Elements of Combinatorial and Differential Topology (Graduate Studies in Mathematics, V. 74)

(x) C An. Then there exists a point Xo E An such that Xo E (xo).

3. The Brouwer Theorem and Sperner's Lemma

85

Proof. Consider the mth barycentric subdivision of the simplex ~n; to each of its vertices Vcr we assign a point Wcr E cI>(vcr). Extending this map by linearity to the simplices of the mth barycentric subdivision, we obtain a continuous map 'Pm: ~ n -+ ~ n. According to the Brouwer theorem, there exists a point Xm for which 'Pm(xm) = Xm . Choose a convergent subsequence {xm.} of the sequence {x m }. We show that Xo = liIDi-+oo xm. has the required property, i.e., Xo E cI>(xo). For each m, take an n-simplex ~~ of the mth barycentric subdivision that contains Xm (if there are several such simplices, we take any of them). Let VO,m, ... , vn,m be the vertices of ~~. Then liIDi-+oo Vj,m. = Xo and Xm = 1:j=o Aj,mVj,m, where Aj,m ~ 0 and 1:j=o Aj,m = 1. We set Wj,m = 'Pm(Vj,m). By the definition of 'Pm, we have Xm = 'Pm(xm) = 1:j=o Aj,mWj,m and, moreover, Wj,m E cI>(Vj,m)' The expression 1:j-o Aj,mWj,m is determined by n+ 1 points in the compact space I x ~ n. Therefore, the sequence {xmo} has a subsequence {xm~} such that the sequences {Aj,m~} and {Wj,m~} converge for all j = 0, 1, ... ,n. We set limi-+oo Aj,m~ = Aj and limi-+oo Wj,m~ = Wj. Then 1:;0 AjWj = Xo. The upper semicontinuity of cI> implies Wj E cI>(xo), because liIDi-+ooV,'m' = Xo, w,'m' E cI>(v3'm')' and limi-+oow"m~to = w 3'. By assumption, the set cI>(xo) is convex. Therefore, Xo = 1:~0 AjWj E cI>(xo), as required. 0 ,

,

,

I

'

"

,

The Kakutani theorem has many applications, but they belong mainly to geometry of convex bodies and mathematical economics, so we do not discuss them in this book.

Chapter 3

Topological Spaces

1. Elements of General Topology 1.1. Hausdorff Spaces and Compact Spaces. A topological space X is said to be Hausdorff if any two different points x, 'JI E X have disjoint neighborhoods U 3 x and V 3 y. This separation property was introduced by F. Hausdorff in the book [51]. The simplest example of a non-Hausgorff space is a space X with trivial (nondiscrete) topology in which precisely two sets are open, X and 0. Exercise 14. Prove that a subspace of a Hausdorff space is Hausdorff. Exercise 15. Prove that different points Xl, X have disjoint neighborhoods Ul, ... ~ Un.

... , Xn

in a Hausdorff space

Non-Hausdorff spaces (in particular, spaces with trivial topology) often arise as spaces of orbits of group actions. Suppose that X is a set and G is a group. An action of the group G on the set X is a map G x X -+ X (which assigns an element g(x) E X to each pair (g,x)) with the following properties: (1) g(h(x)) = (gh)(x);

(2) e(x) =

X,

where e is the identity element of G.

A topological group is a Hausdorff topological space G such that it is simultaneously a group and the maps (g, h) 1--+ gh and 9 1--+ g-1 are continuous. Topology usually deals with continuous actions of topological groups on Hausdorff topological spaces. For a point X E X, the set G(x) = {g(x) EX: 9 E G} is called the orbit of the point x under the action of the group G.

-

87

88

3. Topological Spaces

Exercise 16. Prove that the orbits G(x) and G(y) of any two points x, y E X are either disjoint or coincide. Let X/G be the set consisting of all orbits under the action of G. Assigning the orbit G(x) to each point x E X, we obtain a map p: X --+ X/G. We define a topology on the set X / G by declaring a set U C X / G to be open if and only if p-l(U) is open. The resulting topological space X/G is called the orbit space. Example 1. Suppose that X = 8 1 X 8 1 (this is the. 2-torus), G = JR, and a is a number. For t E G and (ei!P, ei1P ) EX, we set t( ei!p, ei1P ) = (ei(rp+t) , ei (1P+ at». If the number a is irrational, then the topology of the space X/G is trivial. Proof. If a is irrational, then each orbit is everywhere dense, and therefore each orbit meets every nonempty open subset of the torus. 0 Example 2. Let X = Matn(C) be the set of complex matrices of order n, and let G = GLn(C) C Matn(C) be the group of nonsingular matrices. For A E X and BEG, we set B(A) = BAB- 1 • If n ~ 2, then the space X/G is non-Hausdorff. Proof. We consider only the case n = 2. The matrices (~~) and (~l) belong to different orbits 0 1 and O 2 , and all matrices (~1) with s:/:O belong to the orbit O 2 . Since

!~ (~ ~) = (~ ~), the orbit 0 1 cannot be separated from O 2 •

o

Now, we prove some most important properties of Hausdorff spaces. First, note that for any distinct points x and y in a Hausdorff space X I there exists a neighborhood U 3 x such that its closure does not contain y. Indeed, if U 3 x and V 3 Y are disjoint neighborhoods, then U C X \ V. The set X \ V is closed; therefore, U c X \ V, and hence un V = 0. Theorem 3.1. If C is a compact subset of a Hausdorff space X and x E X \ C, then the point x and the set C have disjoint neighborhoods. Proof. Each point c E C has a neighborhood U whose f!losure does not contain x. Such neighborhoods form a cover of the compact space C, which has a finite sub cover Ul, ... , Un. We set V = U?=1 Ui. Then C c V and x f/. V, i.e., V and X \ V are disjoint neighborhoods of the set C and the point x, respectively. 0 Corollary 1. Any compact subset C of a Hausdorff space X is closed.

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Proof. Each point x E X \ C has a neighborhood disjoint from C. This means that the set X \ C is open. 0 Corollary 2. Any two disjoint compact subsets A and B 01 a Hausdorff space X have disjoint neighborhoods. Proof. Each point a E A has a neighborhood whose closure does not intersect B. Since the set A is compact, it has a finite cover {U1,.'" Un} consisting of such neighborhoods. The required neighborhoods of the sets A and B are V = U~=l Ui and X \ V. 0 Problem 26. Prove that any closed subset C of a compact space K is compact. Theorem 3.2. Let I: X -+ Y be a continuous one-to-one map from a compact space X to a Hausdorff space Y. Then I is a homeomorphism. Proof. The image Y of the compact space X under the continuous map I is compact because the preimage of any open cover of Y is an open cover of X (therefore, it has a finite subcover). Any closed subset C of the compact space X is compact; therefore, its image I(C) c Y is compact as well. Since Y is Hausdorff, it follows that I(C) is a closed subset of Y. But I(C) is the preimage of C under 1-1; therefore, the map 1-1 is continuous. 0 Exercise 17. Construct a continuous one-t()-one map of the half-open interval [0,1) onto the circle 8 1 " (In Theorem 3.2, the compactness of X is essential. ) Exercise 18. Prove that a continuous map of the interval onto the square cannot be one-t()-one. In both of the above examples of non-Hausdorff orbit spaces X/G, the groups G are noncompact. Compact groups never cause such troubles. Theorem 3.3. II a compact group G acts on a (Hausdorff) space X, then the orbit space X/G is Hausdorff. Proof. Each orbit G(x) is the image of the compact space G under the continuous map G -+ G x {x} -+ G(x); therefore, G(x) is compact. If G(x) and G(y) are different orbits, then, according to Theorem 3.1, the point x has a neighborhood U whose closure is disjoint from G(y). Thus, p(U) and (X/G) \ p(U) are disjoint open sets which contain the points of X/G representing the orbits G(x) and G(y). 0 Problem 27* ([70]). (a) Prove that for any topological space X, there exists a Hausdorff space X H and a continuous map (T: X -+ X H with the following properties: if Y is a Hausdorff space and I: X ..... Y is a continuous

3. Topological Spaces

90

map, then there exists a unique continuous map fH l XH - Y such that fH q = f· (b) Let XjG be the orbit space from Example 2 on p. 88. Prove that (XjG)H is homeomorphic to cn. (c) Prove that a continuous map f: Matn(C) --+ C such that J(BAB- 1) = A for all B E GLn(C) has a unique representa.tion in the form f(A) = F(CI (A), ... ,en(A)), where Ct (A), ... ,en(A) are the coefficients of the polynomial det (A + >.1) and F: C --+ cn is a continuous function. A topological space X is said to be locally compact if every point x has an (open) neighborhood Ux 3 x with compact closure.

~

X

Theorem 3.4. Let X be a locally compact Hausdorff spae&-.-Then, for any open set U 3 x, there e.cists an open set Us 3 x such that U z is compact and contained in U. Proof. Take an open neighborhood Wx 3 x with compact closure and consider the compact set K that is the closure of K' int(Wz n U). The set C = K \ K' is compact and does not contain z. Therefore, according ~o Theorem 3.1, x and C have disjoint neighborhoods UJ and Uc. The set Uz has the required properties. 0

=

Theorem 3.5. For any compact subspace K of a locally compact Hausdorff space X and any open set U C X containing K, there exists an open subset V of X such that K eVe V c U and V is compact. Proof. For each point x E K, take a neighborhood Ux 3 x such that U x c U and a neighborhood Wx 3 x with compact closure W X ' The closure of Vz = Ux n Wx is compact because it is a closed subspace of the compact space W z. Using the compactness of K, we can choose a finite set of points Xl, •.• , Xn E K such that K c V = VXl U .•. U Vz ,., The set V = V Xl U ..• U V Xn is compact, and V c U Xl U ••. U U zn c: U. [] Let X be a Hausdorff space. The one-point compactijication of X is the topological space X+ = X U { oo} in which the open sets are all open subsets of X and the sets U c X+ such that their complements X+ \ U are compact subsets of X. (It is assumed that 00 is a point that does not belong to X.) We must verify that the finite intersections and arbitrary unions of open subsets of X+ are open. Clearly, the intersection with X of a finite intersection or an arbitrary union of open subsets of X+ is open in X. Suppose that 00 belongs to a finite intersection of sets open in X+. Then the complement of the intersection of these sets is a finite union of compact sets; therefore, it is compact. Now, suppose that 00 belongs to a union of a family of open

1. Elements of General Topology

91

subsets of X+ J Then 00 belongs to some set U from this family. The complement of the union of such sets is a closed subset of the compact set X \ U; hence it is compact. Thus, X+ is a topological space, and X is its subspace. Take an arbitrary open cover U of X+. We show that U has a finite sub cover . The point 00 belongs to one of the sets U E U. The set X \ U is compact; therefore, U contains a finite subcover of this set. Now, we show that if the space X is not only Hausdorff but also locally compact, then X+ is Hausdorff. For this purpose, we must find disjoint open neighborhoods of a point x E X and the point 00. The point x has an open neighborhood Vz with compact closure. The set U = (X \ V:,:) U {oo} is an open neighborhood of 00 disjoint from Vz . 1.2. Normal Spaces. A topological space X is said to be normal if each point of X is a closed set and every two disjoint closed subsets A and B of X have disjoint open neighborhoods U and V. Corollaries 1 and 2 of Theorem 3.1 show that any compact Hausdorff space is normal. Exercise 19. Prove that any metrizable space 1s normal. Urysohn's lemma, which we proved for the space lR" (see p. 56), remains valid for any normal space. Urysohn himself proved it for normal spaces. Theorem 3.6 (Urysohn's lemma). Let A and B be disjoint closed subsets 0/ a normal space X. Then there exists a continuo'U.S function f: X - [0,1] such that f(A) = 0 and /(B) = 1. Proof. Let V be an open subset of a normal topological space X, and let U be a subset of X for which U c V. Then there exists an open W such that U ewe W C V, Indeed, for W we can take an open set which contains the closed set U and does not intersect an open neighborhood of the closed set X \ V. For U = A and V = X \ B, we construct an open set Al such that

(1)

A C Al eX \ B

and Al eX \ B. Then, we insert intermediate sets. Al and A2 so that

(2)

A

c Al

C Al C A2 C X \ B

and the closure of each set is contained in the next set. For the sequence of sets (1), we define the function h: X fl(x)=

o { 1/2 1

ifx E A, ifxEA\\A, if x EX \ AI.

[0,1] by

3. Topological Spaces

For the sequence of sets (2), we define the function

°1/4 hex) =

hIX

...... [0, I} by

if x E A,

if x E A~ \A,

1/2 j.f x EAt \ A~, 3/4 if x E A2 \ All 1 if x E X\A 2.

Then, we construct a third sequence of sets by inserting intermediate open sets between the neighboring terms of sequence (2) and define the function !J(x) for this sequence, and so on. lt is easy to show that 12(x) ~ JI(x), and similarly, In+!(x) ~ In(x). Hence the limit liilln_oo In (x) = I(x) exists. Clearly, I(x}- 0 for x E A and I (x) = 1 for x E B. We must only prove that the function I (x) is continuous. Suppose that at the nth step, a sequence of sets

A

c At

C .. ,

cAre X \ B,

where Ai c AHlI is constructed. (This sequence corresponds to the function In.) We put Ao = intA (intA denotes the interior of A), A-I = 0, an~ Ar+! - X. Consider the open sets AHI \ A.-I for i = O,~, ... , r. Clearly,

X=

r

r

i=O

i=O

U(Ai \ Ai-I) c U(Ai+! \ A-I) ;

therefore, the Ai+! \ Ai-l cover the space X. On each set Ai+! \~-1, the function In(x) takes two values, which differ by 1/2R • Clearly, I/(x) - In(x)1 ~ ~~+!1/2k = 1/2n. For each point x E X, choose an open neighborhood of it in the form Ai+l \ ~~l' ·The image of the open set A+! \ Ai-l is contained in the interval (f(x) - e, I(x) + e), where e > 1/2n. Letting n tend to infinity, we see that the function I is continuous. 0 The Tietze extension theorem can be derived from Urysohn's lemma. Theorem 3.7 (Tietze). Suppose that Y is a normal topological space, X C Y is a closed subset olY, and I: X ~ (-1,1] is a continuous function. Then there exists a continuous function F: Y -+ [-1,1] such that its restriction to X coincides with I.

The proof is the same as in the case of Y = ]Rn (see Theorem 2.4 on p. 57). The only difference is that ]Rn is replaced by Y and, instead of Urysohn's lemma for ]R't, Urysohn's lemma for normal topological spaces is applied. The corollary of Theorem 2.4 remains valid as well.

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93

1.3. Partitions of Unity. Let cp be a continuous function on a topological space X. The support of

= {x EX: ~(x) '" O}.

Let {Ua } be an open cover of a topological space X/A partition of unity subordinate to {Ua } is defined as a family of continuous functions CPa: X -+ [0,1) with the following properties: (1) the family CPa is locally finite, Le., each point x E X has a neighborhood Vex) intersecting only finitely many supports sUPP(CPa); (2) ECPa(x) = 1 for any x E X; (3) SUPP(CPa) C U'a for all a. Sometimes, families {Ua } and {cp.a} with different indices are considered. In this case, it is assumed that for every /3, there exists an a such that supp(cp.a) C Ua . Theorem 3.8 (Stone [123)). Let X be a metrizable topological space. Then, for any at most countable open cover {Ui} of X, there exists a partition of unity {CPi} subordinate to this cover. Proof [37,82). First, consider the case of a finite cover UII ..... , Un. The functions fi(x) = d(x, X \ Ui) are continuous (see the remark on p. 55); therefore, the function F(x) = E~=11i(x) is continuous as well. Each point x E X is covered by some set Ui . We have fi(x) > 0, and hence, F(x) > 0 for all x E X. Let

Then

SUpp(gi)

= {x : fi(x) > F(x)/J..n + I)} c C {x : Ii(x)

Moreover,

~9i(X) ~

t,

~

{x : fi(x) ~ F(x)/(n + In O} c {x: fi(x) > O} = Ui.

(Ii(X) - n

11

F(X»)

= F(x) - n: 1 F(x) =

:~i ?

O.

The required partition of unity consists of the functions n

cpi(X) = 9i(X)/(L9lx i=1

»).

Now, consider the case of a cOl.\ntable open cover Uo, Ul. U2,. time, we define the functions Ii: X -+ [0, 2-i ) by

fi(X)

:b min

{d(x, X \ Ui), 2-i}.

h

•

This

3. Topological Spaces

94

We hav~ fi(X) > 0 for x E Ui and fi(X) = 0 for x ¢ Ui. The function F is defined differently too: we set F(x) = E:o 2-i fi(X). Since {Ui} is a cover, it follows that F(x) > 0 for all x E X. The function F(x) is continuous because E[:..o2- i fi(X) is continuous and, since fi(X) ~ 2-i , for any € > 0, we can choose N such that E:N+l 2-i fi(X) < c. We set 9i(X) = max {f,(x) - tF(x), o}. As in the case of finite covers, we have SUPP(9,) CUi' We prove that the family of functions {9i} is locally finite. Take x E X. The continuity of F implies the existence of a neighborhood Vex) of x such that for some e > 0, F(y) > € at all y E Vex). Choose io for which 2-io < c/3. We have fi(Y) ~ 2-i for any point y E X. Therefore, if y E V(x) and ~ ~ io, then

1 ( It.-i e"" _ . ..... 2 f ,·(x) - -3 F x) -< ~ - -3-

I() -

c.

< o·' 3

-

hence 9i(Y) = o. Finally, we prove that E:09i(X) > 0 for all x E X, i.e., for any point x E X, there exists a number i such that 9i(X) ;> O. Since J;(x) > 0 for some j and fn(x) ~ 2- n , it follows that SUPjEN J;(x) = fio(x) for some io, and fio(X) > O. The definition of F implies 00

00

F(x) = L2-1fi(X) ~ L2-i /i0(X) = 2fio(X)J i=O '=0 Therefore,

. ( ) > ~. ( ) _ 2fio(X) _ fio(x) ..... 0

9'0 x -

JlO X

3"'-

3

'-"" .

The required partition of unity consists of the functions 00

'Pi(X) = 9i(x)1

(L 9i(X»),

o

i=1

1.4. Paracompact Spaces. Suppose that U = {Ua } and V = {Vp} are open covers of a topological space X. We say that the cover V is a refinement of U if each set Vi3 is contained in some Ua' A cover V = {Vp} is said to be locally finite if every point x E X has a neighborhood intersecting only finitely many sets Vp. A topological space X is called paracompact if it is Hausdorff and any open cover U of X has an open locally finite refinement V. The most important property of paracompact spaces is that for any open cover of a paracompact space, there exists a. locally finite partition of u.nity su.bordinate to it. This property of paracompact spaces follows from Theorems 3.9 and 3.10 below (which are of independent interest),. Before proving

1. Elements of General Topology

95

them, we give an example which shows that the class of paracompact spaces is very large. Example. Any set X

c

lRn (with induced topology) is paracompact.

Proof. Consider any open cover {Ua } of the topological space X. For each set UCt , there exists an open set U~ C lRn such that Ua = U~ rr X. Let Xk = {x EX: IIxli < k} for k = 1,2 .... The sets Xk are open in • 00 X, each X k IS compact, X k C Xk+1 for all k, and X = Uk=l Xk. Consider the compact set Xk \ Xk-l' For every point ~ E Xk \ Xk-l, we choose its open neighborhood V; in lRn such that V; c U~ for some a, V; n X = Vz C Xk+l, and Vz n Xk-2 = 0. The open sets V; cover the compact set X1c \ Xk-l; hence there exists a finite family of sets V. which covers Xk \ Xk-l' The union of all such families over all k is a locally finite 0 refinement of the cover {UoJ. Theorem 3.9 (Dieudonne [32]). Any paracompact space is nonnal. Proof. First, we prove that any paracompact space X is regular, I.e., every Dpen neighborhood of a point x E X contains the closure of some other open neighborhood of x. Let U be an open neighborhood of a point x EX. Since X is Hausdorff, for every point y E X \ U, there exist disjoint open sets Uy 3 Y and Wy :3 x. The sets Uy (for all y E; X\U), together with U, form an open cover U of X; it has a locally finite open refinement V. The local finiteness of V means that the point x has a neighborhood W intersecting only finitely many elements of the cover V. Let Vi, ... , Vn be those of them which are not contained in U. Each Vi is contained in UYi for some Yi ~ X\U, We put Z = W

n WY1 n ... n W y .,

and

C=

z.

The set Z is open and contains x because x ~ W and x E WII for any point Y E X \ U. It remains to show that C cU. Consider the union T of all elements of the cover V not contained in U. By construction, TnW

Clearly, Z C

W~

Z

c VIV' ··U Vn

C UYl U··· UUy .,.

therefore,

n T C W n WY1 n ._. n Wy ., n (UY1 U ... U Uy .,) •

By construction, Wy n Uy W

= 0, whence

n WY1 n ... n Wy ., n (UY1 U ... U Uy.,) =

.0;

this means that Z n T = 0 t i.e., Z C X \ T, Since X \ T is closed, we have C= Z C X\T.

3. Topological Spaces

96

Every point of X \ U belongs to some element of V not contained in U. Hence X \ U c T, i.e., X \ T c U (these inclusions are equivalent to X=TUU). Now, we prove that X is normal. Suppose that A and B are disjoint closed subsets of X. Each point a E A belongs to the open set X \ B; therefore, a has an open neighborhood Zo for which Co. = Za eX \ B. The sets Za (for all a E A), together with X \ A, form an open cover U of X; it has a locally finite open refinement V. Let U be the union of all elements of V not contained in X \ A. Then U is an open set containing..A. It remains to construct an open set V containing B and disjoint from U .. We construct it as the union of certain sets Vb over all b E B. Namely, for each point b E B, we choose an open neighborhood Wb 3 b intersecting only finitely many elements of V. Let Yi, .•. , Yn be those of them which are not -contained in X \ A. By construction, we have 1'4 C Za.' where ai E A. We put

Vb = Wb n (X \ Cal) n ... n (X \ Can). The set Vb is open, and b E Vb, because B C X \ Ca. Moreover,

U n Wb c Yl n ... n Y n C therefore, VbnU

= 0.

Thus, if V

ZOI

n· .. n Zan

C Cal

= UbEB Vb, thenB C

n ... n Can;

V and VnU

= 0.

0

Remark 3.1. The reader might have noticed that we repeated twice very similar arguments. Instead, we could prove one general assertion and apply it twice in different situations. Such a proof of Theorem 3.9 was given in [20, Chapter IX, Section 4.4, Proposition 4}. Theorem 3.10. For any open locally finite cover U = {Uo I a E A} of a normal space X, there exists a partition of unity subordinate to this cover. Proof. First, we construct an open cover V = {Va: a E A} of X such that Va C Uo for all a E A. The construction uses transfinite induction~ let us briefly recall what this means (the details, including the proof of Zermelo's theorem, can be found in [134]). A set A is said to be well ordered if it is ordered and any nonempty subset of A has the least element, which precedes all the other elements of A. According to Zermelo's well-ordering theorem, any set A can be well ordered. Suppose that A is a well-ordered set and P is a property such that if all elements preceding some a E A have this property P, then the element a has property P (in particular, the least element of A must have property Pl. Then all elements of the set A have property P. Indeed, if the set of elements of A not having property P is nonempty, then it has a least element ao· All elements preceding ao have property P; therefore, the element ao also has property P, which contradicts its defimtion.

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Suppose that for each a < ao, there exist open -sets Va such that Va C Ua and, for every al < ao, the family {Va: a ~ al}U{Ua : a > al} is a cover of X. It is required to construct an open set Vao such that Vao C Uao and {Va: a ~ ao} U {Ua : a> ao} is a cover of X .. First, we show that {Va: a < ao} U {Ua : a ~ ao} is a cover of X. To this end, we apply the local finiteness of the cover U. Each point x E X belongs to only finitely many sets Uf31 , ... , Uf3n; therefore, we can choose the largest element among f31, ... ,f3n. For definiteness, we assume that f3n > f3i for i = 1, ... , n - 1. If f3n ~ ao, then x E Ua, where a = f3n ~ ao. If f3n < ao, then, by assumption, the family {Va: a ~ f3n} U {Ua : a > f3n} is a cover of X. But x ¢ Ua for a > f3n; therefore, x E Ua::5f3,. Va C Uaao Ua ), together with Uao ' covers the whole space X; therefore, X \ Uao C W. The closed sets X \ Uao and X \ W do not intersect; since X is normal, they have disjoint open neighborhoods Z ::> X \ Uao and T::> X \ W. Clearly,

X \ Uao C Z C Z C X \ T C W. Let Vao = X \ Z. The set Vao has the required properties. Indeed, Vao = X\Z C Uao and Vao UW =X, because Z C W. Now we construct the partition of unity subordinate to the cover U. Instead of U, we take the cover V constructed above, for which Va C Ua ; clearly, it is locally finite. The normality of X implies the existence of an open set Wa such that Va C Wa (: W Q C Ua . By the Tietze extension theorem, there exists a continuous map 9a: X --+ [0,1] for which 9a(X\ W a ) = 0 (Le., supp 9a C W 0: C Ua ) and 9a(Va ) = 1. The sets Va C Va cover X; hence EaEA9a(X) > 0 for each x E X. Since the cover V is locally finite, it follows that the function EOEA9a(X) is continuous. To obtain the required partition of unity, we set 'Pa(x) = 9a(X) (EaEA9o(x)). 0 Corollary of Theorems 3.9 and 3.10. For any open cover of a paracompact space X, there exists a locally finite partition of unity subordinate to it. Proof. Let U = {Ua : a E A} be an open cover of a paracompact space X, and let V = {Vf3 : f3 E B} be its locally finite refinement. Then there exists a --map A: B ~ A such that Vf3 C UA(f3). According to Theorem 3.9, the space X is normal; therefore, Theorem 3.10 implies the existence of a partition of unity {'Pa} subordinate to V. For each a E A, we set 1/;a = EA(f3)=o: 'Pf3. This sum is well defined and continuous because supp 'Pf3 C Vf3 and the cover V is locally finite. Let Co = UA(f3)=o supp 'Pf3. The set C a is closed, being the union of a locally finite family of closed sets. Clearly, Co: C Ua and "pa(x) = 0 for x ¢ Co:. Thus, supp"pa C C a C U Q •

98

3. Topological Spaces

It is easy to verify that the family of sets {Ga } is locally finite. Indeed, for any point x E Xi there exists a neighborhood W intersecting only finitely many elements of the cover Vi we denote them by V,81'" . , V,8,.. The neighborhood W does not intersect Ca if a ~ {A(lh) , .•. , A(Pk)}' Thus, the families of sets {suPPCf',8} and {supp'ljla} are locally finite; therefore,

o We have proved (see p. 93) that for an arbitrary at most countable cover of a metrizable space, there exists a partition of unity subordinate to it. Below we prove a somewhat stronger assertion.

Theorem 3.11 (Stone [123]). Any metrizable space is pamcompact. Proof (see [114]). Let U = {Ua : a E A} be an open cover of a metric space X with metric d. We again use the fact that the set A can be well ordered. For x E X and r > 0, consider the open disk Dx,r ~ {y EX: d(x,y) < r}. For a E A and n EN, we define Va,n as the union of the sets Dx,2-ra over all points x E X satisfying the following three conditions:

{I)

Dx,3.2-ra C Uai

(2) x ~ U,8 for P < ai (3) x ~ V,8J for j < n. We first define the sets Va,n for n = 1 (in this case, only the first two conditions are taken into consideration), then for n = 2, and so on. We start by proving that the sets Va,n cover X. For every point x EX, consider the set B = {P E A : x E U,8}. Let a be the least element of B. Choose n such that Dx,3.2-n C Ua . If x ~ V,8,j for j < n, then x satisfies conditions (1) (3), and therefore x E Va,n. Thus, the point x belongs to some set V,8J with I ::; n. It remains to prove that the cover {Va,n} is locally finite. For a point x EX, consider the set

B

= {P E A : x E V,8,n for some n}.

Suppose that a is the least element of B and let x E Va,n. Choose j E N such that D x,2-i C Va,n' We show that the open set D x,2-1-" intersects only finitely many sets V,8,i. For this purpose, it suffices to prove that this set does not intersect V,8,i for i 2: n + j and intersects at most one V,8,i for i < n j. First, suppose that i ~ n + j > n. The set V,8,i consists of open disks ofradius 2- i centered at points satisfying conditions (1)-(3) .. In particular, it follows from (3) that if y is the center of such a disk, theh y ¢ Va,n' But

+

2. Simplicial Complexes

99

Dx,2~;1 C Vo,n! therefore, d(x, y) ~ 2- j • On the other hand, n+j ~ j+i and i ~ j+1, whence 2- j -"+2-; ~ 2-3, which means that Dx,2-,-.. nDy,2-i = 0.

Now, suppose that i < n+j, p E D x ,2 ;-n nv,8,;, and q E D x ,2-;-n nV;,i' where f3 i "/. For definiteness, we assume that f3 < "/. To obtain a contradiction, it is sufficient to prove that the relations p E V,8,i and q E V..", imply d(p, q) ~ 2-;-n+1. Let y and z be the centers of disks Dy,2-o and D~,2-o such that p E Dy ,2-' C V,8,i and q E D%,2 0 C V-y,;. Condition (1) implies D y,3.2-. c:: U,8, and (2) implies z ¢ U,8' Therefore, d(y, z) ~ 3·2-~, hnd hence

d(p, q) ~ d(y, z) ~ d(prY) .... d(q, z) ~ 3·2-; '"t" 2-i

-

2-; = 2-;

> 2-n - H1 . 0

2. Simplicial Complexes The Euclidean space IR" is the most important example of a topological space. All basic classes of topological spaces (simplicial complexes, CWcomplexes, and manifolds) are constructed by gluing together Euclidean simplices or disks. For purely technical reasons, in homotopic topology, it is more convenient to deal with CW-complexes than with simplicial complexes. The point is that simplicial complexes carry a great deal of geometric ipformation, which is obviously excessive for needs ot topology. Nevertheless, simplicial complexes form a fairly interesting and extensive class of topological spaces, and they are most convenient to use (at least, they are most often used) in geometric topology. A simplicial complex K is a set of simplices in IRn satis7ing the following conditions: • all faces of simplices from K belong to K; • the intersection of any ~wo simplices from K is a face for each of them (for convenience, we assume that the empty set is a face of dimension -1 for any simplex); • any point that belongs to one of the simplices from K has a neighborhood which intersects only finitely many simplices from K. The dimension of a complex K is defined as the maximum dimension of its simplices. A simplicial complex K is said to be finite if it consists of a finite number of simplices. In what follows, w~ mainly consider finite simplicial complexes, To every simplicial complex K correspond& the topological space IKi, which is the union of all simplices of K with topology induced by IR". On p. 81, the barycentric subdivision of a simplex was defined. Taking the barycentric subdivisions of all simplices of K 1 we obtain the barycentric subdivision of the simplicial complex K.

100

3. Topological Spaces

Problem 28. Prove that the simplices in the barycentric subdivision of the simplex A" are in one-to-one correspondence with the ordered sets of vertices of A". 2.1. Rectilinear Cell Complexes. A convex polyhedron of dimension k is defined as a subset of lRk that is determined by a system of linear inequalities of the form Ax ::; b, contains a k-disk, and is contained in a k-disk. A k-dimensional convex polyhedron contained in a k-dimensional (affine} subspace of lR", where n ~ k, is called a rectilinear k-ceU_ A rectilinear cell complex K is a set of rectilinear cells in 1R" satisfying the following conditions: • all faces of rectilinear cells from K belong to K; • the intersection of any two rectilinear cells from K is a face for each ofthemj • every point of the set IKI has a neighborhood intersecting only finitely many rectilinear cells from K (here IKI again denotes the union of all cells of K). Any simplicial complex is a rectilinear cell complex. A rectilinear cell complex K' is called a subdivision of a rectilinear cell complex Kif IKI = IK'I and any cell of K' is contained in some cell of K. The union of all cells of dimension at most l in a rectilinear cell complex K is called the I-skeleton of K; we denote it by KI. If the dimension of K is not less than I, then the I-skeleton of K is an l-dimensional rectilinear cell complex. Theorem 3.12. Two rectilinear cell complexes Kl and K2 such that IK2J have a common subdivision L.

IKII =

Proof. Clearly, the intersection of two rectilinear cells is again a rectilinear cell. Let L be the set of all cells of the form Cl n C2, whe~e Cl is a cell from KI and C2 is a cell from K2. Then L is a rectilinear cell complex, ILl = IKll = IK21, and any cell Cl n C2 from L is contained iIi the cell Cl of KI and in the cell C2 of K2. 0 The following assertion shows that, topologically, rectilinear cell complexes give nothing new in comparison with simplicial complexes. Theorem 3.13. Any rectilinear cell complex K has a subdivision which is a simplicial complex. Proof. We prove this theorem by induction on n = dimK. The rectilinear cells of dimension ~ 1 are simplicesj therefore, for n ::; 1, the assertion of the

101

2. Simplicial Complexes

theorem is obvious. Suppose that we have already constructed a simplicial complex L which is a subdivision of the (m -I)-skeleton of K. Inside each m-cell cm of K, we choose a point M. Consider the simplices such that for each of them, one of the vertices is M and the remaining vertices are the vertices of one of the simplices constituting the boundary of em. As a result, we obtain a simplicial subdivision of K. 0

en

Remark 3.2. For M we can take a vertex of the cell rather than its interior point. Then the constructed simplicial subdivision has the same vertices as the initial rectilinear cell complex. 2.2. Simplicial Maps. Let Kl and K2 be simplicial complexes. A map f: \Kl[- \K2\ is said to be simplicial if the image of any simplex ~l from Kl is a simplex ~2 from K2 and the restriction of J to ~l is linear, i.e.~

f(2: AiVi) = 2: Ad (Vi) ,

(1)

where the Vi are the vertices of ~~, E ~ = 1, and Ai ;::: Q. By definition, the vertices (i.e., O-simplices) of the complex Kl are mapped to vertices of K2. Therefore, the map f determines a map fO: ~ of O-skeletons. On the other hand, according to (1), J is uniquely determined by jO. The map has the following property: if Vo, •.• ,Vn are the vertices of a simplex from Kl, then fO(vo), '" ,jO(vn ) are the vertices of a simplex from K2 (some of the points fO(vo), ... , fO(v n ) may coincide). We refer to maps of 0skeletons with this property as admissible. Each admissible map ~ of O-skeletons determines a simplicial map \K114 \K2 \. We usually denote simplicial maps by Kl -+ K2.

Kf -

r

KP -

Exercise 20. Prove that any simplicial map is continuous. Exercise 21. Prove that the image of the k-skeleton of Kl under a simplicial map is contained in the k-skeleton of K 1. Theorem 3.14. Suppose that f: K - K is a simplicial map, ~' is a simplex from the barycentric subdivision of the complex K, and f(~') = ~'. Then the restriction 011 to ~' is the identity map. Proof. For the simplex ~', there is a unique simplex ~ in K which contains ~, and has the same dimension. Moreover, ~' uniquely determines the ordering of the vertices of ~ under which Vo is the common vertex of ~ and ~', [VO,Vl] is their common edge (or, to be more precise, the edge of ~ containing an edge of ~'), [vo, Vb V2] is their common 2-face, etc. Conversely, any ordering of the vertices of ~ uniquely determines a simplex in the barycentric subdivision.

Since I(~') = ~', we also have 1(l:1) = ~; the map 1 can only permute the vertices of~. If this permutation were not the identity, then it 'would

102

3. Topological Spaces

change the ordering of the vertices of ~, and the new ordering would cor.. respond to a different simplex of the barycentric subdivision, i.e., we would have f(~') :f= ~'. Therefore, the restriction of J to ~ ::> A' is the identity ~

0

2.3. Abstract Simplicial Complexes. From the point of vIew of topology, of interest are the topological spaces 1KI determined by simplicial complexes K rather than the complexes themselves. Each simplicial complex K determines not only the space IKI but also its embedding in ]Rnj this is a superfluous information, which often complicates dealing with simplicial complexes. To get rid of a particular embedding in ]Rn, we define an abstract simplicial complex K as a vertex set {v o } and a family of its subsets, called simplices (a set of k + 1 vertices is a k-simplex); any subset of a simplex from K must be a simplex from K. To each abstract simplicial complex K we assign a topological space IKI as follows. To every abstract k-simplex {ViI' ••• , Vilc+1} corresponds the geometric simplex [ViII' .. ! Vilc+l]' which we treat as a topological space. In the disjoint union of these topological spaces, we identify the respective points of the simplex [VI, ... ,vp] and of the face LVI, . i' , vpJ iI1 the simplex [VI, .• , , Vp, Vp+I, ••• , v q ]. The topology on the coset space )Kl thus obtained is defined as follows: a set U is open in IKI if and only if its intersection with each simplex is open in the topology of this simplex. Suppose that K is an abstract simplicial complex and q: J
2. Simplicial Complexes

103

(t\ t 2 , &,. J~, t2n+l) are the vertices of a (nondegenerate) simplex. If the number of points is precisely 2n + 2, then the volume of this simplex is equal to

1

1

TI

'"

T;n+l[

± (2n + 1)! ., .............. '2'';';'i

# o.

1 T2n+2 • • • T2n+2 Smaller sets can be extended to (2n + 2)-point sets in an arbitrary way.

0

Remark 3.3. Points Xl, ... , Xk in the space]RN are sald to be generic, or in general position, if no m + 1 of these points belong to the same (m -1)dimensional affine subspace for m:::; N. To construct a realization of an dimensional abstract simplicial complex (with vertices VI, ••• , Vk) in ]R2n+l, it suffices to specify generic points Xl, .. J, I Xk in ]R2n+l,.. An explicit construction of generic points was given in the proof of Theorem 3.15; yet another construction is as follows. First, take two different points Xl and X2 in ]RN. Then choose a point X3 not belonging to the straight line XIX2. After that take a point X4 not belonging to the plane XIX2X3, and so on. In this way, we construct a set {x}, •.. , x N +l}' Then, we. draw hyperplanes through all of its N-point subsets, take a point XN+2 belonging to none of these hyperplanes, draw hyperplanes through all N -point subsets of the new set, choose a point belonging to none of them, and so on.

n-

A simplicial sub complex L c K is said to be complete if any set of vertices of L that spans a simplex of the complex K spans a simplex of the complex L. Problem 29. Prove that a simplicial sub complex L c: K is complete if and only if every simplex from K with boundary contained in L is itself contained in L. Problem 30. Let L c K be a simplicial subcomplex, and let L' and K' be the barycentric subdivisions of L and K. Prove that the subcomplex L' C K' is complete. 2.4. Simplicial Approximations. Simplicial maps are much simpler than continuous maps. For example, for any two simplicial complexes K and L, there are only finitely many simplicial maps K -+ L. Nevertheless, any continuous map can be approximated by simplicial maps (although, such an -approximation may require passing from the complexes K and L to their subdivisions). It is most important for homotopic topology that any continuous map of simplicial complexes is homotopic to a certain simplicial map. This fact substantially facilitates studying homotopy classes of maps, but its proof requires a certain ~ffort. Suppose that K and L are simplicial complexes and f: IKI - ILl is a continuous map. For each point X E IKI, consider the point f(x} E ILl. It

3. Topological Spaces

104

is an interior point of precisely one simplex from L. We say that a simplicial map ep: K _ L is a simplicial approximation of the map I if, for each x E IKI, the point ep(x) belongs to the simplex whose interior contains I(x). Theorem 3.16. Any simplicial approximation ep of the map I is homotopic to I. Proof. Let ft(x) be the point dividing the segment with endpoints ep(x) and f(x) in the ratio t : (1- t). Then ft is a homotopy between fo = ep and h=~ 0

When dealing with simplicial approximations, it is more convenient to use another definition of simplicial approximation, which employs the notion of a star. Let K be a simplicial complex; and let ~ be a simplex in K. The star of the simplex ~ is defined as the union of the interiors of all simplices from K containing~. The star of a point x E IKI is the star of the simplex from K that contains x in its interior. The star of a simplex that has maximum dimension is the interior of this simplex. We denote the star of a simplex ~ by st ~ and the star of a point x by st x. Theorem 3.17. A simplicial map ep: K -+ L is a simplicial approximation for a continuous map f: IKI-ILI if and only if f(stv) C step(v) for each vertex v of K. Proof. First, suppose that ep is a simplicial approximation of I and v is a vertex of K. Take x E st v. Consider the simplex ~K with vertex v whose interior contains x and the simplex ~L whose interior contains f(x). On one hand, ep(x) is an interior point of the simplex ep(~K) with vertex ep(v), and on the other hand, it belongs to the simplex ~L. Therefore, ~L => ep(~K) 3 ep(v), which means that f(x) E int ~L cat ep(v). Now, suppose that f(stv) c step(v) for any vertex v of K. Take x E lKI and let Vo, ••• , Vn be the vertices of the simplex ~ from K whose interior contains x. We }lave

0 n

f(x) E f

(

stvi

)

0 n

C

cO n

f(stvi)

step(Vi) = intep(~).

Therefore, ep(~) is the simplex that contains f(x) in its interior. It remains to note that ep(x) E ep(~), because x E ~. 0 Corollary. If ep: K - L and 't/J: L - M are simplicial approximations of continuous maps f: IKI - ILl and g: ILl - IMI, then't/Jep is a simplicial approximation of the map gf.

2. Simplicial Complexes

lU5

Let K be a finite simplicial complex, and let K(n) be its nth barycentric subdivision. Note that the maximal diameter of simplices from K(n) tends to zero 88 n --+ 00 (see p. 81). Theorem 3.18 (simplicial approximation theorem). (a) Suppose that K and L are simplicial complexes, the complex K is finite, and f: IKI --+ ILl is a continuous map. Then, for some n ~ 0, there exists a simplicial map cp: K(n) --+ L which is a simplicial approximation for f.

(b) If, iTt addition, the restriction of f to a subcomplex K1 c K is simplicial, then the simplicial approximation cp can be chosen so that it coincide with f on KI. Proof. (a) The stars of the vertices of the complex L form an open cover of the topological space IL I. The preimage of this cover under f is an open cover U of the compact subset IKI of Euclidean space. According to Lebesgue's theorem on open covers (Theorem 2.6 on p. 59), there exists a number 6 > 0 such that any subset B c IKI of diameter less than 0 is contained in an element of the cover U. Choose n such that the diameter of any simplex from K(n) is less than 6/2. The required simplicial map cp: K(n) --+ L is defined as follows. ,Let v be a vertex of K(n). Then the diameter of st v is less than 6; therefore, the set f (st v) is contained entirely in the star st w of some vertex w of t. We set cp(v) = w (if there are several suitable vertices w, we take anyone of them). We have defined a map of O-skeletons. We must verify that this map is admissible, i.e., cp( VI), •• ! ,CP(Vk) are the vertices of a simplex from L whenever VI,"" Vic are the vertices of a simplex from K(n). Note that vertices VI •••• , Vic form a simplex ~ if and only if n~1 st Vi = st ~ 1= 0. Suppose that VI, ••• ,Vic are the vertices of a simplex from K(n). Then n~=1 st Vi i- 0, and hence n~=1 f(stvi) 1= 0. But n~=1 st cp(Vi) :::) n~=l f(stvi) i- 0; therefore, the vertices cp( vd, ... , cp( Vic) form a simplex in L. By Theorem 3.17, the simplicial map cP: K(n) --+ L is a simplicial approximation of f. (b) Let V be a vertex of KI- Then f(v) = W is a vertex L. IT the subdivision K(n) is sufficiently fine (Le., the number n is sufficiently large), then '-f(stv) C stw for this subdivision, and we can set cp(v) = w. 0 The simplicial approximation theorem enables us to prove the following assertion. Theorem 3.19. Any continuous map f: sn --+ sm, where n < m, is homotopic to a constant map (which takes sn to one point).

3. Topological Spaces

106

Proof. It is sufficient to prove that / is homotopic to a nonsurjective map
tcp(x) - (1 - t)~o CPt(x) = IItcp(x) - (1- t)~oll is a homotopy between cp and the constant map sn ---t -~o E 8 m • The sphere sn can be represented as a simplicial complex K that is the n-skeleton of an (n+ 1)-simplex. Similarly, the sphere gm can be represented as a simplicial complex L. The continuous map /: IK/ ---t ILl has a simplicial approximation cp: K(N) -+ L. The map It' is not surjective because its image is contained in the n-skeleton of the complex L, and it is homotopic to / by Theorem 3.16. 0 Example. Let K be a triangulation of an n-simplex L witli-vertices vo, VI; ••• , V n . Suppose that the vertices of K are labeled by the numbers 0, 1, . J. , n. We define a simplicial map cp: /KI "'-+ 1LI by assigning the vertex Vi to each vertex a E K labeled by i. The map

0 is a fixed number. If k is sufficiently large, then the simplex L = [vo, ... , v n ] is contained inside the simplex lao, ... , an]. Consider the simplicial complex Kl whose vertex set consists of the points ao, ... ,an and the vertices of the complex K (recall that K is a triangulation of L); the simplices of Kl are those of K and all simplices for which one of the vertices is ai and the other vertices are those of a simplex from K contained in the face [va, .• . , Vi, ..• ,Vn]. An example of such a complex Kl for n = 2 is given in Figure 1. We define a simplicial map '1/1; KI -+ L 1 If the simplices ~ i and 'P( ~ i) have the same orientation, then we take the plus sign, and if they have opposite orientations, then we take the minus sign. Note that d ~i C 'P(~i), then the simplex 'P(~') has the same dimension as ~i.

2. Simplicial Complexes

107

Figure 1. The construction of the complex Kl

Figure 2. Orientation of simplices

in such a way that it coincides with

=

3. Topological Spaces

For an arbitrary n, the argument is similar. The signs alternate because the orientations of the simplices [Vio' Vii' ..• ,Vi,,] and [aio' ail' ... ,ai,,] are the same for even k and opposite for odd k. 0 2.5. Nerves of Covers. An arbitrary family U = {Ua } of subsets of a set X determines a simplicial complex N = N(U) whose vertices {va} are in one-to-one correspondence with the sets {Ua}i a set tho,"" Va" is a simplex if and only if Uao n··· n Ua" # 0. 1£ X is a topological space and U is its cover (not necessarily open), then N is called the neroe of the cover U. Example. Let K be a simplicial complex with vertices {va}, and let Ua = st Va be the star of the vertex Va, i.e., the union of the interiors of all simplices containing Va. Then the nerve of the cover {Ua } coincides with K. Proof. Vertices VaO , ... , va" form a simplex stvak = sta k # 0.

a k if and only if st Voo n ... n 0

We say that an open cover U of a space X is contractible if all nonempty finite intersections uaon·· ·nualo are contractible. The nerve of a contractible cover carries a great deal of information on the homotopy structure of X. For example, the following assertion is valid. Theorem 3.21. If U = {Ua } is a contractible locally finite cover of a paracompact space X, then the neroe N = N (U) is homotopy equivalent to X. Proof. The contractibility of the cover and the paracompactness of the space are used at different places in the proof. Thus, first we assume U to be an arbitrary locally finite open cover of an arbitrary space X. Let us construct an auxiliary space Xu. For each nonempty intersection Uao n···n Van = Uao ...an , consider the direct product Uao ...an xa~o ...a.. ' where a~o ...a .. is the simplex with vertices ao, ... , an. In the disjoint union of the topological spaces thus obtained, we identify each point (x, y), where :t E Uao ...an and y E [ao ... OJ ... an] c a~o ... an' with the corresponding point of the space Uoo ...a•... an x a:;;:~.ai a .. by using the inclusion Uao ...Otn C Uao ...a•... an.

...

Step 1. If the space

X

is paracompact, then X

rv

Xu.

Let p: Xu --+ X be the map induced by the natural projections Uao ...an x a~o ...an --+ Uao ... an • This map is continuous, because the cover U is open. Every point of the set p-l(X) can be represented as Ea taxa, where t(1: ~ 0, Eta = 1, and the summation is over all a such that X E Ua; by XQ we denote the point x together with the fact that x E UQ • This sum is finite, because the cover is locally finite, Since X is paracompact, there exists a partition of unity {c,oa} subordinate to the cover {Ua }, i.e., such that supp <Pa C Ua. Consider the map

2. Simplicial Complexes

109

Xu defined by sex) = ~ c,oo(x)xo (if c,oo(x) = 0, then the corresponding term is zero, and if c,oo(x) =1= 0, then x E Uoj by Xo we denote the s: X

-+

point x together with the fact that x E Uq ). Clearly, po s = idx. We must only verify that sop I'V idxu' Suppose that a point x belongs to sets Uao , ... , UOn and does not belong to any other Uo . Then the points y = ~toxo and s(P(y)) = ~ c,ooxo belong to the simplex with vertices x oo ' .•. , X On ' The required homotopy uniformly moves s(P(y)) to y along the segment joining these points. Step 2. H the cover U is contractible, then Xu

I'V

IN(U)I.

Consider the space Xu. Let p: Xu -+ IN(U) I be the map induced by the natural projections Uoo ...On x ~:!o ... o .. -+ ~ao ... a ... Then the preimage of a vertex a is Uo , the preimage of an interior point of an edge [a,.8] is Ua ,{3t the preimage of an interior point of a simplex fa, (3, 1'] is Ua ,{3,'Y' etc. Let us successively contract these preimages to points (first, we contract the preimage Ua of each vertex, then the preimages of all interior points of edges, and so on). As a result, we obtain the space IN(U)I. 0 2.6. Pseudomanifolds. A finite simplicial complex K is called an n-dimensional pseudomanifold if the following conditions hold: • K is homogeneous, i.e., each simplex of K is a face of some simplex;

11.-

• K is unramified, i.e., each (n - 1)-simplex of K is a face of no more than two n-simplicesj • K is strongly connected, i.e., for any two n-simplices ~: and ~b' there exists a sequence of simplices ~f = ~:, ~~, ... , ~~ = ~b in which every two neighboring terms ~f and ~f+l have a common (n - 1)-face. The union of all (n ....,.1)-simplices in an n-pseudomanifold M n such that each of them is a face of precisely one n-simplex is called the boundary of Mn and denoted by aMn. If aMn = 0, then the pseudomanifold Mn is said to be closed. Any (n - 1)-simplex in a closed pseudomanifold Mn is a face for exactly two n-simplices. An orientation of a simplex ~ n C Rn is one of the two possible fami~es of all reference frames in R n of the same orientation with origins at the points of the simplex ~n. For n > 0, each simplex admits exactly two orientations. If a simplex is endowed with an orientation, then this orientation is said to be positive, and the opposite orientation is said to be negative. An orientation of a simplex ~n induces an orientation of its face ~n-l C ~ n as follows. Take a positively oriented frame with origin at a point x E ~n-l such that its first n - l vectors belong to ~n-l and the last

3. Topological Spaces

110

tzi2!2j 1

462

Figure 3. A triangulation of the Mobius band

vector is directed inside ~n. The orientation of ~n-l determined by the first n - 1 vectors is declared to be positive. Is orientable if all of its n-simplices can be oriA pseudomanifold ented in such a way that any two simplices having a common (n - 1)-face induce opposite orientations on this face. It follows from the strong connectedness condition that if a pseudomanifold is orient able , then it admits exactly two orientations. An orient able pseudomanifold Mn with a fixed orientation is said to be oriented.

un

Example. Let us represent the Mobius band as an abstract simplicial complex with six vertices (see Figure 3). Any realization of this abstract simplicial complex, including a realization in IR5 (which exists according to Theorem 3.15 on p. 102), is a nonorientable pseudomanifold. Example. Suppose that M n c IRm is a pseudomanifold (possibly, with boundary). Let us embed IRm into ]Rm+l and take a point a in ]Rm+1 \ ]Rm. The union of all segments of the form [a, x], where:t E Mn, is an (n + 1)dimensional pseudomanifold. It is called the ~uspension of Mn and denoted by I:Mn.

Remark 3.4. The suspension of a usual (topological or smooth) closed manifold M n can be a manifold only if Mn is a homology' sphere. Thus, the class of pseudomanifolds is larger than that of manifolds. On the other hand, if M n is a. pseudomanifold and (Mn)n-2 is its (n - 2)-skeleton, then Mn \ (M")n-2 is a. manifold; thus, any pseudomanifold becomes a. manifold after removing a set of codimension 2. Example. The suspension over the two-dimensional pseudomanifold shown in Figure 4 is a three-dimensional pseudomanifold whose boundary is not a pseudomanifold (the strong connectedness condition is violated). 2.7. The Degree of a Map to a Euclidean Space. Let Mn be an n-pseudomanifold. We say that a map f: M n - ]Rm is simplicial if its restriction to each simplex is affine. Any simplicial map Mn -+ IRm is completely determined by its restriction to the O-skeleton (M")O, and any map (Mn)O _]Rm can be extended to a simplicial map Mn -+ r,

111

2. Simplicial Complexes

Figure 4. The boundary of the suspension of this pseuddmanifold is not a pseudomanifold

Consider a simplicial map I: M" -lR" (the dimensions are equal). We say that a point Y E lR" is a regular value of I if y does not belong to the image of the (n - I)-skeleton of M". In particular, any point y e lR" that does not belong to the image of K is a regular value of I. The set of regular values of I is everywhere dense in lR". Suppose that M" is a compact oriented pseudomanifold, I' MR ........ lRn, is a simplicial map, and y is a regular value of I. The degree 'Of the map I with respect to the point y is defined 88 deg(f,y) =

L

sign J,(x) 1

:&E,-1(1I)

where sign J, (x) is the sign of the determinant of the linear part of the affine map I at Xj thus, sign J,(x) = 1 if I preserves the orientation ofthe simplex .6." whose interior contains x and sIgn J, (x) = -1 otherwise. The sum under consideration is finite because the number of n-simplices in M" is finite and each of these simplices contains at most one point of the set I-I(y). Theorem 3.22. Suppose that M" is a compact oriented pseudomaniloldJ I: M" - lR" is a simplicial map, and Yl and Y2 are regular values 01 I r II Yl and Y2 belong to the same connected component 01 lR" \ 1(8M"), then deg(f, yd = deg(j, Y2). Proof. The image of the (n L 2)-skeleton of M" does not separat~ lR", because no (n - 2)-dimensional subspace of lR" separates lR" (and the union of finitely many (n - 2)-dimensional subspaces of lR" does not separate lR" either). Hence there exists a finite-sided polygonal curve L in lR" with endpoints Yl and Y2 which does not intersect the image of the (n ..... 2)-skeleton of M", does not intersect I(BM"), and intersects the image of the (n -1)skeleton in only finitely many points al, •. , , ak. The set I~ 1 (a;.) contains no points belonging to simplices of dimension ~ n - 2; therefore, 1-1 (a;.) is a finite union of sets each of which either consists of one interior point of an

3. Topological Spaces

112

--r

~ I I I

I

*

*

at.

a

>: J. I I I

I

ai

ai

i

*

b

Figure 5. PBBBing through a critical value

(n - I)-face of Mn or is a straight line segment contained inside an n-face and joining interior points of two (n - I)-faces. By assumption, 1-1 (O-i) n8Mn = 0; therefore, to each ofthe sets (points and segments) forming 1-1(O-i) correspond two n-simplices. Namely, to each interior point of an (n - 1)-simplex correspond the two n-simplices for which the given simplex is a common face, and to each segment inside an n-simplex correspond the n-simplices adjacent to the (n.l... I)-faces which contain the endpoints of this segment. If the images of these simplices have the same orientation, then the number of preimages, as well as the signs of the determinants at the preimages, do not change in passing through the point O-i (see Figure 5a; we mean only the preimages belonging to the two simplices under consideration). If the images..of the simplices have opposite orientations, then two preimages at which the determinants are of opposite signs either appear or disappear (see Figure 5b). Thus, the sum of the signs of the determinants does not change. 0 Theorem 3.23. Let Mfl. be an oriented pseudomanifold, and let I, g: un --+ lRfl. be simplicial maps with coinciding restrictions to 8MR. If Y is a reguiar value of I and g, then deg(f,y) = deg(g,y). Proof. Consider the family of maps It = (1 - t)1 + tg. Clearly, fo = 11 g, and the restriction of It to 8MR does not depend on t. Let X be the connected component of]Rn \ ft(8Mfl.) that cpntains y (X does not depend on t, because ft(8Mn) does not depend on t). The set X is open; therefore, the regular values of each map It are everywhere dense in it. In particular, each map It has a regular value Yt EX. According to Theorem 3.22, deg(/t, Yt) does not depend on the choice of a regular value Yt E X; hence the function 0 with the following properties: for all T E (t - e, t + e) n [0,1], the point Yt is a regular value for the maps fn the preimages f;:l(Yt) contain the same number of points, and the determinants at the respective points have the same sign. Indeed, if Yt is an interior point of the image of an n-silllplex under an

II =

2. Simplicial Complexes

113

1

-L o

2

Figure 6. The supplementary triangulation of a simplex

affine map to ]Rn, then it remains an interior point of the image when the affine map slightly changes. Thus, the function rp(t) is constant on the set Ut = (t - e, t + e) n [0,1]. The family of sets {Ut}~ where f E [0,1], forms an open cover of the interval [0,1]. Considering a finite sub cover of this cover, we see that the function rp(t) is constant on the entire interval [0,1]; therefore, rp(O) = rp(1), i.e., deg(f, yI) = deg(f, yo). 0 Theorem 3.23 readily implies Sperner's lemma (and even its refined version given in Theorem 2.27 on p. 83). We consider bnly the case of twodimensional simplices; this is sufficient to illustrate the general idea of the proof (a detailed proof for n-simplices is given in [130]). We embed a triangulated simplex in a larger simplex and triangulate the new object so as to prevent the appearance of new completely labeled simplices (see Figure 6); this can easily be done. The large triangulated simplex is an oriented pseudomanifold. Consider a simplicial map I from this pseudomanifold to a fixed completely labeled simplex ~ in]Rn (each vertex is mapped to a vertex with the same label). On the boundary, the map I coincides with the identity map; therefore. its degree (with respect to the interior points of ~) is 1. But the degree of I equals the difference between the numbers of completely labeled simplices with positive and negative orientations. By construction, all completely labeled simplices are contained in the given triangulation. 2.8. The Borsuk-Ulam Theorem. In 1933, K. Borsuk [17] proved the following asSertion, which had been conjectured earlier by S. Ulam.

sn sn.

Theorem 3.24 (Borsuk Ulam). Let I: Then I(x) = I(-x) lor some point a; E

--+

]Rn be a continuous map.

The points x and -x are said to be antipodal; for this reason, the BorsukUlam theorem is also known as the antipodal theorem. A map g: sn --+ ]Rn is called odd, or antipodal, if g(-x) = -g(x) for any x. It is easy to see that the Borsuk-Ulam theorem is equivalent to the following statement.

3. Topological Spaces

114

Theorem 3.25. If g:

x

E

sn.

sn

---t ]Rn is an odd map, then g(x)

= 0 for

some

Indeed, if f: sn - ]Rn is an arbitrary map, then the map g(x) = f(x)- f( -x) is odd, and the equality f(x) = f( -x) is equivalent to g(x) = O. Conversely, if g: sn . . ., ]Rn is an odd map, then the Borsuk Ulam theorem implies g(x) = g(-xl for some x ~ 8" .. On the other hand, g(-:z;) = -g(x)j hence g(x) = O. We derive Theorem 3.25 from the following assertion. Theorem 3.26. Let h: Dn - ]Rn be a map such that its restriction to is odd. Then hex) = 0 for some point ~ E Dn.

sn- 1 = 8Dn

To derive Theorem 3.25 from Theorem 3.26, it suffices to represent D n 88 the section of the disk Dn+ 1 (with boundary sn) by a plane passing through its center and define h to be the composition of the projection of D n onto the hemisphere and the map g.

r,

Proof of Theorem 3.26. Instead of Dn, we shall considet the n-cube where I = [-1,1]. This cube is symmetric with respect to the origin. Suppose that the restriction of h: In - ]Rn to 8r is odd and 0 rt her). The set h(In) is compact and, hence, is disjoint from some n-disk centered at O. Let r be the radius of this disk. Since the map h is uniformly continuous, the image of any simplex from a sufficiently fine triangulation of the cube P is contained in a. disk of diameter e <. r. For each vertex V of such a triangulation, we set he(v) = h(v) and extend the map he to the entire simplex by linearity. Every point :t E In belongs to some simplex of the triangulationj therefore, the points hex) and he(x) belong to the same disk of diameter Ef which means that IIh(x) - he (x) II :S e. We can assume that the triangulation of the cube P is symmetric with respect to the origin. In this case, the restriction of he{x) to 8r' is odd. Moreover, the relation IIh(x) - he (x) II :? e implies 0 rt he(P). To obtain a Contradiction, it is sufficient to construct a simplicial map cp: P - ]Rn such that its restriction to 8P coincides with he and deg( cp, 0) is odd. Indeed, on the one hand, deg(he,O) = OJ on the other hand, deg( cp, 0) = deg(he, 0) by Theorem 3.23. Clearly, if cp; P _ ]Rn is an odd map and 0 is its regular value, then deg(cp,O) is odd. Ind~, cp-.l(O) consists of 0 and pairs of the form (x, ~x), and th~ parity of the sum :E ±1 depends only on the number of terms. The required odd map cp is very easy to construct. We take an arbitrary map from the interior vertices of the symmetric triangulation defined

2. Simplicial Complexes

115

in constructing he to lRn such that it takes symmetric vertices to symmetric points. For the vertices v E 8In; we set
=

The followin~ theorem is equivalent to the Borsuk lnarn theorem. Theorem 3.27. Ifm

>n

~

1, then there exists no odd map f:

~

- 8"'.

~ 1, then any odd map 8 m - 8'" is &80 an odd map sn t: IRn+l \ {O} C IRm \ {OJ, and Theorem 3.27 follows from the Borsuk Ularn theorem. Conversely, given an odd map sm - IRm {OJ, it is easy to construct an odd map sm _ sm- 1 , Indeed, if m

>n

8 m -+

l

Remark 3.5. A very simple proof of the Borsuk Ulam theorem is given in [84]. The proof of Theorem 3.26 given above has much in common with [43]. Problem 31*. (a) (Tucker's lemma [126]) Suppose that the n-cube P is triangulated so that the triangulation of its boundary 8P is symmetric with respect to the center; suppose also that the vertices of the triangulation are labeled by the numbers- ±1, ±2, •.. ; ±n in such a way that for any vertex v E 8P, the labels of v and -v add to zero. Prove that this triangulation has a pair of adJacent (i.e., Joined by an edge) vertices such that their labels add to zero. (b) Derive the Borsuk Ularn theorem from Tucker's lemma. Remark 3.6. A purely algebraic proof of the Borsuk U1am theorem for polynomial maps is given in [104]. 2.9. Corollaries and Generalizations of the Borsuk-Ulam Theorem. The Borsuk Ularn theorem has many interesting corollaries. Borsuk included one of them in the sarne paper [17] where he proved the Borsuk-Ulam theorem, although this corollary had already been proved by Lyusternik and Shnirelman [76, p. 26}.

sn

Theorem 3.28 (Lyusternik-Shnirelman). If the sphere is covered by closed sets Flo ... ,Fn +1. then one of these sets contains a pair of antipodal points of the sphere.

3. Topological Spaces

116

Proof. We denote the set symmetric to Fi with respect to the center of the sphere by -Fi. Let us show that if Fi n (-Fi) = 0 for i = 1, ... , n, then

Fn+1 n (-Fn+I)

i= 0.

Applying Urysohn's lemma (see p. 56) to the disjoint closed sets FI and -FI in lRn+1, we can construct a continuous function CPI: sn r-+ [0,1] such that CPI(FI) = and CPI(-FI) = 1. (Urysohn's lemma gives a function I for which I(FI) = --I and I( -FI) = 1; we set CPI = (1 + /)/2.) Similarly, for F2, ... ,Fm we construct functions CP2, ... ,CPn' Consider the map cP: sn -+ lRn defined by cp( x) = (CPI (x), ... , CPn (x». According -to- the Borsuk Ulam theorem, there exists it point Xo E sn for which cp(xo) = cp( -xo). If x E ±Fi for i = 1, ... , n, then CPi(X) - CPi( -x) = ±lj therefore, CPi(X) i= CPi( -x). Thus, Xo ¢ Fi and Xo ¢ (-Fi) = Fi. Hence Xo E Fn+I and -Xo E F n +1. 0

°

n:"l

n:=1

n:=l

Theorem 3.29. Let Fl, .. '1 Fn be measurable subsets 01 lRn. Then there exists a hyperplane which cuts each set Fi into two parts 01 the same volume. Proof. Take x E sn- 1 C lRn. Suppose that the sphere sn-l is centered at the origin. For c E lR, we set 2 IIc(x) = {y E lRn : (y,x) = c}. It is easy to verify that for each vector x E sn-I, there exists a unique

number c E lR such that the hyperplane IIc(x) cuts FI into two parts of equal volume. We set CPI(X) = c. The hyperplane cutting FI into two equal-volume parts is the same for x and -x. Clearly, II-c(-x) = IIc(x), whence CPI( -x) = -c. Let CP2,"" CPn be similar functions for F2, .... Fn and consider the map cP: sn-I -+ lRn - I defined by

cp(X) = (CPn(X) - CPI(X), •.. ,CPn(x) - CPn-I(X»), Clearly, cp(x) = -cp( -x). Therefore, by the Borsuk Ulam theorem, there exists a point Xo E sn- I for which cp(xo) = 0, i.e., CPl(XO) = CP2(XO) = ... = CPn(XO) = c. The hyperplane IIc(xo) has the required property. 0 It is easy to prove that the length of any closed centrally symmetric

curve on the unit sphere sm is at least 211' (any centrally symmetric curve contains two antipodal points, and the length of any arc joining two such points is at least 11'). This assertion has the following generalization. Problem 32* ([13]). Let sn and sm be the unit spheres, and let cp: 8 n --t sm be an odd map. Prove that the n-dimensional volume of
3. CW-Complexes

117

The Borsuk Ulam theorem implies also a nonlinear generalization of the well-known Radon theorem: II a set A c lRn contains at least n + 2 points, then A has disjoint subsets Band C such that their convex hulls have a common point. To prove the Radon theorem, it suffices to consider the case where A consists of exactly n + 2 points; in this case, the theorem can be stated as follows: II I: ~ n+1 --+ lRn is a linear map, then ~ n+1 has two disjoint laces with intersecting images. The nonlinear generalization of this theorem consists in replacing the linear map I by an arbitrary continuous map f. Namely, the following assertion is valid; we state it as a problem. Problem 33* ([12]). (a) Given a nondegenerate (Le., containing an (n+l)disk) convex polyhedron P in lRn +1 and a continuous map I: ap ~ lRn , prove that P has disjoint faces 3 with intersecting images. (b) Prove that if I: a~ n+1 --+ lRn is a continuous map and ~~, ..• , ~~+2 are the n-faces of the simplex ~n+1, then n~!12 I (~f) :/: 0.

3. CW-Complexes In homotopy topology, the CW-complexes, which were introduced by Whit~ head [141], are most convenient in many respects. The CW-complexes are obtained from the closed disks D n by attaching their boundaries aDn = sn-l to each other. For this reason, we start with the general operation of attaching via a map. 3.1. Attaching via Maps. The operation of attaching a space X to a space Y via a map
The natural projection Y -+ YU""X is always injective, while the natural projection X --+ Y U"" X is injective only if the map
3. Topological Spaces

118

Figure T. Attaching via a map

Theorem 3.30. II X and Y are normal topological- spaces, A c X is a closed set, and cp ~ A -+ Y is a continuous map. then the space Y Ucp X is normal. Proof. First, we prove that every point c ~ Y Ucp X is-a. closed set. If E p(X\A) or C E p(Y) \p(A), then p-l(c) consists of one point (belonging to X or Y). If C E p( A), then the preimage of c in Y consists of one point c, and the preimage of c in X is the set cp-l(c), which is closed because the map cp is continuous. Let Cl and C~ be disjoint closed subsets of Y Ucp X. Then the set C = Cl U C2 is closed, and the function ftC rl' 1 which takes the value o on Cl and the value 1 on C2 is continuous. Therefore, it is sufficient to show that any continuous function I; C -+ I I where C is 8 closed subset of Y Ucp X, can be extended to the entire space Y Ucp X. Let C c Y Ucp X be a closed set, and let I: C -+ I be a continuous function. Consider the closed sets Cx = p-l(C) nX and Cy = p-l(C) ny. C

11

On these sets, f determines the functions f x: CX .!. C !... I and Iy: Cy C L I. By the Tietze extension theorem, the function fy can be extended to a continuous function Fy: Y -+ I. On the set A, Fy determines the function I A: A :!.. Y ~ I. The continuous functions f X and I A coincide on Cx nA; therefore, they determine 8 continuous function IXA: ex uA -+ I. Now, it is time to use the closedness of A. We have to construct an extension to Y Urp X of the function iXA, which is defined on Cx U A, where Cx is a closed set. By assumption, the set A is closed; hence CxuA is closed as well. By the Tietze extension theorem, the function f X A can be extended to a function Fx: X -+ I. Moreover, if x E A, then Fx(x) = Fy(cp(x». Hence the functions Fx and Fy determine a function F on YUrpX; it is continuous, because Fx and Fy are continuous. By construction, FtC = f j thus, F is the required extension of f. 0 3.2. Definition of CW-COInplexes. A topologicai space X is called a CW-complex if X = U:O Xi, where XO is a discrete space and XHI is obtained by attaching the disjoint union UaEA D~+1 of (i + I)-disks to Xi

3. CW-Complexes

119

Figure 8. A space Dot homeomorphic to a CW-complex

r'l

s:.

via a continuous map cp: UaEA~ r where = 8IYa+1. Moreover, X must satisfy conditions (c) and (w) specified below. The images of D~+1 and int D~+1 under the natural projection onto XH~ c: X are called, respectively, closed and open cells of dimension i + 1. The conditions (c) and (w) are as follows: (c) each Closed cell intersects only finitely many open cells; (w) a set C C X is closed if and only if all intersections of C with closed cells are closed. Note that if the number of cells is finite" then conditions (c) and. tw) hold automatically. Condition (w) determines the topology of the space X if the dimensions of cells in X are unbounded. The symbols (c) and (w) are abbreviations for "closure finite" and "weak topology." The open cells are pairwise disjoint and cover the space X .. The space Xi is called the i-dimensional skeleton of the CW-complex X. IT a CW-complex X has cells of dimension r. and no cells of larger dimensions, then X is said to be an n-dimensional CW-complex. The natural map X~+l: D~+1 _ XHI c X is called the characteristic map of the cell. Example. Suppose that XO is Sl = 8D 2 with discrete topology, Xl = XO, and X2 = X is obtained by attaching D2 to XO via the identity map Sl _ S1'\( Then the space X 1!latisfies (w) but does not satisfy (c). Example. Let S! be the circle of radius 1/"" centered at (O,l/n). Consider the union X = U~l S! (see Figure 8) with topology induced by ]R2. The natural one-to-one map f: X - Y, where Y is the CW-complex consisting of one O-cell and the cells D! attached to it at both endpoints (n = 1,2, ... ), is not a homeomorphism. Proof. Take a point Xn different from the origin on each circle S!. Let F be the subset of X consisting of the points X n , where n = 1,2, .... It is not

120

3. Topologic81 Spaces

closed, because liII1n-+oo Xn = (0, 0) ~ F. On the other hand, the set I(F) is closed, because it intersects each closed I-cell in precisely one point. 0 One of the most important advantages of CW-complexes is that continuous maps of a CW-complex can be constructed by induction on skeletonsj extending maps defined on the boundaries of cells inside the cells by continuity, we necessarily obtain a continuous map I: X - Y of the entire CW-complex, because j-l(C) is closed if and only if its intersections with all closed cells are closed. Let X be a CW-coJIlplex. A subcomplel: of X is a closed subspace which is a union of open cells. Below, we give some of the most important examples of CW-complexes. The sphere 8 n is a CW-complex with one O-cell and one-n;.ocell. On it is easy to define the structure of a CW-complex with two cells of each dimension from 0 to n by induction: it suffices to attach the northern and southern hemispheres to the equator sn- l G The real projective space lRpn is the quotient of lRn+1 \ {OJ modulo the equivalence relation defined by x '" .xx for all .x E lR \ {OJ. Replacing lR by C, we obtain the definition of the complex projective space Cpn. To every point (Xl, ... , Xn +1) E lRn+1 \. {OJ corresponds the point (Xl: ..• : Xn +1) E lRpnj the numbers Xl, ... I Xn+1 are called the homogeneous coordinates ofthis point in lRpn. For Cpn, the notation is similar. The map (Xl: X2) ....... XdX2 is a homeomorphism between lRpl \ {(I : On and lRlj therefore, lRpl ~ 8 1 • Similarly, Cpl ~ 8 2 •

sn,

sn.

To introduce the structure of a CW-complex on lRpn, consider the map

j: Dn _ lRpn defined by I(x!, ... ,xn ) = (Xl:···; Xn: The image of the boundary 8 n - l

Jl- X~ _ ...

-r

x~).

c Dn is contained in

lRpn-1 = {(Xl: ... : Xn ; Xn+1) E lRpn : Xn+\ = OJ. The map

I

is a.homeomorphism of int Dn to lRpn \ lRpn-ll the inverse map

is given by

(Xl ~ ... : X n +1) ~ (.x-1XlXn+b ... , .x- I Xn Xn+1), where A > 0 and .x2 ::::;: x!+1 (x~ + ... + x~). Thus, JRpn is obtained from lRpn-1 by attaching one cell of dimension n. Similarly, Cpn is obtained from Cpn-l by attaching one cell of dimension 2n. We assume that D 2n = {(Zl, ... , zn) E C n : IZl12 + ... + IZnl2 ~·I}.

3. CW-Complexes

121

,,

Y = f(x)

x Yt

,,

,,

,,

Figure 9. The graph of the function /

Consider the map

J:

D 2n. - . CPR defined by

Jtzl, .. "Zn.) = (Zl: ... -; Zn.~ VI "\I Zl I2

-f"

--1Zn.1 2).

On CP", CPR-I, the inverse map has the form

)Zl : ... : Zn.+1)

1-+

(A-1zIZn.+1,."" ,A-1zn.Zn.+1),

where A > 0 and A2 = 1Zn.+112(lzlI2 + '" + JZn.12). Thus, CPR admits the structure of a CW-complex with cells of dimensions 2i for i = 0,1, ... , n. Problem 34. Prove that Cpa is obtained from 02n. C: CD by identifying each point x of 8D2n = 8 2n.-l with AX for all A E C such that IAI = 1. The same constructions as those used to obtain the CW-complexes 8n., IRPR, and CPR give ~he CW-complexes 8 00 , lRPOO, and CPOO. Problem 35. Prove that the space 8 00 is contractible. CW-complexes resemble simplicial complexes in many respects. It can be even proved that any CW-complex is homotopy equivalent to a simplicial complex (the proof of this assertion is given in, e.g., [99,112]). Bu.t there exist also CW-complexes which are not homeomorphic to simplicial com..., plexes. To construct an example of such a CW...complex, consider the contin-uous function on the interval [ = [0,1] defined by J(x) = xsin(7r/2x) for x >o and J(O) = 0 (see Figure 9); the image of [ under J is the interval [Yb 1]. Consider the map [2 - . 1R.3 given by (x, y) 1-+ (x, xy, f(y» (see Figure 10). In the plane x = 1, its graph coincides with that of f. In the plane x = c, where 0 < c ~ 1, this is the same graph shrank by a. factor of c in the y-direction. Finally, in the plane x = 0, the graph is the segment consisting of the points (0,0, z), where z E [Yh l].

3. Topological Spaces

122

z

y

Figure 10. The graph of the map of the square

a

b

Figure 11. The structure of neighborhoods of the points y.

Consider the CW-complex X whose O-cells are the images of the vertices of the square ]2 and the point (0,0, VI) under this map, I-cells are the images of the sides of the square and the segment of the z-axis between and VI, and 2-cell is the image of the square. It is easy to show that this CW-complex X is homeomorphic to no simplicial complex, Le., X is a nontriangulable CW-complex. Indeed, X is a compact topological space; therefore, a simplicial complex homeomorphic to X cannot have infinitely many vertices. On the other hand, all the points (0,0, Vi), where Yi is the value of f at a point of local maximum or minimwri, must be vertices of any simplicial complex homeomorphic to X because of the structure of small neighborhoods of these points. The two simplest cases of these neighborhoods are shown in Figure Ila. In the other cases, several half-planes are added; the additional half-planes are shown in Figure llb.

°

3.3. Topological Properties. CW-complexes have many good topolog" ical properties. Thus, any CW-complex is Hausdorff (and even normal); in CW-complexes, connectedness is equivalent to path-connectedness; any

3. CW-Complexes

123

CW-complex is locally contractible; any CW-complex is paracompact. In this section, we prove these and other properties of CW-complexes.

Theorem 3.31. Any CW-complex is a normal topological space. Proof. First, we prove that every skeleton xn of a CW-complex X is normal. For n = 0, this assertion is obvious: each point in the discrete space XO is both open and closed. The induction step is Theorem 3.30. Now, we prove the normality of X. Let C c X be a closed set, and let I: C -+ I be a continuous function. The function 1 determines a continuous function 10 on C n Xo, which can be extended to a function Fo on XO. The functions 1 and Fo determine a continuous function II on the closed set (C n Xl) U XO, which can be extended to Xl. Continuing this procedure, we obtain a function F: X -+ I which is continuous on each skeleton and, in particular, on each closed cell. Condition (w) implies the continuity of F. D Problem 36. Prove that any compact subset of a CW-complex intersectljl only finitely many open cells. In CW-complexes, connectedness coincides with path-connectedness; moreover, there is a fairly simple criterion for a CW-complex to be connected.

Theorem 3.32. (a) A CW-complex X is connected il and only il its 1. skeleton Xl is connected. (b) A CW-complex is connected if and only if it is path-connected. Proof. (a) If n ~ 2, then attaching the dis~ D n to the skeleton xn-l via a map sn-l ~ xn-l does not change the number of connected components. Indeed, for n ~ 2, the image of sn-l under a continuous map is connected; therefore, it is entirely contained in one connected component. Moreover, any space obtained by attaching D n to a connected space is connected. Clearly, if the skeletons xn of a CW-complex X are connected for n ~ 1, then X itself is connected, and if the xn are disconnected, then X is disconnected as well. (b) For I-dimensional CW-complexes, there is no difference between connectedness and path-connectedness. Thus, in the proof of assertion (a), we can replace "connectedness" by ''path-connectedness" because the sphere sn- 1 and the disk D n with n ~ 2 are both connected and path-connected. D A topological space X is said to be locally contractible if, for each point x E X and every open set U 3 x, there exists 8. contractible open set V such that x EVe U (the contractibility of V means that the identity map

124

3. Topological Spaces

Figure 12. The construction of the set V~+1

v

-+ V is homotopic to the constant map V -+ x). Local contractibility is very useful in the theory of coverings.

Theorem 3.33. Any OW-complex is locally contractible. Proof. Let X be a CW-complex. We construct a contractible neighborhood V of a given point x by induction on the dimension of the skeletons; this V will also satisfy the condition V C U. Each point E X is contained in a unique open cell int ~ int D';; . The set int e';; n U is open in the topology of xm. Let Vm be the open disk centered at x ofradius so small that V m C int e,;;nU, and let 1["': Vm --+ Vm be a homotopy between the identity map and the map Vm r l Vm that takes the entire disk Vm to the point x. Suppose that n ~ m and the required neighborhood Vn ot x in xn, together with the homotopy 1[', has already been constructed. Let us construct the neighborhood Vn+1 of x in X n+1 and the homotopy 1:+1. Suppose that x: Dn+1 -+ X n+1 is the characteristic map of some cell. Then Vn = X-I (Vn ) is a closed subset of sn c Dn+1 and U' = X-leU) is an open (in the topology of Dn+1) subset of Dn+1~ moreover, V~ C U', because V n C U. The set Vn is compact; therefore, for some E E (0,1), the set

x

e:

V:+1 =

{tv; 1- ~ ~ t ~ 1,v E Vn }

is contained in U' (see Figure 12). The set V~+1 = {tv: 1 -

E


~ 1, v E V~}

V:+I'

is open in Dn+1, and its closure coincides with It is easy to construct a homotopy between the identity map V~+1 -+ V~+1 and the natural projection from V~+1 to V~ treated as a map V~+l -+ V~+1 such that it is the identity on V~ and its image is contained in V~. After constructing

3. CW-Complexes

125

such neighborhoods V~+l and the corresponding homotopies for all (n + 1)cells, we obtain a neighborhood Vn+l of X in Xn+l such 'that V R+1 C Uj in addition, we obtain a homotopy between the identity map Vn+l --+ Vn+l and some map Vn+l --+ Vn+l such that its image is contained in Vn and its restriction to Vn is the identity map. Now, using the homotopy Jr, we can construct the required homotopy fJ'+ 1 . The set V = U~=O Vn is open in X, and the homotopies f? determine a homotopy between the identity map V - t V and the map V - t V taking the entire set V to the point x. 0 Theorem 3.34. Any CW-complex is paracompact. Proof. Let!l = {Ua : a E A} be an open cover of a CW-complex X. We construct a locally finite refinement V = {V,8 : 13 E B} of U by induction on the skeletons xn, where n = 0, 1, .... Namely, at the nth step, we construct an open cover {V,8,n} of xn so that V,8,o c V,8,l C ... and V,8 = LJ~=o V,8,n' Simultaneously, also by induction on n, we construct the index set B: at the nth step, we add the indices Bn which correspond to the sets V,8,n needed to completely cover xn \ X n - 1 . For n = 0, we put Bo = XOj for each 13 E Bo, the set V,8,o consists of one point 13 E Xo. For every 13 E Bo, we choose a(f3) E A such that 13 E Ua (,8)' Suppose that n ~ 0 and the index sets Bo, ... ,Bn , as well as the sets V,8,n for 13 E BoU' . 'UBn , have already been constructed. We construct sets V,8,n+l of two different types. First, for each 13 E BoU' . ·UBn , we extend V,8,n to a set V,8,n+i which is open in Xn+l and lies in Ua (,8). Some part of Xn+l \X n may be not covered by the sets V,8,n+l' To completely cover xn+l, we construct additional sets V,8,n+b where 13 E Bn+b so that each of them is contained in some V'a(j3). We assume that V,8,o = V,8,l = ... = V,8,n = {2} for 13 E Bn+l' We begin with the extension of V,8,n' Suppose that X: Dn+l - t Xn+l is the characteristic map of some (n + I)-cell, Vp,n = X- 1 (V,8,n), and U~(,8} = X- 1 (Ua (,8)' The set U' = U U~(,8) is open in the topology of Dn+lj therefore, Dn+l \ U' is closed. Moreover, sn = aDn+l c U' .. Hence, for some ~ E (0,1), the set {tv: l - c ~ t:::; l,v E sn} is contained in U'. Changing the map X if necessary, we can assume that c = 1/2. We put V;,n+l

= {tv:

1/2 < t:::; l,v E V;,n}.

The set V,8,n+l is defined as the union of all X(V;,n+l) over all (n + I)-cells. Clearly, V,8,n C V,8,n+l C Ua(,8) and V,8,n+l is open in xn+l. Now, let us construct the additional sets V,8,n+l' We again change the characteristic map X in such a way that c = 1/2. This means, in particular,

126

3. Topological Spaces

that if B = {tv ! 0 ~ t < 3/4, 'V E sn}, then the set Dn+l , B is already covered by the sets V;,n+1 extending the V;,h' It remains to cover B. The open sets int X- 1 (UaJ cover the compact set Bj hence we can choose finitely many' indices 01, ••• , Ok such that the sets int X-1(Ua .) cover B. We put Va ..n +l = X(B nX- 1 (Ua for i 1, .•. , k. We construct such sets for all (n+ I)-cells. Clearly, V,a n xn = V,a,nl so V,a is open. Moreover, if (3 E Bn and V,a,n C Ua(,a), then V,a C Ua(,a). Thus, it only remains to prove that the cover V = {U,a : {3 E B} is locally finite. First, we prove (by induction on n) that each Vn = {U,a,n : {3 E Bo U '" U Bn} is a locally finite cover of the skeleton xn. For n = 0, this is obvious. Suppose that the required assertion is proved for the skeletons of dimensions at most n. Take any point x E xn+1. Let X: Dn+ 1 - X n+1 be the characteristic map of the cell containing x. First, suppose that x belongs to the boundary of the (n+ 1)-cell X(Dn+1 ), i.e., x E xn. By the induction hypothesis, there exists a.n open set Wn 3 x in xn which intersects only finitely many sets V,a,n' We put

.»

=

W~+1 = {tv: 3/4

< t ~ l,v E X...1 (Wn )}.

The set W n+1 is defined as the union of X(W~+1) over all (n + 1)-cells containing x. The set Wn+l is open in xn+1" and it is disjoint from all of the sets V,a,n+1 with {3 E Bn+1 (such sets correspond to t < 3/4). Now, suppose that x is inside the (n+ 1)-cell xC D n +1 ). If x is the image of the center of the disk Dn+1, then we put W~+1 = {tV! 0 < t < 1/2,v E S"} and W n +1 = X(W~+1)j the set W n +1 is disjoint from all of the V,a,n+1 with {3 E BoU·' ·UBn • Now, suppose that x is not the image ofthe center. Let x, be the projection of the preimage of x onto S" from the center. By the induction hypothesis, there exists an open set Wn 3 x, in xn which intersects only finitely many sets V,a,n' We put W~+1 = {tv: 0 < t < 1,v E X- 1 (Wn )} and W n+1 = X(W~+1)' The set Wn+1 intersects only those V,a,n+1 with {3 E Bo U •.. U Bn for which Wn n V,a,n # 0. By construction, each (n + 1)cell intersects only finitely many sets V,a,n+1 with /3 E Bn+1' Therefore, Wn+t intersects only finitely many sets V,a,n+1' Now, take a point x inside an n-cell of X. Using the construction described above, we can obtain a sequence of sets Wn C W n+1 C • (. such that x E W m and W m is open in xm for any m ~ n. Moreover, each set Wm intersects only those V,a,m for which Wn n V,a,n # 0. Therefore, the set W = U:=n W m is open in X and intersects only those V,a for which Wn n Y,a,n =F 0. There are only finitely many such sets. 0 3.4. Cellular Approximation. Let X and Y be CW-complexes. A continuous map f: X - Y is said to be cellular if fer) c yn for all n.

127

3. CW-Complexes

Theorem 3.35 (cellular approximation theorem). Suppose that X and Y are CW-complexes, A C X is a subcomplex (possibly, A = 0), and I: X - Y is a continuous map whose restriction to A is cellular. Then there exists a cellular map g:...X - Y homotopic to I; moreover, the homotopy is the identity on A. Proof. We construct the map 9 and the homotopy by induction on the dimensions of cells u;: in X, considering each cell separately and extending maps already defined on the boundaries. of the cells. To construct the map g, it is sufficient to consider each cell e{J in Y, where m > n, and "squeeze" the image of u;: from this cell onto the boundary 8e'; in such a way that the map I does not change outside int e{J. Thus, the problem is as follows. Given a continuous map I: DI1, -4 Y and the characteristic map x~ Dm -+ Y, where m > nand l(sn- 1 ) C Y \ intX(Dm}, construct a continuous map g: DI1, _ Y with the following properties:

(1) if I(x) ¢ int X(Dm), then g(x) = I(x); (2) the map 9 is homotopic to I, and the homotopy is the identity outside intx(Dm)~

(3) g(Dn) C Y \ int X(Dm).

Step 1", There exists a map g: Dn ..... Y such that it has properties (1) and (2) and its image does not contain at least one point y E int X(Dm). Let D;' = {x E ]Rm ; Ilxll ~ ~} (we assume that Dm = Dr).. For 0.( E < 1, the disk D';' is homeomorphic to XeD';') c Y. To abbreviate the notation, we identify D:' with X(~) c Y. On the compact set (D3;4)' the lnap f is uniformly continuous; hence we can choose 6 > 0 such that if x,y E 1- 1(D3;4) Dn and YII < 5, then II/(x) - l(y)1I < 1/4. Consider a ~uffi.ciently fine triangulation bf the disk Dn (identified with an n-simplex), in which the diameter{)f each simplex is less than 6. If the image of a simplex from this triangulation under I intersects SV2"l = 8Dij2' then this image is entirely contained in D3;4 \ Dij". All the simplices of this triangulation are divided into three disjoint classes: (a) the simplices whose images are disjoint from sv;t;

,-1

e

nx -

(b) the simplices whose images are entirely contained in (c) the simplices whose images intersect

SV2"\

.sn;1.

We construct the map 9 and the homotopy separately for each simplex from the triangulation. In case (a), we set g(x) = I(x) for all points of the simplex. In case (b). we set g(v) = I(v) for all vertices of the simplex and extend the obtained map by linearityv For a simplex whose image intersects SV2"l, the situation is most complicated, because the map is already defined

128

3. Topological Spaces

Figure 13. The first step of cellular approximation

on some of its faces (namely, on those satisfying (a) or (b», and we must consistently extend this map over the entire simplex, For the vertices, we set g(v) = f(v). On a I-face, the pmp is either already defined or not. In the latter case, we extend the map from the endpoints to the entire I-face by linearity. If the map 9 is not yet defined on a 2-face, then we define it as follows. The 2-face d 2 can be covered by segments of the form [m, xl, where m is the barycenter of the simplex d 2 and x is a point on the boundary 8d2 . At the boundary points x, the map 9 is already defined. We set gem) = f(m) and extend the obtained map to [m, xl by linearity (see Figure 13). Then, the same construction is performed for 3-faces, and so on. Let die be a simplex in the triangulation of Dn.. Clearly, g(d k ) is con-. tained in the convex hull of f(d k ). In case (c), this convex hull does not intersect Dri4. Indeed, if Yo E f(d k ) n sri2"l, then f(dkl is contained in the disk of radius 1/4 centered at Yo, and this disk does not intersect Dri4. The homotopy ft between f and 9 is defined as follows. If f(x) = g(x)~ then we set hex) = f(x) for all x. If f(x) =I: g(x), then both points f(x) and g(x) belong to the disk D m , and we set hex) = (1- t)f(x) + tg(x). The intersection of Dri4 with the image of 9 is contained in a finite union of affine planes of dimension n < mj therefore, the disk Dri4 contains a point y not belonging to the image of g, as required. Step 2. There exists a map gl! D n

-+

Y with properties (1), (2), and (3).

According to Step l,y the map f can be replaced, by a map go whose image does not contain some point y E int X(Dm). Consider the composition gl of the map go and the projection from y onto the boundary of the disk (see Figure 14). It has property (3) and is homotopic to go; the homotopy between go and gl is defined by gt = (1- t)90 + tgl. 0 3.5. Geometric Realization of CW-Complexes. Let X be a CW-complex. A continuous map i; X -4 Rn is called an embedding if it is a homEl-' omorphism of X onto i(X).

3. CW-Complexes

129

Figure 14. The second step of cellular approximation

Theorem 3.36. For any finite CW-complex X of dimension n, there exists

an embedding of X into lR(n+1)(n+2)/2. Proof. The finite CW-complex X is compact; therefore, by Theorem 3.2 (see p. 89), any injective map X - lRN is an embedding. We prove the theorem by induction on n = dimX. For n = 0, the assertion is obvious. Suppose that we have constructed an embedding in_I: xn-l _lRN for the (n -I)-skeleton xn-l. Applying a translation if necessary, we can assume that ft in-I (Xn-l). Let us define an embedding in~ xn-l _lRN $lRn $lR by setting

°

in(x) = (in-l(X),O,O) E

lRN $lRn $lR

for x E X n - 1. Consider the n-cells Xa(n:!), where a = 1, ... , k. Each point of the disk n:! can be represented in the form tXa, where ~ t ~ 1 and Xa E 8:;-1 = an:!, i.e., IIxall = 1. At the points Xa , we have already defined the values in-l(Xa(Xa )); we denote them by in-l{Xa ) for short. At tXa En:!, we set

°

. (tXa ) = {(O,txa,o:) (2t - l)in-l (Xa), (1 - t)Xa, 20:(1 - t»)

't n

if t ~ 1/2, if t ~ 1/2;

these expressions coincide at t r 1/2. Let us show that the map in is injective. Suppose that in(tlXa) = in(t2Xp). For tl ~ 1/2 and t2 > 1/2, we have In-l(Xp) '" 0. If tI, t2 ~ 1/2, then, obviously, tl = t2, Xa = x{3, and 0: = p. If tI, t2 ~ 1/2, then the equality (1 - tl)Xa = (1 - t2)Xp implies tl == t2 (recall that IIxall = IIxpll == 1); therefore, Xa = xp and 0: = p. 0 Remark 3.7. The bound for the dimension in Theorem 3.36 can be made more precise; namely, any finite CW-complex of dimension n is embedded in ]R2n+1. The proof of this assertion is given in [60].

3. Topological Spaces

130

4. Constructions We have already encountered some constructions on topological spaces, such as direct product, wedge product, and the operation of attaching via a map. In this section, we discuss these and other constructions, as well as relations between them, in more detail. We are also interested in rendering these constructions simplicial {or cellular}t i.e., such that applying them to simplicial complexes (or CW-complexes) yields simplicial complexes (or CW-complexes). 4.1. Cartesian Product~ Recall that if X and Yare topological spaces, then the base of the topology of X x Y consists of the Cartesian products of open subsets of X and Y. Both projections px(x~y) = x and py(x,y) = y are then continuous. A product of two simplices of positive dimensions is not a simplex, but it is a rectilinear cell. Theorem 3.13 (see p. 100) shows that any product of two simplices can always be triangulated, i.e., represented as a simplicial complex~ The remark after this theorem allows us to construct this simplicial complex without employing addition~ vertices. If X and Y are CW-complexes, then the space X x Y can be partitioned into cells in a natural way. Namely, consider the cells
aDf1+q

~

(aDP x Dq) u (DP )( aDq).

Thus, the maps

(Dp+q,sr*q-l) _ (XP xyq,xP

X

yq-l UXP.... 1

X

yq).

The set X x Y is endowed with the product topology. If X and Y are finite CW-complexes, then the partitioning of X x Y into cells described above satisfies conditions (c) and (w), i.e., X x Y is a CW-complex. However, for infinite CW-complexes, condition (w) may not hold. Problem 37. Let sPy sq = (Sp x {*}) u ({*} x sq) c sP x sq. Prove that Sp x sq/sP V sq ~ SJ1+q. 4.2. Cylinders, Cones, and Suspensions. Let I = [0,1]. For a topological space X, the space X x I is called the cylinder over X. The cone over X is defined as the quotient X x I/(X x {1}} modulo the equivalence relation defined by declaring that Xl X {1} X2 ~ {l} for any Xl! X2 EX. The cone over X is denoted by ax. The suspension over X is the quotient space f"V

EX = X x 1/ (X x {1} U.X x {O}) =

ax/(x x {O})..

4. Constructions

131

Exercise 22. J;>rove that CSn ~ Dn+1 and

Esn ~ sn+1t

If X is a CW-complex and A is its sub complex, then X/A is a CWcomplex. Therefore, ex and EX are CW-complexes. Thus, the constructions of cylinders, cones, and suspensions are cellular.

4.3. Joins. The join X * Y of topological spaces X and Y is the quotient of )( >I. I X Y modulo the equivalence relation defined by declaring that (Xl, tl, Yl) "" (X2' t2,Y2) if either tl = t2 = and Xl = X2 or tr = t2 = 1 and YI = Y2· If X, Y c lRn and the segments [x, y], where X E X and Y E: Y, have no common interior points, then the join X * Y has a very simple geometric description: in this case, X * Y is the union of all segments [x, y].

°

Exercise 23. Prove that DP * Dq ~ Dp+q+1. Exercise 24. Prove that DO * X ~ singleton and SO is a two-point set.) Exercise 25. Given

X

ex and SO * X

ESP, Y E sq, and

(x, t, y) is a homeomorphism of SP

1-+

t

~ EX. (Here

DO is a

E [0,1], prove that the map

(cos ~t x, sin ~t

Y)

* sq onto Sp+q+i.

Problem 38. Prove that the natural embedding I ~ X -- X * Y defined by f(x) = (x,O,y) (the point (x,O,y) EX * Y does not depend on y) is homotopic to a constant map. If Clo, ••. , ap, bo, •.. ,bq are points in general t>osition in lRp +q +1, then the join of the simplices with vertices Clo, ••• , lLp and bo, ... ,bq is a simplex with vertices Clo, ••• , lLp, bo, ... ,bq • This remark shows that the join of disjoint abstract simplicial complexes A and B consists of the simplices of the form aU /3, where a is a simplex from A and /3 is a simplex from B. (Recall that a simplex of an abstract simplicial complex is merely a set of vertices.)

Theorem 3.37. If X, Y, and Z are finite simplicial 'complexes, then (X * Y) * Z ~ X * (Y * Z). 'Proof. Consider realizations of X, Y j and Z in lRn such that their vertices {xo:}, {y,B}, and fry} are generic points. Then both spaces (X * Y) * Z and X * (Y * Z) are homeomorphic to the union of all simplices with vertices Xio, ... , xip, Yio' ... , Yjq' Zko, ••• , Zkr , wherep,q,r ~ 0. 0 Suppose that K is a simplicial .complex, pEN, and 2 ~ j ~ p. The deleted join Jf(K) is defined as follows. Consider the p-fold join

3. Topological Spaces

132

JP(K) = K *.. ·*Kj a simplex of JP(K) is an ordered set (O"l," p,O"p) of simplices from K. Among all such sets we select those in which any j simplices have empty intersection. These sets are the simplices of the complex Jf(K). We denote the k-skeleton ofthe simplex /In by Skk /lnj the simplex /In has the natural structure of a simplicial complex. There is a construction based on the deleted join Ji which reduces the proof pf the impossibility of embedding the simplicial complex skn /l2n+2 in ]R2n (Le., the nonexistence of a homeomorphism of skn /l2n+2 onto a subset of ]R2n) tp the Borsuk Ulam theorem. Recall that, according to""Theorem 3.15 on p. 102, any finite n-dimensional simplicial complex can be embedded in ]R2n+1. The impossibility of embedding slen /l2n+2 in ]R2n was proved independently by van Kampen [65] and Flores [38]. Our proof mainly follows [46]. Suppose that K is a simplicial complex and f: K -+ ]R2n is an embedding (we denote the space IKI by K for short). Let CK be the cone over K. The map f determines a map I: CK -+ ]R2n+1 such that its restriction to K is one-to-one and /(K) n I(CK \ K) = 0. Consider the subspace k of C K x C K consisting of the products of all pairs of disjoint simplices in C K such that at least one of them belongs to K. It is easy to construct a homomorphism cp: k -+ Jj(K). Indeed, each point of k has a unique representation in the form (t1x1 +(I-t1)V, t2x2+(I-t2)v), where 'I} is the vertex of the cone CK, Xl and X2 are points belonging to disjoint simplices 0"1, 0"2 C K, and at least one of the numbers t1 and t2 equals 1. We set

cp (t1x1

+ (1 -

t1)V, t2x2 + (1- t2 ») V =

ift1 = 1; { ((X1'~'X2) t). X1.1- ~, X2 If t2 = 1.

On the spaces k and Ji(K), the natural involutions (a, b) +-+ (b, a) and (Xl, t,X2) +-+ (X2' 1 - t, Xl} are defined. The homeomorphism cp commutes with these involutions. For the map we define a map i: k -+ ]R2n+1 by setting i(a, b) = I(a) -/(b). This map anticommutes with the involution, i.e., j(a, b) = - i(b, a). Moreover, i(a, b) =/: 0 for all (a, b) E k. Indeed, suppose that (a, b) E k and I(a) = /(b). Then I(a) = I(b) E I(K), because one of the points a and b belongs to K. Hence both a and b belong to K, because I(K)n I(CK \ K) = 0. Finally, a = b, because the restriction of I to K is one-toone. On the other hand, by assumption, a and b belong to disjoint simplices. Thus, if the deleted join .li(K) is homeomorphic to ,s2n+1 and the natu• :tal involution becomes the symmetry with respect to the center of the sphere under the homeomorphism, then K cannot be embedded in ]R2n. Indeed,

I,

4. Constructions

133

_bO_C a

b

C

-a Figure 15. The deleted join of three points

if there were an embedding of K into ]R2n I then we could construct a map g: 8 2n+1 --+ ]R2n+1 \ {O} such that g(-x) = -g(x) for all x E 8 2n+1, which contradicts the Borsuk Ulam theorem.

Theorem 3.38. The space J?(skn ~2n+2) is homeomorphic to S2,*+1, and the homeomorphism transforms the natural involution into the symmetry with respect to the center of the sphere. Proof.. For n = 0, the proof is seen in Figure 15. The points a, b, and c constitute a copy of sko ~ 2 ; for the other copy we take -a, -b1 and -c.;. To obtain the deleted join, we connect a to -b and -c, and so on. It is easy to verify that the natural involution becomes the symmetry with respect to the center. This construction is generalized to an arbitrary n as follows. For one copy of skn ~ 2n+2 we take the n-skeleton of a simplex with vertices Vo, ... , V2n+2 in ]R2n+2 and place the origin at the barycenter of this simplex. The vertices vo, , .. , V2n+2 can be treated as vectors with sum zero. For the second copy of skn ~ 2n+2 we take the n-skeleton of the simplex with vertices -Vo, ••• , -V2n+2·

The required space Ji(skn ~2n+2) is obtained as follows. In each of the sets vo, ... , V2n+2 and -vo, ... , -V2n+2, we choose n+ 1 points in such a way that all the chosen points together have pairwise different numbers. The convex hull of these points is a (2n + l}-dimensional simplex. If the simplices thus obtained have no common interior point, then their union is the required space. First, note that none of the convex hulls under consideration contains O. Indeed, the equality E .AiVi = 0 can hold only if all numbers .Ai are equal, whereas each convex hull misses one of the vectors Vi. Let e denote the unit vector in ]R2n+2. We show that the ray {Ae: A 1). O} intersects the set under consideration in one point. Renumbering the vectors if necessary, we can assume that e = E O:iVi, where ao ~ a1 ~ . u. ~ a2n+2. Then e = e - an+l EVi = E(ai - a n +1)vi = Ef3ivi, where f3i ~ 0 for o ~ i ~ n, f3n+l = 0, and f3i ~ 0 for n + 2 ~ 1 ~ 2n + 2. We set

3. Topological Spaces

134

fJ' = - L~=o ,8i, ,8"

= L~:!';2 ,8i, and ,8 = ,8'

numbers ,8' and ,8" is nonzero, because e

=/:

+ ,8";

at least one of the O. The point

n ( ,8i) ,8'~ 2n+2 ,8i e Vi ) + -,8',8 L -, t L ,Vi = i=O,8 f3 i=n+2,8 {3 belongs to one of the convex hulls under consideration (if {3' = 0 or (3" = 0, then the corresponding term is assumed to be zero). After renumbering the vectors Vi, we can represent any point y from the convex hull under consideration in the form y = L OiVi, where- ai ~ 0 for o ~ i ~ n, an+l = 0, Oi ~ 0 for n+2 ~ i ~ 2n+2, and L lail = 1. Suppose that some point >.y with >. > 0 belongs to one of the convex hulls. Then >.y = L {3i vi, where L l{3il = 1 and among the numbers f3i there are at most n + 1 positive numbers and at most n+ 1 negative numbers. Clearly,

2n+2 (

~

,8, ) ai - ;

Vi = Y -

>.y

T

= 0;

thus, all numbers Oi - ,8i/ >. are equal to the same number 'Y, i.e., Pi :: >'(Oi - 1'). If 'Y > 0, then f30 5 .. j ~ {3n+l = -~'Y .< 0, and if'Y < 0, then f32n+2 ~ .• ~ ~ {3n+1 = ->''Y > O. This contradicts the assumption that no more than n + 1 of the f3i can be negative and no more than n + 1 of them can be positive. Therefore, l' = 0, i.e., {3i = >'a,. By assumption, >. > 0 and L lail == L 1f3i1 = 1; hence >. = 1, Thus, each ray {>.e : >. > O} intersects the set under consideration in precisely one point. Therefore, this set is homeomorphic to ,s2n+1. Moreover, we have proved that each point has a unique representation in the form L aiVi, where L lai I = 1, at most n + 1 of the numbers ai are positive, and at most n + 1 of them are negative. This means that the (2n + 1)-simplices under consideration have no common interior points. 0 Now, we give yet another example of calculation of deleted joins. Theorem 3.39 (see [118]) • .!f(~") ~ ;n+l(Skj_2 ~p-il). Proof. If n = 0, then ~ n = * (one point), and the simplices of the complex .!f(~n) have the form (O"h ... 'SO"p), where at most j -1 simplices 0"1 are singletons and all of the remaining simplices are empty. Thus, Skj~2 ~p-l = Jl(skj_2 ~p-l).

It is easy to verify that if A and B are arbitrary simplicial complexes, then .!f(A * B) ~ .P,'(A) * .!f(B). Indeed, in the definition of a join, all simplices from A and B are considered different (even if A = B). Clearly, if o,d,n b(3 = 0 for all a and {3, then the intersection of akl U blH ". "-1 ak, U blj is empty if and only if aklj"l' .. n akj ;:::: 0 and bl l n ... n blj = 0~

4. Constructions

1;:$5

Using the homeomorphism /In ~ In+l(/lO), we obtain

Jf(lln) ~ Jf(llo •... * llo) ~ r+i

(Jf(IlO») ~ r+l(Skj_2 llP-1).

0

Corollary 1. .!C(lln) ~ J"+l(sP- 2) ~ 8(n+l)(p-i)-1. Corollary ~. The space Jj(lln) is homotopy equivalent to a wedge of {en + 1)(j - 1) - l)-spheres. 4.4. Symmetric Product. Let X be a topological space. The spnmetric group 8 n acts of the space xn = X X ••• x X by the rule q(X1~' •. ,xn ) = (Xu(l) , ••• ,Xu(n)' The quotient space of xn by this action is balled the nth symmetric product of the space X and denoted by Sfn(Xh. Exercise 26. Prove that Sp2(JR) ~ {(x, y) E JR2 ~ Y ~ o}. Theorem 3.40. spn(82) ~ spn(cp1) ~ epn I Proof. Suppose that (a1: b1), .. , , (an: bn ) E Cp1 and n

n

II(llix - biY) = ~C1~xkyn-k. i=l

k=O

When a pair (lli: bi) is replaced by (Alli: AbiJ, all coefficients Ck are multiplied by An; hence the formula

(a1: b1), ... , (an: bn)) 1-+ (q,: ...Q: en) determines a map Cp1 x ... X Cp1 -+ Cpa. A permutation of (a1 ~ b1), ... , (an: bn ) does not change the point (CO: .•• : en); thus, we have a ~ap h: spn(cp1) -+ CP". Any polynomial in one variable decomposes into linear factors over the field C. Therefore, any polynomial E~=o ckxlcy"'-k where not all the numbers Ck are zero, can be represented in the form II~l (lliX - biY), where for each i at least one of the numbers lli and bi is nonzero. This means that the map h is surjective. This map is also injective because the coefficients ot any polynomial determine its roots up to permutation. Clearly, h is continuous. Thus, h: spn(S2) -+ Cpa is a continuous one-to-one map. The space spn(S2), being a continuous image of the compact space S'- x ... X 8 2, is compact. The space Cpa is Hausdorff because it is a CW-complex. Therefore, by Theorem 3.2 (see p. 89), the map h ls a homeomorphism. 0 Theorem 3.41. (a) spn(c) ~ cn. (b) spn(C \ {O}) ~ cn-1 X (C \ {O}). Proof.. (a) The map b t--tI (1 :J b) is an embedding of C int(). CP~ ~ 8 2, and it determines an embedding of spn(c) intb spn(cpl) ~ CP". To

3. 'lbpological Spaces

each point (bb' h, bn ) E SP"(C) it assigns the coefficients of the polynomial n~l(x - bi); therefore, the image of spn(c) is C n c CP". The map spn(c) _ Cn is a homeomorphism because the map spn(cplj - CP" is a homeomorphism. (A direct proof that the map spn(c} _ cn is a homeomorphism is given in [49].) (b) We must prove that if b1 , ••• , bn E C \ {OJ, then the coefficients of all polynomials of the form n~=1 (x - bi) form a set that is homeomorphic to C n- 1 X (C \ {O}). Clearly, the polynomial xn + Cn_lXn-1 + ... + CO has no zero roots if and only if CO i: O. Hence all but the last coefficient of the polynomial n~1 (x - bi) can be arbitrary. 0 Theorem 3.42. spn(JRP2) t:::: JRp2n. Proof. To each point JRp2 correspond two antipodal points of the sphere 8 2 • The stereographic projection takes a pair of antipodal points of 8 2 to z and _z-1 (or to 0 and 00), i.e., to points ofthe form (a: b) and (-b : a) in Cpl. To every unordered set of n points in

]Rp2

we assign the polynomial

n

(1)

I(x,y) = II(~x-biy)(-bix-~y). i=1

When a point (ai : bi) is replaced by (A~ : Abi) or (-bi : ai), the polynomial I is multiplied by IAI2 or -l~ respectively. Thus, I is determined uniquely up to multiplication by a nonzero real number. It is easy to verify that

(2)

I(-ii,x) = (-l)nl(x,y).

Clearly, every homogeneous polynoniial of degree 2n satisfying (2) can be represented in the form (1) because its roots can be arranged into pairs (a: b), (-b: a). For a polynomial of the form Ek=-n ckxn-kyn+k, relation (2) is equiv. alent to c~ :; (-1)kck. Thus, the polynomial I(x, y) is completely determined by the real coefficient CO and the complex coefficients CI, ••• , Cn. These coefficients are arbitrary (except that they cannot vanish simultaneously). Thus, the space of all polynomials I considered up to multiplication by a nonzero real number is homeomorphic to lRp2n. We have constructed a one-to-one continuous map h : spn(JRP2) lRP2n. The space spn(lRP2) is compact, and lRp2n is Hausdorff; therefore, h is a homeomorphism. 0 Remark 3.S. An interesting discussion of the properties of the homeomorphism h: spn(]Rp2) _lRp2n is contained in [9].

4. Constructions

137

In conclusion, we describe, without proof, the structure of the space spn(81 ). Let 8 1 ';(' D n be the space obtained from I x Dn by identifying the points (O,x) and (1, h(x», where h : Dn --+ D n is the symmetry about a hyperplane passing through the center of the disk (for h we can take any orientation-reversing homeomorphism). Then spn(81 ) ~ 8 1 X Dn-l for odd n and spn(81 ) ~ 8 1 ';(' Dn-l for even n. The proof of this assertion is given in [90j. Exercise 27. Prove that SP2(81 ) is the Mobius band.

Chapter 4

Two-Dimensional Surfaces, Coverings, Bundles, and Homotopy Groups

1. Two--Dimensional Surfaces 1.1. Basic Definitions. Let M2 be a tw
out boundary. Exercise 28. Prove that the tW
Figure 1. A pseudomanifold which is not a surface

-

139

14U

4. Two-Dimensional Surfaces

Figure 2. Invariance of the boundary

A two-dimensional surface with boundary is a topological space_homeomorphic to a two-dimensional pseudomanifold }.(2 in which every nonboundary point has a neighborhood homeomorphic to the open disk D2 and every boundary point a has a neighborhood homeomorphic to

in the latter case, the homeomorphism must take a to (0,0) ED!. To show that the boundary of a two-dimensional surface is well defined, we have to prove the following assertion. Theorem 4.1. If h: M2 _ N 2 is a homeomorphism of two-dimensional surfaces, then h(8M2) = .8N2. Proof. It is sufficient to verify that if a E M2 \ 8M 2, then h(a) ¢ 8N2. By assumption, the point a E M2 \ 8M2 has a neighborhood U homeomorphic to D2. Suppose that b = h(a) E 8N2. Then b has a neighborhood homeomorphic to D!, and the homeomorphism takes ~ to (0,0) ED!. Let W = un h- 1 (V) be a neighborhood of a. We identify U with D2 and V with D! and assume that h is a homeomorphism of W C D2 C ]R2 onto h(W) C D! C ]R2 such that h(O, 0) = (0,0).

V

For sufficiently small ~ > 0, the open set W contains all points Z E ]R2 for which liz - all ~~. Let 8 1 be the circle of radius ~ centered at a (see Figure 2). According to the Jordan theorem, the curve h(81 ) divides the plane ]R2 :::> D! into two connected components, one bounded and one unbounded. On the one hand, the point b = h(a) belongs to the image of the disk bounded by 8 1 , and the image of this disk is the bounded component. On the other hand, the set ]R:' = {(x, y) E ]R2: y <: O} does not intersect D!i therefore, ]R:' does not intersect h(81 ) either, and hence b belongs to the' unbounded connected component. 0

1. Two-Dimensional Surfaces

141

b

a

Figure 3. The structure of a neighborhood of a point on a twodimensional surface

000 1

1

1

a b c Figure 4. A triangulation of the circle

Any point of a pseudomanifold M2 contained inside a 2-simplex has a neighborhood homeomorphic to D2, and any point contained inside a 1simplex has a neighborhood homeomorphic to D2 or D~. Thus, to determine whether a given pseudomanifold M2 is a two-dimensional surface, it is sufficient to consider its vertices. Let v be a vertex of M2. Clearly, the union of all simplices with vertex v in M2 includes m sets of the form shown in Figure 3a and n sets of the form shown in Figure 3b. Any sufficiently small punctured neighborhood of v has n+m connected components. On the other hand, puncturing the sets D2 and D~ does not violate their connectedness. Therefore, the pseudomanifold M2 is a two-dimensional surface if and only if m + n = 1 for any vertex v, i.e., the union of all simplices with vertex v is either of the form shown in Figure 3a or of the form shown in Figure 3b.

1.2. Reduction of Two-Dimensional Surfaces to Simplest Forms. A triangulation of a topological space X is a homeomorphism X -+ IKI, where K is a simplicial complex. The complex K is also called a triangulation of X. As a rule, triangulations are very hard to construct, because every simplex of a triangulation must have no coinciding vertices and different simplices must have different sets of vertices. For example, the partitions of the circle shown in Figures 4a and 4b are not triangulations. A simplest triangulation of the circle is shown in Figure 4c; it has three vertices.

142

4. Two-Dimensional Surfaces

2

sphere

torus 3

3

Klein bottle

projective plane

Figure S. Triangulations of some surfaces

Figure 5 shows triangulations of the simplest two-dimensionJ surfaces (vertices with equal numbers are identified). We t1'egard these triangulations as abstract simplicial complexes (although t according to ';l'heorem 3.15 on p. 102, any two-dimensional abstract simplicial complex can be realized in the Euclidean 5-space). Note that a triangulation of the projective plane cannot be constructed in the same way as those of the torus and the Klein bottle (such a trian-, gulation would contain two different triangles with coinciding vertices; they are hatched in Figur~ 6a). The situation can easily pe improved by taking the secondary diagonal in one of the comer squares-, as shown in Figure 6b~ Looking at Figure 7, it is easy to see that removing a disk from the projective plane, we obtain the Mobius band. Let TA, K2} and p2 be the triangulations of the torus, Klein bottle, and projecti~ plane shown in Figure 5, and let p, q and r be nonnegative integers. We define a two-dimensional surface S2#p~#qK2#r p2 as follows. Consider a triangulation of the sphere S2 so fine as to contain p + q + t disjoint 2-simplices. We remove these simplices and identify p of the resulting triangles with the boundaries of p copies of T2 \ Ll2, where Ll2 is one of the siDJplices in T2; then, we attach q copies of K2 \ Ll2 and r copies of p2, Ll2 in the ~ame manner. Clearly, the two-dimensional surface obtAlned does

143

1.. TwcrDimensional Surfaces

a Figure 6. A triangulation of the projective plane

2

Figure 7. Mobius band

Figure 8. Klein bottle

not depend on the triangulation of the sphere 8 2 (up to homeomorphism). In a similar waYf we can define M2#N2 for any tw()-dimensional pseudo, manifolds M2 and N 2. Clearly, 8V'f:.M2 ~ M2 for any tw()-dimensional pseudomanifold M2. ,

<'

Theorem 4.2. 8 2#2P2 ~ K2. Proof~ The surface S2#2P2 is the cylinder S2 X J to both ends of which Mobius bands are attached. The Klein bottle K2 CaQ also be represented as a cylinder with two Mobius bands attached to its boundaries; the cylinder is hatched in Figure 8. q

4. Two-Dimensional Surfaces

144

Figure 9. The surfaces TJ#P2 and J(2#p 3 are homeomorphic

Proof. The surfaces 'I"l#P2 and J(2#p2 are shown in Figure~---
S2#pT2#qK2#rp2 is homeomorphic to s2#m'I"l or S2#nP2 j for brevity, we denote these surfaces by mT2 ana nP2, respectively (it is assumed that m > 0 and n > 0). Let M2 be a two-dimensional pseudomanifold with v vertices, e edges, and f faces. The Euler chamcteristic of M2 is defined as X(M2) = v-e+ f. TheorelD 4.4. Any closed two-dimensional surface M2 is homeomorphic to m'I"l or np2. The numbers m and n are determined by the relations X(M2) = 2 - 2m and X(M2) = 2 - n. Proof (see [125]). The edges of M2 form a graph. In this graph, we sequentially delete edges; after an edge is deleted, the faces incident to it merge into one domain, which we also call a face. We delete edges in such a way that the graph will remain connected and the number of vertices will not change. After the deletion of an edge, the number of faces either does not change or decreases by 1. At the end, we obtain a maximal tree with v vertices and v -1 edges (and one face). Under the deletions, the quantity number of vertices -

number of edges

+

number of faces

does not increase, and it equals 2 in the end; therefore, v - e + f $; 2. Suppose that there exists a closed two-dimensional surface which gives a counterexample to the theorem. Among all such surfaces we choose those for which the number 2 - v + e - f ~ 0 is minimal and among those, select the surfaces with minimal v. Finally, among the selected surfaces we choose a aurface for which the minimal degree of a vertex is minimal. We denote this surface by M2.

1. Two-Dimensional Surfaces

145

Figure 10. The boundary of the surface

Let A be a vertex of minimal degree pin M2, and let AA1 A 2, AA2A 3 , .. 1., AApAl be the faces incident to it. If p = 3, then either M2 is the surface of a tetrahedron or there exists a closed two-dimensional surface M2 obtained from M2 by removing A and replacing the faces AA 1 A 2, AA2A3, and AA3Al by one face A 1 A 2A 3. If M2 is the surface of a tetrahedron, then M2 ~ 8 2 and x(M2) = 2. Thus, both assumptions contradict the choice of M2j hence p ~ 4. Suppose that for some i, the vertices Ai and ~+2 are not joined by an edge. Then, deleting the edge A~+1 and adding the edge Ai~+2, we obtain a triangulated surface for which the numbers 2 - v + e - ! and v are the same but the minimal degree of a vertex is smaller. This contradicts the choice of the surface M2. Therefore, for each i, the vertices Ai and ~+2 are joined by an edge. The surface M2 contains the edges AAI, AIA3, and AA3 , but it cann~t contain the face AAIAa., because p ~ 4. Let us cut M2 along AA1 , A 1 A 3, and ,AA3 • As a result, we obtain a surface bounded by a graph with six edges and six vertices in which every vertex has degree 2. This graph has no multiple edges; therefore, it either consists of two triangles or is a hexagon (see Figure 10). To the boundary of the obtained surface we attach either two triangles or a (triangulated) hexagon. As a result, we obtain a two-dimensional surface M2. In the former case, we have i = ! + 2, e = e + 2, and v = v + 3, whence (1)

v - e+ i =

v - e +!

+ 2;

in the latter case, we have (2)

v-e+]=v-e+!+1. the quantity 2 .,.... v + e - ! decreases under the passage from

In both cases, M2 to M2; since it is minimal for M2, the assertion of the theorem must

146

4. Two-Dimensional Surfaces

hold for M2, i.e.,

(3) The surface M2 is obtained from }J2 by a fairly simple transformation. In the former case, M2 ~ M2#T'2 or M2 ~ M2#K2 (either a handle or a twisted handle is attached). In the latter case, M2 ~ M2#P2 .. Thus,

(4) It is easy to verify that if X(M2) = 2 - 2m' or X(M2} = 2 - n' (m' and n' are defined by (3», then X(M2) = 2-2m or X(M2) = 2-n (m and n are defined by (4»; it suffices to apply equalities (1) and (2) and the relations X(M2#T'2) = X(M2#K2) = X(M2) - 2 and X(M 2#p 2) = X(M 2}-=_1. 0

1.3. Completion of the Classification of Two-Dimensional Surfaces. To complete the classification of two-dimensional closed surfaces. it remains to prove the following assertion. Theorem 4.5. The sur/aces fP, m~ (m = 1,2, ... ) are pairwise nonhomeomorphic.

1, 2,.~.),

and nP2 (n =

M;

Proof. Suppose that and Mi are two-dimensional closed surfaces and hI M;- Mi is a homeomorphism. We show that X(M;) = x(Mi). There are two graphs on Mi, namely, the graph G2 formed by the edges of and the graph G I , the image of the graph formed by the edges of M;' Suppose that the graph GI has VI vertices, el edges, and II faces on Mi, and the graph G2 has V2 vertices, e2 edges, and h faces on Mi. We can change Gi so that the resulting graph have the same numbers of vertices, edges, arid faces and be piecewise linear, and its edges intersect the edges of G2 transversally. Consider the graph G == GI U G2. Suppose that it has v vertices, e edges, and / faces on Mi. We show that V - e + / = V2 - e2 + h. Consider any face of G2 (i.e., a 2-simplex of the pseudomanifold Mi). Let Va, ea , and /a be the numbers of vertices, edges, and faces of G contained in this face", According to the Euler formula, Va - eo +/0 : : :; 1 (the Euler formula has 2 rather than 1, but it takes into account the unbounded domain, which is absent in the case under consideration). We write Va and e a as Va = t1a+v: and eo = e~ + e:, where v~ and e~ are the numbers of vertices and edges contained in the boundary of the face under consideration and and are the numbers of interior vertices and edges. Clearly, v~ e~, and hence + /a = 1. This means that the deletion of all interior vertices, edges, and faces does not change the Euler characteristic. The equality v~ = e~ implies that the deletion of all vertices belonging to the edges of G2 does n~t change the Euler characteristic either. Thus, V - e + f = V7~ e2 + h.

MI

v: -e:

=

v:

e:

147

1. Two-Dimensional Surfaces

3 2 ~_ _-1-_-711

2 3 Figure 11. The orientation-reversing path on the projective plane

The graph G can be regarded as a graph on the surface M?, therefore, v-e+ f = vl-el +h, whence vl-el +It = v2-e2+ 12, i.e., X(Ml) = X(M~~. So~ the Euler characteristic of a two-dimensional ~urface does not depend o~ ~he triangulation, and to obtain the Euler characteristics for arbitrary triangulations of 8 2 , T2 J and p2, it suffices to calculate tha Euler characteristics for their simplest triangulations. It is easy to show that X(8 2 ) = ~, X(T2) ~ 0, and X(p2) = J for the simplest triangulations. It is also easy to verify that XCM2#N2) = XCM2) t X(N 2) - 2. Therefore, x(mT2) = 2 - 2m and x(nP2) = 2 - n. The Euler characteristics are equal only for the surfaces mT2 and 2mp2. Thus, i~ remains to prove that these surfaces are not homeomorphic (for m ~ 1). The surfaces $2 and T2 with simplest triangulations are orient able pseudomanifolds; hence m~ with SOllle triangulation is an orientable pseudomanifold. On the other hand, th~ surface p2 with simplest triangulation is a nonorientable pseudomanifold. Indeed, there is a closed path on f2 (namely, the path abca shown by the dashed lines in Figure 11) such that after it is traversed, any compatible orientations of the 2-simplices sharing the edge 23 become incompatible. On the surface nP2 (n ~ I} with some triangulation, there is a similar closed path: it has six edges, transversally intersects edges of six simplices, and reverses orientation. This means that nP2 (n ~ 1) is a nonorientable pseudomanifold. It remains to verify that the orientability of a two-dimensional surface is invariant under homeomorphisms; I.e., the orient ability of a surface can be defined without referring to triangulations. Suppose that M2 is a closed two-dimensional surface and -y: 1= [0,11M2 is a. path. We cover M2 by open sets Ui homeomorphic to ]R2. Specifying an orientation at one point {/; E Ui, we thereby specify orientations at all points of Ui; here, by an orientation we mean a direction of going around the point x.

4. Two-Dimensional Surfaces

148

Figure 12. Constructing surfaces by gluing together polygons..

The connected components of the sets 1-1 (Ui ) form an open cover of the compact set I. It has a finite sub cover W b ... , Wn . We assume that 0 E WI, WjnWj+1 :/: 0, and 1 E Wn • Given an orientation at a point x E--nWJ), we can extend it over all points of the set ')'(Wj). Since ')'(Wj) n ')'(Wj+1) :/: 0, we can also extend it over all points of ')'(Wj+I). In this way, we can transfer orientation from ')'(0) to ')'(1) along the path ')'. The result does not depend on the finite subcover of the cover of I by the connected components of the sets ,),-1 (Ui). Indeed, let us identify one of the domains Ui with JR2 and consider the part of the curve "'( contained in Ui JR2. The set Uj n Ui is open in ]R2. The transfer of orientation along a connected component of"'( n (Uj n Ui) by means of Uj gives the same result as the transfer of orientation along ')' in JR2 •1

=

We say that a two-dimensional surface M2 is orientable if the transfer of orientation along any closed path is orientation-preserving, i.e., anyorientation transferred along any closed path coincides with the initial orientation. Clearly, a. pseudomanifold homeomorphic to a two-dimensional surface is orient able if and only if this surface js orient able. This means, in particular, that the nonorientable pseudomanifold nP2 cannot be homeomorphic to the orientable pseudomanifold mTl. 0

Exercise 29. Prove that the surfaces nTl and nP2 can be obtained from a 4n-gon and a 2n-gon by identifying their sides as shown in Figure 12. Problem 39. (a) Prove that there exists a closed curve "'( on the surface nP2 such that nP2 cut along this curve is orientable. (b) Prove that if n is even, then a neighborhood of the curve'} is homeomorphic to the cylinder, and if n is odd, then it is homeomorphic to the Mobius band.

Problem 40. Let M? and Mi be nonhomeomorphic two-dimensional surfaces with boundaries. Can the spaces Mi x I and M? x I be homeomorphic? • 11£ some orientation is specified at one point of the set Uj at all points of this set.

Rl

R2, then it is thereby specified

149

2. Coverings

Figure 13. Curves on a two-dimensional surfa.ce

1.4. Riemann's Definition of the Surface Genus. Riemann defined the gen~ of a closed orient able two-dimensional surface M2 as follows. Suppose that on the .surface M2, there exist p self-avoiding closed curves C1, ..• ,Op such that they are pairwise disjoint and the set M 2 \(Ciu·· .UOp ) is connected, and let p be the maximum possible number of such curves. Then the surface M2 has genus p. We show that the number p thus defined indeed coincides with the genus 9 of the surface M2. It is easy to give an example of curves which shows that p ~ 9 (see Figure 13). It remains to prove that if there exist p curves C1, ..• , Op on the surface M2 such that the complement of their union is connected, then p $ g. Let us cut M2 along the curves 01, . .. ,01" We obtain a connected orientable surface whose boundary has 2p connected components. Pasting each component with a disk, we obtain a closed orient able surface M2 with Euler characteristic XCM2) + 2p = 2 - 29 + 2p. But X(M2) $ 2; therefore, p $ g. Problem 41. Prove that n pairwise disjoint Mobius bands can be arranged on the closed nonorientable surface nP2 ~ while n + 1 cannot.

2. Coverings We have considered fundamental groups and coverings in detail only for onedimensional complexes, but we have defined them for any path-connected topological spaces. Our proofs of properties of fundamental groups and coverings do not use special features of one-dimensional complexes; the only exception is the proofs of the existence and uniqueness theorems for the 1ifting of a path starting at a given point and for the covering corresponding to a given subgroup of the fundamental group of the base. For coverings of arbitrary path-connected spaces, the existence and uniqueness of the lifting of a path is not as obvious as for coverings of onedimensional complexes, but they are fairly easy to prove. Fot each point of a path 'Y, we take the neighborhood involved in the definition of a covering. Since the closed interval is compact, we can choose a finite cover of'Y by such neighborhoods. Using this finite set of neighborhoods and their preimages,

4. Two-Dimensional Surfaces

150

we construct a lifting of ~ with a given starting point. Obviously, this lifting is unique. For coverings corresponding to a given subgroup of the fundamental group, the proof is more complicated. The construction given in the proof of Theorem 1.23 on p. 42 uses essentially the structure of a one-dimensional complex. Moreover, for general spaces, the theorem is false; it is valid only under certain assumptions. Before stating and proving the general theorem, we consider the simplest case of the universal covering of a closed orient able two-dimensional sllrlace. 2.1. Universal Coverings of Two-Dimensional Surfaces. Recall that a covering p: X - t X is said to be universal if 7rl (.x) = o. The torus T2 can pe obtained by identifying the points (x-:t----m,y + n) and (x, y) for all pairs ~f integers m, n. Therefore, the universal covering of the torus has the form p: ]R2 -+ TJ., The simplest construction of the universal covering of the sphere with 9 handles, where 9 ~ 2, uses hyperbolic geometry. Oonsider a regular 4g-gon with angle on the hyperbolic plan~ H21 Let G be the group of motions of the hyperbolic plane generated by the translations such that each of them takes a side of the 4g-gon under consideration, to the opposite side. The images of the 4g-gon under the action of G form Q. tili,ng of the hyperbolic plane. Therefore, the map p: H2 -+ H2/ G ~ is the universal covering of the sphere M; with 9 handles. (The facts of hyperbolic geometry used in this construction can be found in [103].) A description of the geometric structure of the universal covering of which does not use hyperbolic geometry can be found in [69].

;g

M;

M;

Problem 42. (a) Prove that the universal covering space of the plane with several punctures is homeomorphic to ]R2.

]R2

(b) Let ~j = {(Zb~ •• ,Zn) E en I Zi = Zj}, and let I:; d en \ Ui~j .6.i j. Prove that the universal covering space of I:; is homeomorphic to en. 2.2. The Existence of a Covering Space with Given Fundamental Group. Let H be a subgroup of 7rl(X,XO). Before trying ~o construct a covering p; X -+ X for which P*7rl(X,XO) ;::;;; H; we determine the properties which X must have. Let 11 and 12 be paths from Xo to x in X, and let ~h and ::r2 be the liftings of these paths starting at xo· The paths ::rt and ::r2 end at the same point if and only if the class of the loop 1~ 121 belongs to H. This observation suggests the following construction of the space X. Let X be a path-connected space with base point xo EX, and let H be a sub~oup ofthe group 7rl (.~, xo). Consider the set of all paths starting at Xo in X. We say that paths 11 and "Y2 are equivalent if the class of the loop 11121

151

2. Coverings

Xo Figure 14. The path ')'lWW

1')'

belongs to H. The points of the space X are classes of equivalent paths; the topology on X will be defined later. The projection p: X -+ X takes each path '1 to its end. The map p is surjective because the space X is path-connected. This construction does not always give the desired result. But if the space X is locally path-connected and locally simply connected, i.e., for each point x E X and any neighborhood U :3 x, there exists a path-connected mmply connected neighborhood V ~ U, then this construction does give the ~equired result. In what follows, we assume that the space X is locally path-connected and locally simply connected. By a neighborhood of a point i,n X we mean a simply connected path-connected neighborhood. Clearly, such neighborhoods form a base for the topology of X, The topology of the space X. To define a topology ~n X, it is sufficient to specify a base of open sets. Suppose that a point E X and is. neighborhood U C X are such that px E U The point i is a. class of equivalent paths. Let ", be one of the paths (starting at xo) in this class. To the pair U, x we assign the set (U, x) G: X consisting of the equivalence classes of all extensions of '7 by paths entirely contained in U. Clearly, the set (U, x) does not depend on the choice of 'Y. Moreover, this set does not depend on the choice of x in the sense that if X2 E (U, Xl), then (U, X2) = (U, Xl). To prove this, take points Xl = p5h and X2 PX2 and let w be ~ path between them contained in U (see Figure 14). Suppose that 'Y11 is an extension of a path"'l between Xl and Xo bY_-$)me path Jy contained in U. Consider the path 'Ylww-l'Y, which is the extension of the path 'YlW between XQ and $2 by the path w- l ,), contained in U. The paths 'Y1')' and 'Ylww-l'Y are homotopic; therefore, 'Yl'Y I-r+ ,)'lWW-l'Y is a. one-to-one correspondence between (U,Xl) and (U, X2).

x

of

=

For the base of the topology of X we take all sets of the form (U, x). Let us verify that any nonempty intersection of two elements of this base contains a nonempty element of the base. Suppose that x E fj 0 V. where

152

4. Two-Dimensional Surfaces

fJ = (U,Xl) and V = (V,X2). We set W = unv and W = (W;x). fJ n V, and W belongs to the base.

We have

W =

The continuity of the projection p. The preimage of any (connected simply connected) neighborhood U consists of open sets from the base; therefore, it is open. The path-connectedness of X~_ Let x be a point of X, i.e., a class of equivalent paths. In this equivalence class, we choose an arbitrary path 'Y(t) and consider the family of paths 'Y8(t) = 'Y(st), where 0 ~ s ~ 1 and o ~ t ~ 1. Each path 'Y8 corresponds to some point xes) EX. Such points form a path between X(O) = Xo and xC!) = x in the space X. The projection is a local homeomorphism. Let p: (U, x) --+ U be the restriction of p to the set (U, x), where U is a connected simply connected neighborhood. The path-connectedness of U implies the surjectivity of p, and the simple connectedness of U implies the injectivity of p. To prove that p is continuous, consider any connected simply connected neighborhood V cU. Its preimage is (V, x). The image of the group 7I"i(X, xo) under the map p. coincides with H. Let 'Y be a loop in X based at Xo, and let ;:y be its lifting starting at xo. The subgroup P.7I"1(X,XO) consists of the homotopy classes of the loops 'Y for which the paths ;:y are closed. By construction, ;:y is closed if and only if the equivalence class of the path 'Y is xo, i.e., the homotopy class of 'Y belongs to H. 2.3. The Uniqueness of a Covering Space with Given Fundamental Group. The proof of the uniqueness of a covering space with a given group P.7I"1(X,XO) C7I"1(X,XO) does not use the local simple connectedness of X; it only uses the local path-connectedness of this space. The proof is based on the following lemma. Lemma. Suppose that q: Y -+ Y is a covering, f: X --+ Y is a (continuous) map, the space X is path-connected and locally path-connected, and f.7I"1 (X, xo) c:: q.7I"1 (Y, Yo). Then the map f has a unique lifting j: X --+ Y (i.e., there exists a unique map j such that qj = f and j(xo) = yo). Proof. Consider any path 'Y joining the point Xo to another point x in X. The map f takes "I to the path f'Y. Consider the lifting ;:y of f'Y starting at yo. If the required map j exists, then j(x) = y, where y is the end of::Y. Therefore, the map j is unique (if it exists). We must only verify that y does nQt depend on the choice of,. In other words, we must show that if 'Yl and 'Y2 are paths between Xo and x and w is the loop composed of 'Yl and 'Y2, then

2. Coverings

153

the lifting of fw starting at Yo is a closed path in Y, i.e., the class of the loop fw belongs to q*7l'1(Y,YO), which is equivalent to f.7l'1(X,XO) C q.7l'1(Y, Yo). This inclusion holds by assumption. It remains to prove the continuity of the map j. At this point, we use the local path-connectedness of X. Take x E X and let y = j(x). It is sufficient to prove that for any neighborhood W of y, there exists a neighborhood V of x such that V c I-I(W). Consider the neighborhood U of y = q(Y) involved in the definition of a covering. Let U be the path-connected component of q-I(U) nw that contains ii. Since the map f is continuous, f-I(Unq(W» contains some neighborhood V of x. The space X is locally path-connected; therefore, we can assume that the neighborhood V is path-connected. Then I(V) C U, i.e., V C I-I(U) c I-I(W). Indeed, any point Xl E V can be joined to X by a path 'Y contained in V. The image f'Y of'Y is contained in U n q(W); hence the lifting of f'Y is entirely contained in U. This meanS that j(y) = y E fl. 0 Using this lemma, we can easily prove the uniqueness of a ~overing space with a given fundamental group. Indeed, let Pi: Xi -+;X (i ~ 1,2) be coverings of a path-connected locally path-connected space X; suppose that (PI).7l'1(X1,Xl) = CP2)*7l'1(X2,X2). Then there exists a homeomorp~m h: Xl -+ X2 such that P2h = PI and h(XI) = X2. The map h is defined as the lifting of PI, and h -1 is the lifting of P2. It is seen from the proof of the lemma that there exists a unique lifting

j for any path-connected space X. But if X is not locally path-connected, then the map j constructed in the proof of the lemma may not be continuous. This is shown by the following example. Example (Zeeman). Consider the topological space X C ]R2 consisting of a circle, an arc AB, and an infinite set of intervals II, h, ... ; one endpoint of each interval is A, and the other endpoints converge to B (see Figure 15). Let Xl a~d X2 be the topological spaces shown in the same figur.!l' Coverings Pi: Xi -+ X twice wind the circles of the respective spaces Xi around the circle of X I are isometries on thf intervals, and homeomorphically map each arc to the arc of X. We have

(PI).7l'1(XI)

= CP2).7l'1(X2) = 2Z c

Z

= 71'1 (X),

although the map PI has no continuous lifting h for which P2h = Pl. Indeed, such a "lifting" h exists and is unique (provided that the image of one point under h is given), but it is discontinuous at the points P and Q, which belong to Pll(B).

4.

154

Tw~Dimensional

Surfaces

p

Figure 15. The Zeeman example

Problem 43. Prove that an mn-fold covering p: as a composition X~Y~X,

X ...... X

can be represented

where PI is an m-fold covering and P2 is an n-fold covering, if and only if the preimage p-l(x) of some point x E X can be partitioned into m-point sets 11, ... ,In SO that allliftings of any closed path in X which start in the same set Ii end in the same set Ij. The proof of the following assertion is based on the covering technique and uses the Borsuk Ulam theorem (although the Borsuk Ulam theorem can easily be derived from ~t). Theorem 4.6. II m > n ~ 1, then there exists no map g: ntpm inducing an isomorphism 01 fundamental groups.

sm

--+

Rpn

sn

Proof. Suppose that Pm: --+ lRpm and Pn; --+ lRpn are double coverings. Let us construct a map I: --+ for which gPm = Pnf. Take a point Xo E and choose a point Yo in the tw~point set p;;lg(xO)' We join Xo to a point x E sm by a path "1, consider the lifting of gPm"Y starting at Yo, and define g(x) = y, where y is the end of this lifting. The map 9 is well with m ~ 2 is contractible. defined, because any loop in

sm

sm

sn

sm

The equality gPm = Pnl implies I( -x) = ±/(x). The sign is selected as follows. Let am and an be the generators of the groups 71"1 (JR.PI'n) and 7I"1(lRpn) (recall that 7I"l(lRpl) = Z and 7I"l(:lRpm) = Z2 for m ~ 2). Suppose that g.am = ka n (here k E Z2 if n ~ 2 and k E Z if n = 1). Then I(-x) = (-l)k I(x), because ka n is the image of the arc joining x and -x under the map gPm' • Thus, if there exists a map g: lRpm --+ lRpn for which g.am, = ±an, then there exists a map I: 8 m --+ for which I(-x) = -/(x). 0

sn

155

2. Coverings

sn

Note that if there exists a map /: sm -+ for which fe-x) = -/(x), then, for the map g: ]Rpm -+ RP" defined by {x, -x} ...-+ {/(x), - I(x)}, we have g.am = ka n • where k is odd. If n ~ 2, then an E Z2, and g. is an isomorphisni. If n = 1, then g. is a nonzero homomorphism Z2 -+ Z, which does not exist. This argument gives a new proof of the Borsuk Ulam theorem for maps S2 _]R2. 2.4. Local Homeomorphisms. A map f j X -+ Y is called a local homeomorphism if each point x E X has a neighborhood U such that the set I(U) is open tn Y and the restriction of I to U is a homeomorphism. Any covering is a local homeomorphism. Under certain conditions, the converse is also true. We consider only the case of finite coverings. We say that a map f: X -+ Y is proper if the preimage of any compact set under this map is compact. Theorem 4.7 (see [55]). Suppose that X and Y are Hausdorff spaces and ¥ is path-connected. Then any proper surjective local homeomorphism f: X -+ Y is a finite covering. Proof. Take a point Y E Y. The set I-I(y) is discrete, because I is a local homeomorphism, and it is finite, because / is proper. Let I-I(y) = {Xl, , •• , x n }. The points Xi and xj (i :/= j) have disjoint neighborhoods Uij :3 Xi and Uji :3 Xj in the Hausdorff space X. Let Vi. = n#i Uij. Then ~ Ui and Ui n Uj 0 for i =f j. For each point Xi. we choose a neighborhood lti such that the set I(lti) is open in Y and the restriction of I to lti is a homeomorphism. The neighborhoods Wi = U, n lti are pairwise disjoint, and I homeomorphically maps Wt onto a neighborhood of y. We set W = n:=I/(W,). To prove that I is a covering, it suffices to show that the preimage of W is contained in U:=l I(W,), and to show this, it suffices to prove that the number n does not depend on the point y, i.e., the preimages of all points of Y contain the same number of points. We use the path-connectedness of Y. Take points YI, Y2 E Y and consider a continuous path "'(: [0,1] -+ Y joinins 'Y(O) Yl and "'(1) = Y2. We prove that the restriction of I ~o the preimage of "y is a covering. Consider the compact space Y "'([0, 1]) G Y. Let X = I-ICY}, and let j be the restriction of I to X. The spaces Y tmd X are Hausdorff, and j is a proper surjective local homeomorphism. Thus, for any point jj E Y, we can construct open sets Wi in the same way as the open sets W, for Y E Y. We put

x,

=

=

=

156

4. Two-Dimensional Surfaces

To prove that j is a covering, it is sufficient to verify that W is an (open) neighborhood of ii and j-l W C U~1 W;. By construction,

yE W. If j(x)

E

y

E n~l

iCWi )

and

j-1cy)

C

Lr=l Wi;

therefore,

iV, then x ¢ X \ U~=l Wi, i.e., x E U~-l Wi'

It remains to show that the set W is open, Le., j(X\Lh=l Wi) is closed. The space Y is compact, and the map is proper; hence X -= j-l(y) is compact. Thus, the set X \ Lr=l Wi is compact as a closed subset of a compact space. The set i(X \ lh-l Wi) is a compact suhset of the Hausdorff 0 space y; hence it is closed.

i

Theorem 4.7 gives a criterion for a local homeomorphism to be a global homeomorphism. Theorem 4.8 (see [56]). Let X and Y be path-connected Hausdorff spaces. A local homeomorphism I: X -+ Y is a (global) homeomorphism if and only if the map I is proper and the homomorphism I. ~ 11'1 (X, Xo) -+ 11'1 (Y, I(xo» is an epimorphism lor some point Xo E X A

Proof. The "only i£" part of the theorem is obvious. We prove the "if' part. It suffices to verify that the map I is one-to-one. Step 1. The map

I

is surjective.

Take arbitrary points Yo E I(X) and Yl E Y and consider a pa.th I = [0,1} -+ Y joining Yo and Yl. Since the map I is proper, the set l-l(a(l» is compact; hence f(X) n a(l) = I (J~I(a(1») is closed in a(l). On the other hand, the set I(X} is open in Y, because I is a local homeomorphism, and therefore I(X) na(l) is open in a(l). Thus, I(X) na(l) = a(1). In particular, Yl E I(X). 0::

Step 2. The map

J injective.

According to Theorem 4.7, I is a covering. Therefore, I.: 11'1 (X, xo) -+ is a monomorphism, and the number of elements in each fiber is equal to the index of the subgroup 1.11'1 (X, xo) in the group 1I'1(Y, I(xo». By assumption, I. is an epimorphism. Hence the covering I is one-fold, Le., f is a homeomorphism. 0

11'1

CY, J(xo)}

Corollary. Suppose that X and Y are path-connected Hausdorff spaces, 11'1 (Y) = 0, and I: X -+ Y is a local homeomorphism. Then I is a (global) hdmeomorphism if and only if it is a proper map.

3. Graphs on Surfaces and Deleted Products

157

12 3 4 5 6 7 1

3 4 567 1 2 3

Figure 16. The graphs K 6 , K7, and K4 ••

Figure 17. Eliminating the intersections of edges

3. Graphs on Surfaces and Deleted Products of Graphs 3.1. The Genus of a Graph. The graphs K3.3 and K5 cannot be embedded in the plane (see Theorem 1.2 on p. 7). It is easy to understand that a graph can be embedded in the plane ]R2 if and only if it can be embedded in the sphere 8 2 • We might consider embeddings of graphs in other surfaces. For example, the graph K6 can be drawn in the projective plane p2) and the graphs K7 and K4,4 can be drawn on the torus (see Figure 16). Theorem 4.9 (Konig [71]).( (a) Any finite graph G can be embedded in a closed orientable two-dimenswnal surface M2. (b) If the graph G is connected and the surface M2 has minimal genus, then each of the domains into which G divides M2 is homeomorphic to the disk. Proof. (a) Allowing intersections of edges, we can draw any graph on the sphere. The intersection of two edges can be eliminated by attaching a handle to the sphere and letting one edge remain on the sphere and the other go along the handle (see Figure 17). (b) It is sufficient to consider the case where G has no multiple edges. Let Ul, ... ,Um be the domains into which G divides M2. By assumption, the graph G is connected and has no multiple edges; therefore, the boundary of

158

4. Two-Dimensional Surfaces

each domain Ui is homeomorphic to the circle. A domain Ui is contractible if and only if, attaching the disk D2 to Ui 'along the boundary), we obtain the sphere 8 2 . Suppose that one of the domains U, is not contractible. If we remove Ui from M2 and attach D2 in its place, then the graph G will turn out to lie on the surface M2 of genus strictly less than that of M2. This contradicts the minimality of the genus of M2. 0 Remark 1. If a graph can be embedded in an orient able surface M2, then it can be embedded in the nonorientable surface M2#P2. Remark 2. If a graph G divides a. surface M2 into contractible domains, then the genus of M2 is not necessarily minimal. For example, the wedge of two circles admits such an embedding in both the sphere and the torus. The minimal genus of an orientable..surface in which a graph G can be embedded is called the genus of/his graph. The genus of a graph G is denoted by g(G). Theorem 4.10. Let G be a connected graph without loops and multiple edges containing v vertices and e edges. Then g(G) ~

moreover,

e

v

6 - 2" + 1;

il G contains no cycles 01 length 3, then g(G) ~

e

v

4 -"2 + 1.

Proof. Suppose that G is embedded in an orient able surface M2. We can assume that all domains into which G divides M2 are contractible. In this case, we have 2 - 2g(M2) = V - e + I, where I the number of domains into which G divides M2. If the boundary of each domain consists of at least n edges, then nl ~ 2ej therefore,

1(2-v+e (2)) n- 2 v 1-;; =2;te-"2+l.

2 g(G)~g(M)~"2

By assumption, the graph G has no loops and multiple edges. This means that n ~ 3; i.e., (n - 2)/2n ~ 1/6. If, in addition, G has no cycles of length 3, then n ~ 4, i.e., (n - 2)/2n ~ 1/4. 0 Example • g(K) > n Proof. The graph 9

(n-3)(n-4) 12 •

Kn

has

n(n2-1)

( K ) > n(n - 1) ..... n n

-

12

edges; therefore,

_ n 2 -7n + 12 _ (n - 3)(n - 4)

'2 + J -

E~ample. g(Km,n) ~ (m-2l(n-2).

12

-

12

0 .

159

3. Graphs on Surfaces and Deleted Products

Figure 18. The dual gra.ph

Proof. The graph Km,n has m+n vertices and mn edges. :Moreover, Km,n has no cycles of length 3. Hence 9

(K

) > mn _ m+n m,n -

4

2

+

1 = (m ...... 2)(n"T" 2). 4

0

The above bounds for the genera of the graphs Kn and K m •n canno~ be improved; that is, each graph Kn (Km,n) can be embedded in an orientable surface of genus equal to the least integer that is larger than or equal to (n-3I~n-4) (respectively, (m- 2l(n-2». Examples of such embeddings are fairlr complicated, especially for the graphs K'ft. The first examples for Kn were constructed by Ringel, Youngs, and Mayer in [83,107,108]. Examples of embeddings of the graphs Km,n were constructed in [106]. A modern exposition of these constructions is given in [45]. 3.2. Map Coloring. We say that a map on a surface M2 is n-colorable if the vertex of any (loopless) graph embedded in M2 is n-colorable, i.e., can be colored with n colors in such a way that any two vertices joined by an edge have different colors. Visually, the problem of coloring the dual graph, i.e., the graph whose vertices correspond to domains on the surface M2 and edges join vertices corresponding to domains with common edges (see Figure 18), is more obvious. For the dual graph, the coloring must be such that neighboring domains have different colors.

Theorem 4.11 (Heawood [52]). Any map on a closed orientable surface genus 9 > 0 is [7+~] -colombIe.

01

Proof. H e is an edge of a graph G, then any proper coloring" of the vertices of (} is also a proper coloring of the graph G - e. Therefore, adding an edge cannot decrease the number of colors required for coloring the vertices of the graph. Thus, drawing additional edges, we can assume that the graph G divides the surface into contractible triangular domains. In this case, we have 2e(G) 3/(G), and the equality v(G) - e(G) + I(G) == 2 .... 2g

=

M;

160

4. Two-Dimensional Surfaces

implies e( G) = 3v( G) + 6g - 6. Clearly, the sum of degrees of the vertices of G is equal to 2e(G)j therefore, the degree of one of the vertices is at most

2e(G) _ 6

(1)

v(G) , -

+

12(g -1) v(G)

•

Suppose that n is a number such that any graph on a surface of genus 9 has a vertex of degree at most n - 1 (an example of such a number is n = 7 + 12(g - 1)). It is easy to prove by induction on the number of vertices that any graph on a surface of genus 9 is n-colorable. Indeed, if the graph G from which a vertex v of degree at most n - t is deleted is n-colorable, then the graph G itself is n-colorable (there remains at least one color for the vertex v). Let neg) be the minimum number of colors sufficient for coloring an arbitrary map on a surface of genus 9 (the number nCg) is finite because any map on a surface of genus 9 is (7 + 12(g -t»-colorable). Consider a graph G for which this minimum number nCg) is realized and formula (1) holds. Clearly, v(G) ~ nCg); therefore, if 9 ~ 1, then ()

7

n g:5

12(g - 1) < 7 12(g - 1) v(G) - + neg) .

+

(Note that this inequality does not hold for the sphere.) Solving the inequality n(g)2 L 7n(g) :5 12g -12 and taking into account the positivity of neg), we obtain the required result. 0 The inequality n(g)2 - 7n(g) :5 12g - 12 can be rewritten as 9 ~

(n(g) - 3)(n(g) .... 4)

~

12

This inequality is closely related to the inequality 9

(K ) n

> (n - 3)(n - 4) 12

-

.

n"

Indeed. if the graph Kn is embedded in a surface of genus g, then neg) ~ because any (proper) coloring of Kn requires n colors. The examples of embeddings of Kn in orient able surfaces constructed by Ringel and other authors show that

neg)

~ [7+J~+489].

r

Combining these inequalities with the inequalities of Heawood, we obtain

n(g) =

+ J~ + 489 ] •

Recall that Heawood's argument does not apply in the case of 9 = O.

4. Fibrations and Homotopy Groups

161

3.3. Deleted Products of Graphs. The deleted product of a simplicial complex K is the subspace of IK x KI consisting of all products ~~ 1< ~~, where ~~ and ~~ are disjoint simplices in K. The deleted product of K hB$ the natural structure of a CW-complex. It has n 2 - n vertices, where n is the number of vertices of K. A graph G without loops and multiple edges can be viewed as a onedimensional simplicial complex with the same vertex set. Thus, for graphs, deleted products are defined also. If a graph has a pair of disjoint edges, then its deleted product is a two-dimensional CW-complex. In what follows, we assume that the graphs G under considerations have pairs of disjoint edges. Theorem 4.12. (a) The deleted product of a graph G is a closed two-dimensional pseudomanifold (not necessarily connected) if and only if the graph G from which any pair of vertices Vi and Vj joined by an edge is deleted, is a ~et of cycles, i.e., any vertex Vk f/ {Vi,Vj} is incident to precisely two edges not going to Vi or Vj. (b) If the graph G is 2-connected (i.e., it remains connected after the deletion of any vertex), then its deleted product is connected.

rt

Proof. (a) Suppose that a vertex Vk {Vi,Vj} is incident to edgeS VkVPll •• '1' VkVp., where v p1 , •.. " vPII {Vi, Vj}. Then, in the deleted product of G, the edge Vk x ViVj is incident to the faces VkVpa X ViVj, where a = ~, .. '" s. Therefore, the deleted product is a closed pseudomanifold if and only if s = 2 for all triples of vertices {Vi, Vj, Vk}. (b) Vertices Vi x Vj and vp x Vq can be jotned by an edge as follows. If p:f j, we join Vi x Vj with vp x Vj, and then, vp x Vj with vp x v q . To join vertices Vi x Vj and Vj x Vi, we choose a vertex Vk {Vi, Vj} and join first Vi x Vj with Vk x Vj, then Vk x Vj with Vk x Vi, and finally Vk x Vi with Vj x Vi. 0

rt

rt

The conditions of Theorem 4.12 hold for the graphs Ks and K3,3. Therefore, the deleted products of these graphs are connected closed surfaces. Problem 44 ([119]). Prove that the deleted product of the graph K 3 ,3 is a sphere with four handles, and the deleted product of the graph Ks is a sphere with six handles. Problem 45. Prove that the deleted product of a graph cannot be a sphere with an odd number of Mobius bands.

4. Fibrations and Homotopy Groups The notion of a locally trivial bundle with fiber F is a generalization of the notion of a covering. For a covering, the fiber is discrete, and for a locally trivial bundle, it can be an arbitrary topological space. To discuss

162

4. TwcrDimensional Surfaces

properties of bundles, we need the notion of a. homotopy group I which is a. generalization of that of a fundamental group. 4.1. Covering Homotopies. A locally trivial. fiber bundle is a quadruple (E,B,F,p), where E, B, and F are topological spaces and p: E - B ~ a map with the following properties: • each point x € B has a neighborhood U for which p-l(U) ~ U ~ Fi • the homeomorphism U x F the diagram

i'"'-+

p-l(U} is compatible with p, i.e.,

U x F_p-l(UY

~/. u is commutative (here U ~ F 4 U is the projection onto the first factor). The spaces E, B, and F are called, respectively, the total space~ base, and fiber of the bundle, and p is its projection map. The map p: E - B is also called a fibration. Example. The covering p: F = p-l(x), where :x; E X.

X-

X is a locally trivial fibration with fiber

Example. The natural projection p: B x F r+ B is a locally trivial fibration. This fibration is called trivial. Fibrations PI: El - B and P2: & - B are equivalent if there exists a homeomorphism h: El - E2 such that PI = P2h. A fibration equivalent to a trivial fibration is called trivial as well. Theorem 4.13 (Feldbau). Any locally trivial fibration over the cube I" is trivial.

=

Proof, First, let us show that if the cube 1" 1"-1 X I is partitioned into two semicubes I'!.. = 1"-1 x [O,!] and = Ik-l x [!,1] and a bundle is trivial over each of them l then this bundle is trivial over the entire cube lie. In other words, we must prove that any homeomorphisms h±: p-l(I~J - F x Ii compatible with the projection determine a homeomorphism h: p-l(I") _ F x .fk which is also compatible with the projection. Compatibility with the projection means that if y E p-l(x), then hey) = (f,x) for f E F, i.e., the homeomorphism h is a family of homeom9rphisms
Ii

4. Fibrations and Homotopy Groups

It):i

x E I!:. For each point a E I!: n Ii = lie 1 X {!}, we have two homeomorphisms p-l(x) -+ F, namely, 'Pa,+ and 'Pa,-. Consider the homeomorphism 1/Ja = 'Pa,-('Pa,+)-l;. F -+ F. We use it to define homeomorphisms 'Pz for x E Ii as follows. Let a(x) be the orthogonal projection of x E Ii on the partition Ik-l X {!}. We set 'Pz = 1/Ja(z)'Pz,+o This definition is consistent with the definition of CPa,+ and CPa,-, because~ at the point x = a E I k- 1 x {!}, we have 1/Ja'Pa,+ = 'Pa,-('Pa,+)-l'Pa,+ = 'Pa,-. Thus, the family of homeomorphisms 'Pz continuously depends on x and, hence, determines a homeomorphism h: p-l(Ik) -+ F X lie compatible with the projection. Now, the required assertion is easy to prove by contradiction. Indeed, suppose that there is a nontrivial locally trivial bundle over the cube Ik. Cut the cube lie into two semicubes. According to what we have proved, the bundle is nontrivial over one of the semicubes. We cut this semicube, and so on. As a result, we obtain a sequence of parallelepipeds such that their diameters tend to zero and the bundle is nontrivial over each of them. This sequence of parallelepipeds converges to some point Xo. By definition, the point Xo h~ neighborhood over which the bundle is trivial. One of the parallelepipeds is entirely contained in this neighborhood) so the bundle must be trivial over this parallelepiped. This contradiction completes the proof. 0

r

:Let p: E ....... B be a locally trivial fibration, and let f; X -+ B be a map. We say that a map j. X -+ E covers f, or IS a lifting of f, if pj = f. Fibrations have a property similar to the existence of a lifting of a path for coverings. The main difference is that for coverings, the lifting of a path with a given starting point is unique, while for fibrations, only the existence theorem is valid. Theorem 4.14 (covering homotopy). Suppose that p: E -+ B is a locally trivial fibration, X is a CW-complex, X' C X is its subcomplex, and the following maps are given: • a map h: X -+ E; • a homotopy H: X x I -+ B 0/ the map h = ph; • a homotopy ii': X, x I -+ E covering the restriction of the homotopy H to X, x I and extending the restriction of h: X x {O} -+ E to X, x {O}.

Then there exists a homotopy ii: X x I -+ E which covers the homotopy H and is a simultaneous extension for the homotopy ii' and tht! map h~ X x {O} -+ E. Proof. First, ,suppose that the fibration is trivial, j.e., E = B x F and pCb, f) = b. In this case, the map Ii is defined as the prod\lct of two maps,

164

4. Two-Dimensional Surfaces

Figure 19. The projection of the cylinder

to B and to F. The map to B coincides with H; it remains to ·define the map to F. This is done with the help of the following lemma, which is also useful in many other situations.

Lemma (Borsuk). Suppose that X is a CW-complex, X, c X is its subcomplex, and f: X -+ Y is a map. Then any homotopy F': X' x I -+ Y of the map f' = f X' can be extended to a homotopy of the map f.

r

Proof. We extend the homotopy by induction. Suppose that Xo E XO, i.e., Xo is a vertex. If Xo E X', then the map {xo} x I -+ Y is defined, and if Xo f/. X', then we send {xo} x I to the point f(xo). Now, suppose that the homotopy is extended over the skeleton X R , where n ~ O. Then, for each (n + I)-cell, we have a map defined on SR x I and on DR+l x {O}, and we must extend it over D R +I x I. To this end, we embed the cylinder DR+! x I in RR+2 and choose a point 0 on the axis of the cylinder above its top (see Figure 19). Let x -
El = ((e,y) E E x Y: pee) = H(y)} and Pl(X, y) = p(x). It is easy to verify that the induced fibration is locally trivial. Moreover, if ii' is a homotopy on a sub complex X, c DR which covers H, then it determines a homotopy ii~ : X' x I -+ EI by the formula iiHy) = (ii'(y) , y); we have pii'(y) = H(y) because ii' covers H. In a similar way, from the map h: X -+ E, a map hI: X -+ EI is constructed. • The base of the fibration PI is homeomorphic to DR+!. According to the theorem of Feldbau, the fibration over DR+! is trivial. Therefore, as in

4. Fibrations and Homotopy Groups

165

the case considered above, the maps ii~ and hI from Y to El = Y x F are determined componentwise by two maps, to Y and to F; moreover, the map to Y is the identity, i.e., ii~ and hI are determined by maps to F. Thus, we have the map X - t F corresponding to hI and a homotopy of its restriction to X'. Extending this homotopy to the entire space X, we obtain a map iil";..Dn x 1- El. The required covering homotopy ii is the composition of HI and the natural projection EI -+ E. Finally, consider the last case, in which the fibration p: E -+ B and the pair (X, X') are arbitrary. We construct the homotopy by induction. On the O-skeleton, the homotopy is given at some points; at the other points, we define it to be constant. In passing from the (n -' I )-skeleton ~o the nskeleton, we must extend a homotopy given on aDn to D n,. We have already learned how to do this. 0 Problem 46. (a) Prove that if Y c X is a contractible sub complex, then X/Y X. (b) Let Y be a sub complex of a CW-complex X. Prove that X/Y XUCY. f'V

t'V

Problem 47. Prove that an n-connected CW-complex is homotopy equivalent to a CW-complex having precisely one vertex and no k-cells for I ~ k~n.

Problem 48. Prove that if a CW-complex X with one vertex has no k-cells for 1 ~ k $' n, then it is n-connected. Problem 49. Given connected CW-complexes A and B with base points ao and bo, prove that A ... B,.." E(A A B), where A" B = A x BfA V Band A V B = ({ao} x B) U (A x {boll. Problem 50. Given an n-connected CW:"complex X and an m-connected CW-complex Y (both of them finite-dimensional), prove that (a) EX is an (n + I)-connected complex; (b) X A Y is an (n + m + I)-connected complex; (c) X ... Y is an (n + m + 2)-connected complex. Problem 51. (a) Prove that E(8 I x 8 1 ) 8 2 V 8 2 V 8 3 • (b)-Prove that if X and Y are CW-complexes, then E(X x Y) '" EX V EY V E(X 1\ Y). f'V

Problem 52. Prove that a locally trivial fibration p: 8 n -+ B whose base B consists of more than one point is not homotopic to a constant map. It follows from the covering homotopy theorem that for any locally trivial fibration p: E '-t B, a path "y joining points a and b in the base B

166

4. Two-Dimensional Surfaces

induces a map of fibe~ p-l(a) -+ p-l(b), which is defined up to homotopy. First, we construct the map itself. We set Fa = p-l(a). To apply Theo.rem 4.14, we take X = F", X' = 0, h;:: idx, and H(x~t) "Y(t). According to this theorem, there exists a homotopy H: Fax I t-t E that covers the homotopy H and is an extension of the map h~ Fax {O} 4 E. For this homotopy, we have pH(x, t) = "Y(t). Therefore, pH(~, I} 'c=. "Y(I} =: b, i.e., pH(x,l) E Fb. The requ~ed map is defined by x 1-+ H(x, I}. This m~p depends on the choice of H. Let us show that for any two homotopies Ho and HI cO,EStructed for homotopic paths "Yo and "Yl, the mapS-$~ Ho(x, 1) and x,...... HI (x, 1) are homotopic.. We again apply Theorem 4.14. This time, we have two parameters, the path parameter t and the homotopy parameter T, because we have the family of paths ~'l"(t). We take X = Fa )( IT~ X, == Fax {O}UFa X {I}, h(y,T) = y for all T E IT and y E Fa, H(y,r,t)-= "YT(t), H'(y, 0, t) = Ho(y, t), and H'(y, 1, t) = H 1 (y, t). By Theorem 4.14~ there exists a homotopy H(y, r, t) that co~ers the homotopy 11 and is a simultaneous extension for the homotopy H' and the map h: X x {O} -+ E. The required homotopy G: Fax IT -+ F" is defined by G(y,r) = H(y,r,I).

=

4.2. Homotopy Groups. The homotopy groups 7r"(X,xo} are a generalization of the fundamental group 1rl(X, xo). We first define the sets 1r"(X, xo) for n ~ 0, and then endow each, set '"(X, xo) (for n ~ 1) witf a group structure. We mark a point So in the sphere 8" and say that two maps (8",80) -+ (X, xo) are equivalent if they are homotopic (Le., joined by a homotopy ht: 8" -+ X such that ht(so) = XI} for all t ~ [0,1]J. The set 1r"(X, xo) consists of equivalence classes. In particular, the elements of 1ro(X,xo) are the path-connected components of the space X. The maps (8", so) -+ (X, xo) are called n-spheroidsj sometimes, it is convenient t~ represent such maps as (1",81") -+ (X,xo) or (D",8D") -+ (X~xo) (recall that 1"181" ~ D"18D" ~ 8"). To make the set 1rn (X, xo) into a group, we must construct a map 19: (8", so) -+ (X, xo) for any two maps 1,9: (8", so) -+ (X, xo). The construction is shown in the top part of Figure 20; the bottom part of Figure 20 illustrates the same construct jon for maps (1",81") -+ (X,.xo)~ For n ;::: 2, the order in which two maps J and 9 are multiplied does not matter, because the maps 19 and 91 are homotopic. The homotopy is easy to construct by looking at Figure 21. For any map I: (1",81") -+ (X,xo), there exists a map i: (r,8I")-+ (X, xo) such that the product I] is homotopic 'to a constant. The map! can qe defined, e.g., by representing the cube P in the form I" ;:::. [n-l x [-1,1] and setting ](x,s) = I(x,-s). For this ]. we have 1](Xc,s) = 1](;1;, -8)

4. Fibrations and Homotopy Groups

1 - - - -_ _--1

1(:)7

~ x ---g

Figure 20. The product of two spheroids

Figure 21. The commutativity of the homotol7 group

I S

IJ.IW.IlllllWlWWlWIWI..4 -

1 Figure 22. The inverse

elemen~

(see Figure 22). Consider the family of maps

(X 8)

gt,

=

{/~(X'8) II(x,t)

if lsi ~ t, if lsi ~ t.

;..

Clearly, go = II and gl is a constant map. Exercise 30. (a) Prove that if 1 "" h and 9 '" gl, then Ig '" /191(b) Prove that J(gh) '" (fg)h. In considering homotopy groups 1I"n(X,XO) for n ~ 1, the space X is usually assumed to be path-connected. In this case, the groups 1fn(X,JCo) and 1I"n(X, :1:1) are isomorphic, but the isomorphism is not canonical: it dependS on the path from Xo to Xl \ For a given path a between Xo and Jl:l, the

168

4. Two-Dimensional Surfaces

f

...

-~-

X

Figure 23. Changing the base point

isomorphism 1I"n(X,XO) -+ 1I"n(X, Xl) is constructed as follows. Given a map f: (sn, So) -+ (X, xo), consider the map sn -+ sn V I which takes the equator to So and each section of the southern hemisphere by a plane parallel to the equator to a point of the interval I; the southern pole is the base point So, and it is mapped to the free endpoint of I (see Figure 23). The composition of this map and the map sn V I -+ X coinciding with f on 8 n and with it on I is an element of the group 1I"n(X, Xl). This element depends only on the homotopy classes of the map f and the path it (we assume here that homotopic paths have the same start and end points). It is easy to verify that the paths a and a-I induce mutually inverse maps. In particular, if a is a loop at Xo, then a induces an automorphism of the group 1I"n(X,XO). This automorphism depends only on the element of 11"1 (X, xo) represented by the loop a. If each element of 11"1 (X, xo) induces the identity automorphism of 1I"n(X, xo), then the space X is said to be n-simple. If a space X is n-simple, then the groups 1I"n(X,X), where X E X, are canonically isomorphic to each other for this n, and we can denote them by 1I"n(X). Problem 53. Let X be a CW-complex, and let xn be its n-skeleton. Prove that the embedding i: xn -+ X induces an isomorphism i.: 1I"k(xn,XO) -+ 1I"k(X, xo) for k < n and an epimorphism for k = n. To solve the following two problems, the relation 1I"n(sn) = Z (n E N) is needed (see p. 235). Problem 54. Prove that for n ~ 2, the group 1I"n(sn V 81, xo) is a free Abelian group of (infinite) countable rank. Problem 55. Prove that the space sn V 8 1 is not n-simple. 4.3. Exact Sequences for Fibrations. Let p: E -+ B be a locally trivial fibration with path-connected base B. Choose a base point bo E B. The map p induces a homomorphism p.: 1I"n(E, eo) -+ 1I"n(B, bo), where eo E p-l(bo). LeF i: F --+ E be the composition of the homeomorphism F ~ p-l(bo) and the embedding p-l(bo) C E. The map i induces a homomorphism

4. Fibrations and Homotopy Groups

169

-L8 Figure 24. Representation of a map as

&

homotopy

i.: 1I"n(F, eo) - 1I"n(E, eo); here and in what follows, we identify the fiber F with p-I(bo). There is also a third homomorphism 8. : 11"n (B, bo) - 11"n-I (F, eo), which is defined as follows. A map I: (sn, so) - (B, bo) can be represented as a homotopy CPt: (sn-l, so) - (B, bo) between two copies CPo, CPI of the constant map, CPo, CPI: sn- I - bo (see Figure 24). According to the covering homotopy theorem, there exists a homotopy tPt: sn- 1 - E for which <po(sn-I) = eo, tPt(so) = eo, and PtPt = CPt (see Figure 25). Clearly, tPI(sn- 1 ) c p-l(bo) = F, because CPI(sn- 1 ) = boo For 8.1 we take the homotopy class of the map tPI: (sn- 1! so) - (F, eo). We must only verify that this map is well defined, i.e., homotopic maps J and If correspond to homotopic maps tPl and tPl' This can easily be done by afplying the covering homotopy theorem once more. Recall that a sequence of group homomo~hisms il';. G 11';'-1 G '" --+ Gi --+ i-I --+ i-2 --+ ... is said to be exact if Kercpi-I = Imcpi, where Kercpi-I = {g E Gi-I : CPi-1 (g) = O} (this is the kernel of the homomorphism CPi-l) and 1m CPi = {cpi(g) : 9 E Gi } (this is the image of the homomorphism CPi). Theorem 4.15. The sequence 01 homomorphisms

•.• --+1I"n(F,eo) ~ 1I"n(E, eo) ~ 1I"n(B,bo) ..!!!..... 1I"n-I(F,eo) - ' " is exact. Proof~ We must prove six inclusions of the form 1m i. c Ker P., Ker p. c 1m i., etc. We prove them separately. When there is no risk of confusion, we do not mention base points and identify the elements of homotopy groups with the spheroids representing them. (1) Imi. C Kerp•. Any spheroid belonging to Imi. has a representative f: sn ---+> E whose image is contained in F. The map pi is constant; therefore, I E Ker P._

4. Two-Dimensional Surfaces

170

1

5

Figure 25. Lifting of the homotopy

(2) Ker P. C 1m i •. Any spheroid belonging to Ker p. has a representative J; sn - E for which the spheroid p!: sn - B is contractible, For a homotopy H: sn x I - B between p! and the constant map, there exists a covering homotopy ii: sn x I - E between ! and some map h. The map ph is constant, i.e., the image of h is contained in F. This means that II E Imi •. (3) Imp. c Ker8•• Suppose that! = pj, where j: sn - E is a spheroid. The spheroid j can be represented as a homotopy CPt. The map CPl is constant, and hence 8.! = O. (4) Ker8. C Imp•. We represent a spheroid !: sn - B as a homotopy CPt: sn-l _ B and consider a covering homotopy CPt. Suppose that ! E Ker8., i.e., 1,01: sn- 1 - F is homotopic to a constant map. Let at be a homotopy in F between CPl and the constant map. Consider the homotopy

_ {CP2t 1Pt= a2t-l

if t

E[O,!],

iftE[!,l}.

=

Corresponding to this homotopy is a spheroid g: sn-l - F for which 9 pg is homotopic to f. Therefore, J E Imp._ ! (5) 1m 8. c Ker i., Suppose that a spheroid f: sn+ 1 l-+ B is represented as a homotopy CPt: sn - B and 'Pt is a lifting of this homotopy. The map 'Po

4. Fibrations and Homotopy Groups

171

is constant; therefore, the homotopy rpt can be treated as a map Dn+! -+ B. Hence rp1: 8 n -+ F is homotopic in E to the constant map. (6) Keri. elmo•. Let I: 8 n -+ F be a spheroid contractible in E. The projection of a homotopy of the spheroid I to the constant map can be regarded as a spheroid g: 8 n+! -+ B, and o.g = I. 0 Remark 4.1. If the total space E of the fibration is path-connected, then the sets 7ro(E, eo) and 7ro(B, 110) are one-point. In this case, there is a one-t~ one correspondence between 7ro (F, eo) and the coset space 7r1 (B, 110) / 1m p. (the subgroup Imp. C 7rl(B, 110) is not necessarily normal, so the coset space may not be a group). Example. For n ~ 2, 1I"n(81 ) = O. Proof. Consider a covering lR -+ 8 1 . Any covering is a fibration with discrete fiber F, and 1I"n-l(F) = 0 for n ~ 2. Using the exact sequence, we obtain 7rn (81) = 7rn (lRl ) = 0 for n ~ 2. 0 Example. 7r2(82) = Z, and 7rn (82 ) F 11",,(83) for n ~ 3. Proof. In the space C 2 with coordinates z and w, consider the sphere 8 3 given by the equation Izl2 + Iwl2 == 1. On 8 3 , the group 8 1 = {ei
hl(aei'P,bei 1/J)

= (~ei('P-1,h"ei1/J)

and

h2(aei'P,bei1/J)

= (~ei(1/J-'Pj,ei'P).

The exact sequence for the Hopt fibration is

.•. __ 7r2(83) ~ 7r2(82 ) ~ 7r1(81 ) ~ 7r1(S3) ..----. ...

0-

We already know that 7rk(8n ) = 0 for k < n (see Theorem 3.19 on p. 105). Therefore, 1I"2(S2) = 7r1 (8 1 ) = Z. Consider another segment of the exact sequence for the Hopf fibration:

... ~ 7rn (81)

..b..... 7(7)(83) .!:.... 7['n(S2) !!!... 7rn _l(S1) - + ~,,~ .

If n ~ 3, then 7['n(SI) ~ 7['n-1 (SI) isomorphism.

= 0; therefore, p*: 7['n(S3) -+ 7['n(S2) is an 0

4. Two-Dimensional Surfaces

172

- 1/J = const

Figure 26. A meridian and a parallel

Figure 27. The transformation of the solid torus

Now, we study in more detail the geometric structure of the Hopf fibration, in particular, the arrangement of its fibers in 8 3 • We represent Cpl as the union of the two closed sets Df C Ul and D~ C U2 determined by the inequalities a ~ b and a ~. b and consider the homeomorphisms 14: p-l(Dl) -+ Dl x 8 1 defined above. The total space 8 3 of the fibration is obtained by attaching the two solid tori Df x 8 1 and D~ X 8 1 to each other via a homeomorphism between their boundaries. The meridians of the solid torus Dl x 8 1 are determined by the equations 1/J = const and the parallels by the equations cp - 'f/J = const (see Figure 26); we assume that the meridians and parallels lie on the boundary, i.e., alb = 1. Each fiber is determined by ~o equations of the form cp ~ 1/J = const and alb = const. We C8.Il transform the solid torus Lq x 8 1 in such a way that the meridians (which are given by '" = const) remain invariant, while the parallels are transformed into the lines cp const. To this end, we cut the solid torus by a meridian plane and tum the upper part of the torus above the cut through the angle 21r leaving the lower part intact (see Figure 27). Choosing a suitable direction of rotation, we obtain a self-homeomorphism of the soljd torus which takes the parallels to the lines cp = const (see Figure 28); the fibers become linked as shown in Figure 29.

=

4. Fibrations and Homotopy Groups

173

cp = const \ .......~ cp = const

-1/J

N--J;1/J =

const

Figure 28. The image of a parallel under the transformation

Figure 29. The Hopf link

Figure 30. Gluing together solid tori

For the second solid torus D~ x Sl, we construct a similar homeomorphism. Then, we attach the boundaries of the transformed tori D~ x Sl and D~ x Sl to each other by identifying points with the same coordinates cp and t/J; thereby, we identify the meridians of one solid torus with the parallels of the other (see Figure 30). Remark 4.2. The arrangement of the fibers of the Hopf fibration can also be described using the fact that the fibers are the sections of the sphere S3 by the complex lines az = f3w with a, f3 E C. Problem 56. Suppose that S3 is the group of quaternions of unit length and Sl C S3 is its subgroup consisting of the quaternions of the form Xl + iX2. The group Sl acts on S3 by right multiplication. Prove that the map p: S3 ---+ S3/ Sl is the Hopf fibration. Problem 57. In the space ofreal matrices (i~ ~:), consider the sphere given by the equation x~ + x~ + x~ + x~ = 1. Prove that the singular matrices divide this sphere into two solid tori.

174

4.

Figure 31. Elements Oltheset

Tw~Dimensional

71'1 (X,

Surfaces

A, ao)

~1

Problem 58. Let p: s> be thO' HoJf fibration. Prove that D' Up Cp1 = Cp2. (It is assumed that-P~D4.' Problem 59. Prove that there exists no retraction r: Cp2 _ Cpl. (It is assumed that CP1 is naturally embeqded in Cp2.) Problem 60. Prove that the complement of the Hopf link in S3 is homotopy \ equivalent to the torus Tl. The Hopf fibration S3 -t' S. ~h fiber S~ has multidimensional generalizations. One of these generalizations Is as follows. Let us represent s2n+! as the unit sphere in 1R2n+2 ~ C n +1 and identify the points (AZ}, ..• , AZn+!) E s2n+! for all AE C with IAI = 1. As a result, we obtain a map s2n+! - Cpn, which is a fibration with fiber Sl. Taking the quaternions lHI instead of the complex numbers, we obtain a fibration s4n+3 _ JB[pn with fiber S3. Thus, for n = 1, we obtain a fibration S' - S4 with fiber S3. All these fibrations are called Hop/ fibrations as well. 4.4. Relative Homotopy Groups. for n ~ 1 and a pair of spaces X ~ A with base point ao E A, an n-dimensional relative spheroid can be defined as a map I: (Dn,8Dn,80) - (X,A,ao); it is assumed-1;hat I is a map of triples, i.e., 1(8Dn) C A and 1(80) = ao. Relative spheroids 10 and h are said to be homotopic if they are connected by a homotopy such that h(8Dn) C A and/t(so) = ao. The set 7rn(X, A, ao), where n ~ 1, consists of the equivalence classes of relative n-spheroids. This set does not generally admit a group structure. The point is that its elements are represented by paths starting at ao and ending at a. E A, and there is no natural way to compose a path starting at ao from two such paths (see Figure 31). Although 7r1(X,A,ao) is not a group, it has a distinguished element, the class of the constant map. We refer to this element as the zero element . • For n ~ 2, the set 7rn (X, A, ao) can be endowed with the tnructure of a group. To define it, we use the description of relative spheroids as maps of

It

175

4. Fibrations and Homotopy Groups

Xl

Figure 32. The composition of relative spheroids

the cube 1" = {(x!, ... ,Xn) : 0 ~ Xi ~ I}. However, we identify f)" with the quotient space ["'I(ap' \ ]n-I), where ]"-1 is determined by the ~uation Xn = 0, rather than with the cube ]n itself. Thus, a relative spheroid is a map I;.]n '-+ X for which I(al") c A and I(a]n \r-Il=;ao. We define the composition of relative spheroids

I 9 (XI,X2,,,,,Xn ) = { /(2XIIX2, •.. ,xn)

I, g: ]n - X

by

ifO~Xl$!,

g(2XI - 1, X2, ., . , xn)

.

I

If 2

~

xl

~ 1.

In Figure 32, the bold line represents the set of points which the map I 9 takes to the base point ao (for n = 2). For h = 2, the construction used in the proof of the- commutativity Qf the group 1l"n(X) does not apply because a relative spheroid I~ (l rot X does not necessarily take ]1 'C.]2 to the base point (in Figure 32, the set ]1 is shown by the dashed line). But for n ~ 3, any relative spheroid 11 ]n - X maps ap'-l to the base point, and the construction does apply. The follOWIng assertion is often used in dealing with relative homotopy groups. Theorem 4.16. A relative spheroid I: (Dn, aD n? 80) ~ (X, A, ao) represents the zero element 01 the group 1I"n(X, A, ao) if and only if there exists a homotopy ht between the map ho = I and some map hi for which hI(Dn) c A such that all the ht coincide with I on gn-l. Proof. Suppose that there exists a homotopy ht with the properties specified in the statement of the theorem. .oonsider the homotopy gt (s) = hl(I- t)8 + tso). This homotopy joins hi to the
if 0 ~ 11811

< 1 - t12,

if 1 - t/2 ~ 11811

s 1~

176

4. Two-Dimensional Surfaces

Clearly,

ho

= 10 = I, h1(Dn) c A, and ht(s) =

lo(s) for s E

sn- 1.

0

The embedding i: A -+ X induces a homomorphism i.: 1I"n(A,ao) -+ 1I"n(X, ao). Any absolute spheroid (Dn, so) +-+ (X, ao) can be regarded 88 a 1 is mapped to relative spheroid (D n , 8 n - 1 , so) -+ (X, A, ao), because ao E A. Homotopic absolute spheroids are homotopic 88 relative spheroids; thus, we have a homomorphism p.: 1I"n(X, ao) -+ 1I"n(X, A, ao). Finally, there is a homomorphism 0.: 1I"n(X,A,ao) -+ 1I"n-1(A,ao), which takes each rel1 , 80) -+ (X, A, ao) to its restriction 1;g -(sn-l, so); ative spheroid (Dn, from the definition of a homotopy between relative spheroids one can easily see that homotopic relative spheroids are mapped to homotopic absolute spheroids.

sn-

sn-

Theorem 4.17 (exact sequence for a pair). The sequence

01 homomor-

phisms •.. -+

1I"n(A, ao) ~ 1I"n(X, ao) ~ 1I"n(X, A, 0.0) ~ 1I"n-1 (A, ao)

-+ •••

is exact. Proof. (1) Imi. C Kerp•. According to Theorem 4.16, any relative spher-' oid I with f(Dn) c A represents the zero element of the group 1I"n(X,A,ao). (2) Kerp. c Imi.. Suppose that I: X takes In-1 to A and 01" \ ,[T'-1 to 0.0 and there exists a homotopy F: x I -+ X in the class of relative spheroids which joins I to the constant map to ao. The section of the cube r H = I" x I by the hyperplane tXn + (1 - t)XnH = 0, where t E [0,1], is homeomorphic to The restriction of F to this section is a homotopy of absolute spheroids. For t = 0, the section is the face XnH = 0, and the restriction to this face coincides with I. For t = 1, the section if! the face Xn = 0, which is mapped to A by assumption.

r ......

r

r.

(3) Imp. C Kero•. Any absolute spheroid I: D n -+ X maps oDn to ao. (4) Kero. C Imp•. Consider a relative spheroid I: r - X whose restriction to r- 1 is homotopic to the constant map to ao (in the class of maps 1"-1 -+ A). Let gt: 1"-1 -+ A be a homotopy between I t r- 1 and the constant map. Consider the homotopy It: -+ X that coincides with gt on 1"-1 and takes 1,,-1 to ao. By the Borsuk lemma (see p. 164), this homotopy can be extended to a homotopy of I. AI3 a result, we obtain a homotopy in the class of relative spheroids that joins I to a relative spheroid taking to ao. (5) Im 0. c Keri •. If an absolute spheroid I: 1"-1 -+ A is the restriction of a relative spheroid g: In -+ X to /n-1 = 1"-1 X {O} c 1", then 9 can be regarded as a homotopy gt: 1"-1 -+ X in X between the spheroid I and the constant map.

or \

or

or

4. Fibrations and Homotopy Groups

177

So

Figure 33. The homotopy

ii 88 a. rela.tive spheroid

(6) Ked. C 1m 0*. A homotopy 9t: ["-1 -+ X between spheroids 90: ["-1 4 A and 91: In-l -+ ao can be regarded as a relative spheroid 9: In -+ X whose restriction to In-1 coincides with 90. 0 Problem 61. Prove that 1I"n-1(X)

~

1I"n(CX,X) for n

~

2.

The following theorem shows that the exact sequenc1 for a fibration is a special case of the exact sequence for a pair. Theorem 4.18. Suppose thatp: E -+ B is a locally trivialfibmtion, eo E E, = p(eo), and F = p-1(eo). Then the maps p.: 1I"n(E, F, eo) -+ 1I"n(B, bo) are isomorphisms for all n ~ 1.

bo

Proof. Suppose that h: Dn -+ E is a relative spheroid such that the absolute spheroid h = ph represents the zero element of the group 1I"n(B, bo). Then there exists a homotopy H: Dn x I -+ B between h and the constant map (in the class of maps taking gn-1 to bo). By the covering homotopy theorem, there exists a homotopy ii f D n x I -+ E which covers the homotopy H and is simultaneously an extension of h and ii': {80} x I -+ eo E E. The homotopy ii joins the relative spheroid h to a map whose image is contained in F. Such a map is homotopic to the constant map in the class of relative spheroids. Therefore, p. is a monomorphism. Let h: gn -+ B be a spheroid. It can be regarded as a homotopy H: sn-1 X 1-+ B between the constant maps gn-1 -+ bo E B (look again at Figure 24 on p. 169). Let ii: sn-1 X I -+ E be a homotopy which covers H and is an extension of the constant map sn-1 X {O} -+ eo E E and of the homotopy Ii: {so} x I -+ eo E E. This homotopy can be regarded as a relative spheroid h: D n -+ E (see Figure 33). Moreover, h = ph, and hence p. is an epimorphism. 0 4.5. The Whitehead Theorem. The Whitehead theorem asserts that if a map X -+ Y of connected CW-complexes induces isomorphisms of all

178

4. Two-Dimensional Surfaces

homotopy groups, then it is a homotopy equivalence. We start with the special case in which the space Y is one-point. The general Whitehead theorem is easily derived from the relative version of this special case. Theorem 4.19. Suppose that X is a 6W-complex, Xo is a vertex of X, and 11'7'1 (X, xo) = 0 for all n ~ O. Then X is a contmctible space. Proof. Let fo = idx, and let 16 be the restriction of 10 to XO (the 0skeleton of X). The equality 11'0 (X, xo) = 0 means that each vertex Xa E XO is joined by a path 'Ya with xo .. The formula Po(xa) b::::: 'Ya(t) defines a homotopy between f6 and the constant map XO -+ Xo. According to the Borsuk lemma, this homotopy can be extended to a homotopy Fo (x, t) on the entire space X. This homotopy joins the identity map Fo(x, 0) lo(x) to the map Fo(x, 1) = hex) taking X O to Xo., Let I{ be the restriction of h to Xl. The equality 11'1 (Xl xo) = 0 implies the existence of a homotopy F1 between I't and the constant map Xl -+ Xo. Extending this homotopy to X ~ we obtain a homotopy Fl between h and a map h taking Xl to Xo. Similarly, applying the equality 1I'n(X, xo) = 0, we can construct a homotopy Fn between maps In and in+! such that 17'1+1 takes xn to Xo for every n. This completes the proof of the theorem in the case of dim X < 00. If dim X = 00, then a homotopy F between idx and the constant map X -+ Xo can be constructed as follows. We set F(x, t n ) =;: In(x) for tn = 2;;1 and define F to coincide with Fn between tn and tn+!' Clearly, the restrictions of F to xn x I are continuous for all n; therefore, the entire map F is continuous. 0

=

A space A C )( is called a deformation retmct of X if there exists a homotopy It': X ~ X, where t E [0,1], such that It t A = idA for all t, 10 = idx, and heX) c A. For the map r = h: X -+ A, we have ri = idA, and -ir is homotopic to idx (here i: A -+ X is the natural embedding of A into X). Clearly, any deformation retract A of X is homotopy equivalent to X ~ Indeed, for the maps r; X -+ A and i; A -+ X, we have ir ~ idx and = idA.

r.

Problem 62. Let m, n 2: 1. Prove that the space 8"' V sn 0:: (8"' x {xo}) U ({Yo} x sn), where Xo E sn and Yo E 8"', is a deformation retract of the space obtained from 8"' x sn by removing a point not belonging t6 8"' V sn.

Theorem 4.19 has the following relative version. Theorem 4.20. Suppose that X is a path-connected CW-complex, A is its subcomplex, ao is a vertex 01 A, and 1r'n(X, A, ao) 0 lor all n > 1. Then A. 'is a deformation retmct 0/ X. -

=

179

4. Fibrations and Homotopy Groups

Figure 34. The cylinder of a map

Proof. Suppose that n ~ 1 and 9n: X -+ X is a map such that 9n(xn-l) C A and 9n .t A = idA. According to Theorem 4.16, the equality 7rn(X, A, ao) = o implies the existence of a homotopy which is fixed on A and joins the map 9n t Xn u A to a map g~+1: xn U A -+ A. Extending this homotopy to the entire space X, we obtain 9n+1' The map 91 is constructed as follows. We join a vertex Xa E X O \ AO to ao by a path and set 9~ (xa) = ao. Clearly, the map ~: XO U A -+ X is homotopic to idxo uA ' Extending the homotopy to X, we obtain the required map 91. 0 Now, we can prove the Whitehead theorem itself. Theorem 4.21 (Whitehead). Suppose that X and Y are connected CWcomplexes, Xo E X and Yo E Y are their vertices, and a map f: (X, xo) -+ (Y, Yo) induces isomorphisms f.: 7rn(X, xo) -+ 7rn(Y, Yo) for all n ~ 1. Then f is a homotopy equivalence. Proof. Let C, be the space obtained by attaching X x I to Y via the map X x {O} = X L y (see Figure 34). The space C, is called the cylinder of the map f. The identity map C/ -+ C / is homotopic to 9: C / -+ Y c C/, and the homotopy is the identity on y, Therefore, the spaces C/ and Y are homotopy equivalent.

c,

The map f can be represented as the composition X i. ~ Y, where i is the natural embedding X = X x {1} -+ C/. The homomorphisms i.9. and 9. are isomorphisms; therefore, i. is an isomorphism also. Let us write the exact sequence for the pair (C" X): .. , -+

7rn(X) ~ 7rn (C/) ~ 7rn (C" X) ~ 7rn -l(X) ~ 7rn -l(C,)

-+

.~ ••

180

4. Two-Dimensional Surfaces

We have 1rn (C" X) = O. Indeed, Kerp. = Imi. = 1rn (C,), whence p. = O. Moreover, Ima. = Ked. = OJ therefore, = 0, i.e., 1rn (C" X) = Kera. = Imp. = O. According to Theorem 4.20, the space X is a deformation retract of C" i.e., the identity map C, -+ C, is homotopic to r: C, -+ X C C, and the homotopy is fixed on X. For If = rj: Y -+ X, we have IfI = rjgi r(idc/)i = idx and I If = girj g(idc/)j = idy. 0

a.

IV

IV

Chapter 5

Manifolds

1. Definition and Basic Properties A second countable topological space X is called a topological manifold if it is Hausdorff and each point x E X has an open neighborhood homeomorphic to an open subset of IRn , the number n being the same for all points x EX. (The example on p. 117 shows that there exist non-Hausdorff spaces in which every point has an open neighborhood homeomorphic to an open subset of Euclidean space.} According to the Brouwer theorem on the invariance of dimension (Theorem 2.9 on p. 61), the dimension of a locally Euclidean space is determined uniquely. Indeed, suppose that U :3 x and V :3 x are open sets in X homeomorphic to open subsets of IRn and IRm , respectively, where n #: m. Then the open set U n V is homeomorphic to both an open subset of IRn and an open subset of IRm~ which contradicts the invariance of dimension. Let Mn be a topological manifold of dimension n. A pair (U, cp), where U c M n is a connected open set and cp: U -+ cp(U) is a homeomorphism onto an open subset of IRn , is called a chart, or a local coordinate system. If cp(x) = 0, then we say that (U, cp) is a local coordinate system with origin x. A smooth structure on a topological manifold M is a family A of local coordinate systems {( U01, CPQ) : O! E A} with the following properties: (i) the sets UQ cover M; (ii) if UQ n Up#: 0, then CPQCP~l is a smooth map; I

(iii) the family A is maximal in the sense that if (U, cp) is a local coordinate system and the maps cpQcp-l and cpcp;;l are smooth for all O! such that UQ n U #: 0, then (U, cp) E A.

-

181

182

5. Manifolds

To define a smooth structure on M, it is sufficient to specify a family Ao of local coordinate systems having properties (i) and (ii). Indeed, property (iii) can be regarded as the definition of the set of local coordinate systems required for completing Ao. A set of charts covering M and having property (ii) is referred to as an atlas. A topological manifold M endowed with a smooth structure is called a smooth manifold, or simply a manifold. (In mathematics, nonsmooth manifolds are much less frequent than smooth ones.) Example. The topological space lRpn admits a structure of a manifold. Proof. The space lRpn can be covered by open sets Ui = {(Xl: •.. : X n +1) E Rpn: Xi '# O}, where i = 1, ... , n + 1. Consider the homeomorphisms CPi: Ui - lRn given by Xl CPi (Xl:"'; Xn+l ) = ( -t· Xi

'r~

Xi-l Xi+l xn+l) - , --~"r' - - . Xi Xi Xi

We must prove that the maps cpjcp-;l, which are defined on cp;(Ui n Uj), are smooth. Without loss of generality, we can assume that i = 1 and j = 2. Then the set Ui n Uj is determined by the inequality XIXZ # o. Let CPI(Xl •. ~. : Xn+l)

X2 X3 Xn+l) = ( -, -~. ,.,-- = (Yl, Y2,rr' ,Yn)' Xl Xl Xl

Then CP2CP1l(YI! Y2,"" Yn)

= CP2(XI : ••• : xn+d -_

By assumption, Yl

(Xl , X3 , .... ; ~) _ (~ , Y2 , ... , ~) X2 X2 X2 YI YI YI

'# OJ therefore, the map IP2cp11

is smooth.

. 0

Exercise 31. Prove that the space Cpn is a manifold. 1.1. Manifolds with Boundary. The definition Of a manifold with boundary is obtained from that of a manifold by 8llowing the images cp(U) of the charts cP: U -+ cp(U) to be open subsets of the topological space lR+. = {(Xl, ... , Xn) E lRn : Xl ~ O}. To define a smooth structure on a manifold with boundary, we modify property (ii) as follows: (ii) if Ua n Up '# 0, then there exist open (in lR) sets Va J CPa(uan Up) and Vp :J cpP(Ua n Up) and mutually inverse smooth maps lap: Vp - Va ~d fpa: Va - Vp such that the restrictions of these maps to cpP(Ua n Up) and CPa(Ua n Up) coincide with CPaCPi l and cppcp;;l, respectivel~

183

1. Definition and Basic Properties

Let Mn be a manifold with boundary. We say that x E Mn is a boundary point if there exists a chart ep: U -+ ep(U) c ]it. such that x E U and

ep(x)

E

aIR+. = {(Xl, . •• "xn )

E]in : Xl

= b};

by a chart we mean a smooth chart, i.e., a chart from a smooth structure. The boundary of a manifold Mn is the set of all boundary points of Mn. We say that X E Mn is an interior point if there exists a chart ep: U -+ cp(U) such that X E U and cp(U) is open in ]in; we again assume the chart to be smooth. To show that an interior point of a manifold cannot be a boundary point, we need the following theorem. Theorem 5.1 (Inverse function theorem). Suppose that J: U -+ V is a smooth map of open subsets of ]in and its Jacobian matrix (M::) , hal nonzero determinant at some point X E U. Then x has a neighborhood U G U such that the set feU) is open in ]in and the map I t fj J U -4 J(U) is a homeomorphism. Moreover, the map (f t U)-l is smooth. Proof. We write f iIi the form f = (h, ... , In), where h, ... , In are funCI tions on U. Suppose that B is a convex set in the domain of J and y, z E B. On the interval [0,1], consider the function ip,(t) Ji(ty+ (l-t)z). Clearly,

-=

(1)

dep) ~ at· fi(Y) - Ji(Z) = epi(l) - CPi(O) = dt: (ti) = ~ ax: (w,)(Yi - Zi),

E (0,1) and Wi = tiY + (1- ti}Z E B. Now, consider the function J(Wl,"" wnl = det (-~(Wi)) of n 2 real variables (the coordinates of WI, ••• ,Wn E tJ). This function is continuous, and J(x, ... ,x) f. O. Hence there exists a number € such that if a point WI, ... , Wn belongs to the disk D';,e;, then J(WI,"" wn ) f. O. This inequality and formula (1) show that if y, z E D';,t and Y f. Zr then J(y) f. J(z). Thus, the restriction of J to the compact set V';,e; is a one-to-one map to some subset of the Hausdorff space ]in. Such a map is a homeomorphism. Let fj = int D~,e' We have already proved that f r U: fj -4 feU) is a homeomorphism. Nowj let us prove that the set J(U) is open in ]in. Take a point 1.£ E U, We must show that any open disk of sufficiently small radius c~ntered at 1(1.£) is entirely contained in J(U). On the compact set aD;~~ the function cp(y) = II/(y) - l(u)1I attains its minimum. This minimum is positive, because I(u) ¢ l(aD';,e;)' Therefore~ we can choose a, positive number d such that II/(Y) - 1(1.£)11 > 2d for all 'II E ?D';,e. Let ~s show that the open disk of radius d centered at J(u) js contained in I(U). Suppose that 11/(1.£) - zll < d. If y E D';,E' then where

ti

IIf(y) -

zll

~ IIf(y) - f(u)II-lIf(u) -

zll ;> 2d -

d = d.

5. Manifolds

184

Therefore, on the set D~,e' the smooth function .,p(w) = II/(w) - zll2 attains its minimum at an interior point (the values of this function on the boundary are larger than 62 , and its value at 1.£ is smaller than 62 ). Let a be a point of minimum for .,p on the set D~,e' Then

0=

~.,p

VXj

= 2

E ~/i (a)(fi(a) vXj

Zi).

We already know that if a E D~,e1 then J(a,. '" a)

= det (~(a» ¥-

O.

Hence z = I(a) E I(U), as required.

r

It remains to prove the last assertion, namely, that the map (1 U)-l is smooth. Since the determinant det (M:(y» is nonzero for y E U, it follows

,

-

that I-I has smooth partial derivatives on the open set U, Therefore, I-I is smooth on U. 0 Corollary. A point 01 a manilold with boundary is an interior point il and only il it is not a boundary point. Proof. If MR is a manifold with boundary and a point o{ M is both interior and boundary, then there exist smooth mutually inverse maps I: U -.. V and I: V --+ U of open subsets of lRR satisfying the following conditions: (i) 0 E U, 0 E V, and 1(0) = g(O) = OJ (ii) U has an open subset U' 3 0 such that I(U') C lR+.. Since the map I has a smooth inverse in a neighborhood of 0, it follows that det (~(O» ¥- O. Therefore, the point 0 has a neighborhood U C U

r

such that I(U) is open in lRR and I U: U --+ I(U) is a homeomorphism. Thus, I(U U') is an open neighborhood of 0 in lRR. This contradicts the inclusions I~U U') C I(U') C lR+.. . 0

n

r

Remark 5.1. An even stronger assertion is valid: An open neighborhood 01 o in lR+. is never homeomorphic to an open subset o/lRR. The proof of this assertion for n = 2 is contained in that of Theorem 4.1 on p. 140. In the general case, the proof uses the multidimensional analog of the Jordan theorem. A compact manifold without boundary is said to be closed. A subset N C MR is called a k-dimensional submanilold of MR if for an arbitrary point x E N, there exists a chart (U, cp) such that Un N = f/>-1 (lRk n cp(U» (it is assumed that lRk is standardly embedded in lRR, i.e., consists of the points whose last n ~ k coordinates are zero). Exercise 32. (a) Let MR be a manifold without boundary. Prove that a submanifold of MR is a manifold without boundary. • (b) Let MR be a manifold with boundary. Prove that any of its submanifolds is either a manifold with boundary or a manifold without boundary.

1. Definition and Basic Properties

185

Exercise 33. Prove that the boundary of a manifold is a manifold without boundary. Problem 63. Let Mn be a connected manifold, and let Nn be a submanifold of Mn (of the same dimension). Prove that if Nn is closed, then Nn = Mn. 1.2. Maps of Manifolds. Let M rn and Nn be manifolds with smooth structures A = {(Ua,CPa)} and B = {(Vp,1PP)}' A map I: Mrn -+ Nn is said to be smooth if all the maps 1Pplcp;1 are smooth. Each map 1Pplcp;1 is defined on the open set CPa (Ua n 1-1 (Vp)) c lRrn and takes values in lRn '\ If I; urn -+ NR and g: N n -+ Mrn are smooth mutually inverse maps, then the map I is called a diffeomorphism, and the manifolds Mrn and Nn are said to be diffeomorphic. It follows from the inverse function theorem that if manifolds M rn and Nn are diffeomorphic, then m = n. Let I: Mrn -+ Nn be a smooth map between manifolds without boundary, and let x E Mrn. Choose local coordinates near x and I(x) and consider the Jacobian matrix (~(x)). The rank ofthis matrix does not depend on the local coordinate systems. This rank is called the rank of the map I at the point x and denoted by rank I (x). Let I: Mrn -+ Nn be a smooth map. If rank/(x) = m, then the map f is called an immersion at the point x, and if rank/(x) = n, then I is called a submersion at x. A map I that is an immersion (submersion) at all points x E Mrn is called an immersion (submersion). An immersion I that is a homeomorphism of Mrn onto I(Mrn) C N n is an embedding.

,

Theorem 5.2. (a) Any smooth map I: Mrn --+ Nn that is an immersion at some point a has the structure 01 the standard embedding lRrn -+ R rn x lRn - rn in a neighborhood 01 this point. To reduce it locally to such an embedding, it suffices to choose a suitable coordinate system on the image, i. e., in a neighborhood oll(a). (b) Any smooth map I: Mrn -+ Nn that is a submersion at some point a has the structure 01 the standard projection R n x lRrn - n -+ R n in a neighborhood 01 this point. To reduce it ~ such a projection, it suffices to choose a suitable coordinate system on the preimage, i.e., in a neighborhood 01 a. Proof. Let (Ua , cp) and (V!(a) , ,p) be local coordinate systems with origins a E Mrn and I(a) E N n l respectively; we assume that I(Uo.) C Vf(a). We write f in local coordinates, i.e., consider the map j = tfJlcp-l. For this map, the Jacobian matrix J = (~(x)) is defined. If I is an immersion, then the rank of J at the origin equals m, and if I is a submersion, then the rank is n. Therefore, we can choose ilt ... ,irn E {I, .•. , n} (if f is an immersion)

186

5. Manifolds

or iI, ... ,in E {1"." m} (if f is a submersion) such that at the origin,

det

~ ... ~) ( ~l~ ,..,., . . '8/;;' ~ °

or

respectively, Without loss of generality, we can 88Sume that ik = k (resp~ tively, ik = k), The Jacobian matrix J can be augmented to a square matrix-of the form ( ~1 or (~~ ~)) where I is the identity matrix of o:.der In - mi. The augmented matrix is the Jacobian matrix for some map F, namely, for the map

"1)

P(Xb ., • ,xm , Yb ,I. • ~ Yn~) = (fl(X)"., ,jm(x), !m+1(x)

+ YI,~., ,jn(x) + Yn-m)

if f is an immersion, and for the map

P(Xl,"" Xm , Yl,.·" Ym-n) L

= (l(x, Y), Yl.·,·, Ym-n) _

_

~

iff is a submersion; wewriteF(x,y) = f(x)+(O,y) andF(x,y) = (f(x,y),y) respectively. At the origin, the Jacobian determinant of t~e map if is nonzero, th~re­ fore, according to the inverse function theorem, F has a smooth inverse F- i in some neighborhood of the origin. In the case of immersion, we replace the map t/J by p-1t/J (in other words, we change the local coordinate system on the image of f), In the new local coordinates, the map f has the form

x

1-+

1/Jfr.p-l j(x) = P(x 1 0)

1-+

P-1(x,O).

In the submersion case, we replace the map r.p by Pr.p (in other words, we change the local coordinate system on the preimage of f). In the new local coordinates, f has the form

(l(x, y), y)

1-+

p-l(x, y) t-+ 1/Jfr.p-i lex, y),

i.e., (x', y) 1-+ X, where x, = lex, y). It remains to note that for y fixed, the map x 1-+ lex, y) is an epimorphism in a small neighborhood of the origin, because the Jacobian determinant of this map is nonzero. P Using Theorem 5.2, we can easily prove the following assertion. Theorem 5.3. Suppose that J: Mm - t N n is a smooth map, X k is a subma:n,ijold oj Nn, and J is a submersion at each point oj j~1(4k). Then f-l(X k ) is a submanifold of dimension k + m - n in Mm,'J

1. Definition and Basic Properties

187

Proof. Let a E l-t(Xk). In a neighborhood of I(a), we can choose local coordinates :r; = (u, v) so that 'IJ E ]Rk, v E ]Rn-k, and the equation v = 0 determines Xk in this neighborhood. The map I is a submersion at a; therefore, by Theorem 5.2, we can choose local coordinates (x, y) in a neighborhood of a so that :t E ]Rn, y E ]Rm-ri, and the map I has the form (x,y) 1-+ x (i.e., (u,v,y) 1-+ (u,v» in the chosen local coordinates near a and I(a). Clearly, the equation of the set l-l(Xk) in these coordinates is v = O. Therefore, ,-l(Xk) is a submanifold of dimension k + m - n. 0 Theorem 5.3 can be used to prove that certain subsets of manifolds are submanifolds and, thereby, manifolds. Example. The sphere

sn is a manifold.

Proof. Consider the map I: ]Rn+1 -+ ]Rl defined by I(xt, ... , X n +1) = x¥+.~ ·+x~+1' The Jacobian matrix J = (2Xb"" 2x n +1) has rank 1 at each point of 1- 1 (1); therefore, the set 1- 1 (1) = is a submanifold in ]Rn+I. 0

sn

1.3. Smooth Partitions of Unity. Let {lfa, a E.A} be an open cover of a manifold Mn. A partition of unity {'P{3, {3 E B} subordinate to this cover is said to be smooth if all the functions 'P{3 are smooth. Theorem 5.4. (a) For an open cover {Ua,a E A} 01 a maniJold Mn, there exists a smooth partition oj unity {'P{3, {3 E B} subordinate to it. (b) IJ the index set A is no more than countable, then it can be assumed that B = A and supp 'Pa C Ua Jor each a E A. Proof. (a) First, let us construct open sets XII; C Mn (k = 1,2, ... ) such that their closures X k are compact, X k C X,HIt and Mn = U:'I XII;. To this end, consider a countable base of the space Mn and take all open sets with compact closures in this base. We denote the chosen sets by WI, W2, .... They cover the manifold Mn. Indeed, any point x E Mn has a neighborhood U(x) with compact closure. The set U(x) can be represented as a union of elements of the base; tlearly, all these elements have compact c1os~. Therefore, x E Wi for some i. We put Xl = WI' The compact set Xl is covered by the open sets Wi; hence Xl C Wil U .. , U Wi,., where il < i2 < ... < i p • We put X2 = WI U W 2 U Wil U· .. U Wi),>' The sets Xa, X4, . \. are defined similarly. Now, let us construct open sets V{3,1 C V{3,a. We put D~ = {x E lRR : II x II < r}. :For each point z E X II; \ Xk-l, we choose an open set ~,3 such that Vz,a C Ua for some a, Vz,a C Xk+It and Vz,3 n Xk-2 = 0; in addition,

188

5. Manifolds

we require that there exist a chart t/Jz: Yz,3 --+! D!;. The open sets Yz,l = 1jJ;l(Df) cover the compact set XIc C Xk+H hence XIc is covered by finitely many sets of the form Yz,l. Let {V,a,l} be the family of all such sets for all ~; by V,a,3 we denote the sets of the form Yz,3 corresponding to the elements of this family. Note that the index set {,B} is at most countable; in addition, {V,a,l} is a cover of M" and fV,a,3} is a locally finite refinement of {Uo}.l It is easy to verify that the function that takes the value e- 1/ t at t> 0 and 0 at t ~ 0 is smooth. Therefore, the function e~l/(l-t)

-yet)

= {0

if t if t

< 1, ~1

is smooth also. Thus, 7(X) = -y(lIxIl 2 /4) is a smooth function on IR"j it takes positive values at all points of the open disk D~ and vanishes outside D~. Consider the functions

9,a(X) = {7(t/J,a(X))-

o

~ x E V,a,3, If x ¢ V,a,3

and hex) = L,a9,a(X). The function h is smooth, because the cover {V,a,3} is locally finite. The sets V,a,l cover the entire manifold M", and 9,a(X) > 0 for x E V,a,l' Therefore, hex) > 0 for any point x E The functions 'P,a = 9,a/h form the required partition of unity, because supp 'P,a C "P,3 C Uo . (b) Consider the open cover Ul, U2, ... . We have constructed the partition of unity 'P}, 'P2,'" such that for each i, SUPP'Pi C U; for some j = j(i}. Let us define ,:pi as the sum of all functions 'Pic such that supp 'Pic C Ui and supp 'Pic ¢. Uj for j < i. Then each function 'Pic is contained as a term in precisely one ,:pi, and SUPP,:pi CUi. . 0

un.

1.4. Sard's Theorem. To ~tate Sard's theorem, we need the notion of a set of measure zero. We say that a set X C IR" (n ~ 1) is of measure zero if, for any ~ > 0, X can be covered by count ably many cubes such that the sum of their volumes is less than e. The cubes may be open or closed; instead of cubes, disks or parallelepipeds can be taken. It is assumed that in lRo. only the empty set has measure zero. Let M" be a manifold. We say that a set X C M" is of measure zero if the manifold M n can be covered by a countable family of charts 'Pi: Ui -+ ]Rn J where each 'Pi(X n Ui) c IRn has measure zero. The consistency of the two definitions is implied by the following two lemmas. Lemma 1. Any countable union of sets of measure zero in IRn is a set of measure zero. • lEach point &': E Xk-l belongs only to those sets Vp,3 that are contained in X.,+l' The number of such sets is finite by construction.

1. Definition and Basic Properties

189

Proof. Suppose that X = U:l Xi C lRn and each set Xi has measure zero. Covering every Xi by a countable set of cubes such that the sum of their volumes is less than e/2i , we obtain a cover of X by a countable set of cubes for which the sum of volumes is less than e(I/2 + 1/4 + 1/8 +.,~) = e. 0 Lemma 2. II I: lRn - lRn is a smooth map and X C lRn is a set 01 measure ~ero, then f(X) c; lRn is a set 01 measure zero. Proof. The space lRn can be represented as a countable union of cubes; therefore, it is sufficient to consider the case where X is contained in a cube 1". Recall that

II/eu) - lev) II ~

max

Iali (x)llIu - vII

:Z:E(u,tI] Xj

(this follows froni equality (1) on p. 183). Let K = max:z:EI" 1~(x)l. Then IIfCu) - I(v) II ~ Kllu-vll whenever u,v E 1". This means that f takes any disk of radius r contained in 1" to a disk of radius K r. Thus, if a set X is covered by disks such that the sum oftheir volumes is less than E, then leX) can be covered by disks such that the sum of their volumes is less than Kne. 0 Exercise 34. (a) Suppose that m < n and urn c Nn is a submanifold. Prove that the set M m C N n has measure zero. (b) Suppose that m < n and I; M m _ N n is a smooth map. Prove that the set I(Mm ) eN" has measure zero. The proof of Sard's theorem uses the following assertion. Theorem 5.5 (Fubini). IIC C lRn is a compact set and any section olC by a hyperplane 01 the lorm Xn = a has measure zero,then C has measure zero. Proof. If n = 1, then C = 0 by definition. Suppose that n ;::: 2. Since C is compact, we have C C lR,,-l X [a, b]. For t E [a, b], we define sets Ct C lRn - 1 as follows:

Ct x {t} = C n (lRn -

1

x {t}) = C n {(Xl, ••• ,xn )

E

lRn

: Xn

= t}.

By assumption, each set Ct C lRn - 1 has measure zero. Take e > 0 and consider a cover of Ct by open (n-l )-cubes .r:'ll, .r:'i 1 , . •• such that the sum of their volumes is less than E. The set J;'-l ' U~l.r:'J-l is open; therefore, its complement C\ (J;'-l x [a, b]) is closed and hence compact. Thus, we can choose 5 = d(t) > 0 such that the set Or is covered by J;'-l x It for every r E (t - 5, t + 5) = It. Indeed, the function IXn - tl attains its minimum 5 on C\ (J;'-l x It); this minimum is positive, because IXn -tl can vanish only at those points of C that belong to Ct. The number d has the required property, because ifr E It and x E Cn then IXn-rl < d, i.e., x ¢ C\(J;'~l x [a, b]) and

x

E

(J;'-l x [a, b]) n C,. c

.Jt-1 x It.

5. Manifolds

190

The open sets It form a cover, of the interval [a, b]. We take a finite subcover Itt, . .. , It,. of this cover and replace each open set It. = (a, (3) by the interval Ii = [a, (3] n [a, b]. The cubes x Ii cover C, and the sum of their volumes does not exceed eL, where L is the sum of the lengths of the intervals Ii".

.r:::/

=

Lemma 3. Any finite cover of an interval I == [a, b] by intervals Ii [ai, bi] C I has a 8ubcover such that the sum of lengths of the intervals from this subcover is less than 2 (b - a).

Proof. We can assume that the cover 11, ... , In is minimal (that is, it ceases to be a cover of I after any interval Ii is removed). Let us order the intervals h, ... , In in such a way that al :$ a2 :$ ... :$ an. Since the cover is minimal" we have b~ :$ ~ ~ ..• :$ brl • These inequalities an~ the minimality of the cover imply bk < ak+2 (otherwise, the intervals [ak, bk] and lak+2, bk+2] would cover [ak+1, bk+1])' Therefore, the sum of the lengths of 11, 13 , • ~. and the sum of the lengths of 12,14, ... are both less than b - a. 0

According to Lemma 3, the cover II, ... , Ik of the interval [a, b] can be chosen so that L < 2(b ..... a). This completes the proof of the Fubini 0 theorem. Let J = Mm -; Nn be a smooth map. A point x E Mrn is said to be critical if rank J (x) < mine m, n). Otherwise this point is said to be regular. A point y E N n is called a critical value if y = f(x) for some critical point x. Theorem 5.6 (Sard [117]). The set of critical values of any smooth map of manifolds has measure zero. Proof. It is sufficient to prove the theorem for a smooth map f: U or+ lRn , where U <:; lRm is an open set. We prove it by induction on m. For m = 0, the assertion is obvious: if n = 0, then the set of critical values is empty, and if n ~ 1, then this set is a one-point subset of lRn and, hence, has measure zero. Suppose that Sard's theorem is valid for all smooth maps V .... lR", where V c lRm - 1 •

Let 0 be the set of critical points of a map J! [/ -+ lRn , and let Oi be the set of all points x E U at which all partial derivatives of f bf order at most i vanish. Clearly" 0 ~ Cl :;) 02 ::> •• , ~ Step 1. The set

ICo \ 0 1 ) has measure zero.

If n = 1, then 0 1 = O. Suppose that n 2 2. Let c E C \ 01. Then ~(c) i- 0 for some i and j. We can assume that t = j = 1.1 Consider the 3 m~ h: U -+ lRm defined by h(Xl, ..• ,xm) =

Ch(X),X2, .. ~,xm).

1. Dennition and Basic Properties

191

Figure 1. Construction of the map 9

Clearly, the Jacobian matrix of h at c has the form (~(c);), where I is the identity matrix~ Thus, we can apply the inverse function theorem to the map h at the point c. According to this theorem t c has a neighborhood V such that the restriction of h to V is a homeomorphism (see Figure 1~. For the map 9 = /h- l , we have

9(Xt, ... ,xm) = (XI,92(X),.d,9n(X»,

tIi particular, 9 takes the set ({t} xlRm - 1 )nh(V) to {t} xlRn - 1 • Let l be the restriction of 9 to this set. It is easy to show that a. point of ({t} k lRm- l ) n heY) is critical for the map gt if and only if it is i!ritlcal for g. Indeed,

(89i) _(1* 8xj

-

b

{~) t

)

J

By the induction hypothesis, the set of critical points of the map l has measure zero in {t} x lRn~ 1 . This suggests that applying the Fubini theorem, it is possible to show that the set h(V n C) has measure zero (and, hence, the image of V (l C lmder I has measure zero). The Fubini theorem does not apply directly because the set f(V n C) is not compact. But V n C can be represented as a countable union of compact sets; therefore, I(V (l 0) can also be represented as a countable union of compact sets. To each of these sets, we carl apply the Fubini theorem and obtain the desired result. Thus, each point c E C \ C 1 has t neighborhood V such that the set I«C \ C1) n V) C fCC n V) has measure zero. Since 0\ C1 can be covered by count ably many such neighborhoods (instead of we can take an open disk of rational radius centered at a point with rational coordinates such that it contains c is contained in V), it, follows that the set f(C\ Cl), is of measure zero.

y

an4

Step 2. The set I(ek \ Ck_i) h~ measure zero for any k ~ 1.

192

5. Manifolds

The proof of this assertion is similar to that of Step 1. Let c E Cle \ Cle-i' Then 8 ~~;I. (e) t= 0 for some j,jI, ... ,ikH- We fan assume that X31

i

= il = 1.

X'k+l

.

Let w(x)

=

8x ~J~ 32

~(e) = {}x~

Consider the map h: U

4

3k+l

(x), Then w(e} = 0, because e E CIe, and

a"+l Ii

(e} t= O.

{}Xil ••• {}XjA:+l

Rm defined by the formula

h(xl, ... ,xm )

= (W(X),X2, .... ,Xm ).

The rest of the proof is a repetition of that of Step 1. The only difference is that it uses the fact that

h «Cle \ Cki-l) n V)

n ({t} x Rm - 1 )

is a critical point for the map 9 and, hence, for the map gt,. Step 3. The set I(Ck ) has measure zero for sufficiently large k (e.g., for k> (mIn) -1). It suffices to consider the case where U = (0, 1)m- and the smooth map is defined in some neighborhood of the cube [0, l]m, because any open se~ U can be covered by countably many open cubes with this property. Let e E Cle n U. Then the Taylor expansion of I(c + h) has no terms of order lower than k + 1. Hence there exists a constant K such that II/(e + h) - I(e) II ~ Kllhllk+l whenever c + hE If. We divide the open cube U = (O,l)m into lm cubes with side length l/~ and consider the small cubes mtersecting Cle. The image of any such cube under I is contained in a disk of radius K dk+l, where d = v'nll is the maximUIIJ distance between points in the cube. Thus, the set I(U n CIe) can be covered by disks such that the sum of their volumes is at most K'lm(dk+l)" = K"lm-(k+l)". If m <:. (k + l)n, Le,! k > (min) -I, then liIIll-->oo zm-(k+l)" = 0. 0

I

1.5. An Important Example: Grassmann Manifolds. Consider the set G (n, k) whose elements are the k-dimensional subspaces of R". A topology on this set is defined as follows. Let VI, ••.• Vie E R" be linearlr independent vectors. The k-tuples (VI, .... , Vk) form an open subset X in R" x ... X R" = R"k. On the set X J we take the natural topology. The set G (n, k J is a quotient of Xi it is obtained by identifying k-tuples of vectors which generate the same subspaces. We endow G(n, k) with the quotient topology. In other words, a set U c G(n, k) is open if and only if the set of all bases of k-dimensional subspaces contained in U is open in X I

1. Definition and Basic Properties

193

The set G(n, k) can also be described as the quotient of X under the natural action of the group GLk(IR), which is defined 88 follows. We represent each k-tuple (Vh ..... ,Vk) as the matrix (~~.~.::: .. ~~~) whose rows are the 11"'1 ••• 11",.. coordinates of the vectors Vi. The matrix A E GLk(lR) takes this k-tuple to the k-tuple (WI j • • • ,Wk) of vectors whose coordinates form the rows of the matrix W = AV. It is easy to verify that the same topology on G(n, k) is obtained by considering only orthonormal sets of vectors rather than all linearly independent sets. Indeed, the map which takes every k-tuple of linearly independent vectors to the k-tuple obtained by applying the Gram Schmidt orthogonalization procedure is continuous. This approach to the definition of the topology on G(n, k) has the advantage that the space G(n, k) is obtained 88 the quotient of a compact space under the action of the compact group O(k). This implies, in particular, that the space G(n, k) is compact and Hausdorff. Another proof of the Hausdorffness of G (n, k) is 88 follows. Take a point x E lRR and consider the function dx on G(n, k) equal to the distance from x to a subspace II E G(n, k). This function is continuouS. Clearly; if ksubspaces 111 and 112 are different, then the point m can be chosen so that x E III and x ¢ 112. In this case, we have dx (lIl) = 0 and dx (1I2) i= O. Theorem 5.7. The topological space G(n,k) is a manifold (without boundary) of dimension k( n - k). Proof. Take linearly independent vectors Vh"" Vk in a subspace II E G(n, k) and consider the rectangular matrix V = V(II) whose rows are the coordinates of these vectors. For each multi-index I = {il,'" ,ik }, where 1 ~ i l < ... < i k ~ n. let '\tJ be the square matrix formed by the columns of V with numbers il,' .. , ik. Since the vectors VI, ... ,Vk are linearly independent, it follows that there exists a multi-index I for which detVI i= O. Consider the rectangular matrix (VI)-IV. Its columns with numbers iI, ... , ik form the identity matrix of order k. If WI, ••• ,Wk are some other linearly independent vectors in II~ then W = AV, where A E GLk(lR) and WI = AVI. Therefore, ~W.r)-IW

= (VI)-lA-lAV = (VI)-lv.

This means that the rectangular matrix {VI )-1 V depends only on the subspace II and the multi-index I; we denote this matrix by III. For any multiindex J, the square matrix 115 is defined. As mentioned before, III is the identity matrix. The elements of the other columns of III can be arbitrary. For every multi-index I, let UI C G(n,k) be the set-Qfthose subspaces II E G(n, k) for which det VI i= O. The sets UI cover G(n, k). Clearly, they

194

5. Manifolds

open. Indeed, if det lIi :f: 0, then det Vi :f: 0 for any matrix V' whose elements are sufficiently close to those of V-k For each subspace II C tJ~, we construct the matrix III and take the ", - k of its columns whose numbers are not included in the multi-index 1, As a result, we obtain a homeomorphislIl fPJ; Uz ~ JR,k(n-k),. It remains to verify that if 1 and J are two multi-indices. then the map CPJ'P/l, jWhich is defined on the open set CPI(UI n UJ), is smooth. The map 'PJ'P/ 1 can be desc.ribed as follows. 1'9 each point x (:: lRk(n-k) we assign the matr~ III whose n - k FOlumns are filled with ,:the cooJ:'l dinates of x and the remaining co11lIIllUt (with numbers containecl in the multi-index J) form the identity matrix. ';I'hen, we take a matrix V for which (lIi)-lV = III. The final result j.s IIJ = (V.f)-lV = (VJ}-llliIl I . To remove the ambiguity involved in the chollce of V \ we assume that VI is the identity matrix. In this case~ to '= hI we assign the matrix (XJ)-1 X and take its columns with numbers not Included in the multi-index J. By assumption, the matrix XJ is nonsingular on the entire domain; therefore, the resulting map is smooth. 0 ar~

?l

Exercise 35. Prove that the correspondence between subspaces and their orthogonal complements induces a. diffeomorphism G(n, k) ...... G(n, n - k). Exercise 36. Prove that G(n, 1) ~ lRpn-l.

Choose a basis in the space lRn. To each k-dimensional subspace II we assi~ (~) numbers x I, called the PlUcker coordinates, as follows. Take linT early independent vectors Vb • •• ,Vk in II and consider the matrix V whose rows are the coordinates of these vectors. For each multi-index 1~ we se~ x{ = det t;i. There are (~) multi-indices, and hence we obtain (~) numbers. The Plucker coordinates are determined uniquely up tq proportionality, Indeed, whell we take a different basis in II, the matrix V is replaced l;>y :AVi where A ~ GLk(lR)1 and each Plucker coordinate is multiplied by det At Theorem 5.8 (Plucker embedding). The PlUcker coordinates determine an embedding i: G( 'Ti, k) -+ lRP(:)-l t Proof. First, we show that l is a homeomorphism of G(n, k) onto its image i(G(n, k». For this purpose, it suffices to verify that the map i is injective, because any one-to-one continuous map from a compact space to a. Hausdorff space is a homeomorphism. Suppose that VI, •.• , Vk E lRn, are variable linearly independent vectors and el,'~" Cn-A: E lRn are constant linearly independent vectors. Consider the square matrix (~) whose rows are the coordinates of these ~ectors. By

1. Definition and Basic Properties

195

the Laplace theorem, we have det

(6)

=

Laldet~,

where al is a constant depending on the matrix C. Clearly, det (~} = 0 if and only if the intersection of the subspaces spanned by VI,"" vk. and CI, ••• ,Cn-k is nontrivial. Thus, the Plucker coordinates of a k-dimensional subspace IT satisfy the equation ~ alxl = 0 if and only if the intersection of IT with the subspace spanned by CI, ••• , Cn-k is nontrivial. It remains to note that for two different k-dimensional subspaces in IRn, there exists an (n - k )-dimensional subspace such that its intersection with one of the subspaces is trivial and the intersection with the other is nontrivial. Now, we show that i is an immersion. Take a chart Ul' On this chart, we introduce coordinates as explained above. To avoid Gumbersome notation, we consider only the simple example of Ul consisting of matrices ITl .of the form

100 Xl ( 0 1 0 X2 001 Xa

X4)

X5 X6

•

The image of a matrix IT1 under t is the set of all deter-

minants of its 8ubmatrices of order k = 3. Considering only the submatrices in which k ~ 1 columns coincide with those of the identity matrix, we see that in the chart Ul, the map i has the form

( o~ ~0 ~1 :~

X3

::)...-. (1, ±Xl, ±X2,±X3, ±X4, ±X5, ±X6,' .. ). X6

One of the coordinates of the image is 1. This means that the image of the chart Ul is contained in a standard chart of the projective space, Le., locally, i is a map of Euclidean spaces. Obviously, the rank of this JIlap is equal to the dimension of the manifold G(n, k). 0 Remark 5.2. The image of G(n, k) under the PlUcker embedding into IRP(~)-l can be determined explicitly by a system of equations called the PlUcker relationsj see [100, Section 30] for details. By analogy with the manifold of k-subspaces in IRn, we can consider the manifold of complex k-subspaces in en. To distinguish between these manifolds, we call them the real and complex Grassmann manifolds, respectively. Moreover, in the real case, we can consider the oriented Grassmann manifold G + (n, k), whose points are oriented k-dimensional subspaces. In this case, sets of vectors are considered equivalent if they not only span the .same k-subspace but also determine the same orientation on this subspace. Exercise 37. Prove that there exists a double covering G+(n, k) _ G(n, k). Exercise 38. Prove that the Plucker coordinates determine an embedding G+(n, k) - S(~)-l.

5. Manifolds

196

Problem 64. (a) Prove that the manifold G+(n, k) is always orient able. (b) Prove that the real Grassmann manifold G(n, k) is orient able if and only if n is even. Problem 65. Prove that G+(4,2) ~ S2 x S2.

°

Problem 66. Prove that the quadric z?+ .. ·+z! = in Cpn-l is diffeomorphic to G+(n, 2). Moreover, under the diffeomorphism, complex conjugation corresponds to reversing the orientation of the plane. Now, we describe the cell structure of the Grassmann manifolds. For each k-dimensional subspace II C Rn, consider the sequence of a; = dim(II n Ri), where i = 0,1, ... , n. It is assumed that R' consists of vectors of the form (Xl, .•. ,Xi,O, ... ,O). Clearly, ao = 0, tli+1 is equal-to-a; or tli + 1, and an = k. Therefore, to the sequence a; we can assign the Schubert symbol U = (Ul, ••• , Uk), where 1 ~ Ul < ... < Uk $ n, dim(II n R U , ) = j, and dim(II n RUj-l) = j - 1.

Lemma. A subspace II has Schubert symbol U if and only if it contains vectors VI, ••• , Vk such that the matrix whose rows are formed by their coordinates has the form

* ... * * * * ... * * ... * •

_.01

....

'

1

°° °**

......

°* ° °° °0 °° b° °0 °° °° .... * * ° °° ... * ° * * "'" * of the Schubert symbol are the numbers of the columns 1

~ 01 01 . . . . . . . . . . . . . . 01.01 ..

0 01

......

1

01

...

0 * 0 ... the elements Ul, ••• , Uk consisting of zeros and ones; the elements vectors v}, ... , Vk are determined uniquely.

01

f.' r f

0

*

0

~~

. . . . . . . . "" ......

01

••

01.

1 0 ...

are arbitmry. Moreover, the

Proof. The one-dimensional space II n RUl is generated by the vector VI = (x}, ... , XUl! 0, ... ,0). The condition dim(TI n R U l - 1 ) = 0 implies X UI -:/: O. Therefore, we can assume that X UI = 1; in this case, the vector VI is determined uniquely. In the two-dimensional space II n R U 2, the vector VI can be extended to a basis by a vector V2 = (y}, ... , YU 2' 0, ... ,0). The condition dim(II n ]RU2-1) = 1 implies Y U 2 -:/: O. Therefore, we can assume that YU2 = 1. In this case, the vector V2 is determined up to the addition of AVI. The Ulth coordinate of the vector V2 + AVI is YUI + A. We can kill it by choosing suitable A. Now, the vector V2 is determined uniquely. The rest of the proof is similar. 0 • The set of all subspaces with a given Schubert symbol U is called an open Schubert cell and denoted by e(u). Any open Schubert cell e(lt) C G(n, k)

un

1. Dennition and Basic Properties

is homeomorphic to the open disk of dimension d(O') = (0'1 -1) + (0'2 - 2) + .. ~ + (O'k - k). The open Schubert cells are pairwise disjoint and cover the entire Grassmann manifold. The distinguishing feature of any cell e(O') is that the subspaces which it contains have bases VI, ... , VA: such that Vi = (ViI, ... , Vid., 0, ... , 0), where Vid~ > 0. Hence the distinguishing feature of the closure e(O') of such a cell is that the subspaces which it contains have bases VI, ••• , VA: such that Vi = (Vil, ••• , Vid., 0, ... ,0), where VitT. ~ O. Note that the bases Vi, .. " vA: can be assumed to be orthonormal. Our immediate goal is to construct a continuous map Xu: Dci(u} _ e(O') C G(n, k) with the following properties: • the restriction of Xu to int Dd(u) is a homeomorphism onto an open Schubert cell e(O'); • the set XuC8D d(u» is contained in the union of open Schubert cells e(T) for which d(T) < d(O'). We construct such a map by induction on k. If k ~> 1, then the Schubert symbol 0' consists of one element 0'1, and d(O') = 0'1 - i. In this case, the set e(O') consists of all one-dimensional subspaces generated by nonzero vectors of the form (Vn, ..• , Vl Ul , 0, ... , 0), where Vl Ul ~ 0, and we define the map Xu: Dd(u) --+ e(O') as follows. We identify the disk Dd(u) with the hemisphere + ... + X!l = 1, where X Ul ~ 0, and define the image of each point (Xl, .•• ,XUl ) to be the one-dimensional subspace spanned by the vector (Xl, ..• , XU! , 0, ... , 0). Clearly, the restriction of the map Xu thus defined to intDd(u) is a homeomorphism onto e(O') and the set Xu (8Dd(u» consists of the subspaces spanned by nonzero vectors of the form (XI, •.• ,Xu!-l, 0, ... ,0); multiplying such a vector by a nonzero number, we can reduce it to the form (Xl, t •• , X n , 0, ... ,0), where xn = 1 and TI < 0'1. To construct the map Xu for k ~ 2, we need an auxiliary proper orthogonal transformation of R,n with positive determinant which takes a given unit vector u to another unit vector V and leaves all vectors orthogonal to u and V fixed. Such a transformation R(u, v) exists if u # -v, and it is unique. Indeed, it is easy to verify that the transformation defined by

xi

(u+v,x) R(u,v)x=x- 1 ( )(u+v)+2(u,x)v

+

u,V

has the required properties (on the plane spanned by u and VI it acts as a rotation because it takes u to V and V to the vector symmetric to u about v). Thus, the point R(u,v)x continuously depends on u, v, and x. Clearly, if u, V E IT, then the projections of the vectors X and R( u, v)x on II J., coincide. Suppose that the required map Xu: Dd(u) --+ e( 0') is constructed for any

198

5. Manifolds

Schubert symbol a = (al, ... " ak) of length k. Consider a Schubert symbol a' = (aI"." ak,O'k+l) of length k + 1 (as usual, we assume that 1 ::; al < a2 < '" < ak < ak+! ::; n). Instead of the map XO": Dd(O"~ -+ e(er'), we construct a map (which is even a homeomorphism) r.p: e(a) x Dd(O")-cl(O') -+ e(er'). The map XO" will be defined as the composition Dd(u') ~ Dd(O') x Dd(O")-d(O') ~ e(a)

here d(a') - d(a)

x Dd(u')-d(O') L

e(al );

= ak+! .... k-1.

In each subspace IT E e( a), we choose an orthonormal 'Oasis VI, .•• , vk such that Vi = (Vii, ••• , Viu., 0, .•. ,0), where Viu, ~ O. Let el,'" . , en be the canonical basis of lR". We set R

= R(eO',", Vk) 0 ••• 0 R(eO'2 , t12) 0 R( eO'I ,VI).

It is easy to verify that R eO'i = Vi for all i = 1, ... ,k. Indeed, the transformations R(eO'1 , VI), . . . , R(e O'. _1 , Vi-I) leave the vector ecti fixed, because this vector is orthogonal to VI, ••• , Vi-l and eO'I , .... , eo-,_l' The transformation R(eO'1 ,vi) takes eO', to Vi. Finally, the transformations R(eo- +1,vi+!), ••. , R( eO',", Vk) leave the vector Vi fixed because it is orthogonal to VHl, •• , , VI: an4 eUI+1 , ' •• ,eO'Te • We identify the disk DO'Te+1- k- 1 with the set of unit vectors W = (WI"'I' WU ,,+! ' 0, .•. ,0) such that WO'lo+l ~ 0 and WUi = 0 for i = 1, ... , k. We define the map r.p: e(a) x Dd(O")-d(u) -+ e(er') by l

r.p(VI, ... , Vk, w) = (VI. •.. ,vk,Rw).

We must verify that the space spanned by the vectors VII' •• f Vk, R W indeed belongs to e(er'). The vectors W and Rw have equal projections on the orthogonal complement of lRO''"; therefore, R W :;:: (*,,4o . , *, wUTe+h'~ . , wo-It+1 ' O~, .• , O)~ The linear independence of VI, ••• , Vk, Rw follows from, (vi,Rw)::; (Reu"Rw) = (eUi'w) = WO'i ::; 0

and (Rw,Rw) = (w,w) = 1. The map r.p is surjective: its inverse

<,0-1

is defined by the formula


where the orthonormal basis Vln L• ,Vk, Vk+! is constructed in precisely the same way as for the k-dimensional subspace with a given Schubert symbol and R is the orthogonal transformation defined above. It is easy to verify that the vector w = R- I Vk+1 has the required properties; namely, w

= (WI.'" 'lW

Uk

+1,O'T' .,0), where WO'Io+1 = Vk+l'O',"+1 ~ 0;

wO', = (eO'.. w) = (ReO'" Rw) (w,w) = (R-IVk+I.R-IVk+!)

= (Vi, VkH) = 0 for i = Ip"" k; and = (VkH,Vk+1) = 1.

2. Tangent Spaces

The map rp-~ is continuous and by the induction hypothesis, Xu is a homeomorphism between intDd(u) and d(u); therefore, Xu' is a homeomorphism between int Dd(u') and d(u') because int(e( (7) x

Dd(u')-d(u»

= e(u) x int Dd(u')-d(u).

2. Tangent Spaces It is easy to define a tangent vector at a point x E M" in a local coordinate system, but the passage to other coordinate systems involves difficulties. For this reason, there are several definitions of tangent vectors, which are used in different situations. One of the most natural definitions is as follows. A tangent vector at a point x E M" is an object to which a certain vector v = (vI,I" ,v") E ]R" is associated in every local coordinate system (U, c.p) with origin x; in a different local coordinate system (V, 1/1) with origin x, associated to the same tangent vector is the vector W = (WI, •.. ,w") related to v by

(1 ~

Wi

=L

a(1/Ic.p-1 )i (O)Vj.

')

aXj

111 other words, W

is the image of v under the action of the Jacobian n\.atrix

of the transition function 1/Ic.p~1. This definition is consistent because the Jacobian matrix of a composition of two maps is the product ofthe Jacobian matrices of these maps. The major drawback of this definition is that it depends on the coordinate system. An invariant definition can be given in several ways. Tangent vectors as classes of equivalent curves. To a vector v E ]R" we can assign the family of all smooth curves 1': (-1, 1) --+ ]R" for which 'Y(O) = 0 and ~(O) = v. If (u, c.p) is a local coordinate system with origin x E M, then the curve 'Y(t) determines the curve;Y = c.p- 11 on the manifold M"; moreover, ;:yeO) = x. Thus, a tangent vector at a point x E M can be defined as an equivalence class of smooth curves ;Y: (-1,1) --+ M" for which :reO) = x. Curves;Yl and ;Y2 are considered equivalent if d(c.p;yl(9)

dt

I

= d(c.p;y2(t» t=O dt

I

t=O'

in some coordinate system (U, c.p) with origin x. H (V, 'If;) is another coordinate system with origin x, then

d('lj;;Y(t»i dt

I

t=o

=

d('lj;
I

t=O

=

L i

8('lj;
8xj

dt

I. t=O

5. Manifolds

200

This implies, first of all, that the equivalence of curves does not depend on the choice of local coordinates. Second, we see that the coordinates of a tangent vector do indeed obey law (1) under a change of

d(plYU,I t-O

local coordinates. Tangent vectors as differentiation operators. Suppose that (U, ep) is a local coordinate system with origin x E Mn, v E lRn , and / is a smooth function defined in a neighborhood of x. To the function / we assign the number ~i 8(t~. 1) (O)v; and call it the derivative 0/ / in the-direction 0/ the vector field v. In a different coordinate system (V, 'IjJ), v is replaced by the vector w with coordinates Wi = ~j 8(.,p~, 1). (O)Vjj therefore, in the new coordinate system, the function / is assigned the number

L a(f'IjJ-1 ) (0) a('ljJep-1 )i (O)Vj = L 0(fep-1) (O)Vj. ..

'J

ax;

OXj

.

ax}

j

Thus, the number assigned to / does not depend on the choice of a coordinate system. We have associated to each tangent vector v at the point x E M n a linear operator v: coo(Mn) -lR (instead of COO (Mn), we could take COO(U), where U is a neighborhood of Xj the number v(f) depends only on the behavior of the function / in an arbitrarily small neighborhood of x). This operator has the following properties: (i) (Av + J.Lw)(f) = Av(f) + J.Lw(f)j (ii) v(fg) = /(x)v(g) + g(x)v(f).

The second property follows from the Leibniz rule 8~~:) =

gB!; + / at.

Exercise 39. Show that (ii) implies vee) = 0 for any constant function c. We can take properties (i) and (ii), together with the linearity of the operator v, as a definition of the linear space of tangent vectors at the point x E Mn, but we must verify that no "irrelevant" operators arise, i.e., that if v: COO(JRn ) -lR is a linear operator such that v(fg) = /(O)v(g) + g(O)v(f), then v(f) = ~i 3!;(O)Vi for some VI" •. , Vn E lR. For this purpose, we need the following auxiliary assertion. Lemma. Suppose that / E COO(U), where U c JRn is a convex neighborhood of the origin, and /(0) = O. Then there exist functions gl, .•. , gn E COO(U) such that f(x) = ~ xigi(X) and gi(O) = 3fCO). Proof. Clearly,

f(x)

= lex) - f(O) = (1 df(tx) dt = (I LXi af(tx) dt.

10

dt

10

aXi.

201

2. Tangent Spaces

therefore, we can take gi(X)

== f~ a~~:)dt.

o

This lemma immediately implies the required assertion. Indeed, since I(x) - 1(0) = E Xigi(X) , it follows that

v(f) =

L 0 •V(gi) + L gi(O)V(Xi) = L

::i

(O)Vi'

where Vi = V(Xi). The tangent vectors at a point x E Mn form a linear space. This space is called the tangent space at the point x and denoted by TzMn J Exercise 40. Suppose that J = {I E coo(]Rn) : 1(0) = O}, J2 = {E ligi ( Ji,gi E J} (the sum is finite), (JjJ 2)* is the space of linear functions on J j J2, and V is the tangent space at the point 0 E ]Rn. (a) Prove that if V E V and I E P, then v(I) = O. Thus, each tangent vector v can be treated as an element of the space (J j P)* . (b) For l E (JjP)*, let vl(l) = l(l(x) - 1(0». Prove that the operator has property (li) at x = o. (c) Prove that the maps V -+ (Jj P)* and (J j J2)* -+ V constructed in (a) and (b) are mutually inverse.

VI

2.1. The Differential. Suppose that I: M m -+ NY' is a smooth map and v E TxMm is a tangent vector. Then we can define a vector df(v) E Tf(x)N n , For example, if v is given as a curve "Y(t), then df(v) is the curve f(-y(t)). If v is given as a linear operator on smooth functions, then we define c{f(v) as the operator df(v)(ep) = v(ep!) (we can do this because ep E COO(Uf(x) implies epf E COO(Vx The map df ~ TxMm -+ Tf(x)NY' is linear; it is called the differential of the map I at the point x.

».

Exercise 41. Prove that d(l

0

g)

= df 0 dg.

A map f is an immersion (submersion) at a point x if and only if the differential of I at x is a monomorphism (epimorphism). This criterion for f to be an immersion (submersion) is sometimes very convenient. Example. Consider the map I from the space ]Rn2 of all matrices of order n to the space ]Rn(n+1)/2 of symmetric matrices defined by I(X) = XT X. This map is a submersion at all points of the set f- 1(In ), where In is the identity matrix. (Therefore, according to Theorem 5.3, the topological space 1-1(In) = O(n) is a manifold.) Proof. Suppose that U E O(n), Le., UTU = In. Consider a smooth curve of the form ry(t) + U + tA in the space of all matrices. The map f takes it

202

5. Manifolds

to the curve In + t(U AT + Aur) + oCt); therefore, its differential takes A to U AT + (U AT)T. Clearly, any symmetric matrix can be represented as X + X T , and any matrix X can be represented as X = U AT. Thus, the differential of I is an epimorphism at the point U. 0 Exercise 42. Prove that the unitary matrix space U(n) is a manifold. Exercise 43. Prove that the map I: U(n) --+ 8 1 defined by I(U) = det(U} is a submersion. In particular, 1-1 (1) = SU(n) is a manifold.

2.2. Vector Fields. On the set T Mn = UXEMft TzMn, the structure of a manifold can be defined as follows. Let (U, tp) be a local coordinate system on the manifold Mn. To each tangent vector at a point x E Mn we assign the pair (tp(x) , v), where v = (Vb .. p vn ) is the set of coordinatesof this tangent vector in the given coordinate system. As a result, we obtain a one-to-one map Ttp: TU =

U TxMn --+ tp~U) x ]Rn C ]R2n. xEU

The sets TU cover T Mn. Declaring all maps Ttp to be homeomorphisms, we endow T M n with the structure of a topological space. The charts (TU, Ttp) determine the structure of a manifold on this space. The manifold T Mn is called the tangent bundle of the manifold Mn. Problem 67. Prove that the manifold Tsn is homeomorphic to the subset of the complex space en +1 given by the equation zl + ... + z~+1 = 1. Assigning to a tangent vector v E 'J'zMn the point x, we obtain a projection p: T Mn --+ Mn. The map p is smooth. Any smooth map I: Mm --+ N n induces the smooth map dl; 'l'Mm --+ T N n (the differential of I). A vector field on a manifold Mn is defined as a smooth section of the projection p, i.e., a smooth map s: M n --+ TMn such that ps = idMnj it takes each point x E Mn to a vector v E T~Mn. The smoothness of the map s means that in any local coordinate system (U,tp), the vector field has the form L:ailx;'l where the ai are smooth functions on tp(Ui). A point x E Mn is said to be singular for a vector field s: Mn --+ T M n if sex) = O. Example. On the sphere 8 2n+1, there exists a vector field without singular points. Proof. If 1'(t) = x + ty + . .. is a smooth curve on the sphere sm, then the vector y E ]Rm+1 satisfies the equality (x,y) = O. Indeed, II 1'(t) 11 ~ 1 implies IIxl1 2 + t(x, y) + ... = 1; therefore, (x, y) = O. The dimension of the space

2. Tangent Spaces

203

formed by such vectors coincides with that of the tangent space; hence the vectors y E ]Rm+1 for which (x, y) = 0 are in one-to-one correspondence with the tangent vectors at the point X E sm. The (2n + I)-sphere lies in (2n + 2)-space. This means that the coordinates of x can be divided into pairs, x = (UI, VI, •.• , Un+1, Vn+1)' We set y = (-VI, UI, .•• , -Vn+l' Un+1)' As a result, we obtain a vector field without singular points on the sphere S2n+1. 0 Problem 68. Prove that there exist three vector fields on the sphere s4n+3 which are linearly independent at each point x E s4n+3. Problem 69. (a) Prove that the maps I,g: s2n+1 _ s2n+1 defined by J(x) = x and g(x) = -x are homotopic. (b) Prove that the maps J,g: ~n _ s2n defined by I(x) = -x and g(xo, XI, • •• ,X2n) = (-XO, Xl, •.• ,X2n) are homotopic. Theorem 5.9. On the sphere ~n, there exists no vector field tpithout singular points. Proof (see [88]). Suppose that vex) is a vector field on sm without singular points, i.e., v: sm _ ]Rm+1 is a smooth map such that vex) =1= 0 and (x,v(x» = 0 for all x E sm. Replacing vex) by v(x)/lIv(x)l\, we obtain a vector field consisting of unit vectors. Let us extend the map v over ]Rm+1 \ {O} by setting v(rx) = rv(x) for r > 0 and x E sm. For t E ]R, we define the map It: ]Rm+1 \ {O} _ ]Rm+1 by !t(X) = x + tv(x). If \lxll = r, then II/t(x)1I = v'f+t2 r, i.e., I(S;!,) c Sjl+t2 r' where s:' is the sphere of radius r centered at the origin. The Jacobian matrix of the map ft(x) has the form 1+ tJ(x) , where I is the identity matrix and J(x) is the Jacobian matrix of vex). Thus, for small t, the map It satisfies the hypothesis of the inverse function theorem, and the set f(S;!,) is open in sm~ . On the other hand, 1(8;') is comvl+t-r pact and, hence, closed in S~ . The connectedness of sm~ implies vl+t-r vl+t-,.

1(8;') = ~l+t2r' Let 0 <: a < b. Consider the set A = {x E sufficiently small, then !teA)

]Rm+1 :

a ~ IIxli ~ b}. If tis

= {x E ]Rm+1: VI +t2 a ~ IIxli ~ VI +t2 b}.

Therefore, the ratio of the volumes of the sets It(A) and A is (v'f+'t2)m+li The ratio of the volumes of !teA) and A can be calculated differently. First, let us show that for a sufficiently small t, the map It is one-to-one on A. Indeed, all partial derivatives of vex) are uniformly bounded on A (because A is compact); hence there exists a constant csuch that IIv(x)-v(y)1I ~ cllx-yll

5. Manifolds

204

for any x, yEA. Suppose that x, yEA and ft(x) = bey). Then x - y = t(v(x)-v(y», whence IIx-yll ~ cltl·llx-yll. For It I < c- l , we obtain x = y. The Jacobian determinant for ft(x) is equal to det(I + tJ(x» = 1 + to'l(X) + ... +tm+1O'm+I(X), where O'l, •• "O'm+1 are smooth functions. If t is sufficiently small, then this determinant is positive and ft is a hom&omorphism between A and ft(A); therefore, the volume of ft(A) equals ao + alt + ••. + am+1tm+1, where ao is the volume of A and ak

=

f· .. J

O'k(X)dxl

a~lxl

~J·dXm+1'


Thus, (v''I-+t2)m+1 is a polynomial in t of degree m+ 1. This is possible only if the number m + 1 is even. But we have assumed that m =~n, i.e., m+lis~d. 0 Remark 5.3. The nonexistence theorem for vector fields without singular points on the sphere 8 2n has many different proofs. The proof given above is very nonstandard. A more standard proof is based on the Poincare-Hopf theorem (Theorem 5.28 on p. 228); it uses also Theorem 5.40 on p. 244 and Example 2 on p. 250. Problem 70. (a) Given a smooth map f: 8 2n -+ ~n, prove that there exists a point x E 8 2n such that either f(x) = x or f(x) = -x. (b) Prove that any smooth map f: lRp2n -+ 1RP2n has a fixed point. Problem 71* ([2]). A division algebm is defined as a finitEHIimensional real space K with bilinear multiplication J.t: K x K -+ K without zero divisors (i.e., such that if v 1= 0 and W 1= 0, then J.t(v, w) 1= 0) and with two-sided identity element e (this means that J.t(e, v) = v = J.t(v, e) for all v E K}. Prove that if dim K ~ 2, then K contains a subalgebra isomorphic to C. 2.3. Riemannian Metric. To define a Riemannian metric on a manifold MR means to specify a smooth inner product (u, v) on the tangent space TxMn (an inner product (u, v) is smooth if the function f: T MR -+ lR defined by I(v) ~ (v,v) is smooth, or, equivalently, the function (X, Y) is smooth for any smooth vector fields X and Y on Mn). Theorem 5.10. Any manifold Mn admits a Riemannian metric. Proof. Let us cover Mn by a countable set of charts tpi: Ui _. lRn and construct a smooth partition of unity {fd for which supp fi CUi. For x E Ui , we define lion inner product ( . , . )i on TxMn as follows. Let vectors v,w E TzMR have coordinates (VI, .. • ,vn ) and (WI, ... ,wn ) in a local coordinate system (Ui,tpi). Then we set (V,W)i = VIWI + .. ·+VnWn <

2. Tangent Spaces

205

For a point x E Mn and a tangent vectors v, W E TxMn, we define 00

(v,w) = L:!i(X)(V,W)i i-I

(if the value (v, W)i is not defined, then x ¢ Ui and li(X) = OJ in this case, we set li(X)(V,W)i = 0). For fixed x, we obtain an expression of the form .AlAI + .. J + ).kAk, where the .A; are positive numbers with E.Ai = 1 and the Ai are positive definite symmetric bilinear forms. Any such sum is positive definite. 0

2.4. Differential Forms and Orientability. The cotangent space at a point x E Mn is defined as the space of linear functions on TxMnj it is denoted by T; Mn. The set T* Mn = UXEMn T; M n can be made into a manifold in the same way as T Mn. Namely, suppose that (U,
=

L

ail",ilcdxit A··· A dxilc '

il <···
where dXi(V) = Vi (the ith coordinate of'/J in this coordinate system) for all i and v. A manifold Mn is said to be orientable if there exists a set of charts {Ua , <Pa : a E A} which covers M n and has the property that the Jacobian 2The definition of the exterior powers of linear spaces can be found, e.g., in [100]. also [135, 8.4].

See

5. Manifolds

206

determinant of CPacp~l is positive for any a, f3 E A. Such a set of charts is called an orientation atlas. Exercise 44. Prove that the product of any two orient able manifolds is orientable.

=

Problem 72. Let I be a self-diffeomorphism of the manifold M n {x E ]Rn : 0 < IIxll < I} defined by I(tu) = {I - t)u, where 0 < t < 1, for all unit vectors u. Prove that f is orientation-reversing.

un is orientable nowhere vanishing n-form 0 on Mn.

Theorem 5.11. A manifold

if and only if there exists

a

Proof. First, suppose that the manifold M n is orientable. Let-1Q'Otl CPa} be an orientation atlas on Mn. The space of n-forms on an n-manifold is one-dimensional; the basis form won ]Rn is w = dXl/\'" "dx n • Consider the form Wa = (5cpa)w on Ua • If I: ]Rn --+ ]Rn is a smooth map, then (5f)w =.I J(x)w, where J(x) is the Jacobian determinant of I. Therefore; 5(cp,8cp~1)W == ..\w, where ..\ is a positive function on CPa(Ua n U(3). Thus, 5(cp~1)5(cp.B)w = ..\w, and hence w(3 = 5(cp(3)w = 5(CPa)(..\w) = ..\awa, where ..\a(x) = ..\(CPaex» is a positive function on Ua n U.B' We can assume that the cover {Ua } is no more than countable (see Problem 1). Let la be a partition of unity subordinate to the cover {Uet } and such that supp let C Ua for all Q. We set 0 = E lawet. Clearly, the form 0 does not vanish on Mn. Now let us assume that 0 is a nowhere vanishing n-form on a manifold Mn. Take an arbitrary atlas {Uetl CPet} and let W = dXl"" ·/\dxn be the basis form on ]Rn. The form 5(cpet)w, which is defined on Ual is proportional to OJ therefore, 5(cpet)w = /-LaO, where /-Let is a function on Ua • Since the function }.Let does not vanish, it follows that /-Let > 0 or }.La < 0 on the entire set Ua (we assume that Ua is connected). If /-La < 0, then we replace CPa by the composition t/Ja of CPa and the map lRn -+ ]Rn defined by (XI, X2, .. , ) ........ (X2 I Xl, ..... ). Clearly,

5(t/Ja)W = 5(t/Ja)dx l A dX2/\'" A dxn = 5(CPa)dx2 /\ dXl " '" /\ dX n where Va = -/-La> O.

= -}.Lan = vaO,

As a result, we obtain an orientation atlas, because 5(cp(3cp~I)W

= 5(cp~1)5(cp(3)w = 5(cp~1)(/-L(3n) = }.L~l}.L(3w.

0

• On each space N;;Mn, we can define an inner product in the same way as in constructing a Riemannian metric on the manifold Mn. Let};in be the

3. Embeddings and Immersions

207

set of unit vectors in AnMn, The space A:Mfl is one-dimensional; therefore, at each point X there are precisely two unit vectors. Thus, the natural projection p: Mn --+ Mn is a double covering. This covering is called the orientation covering of the manifold Mn, and Mn ts called the brientation covering manifold. This term is explained by the following theorem. Theorem 5.12. (a) A manifold M n is connected if and only if the manifold M n is nonorientable.

(b) The manifold Mn is always orientable. Proof. (a) If the manifold M n is disconnected, then each of its connected components is a nowhere vanishing n-form!l. Conversely, if an n-form n nowh~ vanishes, then !l(x)/II!l(x) II is a connected component ofthe manifold Mn. (b) By definition, each point x E Mn is an n-form !lex) in the tangent space at the point p(x) E Mn. The covering p: Mn -+ Mn induces an iso~rphism of cotangent spaces; therefore, !lex) can be regarded as a form on Mn. 0 Exercise 45. Prove that if 71'1 (Mn) = 0, then the manifold Mn is orient able, The points of Mn can be treated as pairs of the form (point x E M n , orientation of the space TxMn). Lifting a path "y C Mn to the covering manifold Mn corresponds to transferring orientation along the path "y. The transfer of orientation along a path can be interpreted geometrically as follows. Let us cover the path "y by finitely many charts 'Pi: Ui r+ lRn in such a way that each set "y n Uj is connected. Using a chart 'Pit we can define compatible orientations on all spaces TxMn with x E Ui. Thus, if [a, b] t "Y n Ui, then an orientation given at a can be transferred to b. Exercise 46. Give a definition of an orientation covering based on the geometric definition of orientation transfer along a path.

3.

E~beddings

and Immersions

We have defined embeddings and immersions only for manifolds without boundary; thus, we assume that all manifolds considered in this section are without boundary. There is a fairly simple procedure for constructing embeddings of compact n-manifolds in ]R2n+1. We describe it in Section 3.l. Then, in Section 3.2, we apply these embeddings to prove that any closed manifold is triangulable.

208

5. Manifolds

For noncompact manifolds, the construction is quite different. It is based on the fact that if n ~ 2m, then any smooth map Mm 1'"-+ ]Rn can be approx.. imated with arbitrary accuracy by an immersion. We discuss immersions of manifolds in Section 3.3; in Section 3.4, we prove that any n-manifold can be embedded in ]R2n+1 as a closed submanifold. All these embedding and immersion theorems were proved by Whitney [145]. A more delicate argument, also due to Whitney, shows that any n-manifold with n ~ 2 immerses in ]R2n-l, and any compact n-manifold embeds in ]R2n. A modern exposition of thE" proofs of these assertions is contained in [1]. 3.1. Embeddings of Compact Manifolds. In this section, we prove that any compact manifold Mn can be embedded in ]R2n+l. The construction includes two steps. First, we prove that Mn can be embedded in]RN for sufficiently large N. Then, we prove that if Mn embeds in]RN with N > 2n+ 1, then M n embeds in ]RN-l. Theorem 5.13. A compact manifold Mn can be embedded in]RN for sufficiently large N. Proof. On the compact manifold Mn, there exists a finite family of charts 'Pi: U, --+ ]Rn, where i = 1, ... ,k, with the following properties: (i) each set 'Pi(Ui) coincides with the open disk of radius 2 centered at the origin; (li) the preimages of the open unit disk under the maps 'Pi cover Mn; we denote them by~. Let us construct a smooth function A: ]Rn --+ ]R such that

A(Y) = and 0

{1o

if if

1,

lIyll ~ lIyll ~ 2,

< A(Y) < 1 if 2> Ilyll > 1. For this purpose, we take the function a () x

={

o e- 1 / z

if x ~ 0, for x > o.

Then, we set {3(t) = a(x - 1)a(2 - x); the function {3 is positive on the interval (1,2). Finally, we define

'"Y(T) = and A(y) =

12

{3(t)dt /

12

{3(t)dt

'"Y(lIyll).

• Let Ai(X) = A('Pi(X». The map Ai(X)'Pi(X) is defined on the entire manifold M n (if x ¢ Ui, then A;,(x) = 0). It is easy to verify that the map

3. Embeddings and Immersions

209

/: Mn _JR(n+1)k defined by

x

t-+

(Al(X), Al (X)c,ol (x), .•. , Ak(X), Ak(X)c,ok(X))

t::i. If X2 E Vi, then Ai(XI) = Ai(X2) = 1; therefore, the equality Ai(xdc,oi(XI) = Ai(X2)c,oi(X2) is equivalent to c,oi(XI) = c,oi(X2), i.e., Xl = X2. If X2 ¢ Vi, then Ai(XI) = 1, while Ai(X2) < l. The restriction to Ui of the map X t-+ Ai(X)c,oi(X) is an immersion. Indeed, if X E Ui, then ~(x) = 1, and the map x t-+ c,oi(X) is a local diffeomorphism. Hence /: Mn _JR(n+1)k is an immersion. It remains to note that anyoneto-one map from the compact space M n to the Hausdorff space JR(n+1)k is a homeomorphism of Mn onto its image. 0

is one-to-one. Indeed, take Xl E

Theorem 5.14. (a) Let /: M n _JR N be an immersion. 1/ N > 2n, then there exists a hyperplane JRN-I C JRN such that the composition 0/ f and the projection onto this hyperplane is an immersion. (b) Suppose that Mn is a compact manifold and /: un _ JRN is an embedding. If N > 2n + 1, then there exists a hyperplane JRN.... 1 C JRN such that the composition of f and the projection on this hyperplane is an embedding. Proof. (a) The kernel of the projection of JRN onto the hyperplane ~-l orthogonal to a vector v consists of the vectors proportional to v. Therefore, the composition of / and the projection onto ~-l is an immersion at a point x E M n if and only if the vector v does not belong to the image of the map TzMn

-!!!... T/(z)JRN ~ JRN.

To eliminate the zero vector, we consider SN-l instead of JRN. The images of collinear vectors under the map df are collinear, and the image of any nonzero vector is nonzero. Therefore, we can endow Mn with a Riemannian metric and construct a map g: T1M n _ SN-I, where TIMn is the set of unit tangent vectors. It is easy to verify that TIMn is a manifold of dimension 2n-l. Indeed, the smooth function T M n - JR evaluating the squared length of a tangent vector is a submersion at the point 1 E JR, and TIM n is the preimage of lER T-4e map 9 constructed above is smooth; therefore, if 2n-l < N -1, then its image has measure zero. In particular, there exists a vector v E SN-I that does not belong to the image of g. The composition of f and the projection onto the hyperplane JR~-l is an immersion. (b) We have already proved that if N > 2n, then the composition of the map f and the projection onto the hyperplane JR~-l is an immersion for almost all v E SN-I. Let us show that if N > 2n + 1, then the composition

5. Manifolds

210

of I and the projection onto ~-l is one-t
I(x) - I(y) g(x, y) = II/(x) - l(y)U,' Here ~ = {(x,y) E Mn x Mn l x = y} is the diagonal of Mn x un. The map 9 is well defined because if x '" y, then, by assumption, I(x) f: fey). The manifold (Mn x Mn) \ ~ has dimension 2nj therefore, if N > 2n + 1, then the image of g has measure zero. Clearly, if a vector v E SN-l does not belong to the image of g, then the composition of J and the projection onto ~-l is one-t
3.2. Triangulations of Closed Manifolds. Let Mn be a compact manifold without boundary, In this section, we prove, following [26], that Mn is triangulable, i.e., there exists a homeomorphism Mn ..,...... IKI, where K is a simplicial complex. To construct a triangulation, we embed Mn in ]RN. Each point x E Mn in ]RN is associated with two affine subspaces containing it, namely, the tangent subspace Txun and the normal subspace NxMn (the orthogonal complement of TxMn). We say that the sphere of radius r centeted at y is tangent to Mn at a point x if y E NxMn and lIy - xII = r. Lemma. For any closed manilold Mn C ]RN t there exists a number r

>

0

such that any sphere 01 radius less than r tangent to Mn intersects Mn only in the point 01 tangency. Proof. First, consider the case Ml C ]R2. Suppose that the graph y = I(x) of a smooth function J int~rsects the circle x 2 + (y - r)2 = r2 at a point (xo, r - v'r2 - xij) and is tangent to the circle at the origin, I.e., 1'(0) = o. Suppose also that maxtE[O,X]I"(t) = C. Then I'(r) = I"(t)dt $; ex (for .,. E [0, x]) and

I;

.

r - Vr2 -

ro

x~ = I(xo)"= 10 /,(1') dr ~

Cx T' 2

Performing a simple algebraic calculation, we see that if Xo E (0, r], then 2 >_. r-v'r2-x 1 __ ~_~o

Xo2

-

2r '

therefore, C ~ ~. This means that if the radius r is small, then the interval [0, r} contains a point at which the second derivative of I is large; to be more specific; its value at this point is at least 1/r.

J. ,l!.;mbeddings and ImmerslOns

L.L.L

Now, consider the general case M'Il C ]RN. The compact manifold M'Il can be covered by finitely many open sets Ui such that for each x E Ui, the orthogonal projection Pi,:.:: Ui -- Tz,M'Il is a diffeomorphism of Ui onto U:,:r: :::;: Pi,:r:(Ui) and, moreover, Ui is the graph of a smooth map 'Pi,:.:: U:,:r: -- N:r:.M'Il. If the sphere of radius r tangent to M'Il at x E Ui intersects U, at a point different from x, then the inequality C ~ l/r proved above implies a certain inequality for the second partial derivatives of 'Pi,:r:. Using it, we can find a lower bound for the radii of the tangent spheres intersecting Ui . If a tangent sphere has radius smaller than the minimum of these bounds over all i and is tangent to M'Il at a point x E Ui, then it does not intersect Ui (but it may intersect Uj for j :f i). Suppose that there exists a sequence of spheres which are tangent to M'Il at points Xl, X2, ••• , intersect M'Il at other points 1/t, Y2, ••• , and have radii rk -- O. We can assume that Xl, X2, ••• E Ui and Xk -- X E Ui (otherwise, we take a subsequence). Moreover, we can assume that all the rk are smaller than the minimum bound mentioned above; then none of the points Yk belongs to Uf.. On the other hand, IIXk - Ykll ~ 3rk -- 0 in ]RN. Therefore, Yk -+ x E Ui, contradicting the fact that all limit points of the sequence Yk must belong to the closed set M'Il \ Ui. 0 For each p > 0, we can choose points at, ... ,am E Mrl such that the open sets q'll(ak' p) = M'Il () {y E]R'Il, lIak '- yll < p} cover Mn. Take the number r from the lemma and choose a number p < r /2 so small that each set qn (ak' p) is homeomorphic to int D n and any straight line containing two points of the set ern (ak' p) makes an angle no larger than 7r/4 with the subspace Ta,.Mn. Then the orthogonal projection O''Il(ak' p) -t Ta,.M n is a homeomorphism onto its image. The sets ck = {x E Mn : IIx - akll ~ IJx - a.;.1I7i = 1, ... ,m} cover Ar'. Moreover, c~ c ern(ak,p), because if x E Mn and IIx - akll > p, then IIx - ai II < p for some i. Each set ck is the intersection of the manifold M n with the convex subset of ]RN determined by the inequalities IIx - ak n ~ IIx - a.;. II , where i = 1, ... , m. Consider the hyperplane Lki given by IIx - akll = IIx - aill. If Lki intersects Na,.Mfl in some point y, then the sphere ofradius lIy - akll centered at Y is tangent to Mn at ak and intersects M n in lli; therefore, (1)

Let 'T be a .ray emanating from ak in the space TfI4.un. This ray and the subspace NakMn generate a half-space fl.,. of dimension N - n+,1. The half-space H-,; intersects ern (ak' p) C M n in some curve 'Y; the projection of'Y on TakMn lies on the ray 'T. The curve 'Y intersects at least one of

212

5. Manifolds

Figure 2. A "bad" set

ct

Figure 3. The section by a balf-plane contained in HT

the hyperplanes Lki. Let us show that "y intersects this Lki in precisely one point and is not tangent to Lki. (If the curve 1 intersected Lki in two points, then the set c~ might be of the form shown in Figure 2.) Suppose that "Y is tangent to Lki or intersects it in two points. Let I be the tangent or the line passing through the two intersection points. Both intersection points belong to "Y C un(ak,p)j therefore, by assumption, the line I makes an angle of 0 ::; 7r I 4 with Talc Mn (for the tangent line, this assertion is proved by passage to the limit). Since the line I intersects un(ak' p), the distance from ale to I is at most p. Taking into account the fact that 0 ::; 7rI 4, we obtain lIak -

yll ::; pi coso::; pi cos(7r/4) = V2p

(see Figure 3), which contradicts (1). Thus, each curve 'Y intersects Lki in at most one point. This enables us to construct a homeomorphism between c~ and the convex polyhedron in TakM" determined by the same hyperplanes Lki as c~ (see Figure 4). (Note that no hyperplane Lki intersecting c~ is parallel to TalcM", because such a hyperplane would intersect NalcM" in a point y for which lIak-yll ::; p. And, of course, Lki cannot intersect c~ and TalcMn on different sides of the point al-.) The homeomorphism transfers the combinatorial structure of the convex polyhedron to C~j the i-faces of ck have an invariant description based on

3. Embeddings and Immersions

213

Figure 4. A set cj! and a convex polyhedron

the intersections of the hyperplanes Lki. This allows us to triangulate all the c~ face by face: first we triangulate the 2-faces, then the 3-faces, and so on. A compact manifold Mn with boundary 8Mn can be triangulated by the same method. Consider the closed manifold M'" obtained from two copies of Mn by identifying the respective points on their boundaries. We embed M'" in jRN, apply the lemma to }In and 8Mn, and choose a number r > 0 such that the assertion of the lemma holds for both manifolds. Then, we choose numbers Pl < r /2 and P2 < r /2 which can be used as P in triangulating M n and 8Mn, respectively. For P = min{Pl' P2}, we first choose points al..t..:.'" am E 8Mn and then supplement them by am+l, ;am+k E Mn c Mn. .J• •

3.3. Immersions. In this section, we prove the following assertion: Any manifold Mn (not necessarily compact) can be immersed in jR2nj moreover, if 2m :::; n, then any smooth map f: Mm -4 jRn can be approximated with arbitrary accuracy by an immersion. First, we calculate the dimension of the set of matrices of a given rank. Let Mn , m be the set of all matrices (~~.~.::: ...~-~~) with real coefficientsj such ani .p. Gntn matrices are in one-t~one correspondence with the linear maps jRm -+ jRn. The set Mn,m is naturally identified with jRmn. Let Mn,m,k be the subset of Mn,m = jRmn consisting of all matrices of rank k. Theorem 5.15. If k:::; min(m,n), then Mn,m,k is a manifold of dimension

k(m+n-k). Proof. Take any element 'Of Mn,m,k. Without lost of generality, we can assume that this element has the form (~ g~ where Ao is a nonsingular matrix of order k. IT € > 0 is sufficiently small, then any matrix A of order k such that the absolute values of all elements of the matrix A - Ao are smaller than e is nonsingular .. It is easy to verify that for such a matrix A, we have

),

(~ ~)

E Mn,m,k

{::=:>

D

= GA.-lB.

5. Manifolds

214

Indeed, rank

(~ ~) = rank [ (_~~-l In~k) (~ ~)] = rank (~ D _ gA-1B)'

The rank of the last matrix coincides with the rank of A if and only if

D-CA-1B=O.

Take a sufficiently small neighborhood U of (~~ ~~) E Mn,m,k C lRmn and consider the map cp: U - lRmn defined by

(~ ~) ~ (~

D-gA-IB)'

This map is invertiblej its inverse has the form

(~ ;)~(~ X+gA-IB)' Moreover, un Mn,m,1e = cp"-1 (lRk(m+n-k) n cp(U)) , where lRk(m+n-k) C lRmn IS the subspace consisting of the matrices of the form (~~). 0 Now, consider the local situation where Mrn is an open subset of lRm. Recall that "almost all" means "all except those from a set of measure zero.n Theorem 5.16. Suppose that U C lRm is an open set, f: U _ lRn is .a smooth map, and n ~ 2m. Then, for almost every linear map A: lRm _ lRn ~ the map g: U _lRn defined by g(x) = f(x) + Ax is an immersion. Proof. Consider the map Fk: Mn,m,k xU - Mn,m defined by Fk(X, x) == X - df(x). According to Theorem 5.15, the dimension of the manifold Mn,m,1e x U equals k( n + m - k) + m. The function k( n + m - k) monoton· ically increases in k for fixed m and n provided that k < (m + n)/2~ Thus, if k :$ m - 1, then k < m < 3m/2 ,$ (m + n)/2 and

k(n+ m - k) +m:$ (m -l)(n+ 1) +tn = (2m-n)

+mn~

1.

By assumption, 2m < nj therefore, dim(Mn,m,k x U) -< dim Mn,m~ and the image of FI!; has measure zero. This means that the set of linear maps of the form X - dl(x), where X E Mn,m,1e (k = 1, ... , m - 1), has measure zerOj Le .., for almost all linear maps A and all x E U'J the matrices A + df (x) have rankm. 0 Now) we are ready to prove the main theorem. Theorem 5.17. Suppose that n ;:::: 2m and 11 Mm - lRn is a smooth map . .'l;hen, for any c > 0, there exists an immersion g: Mm _ lRn such that II/(x) - g(x) II < c lor all x E Mm.

3. Embeddings and Immersions

21::>

Proof. Take a countable family of triples of open sets Ui,l C Ui ,2 C Ui,a such that the sets Ui,l cover Mm and for each i we have Uj.,k = c,o;l(Dr), where = {x lRm : IIxll ~ k} and c,oi: Ui,a -+ c lRm is a smooth chart; in addition, we require that the cover {Ui,a} be locally finite. a We construct the desired map 9 by induction, successively replacing the map li-l constructed at the (i -1)st step by a map Ii such that

Dr

E

Dr

(i) II/i(x) - li-l (x) II < e/2i for all x E Mmj (ti) the rank of Ii on the set U.,l equals m; (iii) the map Ii coincides with li-l outside Ui,2; (iv) the rank of Ii equals m at all points of the set Gj.

= U,,2n( U~:~Uj,l)'

Setting 10 = I and g(x) = lilli.....co li(x), we obtain the required map; it is smooth because the cover {U.,a} is locally finite, We proceed to constructing the maps Ii. For this purpose, we need a smooth function A: lRm ,..... lR such that

,\() = {1 y

0

1,

if nyll ~ if lIyll 2: 2.

(The construction of such a function was described on p. 208.) The map Ii will have the form Ii (x) li-1 (x) + ,\( c,oi (x) )Ac,oi (x). Obviously, all maps of this form have property (iii). We can perform the construction in the local coordinates associated with the chart 'Pi :.. Ui,a r+ lRm. In other words, we can assume that Ii is a map from c lRm to lRn and li(Y) = li-l(Y) +'\(Y)AY· If x E Ut,lt then y = c,oi(X) E 15ft; therefore, ,\(y) = 1. In this case, we have li(Y) = li-l(Y) + Ay. According to Theorem 5.16, the map Ii has rank m at all points for almost every A. ThIs ensures (i) and (ii). It remains to achieve (iv).

=

Dr

For each t, the set q = Ui,2'n (U~:~ Ui,l) is compact, and li-l has rank m at all of its points. Therefore, if all elements of a matrix A are sufficiently small, then the map li(Y) = li-l (y) + ,\(Y)AY has rank m for all y E OJ. (the function aCyl = minA: rank/.(y)~m max: lai;1 attains its minimum on the set Gi). d 3.4. Embeddings of Noncompact Manifolds. In this section, we sho~ that any n-manifold can be embedded in lR2n+1 as a closed submanifold. The proof is valid for both compact and noncompact manifolds, but we have already given a simple proof for compact manifolds. We say that an immersion I: M m -- Nn is one-to-one if the map M m -+ I(Mm) iSt Qne-.to-one. If the manifold Mm is compact~! then any 3Such open sets were constructed in the proof of Theorem $.4.

216

5. Manifolds

Figure 5. A one-to-one immersion which is not an embedding

one-to-one immersion of this manifold is an embedding. But for noncompact manifolds, this is not so (see Figure 5).

Theorem 5.1S. Suppose that n ~ 2m + 1 and I ~ Mm -+ lRR is an immersion. Then, lor any e > 0, there exists a one-to-one immersion g: Mm -+ lRR such that II/(x) - g(X) II < e lor all x E Mm. "'----

Proof. As in the proof of Theorem 5.17, we construct the desirea map 9 by successive approximations. We define open sets U i ,l C Ui ,2 C Ui,3 and a smooth function ).: lRm -+ lR with the same properties as in Theorem 5.17 but require in addition that the restriction of I to Ui,3 be one-to-one (this can be achieved because any immersion is locally one-to-one). This time, we seek maps Ii in the form li-l (X)+).(rpi(X»)Vi, where the Vi E lRR are constant vectors. We want the vectors Vi to be sufficiently small, so that IIVili < e/2i . The maps h and li-l differ only on the compact set U i ,2; therefore, if li-l is an immersion and Vi is sufficiently small, then h is an immersion also. The equality li(X) = li(Y) is equivalent to li-l(X) + ).(rpi(X»Vi = h-l(Y) + ).(CPi(Y»Vi, and it implies that ( ) 1

li-l (x) - li-I(Y) Vi = - ).(cpi(X» _ A(rpi(Y»

(provided ).(rpi(X» =1= )'(cpi(Y»))· Let N be the open subset of Mm x Mm consisting of all pairs (x, y) for which ).(rpi(X» =1= ).(rpi(Y». Consider the map N -+ lRR defined by the right-hand side of (1). The manifold N has dimension 2m < nj therefore, the image of this map has measure zero. This means that we can choose an arbitrarily small vector Vi which satisfies (1) nowhere on N. For this vector, the equality li(X) == li(Y) implies h-l(X) = li-I(Y) and A(cpi(X» = ),(cpi(Y». In particular, if x E Ui,l and li(X) = /iCy), then ).(cpi(X» :::;; ).(rpi(Y» = 1, which means that Y E Ui,lol Since the restriction of li-l to Ui,l is one-to0Il:e, we have x = y. Summarizing, we conclude that the restric~ion of h to Uj=l Uj,l is one-to-one. q As we will see, the obstruction to a one-to-one immersion I being an embedding is that the sequence {/(xn)} may converge even when the sequence {xn} has no limit points. We denote the set of limits of all such sequences {/(x n )} by L(f).

3. Embeddings and Immersions

217

Exercise 47. Prove that if I is a map from a compact manifold to 1R1'l, then

L(I) =

0.

Theorem 5.19. A one-to-one immersion and only il L(f) n I(M m ) = 0.

I:

Mm

--+

lRl'l is an embedding il

Proof. A on~to-one immersion I: M m --+ lRl'l is an embedding if and only if the map 1-1 : J(M m ) --+ M m is continuous. First, suppose that ,-1 is continuous. Then the relations limk-+oo I(xk) = y and y E I(Mm) imply limk_oo Xk = 1-1(y). Therefore, L(f) n I(Mm) =0.

Now, suppose that 1-1 is discontinuous. Then there exists a point y E I(MI'l) and a sequence Yk --+ Y such that the sequence {Xk = J- 1(Yk)} does not converge. The sequence {Xk} cannot have limit points different from x = 1-1 (y). Indeed, if xk. --+ x' ::f x, then Yko --+ I(x') ::f y. Therefore, the sequence {Xk} contains a subsequence having no limit points. 0 Now, we can prove the main theorem.

Theorem 5.20. For any manilold M m I there exists an embedding J: Mm --t 1R1'l, where n = 2m + 1. M oreaver, there exists an embedding such that the set I(M m ) is closed in 1R1'l. Proof. First, let us show that there exists a smooth function h: Mm --+ lR for which L(h) = 0. Take the same sets Uj,l C Vi ,2 c Ui,a, smooth function ..\: lRm --+ lR, and charts N, and h(Xk) ~ i > N. Hence the sequence {h(Xk)} has no limit points. Consider the map 12: Mm --+ lR2m+1 defined by 12 (x) = (h(x),O, ..• ,0). According to Theorem 5.17, for any e ::> 0, there exists an immersion fa: Mm --+ lR2m+1 such that jlh - lall < e, and by Theorem 5.18 there exists a on~to-one immersion I: U m ..-+ lR2m+1 such that llfa -- JII < E. We show that L(f) = 0 (for all e). Suppose that a sequence {Xk}, where X/c E Mrn, has no limit points. Then, for any positive integer N, we can choose keN) such that h(Xk) > N whenever k ~ keN). Therefore, the sequence {/(xk)} has no limit, because II/(xk) - 12 (Xk) " < 2c:.

218

5. Manifolds

It remains to verify that the set f(Mm) is closed. This following lemma.

is implied by the

Lemma. The set f(M m ) is closed in an if and only if L(f) C J(M m). '\

Proof. First, suppose that f(Mm) is closed in an and y E L(f). Then y = limk~oo f(Xk), where Xk E Mm. and hence y E f(Mm). Now, suppose that L(f) C f(M m) and a point y belongs to the do. sure of f(Mm). Then there exists a sequence {Xk} C Xk E Mm such that /(Xk) - y. H the sequence {Xk} has a limit point ~, then it has a subsa. quence Xki - x, and y = liIlli-+oo f(xk.) = f(liIlli-+oo xk.) = f{x) E f(Mm). H the sequence {Xk} has no limit points, then y E L(I) c f(Mm). 0 For the embedding f constructed above, the set L(f) is empty, and hence f(Mm) is closed. 0 3.5. Impossibility of Certain Embeddings. In this section, we prove that no closed nonorientable manifold of dimension n can be embedded in an+! . The proof uses fairly obvious assertions concerning transversality and general position, which we do not prove rigorously. We only give the definition of transversality.

Suppose that X and Y are smooth manifolds and W C Y is a submanifold of Y. We say that a smooth map f: X -+ Y is transversal to the submanifold W at a point x E X if one of the following conditions holds: (a) f(x) ¢ W; (b) f(x) E W and TJ(x)W + (df)x(TxX) = T.(~)Y"l If a map I is transversal to W at all points x EX, then we say that is transversal to W 4 Example. H dim X + dim W if and only if f(X~ n W = 13.

< dim Y I then

I: X

-

f

Y is transversal to W

Theorem 5.21. Suppose that M n is a manifold without boundary (not nec· essarily compact) and /j Mn _ ,Nn+! is an embedding into a simply connected manifold N n+l such that f(M n ) is closed. Then M n is orientable. Proof (see [116]). Suppose that the manifold un is nonorientable. Let 'Y be a curve on Mn such that the transfer along this curve is orientationreversing. Then the transfer along'Y reverses the directions of normal vectors to Mn. H the length of a normal vector transferred along 'Y is constant and sufficiently small, then the curve described by the end of this vector does not intersect Mn, Using this nonc1osed curve, we can easily construct a smooth closed curve ::y which transversally intersects Mn in one point. We show that in fact such a curve does not exist.

3. Embeddings and Immersions

219

The contraction of the curve ::y in the space Nn+1 determines a map

g: D2 -+ NR+l whose restriction to aD2 coincides with::y. We can assume that 9 is smooth. Slightlx. varying f and g, we bring feAr) and g(D2) into general position. For n ;?: 3, the generic disk g(D2) is self-avoiding; therefore, the intersection of f(Mn) and g(D2) consists of closed curves and arcs of curves whose endpoints are different points of::Y; these curves and arcs are self-avoiding and pairwise disjoint. If n = 2, then the self-intersections of the disk may be impossible to remove by slightly changing g, but they can be made double and triple; the double points sweep out some curves, and the triple points are isolated. In general position, the submanifold f(MR) does not pass through the triple points of the disk g(D 2 ). Thus, the intersection ot f(Mnl and q(D2) again consists of closed curves and arcs, but this time, the curves may transversally intersect each other and have transversal se1£intersections. However, the number of intersection points of these curves with the curve ::y is again even. This is sufficient to obtain a contradiction, becaUSe Mn intersects ::y in precisely one point. 0 Corollary. No closed nonorientable manifold of dimensio". n can be t(mbedded in lRR+l. Thus, any closed two-dimensional surface embedded in S3 is orientable. Using this fact, we can obtain a complete description of all closed two-dimensional surfaces which can be embedded in Rpa. Clearly, lRp2 is one of them. To the surface lRp2 embedded in lRp3 we can attach any number of handles. Thus, any closed nonorientable surface with odd Euler characteristic can be embedded in lRp3 • Theorem 5.22 (see [211). No closed nonorientable two-dimensional surface with even Euler characteristic can be embedded in lRp3 . Proof (see [30]). Suppose that M2 is a closed nonorientable surface embedded in lRp3. We want to prove that its Euler characteristic X(M2) is odd. Let q ~ S3 -+ S3 be the antipodal involution (it is defined by q(x) == -x), and let p: S3 -+ lRp3 = S3 / q be the natural projection. Take the equatorial sphere S2 in S3, We assume that .lRp2 = p(S2) C JR.P3. Slightly moving M2 if necessary, we can assume also that it intersects lRp2 transversally. If M2 n lRp2 is disconnected, then by attaching several handles to M2, we can Construct a new two-dimensional surface N2 for which N 2 n lRP2 is connected. Clearly, the surface N2 is nonorientable and X(M2) == X{N2) (mod 2). In what follows, we assume that M2 n IRp2 is connected. Let us show that p-l(M2) n S2 is connected as well. Since the surface p-l(M2) is embedded in S3, it is orient able (by the corollary of Theorem 5.21). The quotient of p-l(M2) under the antipodal involution q. is a nonorientable two-dimensional surface; therefore, the restriction of q to

5. Manifolds

220

Figure 6. The surgery

p-i(M2) is orientation-reversing. On the pther hand, (T preserves the orientation of the sphere S3. Hence (T cannot take connected components of S3 \p-l(M2) to themselves. Each connected component of s2 \ Jf'. . I{M2 ) is contained in a connected component of S3 \p-l(M2); thus, (T cannot take connected components of s2 \p-l(M2) to themselves either. It follows that the number of connected components of s2 \p-l(M2) is even. By assumption. M2n1RP2 is connected. Therefore, p-l(M2)nS2 has one or two connected components, i.e., S2 \p-l(M2) has two or three connected components. But we have proved that the number of connected components of S2\p-l(M2) is even. Hence S2\p-l(M2) has two connected components. This means that p-l(M2) n S2 is connected, Le., p-l(M;2) n S2 ~ 8 1 .. In what follows, it is convenient to assume that JR.P3 is obtained from D3 by identifying antipodal points of the sphere SJ = aD3. In this case, the restriction of p: D3 -1Rp3 to S2 is a. homeomorphism, while the restriction of p to S2 remains a double covering. Let D3 ±: Dk = {x E R3 J IIxll ::5 R}. We can choose e > 0 such that the intersection of the closure of D~ \ Dk_e with M2 is homeomorphic to the product of p~I(M2) n S2 ana the interval {R - e, Rj, Le., to the cylinder SI x I. In JRP3, this cylinder transfo~ into ~ Mobius band. Let I;~ be the sphere 8D~_E; it intersects M~ in a circle. We cut I;2 and M2 along this eircle and glue together the four pieces thus obtained into two closed surfaces as shown in Figure 6. Namely, we attach one half of the -sphere ~2 to the Mobius band and obtain RP2. The other half of I;2 is attached to the remaining part of M2. As 8. result; we obtain an orient able (we again apply the corollary of Theosurface, because it is embedded in rem 5.21). Suppose that this surface has 9 handles. Then X(M2) + X(I;2) = X(Rp2) + 2 - 2g, and hence X(M2) X(RP2) = 1 (mod 2). 0

na \

na

=

4. The Degree of a Map 4.1. The Degree of a SUlooth Map. Let f: MR --t NR be a smooth Illap of manifolds of the same dimension n. We assume that the manifolds MR

4. The Degree of a Map

221

and N n are closed and oriented. According to Sard's theorem, the map I has a regular value y E NR. Let x E I-I(y). Since dl(x): TzMn - T1I Nn is an isomorphism, we can choose local coordinates near x and y which determine orientations compatible with those of the manifolds Mn and N n . Let signJ/(x) denote the sign of the Jacobian of I at the point x. The degree of the map I with respect to the point y is deg(f, y) =

L

sign J/(x).

zE/-l(1I)

This sum is defined because the set I-I(y) is finite. Indeed, suppose that I-I(y) contains infinitely many different points. Since the manifold un is compact, there exists a sequence of pairwise different points Xi E I-I(y), where i E N, which converges to Xo. We have I(xo) = Yj by the inverse function theorem, the point Xo has a neighborhood U' in which I is a homeomorphism. In particular, (U \ {xo}) n 1-1 (y) = 0. This contradiction shows that I-I(y) is finite. We assume that if the set I-I(y) is empty, then deg(f, y) = o. J

Example. Let Sl = {z E C :t Izl = 1}. Consider the map I~ Sl _ Sl defined by I(z) = zn, where n E Z. If n =f 0, then deg(f,w) = n for any point W E SI. (If n = 0, then the irregular point w = 1 is excluded from consideration. ) Let I, g: M m - N n be smooth maps. We say that I and 9 are smoothly homotopic if there exists a smooth map (homotopy) F t Mm x I .,..... N n such that F(x, 0) = I(x) and F(x, 1) = g(x) for all x E urn.

un -

Theorem 5.23. Suppose that I, g: NR are smoothly homotopic maps 01 closed oriented manifolds and y E Nn is a regular value for both maps. Then deg(f,y) = deg(g,y). Proof. Let 1-1 (y) = {Xl,"" Xk}. The points X}, ... , Xk have pairwise disjoint neighborhoods UI,"" Uk which I diffeomorphically maps to neighVi \ J{M \ U Ui). This borhoods Vi, ... , Vk of y. Consider the set V = set is open and contains y. The preimage of each point y' E V consists of exactly k points ~, ... ,x~, and signJ,{~) = sign Jf{x). Therefore, deg(f, y) = deg(f, y'). Consider a similar neighborhood of y for the map 9 and,take the intersection of these two neighborhoods. We obtain an open set Way such that any point .t E W is a regular point for I and gj moreover, deg(f, y) = deg(f, z) and deg(g, y) = deg(g, z) for z E W. It follows from Sard's theorem that the map F: Mn X I ...... JV'I has 8 regular value z in the open set W. Let us show that deg(f, z) = deg(g, z). According to Theorem 5.3 (see p. 186), the set F-I{z) is a one-dimensional submanifold in Mn xl, The connected components of this set are circles

n

5. Manifolds

222

Figupe 7. The manifold ,-I(Z)

Figure 8. The orientation of the maItifold F- 1 (z)

and segments; the endpoints of the segments may belong to one of the sets Mn x {O} and Mn x {I}, or they may belong to different sets (see Figure 7). The manifold F-I(z~ can be oriented as follows. First, we orient the manifold Mn x I. Then, we choose positively oriented local coordinate systems near w E F~l (z) and z so that the map F has the form (Xl, ... ,Xn , X n +1) ...... (Xl, ... ,Xn ) in these local coordinates. The orientation of F-l(i) at the point w is determined by the direction of the coordinate X n +!. We are interested only in those connected components of FTl(z) that are segments. H the endpoints of such an oriented segment belong to different sets Mn x {O} and M n x {I}, then the orientations at these endpoints have the same sign with respect to the orientation of I, and if the endpoints belong to the same set, then the orientations at its endpoints have opposite signs (see Figure 8). The signs of the Jacobian determinants of the maps 1 = F t M n x {O} and 9 = F Mn x {I} at each point w E p-l(z) are completely determined by the sign of the orientation of the manifold F-l(z) at w with respect to the orientation of I. Therefore, the endpoints of an oriented segment correspond to either two points at which the Jacobians of 1 and 9 ha.ve the same sign or two points at which the Jacobian of the same map (I or 9) takes values of opposite signs. This implies deg(j, z) = deg(9, z). 0

r

4. The Degree of a Map

223

Consider an oriented manifold of the form W n +1 = MfI, X. I. The orientation of wn+! induces opposite orientations on the manifolds Mn x {O} and Mn x {l}. Thus, Theorem 5.23 is a special case of the following assertion. Theorem 5.24. Suppose that wn+! is a compact orienteil manifold with boundary awn+! (endowed with the induced orientation), N R is a closed oriente4 manifold, f: W n +! -+ Nn is a smooth map, and 11 is a regular 'Value 0]1 t awn+!. Then deg(f t awn+t, 1/) = o. The ,Proof of Theorem 5.24 is similar to that of Theorem 5.23. Note the orient ability assumption is essential. For example, the pegree of the restriction to ~he boundary of the projection of the Mobius band onto its median circle is nonzero (it equals ±2). But when degrees are considered modulo 2, Theorem 5.24 holds also for nonorientable manifolds W n +!. tha~

Theorem 5.25. Suppose that f: M n -+ Nn is a smooth map of closed oriented manifolds, the manifold N n is connected, and X, 11 E NR are regular 'Values lor f. Then deg(f, 11) = deg(j, x). Proof. Let h: Nn -+ Nn be a diffeomorphism. A point 3: E N n is a regular value for f if and only if the point hex) is a regular value for hf~ I~ follows directly from the definition of the degree of a map that if h is orientationpreserving, then deg(f, x) = deg(hf, h(x». Therefore, it suffices to prove the existence of a diffeomorphism h; JVn -+ Nfl, with the following properties: (a) h is orientation-preserving;

(b) hex) = y; (c) the map hf is smoothly homotopic to!~ Indeed, if such a diffeomorphism exists, then the point y is a regular value for the homotopic maps J and hf, and therefore deg(hf,y) = deg(f,y). Diffeomorphisms ho and hI are said to be isotopic if they are smoothly homotopic and all the intermediate maps ht are diffeomorphisms. Clearly, any diffeomorphism (of orient able manifolds) isotopic to the identity diffeomorphism is orientation-preserving. Thus~ it remains to prove the following lemma. Lemma (homogeneity of manifolds). Let Nn be a connected manifold without bo,undary. Then, for any two points x, y E N n , there exists a diffeomorphism h: N n -+ NR that is isotopic to the identity and takes x to y. Proof. Let A: ]Rn l""'"'+ lR be a smooth function such that A(X) > 0 for IIxll < 1 and A(X) = 0 for IIxl[ ~ 1. Consider the differential equation : = >.(x)c, where c E lRn is a fixed vector. Let Ft(x) be a solution to this equation with initial condition Fo(x) = x. Clearly, Ft+s = Ft 6 Fs; therefore, Ft is a diffeomorphism isotopic to the identity. The map Ft leaves all points outside

224

5. Manifolds

the unit disk fixed and shifts all points inside the unit disk in the direction of the vector c. Suppose that IIxll < 1 and lIyll -<; 1. We set c = y ..... x. For some t > 0, the diffeomorphism F, takes x to y and leaves all points outside the unit disk fixed. Using this construction 1 we can obtain the required diffeomorphism h:< Nn -+ Nn in the case where the points x, y E Nn belong to the same chart cp: U -+ lRn , where cp(U~ is the open unit disk. We say that points x, yEN" are equivalent if there exists a diffeomorphism which is isotopic to the identity and takes x to y. The above considerations show that the equivalence classes are open sets. But the connected manifold Nn does not admit a nontrivial decomposition into a union of pairwise disjoint open sets. This means that there is only one equivalence class. 0 According to Theorem 5.25, if M" and N n are closed oriented manifolds and N" is connected, then, for any smooth map J: Mn -+ N", the number deg(J, x) does not depend on the regular value x. This number is called the degree of the smooth map 1 and denoted by deg I. Remark 5.4. For closed (but not necessarily orient able) manifolds Mn and N", the degree modulo 2 can be considered (for nonorientable manifolds, the Jacobian determinant has indefinite sign, but -1 = 1 (mod 2)). For the degree modulo 2, Theorems 5.23-5.25 remain valid. Problem 73. Let M2 be a sphere with 9 handles, where 9 ~ 1. Prove that any smooth map I: 8 2 -+ M2 has degree zero. Problem 74. Prove that deg(Jg) = (degf)(degg). Problem 75. Let P(z) be a polynomial of degree n. Prove that the map C -+ C defined by z 1-+ P(z) can be extended to a smooth map Cp1 -+ Cpl. Calculate the degree of this map. Problem 76. Let R(z) be a simple fraction of two polynomials of degrees m and n. Prove that the map defined by z 1-+ R(z) can be extended to a smooth map Cp1 -+ Cpl. Calculate the degree of this map. Problem 77. Given a map I: S" -+ S", consider the map 'E/: 'ES" -+ 'ES" which takes (x, t) to (J(x), t) for all (x, t) E S" x {t} and all t. Prove that degl = deg'Ef. Problem 78. Let 8 2n- 1 be the unit sphere in the space en with coordinates (r1 ei 'Pl, ... ,rnei'Pn). Calculate the degree of the map f: 8 2n- 1 -+ S2n-l defined by (rl ei'Pl,. •.• ,rnei'Pn ) 1--+ (rle ik1 'Pl, • •• ,rneikn'Pn)~

where kl' ... ! k n are integers.

4. The Degree of a Map

225

Problem 79. Is the map I: SO(n) - SO(n) (n ~ 2) defined by I(A) = A2 homotopic to the identity map?

4.2. The Index of a Singular Point of a Vector Field. Suppose that M" is a manifold without boundary and v: M" - T M" is a smooth vector field on M". A point x E M" is singular for v if vex) = o. A singular point x is said to be isolated if it has a neighborhood containing no other singular points. Let U be an open subset in R.", and let v: U _ R." be a smooth vector field with an isolated singular point Xo E U. For sufficiently small r > 0, the disk II x - Xo II ~ r contains no other singular points. Consider the map of the sphere IIx - xoll = r to the unit sphere defined by x ~ v(x)/lIv{x)lI. The degree of this map is called the index of the singular point Xo. Clearly, the index of a singular point is an integer, and it continuously depends on r (provided that the singular point Xo under consideration is the only singular point in the disk IIx - xoll ~ r)j therefore, the index does not depend on ,. at all. The index of an isolated singular point Xo E M" of a vector field v can be defined as follows. Consider a smooth chart cp: U - R.", where Xo E U and

5. Manifolds

226

Proof. For any a E ]Rn.. the map ~ t-+ a + I(x) is a diffeomorphism. Ther~ fore, the maps It(x) = (t-l)/(O)+ I(t) are diffeomorphisms for all t. Moreover, h = I and 10(0) = O. Thus, we can assume that 1(0) = O. According to the lemma on p. 200, I can be represented in the form I(x) = t:Xi9i(X~, where 91, ... ,9n are smooth maps and 9i(0) = G!;(O). Setting

F(x, t)

= x19l (tx) +,.,. •+ Xn9n(tX),

we obtain an isotopy between

F(x, 0) =

I

Xl

and the linear transformation

81 al -a (0) + · '1+ Xn -a (0). Xl Xn

It remains to prove that any orientation-preserving linear transformation is isotopic to the identity transformation. Each matrix with positive determinant can be represented as SU, where S is a symmetric positive defi.. nite matrix and U is an orthogonal matrix with positive determinant. There exists a basis in which the transformation S has diagonal matrix with positive diagonal elements, and there exists a basis in which the transformation U has block-diagonal matrix with diagonal elements 1 and (-C::Y' The isotopies between the transformations S and U and the identity transformation are constructed in the obvious way. 0

=:).

We proceed to prove Lemma '1. First, suppose that the diffeomorphism I: ]R" ...... ]Rn preserves orientation. Let It be an isotopy between I and the identity map. The index of the singular point It (Yo) of the vector field dft (w) does not depend on t; therefore, the index corresponding to t = 1 coincides with that corresponding to t = O. But this is exactly what we need. Now suppose that the diffeomorphism f reverses orientation. Consider the symmetry S(Xi,X2, ... ,xn ) = (-X1,X2, .•. ,xn ) about the hyperplane Xl = O. The diffeomorphism sl preserves orientation. Therefore, it is sufficient to show that the indices of the singular points Xo and s(xo) of the vector fields w and ds(w) coincide. If w(x) = (WI' W2,"" wn}t then ds(w(s(x») = (-WI, W2,"., wn ) = sw(x).

Hence the index of the singular point Xo of w is equal to the degree of the map W: 8"-1 _ 8"-1 defined by W(x) = w(x)/lIw(x)lI, and the index of the singular point s(xo) of ds(w) is equal to the degree of the map W" = sWs-l. We have degs '= degs- 1 = ~1 and, according to Problem 74, degW' = (degs)2degW degW. j

=

Problem 80 (Poincare). Suppose that the trajectories5 of a vector field "? with one singular point in the plane are tangent to some circle (J internally • 5The trajectOf'J/ of 8, vector field is a cw-ve -yet) such that the velocity vectOll ~(to) is equal to v(-y(to for all to·

»

4. The Degree of a Map

227

at i points and externally at e points, and C encloses the singular point. Prove that the index of this singular point is equal to 1 + (i.,.... e)/2. Problem 81 *. Let ! be a smooth function in the plane with coordinates (Xl, X2). Prove that the index of an isolated singular point of the vector field v = grad! = (8fI8xl,8fI8x2) can take the values 1, 0, Ll, -2,. ... and cannot take other values. Suppose that a manifold M n is embedded in ]RN and 'I(J: rJ -+ Mn C ]RN is a diffeomorphism of a domain U c ]Rn onto a domain !p(uj c Mn, Let x = (Xl, .•• ,Xn) E U~ Then the vectors ~(x) = B£(x) form a basis in the space T!p(:.;)Mnj hence v(!p(x» = ~ vi(x)ei(x), where the 'Vi are smooth functions. Each vector ej(x) is determined by the curve rp(Xlt .•. ,Xj+t, .. ,. ~Xn). The map v takes it to the curve ~ Vi( ... ,Xi +t,.,. )ei(,rt ,Xi + t1 t •• ). The tangent vector to this curve is ~8v.

L.J 8 '. (x)~(x)

. x,

~

~

8~

+ L.J vi~x)-8 (x). Xj ...

IIi particular, if rp( x) is a singular point of v I then this tangent vector belongs to the space spanned by el (x), ... , en (x). This means that the tangent spaces at the singular points of the vector field v are invariant under the map dv. A singular point y of a vector field v is said to be nondegenemte if the linear operator dv: TyMn -+ TyMn is nondegenerate.

Theorem 5.26. Any nondegenemte singula1t point y of a vector field 11 is isolated and hal} index ;il; the sign of the index coincides with that of the determinant of the operator dv T TyMn J TlI Mn r Proof. Choose local coordinates with origin y. We shall consider vasa map from IRn to ]Rn. By assumption, the Jacobian determinant of this map does not vanish at the origin; therefore, by the inverse function theorem, there is a neighborhood U of the origin such that v is a diffeomorphism of U onto its image. (This implies, in particular ~ that v has exactly one liin~ gular point in U.) Identifying the neighborhood U and its image with IRn, we obtain a diffeomorphism v: IRn -+ ]Rn. This diffeomorphism preserves orientation if and only if det(dv) > O. According to Lemma 2 on p. 225, any orientation-preserving diffeomorphism IRn -+ ]Rn is isotopic to the identity diffeomorphism. Thus, if v preserves orientation, then the singular point has index 1. If the diffeomorphism 11 reverses orientation, then it is isotopic to a symmetry about a hyperplane. In this case, the map sn-l - t sn- 1 has degree -1, and the index of the singular point is -1. [J

228

5. Manifolds

One of the most important properties of vector fields with isolated singular points on closed manifolds is that the sum of indices of their singular points is constant. To prove this, we need the following assertion, which is also used in many other theorems. Theorem 5.27 (tubular neighborhoods). Suppose that Mn is a closed manifold and f: M n - lRm is an embedding. Let Me be the set of all points in lRm which are at distance at most e from f(M n ). Then the numbere > 0 can be chosen so that each point y E Me has a unique representation y = x + where x E Mn and flTzMn.

e,

e

Proof. Let N be a set of pairs (x, e), where x E Mn and is a vector orthogonal to TzMn C lRm. We can define the structure of an m-manifold on the set N as follows. We introduce local coordinates (UI,"" Un) on Mn and choose an orthonormal system of vectors CI,"" em-n orthogonal to TzMn at each point x of this local coordinate system; we assume that the vectors ei smoothly depend on x. To each pair (x,e) we assign the coordinates (UI, ... , Un, 6,.··, em-n) , where e eici + ... + em-nem-n' In these coordinates, the map F defined by (x, x + has the form

=

e) . . . .

(UI, ... ,un,6, ... ,em-n)

xCUI, ••• ,un) E Mn

e

....... x(Ult· .. ,Un) + €leI + ... + em-nem-n

E lRm ,

lRm is the point of M n with local coordinates (UI, .. ' ,un). The Jacobian matrix of this map is

where

C

aek ' .•. , en + "'" aek ' el, .•. ,em-n ) , ( el + "'" ~ ek aUI ~ ek aUn where the ~ = ~ (i = 1, ... ,n) are the basis vectors of TzMn. In this expression, each vector is represented by the column of its coordinates. The vectors el, ... , en form a basis in the space TzMn, and the vectors CI, .•. ,Cm-n form a basis in its orthogonal complement. Therefore, for = 0, the map (x, x + is locally one-to-one. Since the manifold M n is compact, we can choose e > 0 such that the restriction FE of F to the set Ne = {(x, e) EN: lIell $ c} is one-to-one. The map FE: Ne - lRm is a oneto-one immersion of a compact manifold; therefore, FE is an embedding. In particular, Me = Fe(NE) is a compact manifold with boundary. Moreover, each point y E ME has a unique representation in the form y = x + where x E Mn and e..lTzMn. 0

e

e) . . . .

e

e,

Theorem 5.28 (Poincare-Hopf). The sum of the indices of singular'points is the same for all vector fields with isolated singular points On a closed manifold Mn. Proof. Consider any embedding f: M n -+ lRm. First, suppose that v is a vector field with nondegenerate singular points on Mn (the case of degenerate singular points is considered at the end of the proof). We use the

4. The Degree of a Map

229

notation from the proof of Theorem 5.27. The vector field v is extended over Me as follows. We represent each point Y E Me as y = X + and set v(y) = v(x) + Clearly, v(x).le and = 0 if and only if y EM". Therefore, the vector field v has the same singular points as v. It follows from Theorem 5.26 that the singular points of v have the same indices as the singular points of v (the operator dv is obtained from dv by adding the identity map as a direct summand).

e

J

e.

e

Lemma. The sum of the indices of singular points of the vector field equal to the degree of the map aMe --+ 8"'-1 defined by y = x + 1--+ particular" this sum does not depend on V.

v is

e e.

In

e,

Proof. The tangent space TS/(aMe ) , where y = x + is the hyperplane orthogonal to The vector vex) belongs to this hyperplane; therefore, (v(y), e) = (e,e) > O. For t E [0,1] and y € aMI$I we set Wt(Y) = tv(y) + (1 - t)e. We have (Wt(y),e) = t(v(y),e) + (1 - t)(e, e) > OJ in particular, Wt(Y) =I O. Hence the degree of the map aMe --+ 8"'-1 defined by y 1--+ Wt(y)/II Wt(Y) II does not depend on t. Thus, we must prove that the degree of the map aMe --+ sm-l defined by

e.

(1)

Y 1--+ v(Y)/lIv(y) II

is equal to the sum of the indices of singular points of the vector field v(y). Removing small disks Dr, ... ,Dr containing singular points from the manifold Me, we obtain a manifold M~ with boundary aMe U ~1 tJ··· U ~-1. The orientation of s;n-l induced by the orientation ot M~ is oppo1 is site to that induced by the orientation of This means that if oriented as the boundary of the manifold M~, then the degree of the map S;n-l --+ s;n-l defined by (1) is equal to the index of the ith singular point taken with opposite sign. Formula (1) defines a smooth map of the manifold M~; therefore, according to Theorem 5.24, the restriction of this map to the boundary has degree zero. Hence the degree of the map aMe --+ sm-l is equal to the sum of the degrees of s;n-l --+ s;n-l taken with opposite signs, i.e., to the sum bf the indices of singular points. 0

Dr.

sr-

It remains to consider the case in which the vector field v has degenerate singular points. We change v only in small neighborhoods of degenerate singular points; thus, we can assume that v is a vector field on an open subset of lRn. Let Yo be an isolated degenerate singular point of v, and let A: ~.n --+ [0,1] C lR be a smooth function taking the value 1 on an open set U 3 Yo and vanishing outside an open set V. We assume that V is sufficiently small for its closure V to contain no singular points different from Yo- Let vo be a regular value of the map v: U --+ lR". We put v' (y) = v(y) - A(y)VO.

5. Manifolds

230

Figure 9. The trajectories of a. vector field

aD

the sphere with 9 handles

Suppose that the minimum value of the function IIv(y)1I on the compact set V\U is 6 > O. The regular value '110 can be chosen so that 11'11011 < 6 (because the set v(U) contains the vector v(yo) = 0); then v'(y) =f 0 for aJlll E V\ d. IT y E U, then A(Y) = 1 and v'(y) = v(y) - '110. Th(>refore, the singular points of v' belonging to U arE! preimages of the regular value '110 of the map 't1: U - Rnj all these singular points are nondegenerate. It remains to note that both the index of the singular point Yo of v and the sum of the indices of the singqlar points of v' that belong to U are equa.'I. to the degree of the map au - sn- 1 defined by y 1-+ v(y)/lIv(y)lI. 0 Example. The sum of indices of any vector field (with isolated singular points) on the sphere with 9 handles is equal to 2 - 29. Proof. On the sphere with 9 handles, there exIsts a vector field with two singular points ofindex 1 and 29 singular points of index -1 ~see Figure 9). 0 Theorem 5.29. Suppose Mn and Nn are closed manifolds and p: Mn _ N n is a smooth k-fold covering .. If the sum oj indices of any vector field with isolated singular points on Nn equals X, then the sum of indices of any vector field with isolated singular points on M" equals kX. Proof. Le~ v be a vector field on N n ~ Since the covering p is a local diffeomorphism, we can define the vector field v == d(P-l)(v) on Mnl w~ mean that each vector v(y} is equal to d(P-I)(v(P(y»), where p-l is the inverse map to the projection of a neighborhood of y onto Q. neighborhood of p(y), To each singular point of v correspond k singular points of v with the same index. 0 Corollary. The sum of indices oj any vector field with isof.ated singular points on the nonorientable surface nI;'2 is. equal ~o 2 - n. Proof. The orientation covering pf the surface nP2 is (n - l)T'i.

0

. .Remark 5.5. A vector field of index XCM2) can be constructed directly from a triangulation of a two-dimensional surface M2 as follows. We take

4. The Degree of a Map

Figure 10. Construction of a

231

vectOl'

field from a triangulation

the barycentric subdivision of the given triangulation and construct a. vec~ tor field on its 1~skeleton such that it exits the vertices corresponding to the centers of faces and enters all vertices of the initial triangulation. This veer tor field can be extended to a vector field on M2 (see Figure ~O). The only singular points of this vector field are the vertices of the barycentric su~ division. The singular points corresponding to the vertices and faces have index 1, and the singular points corresponding to the edges have index - L This construction can be generalized to n- manifolds; in this case, the vector field on the I-skeleton of the barycentric subdivision is constructed so that it is directed from centers of k-faces to centers of {-faces for k> l. Theorem 5.30. The sum of the indices of a vector field with isolated sinflular points on a closed manifold of odd dimension is zero. Proof. By the Poincare-Hopf theorem, the sums of the indices of singular points of vector fields v and -v are equal. Therefore, it suffices to prove that if a map f: sn-l -+ sn- 1 has degree d, then the map -I has degree (_1)nq, or, in other words, that the degree of the map x 1--+ -x equals (_l)n. But this map is the composition of n maps of the form

C···, Xi-bXi, Xi+b.") ....... C···, Xi-l, -Xi, Xi+l,~")' each of which has degree -1.

tJ

4.3. The Hopf Theorem. We h~ve already proved that if maps I, gr Mn ..... Nn are smoothly homotopic, then deg I = deg g (this follows from Theore~5.23 and 5.25). Hopf [58] proved that if N n = sn, then the converse is true. Theorem 5.31 (Hopf). Let Mn be a closed oriented connected manifold. Ca) If the degrees of smooth maps I, g: Mn -+ S" are equal, then these maps are homotopic. (b) For any integer m, there is a smooth map I: M n -+ sn of degree m.

5. Manifolds

232

Proof. (a) Many of the arguments used below apply only in the case of n ~ 2; for n = 1, they must be changed, although not substantially. We consider only n ~ 2, because the proof for maps SI --+ SI is fairy simple.

sn

First, consider the simplest case in which there is a regular point Yo E such that the sets 1-1 (Yo) and g-l(yO) consist of Ideg II = I deggl elements. The map 9 is homotopic to a map gl such that I- l (yo) = g1 1(yo); the signs of the Jacobian determinants of / and 91 at the preimages of Yo coincide. Indeed, it suffices to show that if {all ... ,ak} and {bll ... ,bk} are sets of pairwise different points of a connected manifold without boundary, then there exists a diffeomorphism of this manifold which is isotopic to the identity map and takes each bi to~. A diffeomorphism taking bl to al exists by the lemma on the homogeneity of manifolds (see p. 223). To obtain the required diffeomorphism, we remove the point al from the manirOId~~ apply the lemma on the homogeneity of manifolds to the resulting manifold (it is connected if n ~ 2), and so on. In what follows, we assume that /-1 (yo) = g-1 (yo) = {all ... , ak} and the Jacobians of the maps I and 9 at the points aI, ••• ,ak have the same sign. Choose pairwise disjoint neighborhoods Ui 3 ~. The set /(M" \ U~=1 Ui) does not contain Yo; therefore, the map / is homotopic to a map h with the following properties:

• h (Mn \ U~=1 Ui )

sn;

= Yl, where Yl is the antipode of Yo on the sphere

• for each i, the map Vi C Ui of~.

h

coincides with / in some neighborhood

If the neighborhoods V; are sufficiently small, then the restriction of I to each of them is a diffeomorphism. Therefore, applying an additional homotopy, we can transform the map h into a map h whose restriction to each V; is a diffeomorphism of V; :::::: IRn onto {Yl} :::::: IRn. According to Lemma 2 on p. 225, any two orientation-preserving (or orientation-reversing) diffeomorphisms IRn --+ lR.n are isotopic. Therefore, the maps / and 9 are homotopic.

sn \

To complete the proof, it remains to consider the case where the preimage of Yo contains points at which the Jacobians have opposite signs. In this case, the homotopy is constructed as follows. For each pair of points (~, bi), where ai E j-l(yO), bi E g-l(yO), and the Jacobians of / and 9 have the same signs at ai and bi , we perform the above construction and obtain a tube Ui X I in C Mn x I j on this tube, the required map to sn is defined as above. We join pairs of the remaining points from the preimages of Yo (under j and g) at which the Jacobians have opposite signs by similar tubes (see Figure 11). The tubes can easily be rendered pairwise disjoint (this is obvious

4. The Degree of a Map

+

233

+

Figure 11. Elimination of preimages with Jacobians of opposite signs

sn

for n ~ 2). On the new tubes, the map to is constructed in the same way as on the old ones. The complement of the tubes is mapped to one point YI. (b) We choose pairwise disjoint open sets UI , ••. , Ujml C Mn and map them diffeomorphically to {YI}, where YI is a fixed point. The diffeomorphisms are chosen so that the signs of their Jacobian determinants coincide with that of m. The remaining part of the manifold is mapped to the point YI (if m = 0, then the entire manifold M n is mapped to YI). 0

sn \

Remark 5.6. A simplicial map of oriented pseudomanifolds also- has a degree, and for these degrees, an analog of the Hopf theorem is validj such an analog was proved in [105]. Problem 82. Prove that if M n is a closed nonorientable connected manifold, then smooth maps f, g: M -+ are homotopic if and only if their degrees are congruent modulo 2.

n sn

4.4. Approximation of Continuous Maps. Any continuous map Mm -+ NR of closed manifolds can be approximated by a smooth map with an arbitrary accuracy. In this section, we prove this assertion by using an embedding of N n in Euclidean space. Another approach is discussed in [54]. We start with approximation of maps M m -+ lRn. A map to lRn can be approximated coordinatewisej therefore, it is sufficient to consider the case n=l.

Theorem 5.32. Suppose that Mm is a closed manifold and f: M"" -+ lR is a continuous function. Then, for any e > 0, there exists a smooth function g: Mm -+ lR such that if(x) - g(x)1 ~ E for all x E Mm .. Proof. Every point x E M"" has an open neighborhood Ux such that If(x) .- f(y)1 ~ e for an Y E Ux. Take a cover {U21 l ' . . . ; Ux ,.} of M n by such neighborhoods and consider a smooth partition of unity {Ai} subordinate to this cover. Let g(x) = f(XI)AI(X) + ... + f(xk)Ak(x), The identity

5. Manifolds

234

E ~(x) =

1 implies

I(x) - g(x)

= I(x) -

L

f(xi)Ai(X) =

L

Ai(X) (J(x) - I(xi»'

If x ¢ Uz., then ~(x) = O. If x E Uz" then I/(x) - l(y)1 ~ e. In both cases, IAi(X) (J(x) - I(xi)) I ~ eAi(X); therefore, I/(x) - g(x)1 ~ EeAi(X) = e. 0 Using the tubular neighborhood theorem (Theorem 5.27 on p. 228), we can construct a smooth approximation of a map I: Mm _ Nn as follows. Suppose that N n c ]RN and gl: Mm _ ]RN is a smooth map for which IIgl(X) - l(x)1I < e, where e is the same as in the tubular neighborhood theorem. Then gl(X) = g(x) + e(x), where g(x) E Nn, e(x).lTg(z)Nn, and lIe(x)1I ~ e. Here g: Mm ..... N n is a smooth map and IIg(x) --i{x)1I 5

Ilgl(X) - I(x) II

+ lIe(x) II < 2e.

Theorem 5.33. Let Mm and Nn be closeJ manifolds. Then (aj any continuo1J.S map f: Mm - Nn is homotopic to a smooth map g: M m - Nn i (b) any two homotopic smooth maps I, g: Mm - ~ are smoothly homo~ topic. Proof" (a) Consider an embedding N n 1'"'t ]RN and take E > 0 from the tubular neighborhood theorem. Let g: Mrn _ Nn C ]RN be a smooth map for which IIf(x)-g(x)1I < e. Then the segment with endpoints I(x) and g(x} is contained entirely in the tubular e-neighborhood. We take the orthogonal projection on Nfl of the point which divides this segment in the ratio i: (1 - t). (b) Two copies of the interval [0,1] can be glued together so as to form the circle 8I" Thus, a homotopy between maps I and 9 can be treated as a map M rn x 8 1 - N n . Approximating this continuous map by a smooth map, we obtain a smooth homotopy between II and g1, where II and gl are approximations of I and g. The fact that the maps I and II (as well as 9 and gl) are smoothly homotopic can be proved by using the construction of

W.

0

The theory of homotopy groups considers maps which take base points to base points. We can pass from arbitrary smooth approximations to approximations with this property by using the lemma on the homogeneity of manifolds (see p. 223). It is seen from the proof of this lemma that if a point y belongs to a small neighborhood of a point Yo. then it is possible to construct a diffeomorphism Nn -+ N n taking y to YO in. such a way that it smoothly depends on y. This allows us to pass from a smooth approximation of a homotopy H, Mm x I _ Nn for which H(xo, t) Yo to a smooth homotopy H' for which H'(xo, t) Yo.

=

=

4. The Degree of a Map

235

Now t we can prove that 7rn(sn) = Z for n ~ 2. We derive this from the Hopf theorem. We have to prove that if maps I, 9 ~ STl, --+ taking Xo td Yo are homotopic t then they are homotopic in the class of maps taking Xo to Yo. The required homotopy is constructed as follow6. According to the above considerations t we can assume that the maps I and 9 and a homo~opy H between them are smooth. The path H(xo, t) is contained ~n some contractible open set U C sn (this is not so for n = 1). Let F(x, t) be a homotopy in the class of maps taking Yo to Yo which joins the identity map sn --+ sn to a map sending U to Yo. The homotopies
sn

Problem 83. Prove that if the indices of singular points of some vector field on a closed manifold Mn sum to 0, then there exists a vector field without singular points on Mn. Let P Mn be the quotient of the space of nonzero tangent vector~ to a manifold Mn by the equivalence relation v "1 AV (A is an arbitrary nonzero number), and letp: PMn --+ Mn be the natural projection. The space PMn has a natural structure of a manifold. The smooth sections of the projection p are talled line element fields on the manifold Mn. In other words, any line element field on Mn determines a one-dimensional subspace of T:eMn a~ each point x E Mn, and these subspaces smoothly depend on x. Problem 84. Prove that the existence of a line element field on a closed manifold Mn is equivalent to the existence of 8 vector field without singular points. 4.5. The Pontryagin Construction. The Hopf theorem suggests an interpretation of the elements of the group 7rn (sn) in the language of framed manifolds. The Pontryagin construction generalizes this interpretation tQ the groups 1I"n+k(sn) with k ~ 0 and n ~ 2. A smooth closed submanifold ~ c R n +k is said to be framed if, at each point x E Mk, an orthonormal set of vectors VI (x), .•. , Vn (x) orthogonal to T:eMk and smoothly depending on x is given. The manifold Mk is not required to be connected; it may consist of several connected components of the same dimension k. It is assumed that the empty set is a framed manifold of ~rbitrary dimension. Two framed manifolds M~ and Mf are framed cobordant if there exists a submanifold Wk+l in lRn +k +1 wit:q the following properties: (a) Wk+l is contained in the strip 0 ::; xn+k+1 ::; 1;

236

5. Manifolds

Me

Mf, =

(b) the boundary of Wk+l consists of the manifolds and and these manifolds are contained in the hyperplanes Xn +k+1 0 and Xn+k+ 1 = 1, respectively; (c) Wk+1 approaches these hyperplanes orthogonally, i.e., at each point of the boundary, the tangent space to Wk+l is orthogonal to the hyperplanes; (d) the smooth families of orthonormal sets of vectors given on

Me and

Mf can be extended to a smooth family of orthonormal sets of vectors on Wk+l.

The set of classes of framed cobordant manifolds of dimension k in ]Rn+k is denoted by n~(n+k). The set n~(n+k) admits the structure of an Abelian group. The sum of two elements is the class of the union of their representatives contained in different half-spaces ]R~+k and lR~+k. The zero element is the class containing the empty set. The inverse element is obtained by reversing the orientation of the orthonormal basis (e.g., by changing Vl(X) for -Vl (x». The same argument as the one used in the proof of the Hopf theorem to eliminate preimages with Jacobians of opposite signs shows that this indeed gives an inverse element. A framed zero-dimensional submanifold in lRR is a set o£ m+ points at whlch positively oriented bases are given and m_ points at which negatively oriented bases are given. The class of such a framed manifold in n~(n) is determined by the number m+ - m_. The Hopf theorem establishes the isomorphism n~(n) ~ 1I"n(S"') for n ~ 2. (To be more precise, the isomorphism holds for n = 1 as well, but the arguments used tor n > 1 do not apply to n = 1.) Theorem 5.34 (Pontryagin). For k ~ 0 and n 2:: 2, the group n~(n + k) is isomorphic to 11"n+k (sn). Proof. To a framed manifold Mk C ]Rn+k we can assign a map I: sn+ k --+ S'" as follows. According to the tubular neighborhood theorem (Theo"rem 5.27 on p. 228), there exists an e > 0 such that if lIall < e, then the map Mk X ]Rn --+ ]Rn+k defined by (x, a) ..... x + ~ tliVi(X) is a homeomorphism of Mk x D~ onto the e-neighborhood of Mk in ]Rn+k. Suppose that Xo E sn+k and Yo E S'" are base points. We identify sn+k \ {Xo} with ]Rn+k and sn \ {yo} with D~. Then, we map all points of ]Rn+k not contained in the e-neighborhood of Mk to Yo. identify the e-neighborhood of Mk with Mk x D~, and take its projection to D~ = sn \ {yo}. If and Mf are framed cobordant manifolds, then, applying a similar construction to Wk+1 , we obtain a map sn+k X I --+ sn which is a homotopy between the maps 10, h: sn+k --+ sn. Now let us assign a framed submanifold Mk C Rn+k to a map I: sn+ k --+ sn. We replace the continuous map 1 by a smooth map g homotopic to it,

Me

4. The Degree of a Map

237

take a regular value Y1 E 8 k different from the base point Yo = 9 (xo), and let Mk = g-1(Yl). The manifold Mk is framed as follows. At each point Y1 E lRn = sn \ {yo}, we take an orthonormal basis el , " f ' en and let Vi(X) be the normal vector to Mk at x for which dg(vi(x)) =~. It can be proved in the same way as for the degree that smoothly homotopic maps deter... mine framed cobordant manifolds and the equivalence class of Mk does not depend on the choice of the regular point Y1. The maps of the groups n~ (n + k) and 1l'n+k (8n ) constructed above are mutually inverse and respect the group operations. 0 Problem 85. Prove that the Hopf fibration p: 8 3 ....... 8 2 is the generator of the group 1l'3(82 ) ~ Z and describe the corresponding framed manifold in

nM3). 4.6. Homotopy Equivalent Lens Spaces. Take an integer p > 1 and integers qh •.. , qn, where n ~ 2 is coprime to p. The unit sphere 8 2n - 1 C en admits the action of the group Zp defined by o"(ZI,.

t.

I

zn) = (exp(21l'iqdp)zI, "-J exp(21l'iqn/P)Zn) ,

where 0" is a generator of Zp. The quotient under this action (which has no fixed points) is a manifold. We denote this manifold bY' Lp(qI, .•. , qn) and call it the lens space. The map 1l': 8 2n- 1 --+ Lp(q1. ... ,qn) is a p-fold covering with automorphism group Zp. Therefore, 1l'1 (Lp(qI, ... , qn») = Zp. If a number k is coprime to p, then Lp(qI, ... , qn) = Lp(kqI, ... , kqn), because in the group Zp with generator 0", the element kO" is also a generator. For n = 2 (i.e., for 3-manifolds), we obtain Lp(ql, q2) = Lp(l, ql 1 q2)i thus, any three-dimensional lens space has the form Lp(l, q). In the topology of 3-manifolds, the spaces Lp(l, q) are usually denoted by L(p, q). Theorem 5.35. Suppose that ql ... qn = ±knqi···
where q1··· ~ = knqi .•.
j( r1 ei 'Pl, •• , ,Tnei'Pn) =

(Tl eik1'Pl •••• ,Tneikn'Pn).

238

5. Manifolds

According to Problem 78, j has degree kl i ~. k n • The condition kjqj == qj (mod p) means that induces a map of quotient spaces I: L - L', where t = L p(ql, ... ,qn) and L' = Lp(tli, ... ,t/n). Indeed, i takes the point with coordinates r jei'PJ e(2Triq,p) to the point with coordinates r jeik1'P1 e(21rtq;p) , because k}qj == qj (mod p). Thus, points equivalent with respect to the map u are taken to points equivalent with respect to the map u'.

i

The degree of I is equal to the degree of Consider the maps

L

-+

Lv 8 2n-

i, i.e., deg I = kl ... kn.

LV 8 2n-

1 idV(deg d) I

1 !V1r'1

L'

defined as follows. The first map contracts the boundary of a small disk in L to a point. The second map is the identity on L and a map 8 2n - 1 ~--s2n-l of degree don S2n-l. The third map coincides with I on L and with the canonical projection '!r': S2n-t -+ 1/ on 8 2n 1. Le~ 9 ~ L -+ L' be the composition of these maps. It follows directly from the definition of the degree of a map that degg = deg/+dp = k 1 ··· kn+dp. But kl'" k n == q~ql1 ... q~q;1 == ±1 (mod p)j therefore, d can be chosen so that degd = ±1. In what follows, we assume d to have this property. Let g'; L' -+ L be a map defined similarly. The maps 9 and g' are the required homotopy equivalences. According to the Whitehead theorem (Theorem 4.21 on p. 179), it is sufficient to verify that the homomorphism g.: '!rn(L) -+ '!rn(L') is an isomorphism for all n ~ 1. The sphere S2n-l is simply connectedj hence, for n = 1, it suffices to show that the map f.: '!rl(L) -+ '!rl(L') is an isomorphism. Consider the commutative diagram

8 2n- 1

!1r

j

~

8 2n-

t

I

L

/'

1 ..-----. 82n-~

)0

!r

I

1r '

L'

)0

If follows directly from the definitions that

L

i'i is the identity mapj therefore,

I' / is the identity- map as well. Thus, f.: '!rl(L) -+ '!rl(L') is an isomorphism. Now, suppose that n ~ 2. Using the universality of the covering Ir: 8 2n- 1 -+

L, we construct the commutative diagram

8 2n- 1 !1r

L

j

__

8 2n- 1

!

!1r' )0

L'

The map g, as well as g, has degree ±1. Therefore, by the Hopt theorem, 9 is homotopic to either the identity map or the symmetry about the equatorial

239

5. Morse Theory

hyperplane. Hence the maps 9.: ll'n(s2n-l) -+ ll'n(S2n-l) are isomorphisms for all n. The map 9.: ll'n(L) - ll'n(L') is an isomorphism also, because the maps 1r.: ll'n(S2n-l) - ll'n(L) and 1r~: 1rn (S2n-l) - ll'n(L') are isomorphisms for n ~ 2. 0 For the three-dimensional lens spaces L(p, q), Theorem 5.35 is stated as follows. Theorem 5.36. If q = ±k2 q' (mod p), then th~ lens spaces L(P, q') are homotopy equivalent.

L(P, q) and

Indeed, we have L(p, q) = Lp(l, q), I.e., ql = 1 and q2 = q. Therefore, the equalities q = ±k2 q' and qlq2 = ±k2 tli th are equivalent.

5. Morse Theory 5.1. Morse Functions. Suppose that Mn is a manifold without boundary and I: M n - lR is a smooth function. A point x E un IS critical for I if and only ifrank/(x) 0, i.e., the map d/: TzMn -lR identically vanishes, In local coordinates (Xl,'" ,xn ), this means that (x) = 0 for i = ~, v . In. A critical point X of a function I is said to be nondegenemteif its Hessian matrix (a!2/xj (x)) is nonsingular. This definition does not depend on the choice of local coordinates because when the local coordinates are changed to (yh .... j ,Yn)1 the Hessian matrix transforms as

=

821y/x )) ( 8Yi 8

3!;

= JT

21 (8 ) 8Xi8x/x)

J,

where J = (~). A smooth function I: Mn. -4 lR is called a Morse function if all of its critical points are nondegenerate. Recall that the index of a quadratic form 'E aijXiXj with symmetric matrix (aij) is defined as follows. Making a change of variables (over the field lR), we can write the quadratic form as -y'f. - ... - Y: + Y~+l + ... + y~. The number q is called the index ot such a quadratic form. The index of a quadratic form can also be defined as the maximal dimension of a subspace on which the form is negative definite. The index of a nondegenerate critical point x of a function I is defined as the index of the Hessian matrix of I at the point x. Exercise 48. Let I(x) = -x~ - ... - x~ + x~+l point Xo = (0, _.. ? 0) is critical and has index q.

+ ... + x~.

Prove that the

Theorem 5.37 (Morse lemma). In a neighborhood of a nondegenemte crit~ ical point of index q, there exist local coordinates with origin at this critical

5. Manifolds

240

point such that the function / has the form /(X1,' ~. ,Xn ) = /(0) - X~ - ••• ~ X~ + X~+1 + ... + x~ in these coordinates.

Proof. We can assume that /(0) = 0 and local coordinates are defined in a convex neighborhood of 0 in lRn. According to the lemma on p. 200, there exist smooth functions 91, ... ,9n such that lex) = ~Xi9i(X) and 9i(0) = 31;(0). By assumption, the point 0 is critical, Le., 31;(0) = O. Applying the same lemma, we obtain /(x) = ~xixjhi';(X), where hij(O) = ~ /PI (0), i.e., ., (hij(O» is the Hessian matrix of / at the critical point. Changing hij(X) for (~j(x) + hji(x», we can assume that the matrix ~j(x) is symmetric, and performing a linear transformation of coordinates, we can also assume that hll (0) =F O. Finally, decreasing the coordinate neighborhood if necessary, we can assume that hll(X)/hll(O) > 0 for all x from this neighborhood. We set

!

Yl

h12(X)

= Xl + hll(x)x2 +, .. +

hln(x) hll(x) Xn

and Yi

.

= Xi for ~ ~ 2.

By the inverse function theorem, (XI, ... , x n ) H (YI, .. " Yn) is a diffeomorphism (possibly, in a smaller neighborhood). It is easy to verify that

2: XiXj~j(X) = hll(X)Y~ + 2: Yiyjhij(x). iJ~2

We make the changes Zl = Yl Jlhll(X)1 and Zi = Yi for i ~ 2, apply similar transformations to the quadratic form in n - 1 variables, and so on. 0 Corollary. Any nondegenemte critical point is isolated. We prove that Morse functions exist on any manifold. We give two different proofs, each with its own advantages, The first proof shows that any smooth function can be turned into a Morse function by a small change; here a "small change" means a small change of the first and second derivatives. The second proof is constructive. Moreover, it shows that there exists a Morse function / for which all sets of the form {x E Mn : fex) ~ c} are compact; this property is useful when dealing with noncompact manifolds. Theorem 5.38. On any manifold Mn, a Morse function exists. First proof. Let g: M n --+ lR be an arbitrary smooth function (e.g.i a constant). We shall construct a Morse function I by successively changing the function 9, as in the proof of Theorem 5.17 (see p. 214). Take the same domains Ui,l C Ui,2 c Ui,a, charts
5. Morse Theory

241

c

lRn is an open set and f: U - lR is a smooth function. Then, [or almost all linear functions lRn - lR, the function

Lemma 1. Suppose that U

f

+A

Ai

has only nondegenerate critical points.

Proof. Consider the map F: U _ lRR defined by

F(x) =

af (x), ..• , aX af ( aXI

R

(x)

)

•

A point Xo is critical for F if and only if the Hessian matrix of f at Xo is degenerate. Therefore, Xo is a degenerate critical point for a function of the form f(x) .,.... alXl - ..• - anXn if and only if F(xo) = (ab ••. , an) and Xo is a critical point for F,. i.e., if (al,"" an) is the image of a critical point of F. It remains to apply Sard's theorem. 0 It follows from Lemma 1 that if 9i-l is a smooth function on a manifold MR, then there exists a linear function A(x) = alxl + ... + anXn with arbitrarily small coefficients ai such that the function gi(Y) = 9i-l(Y) + '\(cpi(y))A(CPi(Y)) has no degenerate critical points on the set Ui,l (Lemma 1 should be applied to U = Ui,2 ~ Ui,l; note that '\(CPi(Y» = 1 for all Y E Ui,l). We have learned how to correct the function 9i-l on the set Ui,l. Now, let us learn how to do this without loss of what has been achieved at the preceding steps. Namely, suppose that a function 9i-1 has no degenerate critical points on the set Ui,l; we want gi to have no degenerate critical points on this set as well. The function gi-l is changed only on the compact set Ui,2, and it has no degenerate critical points on the compact set Ui,2n ( U;:~ U i,l)'

u;:l

Lemma 2. Suppose that f, g: U - lR are smooth functions on an open set U C lRR and f has no degenerate critical points on a compact set K cU. Then there exists a number e > 0 such that if all the first and second partial

derivatives of the function f - 9 at all points of K are less than e in absolute value, then 9 has no degenerate critical points on K. Proof. The function

vanishes only at the degenerate critical points of f; therefore, the minimum value 6 of F on the compact set K is positive. If e it sufficiently small, then

~ L...J

af )2 -L (Og)2 ( -OXi -OXi <8/2

and [det

(0:i £Xi) 2

r-

[det

(O!:Xi)

r

< 8/2;

5. Manifolds

242

hence

L

(::J + 1

[det

(O!:Xi)

r

> 0,

which means that 9 has no degenerate critical points on K.

o

If the numbers ai, .,. . ,an are sufficiently small, then the first and second derivatives of the function ,x(X) (alxI +_ .. + dnxn) are small also. Thus, the required function gj can be constructed by applying Lemma 2. 0 Second proof. We embed the manifold Mn in ]Rm; take a point a E ]Rm, and define f(x) = IIx - all 2 for x E Mn. Let Ul, ••• , Un be local coordinates on the manifold Mn, and let Xi(Ut,. I , I Un) (i = 1, ..• , m) be the coordinates of the point (UI, •.. ,un) in ]Rm. The functions Xi are smooth; therefore, the function f is smooth also. Our purpose is to choose a such that all critical points of f are nondegenerate. Clearly,

at = 2(Xi - ~) and ar.l:t = 20ij. Therefore, of Lm -0/-OXk Lm (xk~ak)----'" OXk --= =2 ~Ui

k=1

OXk OUi

k=l

OUi

and

o2f ~ 02f OXk OXl ~ 0/ 02Xk )OUiOUJ~ = k,l=l L..J OXkOXl OUi OUJ- + k=1 L..J OXk o,J,-OUl:J

~ (OXk OXk

= 2 Lk-l

o"~ oU¥J "'$

+ (Xk -

Q2Xk ) ak) au-ou- . ,

.,

vectors ei = ~! ••• , ~~), where i = 1, ..... , n, form a basis in the tangent space Ta;Mn; hence a point x E Mn is critical for f if and only if the vector = x - a is orthogonal to Ta;Mn, and this critical point is degenerate if and only if the matrix with elements gij + (e,lij), where gij = (ei' ej) and lij = (l:~" ! is degenerate. Here gij and laj depend on the point x E Mn (and on the local coordinate system). On p. 228, we considered Fhe m-manifold N consisting of pairs (x, e), where x E Mn and is ~ vector orthogonal to TzMn c lRm. Let us show that a point (x, e) E N is critical fot the map (x, e) f-+ x - eE JR"I if and onl;r if the matrix with elements gij + (e, lij) is degenerate. This would imply that the function f(x) = Ux - all 2 on the manifold M'n has a degenerate critical point if and only if a is a critical value for the map {x, e) f-+ x - e, and, by Sard's theorem, f(x) = IIx-all 2 is a Morse function for almost every a E ]Rm. We have already calculated the, Jacobian matrix of the map (x, e) f-+ x+e (see p. 228). The Jacobian matrii J of the map (x,e) 1-+ x......-~ is obtained

t

The

e

... £i/;!, ),

e

5. Morse Theory

243

by similar calculations:

(e + L {k ::~ ,..•., en + L ek ;:: "-~b -em-n) . l

I

i .. ,

The vectors el, ... , en form a basis in the space T:z;Mn, and the vectors EI, ••• ?em - n form a basis in its orthogonal complement. Therefore, the matrix A = (el, •.. , en, n) is nonsingular, which means that the rank of J is equal to that of

ell ... ,em

ATJ = ((9i (ei,Eek~)) j -

*

0) .

1m n

a!1u,.

It remains to verify that -(ei,Eek~) = (e,lij), where Ii; = This 2 8 x ) = - (~81l:)' {)~ (ek, ~ 8x ) = 0 . equal1' t y '18 eqw'valent t 0 (ek, ~ iju..' ~ , I.e" &. v .....vu, 1 ~~ VUi

I! = fi are orthogonal, and hence (eAa ei) = 01 I

But ihe vectors e1c and

0

Remark 5.7. If a manifold Mn is embedded in am as h closed subset and I(x) = IIx - a1l 2 , then all sets of the form {x E Mn\/(x):s; c} are compact. A Morse function I is said to be regular if it takes different 'values at different critical points. Theorem 5.39. On any closed manilold Mn, a regular Morse function exists. Proof. Let Xl, ... ,Xn be the critical points of a Morse function I: Mn - lR. Choose pairwise disjoint neighborhoods Ui 3 Xi and their open subsets Vi 3 Xi for which there exist smooth functions I.f'i: M n - IR taking the value 1 on the set V; and identically vanishing outside Ui • Consider the function g(x) = I(x) + ell.f'l(X) +". + Ekl.f'k(X). The minimum value of the function +t. i+ on the compact set Ui \ Vi is positive; therefore, if ei is sufficiently small, then g(x) has no critical points belonging to fli \ \ti. The function 9 is a regular Morse function if the numbers el, ... ,ek are sufficiently small and the numberS g(Xi) = I(x.) +ei are pairwise different. 0

fit 12

1£t1 2

5.2. Gradient Vector Fields and the Operation of Attaching Handles. Let I be a smooth function on a manifold Mn. If Mn is endowed with a Riemannian metric, then the function I determines the gradient vector field grad I characterized by the property that (grad I. v) = v(f) for any smooth vector field v on the manifold Mnj here v(f) is the derivative of I in the direction of the vector field v. If M n = ]Rn and the Riemannian metric is determined by the canonical inner product, then grad I = (it,;·· If This readily implies that the singular points of the vector field grad f coincide with those of the function f, and nondegenerate singular points correspond to nondegenerate critical points.

£t).

244

5. Manifolds

Theorem 5.40. For any Riemannian metric and smooth function f on a manifold Mn, the index of a non degenerate singular point Xo 0/ the vector field grad f is equal to (-I)t, where i is the index 0/ the critical point Xo 0/ /. Proof. First, we show that the index of a singular point of the vector field grad / does not depend on the choice of a Riemannian metric. Let (v,w)o and (v,wh be two Riemannian metrics on Mn. Then the formula (v, w)t = t(v, w)o+(I-t)(v, wh, where t E [0,11, defines a continuous family of Riemannian metrics. The indices of singular points of the-vector fields grad / (defined for these Riemannian metrics) continuously depend on t and take integer values; therefore, they do not depend on t. For the function /(x) = -x~ - •.• - x? + xl+! + ... + x; on the space lRn with canonical inner product, the vector field grad f has the form 2( -Xl, ... , -Xi,Xi+1,"" x n ). The origin is a singular point of index (_l)i for this vector field (see the proof of Theorem 5.30 on p. 231). P Corollary. Let f be a Morse function on a closed mani/old Mn. Then the alternated sum L:=l (-l)ic" where c, is the number of critical points 0/ index i, does not depend on the choice of f. Proof. The alternating sum under consideration is equal to the sum of the indices of singular points of the vector field grad f, and the sums of the indices of singular points are the same for all vector fields on a given closed manifold. 0 The topological structure of a closed manifold M n is largely determined by the set of indices of critical points for a regular Morse function f. We give precise statements below. The main objects of study in Morse theory are the sets Ma = {x E Mn : f(x) $ a} and the level surfaces f-l(a). Morse theory studies what happens to them when a passes through critical values. Note that if a is not a critical value, then Ma is a manifold. Theorem 5.41. Suppose that an interval [a, b] contains no critical values of a Morse function f on a closed manifold Mn. Then the manifolds Ma and Mb are diffeomorphic; in particular, the level surfaces f-l(a) and J-I(b) are diffeomorphic. Moreover, the manifold f- l ([a, bD is diffeomorphic to f-l(a) x [a, b]. Proof. Choose e > 0 such that the interval [a - e, b + e] contains no critical values of f. Let A(S) be a smooth function which takes the value 1 at • S E [a, b] and vanishes at 8 ¢ [a - e, b + e]. If f(x) E [a - e, b + e], then the vector field gradfIII grad fll2 exists (we assume that the manifold Mn

5. Morse Theory

245

is endowed with a Riemannian metric). Consider the vector field ( ) _ >.(J(x» grad! II grad 1112 '

vx -

which is defined on the entire manifold Mn. Note that v(x) = 0

i~

f(x) ¢

[a-e,b+e]. The smooth vector field v on the compact manifold Mn determines the scrcalled integral curves, i.e., curves 'Y(x, t) for which "'r(x, 0) x and 8-r ,t) = v(-y(x, t». The latter equality means that the tangent veCtor at the point 'Y(x, t) to the curve WeT) = 'Y(x, t+T) is v (-y(x, t)). In other words, if g: Mn _ IR is an arbitrary smooth function, then, at the point x E M n , the operator v assigns to this function the number 8g <W,t» Thus,

=

bt

It=O'

I

= v(g) = (v, grad g). at t=o Taking the initial function I for g, we see that if I(x) E [a, b], then

og(-y(x,t))

(1)

81('Y(x,t»

at

I

t=O

= (

grad

I

II grad III

2'

grad

I)

=1.

=

Consider the map 'Pt: Mn - Mn defined by 'Pt(x) 'Y(x,t). It has the properties 'Po = idMn and 'Pt+a = 'Pt'Ps. Therefore, 'Pt is a diffeomorphism. Formula (1) shows that 'Pb-a(Ma ) = Mb. A diffeomorphism between the manifolds I-I (a) X [a,b] and 1-1 ([a, b]) is defined by (x, t) ~ 'Y(x, t - a). D Using Theorem 5.41, we can describe the topological structure of a closed manifold in the case where this manifold admits a Morse function with exactly two critical points (a maximum and a minimum). Theorem 5.42. Suppose that on a closed manilold Mn, there exists a Morse function I having exactly two critical points. Then the manilold is 6 homeomorphic to sn.

un

Proof. Let Imax and !min be the maximum and minimum values of the function J. According to the Morse lemma, there exists a local coordinate system with origin at the point of maximum in which the function I has the form I(:¥;t, • .• ,xn ) = Imax - x~ - ..• - x~. Therefore, we can choose 5 > 0 so that the level surface I-I (Jmax - e) is diffeomorphic to 8 n - 1 and the inequality lex) 2:: fmax - e determines a manifold diffeomorphic to Dn. We assume that e is such that similar conditions hold at the point of minimum. 6Milnor [85) showed that for n ~ 7, M" may not be diffeomorphic to sn. This means that for n ~ 7, the sphere S" may admit nonisomorphic (or nonstandard) differentiable structures. There are exceptions; for example, the sphere S12 admits no nonstandard smooth structures.

5. Manifolds

246

The function I has no critical values between the points Imin + e and I max - e. Therefore, according to Theorem 5.41 t the preimage of the interval [/miD + e, Imax - e] is diffeomorphic to 8"'-1 X I. Hence the manifold Mn is obtained from 8"'-1 x I by attaching two copies of Dn via some diff~ morphisms CPl: 8"'-1 _ 8"'-1 and CP2: 8"'-1 _ 8"'-1 of the boundaries. It is easy to show that such a manifold is homeomorphic to Indeed, if CPl = CP2 = ids n ... l , then this assertion is obvious. It remains to show that any diffeomorphism cP: 8"'-1 ~ 8"'-1 can be extended to a homeomorphism ~: Dn _ Dn. For x E Dn, we set

sn.

~(x) = {XCP(~/IiXII) ~ x=/: 0, o if x = O. At the point 0, the map

~

is continuous but not differentiable.

o

Now consider the case in which there is precisely one critical point between the level surfaces I-l(a) and I-l(b).

Theorem 5.43. Suppose that Xo is a nondegenemte critical point 01 index i 01 a smooth function I and the intenJal [a, b] = [/(xo) rE,f(xo) +e] contains no images 01 other critical points. Then the space M" ~ homotopy equivalent to the space obtained from Ma by attaching a disk Di via ~ map aD' - Mo.. Proof. Theorem 5.41 allows us to assum~ the number: e to be arbitrarily small: the manifolds Ma and Mb can change their structures only when a critical value of I is passed through. Applying the Morse lemma, we choose local coordinates with origin Xo such that in these coordinates we have f(Xl, ... , x n ) = f(xo} x~ - ... - x~ + x~+1 + ("i + x~. In the chosen local coordinate system. the intersection ofthe level surface f(x) = f(xo)-e with the linear subspace generated by the first i coordinates is the i-disk x~ .. . +x~ ~ ej we denote it by D~. The coordinates of the points of the surface f(x) = f(xo)+e which are projected to this disk satisfy the inequality x~+1 + .• '+x~ ~ 2£; we denote the corresponding (n-i)-disk by D~-i (in Figure l~, the c~ of n = 2 and j = 1 is shown). We assume that f: is so small that D! >{ D'2;' is contained entirely in the chosen coordinate neighborhood JJ Cl Xo. Let us construct a deformation retraction r: Mb ~ A, where A = Ma U D! (and Mci n JJ! aD!). To simplify the notation, we write coordinates (Xl,."'X n ) in the form (x_,x+), where x ..... F (Xl,,,"'X,) and x+ = (Xi+l, JU., xn). On the set D! x D'2;;' n Mb, the map'" is defined as follows. Suppose that u is the vector field that takes the value (0, -x+) at each point (x_,x+) ED! x D'!le-i n Mb' Then r sends x to the endpoint of the closure of the integral curve of u passing through x (see Figure 12)0/. T-

+

=

247

5" Morse Theory

Figure 12. The deformation retraction

Outside the coordinate neighborhood U, the map r is defined similarly by using the vector field v(x) = gradf(~). Note that it x E U, ~hen vex} = ~x_, -x+). To define the map r on the entire set Mb, we need to construct a vee.. tor field which coincides with u on D! x D~-i and, wi,th v outside U'P Let A: Mn - [0, I] be a smooth function vanishing on D~ X D~-i and taking the value 1 outside U. For x E U, we set w(x) = (A(X)X_, -x+), and for x ¢ U, we set w(x) = vex). On the integral curves of the vector field w, the absolute values of the coordinates x+ decrease; therefore, all these integral curves reach the level surface f (x) = f (xo) - e (we mean the integral curves which start in Mb outside Ma and outside D x D~-i).

-1

f

A homotopy between idMb and r: Mb -' A C Mb is constructed as follows. For each point x E Mb, we consider the segment of the integral curve a from x to rex) and take the point dividing it in the ratio t ~ (1 - t). Theorems 5.41 and 5.43 imply the following important theorem, which relates the indices of critical points of Morse functions to CW-structures on manifolds. Theorem 5.44. Suppose that f is a Morse function on a closed manifold Mf'I. which has ct critical points of index i (i = 0,1, ... , n). Then Mn is homotopy equivalent to a CW-Complex with Ci i·cells. Proof. Suppose that Xo is a critical. point of index i and the interval [a, b] = [f(xQ~ - E, f(xo) + e] contains no images.of other critical points. It is suffi.. cient to show that if the space Ma is homotopy equivalent to a CW-complex X, then the space Mb is homotopy equivalent to the space X U x Di tJ where x: 8'--1 = aDi _ X is a cellular map.

According to Theorem 5.43, the space Mb is homotopy equivalent to Ma Urp D i , where
248

5. Manifolds

hcp: Si-l --+ X. It remains to prove the following two assertions (the space

Y in the lemma plays the role of Ma). Lemma. (a) If maps Xo, Xl: Si-l --+ X are homotopic, then the spaces X Uxo Di and X UX1 Di are homotopy equivalent.

(b) If cp: S,,-l then the spaces Y

--+ Utp

Y is a map and h: Y -- X is a homotopy equivalence, Di and X Uhtp Di are homotopy equivalent.

Proof. (a) Let Xt be a homotopy between XO and Xl. Consider the map Xo = X Uxo Di --+ X UX1 Di = ' x l defined as follows. -For x E X, we set o(x) = x. Each point 1.£ E Si-l is identified with Xo(u) in the space Xo; therefore, the equalities 0(1.£) = o(Xo(u)) = Xo(u) must hold. For t E [0,1/2] and 1.£ E Si-l, we set o(tu) = 2tu. Each point 1.£ E; Si-l is identified with Xl(U) in the space Xl; hence we must define 0 so that 0(1.£/2) = u = :n(u). To ensure that 0(1.£/2) = Xl(U) and 0(1.£) = Xo(u), we set o(tu) = X2-2t(U) for t E [1/2,1] and 1.£ E Si-l. 0:

A map (3: Xl --+ Xo is defined similarly. In this case, the equalities (3(u/2) = Xo(u) and (3(u) = Xl(U) must hold, and we set (3(tu) = X2t-l(U) for t E [1/2,1] and 1.£ E Si-l. It remains to verify that {30 ,.., idxo and 0{3 ,.... idxl' Direct calculations give if t E [0,1/4], 4tu { (3(o(tu» = X4t-I(U~ if t E [1/4,1/2], if t E [1/2,1]. Informally, a homotopy joining this map to the identity map can be described as follows. We uniformly expand the interval [0,1/4] to [0,1] and leave the map linear on this interval. Simultaneously, on the remaining part of the interval, we replace Xk(t) by X(l-s)k(t), where 8 E [0,1] (at the endpoints, the map xo is given; we want to obtain it at 8 = 1). A homotopy between 0{3 and the identity map is constructed similarly. (b) Let g: X --+ Y be a homotopy inverse to h. Consider the maps H: Y Utp Di -+ X Uhtp Di and G: X Uhtp Di -+ Y Ughtp Di defined by the conditions H tY = h, H Di = id D, and G X = g, G Di = idDi. Since gh '" idy , we have ghcp '" cpo Therefore, according to the asser~ tion (a) proved above, there exists a homotopy equivalence 0: y Ughtp Di --+ Y Utp Di. We show that the map oGH is homotopic to the identity. By construction, o:GH (y) = y for y E Y and

X2-2t(U)

r

o:GH(tu)

= {2tu X2-2tCP(U)

r

r

if t E [0,1/2] and 1.£ E S~-l, if t E [1/2,1] and 1.£ E S,-I;

here Xt is the homotopy between gh and idy.

5. Morse Theory

249

The construction of a homotopy between aG H and the identity map largely repeats the construction from (a). The main difference is that this time, there are no constraints on the image of the right endpoint of the interval [0,1]. Hence for the map of the interval remaining after the expansion of [0,1/2] we can take the initial map of some left segment of the interval [0,1]. Formally, this map can be defined as 2

'tfJs(tu) = { 1+s

tu

X2-2t+sCP(U)

ift E [0

'

lli]

and u E Si-l

2

.

'

if t E [!{=!,1] and u E 8'-1.

The linear function 2-2t+s is chosen because it takes the value 1 at t = !{=!. Thus, aGH id. Similarly, {3HG '" id, where {3 is the homotopy inverse of a. Therefore, I'V

GHa '" {3a(GHa) = {3(aGH)a '" {3a '" id. We obtain

HaG", ({3HG)HaG = {3H(GHa)G '" {3HG,.., id. The relations G H a id and HaG '" id mean that the maps If and aG are homotopy inverse to each other; in particular, H is a homotopy equivafV

0

~~

Problem 86. (a) Suppose that Y is a contractible sub complex in a CWcomplex X, Le., the embedding Y -+ X is homotopic to a constant map. Prove that X/Y rv X V E¥'. (b) Suppose that m < n and the sphere gm is canonically embedded in n 8 • Prove that sn / 8 m '" sn V gm+l. 5.3. Examples of Morse Functions. Torus. Let us represent the torus T'" as the quotient space IRn /27rzn. Then the smooth functions on T'" are the smooth functions of n variables that are 27r-periodic in each variable. Example 1. Let Cl, ••• , en be real numbers. The function l(xI, ... , Xn) = Cl sin Xl +.. '+en sin Xn is a Morse function on T'" if and only if ct, ... , en i= 0. This function has (~) critical points of index k.

l!t

°

Proof. The equalities = ... = if;; = imply Cl cos Xl = ... = encos x n = 0. If Ci =f:. 0, then Xi = ±~ + 2m7r. Clearly, if Ci = 0, then the function I has a nonisolated critical point. If CI, ... ,en i= 0, then we have

(a::a:j) =

diag( -CI sin XI, ••• , -en sinxn)

= diag(cICl. ... , enen)

250

5. Manifolds

at this critical point; here Xi = -Ei! + 2m'lr J The index of the critical point is equal to the number of those i for which EiCi <: O. The eritical points of index k are determined by k numbers i for which EiCi < O. 0 Note that if 8 1 is represented as a CW-complex with one O-cell and 1.., cell, then the torus rn 8 1 X .,. >< 8 1 is 8. CW-complex with (~) k-eells. The Morse function from Example 1 determines precisely this partition of the torus into cells.

=

Sphere. We represent the sphere sn as the submanifold in lRn determined by the equation x~ + .. i + x!+1 == 1. Example 2. The function I(X1, . .• I Xn+1) = xn+1 is a Morse function on sn, and it has two critical points, one of index a and one of index n. Proof. The sphere 8 n can be covered by 2(n + 1) charts, each of which is determined by the inequality Xi > a or Xi < 0; the local coordinates for such a chart are (Xl, ..• , Xi-I, Xi+l,- rt.,xn+1)' If i:l 1i+ 1, then the function I is smooth and has no critical points on the corresponding chart. On the chart Xn+1 > 01 I has the form v'1~ x~i here x~ + + x! < 1 and the positive value of the root is taken. A simple calculation shows that ~ 8/ = ..• = 8/ = 0 only at the point ~

xr ... -

0

••

-in.

(0""10); at this point, the matrix (a!21.,) is equal to Thus, the point (Xl, ..• , X n , X n+1) = (0, ... ,0,1) F 8 n has index n. {XI,"pXn) =

A ~imilar calculation for the chart xn+~ < 0 shows that at the point (0, ... , 0, -1), the Hessian matrix is In, i.e .... this point has index O. 0 Example 3. Let Cb ..• , Cn+1 be real numbers. The function I(XI, ... , Xn+1) + ... + Cn+1 x!+1 is a Morse function on 8 n if and only if the numbers Cl, ••• ~ Cn+1 are pairwise distinct. This Morse function has two critical points of each index 0,1 12, ... ,n. = CI x~

Proof. On the. charts Xn+1

1 --

> 0 and Xn +1 < 0, the function f

CIX12' +"r + CnXn2 +<:n+1 (1 - xl,2 -'" -

=

(C1 -

has the form

2) Xn

Cn+1)x~ +,. .. + (en - Cn+1)x! + Cn+l'

Such a function has only one critical point, namely, (Xb~ or' xn) = (0, ... ( 0)1 This critical point is nondegenerate if and only if all the numbers CI, • en are different from en+1; the index of a nondegenerate critical point is equal to the number of those Cl, "0, Cn which are smaller than en+1. The points '(~l'" . , x,n) = (0, ... ,0) on the charts Xn+1 > 0 an~ xn+l <; 0 correspond to the pomts (Xl, .•. ,Xn , x n +1) = (0, ... ,0, ±1) E 8 . ~ 0

•

,

5. Morse Theory

251

°

Performing similar calculations for the other charts Xi > and Xi < 0, we see that -critical points are nondegenerate if and only if the numbers CI, .•• , Cn+l are pairwise distinct. Clearly,. for each k =. 0,1, I,., n, there exists precisely one i for which k of the numbers CI,.··, Ci-l, Ci+l,"" Cn+l are less than Ci. 0 Real projective space. We represent the real projective space lRpn as the manifold obtained from the sphere 8 n by identifying X and -x. Let I pe a Morse function on such that f ( -x) = f( x). The~ the function I can be regarded as a function on lRpn, ~d it remains a Morse function on lRpn. Moreover, each critical point of f as a function on lRpn corresponds to two critical points (with the same index) of J as a function on

sn

sn.

The function f(xI, .•• , Xn+l) = clxi + ... + Cn+lX~+l does have the tequired property; therefore, the following assertion is valid. Example 4.,. The function f(x}, ..• , Xn+l) = clxf+· . +Cn+lX~+l is a Morse function on Rpn if and only if the numbers Ct,., ••• , Cn+! are pairwise distinct. This Morse function has one critical point of each of the indices 1 1,2, ... In.

°

Complex projective space. We specify points of the complex projective by homogeneous coordinates (Zl : .,. ; zn).

spac~. CP"

Example 5. Let c},.t., Cn+l be pairwise distinct real numbers. Then the function f(zl ;' .' Zn+1) = cllz11 2 + ... + Cn+1I Zn+11 2 }

(j

IZ I1 2

+o'J'+IZn+11 2

is a Morse function on Cpn. This Morse function has one critical point of each of the indices 0, 2, ... , 2n. Proof. The manifold Cpn can be covered by n + 1 charts, each determined by an inequality of the form Zk =/: 0. On the chart Zn+1 -=1= 0, we define local coordinates by Wk = Zk/Zn+1 for k = 1, ... ,n (each complex coordinate Wk corresponds to ~wo real coordinates). In these local coordinates, the function f has the form

f(

Wh'"

,Wn

) - cll w l1 2 +." + enl wnl 2 + Cn+1 IWIJ2 + ...... + Iwnl2 + 1 . -

It is more convenient to pass to the coordinates Uk = IlwII2 = IWl12 +~ ... + Iwn l2, Using the equalities

Wk/ vllwll2

+ 1, where

1 IIwW~ I 12 2 IIwII2 + 1 = 1 - IIwll2 + 1 :::; 1 - 1.£1 -,. .. - lUnI !

we obtain f(ub"" un) = (cl - Cn+l)IUlI2 + ..• + (Cn - CA+1)lunI2 + Cn+l·

5. Manifolds

252

Moreover, Uk = Xk + iYk and IUkl2 = x~ + y~. The rest of the proof is as in Example 3. In the complex case, the indices of critical points are twice the indices in the real case because IUkl2 = ~+~. 0 The manifolds SO(n) and U(n). Our construction of Morse functions on SO(n) and Uen) is borrowed from [133]. First, we describe the structure of the tangent spaces to SO(n) and U(n) at a point X (by definition, XX· = In is the identity matrix, where X· = XT; in th~ real case, complex conjugation can be omitted). If is a tangent \lector, then the matrix X + must satisfy the relation (X te)(X· + te*) = In up to terms of order t; therefore, eX* + Xe* = o. At the point X = In, this condition takes the form + = 0; i.e., the matrix is skew-Hermitian (skew-symmetriEfil the real case) at this point. The dimensions of the spaces of skew-symmetric and skew-Hermitian matrices are easy to calculate; they coincide with those of SO(n) and U(n). Clearly, for the matrix { = eX, the equality {X*+X{* = 0 is equivalent to + = O. Hence any matrix for which eX* + Xe* = 0 belongs to the tangent space at X.

e

te

+

e e*

e

e e*

e

Lemma. Let 17 be a matrix 0/ order n, i.e., a vector in ]Rn?l. Suppose that the space of matrices in endowed with the inner product (A, B) = tr(AB*). Then the orthogonal projection of the vector 17 on the tangent space to SO (n) or U(n) at a point X is !(17 - X17* X). Proof. First, consider the case of X = In. In this case, the orthogonal complement of the skew-Hermitian (skew-symmetric) matrices consists of Hermitian (symmetric) matrices. Therefore, it suffices to represent the matrix 17 as a sum of Hermitian and skew-Hermitian matrices: 11 = !(17 + 17*) + !(1] -17*)· The matrix !(1] -1]*) is the projection of 17 on the tangent space, and !(1] + 1]*) is the projection of 17 on the orthogonal complement of the tangent space. For an arbitrary point X, we proceed as follows. First, we translate the vector 1] to the point In: 17 1-+ 1]X- 1 = 1]X*, Then, we find the projection of 17X* on the tangent space. It is equal to !(17X* - X1]*). Finally, we return to the initial tangent space: !(1]X* - X1]*) 1-+ !(1]X* - X1]*)X = !(1] - X17* X). [) Example 6. On the space SO(n) or U(n), consider the function fA(X) = Retr(AX), where A is a given matrix. A point X is critical for this function if and only if A* = XAX. ,Proof. We have fA(X) = ~pq(apquqp-{3pqVqp), where apq = a pq +i{3pq and xpq = upq + ivpq; hence gradf = A* is a constant matrix.

5. Morse Theory

253

The critical points are the points for which grad I has zero projection on the tangent space. According to the lemma on p. 252, this projection is equal to ~(A* ~ XAX). 0 Example 7. Let A = diag(al, ... , an), where 0 ~ a1 IA(X) is a Morse function on SO(n) (or on U(n)).

< .•. < an.

Then

Proof. According to Example 6, a point X is critical if and only if A* = XAX. In this case, A = (XAX)* = X-1AX-1. Therefore,

A2

= (XAX)(X- 1AX- 1) = XA 2 X- 1,..

i.e., the matrices A2 and X commute. Moreover, A is a diagonal matrix with different eigenvalues. Therefore, X is a diagonal matrix (see [100, Problem 39.1a)). By assumption, X is unitary, whence X = diag(±1, ... , ±1)j in the real case, the product of diagonal elements equals 1. We have found the critical points of the function IA. Now, we must verify that all of them are nondegenerate.

e

e* e

Let be a tangent vector at the point In (i.e., + = 0), and let Xo = diag(e1, ... ,en), where ei = ±1, be a critlcal point. Consider the map {' ~ Xoe{ j for the local coordinates near Xo we take the superdiagonal and diagonal elements of the matrix In these coordinates,

e.

IA(X) - IA(XO) = Retr A(X - Xo) = Retr AXo{e{ - In)

= Retr Axo(e + ~e2 +.~.) = Re (

L apepepp + ~ L apepepjejp + ... ) . p

p

e

The numbers pp are purely imaginary; therefore, the real part of the first sum vanishes. Moreover, epj = -ejp; hence all the terms in the second sum are real, and IA(X) - IA(XO) = - E1~p~q~n(apep + aqeq)lepq l2 + .1 j . By assumption, aq > ap ~ 0 for q > Pi therefore, sign(apep + aqeq) = sign(aqeq) for q ~ p. Thus, if P ::; q, then the sign of the term with lepq l2 is opposite to that of eq. In particular t the quadratic part of IA(X) contains the squares of all coordinates. D Grassmann manifolds. We use the notation and the properties of Grassmann manifolds from Section 1.5. Of special importance for our purposes in this example is the Schubert symbol 0' = (0'1, ••• , Uk), where 1 ::; 0'1 < . ",. < Uk ::; n, and the number d(u) = (0'1 - 1) + (0'2 - 2) +" .. + (Ulc - k), which is equal to the dimension of the open Schubert cell e(u). The Schubert symbol is a special name for the multi-index used to describe the cell structure of a Grassmann manifold. Each Schubert symbol

5. Manifolds

254

is associated with the PlUcker coordinate Xu- The set of Plucker cobrdinates of a subspace II E G(n, k) is a point in the projective space lRP(~)-l. Let eu be the point of this projective space that corresponds to the u·axisj we have eu = (0 r 0 : ... : 0 r 1 : 0 ~ ... : 0), where 1 occupies the position corresponding to the multi-index u. Clearly, the point eu belongs the image of the open Schubert cell e(u). U

-tq

Example 8. Consider the function lex) =. LCaX! on the space lRP(:)-l (it is assumed that Cu E lR and Ex! = 1). Let i: G(n,k) -+ lRP(~)-l be the PlUcker embedding. Then the numbers Cu can be chosen so that Ii is a Morse function on G(n, k) with critical points j l(eD') of index d(u). (The choice of the numbers Cu is constructive; it is described in the course of the proof.) Proof. H the numbers ~ are l>airwise distinct, then, according to Example 4, the function I is a Morse function on lRP(~)-1 with critical points eu • Therefore, all points i-l(eu ) are critical for Ii. Let us find out how the function fi behaves in a neighborhood of the point i- 1 (eu)' The multi-index u corresponds to the chart Uu on the manifold G(n, k), and the point i- 1 (eu ) is the origin of this chart. In these local coordinates, the PlUcker embedding has th8r following structure. The Plucker coordinate X T is equal to the determinant of the matrix formed by the columns Tl,' .• ,Tk of the matrix in which the columns Ul, ~ .. ,Uk constitute the identity matrix and the remaining columns are filled with the numbers Yt, ••• ,Yk(n-k)' Clearly, X T is a homogeneous polynomial in the variables Yb • •• ,Yk(n-k) of degree equal to the number of columns Tl, ••• , TIc different from the columns ul,".' Uk (we have Xu =r 1). In calculating the Hessian matrix of the function E Co-:;X at the origin, we are interested only in the lin;,;.,. ear polynomials X-ro. We say that Schubert symbols u and T (of length k) are neighboring if they have precisely k -1 common elements, It is easy to verify that if u and T are neighboring Schubert symbols, then X T = ±Yi. Moreover, for each index i, there is exactly one Schube~ symbol T(i) neighboring u. At the origin" the function f takes the value Ca, and

I()y

Cu

-

Cu

-+ ECrx~(y) _

1 + Ex~(y)

Cu

_ E(Cr - Cu)x~(y) 1 + Ex;(y) i

-

Therefore, the quadratic form that ~pproximates the difference fey) - Cu il:! E~~~-k) (Cr(i) """Cu )x~(i) (y); here the summation is over the Schubert symbols T( i) neighboring u. H the numbers Cr are pairwise distinct, then the critical point i-I (e a ) is nondegeneratej its index equals the number of the Schubert symbols T(i) (neighboring u) for which Cr(i) < CuJ We order the Schubert symbols by declaring that 7 <0 fT if 'T'k = Uk, Tk-I = uk-Il Ti+l = Ui+lt and T~ < Ui· Take constants ~ such that r';'

255

5. Morse Theory

<

for t

< u.

It. Is easy to verify that for each Schubert symbol u, there are d( u) neighboring Schubert symbols .,. smaller than u. Indeed, consider an element O'i of U. To obtain a Schubert symbol smaller that u, we should replace it by any positive integer smaller than Ui and different from U1, ••• , Ui-1; the number of such integers equals u; ~ i( Thus, for the chosen numbers Crr, i- 1 (eD') is a nondegenerate--critical point of index d(u). It remains to show that the function Ji has no other critical pointS. Consider any k-dimensional subspace II; let u be its Schubert symbol. In II we take the basis VI, ..... , Vk, where V, = (Vilt"'" ViO'.-l'~' 0, ... t ,0) and ViD', = 0 for j < i. We say that a vector Vi has a nonzero coordinate if ViB # 0 for some 8 < Ui (the coordinate ViC1~ = 1 is ignored). We assume that II # i- 1 (eD')' i.e., at least one of the vectors VlI'" ,Vk has a nonzero coordinate. Choose such a vector Vi with largest possible i (this will be convenient in what follows). C-,I

Crr

Consider the subspace lIt spanned by the vectors VIt. '" and the vector viet)

= «1

Vi-I, Vi+l, J1 ., Vk

+ t)Vil,"" (1 + t)ViD'.-l, 1,0, r u ~O).

Clearly, 1£ II is a critical point of the function Ji, then t = () is a critical point of the function

= 2:r Crx~(t)/ 2:1' x~(t)j hence

(Epcpx~xp) (Erx~) i

I

..-I

(Epcpx~) (Erx~xr)

(Erx~)

2

'

where Xr = Xr(O) and x~ = x~(O). The situation is the simplest in the case i = k. In this case, the matrix \t;.(t) b,as the following structurer depending on Tk. If Tk < Uk, then the last row of Vr(t) is obtained from the last row of Vet) by multiplying all of its elements by 1 + t. In this case, we have Xr(t) = (1 + t)xr(O) and x~(t) = x~(O). If Tk = Uk, then the last column of Vr(t) consists of the elements 0, ... ,0,1. In this case, xr(t) = const and x~(t) = O. If Tk > Uk, then the last column of Vr(t) is zero. In this casej Xr(t) = O.

5. Manifolds

256

Thus, x~ "" 0 only if 1'k < Uk, and xp "" 0 only if Pk ~ Uk' Therefore, for Schubert symbols T T, 1'k and P = p, Pk (here T and p are shorter Schubert symbols), the nonzero terms in the nwnerator of the expression for cI(O) take the form

=

( L CP'PIoX~'P,,) (L x~,7lc + L X~,u,,) ,7lc
'T

- ( L cp,p"x~,p" + L c,.,U"X~,UfI P,PII:
Canceling the common part

L

'T

Lp,PfI
L X~'7lc).

) ( \.,., 7'fI
.

Cp'Pflxlp"x~,7'fI' we obtain

c,.,UIo)X~'PflX~,Ul'"

The constants Cu are chosen so that Pk < Uk implies Cp,Plo < c,.,UIo' Therefore, the equality 11"(0) = 0 implies X~'PflX~,UIo = 0 for all T and p, Pk < Uk. But if T, Uk = 0', then X'T,UIo = Xu = 1. Hence Xp,PIo = 0 for all p, Pk < Uk. On the other hand, if the Schubert symbol P consists of j and 0'10 i •. , Uk-lo then xp = ±Vkj; this contradicts the assumption that one of the numbers Vkj with j < Uk is nonzero. Now assume that i < k. Recall that the nwnber i is maximal in the sense that the vector Vi has a nonzero coordinate and the vectors Vi+!, • ,. I Vk have only zero coordinates (we ignore the coordinates VjUj = 1). Thus, X'T(t) "" 0 only if Ti ~ Uh Ti+1 = Ui+lJ ..• , Tk = Uk, and x~(t) "" 0 only it Ti < Ui, Ti+1 = Ui+lJ' •• , 1'k = Uk; we have x~(O) = X'T(O). We write the Schubert symbol T

= (TlJ.' • ,Ti, Ti+1, •.• ,Tk) = (71, •.• ,Ti, Ui+l,.' . ,Uk)

in the form T, Ti, where T = (Tt, .•. , Ti-l); we ignore the common part Ui+1, ••.• Uk. In this notation, the nonzero terms in the numerator of the expression for 11"(0) have almost the same form as above; the only difference is that the inequalities 1'k < Uk and Pk < Uk are replaced by Ti < Ui and Pi < Ui, respectively. 0 Remark 5.8. The first explicit construction of a Morse function on a Grassmann manifold was given in [147]. Our exposition follows [48]; see also [3]. A simpler construction of a Morse function on a Grassmann manifold was suggested in [133], but it uses properties of commutators of vector fields.

Chapter 6

Fundamental Groups

In Section 2.1, we defined the fundamental group 1I"1(X,XO) of an arbitrary path-connected space X with base point Xo and proved the basic properties of fundamental groups. In this section, we use these properties to calculate the fundamental groups of some particular topological spaces.

1. CW-Complexes The same argument as the one used in the proof of Theorem 1.15 on p. 30 proves that any finite connected CW-complex is homotopy equivalent to a CW-complex with only one vertex (O-cell) eO. In what follows, when calculating the fundamental group of a CW-complex X, we always assume that it has exactly one vertex, ear We also assume that eO is the base point (when the base point changes, the fundamental group is replaced by an isomorphic group; the isomorphism is induced by the map w J-+ a-Iwa, where a is a path between the base points). 1.1. Main Theorem. If the cell structure of a space X is specified explic.o. itly, then the fundamental group of X is easy to calculate. Namely, the l-cells correspond to generators of the group, and the 2-cells correspond to relations. We explain this in more detail. The I-skeleton Xl is a wedge of circles; therefore, 1r1(XI. eO) is the free group with generators a!, ..... , ak, where_~ is the number of I-cells in X. The characteristic map of the ith 2-cell X~! D2 --+ X induces a map f3i: 8D 2 /-it Xl eX. Choose a base point 8 in 8 1 . The map f3i corresponds to an element of the fundamental group 1rl(Xl,f3i(S». To this element we can assign an element of the group 1rI(X1,eO) by choosing a path from f3i(S) to eO. Thus, to the ith 2-cell we

-

257

6. Fundamental Groups

258

s

Figure

t

~e homoto~

assign an element of the group 71"1 (Xl , eO). We denote it by the same symbol Pi; it is determined up to conjugation. It is easy to verify that the map Pi: 8 1 -+ X corresponds to the identity element of the group 71"1 (X, Pi (8»; the homotopy is shown in Figure 1. The natural embedding Xl -+ X takes the generators aI,,,, .. , aA: E 7I"l(X 1,eO) to elements aI,J •• ,ak Ej: 7I"l(Xt eOl, The element Pi is a word 1 , The corresponding word bi in the alphabet 1 , •.• , in the alphabet t , •.• , atl is the identity element of 71"1 (X, eO); thus, we obtain the relation bi = 1, where bi is a word in the alphabet atl, ••• ,at l • The relations bi = 1 and abia- 1 = 1 are equivalent; therefore, the ambiguity in the choice of Pi is inessential in this respect.

at

at

at

Theorem 6.1. The group 71"1 (X,.e°) is genemted by the elements at. u'. , aI;, and any relation in this group reduces to the relations br, .•. ,b,. (This means that if a word P in the alphabet ~ a~l corresponds to word b in the alphabet atl,J •• ~, a~l that represents the identity element of 71"1 (X, eO), then f3 E N j where N is the minimal normal subgroup in 71"1 (xl, eO) containing the elements PH . L , p,. )

atl ... ,

a

Proof. We represent 8 1 as a CW-complex "With one O-ce11 and one 1-cell; the O-cell is the base point 8. According to the cellular approximation theorem, any continuous map (81,8) ...... (X, eO) is homotopic to a cellular wap (81,8) -+ (Xl, eO). 'fperefore, the embedding .xl M x, induc~ an epimorphism 71"1 (Xl, eO) __ 71"1 (X, eO) of fundamental groups; Le., the elements al, ... , al; generate the group ;1 (X, eO)"} The same cellular approximation theorem implies that the embedding X2 -+ X induces an isomorphism of fundamental groups. Indeed, since the embedding Xl -+ X induces an epimorphism of fundament at grou~ it suffices to verify that if two loops in X~ (based at eO) are homotopiC in the space X J then they are homotopic in the space. X2. A homotopy in X between two loops in Xl is a map H ~ 12 ..-.4{ X. Let us represent the square 12 as a CW-complex with four vertices, four edges, and one 2-cell. The restriction of H to 812 is a cellular map; therefore, H Uihomotopic to

1. OW-Complexes

259

Figure 2. The covering PI

a cellular map H': J2 .- X2, and the homotopy is fixed on {)I2. The map H' is the required homotopy in the space X2. It remains to verify that the kernel ofthe homomorphism 7r1 (Xi teO) -+ 7r1(X2,eO) induced by the embedding Xl -+ X2 is the group N. Consider the covering PI: Xl - XI corresponding to the subgroup N C!' 7r1(Xl,eo1, i.e., such that Pt.7rI(.X\iQ) = Nt where eo E p-l(eo). The covering PI in the case where X2 is the torus and Xl is its I-skeleton (the wedge of two circles) is shown in Figure 2. The group N is normal; therefore, the covering PI is regular. According to the definition of PI, the lifting of each Pi:' (Sl,s) _ (XI,eO) starting at ifJ is closed. It follows from the regularity of Pl tha11 any lifting Pi:f ~f A starting at E Pll(eO) is closed as well. Therefore, we can assume that Pi} is the characteristi~ map [Jij: aD} ....... Xl. Attaching the 2-cells to Xl via the maps [Jij, we obtain a CW-complexX 2 , and the covering PI can be extended to a covering P2: X2 -+ X2.

eJ

Consider any loop w; (Sl,s) -- eX:\eOl contractible in X 2 • The contractibility of w implies the c10sedness of its lifting to X2 t moreover., the lifting of w is entirely contained in Xl. Therefore, the loop w is the projection of a loop in Xl based at ~. Hence w is an element of the group N, as required. 0 Corollary. The group '7fl(X, eO) is isomorphic to 7rl(X2,eO), i.e., the fundamental group of a CW-complex is completely determined by the 2-skeleton of this complex. 1.2. Some Examples. Example. 7i"l(CP") = O. Proof. The 2-skeleton of C..pn is Cpl ~ S2,

o

Example. The group 7r1 (nT2) has generators ai, b1, ' .. , an, bra and one rela.1 · 1 t Ion i=l a~'b·~ai-lbi ) =.

rrn (

6. Fundamental Groups

260

o

Proof. Look at Figure 12 on p. 148. Example. The group a~ ... a~ = 1.

7rl (nP2)

has generators

al ........,

lln and one relation

We have already proved that the surfaces S2, TJ. , 2T2, ... , p2, 2p2, ... are not pairwise homeomorphic (see Theorem 4.5 on p. 146). The use of fundamental groups makes the proof very simple. Recall that the commutator subgroup of a group G is defined as the subgroup G' consisting of all products of the form aha Ib- l , where a, bEG. The identity xaba-lb-1"Yx - 1 = xax-la-la(xb)a~l(xb)-lx"Yx-l shows that the subgroup G' eGis normal. Clearly, the quotient group

GIG' is commutative, Exercise 49. (a) Prove that if G = 7rl(nT2), then GIG' 9! z2n..,. (b) Prove that if G = 7rl(nP2), then GIG' 9! zn-l E9~. It is easy to verify that the groups Znl EB Z~l and zn:l, E9 {O, I}, are isomorphic if and only if nl = n2 and £1;:::: f2.

Z~2,

where

£j E

Problem 87. (a) Prove that any subgroup of finite index in the group 7rl(nT2) is isomorphic to 7r1(mTJ.) for some m; moreover, m -1 is divisible by n-1. (b) Prove that any subgroup of finite index in the group 7r1(nP2) is isomorphic to either 7r1(mP2), where m - 2 is divisible by n - 2, or 7r1(mT2), where 2(m - 1) is divisible by n - 2. Example. The fundamental group of the surface nT2 from which Ie ~ 1 disks are removed is the free group of rank 2n + k - 1. Proof. The fundamental group of the space under consideration is generated by aI, bI,' .. ' lln, bn and C}, ••• , Ck and the relation n

C! ... Ck

II(aibia;-lb;l) = 1 i=l

(see Figure 3). The element Ck can be expressed in terms of the remaining elements, which are algebraically independent. 0

Corollary. The surface nT2 from which k ~ 1 disks are removed is homotopy equivalent to the wedge of 2n + k - 1 circles. (This assertion can easily be proved directly by constructing a deformation retmction.) Example. The fundamental group of the surface nP2 from which k ~ 1 disks are removed is the free group of rank n + k - 1.

261

1. CW-Complexes

Figure 3. A sphere with handles and holes U~ing

the properties of fundamental groups, we can give yet another proof that the circle 8 1 = aD 2 is not a retract of the disk D2. Namely, suppose that there exists a retraction r: D2 --+ 8 1 . Then the composition A .!.... X ~ A is the identity map; therefore, it induces the Identity homomorphism of fundamental groups. But the composition ?rI(A) ~ ?rl(X) .!::.. ?rl(A) cannot be the identity, because ?rICA) = Z and ?rl(X) = o. A more delicate algebraic argument proves the following assertion. Recall that on p. 74, we proved the general theorem that the boundary of a compact manifold is never a retract of this manifold. Problem 88. Let X be a Mobius band with boundary A. Prove that A is not a retract of X. Problem 89. (a) Let X be the torus r2 from which the open disk D2 is removed. Prove that A = aD 2 is not a retract of X. (b) Prove the same assertion for a sphere with 9 handles instead of T2. Problem 90. Let X be a closed nonorientable surface from which the open disk D2 is removed. Prove that A = aD2 is not a retract of X.

Now we consider a more complicated example of calculating the fundamental group. The main difficulty is in describing the cell structure of the space.

M:

Example (see [41]). Let be the manifold of unit tangent vectors to a sphere with 9 handles. Then the group ?rl (M:) has generators aI, . .. ,ag , bI, .•. -,-bg,c and defining relations aiC =~, bic = cbi, 11f=I(aibiailbil) =

c2- 2g •

Proof. Consider any vector field with isolated singular points on the manifold = gT2. We can assume that all singular points of this vector field are inside the disk D2 c The indices of the singular points of a vector sum to 2-2g; therefore, the degree of the map aD 2 ....... 8 1 defined field on

M; M;

M;.

6. Fundamental Groups

262

Figure 4. The vector field

t1

by X 1-+ v(x)/lIv(x)1I is equal to 2 - 2g. The map aD2 --+ Sl1S-homotopi~ to any other map of the same degree. The homotopy can be represented as a vector field on the annulus '[)2 \ D~, where D~ is a disk concentric to D2. Thus, we can construct a vector field on D~ which consists of unit vectors; moreover, we can assume that on 8D'J, this vector field is as follows: traversing a. small arc, the vector vex) makes 2 - 2g rotations, and outside this arc, the vector field is constant. We extend this vector field inside D~ along the radii (and leave it undefined at the center of the disk). Using this vector field v, we define characteristic maps from 1- and 2-cells to as follows. We represent the two-dimensional surface as a 4g-gon with sides glued together. We assume that the center of the disk D~ corresponds to the vertices of this polygon (recall that all these vertices are glued to one point) and the arc of aD'; outside of which the vector field v is constant is contained entirely inside one of the angles of this polygon (see Figure 4). Lei Vo be the constant value of v outside the arc specified above. For the O-cell of M: we take the pair (center of D~, vector vo). For the open 1-cells we take the following sets:

M; \

M:

M;

• the pairs (center of • the pairs (point the 4g-gon.

X,

D:, any unit vector w I: vo); vector v(x», where

X

is inside the ith side of

Clearly, the continuous extensions of the corresponding characteristic maps (0,1) - t take the endpoints of the interval [0, l] precisely to the O-cell.

M:

Now, we construct the 2-cells. We take a homeomorphism of the interior of the 4g-gon onto the interior of the (4g + 1)-gon obtained by blowing up one vertex (namely, the vertex corresponding to the arc on which the vector field is nonconstant)j the blow-up is shown in Figure 5. The vector field V naturally determines a map from the closure of the (4g + 1 )-gon to M;.

1. CW-Complexes

263

.....

L ________ _

Figure 5. The blow-up

This map takes all vertices to the O-cell, each interior point x to the pair (x, v(x»~ ~d each point of the blown up side to the uniquely qetermined limit tangent vector at the center of D~; the remaining sides are mapped to the corresponding I-cells. In addition to this 2-cell, we consider the 2-cell consisting of all unit 'vectors at the points of one of the sides of the 4g-gon. It is easy to verify that the complement of the union of ali closed 2-cells is an open 3-cell. Indeed, this complement consists of the unit tangent vectors at the interior points of the 4g-gon, and at each point x, all vectors different from vex) are taken. Such a set is homeomorphic to the direct product of an open 4g-gon and (0,1). 1 The generators al, •. ~ ,ag , bl, _•. , bg correspond. to sides of the 4g-gon. The generator c corresponds to the I-cell over the center of The relation ITf=l(aibia;lb;l) = c'2- 2g is determined by the (4g+ 1)-gonal cell (the map from the blown up side to the I-cell c has degree 2 - 2g). The relations ~ca;lc-l = 1 and. bi cb;lc- 1 = 1 are determined by the 4-gonal cells. 0

D:,

Problem 91. Prove that tangent vectors to 8 2 .)

M3

~ lRp3. (Here

M3

is the manifold of unit

1.3. The Fundamental Group of SO{n). To calculate the fundamental group of a space, it is not necessary to know the cell structure of this spacej in many cases, the exact sequence of a fibration is sufficient. Example. For n ~ 3, 7l"l(SO(n») = Z2' The generator of the group 7l"1(80(n» is the path consisting of all rotations about a fixed axis. Proof. Take a unit vector e in lRn +1 and consider the map SO(n+ 1) -+ sn which assigns the vector Ae to each matrix A. This map is a locally trivial fibration with fiber SO(nJ. We write the exact sequence for this fibration: 7l"2(sn)

-+

7l"l(SO(n»)

-+

71"1 (SO(n

+

l»

-+

7l"l(sn).

If n ~ 3, then tr2(8n )= 7l"l(sn) = OJ therefore, 7l"1(80(n» ~ 7l"1(80(n + 1».

Moreover, 80(3) :::::: lRp3 (see the solution of Problem 91) and 1f"1(lRp3)

=~.

264

6. Fundamental Groups

Let us represent lRp3 as the disk n3 with boundary antipodes identified. The homeomorphism SO(3) ~ lRp3 takes the rotations about a fixed axis to some diameter of this disk. Each path-diameter of 3 corresponds to a nonzero element of the group 11'1 (lRp3), because the lifting of this path under the covering S3 --+ lRp3 is nonclosed. The embedding SO(n) --+ SO(n + 1) maps rotations about a fixed axis to rotations about a fixed axis, and it takes the generator of the fundamental group to the generator of the fundamental group. 0

n

The Kakutani theorem. The relation 1I'1(SO(3» = Z2 hlWthe following consequence. Theorem 6.2 (Kakutani [64]). Let S2 = {u E lR3 ; IIxll = 1}, and let f: S2 --+ lR be a continuous map. Then there exists an orthonormal basis Ul, U2, U3 in lR 3 such that f(Ul) = f(U2) = f(U3)' Proof. Take an orthonormal basis el, e2, es and consider the map cp: SO(3) --+ lR3 defined by cp(A) = U(Ael), f(Ae2), f(Ae3)). We must prove that at least one point cp(A) belongs to the line I determined by the equation x = y = z. Suppose that cp(SO(3» does not intersect this line. Let "p be the composition of cp and the projection onto the plane x+y+z = 0 (this plane is orthogonal to 1). By assumption, the origin does not belong to the image of "pi therefore, we can consider the composition of "p and the projection from the origin onto the unit circle Sl in the given plane. As a result, we obtain a map ifi: SO(3) --+ SI. Consider the restriction of 1[; to the subgroup Sl.....-+ SO(3) consisting of the rotations about I. Let us find out what the curve "p(SI) is. Let t be the parameter on the circle S1 (the rotation angle about the axis I). The rotation through the angle t = 211'/3 takes the vectors el, e2, and e3 to e2, e3, and el, respectively. In the plane x + y + z = 0, this corresponds to the transformation (x, y, z) 1-+ (y, z, x). It is easy to show that this transformation is the rotation through 211'/3. Indeed, the cosine of the angle between the vectors (x, y, z) xy-(x+Yr~ 1 There£ore, the h and ( y, z, x ) , were z -- -x - y, equals a;2+ y2_(.:c+y)2 - -2"' structure of the curve ¢(~l) is as follows. As t varies from 0 to 211'/3, the curve goes from the point (xo, Yo, Zo) to the point (Yo, zo, xo). The next segment of the curve, corresponding to t between 211'/3 to 411'/3, is obtained by rotating the preceding one through ;l1l'/3. The remaining p~ of the curv~ is this segment rotated through 211'/3. The curve ifi(8 1 ) has siInilar structure. As t changes from 0 to 211'/3, the point 1jj(t) rotates through +2k1l'l where k is an integer. Thus, as t changes from 0 to 211', _the point 1jj(t) rotates through 211'(3k + 1) =/:- O. This means that the map ¢: SO(3) --+ 8 1 induces .a nonzero homomorphism 1I'1(SO(3» --+ 11'1(81). But any homomorphism Z2 --+ Z is zero. 0

2;

1. CW-Complexes

265

Corollary. Any bounded closed convex subset K of R3 ron be inscribed in a cube. Proof. For each unit vector u E R3, take the two supporting planes of K orthogonal to u. Let !(u) be the distance between these supporting planes. Applying the Kakutani theorem to f, we obtain the required result. 0 The spinor group. H n ~ 3, then 1r1(SO(n» = Z2. This means that SO(n) is doubly covered by a simply connected manifold. We denote this manifold by Spin(n). This is a group, as well as SO(n). The group Spin(n), which is called the spinor group, can be constructed as follows. The Clifford algebra d n is the associative algebra with identity generated by elements e1 t ••• , en satisfying the relations e~ = -1 and eiej + ejei = 0 for i #: j. Let Rn be the linear subspace of Cn spanned by the vectors eI, •.. , en. It is easy to verify that (Exiei)2 = - L:x~; therefore, all nonzero elements of R n C Cn are invertible. In particular, all elements of the unit sphere 8'1 .... 1 C lRn C Cn are invertible. In the multiplicative group of invertible elements of Cn, the elements of the unit sphere sn- 1 generate a certain subgroup; this subgroup is denoted by pin(n). The group Spinen) is a part of the group pin(n). To describe this part, we use the decomposition of the linear space Cn into the direct sum d,! ffi C~, where ~ is the subspace generated by all products of the form ejl ... ehle+i; the spaces d,! and C~ intersect only in zero, because the application of the relations = -1 and ~ej + ejei = 0 cannot change the parity of the number of generators in the product. The group Spine n) is the part of pinen) that is contained in d,!. In other words, the group Spin(n) consists of the products u1u2"'U2k, where Ui E sn-1. To construct a homomorphism Spin(n) --+ SO(n) that is a double covering, we need the involutive anti-isomorphism of the algebra Cn defined by ~l ••• ~" 1--+ fi" . • . eil; this map leaves the elements e~ and eiej + ejei fixed, and therefore this is indeed a self-map of Cn. We denote it by u 1--+ u*. Clearly, (u*)· = u and Cuv)* = v*u*; moreover, u ERn implies u* = u. To each element u E pine n) we assign the map !p(u): R n --+ Rn defined by !pCu)x = uxu*. First, we must verify that the element uxu* indeed belongs to Rn, i.e., can be represented as a linear combination of el. ... , en. It is-suflicient to consider the case u E sn- 1 , Suppose that 1.£ = EUi~ and x = L:xiei. Then ux = - EUiXi+ L:i#j UiXj~ej. The product uxu* = uxu has a, part that belongs to R n for obvious reasons. The remaining part is of the form E ~xjUk~eje", where the summation is over the triples of pairwise distinct numbers. But eiejek = -ekejei; hence this sum vanishes. Next, we 'verify that !pCu) is an orthogonal transformation. First, note that if 1.£ = U1" 'uk, where Ui E sn-1, then u*u = uu* = (_l)k~ Moreover,

e1

266

6. Fundamental Groups

as mentioned before, (EX,ei)2:::: ~ Ex~ = +-lIxI[2. Therefore,lIc,o(u)xIl 2d -uxu·uxu· = -( _1)kux 2u· = (-I)kll x I[2uu• = IIx1l 2 • Theorem 6.3. The map c,o: pin(n) - O(n) is a group epimorphism. More" over, cp-1(SO(n)) = Spin(n) and Kercp f Spin(n) = -l}.

p,

Proof. Clearly, cp(uv)x = uvx(uv)* = uvxv·u* = cp(u)cp(v)x. Therefore, cp is a group homomorphism. We show that if u E 81'-1, then cp(u) is the symmetry with respect tQ the hyperplane orthogonal to~. We represent f ~ AU+W{ where wl:u. We must prove that cp(u)x = -AU+W. Bu~ cp(u)x = 'If(AU+W)U· = UAUU" +uwu· F -AU + uwu·. It remains to show that uwu· = w (u· = U in the case pnder consideration). If E UiWi = 0 and ~ u~ = i, then uw = Ei;l:j UiWjeie~ = - Ei;l:j WjUiejei = -wu and uwu· = -wuu* = w. Any orthogonal. transformation can be represented as a composition of symmetries about ~yper­ planes; therefore, cp is an epimorphism. Clearly, an orthogonal transformation is orientation-preserving if and only if it is a composition of an even number of symmetries about hyperplanes. Therefore! cp-1(SO(n» = Spin(n). Suppose that u E Spin(n) and and hence UfiU·U = fiU. But u·u = 1. for u E Spin(n). Therefore~ Ufj = eiU. Conversely, if ue; = fiU for all i and U E Spin(n), then lJ'(u)x = x for all f. It is easy to verify that U = E ail ... i2/ceh •.. ei210 implies fiue~1 = -fiuei = E(-ly(i)a;1 ...i 2 /cfi1 ·· ·e~/c, where e(i) = 0 for i ¢ {iht .... ,i2k} and e(i) = 1 for? E {il"~i,i2k}' Thus, ifu E q:! and fiU 7 uei for.all i, then u::::; (V 1, where A E R. Therefore, Kercp f Spin(n) F {I, -1}~ 0 Thus, the group Spin(n) is a double covering of SO(n). It remains to show that this covering is nontrivial, i.e., the space Spin(n) is connected. It suffices to verify that Spin(n) contains a path joining 1 and .... 1. Such a path is given by (e1 cost + e2 sin t)(e1 cos t - e2 sin t) = - cos 2t - e1e2 sin 2t,

where t E [0, 'Tr /2].

2. The Seifert-van Kampen Theorem 2.1. Equivalent Formulations. Suppose that a path-connected topological space X is the union of path-connected sets Ul and U2 with pathconnected intersection Uill U2. Choose a point Xo E UI n U2 and consider the fundamental groups 'TrI(U! n U2,XO), 'Trt(Ul,XO), and 1I"1(U2,XO)... Suppose that these groups are defined by sets of generators S, Su, and S2 and

2. The Seifert van Kampen Theorem

267

Xo

Figure 6. The counterexample

relations R, Rl, and R2. For brevity, we use\;he same notation for maps of topological spaces and for the maps of fundamental groups that they induce. The embedding .,pi,: Ui, -+ X takes each element 8i E Si to .,pi8i, E 7rl(X,XO). Thus t 7rl(X,XO) contains sets (alphabets) .,plSI and .,p2S2 such that every element of the image of .,pi: 7rl (U;" xo) -+ 7rl (X, xo) can be written as a wor~ in the alphabet .,piS~! Now, take an element 8 E S and consider its image in 7rl (Ui" xo) under the embedding IPi: Ul n U2 -+ Ui. The element IPi8 can be represented as a word in the alphabet Sit and the element .,piIPi8 can be represented as a word in the alphabet .,piS,. In th~ group 7rl(X,XO), the relation .,pI IPl S = .,p21P2S holds, because .,pI IPl = .,p2IP2 (both of these compositions coincide with the embedding of Ul U~ into X). Thus, the group 7r1 (X, xo) contains the elements .plS1 and 1/J2S2, and these elements are related by .,plRI, 1/J2R21 and .p1IPIS = 1/J2IP2S, where S E S. Consider the group G with generators .plSI and 1/J2S2 and defining relations .,plRI, .,p2R2, and .,plIPIB = tP2IP2S, where S E S. It is natural to expect that G is isomorphic to 1[l(X,,:rO). This is not so without additional assumptions; there exists an example for X = SI (see Figure 6). But under certain assumptions about the path-connected spaces under consideration, this assertion becomes true. For example, in 1931, Seifert [120] proved that it is true in the case where Ul and U2 are sub complexes of a simplicial complex, and in 1933, van Kampen [67] proved that it is true if Ul and U2 are open in X.

n

The theorem of Seifert (for CW-complexes) is an obvious consequence of Theorem 6.1 on p. 258 with only one additional assumption: the generators of the fundamental groups are the I-cells, and the relations are the 2-cells. Thus, before proceeding further, we give an invariant definition of the group G, which does not depend on the choice of generators and relations. Let Go. Gl, and G 2 be groups defined by sets of generators S, SIt and

268

6. Fundamental Groups

S2 and sets of relations R, Rl, and R2. Suppose that G is a group,

is a commutative diagram of homomorphisms, and G' is another group for which the similar diagraIll

holds. We show that in this case, there exists a unique homomorphism u~ G -+ G' for which UtPi = tP~. Let F be the free group on generators "/JISl and tP2S2, and let N be its minimal normal subgroup containing the words tPlRI, tP2R2, and tPI'PIS(tP2'P2S)-1, where S E S. For each element Si E Si, we set UtPiSi = tP~Si' Any map from the generators of the free group F to an arbitrary group can be uniquely extended to a homomorphism; thus, we obtain a homomorphism u: F -+ G' . If u(N) = 1, then the homomorphism u induces a homomorphism u: G -+ G' for which UtPi = 1/It.. Therefore, it suffices to verify that u(N) = 1 (the homomorphism u is unique because the elements tPlSl and 1/J2S2 generate the group G). If Ti E ~, then UtPiTi = tP~Ti = 1, because the word ri is the identity element of Gi. If s E S, then UtPi'PiS = .,p~'PiS; hence the equality ~'Pl = p' = tP~'P2 implies UtPI'Pt S = UtP2'P2S. Consider all commutative diagrams of homomorphisms

~/'" Go I{J2 ""

Gt

~l G

G2 /1/12

with fixed homomorphisms 'Pt and 'P2. The group G in these diagrams is called the amalgam of the groups G I and G2 with respect to the group Go (and the homomorphisms 'PI and 'P2) if it has the universal property mentioned above: for any other group G' from a similar commutative diagram, there exists a unique homomorphism (J': G -+ G' such that utP" = tPi. To obtain an invariant definition of the group G, it remains to show that the amalgam is unique up to isomorphism.

2. The Seifert van K8IDpen Theorem

269

Let G and G' be two amalgams with respect to the same groups. Then there exist homomorphisms u: G -+ G' and u': G' -+ G for which u'I/Ji = .,p~ and u'.,p~ = .,pi' Consider the homomorphism u'u! G ...., G. It has the property that u'U.,pi = u'.,p~ = .,pi. By assumption, there is only one homomorphism G -+ G with this property. On the other hand, the identity homomorphism does have this property; therefore, u'd = ide. Similarly, qU' ::;; ide, f and hence u is a group isomorphism. 2.2. The Proof. Theorem 6.4 (van Kampen). Let Ul and U2 be open path-connected subsets of the space X = Ul U U2 such that their intersection Ul n U2 is path-' connected. Consider the commutative diagram

"

11"1 (Ul)

11"1 (Ul

~

n U2)

~

.1.

11"1 (X)

~

11"1 (U2 )

induced by the embeddings (the fundamental groups are with respect to the same base point Xo E UI n U2). Then the group 1I"I(X) is the amalgam of 1fl CUI) and 11"1 (U2) with respect to the group 11"1 (UI, n U2 ) and the homomo~ phisms CPI and CP2. It is more convenient to prove the theorem of van Kampen in terms of generators and relations rather than in terms of amalgams; we have explained above that the two statements are equivalent.

Proof (see [31]). Consider the commutative diagram 11"1 (Ut)

11"1 (UI

/'"

n U2)

.1.'

~

H

~1I"I(U2)~ 2

We must prove that there exists a unique homomorphism (1': 1I"1(X) \"'""+ H for wJIich U.,pi = .,p~. The uniqueness of u is easy to derive from the following assertion. Step 1 (generators). The images of the groups 1I"I(Ud and 1I"2(U2) under the homomorphisms .,pI and .,p2 generate 11"1 (X). Consider an arbitrary loop a: [0,1] -+ X based at Xo. The interval [0,1] is covered by two open sets, a-l(UI) and a- I (U2). Let 5 be the Lebesgue

6. Fundamental Groups

270

number of this cover. Choose points 0 = tl < t2 < •. / < tn+1 == 1 such that tk+1 - tk < 6. The image of each interval [tk' tk+1} is entirely contained in one of the sets Ul and U2. The loop 0: is the composition of (not necessarily closed) paths O:t, a2, .. J ,Oni where each ak is the restriction of at to the interval [tk' tk+1]' To represent d as BUch 8. composition of loops, we join each point a(tk) E Ui to Xo by a path Pk contained in Ui; if O(tk) e U1 n U2, then the path Pk must be contained in Ul n U2. If the path Ok is contained entirely in Ui, then the loop p;;lakPk+l is contained entirely in Ui as well; therefore, the class of the loop p;;lakPk+1 is contained in the image of the group 1I"1(Ui) under the homomorphism ..pi . . Moreover, a is the composition of the loops alf32, PilasP31 I" *1', p:;;~la~lf3n, p;;lOn. Now it is easy to prove that the homomorphism er is unique.--Indeed, let us represent an element a E 11"1 (X) in the form a = Ilk:l 'l/Ji(k)'Yk, where 'Yk E 11"1 (Ui(k». The equality er'I/Ji = 'I/J~ implies er'I/Ji(k)'Yk = 'I/J~(k)'Yk' Therefore, n

(1)

ero: =

II 'I/J:(k)'Yk. k=1

Relation (1) completely determines the homomorphism er. It remains to verify that this homomorphism is well defined, l.e., a = Ilk=l'I/J~(k)'Yk depends only on a and not on the representation a Ilk=I'I/Ji(k)'Yk. For this purpose, it suffices to prove the following assertion.

=

Step 2 (relations). Suppose that 'Yk E 1I"1(Ui (k» and Ilk=I'I/Ji(k)'Y1c Ilk=l'I/J~(k)'Yk = l. The equality Ilk=l 'l/Ji(k)'Yk = 1 means that there exists a map with the following properties:

= 1; then

I: [2 -+ X

• the restriction of I to the bottom edge of the square P represents the class of the loop Ilk=l 'l/Ji(k)'Yk; • the map I takes the other edges of [2 to the base point Xo (see Figure 7). Let 6 be the Lebesgue number of the cover of [2 by the open sets 1-1 (Ul) and /"..-1 (U2). We partition [2 into rectangles by vertical and horizontal intervals in such a way that the diagonal of each rectangle is less than O. In the set of such intervals we include all vertical intervals dividing the bottom edge of the square into n equal intervals (it is assumed that each of these equal intervals corresponds to one of the paths 'Yi; in particular, their endpoints are mapped to the base point xo). • By construction, the image of each rectangle is contained in Ul or U2. Let a be a vertex of one of the rectangles. We connect the point I(a) to the base

2. The Seifert van Kampen Theorem

-t-1~:

271

-J--

-"'--1'"--~I lit

-T-'--I -r-

-+-~r-: -~-

-+--1--1 -to1 1 I I -T-"'+-I -r-

Figure 7. The map of the square

f(U34)

Figure 8. A rectangle from the partition and its image

Figure 9. The map of the disk

point Xo by a path Wa so that if a E Vi, then Wa c: Ui (in particular~ if a E U1nU2, then Wa C UlnU2) and if a = xo, then Wa = Xo (the path is constant). Consider one of the rectangles in the partition. Let 012, 023, 034, 041 be its sides (see Figure 8). It is easy to show that the loop 1312132313341341, where f3pq = w;1 f(opq)wq, is contractible in the space X. Indeed, it suffices to prove that the corresponding map 8 1 --+ X can be extended to a map D2 --+ X'. The required map D2 --+ X can be constructed as follows. First, we send D2 to a rectangle with segments attached to its vertices (see Figure 9). To the rectangle we apply the map fl and the segments are mapped to X by using wp. For the element of 1rl(X) represented by the loop f3pq we use the same notation f3pq. By construction, the loop f3pq IS contained in Ul or U2; hence f31Hl = tPif3i' where f3i E 1rl (U.). We set f3~ tP~f3i E H. We must verify

=

6. Fundamental Groups

272

lIn-I

1I2

I I I1 lID Figure 10. The products II.

that the element /3~ is well defined (the loop /3pq may lie in UI n U2). H /3pq c UI n U2, then /3pq = .,p1/31 = 'l/J2f32, where /31 = 'PI/3 and f32-~_'I?2/3 for some /3 E 7r1(UI n U2). We have to prove that 'I/J~/31 = 'I/J~I32. By assumption, 'I/J~ 'PI = 'I/J~'P2; therefore, 'l/JiJ31 = 'I/J~ 'PI/3 = 'I/J~'P2/3 = 'I/J~/32' By construction, one of the sets U1 and U2 contains not only the loop /3pq but also the four loops /312, /323, /334, and /341; hence /3pq = 'l/Ji~pq, where ~pq E 7r1 (Ui) and i is the same for all the four loops. The proof of the contractibility of the loop ~12~23~34~41 is similar to that of contractibility of /3121323/334/341 (with X replaced by Ui). Thus, /3~2/3~3/3~~1 = ('I/J~~12)('I/J~~23)('I/J~~34)('I/J;~41)

= 'I/J;(~12~23~34~41) = 'I/J~(I) =

1.

Let us summarize. To each of the (directed) edges of the rectangles into which the square P is partitioned we have assigned an element of the group H, and the elements assigned to the edges of the same rectangle satisfy the relation /3b/3~3/3~/341 = 1. We must prove that the product of the elements of H that correspond to the bottom edge of the square is equal to 1. We denote this product by ITo (see Figure 10). Consider the product of all relations of the form /3b/3~3/3M/341 = 1 over the rectangles adjacent to the bottom edge of the square. We multiply the relations in such a way that the common sides of the rectangles are included with opposite orientations. AB a result~ we obtain ITo = TIl, where III is the product of the elements of H that correspond to the next-to-bottom horizontal segment. (The elements corresponding to the two extreme vertical sides are not canceled, but these sides are mapped to the base point Xo and, therefore, correspond to the identity element of the group.) Similarly, we obtain III = II 2 , ... , IIn -1 = TIn. But the upper horizontal segment is mapped to the base point Xo; hence lIn = 1. 0 Corollary. lIn ~ 3, then 7rl(Mn) point 01 the manifold Mn,

~

7rl(MR \ {x}), where x is an arbitrary

Pl"oof. Take a neighborhood of x homeomorphic to lRR for U1 and an arbitrary point of the set U1 \ {x} for the base points in the spaces Mn and

2. The Seifert van Kampen Theorem'

273

Figure 11. The sets Ul and U2

Mn \ {x}. The set {x} is closed; hence U2 = Mn \ {x} is open. Clearly, UlnU2 = IRn\{x} '" sn- l . By assumption, n ~ 3; therefore, 11"1 (UlnU2) = 1. Moreover, 1I"1(Ul) = 1. Thus, the groups 1I"1(Mn) and 1I"1(Mn \ {x}) have the same sets of generators and relations.

0

Remark 6.1. This assertion can be proved without employing the van Kampen theorem. Indeed, consider a triangulation K of the manifold Mn. We can assume that x is inside some simplex ~ n of this triangulation. Then the space M n \ {x} is homotopy equivalent to the simplicial complex obtained from K by deleting ~ n. If n ~ 3, then the deletion of an n-simplex does not affect the 2-skeleton. 2.3. Knot Groups. A knot is the image of the circle 8 1 under a continuous map I: 8 1 - IR3, and the group of a knot K is defined as the group 1r'l(IR3\K,xo), where Xo is an arbitrary point ofIR3\K. A knot K is said to be polygonal if I is a piecewise linear function of the parameter !P on the circle 8 1 = {ei
11"1 (IR3 \

8 1)

= Z.

Proof. For Ul and U2 we take the open subsets of IR3 \ 8 1 shown schematically in Figure 11. Clearly, Ul n U2 8 1 V 8 1 and Ui '" 8 1. We choose generators a and b in the group 11"1 (Ul n U2) and generators al and a2 in the groups 11"1 (Ul) and 1l"1(U2) as shown in Figure 12; we impose no relations on al and a2. Each homomorphism !Pi: 1I"1(UI n U2) - 1I"1(Ui ) takes both elements a and b to the same element ai. Therefore, the group 1l"1(IR3 \ 8 1 ) has generators 1Plal and 1P2a2 and defining relation 1Plal = 1P2a2. As a result, we obtain a group on one generator a with no relations. 0 /"OJ

6. Fundamental Groups

274

,, I

Figure 12. The choice of generators

Figure 13. A knot diagram

Problem 92. Prove that :IR3

\

81

,..,

s2 V 8 1 .

For any smooth knot, we can choose a plane such that the projection of the knot on this plane satisfies the following conditions: • the projection of no tangent to the knot degenerates into a point; • no point is the projection of more than two different points of the knot; • the set of crosses (Le., the points in the plane to which two different points of the knot are projected) is finite, and the projections of the tangents to the knot at the two points corresponding to a cross do not coincide. For a polygonal knot, a projection plane with similar properties can be chosen. The analog of the tangents to a smooth knot is the straight lines containing the edges of the polygonal knot. A diagram 'of a knot is its projection on a plane satisfying the conditions specified above. For each cross, the diagram must indicate which branch of the knot passes over the other (see Figure 13). Instead of the knot K, consider any knot K' which coincidE'S with some diagram of K everywhere except in small neighborhoods of the crosses; at each cross, one of the branches passes above the plane of the diagram and the other remains in this plane (see Figure 14). Clearly, the groups of the . knots K and K' are isomorphic because the spaces :IR3 \ K and :IR3 \ K' are homotopy equivalent. In calculating knot groups, we assume that the knots

2. The Seifert van Kampen Theorem

275

Figure 14. A neighborhood of a cross

V2

- ......-- -----

--~-

VI

Figure 15. The sets U1 and U2

are embedded in space as K'. We consider only one simple example, namely, calculate the group of the trefoil knot, which is shown in Figure 13. The group of any other knot can be calculated by the same method. Example. The group of the trefoil knot is defined by generators x and 11 and the relation xyx = yxy.

Proof. For UI and U2 we take the open subsets of ]R3 \ K shown schematically in Figure 15. The set Ui is obtained from the half-space by removing n arcs, where n is the number of crosses in the knot diagram. It is easy to verify that the projections of these arcs on the plane of the knot diagram are pairwise disjoint; therefore, Ui is homotopy equivalent to the wedge of n circles. Clearly, Ul n U2 is homotopy equivalent to the wedge of 2n circles. Take the generators of the fundamental groups 11"1 (Ul), 1r1(UI n U2), and 1I"1(U2 ) shown in Figure 16 (for the group 1I"1(UI), in addition to its generators, the loop 'PIX2 and a loop homotopic to it are shown). Clearly, 'PIXI = x. Looking at Figure 16, we see that 'PlX2 = yzy-l. In U2, we have--the simpler relation h n 0 hn-l o· • . 0 ho = h is defined, and the map h is continuous. Clearly, the map h: Bo --+ h(Bo) is one-to-one, and hence h is a homeomorphism (see Theorem 3.2 on p. 89). Thus, the space X = n~o Xn = h(Bo) is homeomorphic to the disk D3. Let us calculate the fundamental group 7l"1(lR3 \X). We set Yn = lR3 \Xn . Then Yo C Yl C··, and lR3 \X = u~=o Yn' Clearly, 7l"l(YO) = Z because Xo

6. Fundamental Groups

278

Figure 19. The construction of the space Xl

A Figure 20. A disk from which two tubes are removed

is the solid torus standardly embedded in 1R3 • Let us show that 71"1 (Y1) = F2 is a free group on two generators a1 Jlnd ~2 and the embedding Yo - Yl induces the monomorphism 7I"l(YO) - 7I"l(Yi) that takes the generator a of the group 71"1 (Yo) to the commutator [ai, a2]. We represent Y1 as the union Yo U A, where A is the space shown in Figure 20. To apply the van Kampen theorem, instead of Yo we must take Yo = Yo U Z, where Z is a small neighborhood of the cylinder 8 1 X I, the lateral surface of the set A. But in the situation under consideration, this is inessential from the homotopy point of view. It is easy to see that the space A (a disk from which two linked handles are removed) is homeomorphic to a disk from which two rectilinear handles are removed. l'herefore, the fundamental group 7I"l(A) is a free group on two generator!:! a1 and a2, which are represented by loops thrown upon the handles removed. The generator of the group 7I"l(Yo) is represented by the loop a thrown upon the removed handles as shown in Figure 21. This loop is homotopic to the composition al a2a}1 a2"l. Therefore~ the group 7I"l (YJ.) is generated by a, ab and a2, which are related by a = tab a2]. Similarly, we can obtain Yn +1 from Yn by attaching 2'1 copies of the set A. Ai; a result, arguing by induction, we conclude that 7I"l(Yn+1) is a free group on 2n+ 1 generators, and the homomorphism 7I"l(Yn ) -+ 7I"l(Yn+1) induced by the embedding Yn - Yn +1 is a monomorphism. Any loop in 1R3 \ B is compact; therefore, it is contained in some set Yn' The image of a homotopy between two loops is compact also. 'rherefore, the

3. Fundamental Groups of Complements of Algebraic Curves

279

Figure 21. The homotopy of the loop a

group 71'1 (1R3 \B) is isomorphic to G = U~=O 71'1 (Yn ). The fundamental group of the space lR.3 \ D3 is trivial, while the group G contains, e.g., a free group on two generators. Thus, the space lR.3 \ X is not homeomorphic to lR.3 \ D3. Note that the commutator .quotient of G is trivial, because each gen~ ator is the commutator of two other generators.

3. Fundamental Groups of Complements of Algebraic Curves

A plane algebraic curve C is the subset of CP2 given by an equation pf the form P(x, y, z) = 0, where P is a homogeneous polynomial of degree n ~ li the number n is called the degree of the curve C . .A curve C is said to be reducible if P = P 1P2, where PI and P 2 are homogeneous polynomials of positive degrees. A point of a curve C is singular if 9Iad P = 0 at this point. The first groups of the form 71'1 (Cp2 \ C) were calculated in [66,148 150]; since then, they have been studied by many authors. We start with several examples, in which C is a union of complex lines in Cp2. This corresponds to the situation where the polynomial P decomposes into linear factors. I

3.1. The Complement of a Set of Complex Lines. In this section, we calculate the group 71'1 (CP2 \ U~lli)' where 11 , ••• , In are complex lines in Cp2. This group depends not only on the number of the lines but also on their arrangement. We consider several examples of arrangements. Note that removing one line turns Cp2 into C2j therefore,

where C 2 = Cp2 \ In and ii = 4 n C2. Thus, calculating the fundamental group of the complement of n lines in CP2 reduces to calculating the fundamental group of the complement of n - 1 (complex) lines in C2.

6. Fundamental Groups

280

Example. If lines h,~. ...,In E Cp2 pass through one point, then n

11"1 (Cp2 \

Uli) = F -1 n

i=l (the free group of rank n -1). Proof. The lines £1 •..• , £n-1 do not intersecf in the space C 2 = Cp2 \ 1n; hence we can choose complex coordinates z and w in such a way that £i = {(z,w) : w = Ci}, where Ci is a constant. Therefore,

C2 \ Example. If lines

n 1

n-l

~l

i=l

U"liNC\{CbLJ,Cn_1} ..... VS,. 1

h, l2, 13

o

E Cp2 intersect in three distinct points, then

11"1 (Cp2 \ U~=lli) = Z EB Z. Proof. In the space C2 = Cp2 \ l3, the lines £1 and £2 intersect at some point; therefore, we can choose complex coordinates z and w in such a way that these lines are determined by the equations z = 0 and w = O. The origin does not belong to C2 \ (£1 lJ £2); thus, applying the projection onto S3 = {(z,w) E C2 : Izl2 + Iwl 2 = 1} along the real rays from the origin in C2 = ]R4, we can show that the space C 2 \ (4 U £2) is homotopy equivalent to S3 n (C 2 \ (i1 U £2))' The latter space is the sphere S3 from which the two circles given by the equations z = 0 and w = 0 (on S3) are removed. Puncturing a 3-manifold does not change its fundamental group (see p. 272). Therefore, we can remove one point from S3 and pass to the space ]R3 \ (K1 U K 2 ), where K1 and K2 are circles. (If Kl, •. . ,Kn are palrwise disjoint images of circles under ho~eomorphisms Sl --+ ]R3, then the set Kl U .. ·UK" C]R3 is called an n-component link.) The fundamental group of the space ]R3\(K1UK2) is calculated in exactly the same way as the fundamental group of a knot (see the example on p. 275). To construct the diagram of the link K1 UK2, consider the stereographic projection of the sphere S3 onto the subspace Rew = 0 from the point (0,1). The projection of Kl is a circle centered at the origin in the plane 1m w = 0, and the projection of K2 is the line z = O. Therefore, the diagram of the link Kl U K2 is as shown in Figure 22. The group 11"1 (]R3\ (K1 UK2)) is defined by generators a and b and the relation b = aba-I, i.e., ab = ba (the second cross gives the same relation). 0 Now, let us explain geometrically how the relation ab = ba arises when we go around the intersection point of two complex lines. The element a is determined by the unit circle in the plane w = wo, and the element b is .qetermined by the unit circle in the plane z = ZOo We traverse the circle b and simultaneously translate the circle a (see Figure 23). Formally, such

3. Fundamental Groups of Complements of Algebraic Curves

281

Figure 22. The diagram of the link Kl U K2

Figure 23. Going around the intersection point of two complex lines

a translation can be described as follows. We assume that Zo = Wo = 1. Suppose that the point x(t) = (1, eit) moves uniformly on the circle b. To each moment of time t we associate the loop that first goes from Xo = (1,1) to x(t), then passes along the circle (e iB , eit ), where t = const, and finally returns from x(t) to Xo. After the whole circle b is traversed (Le., at t = 211"), we obtain the loop bab- 1 j this loop is homotopic to the initial loop a. Problem 94. Given a link L = K1 u· .. U Kn whose diagram consists of n disjoint circles (such a link is called the trivial n-component link), prove that the space ]R3 \ L is homotopy equivalent to the wedge of n copies of the space 8 2 V 8 1 • Problem 95. Given a link L = K1 U K2 with diagram shown in Figure 22, prove that ]R3 \ L T'l V 8 2 • f'.J

Example. Let it. ... ,in be pairwise distinct lines in C 2 passing through the origin. Then the group 11"1 (Cp2 \ Lr=1 ii) is defined by generators aI, ~ •. , an and the relations

6. Fundamental Groups

282

Figure 24. The diagrams of two links

Proof. The case n = 2 was considered in the example on p. 3.1. We consider only the cases n = 3 and n = 4. As in the above-mentioned example, calculating the required fundamental group reduces to calculating the groups of the links shown in Figure 24 (the circles are pairwise linked). For n = 3, the inner crosses giv~ the relations bi = lli+2lliaii2 and the outer crosses give the relations ai ai+1biai';1' Expressing the generators bI, ~, and ba in terms of all a2, and a3 with the use of the former relations and substituting the obtained expressions into the latter relations, we see that the required group has generators all a2, and a3 and defining relations ala2 a 3 = a2a3a1 = a3 a l a 2· For n = 4, we obtain the relations bi ai+1lliai11' Ci = ai+2biaii2' and oJ:d: ai-1Ciai":1' First, we express the generators {bi} in terms of {ail, then we apply the relations obtained to express {cd in terms of {lli}, and finally we substitute the resulting expressions into the relations ai = ai-1Ciai!1' As a result, we obtain the relations a1a4a3a2 = a4a3a2a1 = a3a2a1a4 !:: a2a1a4a3. It remains to set 01 = al, 02 = a4, 03 = a3, and 04 = a2. 0

=

=

3.2. The van Kampen Theorem. In the paper [148] published in 1929, Zariski proposed a method for calculating the group 1I'l(CP2 \ q). Soon after, he revealed that his proof used a conjecture which he was not able to prove (if an infinite group is defined by finitely many generators and relations, then the intersection of all of its subgroups of finite index is trivial)l In 1951, this conjecture was disproved. Another method for calculating the group 1I'l(CP2 \C) was suggested by van Kampen [66]. The theorem on the structure of the group 11'1 (CP2 \ C) that he proved is closely related to his other theorem discussed on p. 269 ·(Theorem 6.4). Both theorems were published in the same issue of the American Journal of Mathematics.

3. Fundamental Groups of Complements of Algebraic Curves

283

Our exposition of the van Kampen theorem on the group 1i"1(CP2 \ C) follows largely [27]. Suppose that a curve C is determined by the equation P = 0, where P = P1 P2. Let Cl and C2 be the curves given by PI = 0 and P2 = O. The equality grad P ;::: PI grad P2 + 1'2 grad PI implies that the singular points of the curve C are the intersection points of the curves Cl and C2 and the singular points of Cl and C2' We assume that C is a curve of degree n given by the equation P = 0, where P = PI ... Pic and the polynomials Pi are irreducible and pairwise distinct. Let Ci be the curve P; = O. The singular points of C are the singular points of the curves C, and the intersection points of pairs of these curves. Suppose that Xo E Cp2 \ C is a base point. We show that almost all lines passing through Xo intersect the curve C in exactly n distinct points ("almost all" here means "all but finitely many"). The number of intersection points of pairs of curves Cl, .... , Cle is finite; therefore~ it suffices to consider only straight lines not passing through these points. Such a line intersects the curve C in n points if and only if it intersects each curve C. in ~ points, where ni is the degree of the curve Ci. Therefore, it is sufficient. to prove the required assertion for an irreducible curve Oi, A straight line intersects the curve Cf, in fewer than ni points if it is tangent to Cj. or passes through its singular point. The number of tangent lines to the irreducible Curve Ct that can be drawn through a given point is finite (it does not exceed the degree of the dual curve Cn, and any irreducible curve has only finitely many singular points. Choose a line I passing through Xo and intersecting C in exactly n distinct points. Take a point a on l which is different from Xo and does not belong to C. Let 10 , h, ... , ls, be the straight lines that pass through a and intersect C in fewer than n distinct points. We choose the point a so that each of the lines lo, ll •... , Is contains precisely one singular point or point of tangency. Finally, we draw a line m different from I through the point Xo and consider the points ao, at, . •.• as at which the lines 10, ... , Is intersect m (see Figure 25). We choose m in such a way that it does not pass through the intersection points of C with the lines Ii, i.e., so that the points ai dd not belong to C. We set E = Cp2 \ (CUloU". Ul s ) and B = m \ {ao, ... ,as }. The projection from the point a onto the line m induces a map p: E --+ B. Lemma. The map p: E --+ B is a locally trivial fibration with fiber F omorphic to C with n punctures.

home~

Proof. We show that for any point x E B, there exists an open set U 3 x such that the map p is a trivial fibration over U. We take any point x' =F x on

6. Fundamental Groups

284

ao

Xo I

Figure 25. The projection of the curve

aa m

d

onto the line m

m and remove the line ax' from Cp2. On the remaining set, which is homeomorphic to C 2 , we take the line m and one of the lines passing through a in CP2 for coordinate axes. In these coordinates, the projection from a onto m has the form (z, w) 1-+ (z,O). If the neighborhood U C B is sufficiently small, then the branches of the curve C over it are sufficiently well approximated by the lines w = (kiZ + A, where i == 1, .•. , n. The lines z = const are not included, because we have eliminated the tangents to C. The lines Wi = (kiZ"+ A have no intersection points over a sufficiently small domain U, because the curve C has no singular points over it. Hence the branches of the curve C over a small domain U C B are sufficiently well approximated by the lines w = Ci, where i = 1" .• , nf "sufficiently well" means, in particular, that the set p-I (U) is homeomorphic to U x (C \ {CI, ..• , en}), and the homeomorphism is compatible with p. 0 The space B = m \ {o.o, ... , as} is homotopy equivalent to the wedge of circles. The group 1f"l(B,xo) is generated by loops hl, •.. ,hs , each enclosing exactly one of the points ai, where i = 1, ... , s; there are no relations between these generators, but if we add a loop ho enclosing ao, then the relation huhl ... h5 = 1 will arise. 8

For generators of the group 1I"1(F,xo) we take loops 91. .•. ,sIn each of which encloseS exactly one of the punctures in C. Moreover, we need the composition of these loops in Cpl to be homotopic to a loop enclosing a (the fiber F is the complex projective line Cpl from which the point a and the n intersection points of this line with the curve C are removed). In the fundamental group of the space with a glued back in, the relation 91 ... 9n = 1 arises. To calculate the group 1f"l(E,xo), we use the exact sequence for the fibration under consideration:

1I"2(B,xo) ~ 1f"l(F,xo) ~ 1I"1(E,xo) ~ 1I"1(B,xo) ~ 'lTo(F,xo).

3. Fundamental Groups of Complements of Algebraic Curves

285

Since the fiber F is connected, we have 1I"0(F, xo) = 1. Moreover, the space B is homotopy equiValent to a wedge of circles, and the universal covering space of a wedge of circles is contractible. Therefore, 1I"2(B, xo) = 1. As a result, we obtain 1-+ 1I"1(F,xo) ~ 1I"1(E,xo) ~ 1I"1(B,xo)

-+

1.

The map of the group 11"1 (F, xo), which is freely generated by gI, •• . ,gn, to xo) is a monomorphism. Hence we can identify 11"1 (F, xo) with the subgroup G = i.1I"1(F,xo) C 1I"1(E,xo). The group 11"1 (B,xo) is freely generated by hI, ... , hs. In the case under consideration, the space B is contained in Ej therefore, 11"1 (E, xo) has elements represented by the same loops as the elements hl, .•. ,hs of the group 1I"1(B,xo)j we use the same notation for these elements of 11"1 (E, xo). The fact that the group 1I"1(B,xo) is free substantially facilitates the calculation of the group 11"1 (E, xo). Indeed, for a free group, any map of its free generators to a group can be (uniquely) extended to a group homomorphism. Hence there exists a unique homomorphism cp: 11"1 (B, xo) -+ 11"1 (E, xo) for which cp(hi ) = hi. The subgroup H = cp1l"1(B,xo) C 1I"1(E,xo) is isomorphic to 1I"1(B, xo), because P.cp = i
Therefore, the group 1I"1(E,xo) is completely determined by the groups G and H and the action of H on G defined by h(g) = hgh- 1 E G. Hence the group 1I"1(E, xo) is defined by the generators gl,"" gn. hI, •.. , hs and the relations hj9ihjl = 'l/Jij(gI,oJ<,gn)" where 'l/Jij(gI,"'pgn) is the expression of hjgihjl E G in terms of the generators g1, ••. , gn' We have calculated the group 11"1 (E, xo). The next step is the calculation of the group 11"1 (E', xo), where

E'

= Cp2 \

(C U lo) = E U (E' n (h U··· U Is» .

The set E' n (II U .• U ls) is a submanifold of codimension 2 in the manifold E'; therefore, the embedding E -+ E' induces an epimorphism 1I"1(E,xo) -+! 11"1 (E',xo). Indeed, any loop in E' is homotopic to a loop disjoint from h, ~ .. , Is. Hence the group 11"1 (E', xo) has the same generators g1, 1 •• , gn" hI, ... , hs ; to the relations inherited from 11"1 (E, xo) some new relations may add. For example, the loops ~ are contractible in E' (see Figure 26), which gives the new relations ~ = 1 for i = 1, •.. , s. We show that no other new relations arise. Consider any homotopy between a loop 'Y and the constant loop Xo in E'. We can assume that the loop 'Y is smooth and avoids the lines 11, • •. • ls' First, we replace the homotopy under consideration by a smooth 0

6. Fundamental Groups

286

Figure 26. ,A contra.ctihle loop

Figure 27. A standard loop

homotopy, and then we slightly change the smooth homotopy so as to obtain a homotopy \If such that the points al, •.. ,a8 are not critical for the map p\Jt, where p is the projection from the point a onto the line m. The map WI P. I.o.t E' has the property that the preimage of II U .-•. U I8 consists of finitely many interior points of the square [2 (the boundary points of [2 are mapped to Xo or to some other points of the loop 'Y; none of their images belongs to Il,l'" ,la)' The main difficulty is that each loop in U \ li' where U is a sufficiently small neighborhood of E Ii \ C, must be replaced by a standard loop lying on a circle centered at ai on the complex line m \ {ao} (see Figure 27) by means of a homotopy in the space E; meanwhile, we consider homotopies of loops in the class of all maps 8 1 ..... E, Le., we allow homotopies to move the base point. To construct such a homotopy, take a path k(s) on the complex line li \ {a} that joins the points a~ and Ui and avoids the point li n C (see Figure 28). The required homotopy is constructed as follows. In the space C 2 = Cp2 \ lo, the projection from a onto the complex line m \ {ao} is written as (z,w) 1-+ (i,O) in some coordinates. Moreover, ttt = tZo,O) and a~ = (zo, wo) in these coordinates. The loop in U\li defined by "Y'(t) = (%(t), w(t) 'iii homotopic to the loop Y'(t) = (z(t),wo). Suppose that 1" is sufficiently small and the minimum distance p from the path k (s) to the points Ii n C

a,

3. Fundamental Groups of Complements of Algebraic Curves

287

Figure 28. A path from lit to a~

Figure 29. Homotopic loops

is sufficiently large, namely, p> max Iz(t) - zol. Then the fOl"mula 'h(t) = (z(t), k(8)) defines a homotopy in E between the loop ')'0 = '"(' and a loop 'Yt which is contained entirely in the complex line m \ {ao}. Moreover, the loop 1'1 lies in a small neighborhood of the point ~ and does not pass through this point. Such a loop is homotopic to a loop lying on a circle centered at ~. We can pass from homotopy in the class of all loops to homotopy in the class of loops with fixed base point Xo as follows. Suppose that 'Ys(t) is a homotopy between loops 'Yo(t) and 'Yl(t), /-L(T) is a path from Xo to 1'0(0), and V(8) = 'Ys(O) is a path from 1'0(0) to 1'1(0). Then the loops /-L'YO/-L .... l and /-LV"flV-1/-L-l are homotopio (see Figure 29), We return to the homotopy W constructed on p. 286. In the space E, any loop l' contractible in E' is homotopic to the loop /-Lk'Yk/-Lj/ , where 'Yk is a loop in a small neighborhood of a point of the line liCk); the homotopy is constructed as shown in Figure 30. In the language of group generators and relations, this means that any word representing the identity element of the group 1fl (E', xo) can be reduced to the form /-Lk'Yk/-L;; 1 by using only the r~ations between the elements of 71"1 (E, xo); here we assume that both groups are defined by the same generators (specified before). We have shown that the loop /-Lk'Yk/-L;;1 is homotopic in E to the loop /-Lkhr(k) (/-Lkt\ where hiCk) is a loop from the set of generators {hI. ..• , h8} and,. is an integer. This means that the word /-Lk/k/-L;;1 can be reduced to the form /-Lkhr(1c)(/-Lk)-1 by using only the relations between the generators of 1fl (E, xo). Finally, the relation hiCk) = 1 implies /-Lkhr(k)(/-Lk)-1 = 1. This means that if a word l'

n

n

288

6. Fundamental Groups

Figure 30. The construction of a homotopy

represents the identity element of 7I"1(E',XO), then the equality 'Y = 1 follows from the relations between the elements of 1I"1(E,xo) and the relations hI = 1, ... , hs = 1. The last step is the calculation of the group 11"1 (Cp2\C), where CP2\C = E' U (lo \ C). As mentioned earlier (see p. 284), after adding the point a E 10 \ C, the relation gl ... gn = 1 arises. The same argument as that used at the preceding step shows that there are no other relations. We now state the final result. Recall that the words 'l/Jij(gl, ... , gn) were defined on p. 285. Theorem 6.6 (van Kampen [66]). Suppose that C is a curue 0/ degree n in Cp2, a E Cp2 \ C, and lo, h, ... , ls are straight lines each 0/ which passes through a and is either tangent to the curue C at one point or passes through one singular point o/C. Then the group 11"1 (Cp2 \ C,xo) is defined by generators gl, ... ,gn and the ns + 1 relationS gl··· gn = 1, gi = 'l/Jij(g1, ... , gn), where i = 1, ..• , nand j = 1, ... , s. 3.3. Applications of the van Kampen Theorem. The van Kampen theorem gives an algorithm for calculating the groups 'Il"I(CP2 \ C,xo). Its most difficult part is constructing the expressions 'l/Jij(g1, ..• , gn). For this reason, we start with a more detailed discussion of their geometric meaning.

Recall the definition of the elements 'l/Jij (gl, •.• ,gn) E 11"1 (F, xo), where F is the fiber over the point Xo. The fiber F is the plane C with n punctures. Let gl, ..• , gn be loops, each going around one of these points (all in the same direction). The base B is C with s punctures; the loops hI"'" hs are defined similarly to g1, ..• ,gn' The loop hj9ihjl is homotopic to a loop lying in the fiber F. The element h j gi hjl can be represented as a word in the alphabet gl, ... ,gn; this word is 'I/J~j(g}, ... ,gn)' The homotopy of h j gi hjl is described as follows (see Figure 31). We take the loop gi and translate it along hj (so that at each point of hj, a loop in the fiber over this point is obtained). After going around the point aj, we return to the fiber over Xo and obtain a loop in this fiber. This new loop is 'l/Jij(g1, ..• , gn).

3. Fundamental Groups of Complements of Algebraic Curves

289

Xo

Figure 31. The homotopy of the loop h,9.h;1

a

Figure 32. Tangents to the conic

Example. Let C2 be a nondegenerate conic in Cp2 (e.g., given by the equation z~ + z~ + z~ = 0). Then 1l"1(CP2 \ C2) = Z2. Proof. Through each point a ¢ C2 pass exactly two tangents to the conic C2 (see Figure 32). The group 1l"1(CP2 \ C2) is defined by the generators 91 and 92 and the relations 9192 = 1 and 9i = .,pi! (91, 92), where i = 1,2. Instead of going around a tangent to the conic, we can consider a simpler situation, namely, going around the complex line z = 0 tangent to the curve z = w 2 in C2. If z = ei
Example. Let Cn be the curve in CP2 determined by the equation z~ + z!1 = o. Then 1l"1(CP2 \ Cn) = Zn.

z? +

Proof. In CP2, the projection onto the real plane Z3 = 0 from the point a = (0 : 0 : 1) is given by (Zl : Z2 : Z3) ....-.+ (Zl : Z2 : 0). A point (Zl : Z2 : 0) has n preimages on the curve Cn if and only if z? + z~ =f: o. The points for which

290

6. Fundamental Groups

FigllTe 33. Going around the

~rigin

Z1 : Z2 = -Ck, where c~ = 1, correspond to tangent lines; moreover, these tangents must have multiplicity n (at each point of tangencY,...R branches must merge). Going around a tangent of multiplicity n is essentially the same as going around the origin In the plane w = 0 for the algebraic function w(z) defined by z = w"l it results in the rotation of the real plane z = 0 through an angle of 27r/n and a cyclic permutation of the branches. Thus, we obtain the relations 91 = 92; 92 = 93, ..• ,9n-1 = 9n, 9n = 91. We also have the relation 9192 ... 9ft = 1. As a result, we obtain a group with defining generator 9 and relation 9ft = 1. 0

Problem 96* ([94]). Let p and q (p, q ~~) be coprime integers, and let Cp,q be the curve in CP2 given by the equation

(z{' + ~)q + (z~ + ~~)P :;: O. Prove that the group 11"1 (Cp2 \ Cp,q) is defined by two generators a and b and the relations aP = 1 and bq = 1. I

Hints and Solutions

1. Let {l-i} be a countable base of X, and let I be the set of those indic~ i for which l-i is contained in some Ua, To every i E I we assign one set Va(i) containing l-i. Let us show that {Ua(i)} is a covel," of X. Each point x E ~ is contained in some set Ua . The set Ua can be represented as the union of l-i; therefore, x E l-io C Ua for some io. Clearly, io E I and x E Ua(io)' 2. The set sn+m- 1 \ sn-1 consists of the points (Xl, •. qXn+m) E n lR + m for which x~ + .. , + x~ < 1 and x~ + .. 1 + X~+m = 1. 1'0 each point (X!,"" Xn+m ) E sn+m-1 \ sn-l we assign (yt, .. ,,, Yn+m),where Yl = Xl,···,Yn = Xn , Yn+l = xn+l/a"'.,Yn+m = xn+m/a, and a ~ YlClearly,

xi .- ... - xa. 2

Yn +!

+ ... + Yn2+m

=

2 + x n+l'" 1 2 -Xl -

2 + xn+m 2 == 1; .• '1-X o

thus, we have defined a homeomorphism sn+m-l \ sn- l ---+ D n x sm- l " where Dn is the open unit disk; it is homeomorphic to ]Rn. 3. Suppose that the set A U B can be represented as the union of nonempty disjoint sets Xl and X2 each of which is both open and closed. Take a point x E An B belonging to Xl. The sets Al = A n Xl and A2 = An X2 are open and closed in A, and Al =J 0, because x E AI. Therefore, Al = A and A2 = 0, Similarly, B n Xl = Band B n X 2 = 0. Thus, Xl = AUB and X2 = 0. 4. Consider the function F: K ---+ lR defined by F(x) = p(x,jex)). Since K is compact, this function attains its minimum at some point Xo. If f(xo) = 0, then Xo is fixed. Suppose that F(xo) = d > O. Then F(J(xo)) = p(J(xo), /(/(xo») < p(xo,/(xo)) = d, which contradicts the assumption.

-

291

292

Hints and Solutions

Figure H.t. Embeddings of the graphs Ka,a and K5

5. Yes, this is possible. The embeddings are shown in Figure H.1. 6. Consider a straight line parallel to none of the lines joining a vertex of one cycle to a vertex of the other cycle. A translation of one of the cycles along this line does not change the intersection number. Indeed, each vertex of a cycle is incident to exactly two edges; therefore, as the vertex passes though an edge, the number of intersection points changes by ±2, and its remainder after division by 2 does not change. When the cycle under consideration is moved far enough to become disjoint from the other cycle, the intersection number vanishes. 7. (a) First, suppose that the vertices of the graph are fixed and only the arrangement of the edges may change. Two positions of the same edge form a cycle. According to Problem 6, the intersection number of this cycle and the cycle constituted by the nonadjacent edges is O. Therefore, the selfintersection number does not depend on the position of one edge, and hence on the positions of all other edges, The self-intersection number does not depend on the arrangement of the vertices either, because, for any n points, there exists a homeomorphism of the plane which takes them to any other n points. (b) The graphs K3,3 and K5 have the property specified in (a); therefore, for each of these graphs, the self-intersection number does not depend on how the graph is drawn in the plane. Hence we can calculate the selfintersection number for any plane image of the graph. It equals 1 for both graphs. In particular, there is always a self-intersection point. 8. (a) It is sufficient to prove that any cycle C in the graph G has even length. The cycle C encloses several faces. Removing an interior face that shares an edge with C, we replace a part of C consisting of nl edges by a part consisting of n2 edges, where nl + n2 is the number of edges in the removed face, which is even. Thus, such a transformation does not affect ·tha parity of the cycle length. Several such transformations give a cycle bounding one face. Its length is even.

293

Hints and Solutions "

(b) Let us choose one point in each domain and join the points belonging to domains adjacent along arcs by edges. As a result, we obtain 8. graph G in which every face contains an even number of edges. (The faces correspond to the self-intersections points of the curve '1; to a point of multiplicity k correspond faces with 2k edges.) 9. Any face of the graph Ks must contain at least three edges. Thus, if Ks were planar, then the inequality e ~ 3v - 6 would hold; on the other hando ~ ~ 10 and v = 5 for Ks. In the graph K3,3~ any face must contain at least four edges, which implies e ::::; 2v - 4 for a planar graph, while for K3,8, we have e = 9 and v = 6. 10. The required homotopy is defined by

ht(A B) = (A ,

0

0) (c~t 1

sint) -smt cost

(1 B0) (cost 0

sint

-sint)

cost "

where 0 ~ f ~ 7r/2. 11~ Let X be a path-connected space. Suppose that X can be represented as the union of two nonempty disjoint open sets A and B I Take points a E A and bE B and consider a path f; [0,1] r--+ X for which j(O) = a and f(1) = b. The interval [0,1] is the union of the nonempty disjoint open sets Ao = f'"-l(A) and Bo = f-I(B). We know that 1 E;' Bo. Let to be the least upper bound for the numbers from Ao. If to E A o, then Ao cannot be open, and if to E' BOj then Bo cannot be open. 12. Each path-connected component of X )s open~ Indeed, every point x E X is contained in an open path-connected set U. Clearly, U is contained in th~ path-connected component of x. Thus, any path-connected component is a union of open sets. The complement of a path-connected component is the union of the other path-connected components, which are open. Therefore, every pathI connected component IS both open and closed. 13. For each point x E U, we can choose an open ball centered at x and contained in U. Each ball is path-connected. By Problem 12, U is path-connected. 14. Each of the sets Xl and 1S path-connected and, therefore, connected. Let A be a elopen (i.e., both open and closed) set in X. Then Al =--XI n A is a clopen subset of Xl. Hence Al = Xl or 0. Similarly, AnX2 = X 2 or 0. Thus, only Xl and. X2 can be nontrivial clop en subsets of X. But the set X2 is not closed, because its closure contains Xl- Therefore, the space X is connected. Let us prove that X is not path-connected. Consider any continuous path "'(: [0,1] -+ X for which 'Y(O) E Xl· Let to be the least upper bound for all t such that 'Y(t) E Xl. By continuity, 'Y(to) E Xl, i.e., "'((to) = (0, yo).

X4

294

Hints and Solutions

Choose ~ > 0 such that if It-tol < e~ then 'Y(t) = (x, Y), where Iy-yol t'C 1/2. Consider the projection on the Ox-axis of the intersection of X with the set defined by Iy - yol < 1/2. Between the origin and any other point pf the projection, there is a point not belonging to the projection. Therefore" if It - tol < e, then 'Y(t) E Xl, and the path "Y(t) is entirely contained in Xl> 15. It is sufficient to prove that in the spaces under consideration(each matrix can be joined by Q. path to the identity matrix. In. Any nonsingular matrix A can be represented as A t::::I SU} where S is a symmetric positive definite matrix and U is an orthogonal matrix. Let V be an orthogonal matrix such that S = VDV-l."where D is a diagonal matrix with positive diagonal elements! Then A.= VDW, where W CI vr l U is an orthogonal matrix. The path V[(l ..... t)D+tI}W joins A with the orthogonal matrix VW in GL(n). Moreover; if detA > 0, then det~VW)'''-O, Now, let us prove that any matrix U E SO(n) can be joined with the identity matrix I by a path in SO(n). In some orthonormal basis, the matrix U is the direct sum of the identity matrix and matrices 01 the form ~ The required path can be constructed by considering the family of matrices (=:~ -~~), where t E [0,1]. ;For the unitary matrices, we can use the fact that any unitary matrix is diagonal in some orthonormal basis and the diagonal elements have the form ei
(=-::an:).

16. (a) Suppose that the covering p: Kn - G is of degree ~rn.. ';['hen the preimage of any vertex v of the graph d consists of 2m yertices Vb H '<' V2m,' and these vertices generate a subgraph K 2m with m(2m - ~) edges ~n Kn. Each of these edges is projected to a loop based at v. Suppose that the number of the loops thus obtained equals l. The preimage of every loop consists of 2m edges; therefore, 2ml = m(2m - 1), i.e., I = (2m - 1)/2, which is impossible. (b) The covering p: K 2m+l -+ G, ~here G is a graph ~onsisting of one vertex and m loops, is of odd degree. 17. Suppose that a map f is null-homotopic. Let Xo E S1 bl( the base point and Yo = f(xo). Consider the loop wet) = Xo exp(211"it) at Xo. The map f takes this loop to a contractible loop wet). Take a point ZO E ~ for which exp(211"iZO) = Yo and consider the lifting nCt) of w(t) starting at ZO0 The loop wet) is contractible; therefore, the path net) is closed. This means that the formula f2(w(t» = net) defines a map f2: SI -+lR. It remains to set bet) = expC211"it).

Hints and Solutions

x::><::::x::::x - C>O Figure H.2. A covering with automorphism group Z

Figure H.3. A covering with automorphism group Z

Figure H.4. A covering with automorphism group

Figure H.5. A covering with automorphism group Z2

ez

z..

e Za

18. The required coverings are shown in Figures H.2 H.5; each of the figures, except the first, shows only the covering space. 19. Consider the wedge of rkG circles, i.e., a one-dimensional complex with one vertex and rk G edges. Let us construct a covering of this complex whose automorphism group H is a subgroup of G. The group H is isomor· phic to the fundamental group of the covering space X. The space X is homeomorphic to a wedge of circles; therefore, the group H is free. The covering under consideration is k-fold; hence X has k vertices and k(rk G) edges. Any maximal tree in X contains k -- 1 edges; therefore, contracting a maximal tree to a point, we obtain a one-dimensional complex with one vertex and k(rkG) - (k -1) :i:::: (rkG -1)k+ 1 edges. This number is equal to the rank of H.

296

Hints and Solutions

Figure H.6. The covering space

20. To construct a monomorphism Fn -+ F2, it suffices to construct a covering for which the base is S~ V Sl and the covering space is homotopy equivalent to Sf V .•. V S~. This covering can be constructed,--e.g., as follows. Let us uniformly arrange circles SJ, ... , S~ on Sf (see Figure H.6); the space thus obtained is homotopy equivalent to Sf V··· V S~. Consider a map which is an (n - I)-fold covering of S~ on sl and identically maps each of the remaining circles SJ, ... ,S~ onto In the algebraic language, the map Fn -4 F2 is described as follows. Let Xl, ••• , Xn be generators of the group Fn , and let a and b be generators of F2. Then Xl 1-+ an-l , X2 1-+ b, X3 1-+ a 00- 1 , X4 1-+ a2L_~2 va- ,.;1., Xn 1-+ an-2ban-2 . For the group Foo, the map is Xk 1-+ akoo- k •

st.

21. No, this is not true. For example, if A

C

= {O}, B =

[0,1], and

= {I}, then dCA, B) = deB, C) = 0 and dCA, C) = 1.

22. Suppose that dH(A, B} = f3 and dH(B, C) = "y. Take e > O. For any point a E A, we can choose a point b E B such that lIa-bll ~ f3+e. For bE B, we can choose a point e E C such that lib - ell ~ 'Y + e. Therefore, lIa - ell ~ f3 + 'Y + 2c:. Similarly, for a point e E C, we can choose a point a E A such that lIa - ell ~ f3 + 'Y + 2c:. 23. Let e > O. For each point X E I(A), consider the set Uz = An 1- 1(D;,e/2)' where D;,e/2 is the open disk of radius £/2 centered at x. Such sets form an open cover of the topological space A. Let b > 0 be the Lebesgue number of this cover. If at, a'l E A and lal - a21 < b, then aI, a2 E Uz for some point x, and the points f(al) and f(a'l) belong to the open disk of radius e/2 centered at x, i.e., If(al) - f(a2)1 -< e. 24. Let r: A -+ X be a retraction. If f = A --4 Y is a continuous map, then fr is an extension of I to X. On the other hand, if each continuous map I: A -+ Y can be extended to X, then the identity map idA: A -+ A can be extended to r: A -+ X. This r is the required retraction. 25. Let f. A -+ A be a continuous map. According to Problem 24, this map can be extended to a map F: X -+ A eX. By assumption, F has a fixed point Xo· We have Xo = F(xo) E A and f(xo) = F(xo) -. Xo.

Hints and Solutions

297

26. Let {Ua } be an open cover of C. Choose open sets U~ in K for which Ua :;:: U~ n C. The sets U~, together with the open set U = K \ C, form a cover of K. This cover has a finite sub cover Ul 7"'" U~, U. Clearly, the sets Ul, ."' , Un cover C. 21. (a) Let us introduce an equivalence relation on X~ We declare that Xl X2 if the images of Xl and X2 under any continuous map from X to a Hausdorff space coincide. Then XH = XI""', and (7 is the natural projection of X onto XI-. (b) Consider the surjective map Matn(C) - C n that assigns the coefficients of the polynomial det(A + AI), where I is the identity matrix, to each matrix A. This map is constant on the orbits; therefore, it induces a map c: X/G - cn. The map c corresponds to a surjective map dl: (X/G)H IV

cn

J

1 If A and B are diagonal matrices, then dl (A) = dl (B) if and only if A and B belong to the same orbit. ' Any orbit contains an upper triangular matrix

A

=

(~.l

. :::... ~.) .

o ...

An

Consider the diagonal matrix .6. m = diag(!, m, m 2 ,., • ,mn -

l ).

We have

lim .6.m A.6.;1 = diag(Alt ..• , An).

m-+oo

Since the space (X/G)H is Hausdorff, it follows that the matrices A and diag(A!, ... , An) represent the same point in this space. (c) The map f induces a map f; X/G - C. To the map f corresponds a map F: (X/G)H = c n - C. 28. Let vo, VI, ••• , Vn be the vertex set (ordered somehow) of the simplex .6.n. Corresponding to it is the simplex in the barycentric subdivision which is determined by the inequalities Xo ~ Xl ?: ... ~ Xn in barycentric coordinates. The vertices of this simplex are vo, the barycenter of [vo, VI], the barycenter of [VO,VI,V2], ' " ? ' 29. Obviously, any complete 8ubcomplex does have the required property. --Suppose that each simplex of K whose boundary is contained in L is itself contained in L. Take a simplex from K such that all of its vertices belong to L. Then all the edges of this simplex belong to L, and therefore all of its 2-faces belong to L, etc. 30. Each simplex .6.n of K' is uniquely determined by a set of simplices 0'0 ~ 0'1 ~ ' ! ~ ~ Un from the complex K (the vertices of.6." are the barycenters of these simplices). Suppose that all the vertices of .6.n belong to L'l

298

Hints and Solutions

Then, in particular, the barycenter of Un belongs to L'. This means that the simplex (Tn itself belongs to L. The simplex An is one of the simplices in the barycentric subdivision of (Tn; therefore, it belongs to L ' . 31. (a) [140] We assume that I = [-1,1]. Then consists of the points (x}, .• . ,xn ) such that IXil ~ 1 for all i and Xi = ±I for some i. We set

ar

(r)~-l = {(Xh

o .,

,x

n ) E r ; Xi

= +1 for some i}.

It is easy to verify that a«F)~-l) consists of the points (x.l.,2 .•• ,xn ) E F such that Xi = +1 and Xj = -1. Let us define {r)~""2 as the union of the (n - 2)-faces of the cube determined by the relations Xi = +1 and Xj = -1 for i < j. Similarly, we define (r)~-j as the union of the (n - j)-faces of the cube determined by x a1 = +1, x a2 = -1, x~ = +1, ... , Xaj-=f-I)i+l for some 1 ~ al < a2 < a3 < 0" < aj ~ n. It is easy to show that a«F)~-j) = (F)~"'j-l U (_(F)~-j"'l). Let Sk(io, i}, ... , ik) be the number of k-simplices with labels io, ... , ik in (In)i. We derive a relation between Sl(i,j) and So(i) by calculating in two different ways the number N(i) of pairs, each consisting of a I-simplex belonging to (In)~ and its vertex with label i ~ 1. First, consider the sum over the I-simplices from (r)~. We obtain

(1)

N(i)= 2Sl(i,i)+Sl(i,-i)+

L

(Sl(i,j)+Sl(i, .... j»)j

j#iJ~\

we have Sl(i, -i) = 0 by assumption. Now, consider the sum over the vertices of the triangulation which belong to (In)~. We obtain

(2)

N(i)

= 2K+ So (i) + SoC -il,

where K is the number of interior vertices in (F)~ labeled by i. Indeed, the boundary of (F)~ consists of (F)~ and -(F)~, and if a vertex v E (F)~ has label k, then the vertex -v E -(F)t has label -k. Comparing (1) with (2) and summing over i between 1 and n, we obtain

L l$i<j$n

n

(Slei, -j) + SI(-i,j»

=L (So (i) + So(-i»

(mod 2)~

i=l

the terms of the form Sl (i, j) are canceled because each of them occurs twice, once in the expression for i and once in the expression for j. Clearly, E(So(i)+So(-i» = 1 because (F)~ consists of one point (+1,-1.+1""..). The further calculations are modulo 2. Let us count the pairs in (P);', e&ch consisting of a 2-simplex and its I-face with labels i and -j, where 1 ~ i < j. First, we count them by taking a SUll). over the 2-simplices.

Hints and Solutions

299

We obtain (modulo 2)

~

(S2(i , -j, k)

+ S2(i(-i, -k)).

k#i,j,k~l

Summation over the 1-simplices with labels i and r j gives

Sl(i,-i) +Sl(~i,j); the interior I-simplices cancel each other modulo 2, and only the boundary simplices remain. tet us equate the obtained expressions to each other, take the sum over all pairs i < ;, and reduce the result modulo 2. We obtain

:E

(S2(io,-iI, i 2)+S2(-io,iI,-i 2))

=

l~io
:E (Sl(i,-;)+Sl(-i,j)) :E (So (i) + SoC -i)) = i.

l~i<;

=

l~i~n

Continuing the calculations in a similar way, we come to (3)

Sn-I{I, ~2, .•. , ±n) + Sn-l( ~1, 2, .•. ~ =Fn)

=1

(mod 2).

Every time we use the fact that if two neighboring numbers iQ and ia+ I are ofthe same sign, then the expression Sk(io, ••. , i a , ia+!, •.. , ik) occurs twice. To obtain a contradiction, we count the pairs consisting of n-simplices and their {n-l)-faces with labels 1, -2,3, ... , ±n in the entire cube In. The vertex opposite to such a face cannot be labeled by -1,2, -3, ... , =Fn because if it were, then we would have an edge with labels i and -i. Thus, precisely one of the labels 1, -2, 3, ... , ±n occurs twice. Therefore, the required number of pairs is even (we count the pairs by summing over the n-simplices). On the other hand, counting the same pairs by Bumming over the (n - 1)simplices with labels 1, -2, 3, ;. ±n, we obtain the expression on the lefthand side of (3), which is odd. (b) Suppose that there exists a continuous map I: 1" - 8In that takes antipodal points of 81" to antipodal points. Consider a sufficiently fine triangulation In such that it is symmetric on 8In and if VI and V2 are adjacent vertices, then II/(VI) - 1(V2) II < 2; in particular, the images of adjacent vertices cannot belong to opposite faces of the cube. We label each vertex V E In of the triangulation by the number of the face of 1" that contains I(v); it is assumed that opposite faces have numbers i and -i. According to Tucker's lemma, some adjacent vertices of the triangulation are labeled by i and -i. This contradicts the definition of this triangulation. 32. The Borsuk-Ulam theorem implies n ~ m. Take an arbitrary (m + 1- n)-dimensionallinear subspace II of lRm+1 ::J sm. Let us show that the set II n !p(S") contains at least two points. Let II...l be the orthogonal complement of II, and let p: Jim+! - Ill. be the orthogonal projection.

i,

Hints and Solutions

300

Then the map pocp: 8 n - t IT.!. ~]Rn is odd, and hence, by the Borsuk Ulam theorem, there exists a point x E sn for which p(cp(x)) = 0, i.e., cp(x) E IT. We have cp( -x) = -cp(x) E IT and cp(x) :F 0 because cp(x) E sm. Almost all (m + 1 - n )-dimensionallinear subspaces of]Rm intersect any fixed (n+ 1)-dimensionallinear subspace in a straight line; i.e., they intersect the sphere sn standardly embedded in ]Rn+l C ]Rm in exactly two points. Thus, almost all (m + 1 - n )-dimensional subspaces have at least as many intersection points with cp(sn) as with sn. Therefore, the n-dimensional volume of cp(sn) is not less than that or sr'. To show th~ we introduce an invariant measure I' on the set G (m, m + 1 - n) of all (m + 1 - n)dimensional linear subspaces of ]Rm and, for each X c 8 m, consider the integral fG(m,m+l-n) IX n ITI dl" If the set X has n-volume, then this integral is finite and proportional to the n-volume of X. 33. (a) The direct application of the Borsuk Ulam theorem does not give the desired result because opposite rays emanating from an interior point of the polyhedron P may intersect oP at points that belong to intersecting faces. We must consider the symmetrized polyhedron P, i.e., the set Q = {x = z", w :- Z, 111 E

Pl.

Clearly, Q is a convex polyhedron symmetric with respect to the origin. We define a map h: Q -+ P as follows. Take.z; = (Zl, .•. , Zn+l) and ~t = (z~, ••• , ~+l). We say that ~ > z' if Zl = zip" n Zk-l :;::: z~~J: and Zk > z1 {possiblYl k = 1), We define hex) = max{z:

X

= Z - W, where Z,W E

Pl.

Geometrically, the map h can be described as follows. If x = Z 4. w, then Z = x + w E ~ + Pi' thus, hex) is the (unique) maximal point of the set P n (x + P). To find this maximal point, we first determine all points for which the coordinate Zl is maximal, then choose the points with maximal coordinate Z2 among them, etc. It is easy to show that x = hex) - h( -,x). Indeed, if x = hex) I Wo, then Wo = hex) - x E P n (x + P) ~x = (-x + P) np. Moreover, since hex) is the maximal point of P n (x + P), it follows that Wo is the maximal point of (-x + P) n P, Now let us prove that the map h is continuous. Suppose that Xn E Q and liilln--+oo Xn = x E Q. Let us represent Xn as Xn = Zn - W n , where Zn = h(xn). Take an arbitrary convergent subsequence Zn,. The subsequence W n • converges also. We set z = lilIli--+oo Zn. and w = liIIli--+oo w"'''' Then x = ~-w, whence z ~ hex). Suppose that Z < hex). Consider the points z' ;:::::: Zn. + c:(h(x) -z) and w' = w ni +c(h(-x)-w), where ~ > o. Since hex) -he -x) = x = z-w, it follows that z' -w' = Zn, -Wn. = x",. Clearly, Z > :Zni-J because

Hints and Solutions

301

hex) - z > 0 and € > O. Let us show that the numbers e > 0 and ni can be chosen so that z', w' E P. For z E PI we can find 5 > 0 such that if II z - v II ~ 5 and v is a point on the ray from z to t E P, then v E P. Let C6 be the set of all such points v. Take c > 0 such that cllh(x) - zll ~ 5/2 and choose 'ni for which 1IZn. ..... zll ~ 6/2. The points Zn. and z + c(h(x) - z) belong to C 6/ 2 • Since the set C6/2 is convex, it follows that the midpoint of the segment joining these points belongs to C6/2 also, whence z, = z~+c(h(x)~z) E C6 C P. Similarly, w' E P. But if Z't w' E P and z' -w' = x np then z, ~ h(xn.) = Zn., which contradicts the inequality z' > z~. This contradiction shows that z = hex). Thus, any convergent subsequence of the sequence {Zn} converges to hex). The set P, which contains the points Zn, is compact; therefore, only finitely many points Zn can be outside an arbitrarily small neighborhood of hex). Hence liffin-+co Zn = hex), i.e., the map h is continuous. Suppose that a 1= 0 is an arbitrary vector and maxzeQ(a,x) = (a,xo). The equality Xo = h(xo) - h( -xo) implies (a,h(xo»

+ (. . . a,h(-xo» = (a,xo) = max(a, x) xeQ = max (a, z - w) = max(a, z) - max( -a, w). ~weP zeP weP

Therefore, (a,h(xo)) = maxzep(a,z) and (-a,h(-xo)) = maxwep(-a,w). This means that the points h(xo) and h( -xo) belong to two different supporting hyperplanes of the polyhedron P. In particular, the pointg h(xo) and h( -xo) belong to disjoint faces of P. Now we can apply the Borsuk Ulam theorem to the map g(x) = !(h(x». It says that there exists a point Xo E aQ for which g(xo) = g( -xo). The point Xo belongs to some supporting plane of the polyhedron Q; hence there exists a vector a 1= 0 for which maxxeQ(a, x) = (a, xo). The points z = h(xo) and w = h( -xo) belong to disjoint faces of the polyhedron P, and fez)

=

f(h(xo»

= g(xo) = g(-xo) = !(h(-xo» = few),

as required. (b) Let B and C be disjoint faces of the simplex ~n+1 such that feB) n f(C) i= 0. The face ~f contains all but one vertex of the simplex ~n+1. This vertex cannot pelong to both laces B and Cj therefore, B C ~~ or C C ~f (or B, C c ~f), which means that f(~f) ~ feB) n f(C). Remark 1. For linear maps f: a~n+l __ ]Rn, assertion (b) is a special case of the Helly theorem. But this special case is the induction step in the inductive proof of the theorem. Thus, it is essentially equivalent to the Helly theorem. Remark 2. Another solution of Problem 2.9 and its generalization are given in [75].

302

Hints and Solutions

34. Consider the characteristic map f: D 2n -+ Cpn defined on p. 121. Its restriction to int D 2n is a homeomorphism of int D2n onto Cpn \ Cpn-l. Clearly, the specified identification of points of the sphere s2n-l turns the sphere into Cpn-l. 35. We can assume that Soo consists ofthe points x = (Xl, X2, . .. ) E ]Roo such that they have only finitely many nonzero coordinates and Ext = 1. Consider the maps rp(x) = (0, XI, X2,' .. ) and ht(x) = (1 - t)x + trp(x). It is easy to show that ht(x} #: for X #: O. Therefore, the formula X 1-+ ht(x)/IIht(x)II defines a homotopy between the identity map idsoo and rp t Soo. Let 9t(X) = (1-t)rp(x) + (t,O,O, ... ). Again, 9t(X) #: for X #: 0. Therefore, the formula x 1-+ 9t (X)/II9t (x) II defines a homotopy between the map rp t soo and the constant map to the point (1,0,0, ... ). 36. Let K be a compact set. For every open cell int e~ which K intersects, choose a point x~ E K n int e~. We must prove that the set T == {x~l is finite. By condition (c), a closed cell may intersect only finitely many open cells; therefore, the intersection of any set T' c T with any closed cell is finite and, hence, closed. By condition (w) (see p. 119), any set T' C T is closed, which means that T is discrete. On the other hand, the set T is compact, as a closed subset of a compact space. It remains to J).ote that all discrete compact sets are finite. 37. The sphere sn can be represented as a CW-complex with one O-cell and one n-cell. Hence SP x can be represented as a CW-complex with cells ot dimensions 0, p, q, and p + q. The cells of dimensions 0, p, and q form a sub complex SP V Contracting this sub complex to a point, we obtain a OW-complex with cells of dimensions 0 and p + q, i.e., a (p + q)-sphere. 3S. ~t Itex) = (x, t, Yo), where Yo is a point of Y. Then 10 = 1 and h takes X to the point Yo. 39. (a) If n is even, then nP2 ~ n22T2#2P2, and if n is odd, then nP2 ~ n21T2#P2. Therefore, it suffices to consider the surfaces 2p2 and p2, for which the required curves are constructed in an obvious way. (b) We must prove that if cutting along a closed curve renders the surface nP2 orientable, then the boundary of the cut surface has two components for n even, and one component for n odd. Such a cutting does not change the Euler characteristic of the surface. If the boundary has two components, then we attach the handle SI X I ~ and if it has one component, then we attach the disk D2. In both cases, we obtain a closed orient able surface (with even Euler characteristic). In the former case, the Euler characteristier does not change, and in the latter, it increases by 1.

°

°

sq

sq.

.

Hints and Solutions

303

Figure H.T. The surfaces

Ml and Ml

Figure H.8. The universal covering of the plane with two punctures

Mr

40. Yes, Mi X I and X I can be homeomorphic. The surfaces Ml and Mi shown in Figure H.7 are not homeomorphic because the boundary of Ml has three connected components and that of Mi is connected, but the spaces Mi x I and Ml x I are homeomorphic because the "handle" can be dragged along the dashed line. 41. The surface nP2 can be represented as the sphere 8 2 in which n disks are replaced by n disjoint Mobius bands. Suppose that there are p disjoint Mobius bands on the surface nP2. Let us cut the surface along the boundaries of these Mobius bands and paste the holes with disks. As a result, we obtain a closed surface M2 with Euler characteristic x(nP2) + p = 2 - n + p. But X(M2) :5 2; therefore, p ~ n. 42. ( a) The universal covering or the plane with two punctures is shown in Figure H.8. Clearly, the universal covering space is homeomorphic to the plane. For the plane with an arbitrary (finite) number of punctures, the proof is similar. (b) Let fal .. .a..; C -+ C\ {al..~ . . ; an} be the universal covering. Consider the map (WI, ••• , wn t-+ (Zb.'" Zn), where

1

Zl

= WI,

Z2

= / zl W2,

This map is a covering

Z3

en -+ I:.

= Jztz':1. W 3,

Z4

= !ZlZ2Z3W4, .. . ~

304

Hints and Solutions

43. If a covering p can be represented in the form p = P2PI, then the sets h = Pl 1 (Yl), ..• ,In = Pl 1(Yn), where {YI, •.• ,Yn} = p;I(X), are as required. Indeed, if "( is a closed path in X, then its lifting to Y joins some points Yi and Yj. Therefore, the lifting of "( to X joins points of the sets Ii and I j •

Now assume that II = {tn, ... , tIm},"', In = {t n1,"" tnm} is a partition of p-1 (x) with the properties specified in the statement of the problem. Take a point Xl EX. Let "( be a path from X to XI, and let Jk be the set of the ends of those liftings of"Y which start in Ik. The enumeration-of the sets Jk depends on the path ,,(, but these sets themselves do not depend on "y. Indeed, suppose that one lifting of y starting in Ik ends in Jk and another lifting starting in Ik ends in J" where I =/: k. Then one lifting of the closed path y,,(-l starting in Ik ends in Ik and another lifting starting in Ik ends in lz, where l =/: k. This is impossible. The covering PI; X --+ Y maps each set Jk to one point. The construction of the covering P2 is obvious. 44. It is easy to calculate the Euler characteristic of the deleted products of the graphs K3,3 and Ks. All of their faces are quadrangular, and each edge belongs to exactly two faces. Therefore, 2e = 4/. The number v of vertices in the deleted product equals n 2 - n, where n iJ the number of vertices in the graph. Hence v = 30 for the deleted product of the graph K 3 ,3 and v = 20 for that of Ks. The number / of faces is equal to the number of ordered pairs of disjoint edges. Therefore, / ::z 36 for the deleted product of K3,3 and J = 30 for that of Ks. It remains to show that the deleted products of K3,3 and Ks are orientable. This can be verified directly, but such a verification is fairly tedious, because one surface consists of 36 quadrangular faces and the other COIlBists of 30 faces. It can be obviated as follows. The deleted product of a graph is naturally embedded in the deleted join of this graph. Indeed, each pair of nonadjacent edges in the deleted join corresponds to a tetrahedron, and each pair of nonadjacent edges in the deleted product corresponds to a parallelogram; this parallelogram can be regarded as a section of a tetrahedron (see Figure H.9). The deleted joins of the graphs K3,3 and Ks are homeomorphic to 8 3 • For the graph Ks, this assertion is a special case (for n = 1) ofTheorem 3.38 on p. 133. For the graph K3,3, it easily follows from the observatio~ that the deleted join of a join is a join of deleted joins (see the proof of Theorem 3.39 on p. 134). Indeed, the graph K3,3 is the join sko a 2 *sko ~2; therefore, J~(K3,3) = J?(sko a 2 *sko ~2) = .Ji(sko ~2)*Ji(sko ~2) ~ 8 1 *81 ~ 8 3 because J~(sko~2) ~ 8 1 .

Hints and Solutions

305

Figure H.9. The embedding of the deleted product in a deleted join

Thus, the deleted products of the graphs K 3 ,3 and K5 can be embedded in 8 3 • On the other hand, no closed nonorientable surface can be embedded in 8 3 (see the corollary of Theorem 5.21 on p. 218). 45. The map ~~ x ~~ -+ ~~ )( A~ determines an involution on the deleted product; this involution has no fixed points because the simplices ~~ and ~~ are disjoint. The Euler characteristic of any fixed-point-free involution is even.

tw~dimensional

surface that admits a

46. (a) The map idy is homotopic to the map Y -+ Yo. Consider the map 10 = idx on X and a homotopy between the maps idy and Y -+ Yo on Y. Extending this homotopy to a homotopy It of 10, we obtain a map h: X -+ X such that it is homotopic to idx and /I (Y/) = Yo. This map induces a map q: X/Y -+ X for which qp = /I, where p: X -+ X/Y is the canonical projection. Thus, qp = /I '" idx. It remains to show that pq'" idy. By construction, It(Y) c Y. Therefore, each map It determines a map 9t~X/Y -+ X/Y; we have 90 = idy and 91 = pq. (b) In the CW-complex X LJ CY, the sub complex CY is contractible; therefore, xu CY '" (X u CY)/CY = X/Yo 47. We prove the required assertion by induction. Suppose that X is an n-connected CW-complex with exactly one vertex and no k-cells, where 1 ~ k ~ n - 1 (for n = 0, no assumptions are made). We must "annihilate" the n-cells of X. Let cp: gn -t X be the characteristic map of some n-cell of the complex X. Since X is n-connected, it follows that the map cp can be extended to a map cp: Dn+1 -+ X. (For n = 0, we assume that cp: SO -+ X takes one point of SO to a fixed vertex Xo and the other point to a given vertex Xi of X; then cp is a path from Xo to xd We think of sn as the equator of the sphere Sft+1 = 8Dn +2 , and of D n +1 as a half of the sphere sn+1 (see Figure H.1O). Attaching Dn+1 to X via cp, we obtain a CW-complex X U Y, which is homotopy equivalent to X (the complex Y for n = 0 is hatched in Figure H.ll). The complex Y has the contractible sub complex Y' which corresponds to the "upper" hemisphere D n +1 (this

Hints and Solutions

306

Figure H.l0. The disk D"+1

Figure H.ll. The complex Y

sub complex is shown by bold lines in Figure H.~i). According to Problem 46, (X U Y)jY' .v X Y .v X. Clearly, the CW-complex (X U Y)jY' has no n-cells (and cells of smaller positive dimensions). For n = 0, this construction gives a CW-complex with one vertex. 48. According to the cellular approximation theorem, any map sn - X is homotopic to a map sn - xn eX; in the case under consideration, the n-skeleton xn consists of one point. 49. The space E(A 1\ B) is obtained from A * B by contracting two cones CA and CB with common generatrix {ao} x {bo} x I to a point (Figure H.12). Clearly, the space CAUCB is contractible (the cones can be contracted to points one after another). Therefore, according to Problem 46,

G

A * B '" A

* B/(CA U CB) ~ E(A 1\ B).

50. (a) According to Problem 47, we have X '" X') where X' is a CW-complex with one vertex Xo and no k-cells for 1 ~ k ~ n. Clearly, EX '" EX'. The complex EX' /Exo '" EX' has no cells of dimensions between 1 and n + 1\ By Problem 48, it is (n + I)-connected. (b) We can assume that X and Y have no cells of positive dimensions greater than or equal to 11. and respectively. Then the cells of positive dimensions in the CW-complex X x y that do not belong to X V Y are products of cells qP x qq, where p ~ n + 1 and q ~ m + l~ the quotient

m,

;:sur

Hints and Solutions

Figure H.12. The two cones

of X X Y modulo X V Y contains only the O-cell and such product cells. Therefore, X 1\ Y has no k-cells for 1 ~ k ~ n + m + L (c) According to Problem 49, we have X * Y f'V E(X J\Y). It remains to apply (a) and (b). 51. Assertion (a) is a. special case of (b). Thus, we prove (b). The join X * Y contains the distinguished subspaces X and Y j let us attach C X and CY to them. Contracting each of the cones ex and CY to a point, we obtain E(X X Y). These cones are contractible subspaces; therefore, X * Y U CX U CY '" E(X X Y). Let Xo E X and Yo E Y be base points. Consider the subspace Z of X * Y which consists of {xo} * Y and X {yo}. The space Z is contractible because the space obtained by contracting the segment [xo, yoJ to a point in Z is homotopy equivalent to the wedge of two cones. Clearly, contracting the subspace Z to a point in X*YUCXUCY, we obtain EXVEYVE(X I\Y). 52. Suppose that there exists a homotopy ht: ~ --+ B such that ho = p and hI takes sn to a point bo E B. The map ids.. : sn -+ sn covers p = hoi therefore, for the homotopy ht, we can consider the lifting let: sn -+ sn, where ko = idsn and kl takes sn to p-I(bo). Since B consists of more than one point, it follows that p-I(bo) #- 0, and the map kl : sn -; sn is homotopic to a constant. Thus, the map ids.. is homotopic to a constant, which is impossible. 53. By the cellular approximation theorem, any map Sir. -+ X, where k ~ n, is homotopic to a map Sir. -+ xn C X j moreover, we can assume that the homotopy leaves the point Xo E xn fixed. For k ~ n - 1v any map Sir. x I -+ X that is a homotopy between two maps /0,!L: Sir. -4 X is

*

sn \

Hints and Solutions

308

Figure H.13. The universal covering space

homotopic to a map 8 k x 1 -- xn eX, and it coincides with 10 and II, respectively, for t = 0 and t = 1. 54. For n ~ 2, the universal covering space of 8 n V 8 1 is the line 1R to which n-spheres are attached at integer points (see Figure H.13). This space is homotopy equivalent to the wedge of count ably many n-spheres. For n ~ 2, the nth homotopy groups of the base and bf the covering space are isomorphic. 55. The group 7I"n(sn V Sl, xo) is the free group oli oountably many generators l:tk (k E Z). The action of the generator of the fundamental group 11"1 (sn V 8 1 , xo) takes l:tk to l:tk±l; thus, this action is nontrivial. 56. Any quaternion q can be represented in the form q

= Xl + X2 i + X3J + x4 i j = Xl + X2 i + (X3 + X4)j =

Zl

+ Z2J.

The group 8 1 acts on 8 3 by the rule Z(Zl + z2j) = ZZ1 + zZ2j. This action coincides with the one used to construct the Hopf fibration. 57. The change of variables Xl = Ul + U2, X4 = Ul - U2, X2 = U3 + U4, determines a homeomorphism of the sphere under consideration onto the sphere u~ + u~ + u~ + ul = and transforms the equation XlX4 - X2X3 = 0 into u~ - u~ - u~ + ul = 0, i.e., u~ + u~ = u~ + ul 58. As explained on p. 120, Cp2 can be obtained by attaching D4 = {(ZbZ~) E C 2 : IZll2 + IZ212 ~ 1} to Cpl {(Zl : Z2 : 'z3) E Cp2 : Z3 = O} 3 via the map I: 8 -- Cpl defined as I(ZhZ2) (Zl ; Z2). But this map 1 coincides with p. 59. It follows from Problem 58 that any retraction r: Cp2 __ Cpl determines a map r: D4 -+ Cpl such that rex) = p(x) for x E 8 3 , Therefore, r is an extension of p to D4; thus, p is homotopic to a constant map. But the map p induces an isomorphism P.: 11"3(83 ) -- 11"3(82 ), where 11"3(83 ) = Z (see p. 235); hence it cannot be homotopic to a const81Jt map. • 60. Let us represent S3 as a union of two solid tori. The complement of the Hopf link in 8 3 is obtained by removing the central circles of these solid X3 = U3 -.. U4

!

=

=

309

Hints and Solutions

tori. It is easy to construct a homotopy between 8. solid torus without the central circle and its boundary T2. The two solid tori under consideration have common boundary. 61. The required isomorphism follows from the exact sequence for a pair,

because the cone ex is a contractible space. 62. Let us represent sm and sn as Jffl/8Jffl and 1"/8r, respectively. Thereby, we represent sm X sn as the cube In+m in which some points of 8In+m are identified; the identification turns 8r+m into sm V sn. Let a E l"+m \ 81"+m. It is clear how to construct a deformation retraction of In+m \ {a} onto 8r+m. This gives a deformation retraction of the punctured product sm x sn onto sm V sn. 63. The set Nn c M n is closed (because N" is compact) and open (because Nn has no boundary and the manifolds N n and Mn have equal dimensions). Therefore, it coincides with Mn. 64. (a) The manifold G+(n, 1) is diffeomorphic to sn-l, and 0+(3,2) R1 G+(3, 1). In what follows, we assume that k ~ 2 and n ~ 4. Let us calculate 1rl ( G (n, k)) by using Theorem 6.1. We are interested only in the 2-skeleton of G(n, k). The 2-skeleton includes Schubert cells of the forms (A ~ ~ 8), 1 A~ g), and (A ~ ~ ~); it is assumed that the remaining part of the matrix consists of zeros and ones, and so it is of no interest to us. Let us show that both 2-cells are attached to the I-cell in precisely the same way as in IRP2. 1000) (1000) d A s x, y -. 00, ( O:r: til 0 x/til 0 an

=

(X y

1 0 0) 0 1 0 '"

(X

1 0 0) 0 -y/x 1 0

"-+

(1 0 0 0) 0 -y/l: 1 0 .

In both cases, any antipodal boundary points of the 2-cell are attached to one point of the I-cell.

Thus, the group 11"1 ( G (n, k») is defined by one generator a and the relation a 2 = 1 (the two 2-cells determine the same relation). Hence 11"1 (G(n,

k)) = Z2-

The !lpace G+(n, k) doubly covers G(n, k); hence 1!"1 (G+(n, k)) = O. This implies the orient ability of G+(n, k). (b) Since k < n, for any two sets of k orthonormal vectors in IRn , there exists an element of SO(n) which transforms one of them into the ()ther; therefore, the group SO(n) acts transitively on G(n,k), i.e., any k-dimensional subspace Ilk of IRn can be transformed into any other kdimensional subspace by an element of SO(n).

Hints and Solutions

310

The stationary subgroup of each point nk E G(n, k), which consists of the transformations taking nk to itself, is isomorphic to (O(k) >< O(n -"I k» n SO(n). Indeed, any orthogonal transformation taking nk to itself is the direct sum of an orthogonal transformation of nk and an orthogonal transformation of (nk).L. Choose an orientation at a point nk E G(n, k) and try to extend it to the whole manifold G(n, k) by means of the action of SO(n). This can be done if the action of the entire stationary subgroup does not change the orientation at the point ITk. Take UI ~ O(k) and U2 E O(n -. k). The pair of matrices (UI' U2 ) acts on the matrix (lk' X), where lk is the identity matrix of order k and X is a k x (n - k) matrix, as follows, (Ik, X) 1-4 (Ull XU2) I'\J (lk' UI I XU2). This determines a map X...,..J UI I XU2 on the space of k x (n .... k) matrices X. Take elements of X as coordinates. In these coordinates the map has matrix ® (UII)T;; A, with determinant (detU2 )k(detU 1 )n-k (see (100]). Consider the case in which the matrix Ul ffiU2 belongs to SO(n), i.e., has determinant 1. This means that det Ul = det U2 = ±1. If det U1 = det U2 ::± 1, then detA;:= 1. If detUl -= detU2:;: \1,.. then detA =+I (_I)k( .... I)n-k = ( 'T l)n. Therefore, if n is even, then the stationary subgroup preserves ori· entation, and if n is odd, then some elelllents pf the stationary subgroup change ~rientation. Now, for odd n, we can easily construct a loop such that the orientation changes when transferred along it. Namely, in the connected group SQ(n), take a path -yet) from In to Ul ffi U2, where Ul E O(k) and U2 E O(n - k) are matrices with determinant -J, Tl1e loop -y(t)IIk in G(n, k) determined by this path has the required property. 65. To each pair of vectors (vu, V12, V13, vt4) and (vall V22, V23, V24) we assign the six numbers Xij = I~! ~! where i < j. These numbers are the Plucker coordinates of the plane spanned by the given vectors; They are connected by the single Plucker relation

uI

t

I,

(1)

Thus, it suffices to prove that the intersection of the hypersurface (1) with the 5-sphere L: X~j = 1 is diffeomorphic to 52 x 8 2 • Let us introduce the new coordinates Yl, ••• ,Y6 defined by 2X12 = Yl +Y4, 2 X34 = Yl - Y4, 2X23 = Y2 + Y5, 2Xl4 :;: Y2 ,. Y5, 2X13 = Y3 + Y6, and 2X24 = Y3 - Y6· In these coordinates, equation (1) takes the form Y~ + Y~ + Y~ = Y~ + Y~ + Y~, and the equation of the 5-sphere is L: Y~ = 2. 'rhe obtained system of equations is equivalent to + y~ + y~ ::;: 1, Y~ + Y~ f. Y~ = 1. This system determines 8 2 X 8 2 •

yr

Hints and

~olutions

oJJ.J.

66. Let n2 E G+(n, 2) be an oriented plane in ]Rn. In this planet take vectors VI and V2 such that IVII = IV21, vl.iv2' and the basis VI,V2 has positive orientation. If the plane rr 2 is identified with C, then the pair VI, V2 is determined up to multiplication by a nonzero complex number. To each pair of vectors VI, V2 we assign the vector VI + iV2 in en, and to this vector we assign the corresponding point in epn-l. Thereby, we define a.one-to-one map from G+(n,2) onto a subset of Cpn-l. Let us show that this set is the quadric z~ + ... + z~ = o. Suppose VI = (Xb ••. , x n ) and V2 = (YI, ... , Yn). Then 1:(Xk + iYk)2 = E x~ - 1: y~ + 2i 1: XkYk = 0, i.e., the point VI + iV2 belongs to the quadric under consideration. Conversely, if Zk = Xk + iYk and 1: z~ = 0, then the vectors Vl = (Xh.' f' Xn) and V2 = (YI, ..... Yn) have equal lengths and are orthogonal. Under complex conjugation, the basis VI, V2 is replaced by VI, -V2j the plane remains invariant, but its orientation changes. 67. Each point of T 8"' is determined by two vectors x, Y E ]Rn+1 for which (x, x) = 1 and (x, y) = O. To this pair of vectors we assign the point x( + IIYII2) + iy E n+1. Let Zk = Xk( + IIYII2) + iYk' Then

VI

L(Xk(

e

Vl

VI + IIYII2) + iYk)2 = (1

+ IIYII2) L x~ + 2iV1' + IIYII2 L

XkYk -

z::

Y~ = 1

because 1:x~ = 1 and 1:xkYk = O. Conversely, take a point u + iv E cn+~ for which 1:(Uk + iVk)2 = 1, i.e., 2 IIull - IIvII 2 =- 1 and (u, v) = O. To the point u + iv we assign the pair of 2 vectors x = v Il 2 and Y = v. We have IIxII = 1!"~iI2 = 1 and (x, y) =

V'1:'lI

O.

68. Let Vo = (XlIX2,X3,X4) E 8 3. Consider VI = (-X2,XI,-X4,X3), V2 = (-X3,X4,XlI-X2), and V3 = (-X4,-X3,X2,Xt). We have (Vi,Vj) = 0 for i =1= j. Therefore, VI, V2, and v~ are pairwise orthogonal unit vectors tangent to 8 3 at Vo. ~ The three linearly independent vector fields on 8 4n+3 are constructed similarly: we divide the coordinates into n + ~ quadruples and perform the same operations on each quadruple. 69. (a) Let vex) be a vector field without singular points on 8 2n+l... We can assume that IIv(x)1I = 1. Consider H(t,x) = (COS7rt)x + (sin7rt)v(x). We have IIH(t,x)1I = 1, i.e., H(t,x) E 8 2n+l. Moreover, H(O,x) = f(x) and H(l, x) = g(x).

Hints and Solutions

312

(b) Suppose that X2k+1(t) = X2k+1 COS1l"t + X2k+2 sin 1I"t and X2k+2(t) = X2k+1Sin1l"t + X2k+2COS1l"t. Let H(t,x) = (-XO,Xl(t), ... ,X2n(t)). Then H(O, x) = g(x) and H(l, x) = f(x). 70. (a) Consider the projection of the vector f(x) to the tangent space at a point x E San. H f(x) i= ±x for all x, then this projection is a vector field without singular points on s2n; such a field does not exist. Assign(b) Every point x E ]Rp2n can be regarded 88 a pair ±x E ing the pair ±f(x) to each of the points ±x, we obtain either two maps il,2= s2n -+ s2n (then i2 = -il) or one map i: s2n -+ 81'\- where the space s2n doubly covers S2fT. The latter is impossible because 11"1 (S2n) = O.

san.

If the map f has no fixed points, then the map il h88 the property that il(X) i= ±x for all x E S2n. According to (a), there is no such map. 71 (see [97]). We can assume that K is identified with JRn 88 a linear space. First, let us prove that the number n is even. Consider a path "Y(t) joining the points e and -e and not passing through 0 in JRn . Corresponding to each point "Y(t) of this path is a nondegenerate linear transformation Ay(t): x 1-+ JL(x,"Y(t)). To the points e and -e correspond the transforma.-

=

tions In and -In with determinants detIn 1 and det(~In) = ( ..... 1)n. Suppose that n is odd. Then det( -In) = -1. On the other hand, det(Ay(t» i= 0 for all t. Therefore, if det(Ay(t» > 0 at the starting point of the path, then det(Ay(t» > 0 at the end point. We have obtained a contradiction. To every vector v E IRn \ {O} we can assign a vector f(v) E JRn \ {O} for which JL(v,f(v)) = e and, moreover, v = ae if and only if f(v) = f3e. Thus, f is a self-homeomorphism of JRn \ {tel, where t E JR. It takes each ray tv, t > 0, to the ray tf(v), t > 0, because f(tv) = t- 1 f(v) for t =1= O. Therefore, we can consider the map i: sn-l \ {±e} -+ sn-l \ {±e} that takes each point v to the intersection point of the ray tf(v), t > 0, with the sphere sn-l. We use this map to define the map g: sn-2 -+ sn-2 that takes each point v to the intersection point of the sphere sn-2 formed by the unit vectors orthogonal to e with the great circle passing through ±e and i(v). The number n - 2 is even; therefore, according to Problem 70, we have 2. This means that f(v) = av+{3e, where a, {3 E g(v) = ±v for some v E 1R, and a i= O. By definition, e = JL(v,f(v)) = JL(v,av+{3e) = aJL(v,v)+{3v, whence JL( v, v) = a-Ie - a-I {3v. Thus, the subspace spanned by the vectors e and v is a subalgebra. This subalgebra is associative and commutative, and it has a two-sided identity element and no divisors of zerOj thus, it is a field. But any field which has dimension 2 as a real space is isomorphic to C. 72. In a small neighborhood of any point of the sphere sn-l = {x E JRn: IIxll = 1/2}, the map f is (up to linear terms) a symmetry about the 1J.yperplane tangent to Sn-l at the given point.

sn-

Hints and Solutions

73. The map This implies deg I

313

I

is homotopic to a constant map because 7r2(M2) = O. = O.

74. Suppose that Xo is 8. regular value of the map Ig, 1-1(xo) = {al, ••• ,ak}, g-l(lli) = {bil, .. ~,bil(i)}' Ei = signJ,(lli), Eij = signJg(bij). Then El + •.. + Ek = deg I and Eil + , .. + Cil(i) = deg 9 for all i = 1, ... , k. Therefore, deg(Jg) = EEiEij = EEi(degg) = (deg/)(degg). 75. Let P(z) anzn + an_lZn..-l + ... + ao, where an 1= O. The map z ....... P(z) extends to the map P: Cpl -+ Cpl defined by

=

(z ~w) ....... (an zn + an_lZn-1W + ... + aown : wn). Let (uo : 1) be a regular value of P. The degree of P is calculated as follows. First, we take the preimages of the point (uo :- 1), i.e., the points %0, ••• , Zk such that P(Zj) = uo. Regularity implies P'(Zj) 1= O. Therefor~ the polynomial P(z) - uo has no multiple roots, Le.; k = n" Next, we determine the sign of the Jacobian determinant for the map z ....... P(z) at each point Zj. 1£ P' (Zj) == a + bi, then the Jacobian matrix of this map at Zj is (~b!)' Its determinant equals a 2 + ~ > 0; therefore, the Jacobian determinant is positive at all points Z1, ••• ,%n. Hence the degree of the map is equal to n, 76. Suppose that R(z) = P(t)/Q(t), where degP =: m and degQ = n. Then the required smooth extension Cpl -+ Cpl is defined by

(z: w) ....... (wm+np(z/w): wm+nQ(z/w)). As in Problem 75, the Jacobian determinant is positive at all the preimages of regular values; thus, we only have to determine the number of preimages. The generic equation P(z) = cQ(z) has max{m,n} roots. 77. We regard ~sn ~ sn+1 as a smooth manifold. Take the equatorial sphere sn = snx {1/2} in sn+1 and choose a regular value Xo for I on it; this value is also regular for the map ~f. Moreover, if J is the Jacobian matrix of I at the point Yo E 1-1(xo), then (t~) is the Jacobian matrix of ~/. 78. Let us show that deg I = kl ... kn. 1£ one of the numbers kl' ... , len is zero, then this assertion is obvious. In what follows, we assume that kl ... kn 1= O. The preimage of any point (rlei1/Jl, ••• , r n ei 1/J.. ) consists of the Ikl ... kn I points (rle i (1/Jl +21rlt)/k1 , ••• ,rn ei (1/J.. +21rln )/kn),

where 0 ~ Ii ~ Ikll-l. Clearly, the Jacobian determinant of I has the same sign as kl •.. k n . 79. It follows from I(-A) = I(A) that the degree of I is even. Therefore, the map f is not homotopic to the identity map. 80. We can assume that C is the unit circle. Instead of a vector field v on C, it is more convenient to consider the vector field w obtained by

314

Hints and Solutions

rotating the vector vat each point (cos so, sin SO) through .....
=

At the points of tangency of the circle C with the integral trajectories, the vector w is normal to the Ox-axis. It is easy to show that at the points of external tangencY1 w rotates in the direction opposite to that in which the circle is traversed. This means that the Jacobian determinant of the map that takes each point of C to the vector w IIot this point is negative at the points of external tangency. At the points of internal tangency, ""the Jacobian is positive. The points of external and internal tangency are the preimages of the two points corresponding to the ~o directions normal to the Ox-axis. Therefore, one point has (i~e)/2 preimages (with signs taken into account)! 83. The case n = 1 is obvious; thus, we assume that n ~ 2. Consider a vector field v with nondegenerate singular points on M71 ~ The singular point~ can be divided into pairs consisting of points whose indices have opposite signs. Let us show how to decrease the number of singular points (if they exist) by two. Take two singular points x+ and x_ with indices 1 and .-or 1. Let '"Y be a path between them not passing through other singular points, and let '"Yt! be the e-neighborhood of '"Y. If e is sufficiently small, then '"YI& R:S D71 and '"Ye contains no singular points except x+ and z~! We can assume that the set '"Ye is covered by one chart, and all the vectors vex), where x E 8'"f.e, have unit length. Consider the map &Ye _ 8 71. . . 1 defined by x 1-+ vex). The degree of this map is equal to the sum of the indices of the singular points x+ and x_, Le., vanishes. Therefore, the map Ott! - 8 71- 1 is homotopic to a constant map. The homotopy can be regarded as a vector field w on '"Ye: which consists of unit vectors and coincides with the initial vector field v on &Yeo Consider the vector field that coincides with v outside '"Ye and with w on '"Ye. This vector field has two fewer singular points than v. In this way, all singular points can be killed. 84. Clearly, if M71 admits a vector field without singular points, then it admits also a line element field. Suppose that there is a line element field on M71. Let us endow M71 with a Riemannian metric and take both unit vectors in each of the I-dimensional subspaces determined by the field. The set of all such vectors is a closed manifold M 71 , which doubly covers M71 and is endowed with a vector field without singular points. (The manifold M71 is either connected or has two connected components diffeomorphic to M71.) Let X be the sum of indices of the singular points of the vector field on M71. According to Theorem 5.29 on p. 230, the indices of singular points of the vector field on M71 sum to 2X. But M71 admits a vector field without singular points; therefore, 2X = 0, whence X = O. Now, the existence of a ~ector field without singular points on M71 follows from Problem 83.

315

Hints and Solutions

85. In Example 4.3 on p. 171, it was shown that the Hopf fibration pt S3 4 8 2 generates the group 1r3(82) OJ! Z. Let us describe the corre-sponding framed manifold in n~(3). Take two near points on the sphere 8 2 • Their preimages are two circles forming a Hopf link. Therefore, the corresponding framed manifold is the circle 8 1 with framing defined as follows. Consider a Hopf link in IR3 such that one of its components is the circle 8 1 under consideration (we assume that it is standardly embedded in IR3), the second component lies on the boundary of the e-neighborhood of 8 1 , and each disk of radius orthogonal to 8 1 intersects the second component in exactly one point. The end of the vector el of the framing moves on the second component of the Hopf link (lIelli = t); the vector e2 lies in the plane normal to 8 1 and is orthogonal to e:l.. 86. (a) The contractibility of the Cone CY implies X/Y == (X U CY)/CY '" XUCY for any sub complex Y eX. It is easy to show that the proof of Lemma 5.2 remains valid for Y instead of Si-l and CY instead of Di. Thus, if f, 9 ~ Y -+ X are homotopic maps, then X UI CY ~ X U g CY.. By assumption, the embedding f; Y --+ X is homotopic to the constant map g: Y -+ Xo EX. But XuI Cy = XU CY and X Ug CY = X V I:Y. (b) The equatorial sphere 8 k is contractible in 8 H1 • Therefore, for n > ffl, the sphere 8 m canonically embedded in 8 n is contractible in sn, It remains to apply (a). 88. It is easy to see that the map i.: 1rl(A) --. 1rl(X) is given by a ~ 2a, where a is the generator of the groups 1rl(A) and 1rl(X). Therefore, r.i. cannot be the identity map. 89. (a) It is seen from the standard representation of the torus T'l as a square with sides glued together that the map i.: 1rl(A) -+ 1rl(X) is defined by a t-+ o.{3o.-l{3-l, where a is the generator of 1rl(A) and 0. and {3 are generators of the free group 1rl(X). Thus, for the commutator subgroups of the fundamental groups, we obtain the zero map. Therefore, the map r.i. cannot be the identity. (b) In this case, the map i.: 1rl(A) -+1rl(X) is defined by

e

a 1-+ 0.1 {310.1-l{3-l 1 ... o.g {3gag-1{3-1 g., and the map of commutator subgroups is again zero. 9(}. The map of commutator subgroups that is induced by i.: 1rl(A) --+ 1rl(X) has the form a 1-+ 20.1 + 20.2 + ... + 20.g • Therefore, r.i. cannot be the identity. 91. The tangent vector at x E sn- 1 C IRn is orthogonal to x. Therefore, each point of the manifold is an ordered pair of orthogonal unit vectors el, e2 E JR3. This pair admits a unique extension to a positively oriented orthonormal basis eI, e2, e3. Therefore, ~ 80(3).

M8

M8

316

Hints and Solutions

The homeomorphism 80(3) ~ lRps is established as follows. Any transformation from 80(3) has an eigenvector; hence it is a rotation through an angle cp about an axis I passing through the origin. To each vector e3 E lRs of length cp (0 < cp ~ 1r) we assign the rotation through the angle cp about the axis es in the direction for which the basis el, e2, es, where el is a vector orthogonal to e3 and e2 is the image of el under the rotation, is positively oriented. The zero vector is assigned the identity transformation. We have established a correspondence between the points of a disk D3 of radius 1r and the transformations from 80(3). Any two antipodal points of the disk correspond to the same transformation. 92. The space 1R3 ,SI is homotopy equivalent to S2 U I, where I is a diameter of the sphere S2. Let It be an arc on the sphere joining the endpoints of I. Then S2 U I "'I (S2 U I) /11 f'J S2 V 8 1 • 94. The space lR3 \ L is homotopy equivalent to the wedge of n copies of the space ~ \ SI, where SI C D3 is the standardly embedded circle (trivial knot). According to Problem 92, we have D3 \ Sl N S2 V SI, 95. In discussing the properties of the Hopf fibration, we have represented the sphere 8 3 as the union of the two solid tori Tl = D? X 8 1 and T2 = D~ X 8 1• The circles {O} ~ SI in these solid tori form the link under consideration. Since lR3 = 8 3 \ *, it follows that the space 1R3 \ L is obtained from the solid torus Tl by removing the circle {O} x 8 1 and one point. 8uch a space is homotopy equivalent to TJ V S2.

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[4] P. S. Alexandroff, Uber atetige Abbildungen kompakter Riiume. Math. Ann. 96 (1927), 555-571.

[5] K. Appel and W. HaIren, Every planar map is four colorable. I. Discharging.lllinois J. Math. 21 (1977), 429--490. [6) _ . Every planar map is four colorable. Amer. Math. Soc., Providence, Rl, 1989. [7) K. Appel, W. HaIren, and J. Koch, Every planar map is four colorable. II. Reducibility. Illinois J. Math. 21 (1977), 491-567. [8] D. Archdeacon and J.

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-

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R2n

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proof 0/ the

Borsuk antipodal theorem. J. Math. Anal.

[44] A. Gramain, La thlorime de

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Index

O-cell of a graph, 5 I-cell of a graph, 5 abstract simplicial complex, 102 action of a group, 87 admissible map, 101 Alexander horned sphere, 277 Alexandroff theorem, 61 algebra Clifford, 265 division, 204 algebraic curve plane, 279 reducible, 279 amalgam, 268 antipodal map, 113 points, 113 theorem, 113 approximation cellular, 126 simplicial, 104 atlas, 182 orientation, 206 attaching via a map, 117 automorphism group of a covering, 38 automorphism of a covering, 38 Balinski theorem, 22 barycentric coordinates, 81 subdivision, 81, 99 second,82 base of a bundle, 162 of a covering, 35

of a topology, 1 point, 30 Borsuk lemma, 164 Borsuk Ulam theorem, 113, 154 Bott Whitney polynomial, 51 boundary of a manifold, 183 of a pseudomanifold, 109 boundary point, 183 bridge, 54 Brouwer dimension invariance theorem, 60 fixed point theorem, 72 bundle locally trivial, 162 tangent, 202 Cantor set, 61 cell closed,119 open, 119 open Schubert, 196 rectilinear, 100 cellular approximation, 126 construction, 130 map, 126 cellular approximation theorem, 126 characteristic Euler, 144 map of a cell, 119 chart, 181 chromatic polynomial, 48 Clifford algebra, 265 closed cell, 119

-

325

326

manifold, 184 pseudomanifold, 109 set,l tw
Index

covering space, 35 univeI'llal, 43 critical point, 190 nondegenerate, 239 value, 190 COOIIII,274 curve algebraic plane, 279 reducible, 279 integral,245 Jordan, 63 regularly homotopic, 67 CW-complex,118 n-dimensional,119 nontriangulable, 122 cycle, 5 cylinder, 130 of a map, 179 deformation retract, 1:78 degree of a covering, 35 of a map, 111,221 modulo 2, 224 of a smooth closed curve, 66 of a IImooth map, 224 of a vertex, 5 of an algebraic curve, 279 deleted join, 131 product, 161 derivative of a function in the direction of a vector field, 200 diagonal,210 diagram of a knot, 274 diameter of a set, 59 dichromatic polynomial, 53 diffeomorphic manifolds, 185 diffeomorphism, 185 differential form, 205 of a map, 201, 202 dimension of a simplicial complex, 99 topological, 59 discrete space, 3 topology, 3 distance, 3 between sets, 55 from a point to a set, 55 Hausdorff, 50 division algebra, 204 dual graph, 12

Index

edge multiple, 5 of a graph, 5 embedding, 128, 185 Pliicker, 194 Euler characteristic, 144 formula for convex polyhedra, 14 for planar graphs, 15 exact sequence, 169 for a fibration, 168, 177 for a pair, 176 face of a planar graph, 13, 15 Feldbau theorem, 162 fiber of a bundle, 162 of a covering, 35 fibration, 162 Hopf, 171, 174 induced, 164 trivial, 162 field line element, 235 vector, 202 gradient, 243 finite simplicial complex, 99 fiVEK:Olor theorem, 16 fixed point, 72 Brouwer theorem, 72 four-color theorem, 16 framed cobordant manifold, 235 framed manifold, 235 free group, 40 Fubini theorem, 189 fundamental group, 32 general position, 103 generic points, 103 genus of a graph, 158 of a surface, 149 gradient vector field, 243 graph, 5 k-connected, 20 complete, 7 connected, 5 deleted product, 161 dual, 12 planar, 5 maximal, 13 graph invariant, 47 polynomial, 47 Grassmann manifold complex, 195 oriented, 195

327

real, 195 group defined by generators and relations, 44 free, 40 fundamental, 32 homotopy, 166 of a knot, 273 spinor,265 topological, 87 Hausdorff distance, 56 space, 87 Heawood theorem, 159 Helly theorem, 301 Hessian matrix, 239 homeomorphic spaces, 2 homeomorphism, 2 local, 155 homogeneous complex, 109 coordinates, 120 homotopic maps, 29 relative spheroids, 174 homotopy, 29 equivalent spaces, 29 group, 166 smooth, 221 Hopf fibration, 171, 174 theorem, 231 horned sphere, 277 image of a homomorphism, 169 immersion, 185 one-to-one, 215 incident vertex and edge, 5 vertex and face, 26 index of a critical point, 239 of a quadratic form, 239 of a singular point, 225 induced fibration, 164 topology, 2 induction transfinite, 96 integral curve, 245 interior, 1 point of a manifold, 183 internally disjoint path, 20 intersection number of two graphs, 7 invariant graph, 47 polynomial, 47 topological, 52

328

'nItte, 53 inverse function theorem, 183 isolated singular point, 225 isomorphic graphs, 47 isotopic diffeomorphisms, 223 join, 131 deleted, 131 Jordan curve,63 curve theorem, 63, 77 piecewise linear, 6 Konig theorem, 157 Kakutani theorem, 84, 264 kernel of a homomorphism, 169 knot, 273 diagram, 274 group, 273 polygonal, 273 smooth,273 toric, 277 trivial, 273 Kuratowski theorem, 8 Lebesgue number, 59 theorem on dOlled covers, 59 on open covers, 59 Lefscbetz formula, 106 lemma Borsuk,l64 Morse, 239 on homogeneity of manifolds, 223 Sperner's, 81, 106, 113 'nIcker's, 115 Urysohn's, 56, 91 lens space, 237 lifting of a map, 163 of a path, 35 line element field, 235 link,28O trivial, 281 local coordinate system, 181 homeomorphism, 155 locally compact space, 90 locally contractible space, 123 locally finite cover, 94 locally trivial fiber bundle, 162 loop, 5, 32 Lyusternik Shnirelman theorem, 115 manifold, 182

Index

dOlled,l84 cobordant, 235 diffeomorphic, 185 framed,235 framed cobor
329

Index

one-point compactification, 90 one-to-one immersion, 215 open cell, 119 Schubert cell, 196

set,l orbit, 87 space, 88 order of a cover, 59 orientable manifold, 205 pseudomanifold, 110 surface, 148 orientation atlas, 206 covering, 207 covering manifold, 207 negative, 109 of a simplex, 109 positive, 109 oriented Grassmann manifold, 195 pseudomanifold, 110 origin of a loca.1 coordinate system, 181

polygonal knot, 273 polynomial Bott Whitney, 51 chromatic, 48 dichromatic, 53 graph invariant, 41 invariant, 47 Tutte, 54 Pontryagin construction, 236 theorem, 236 positive orientation, 109 product deleted, 161 symmetric, 135 topology, 4 wedge,30 projection map, 162 proper map, 155 pseudomanifold, 109 closed,109 orientable, 110 oriented, 110 quotient space, 4

paracompact space, 94 partition of unity, 93 smooth,187 path-connected space, 29 Peano theorem, 62 piecewise linear Jordan theorem, 6 Pliicker coordinates, 194 embedding, 194 relations, 195 planar graph, 5 maximal,13 plane algebraic curve, 279 Poincare-Hopf theorem, 228 point base, 30 boundary, 183 critical, 190 nondegenerate, 239 fixed,72 interior of a manifold, 183 regular, 190 singular isolated, 225 nondegenerate, 227 of a vector field, 202, 225 of an algebraic curve, 279 points antipodal, 113 generic, 103 in general position, 103

Radon theorem, 117

rank of a free group, 40 of a smooth map, 185 real Grassmann manifold, 195 real projective space, 120 realization of an abstract simplicial complex, 102 rectilinear cell, 100 cell complex, 100 reducible algebraic curve, 279 refinement of a cover, 94 regular covering, 36 Morse function, 243 point, 190 space, 95 value, 111 regularly homotopic curves, 67 relative spheroid, 174 homotopic, 174 retract, 72 deformation, 178 retraction, 72 Riemannian metric, 204 Sard's theorem, 190 Schubert symbol, 196 neighboring, 254

Index

330

second countable space, 2 Seifert van Kampen theorem, 269 self-intersection number of • graph, S. semicontinuity upper, 84

set Cantor, 61 closed,1 of labels complete, 80 of measure zero, 188 open, 1 well-ordered, 96 simplicial approximation, 104 theorem, 10.5 complex, 99 construction, 130 map, 101, 110. simplicial complex abstract, 102 finite, 99 simply connected space, 33 singular point isolated, 225 of a vector field, 202, 225 nondegenerate, 227 of an algebraic curve, 279 skeleton of a complex, 100 of a OW-complex, 119 smooth homotopy, 221 knot, 273 manifold, 182 map, 185 partition of unity, 187 structure, 181 space n-simple, 168 compact, 3 locally, 90 connected, 3 contractible, 29 locally, 123 cotangent, 205 covering, 35 discrete, 3 Hausdorff, 87 lens, 237 locally compact, 90 contractible, 123 metric, 3 metrizable, 3 normal,91 • of orbits, 88 paracompact, 94 path-connected, 29

projective complex, 120real, 120 regular, 95 second countable, 2 simply connected, 33 tangent, 201 topological, 1 spaces

homeomorphic, 2 homotopy equivalent, 29 Sperner's lemma, 81, 106, ~ sphere Alexander homed, 277 spheroid, 166 relative, 174 spinor group, 265 star of a point, 104 of a simplex, 104 Steinitz' theorem, 23 Stone theorem, 93, 98 strongly connected complex, 109 subcomplex, 120 complete, 10.3 subdivision barycentric, 99 second,82 of a rectilinear cell complex, 100 subgraph,8 submanifold, 184 submersion, 185 support of a function, 93 surface orientable, 148 twtHIimensional closed,139 with boundary, 140 without boundary, 139 suspension, 110, 130 symmetric product, 135 tangent bundle, 20.2 space, 201 vector, 199 theorem AlexandrofI, 61 antipodal, 113 Balinski, 22 Borsuk UlaID, 113, 154 Brouwer on fixed point, 72 on dimension invariance, 60 cellular approximation. 126 Feldbau, 162 five-color, 16

Index

four-color, 16 Fubini,189 Heawood, 159 Helly,301 Hopf,231 Jordan curve, 63, 77 piecewise linear, 6 Konig, 157 Kakutani, 84, 264 Kuratowski, 8 Lebesgue on closed covers, 59 on open covers, 59 Lyustemik-8hnirelman,115 Menger Whitney, 21 on covering homotopy, 163 on inverse function, 183 on simplicial approximation, 105 on tubular neighborhoods, 228 Peano, 62 Poincare-Hopf, 228 Pontryagin, 236 Radon, 117 Sard's,l90 Seifert-van Kampen, 269 Steinitz', 23 Stone, 93, 98 Tietze, 57, 92 van Kampen, 269 Whitehead, 179 Zermelo's,96 Tietze theorem, 57, 92 topological dimension, 59 group, 87 invariant, 52 manifold, 181 space, 1 topology discrete, 3 induced, 2 induced by a metric, 3 nondiscrete, 87 product, 4 trivial, 87 toric knot, 277 total space of a bundle, 162 trajectory of a vector field, 226 transfinite induction, 96

331

transversal map, 218 tree, 15 maximal,32 trefoil, 275 triangle inequality, 3 triangulation, BO, 210 of a topological space, 141 trivial fibration, 162 knot, 273 link,281 topology, 87 tubular neighborhood theorem, 228 Thcker'.lemma, 115 Thtte invariant, 53 polynomial, 54 two-dimensional surface closed,139 with boundary, 140 without boundary, 139 universal covering, 43, 150 covering space, 43 unramified complex, 109 upper semicontinuity, 84 upper semicontinuoUB map, 84 Urysohn'. lemma, 56, 91 value critical, 190 regular, 111 van Kampen theorem, 269 vector field,202 gradient, 243 tangent, 199 vertex of a graph, 5 wedge, 30 product, 30 well-ordered set, 96 Whitehead theorem, 179 Whitney number, 69

Zeeman example, 153 Zermelo's theorem, 96

Titles in This Series 74 73 72 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36

V. V. Prasolov, Elements of combinatorial and differential topology, 2006 Louis Halle Rowen, Graduate algebra: Commutative view, 2006 R. J. Williams, Introduction the the mathematics of finance, 2006 Sean Dineen, Probability theory in finance, 2005 Sebastian Montiel and Antonio Ros, Curves and surfaces, 2005 Luis Caffarelll and Sandro Salsa, A geometric approach to free boundary problems, 2005 T.Y. Lam, Introduction to quadratic forms over fields, 2004 Yuli Eidelman, Vitali Milman, and Antonis Tsolomitis, Functional analysis, An introduction, 2004 S. Ramanan, Global calculus, 2004 A. A. Kirillov, Lectures on the orbit method, 2004 Steven Dale Cut kosky, Resolution of singularities, 2004 T. W. Korner, A companion to analysis: A second first and first second course in analysis, 2004 Thomas A. Iveyand J. M. Landsberg, Carlan for beginners: Differential geometry via moving frames and exterior differential systems, 2003 Alberto Candel and Lawrence Conlon, Foliations II, 2003 Steven H. Weintraub, Representation theory of finite groups: algebra and arithmetic, 2003 Clklric Villani, Topics in optimal transportation, 2003 Robert Plato, Concise numerical mathematics, 2003 E. B. Vinberg, A course in algebra, 2003 C. Herbert Clemens, A scrapbook of complex curve theory, second edition, 2003 Alexander Barvinok, A course in convexity, 2002 Henryk Iwaniec, Spectral methods of automorphic forms, 2002 Ilka Agricola and Thomas Friedrich, Global analysis: Differential forms in analysis, geometry and physics, 2002 Y. A. Abramovich and C. D. Aliprantis, Problems in operator theory, 2002 Y. A. Abramovich and C. D. Aliprantis, An invitation to operator theory, 2002 John R. Harper, Secondary cohomology operatiOns, 2002 Y. Eliashberg and N. Mishachev, Introduction to the h-principle, 2002 A. Yu. Kitaev, A. H. Shen, and M. N. Vyalyi, Classical and quantum computation, 2002 Joseph L. Taylor, Several complex variables with connections to algebraic geometry and Lie groups, 2002 Inder K. Ran&, An introduction to measure and integration, second edition, 2002 Jim Agler and John E. MCCarthy, Pick interpolation and Hilbert function spaces, 2002 N. V. Krylov, Introduction to the theory of random processes, 2002 Jin Hong and Seok-Jin Kang, Introc:luction to quantum groups and crystal bases, 2002 Georgi V. Smirnov, Introduction to the theory of differential inclusions, 2002 Robert E. Greene and Steven G. Krantz, Function theory of one complex variable, thil;d edition, 2006 Larry C. Grove, Classical groups and geometric algebra, 2002 Elton P. Usu, Stochastic analysis on manifolds, 2002 Hershel M. Farkas and Irwin Kra, Theta constants, Riemann surfaces and the modular group, 2001 Martin Schechter, Principles of functional analysis, second edition, 2002

TITLES IN THIS SERIES 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1

James F. Davis and Paul Kirk, Lecture notes in algebraic topology, 2001 Sigurdur Helgason, Differential geometry, J..ie groups, and symmetric spaces, 2001 Dmitri Burago, Yurl Burago, and Sergei Ivanov, A course in metric geometry, 2001 Robert G. Bartle, A modem theory of integration, 2001 Ralf Korn and Elke Korn, Option pricing and portfolio optimization: Modern methods of financial mathematics, 2001 J. C. McConnell and J. C. Robson, Noncommutative Noetherian rings, 2001 Javier Duoandlkoetxea, Fourier analysis, 2001 Livlu I. Nicolaescu, Notes on Seiberg-Witten theory, 2000 Thierry Aubin, A course in differential geometry, 2001 Rolf Berndt, An introduction to symplectic geometry, 2001 Thomas Friedrich, Dirac operators in Riemannian geometry, 2000 Helmut Koch, Number theory: Algebraic numbers and functions, 2000 Alberto Candel and Lawrence Conlon, Foliations I, 2000 Gunter R. KrsUBe and Thomas H. Lenagan, Growth of algebras and Gelfand-Kirillov dimension, 2000 John B. Conway, A course in operator theory, 2000 Robert E. Gompf and Andru I. Stipslcz, 4-manifolda and Kirby calculus, 1999 Lawrence C. EVIlIlS, Partial differential equations, 1998 Winfried Just and Martin Weese, Discovering modem set theory. II: Set-theoretic tools for every mathematician, 1997 Henryk Iwaniec, Topics in classical automorphic forms, 1997 Richard V. Kadison and John R. Rlngrose, Fundamentals of the theory of operator algebras. Volume II. Advanced theory, 1997 Richard V. Kadison and John R. Ringrose, Fundamentals of tho theory of operator algebras. Volume I: Elementary theory, 1997 Elliott H. Lieb and Michael Loss, Analysis,I997 Paul C. Shields, The ergodic theory of discrete sample paths, 1996 N. V. Krylov, Lectures on elliptic and parabolic equations in Holder spaces, 1996 Jacques Dixmier, Enveloping algebras, 1996 Printing Barry Simon, Representations of finite and compact groups, 1996 Dino Lorenzini, An invitation to arithmetic geometry, 1996 Winfried Just and Martin Weese, Discovering modem set theory. I: The basics, 1996 Gerald J. Janusz, Algebraic number fields, second edition, 1996 Jenz Carsten Jantzen, Lectures on quantum groups, 1996 Rick Miranda, Algebraic curves and Riemann surfaces, 1995 Russell A. Gordon, The integrals of Lebesgue, Denjoy, Perron, and Henstock, 1994 William W. Adanu; and Philippe Loustaunau, An introduction to Grebner bases, 1994 Jack Graver, Brigitte Servatius, and Herman Servatius, Combinatorial rigidity, 1993 Ethan Akin, The general topology of dynamical systems, 1993

Modern topology uses very diverse methods. This book is devoted largely to methods of combi natorial topology, which reduce the study of topological spaces to investigations of their partitions into elementary sets, and to methods of differential topology, which deal with smooth manifolds and smooth maps. Many topological problems can be solved by using either of these two kinds of methods, combinatorial or differential. In such cases, both approaches are discussed. One of the main goals of this book is to advance as far as possible in the study of the properties of topological spaces (especially manifolds) without employing complicated techniques.This distinguishes it from the majority of other books on topology. The book contains many problems; almost all of them are supplied with hints or complete solutions.

ISBN 0- 8218- 3809- 1

9 780821 838099

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