Manifolds and Differential Geometry (Graduate Studies in Mathematics 107)

Special case 2: Assume again that M = ~n. Let K c 0 be as in the hypotheses. For each point p E K let Up be an open ball centered at p and contained in O. Let Kp be the closed ball centered at p of half the radius of Up. The interiors of the Kp's form an open cover for K and so by compactness we can reduce to a finite subcover. Thus we have a finite family {Ki} of closed balls of various radii such that K c UKi' and with corresponding concentric open balls Ui cO. For each Ui, let
13 (x)

= 1-

II(1 -


General case: From the second special case above it is clear that we have the result if K is contained in the domain U of a chart (U, x). If K is not contained in such a chart, then we may take a finite number of charts (Ul,Xl), ... , (Uk,Xk) and compact sets K 1 , •.. , Kk with K c U7=lKi , Ki CUi, and UUi C O. Now let
13 = 1 -

II (1 -


o

i=l

Let [fl E C~(M,~) (or E C~(M,q) and let f be a representative of the equivalence class [fl. We can find an open set U containing p such that U is compact and contained in the domain of f. If 13 is a cut-off function that is identically equal to 1 on U, and has support inside the domain of f, then 13 f is smooth and it can be extended to a globally defined smooth function that is zero outside of the domain of f. Denote this extended function by (13 f)ext· Then (13 f)ext E [fl (usually, the extended function is just written

30

1. Differentiable Manifolds

as (3J). Thus every element of C:r(M, JR) has a representative in COO(M, JR). In short, each germ has a global representative. (The word "global" means defined on, or referring to, the whole manifold.) A partition of unity is a technical tool that can help one piece together locally defined smooth objects with some desirable properties to obtain a globally defined object that also has the desired properties. For example, we will use this tool to show that on any (paracompact) smooth manifold there exists a Riemannian metric tensor. As we shall see, the metric tensor is the basic object whose existence allows the introduction of notions such as length and volume.

Definition 1. 70. A partition of unity on a smooth manifold M is a collection {'POI.} OlEA of smooth functions on M such that (i) O:S 'POI. :S 1 for all a. (ii) The collection of supports {supp( 'POl)}OlEA is locally finite; that is, each point p of M has a neighborhood Wp such that Wp n supp( 'POI.) = 0 for all but a finite number of a E A. (iii) L:OlEA 'POl(P) = 1 for all p E M (this sum has only finitely many nonzero terms by (ii)). If 0 = {OOl}OlEA is an open cover of M and SUPP('POl) C 001. for each a E A, then we say that {'POl}OlEA is a partition of unity subordinate to 0= {OOl}OlEA.

Remark 1.71. Let U = {UOl}OlEA be a cover of M and suppose that W = {W,B} ,BEB is a refinement of U. If {'I/J,B} ,BEB is a partition of unity subordinate to W, then we may obtain a partition of unity {'POI.} subordinate to U. Indeed, if f : B -+ A is such that W,B C Uf(,B) for every (3 E B, then we may let 'POI. := L:,BEf-1(0l) 'I/J,B.

Our definition of a smooth manifold M includes the requirement that M be paracompact (and Hausdorff). Paracompact Hausdorff spaces are normal spaces, but the following theorem would be true for a normal locally Euclidean space with smooth structure even without the assumption of paracompactness. The reason is that we explicitly assume the local finiteness of the cover. For this reason we put the word "normal" in parentheses as a pedagogical device.

Theorem 1.72. Let M be a (normal) smooth manifold and {UOl}OlEA be a locally finite cover of M. If each U01. has compact closure, then there is a partition of unity {'POl}OlEA subordinate to {UOl}OlEA. Proof. We shall use a well-known result about normal spaces sometimes called the "shrinking lemma". Namely, if {UOl}OlEA is a locally finite (and

1.6. Coverings and Discrete Groups

31

hence "point finite") cover of a normal space M, then there exists another cover {VaJaEA of M such that Va C Ua . This is Theorem B.4 proved in Appendix B. We do this to our cover and then notice that since each Ua has compact closure, each Va is compact. We apply Lemma 1.69 to obtain nonnegative smooth functions 'l/Ja such that supp 1Pa C Ua and 1Pa IVa == 1. Let 1P := LaEA 1Pa and notice that for each p EM, the sum LaEA 1Pa (p) is a finite sum and 1P(p) > 0. Let 'Pa := 1Pa/1P· It is now easy to check that {'Pa}aEA is the desired partition of unity. 0 If we use the paracompactness assumption, then we can show that there exists a partition of unity that is subordinate to any given cover.

Theorem 1.73. Let M be a (paracompact) smooth manifold and {Ua}aEA a cover of M. Then there is a partition of unity {'Pa} aEA subordinate to {Ua}aEA. Proof. By Remark 1. 71 and the fact that M is locally compact we may assume without loss of generality that each Ua has compact closure. Then since M is paracompact, we may find a locally finite refinement of {Ua}aEA which we denote by {Vi}iEI. Now use the previous theorem to get a partition of unity subordinate to {ViLEI. Finally use remark 1.71 one more time to get a partition of unity subordinate to {Ua}aEA. 0 Exercise 1.74. Show that if a function is smooth on an arbitrary set ScM as defined earlier, then it has a smooth extension to an open set that contains

S. Now that we have established the existence of partitions of unity we may show that the analogue of Lemma 1.69 works with K closed but not necessarily compact: Exercise 1. 75. Let M be a smooth manifold. Let K be a closed subset of M and 0 an open set containing K. Show that there exists a smooth function (3 on M that is identically equal to 1 on K, takes values in the interval [0,1]' and has compact support in O.

1.6. Coverings and Discrete Groups 1.6.1. Covering spaces and the fundamental group. In this section, and later when we study fiber bundles, many of the results are interesting and true in either the purely topological category or in the smooth category. Let us agree that a CO manifold is simply a topological manifold. Thus all

32

1. Differentiable Manifolds

°

relevant maps in this section are to be C r , where if r = we just mean continuous and then only require that the spaces be sufficiently nice topological spaces. Also, "Co diffeomorphism" just means homeomorphism. In the definition of path connectedness and path component given before, we used continuous paths, but it is not hard to show that if two points on a smooth manifold can be connected by a continuous path, then they can be connected by a smooth path. Thus the notion of path component remains unchanged by the use of smooth paths.

Definition 1.76. Let fa : X ---t Y and h : X ---t Y be C r maps. A C r homotopy from fa to h is a C r map H : X x [0, 1] ---t Y such that

H(x,O) = fo(x)

and

H(x, 1) = h(x) for all x. If there exists such a C r homotopy, we then say that fa is C r

homotopic to h and write fa ~ h. If A c X is a closed subset and if H(a, s) = fo(a) = h(a) for all a E A and all s E [0,1]' then we say that

fa is C r homotopic to h relative to A and we write fa ~ map H is called a C r homotopy.

h

(reI A). The

In the above definition, the condition that H : X x [0, 1] ---t Y be C r for r > can be understood by considering X x [0, 1] as a subset of X x lR. Homotopy is obviously an equivalence relation.

°

Exercise (reI A) if H(x, s) = H(a, s) =

1. 77. Show that fa : X ---t Y and h : X ---t Yare C r homotopic and only if there exists a C r map H : X x ffi. ---t Y such that fo(x) for all x and s ::; 0, H(x, s) = h(x) for all x and s 2: 1 and fo(a) = h(a) for all a E A and all s.

At first it may seem that there could be a big difference between Coo and CO homotopies, but if all the spaces involved are smooth manifolds, then the difference is not big at all. In fact, we have the following theorems which we merely state. Proofs may be found in [Lee, John].

Theorem 1.78. If f : M ---t N is a continuous map on smooth manifolds, then f is homotopic to a smooth map fa : M ---t N. If the continuous map f : M ---t N is smooth on a closed subset A, then it can be arranged that

coo

f :::: fa (relA). Theorem 1.79. If fa : M ---t Nand h : M ---t N are homotopic smooth maps, then they are smoothly homotopic. If fa is homotopic to h relative to a closed subset A, then fa is smoothly homotopic to h relative to A. Because of these last two theorems, we will usually simply write f :::: fa instead of f

~ fa, the value of r being of little significance in this setting.

1.6. Coverings and Discrete Groups

33

Figure 1.5. Coverings of a circle

Definition 1.80. Let M and M be CT manifolds. A surjective CT map P : M -+ M is called a CT covering map if every point p E M ha~ an open connected neighborhood U such that each connected component Ui of p-l(U) is CT diffeomorphic to U via the restrictions plui : Ui -+ U. In this case, we say that U is evenly covered by p (or by the sets Ud. The triple (M, p, M) is called a covering space. We also refer to the space M (somewhat informally) as a covering space for M. Example 1.81. The map IR -+ Sl given by t r---t eit is a covering. The set of points {e it : () - n < t < () + n} is an open set evenly covered by the intervals In in the real line given by In := (() - n + 2nn, () + n + 2nn) for n E Z. Exercise 1.82. Explain why the map (-2n, 2n) -+ Sl given by t not a covering map.

r---t

eit is

Definition 1.83. A continuous map f is said to be proper if f-l(K) is compact whenever K is compact. Exercise 1.84. Show that a CT proper map between connected smooth manifolds is a smooth covering map if and only if it is a local CT diffeomorphism. The set of all CT covering spaces is the set of objects of a ~tegory. A morphism between CT covering spaces, say (Ml' PI, M l ) and (M2' P2, M 2), is a pair of CT maps (1, f) such that the following diagram commutes:

j

Ml~M2

PI!

f

!P2

Ml~M2

This means that f 0 Pl = P2 0 f. Similarly, the coverings of a fixe~pace M are the objects of a category where the morphisms are maps : Ml -+ M2

1. Differentiable Manifolds

34

that make the following diagram commute:

meaning that PI = P2 0 0 and suppose that p : M --+ M is a CO covering map with M paracompact. Then there exists a (unique) C r structure on M making p a C r covering map. Proof. Choose an atlas {(Uo:, XO:)}O:EA such that each domain Uo: is small enough to be evenly covered by p. Thus we have that p -1 (U0:) is a disjoint union of open sets U~ with each restriction PIUi a homeomorphism. We Q

now construct charts on M using the maps

Xo: 0

PIUi defined on the sets U~ Q

1.6. Coverings and Discrete Groups

35

(which cover M). The overlap maps are smooth since if U~ (Xa 0

=

~luJ 0

Xa 0

(Xf3

0

n UJ # 0 then

~IU~)-1

~Iu~ 0 (~Iu~) -1 0 x~1

= Xa 0 Xf3-1 . We leave it to the reader to show that M is Hausdorff if M is Hausdorff.

D

The following is a special case of Definition 1. 76. Definition 1.87. Let a : [0,1] -7 M and /3 : [0,1] -7 M be two CT maps (paths) both starting at P E M and ending at q. A CT fixed endpoint homotopy from a to /3 is a family of CT maps Hs : [0,1] -7 M parameterized by s E [0,1] such that

1) H: [0,1] x [0,1] -7 M defined by H(t, s) := Hs(t) is CT; 2) Ho = a and HI = /3; 3) Hs(O) = P and Hs(l) = q for all s E [0,1]. Definition 1.88. If there is a C T homotopy from a to /3, then we say that a is CT homotopic to /3 and write a c::= /3 (CT). If r = 0, we speak of paths being continuously homotopic. Remark 1.89. By Theorems 1.79 and 1.78 above we know that in the case of smooth manifolds, if a and /3 are smooth paths, then we have that a c::= /3 (CO) if and only if a c::= /3 (CT) for r > 0. Thus we can just say that a is homotopic to /3 and write a c::= /3. In case a and /3 are only continuous, they may be replaced by smooth paths a' and /3' with a' c::= a and /3' c::= /3.

It is easily checked that homotopy is an equivalence relation. Let P(p, q) denote the set of all continuous (or smooth) paths from p to q defined on [0,1]. Every a E P(p, q) has a unique inverse (or reverse) path a+- defined by a+-(t) := a(l - t). If PI, P2 and P3 are three points in M, then for a E P(PI, P2) and /3 E P(p2,P3) we can "multiply" the paths to get a path a*/3 E P(PI,P3) defined by a(2t) for t < 1/2, { a * /3(t) := /3(2t-1) for 1/2 :::; t :::; 1. Notice that a*/3 is a path that follows along a and then /3 in that order. 5 An important observation is that if al c::= a2 and /31 c::= /32, then al * /31 c::= a2 */32.

°: :;

SIn some settings, it is convenient to reverse this convention.

36

1. Differentiable Manifolds

The homotopy between al *(31 and a2 *(32 is given in terms of the homotopies Ha : al ~ a2 and H(3 : (31 ~ (32 by

Ha(2t, s)

for 0 :::; t

< 1/2,

H(t, s) := {

and 0 :::; s

< 1.

H(3(2t - 1, s) for 1/2 :::; t < 1, Similarly, if al ~ a2, then ai ~ at. Using this information, we can define a group structure on the set of homotopy equivalence classes of loops, that is, of paths in P(p, p) for some fixed p E M. First of all, we can always form a * (3 for any a, (3 E P(p, p) since we always start and stop at the same point p. Secondly, we have the following result.

Proposition 1.90. Let 1fl (M, p) denote the set of fixed endpoint homotopy classes of paths from P(p, p). For [aJ, [(3] E 1fl (M, p), define [a]· [(3] := [a*(3]. This is a well-defined multiplication, and with this multiplication 1fl (M, p) is a group. The identity element of the group is the homotopy class 1 of the constant map 1p : t f--7 p, the inverse of a class [a] is [a+---]. Proof. We have already shown that [a] . [(3] := [a must also show that 1) For any a, the paths a constant map 1p.

0

a+--- and a+---

* (3]

is well-defined. One

a are both homotopic to the

0

* a ~ a and a * 1p ~ a. 3) For any a,(3,"( E P(p,p), we have (a * (3) * "( ~ a * ((3 * "(). 2) For any a E P(p,p), we have 1p

Proof of 1): 1p is homotopic to a

a+--- via

0

a(2t) for { H(t, s) = a(s) for a+---(2t - 1) for

0:::; 2t :::; s, s :::; 2t :::; 2 - s, 2 - s :::; 2t :::; 2,

where 0 :::; s :::; 1. Interchanging the roles of a and a+--- we also get that 1p is homotopic to a+--- 0 a. Proof of 2): Use the homotopy

H(t s) = { a( l!st) for 0 :::; t :::; 1/2 + s/2, , p for 1/2+s/2:::;t:::; 1. Proof of 3): Use the homotopy

a(l!st) H(t, s) = { (3(4(t _ Its)) "(( 2~s (t

-

2t

s ))

for 0 < t < 1+s 4 ' for 1+s < t < 2+s 4 4 ' for 2+s < t < 1. 4 -

D

The group 1fl (M, p) is called the fundamental group of M at p. If desired, one can take the equivalence classes in 1fl(M,p) to be represented

1.6. Coverings and Discrete Groups

37

by smooth maps. If "( : [0,1] --t M is a path from p to q, then we have a group isomorphism 7rl(M,q) --t 7rl(M,p) given by

[a]

r---+

[r * a

* ,,(<-].

It is easy to show that this prescription is a well-defined group isomorphism. Thus, for any two points p, q in the same path component of M, the groups 7rl(M,p) and 7rl(M,q) are isomorphic. In particular, if M is connected, then the fundamental groups based at different points are all isomorphic. Because of this, if M is connected, we may simply refer to the fundamental group of M, which we write as 7rl (M).

Definition 1.91. A path connected topological space is called simply connected if 7rl(M) = {1}. The fundamental group is actually the result of applying a functor (see Appendix A). Consider the category whose objects are pairs (M,p), where M is a C r manifold and p is a distinguished point (base point), and whose morphisms f : (M,p) --t (N, q) are C r maps f : M --t N such that f(p) = q. The pairs are called pointed C r spaces and the morphisms are called pointed C r maps (or base point preserving maps). To every pointed space (M,p), we assign the fundamental group 7rl(M,p), and to every pointed C r map f : (M,p) --t (N, f(p)) we may assign a group homomorphism 7rl(f) : 7rl(M,p) --t 7rl(N, f(p)) by

7rl(f)([a]) = [f 0 a]. It is easy to check that this is a covariant functor, and so for pointed maps f and g that can be composed, (M, x) (N, y) !4 (P, z), we have 7rl(g 0 f) = 7rl (g) 0 7rl (f).

-L

Notation 1.92. To avoid notational clutter, we will often denote 7rl(f) by f#· Definition 1.93. ~et p : ]l.L--t M be a C r covering and let f : P --t !,f be a C r map. A map f : P --t M is said to be a lift of the map f if P 0 f = f. Theorem 1.94. Let p : M --t M be a C r covering, let "( : [a, b] --t M be a C r curve and pick a point y in p-l(-y(a)). Then there exists a unique C r lift ;:y : [a, b] --t M of"( such that ;:Y(a) = y. Thus the following diagram commutes:

38

1. Differentiable Manifolds

Figure 1.6. Lifting a path to a cover

If two paths a and f3 with a(a) = f3(a) are fixed endpoint homotopic via a homotopy h, .!hen for a given point yin p-1(,(a)), we have the corresponding lifts ii and f3 starti"!:Jl at y. In this c~se, the homotopy h lifts to a fixed endpoint homotopy h between ii and f3. In short, homotopic paths lift to homotopic paths.

Proof. We just give the basic idea and refer the reader to the extensive literature for details (see [Gre-Hrp]). Figure 1.6 shows the way. Decompose the curve, into segments that lie in evenly covered open sets by using the Lebesgue number lemma. Lift inductively starting by using the inverse of p in the first evenly covered open set. It is clear that in order to connect up continuously, each step is forced and so the lifted curve is unique. A similar argument shows how to lift the homotopy h. A little thought reveals that if p is a CT covering, then the lifts of CT maps are CT. 0 There are several important corollaries to this result. One is simply that if a : [0,1] -+ M is a path starting at a base point p E M, then since there is one and only one lift ii starting at a given pi in the fiber p -1 (p), the endpoint ii(l) is completely determined by the path a and by the point pi from which we want the lifted path to start. In fact, the endpoint only depends on the homotopy class of a (and the choice of starting point pi). To see this, note that if a, f3 : [0,1] -+ !:! are fixed endpoint homotopic paths in M beginning at p, and if 0: and f3 are the corresponding lifts with 0:(0) = ,8(0) = pi, then by the second part of the theorem, any homotopy h t : a ~ f3 lifts to a unique fixed endpoint homotopy h t : 0: ~ f3. This then ~

~

1.6. Coverings and Discrete Groups

39

/3(1).

implies that a(l) = Applying these ideas to loops based at p EM, we will next see that the fundamental group 7rl(M,p) acts on the fiber 8J-l(p) as a group of permutations. (This is a right action as we will see.) In case the covering space M is simply connected, we will also obtain an isomorphism of the grou~I(M,p) with the deck transformation group (which acts from the left on M). Before we delve into these matters, we state, without proof, two more standard results (see [Gre-Hrp]): Theorem 1.95. Let 8J : M -+ M be a C r covering. Fix a point q E Q and map with cP(q) = 8J (is):.... If Q is a point is E M. Let cP : Q -+ M be a connected, then there is a0!Yost one lift cP : Q -+ M of cP such that cP(p) = is· If cP# (7rl (Q, q)) C 8J# (7rl (M, is)), then cP has such a lift. In particular, if Q is simply connected, then the lift exists.

C!:

Theorem 1. 96. Every connected topological manifold M has a CO simply connected covering space which is unique up to isomorphism of coverings. This is called the universal cover. Furthermo!!;, if H is any subgroup of 7rl(M,p), then there is a connected covering 8J : M -+ M and a point is EM such that 8J#(7rl(M,iS)) = H. If follows from this and Theorem 1.86 that if M is a C r manifold, then ther!.. is a unique C r structure on the universal covering space M so that 8J: M -+ M is a C r covering.

Since a deck transformation is a lift, we have the following corollary. Corollary 1.97.l;et 8J: M -+ M be a C r covering map and choose a base point p EM. If M is connected, there is at most one deck transformation cP that maps a given PI E 8J-l(p) to a given P2 E 8J-l(p). If M is simply connected, then such a deck transformation exists and is unique. Theorem 1.98. If M is the universal cover of M and 8J : M -+ M is the corresponding universal covering map, then for any base point Po EM, there is an isomorphism 7rl(M,po) ~ Deck(8J).

Proof. Fix a point is E 8J -1 (po). Let a E 7rl (M, po) and let 0: be a loop representing a. Lift to a path a starting at is. As we have seen, the point a(l) depends only on the choice of is and a = [0:]. Let cPa be the unique deck transformation such that cPa(iS) = a(l). The assignment a I-t cPa gives a map 7rl(M,po) -+ Deck(8J). For a = [0:] and b = [,B] chosen from 7rl(M,po), we have the lifts a and /3, and we see that cPa 0 /3 is a path from cPa (is) to cPa(/3(l)) = cPa(cPb (is)). Thus the path -;;; := a * (cPa 0/3) is defined. Since

40

p

0

1. Differentiable Manifolds

<Pa = p, we have p

0

~=

p

0

[a * (<Pa (3) ] 0

= (p 0 a)

* (p 0 (<Pa 0 (3) )

= (p 0 a)

* ((p 0 <Pa) 0 (3)

= (p 0 a) * (p 0 Since

0:

(3)

= 0: * (3.

* (3 represents the element ab E 1Tl(M,po), we have <Pab(]i) = ~(1) =

<Pa(
0


It is easy to see that the map a f-t <Pa is onto. Indeed, given f E Deck(p), we simply take a curve ~ from p to f(P), and then we have f =
Finally, if <Pa = id, then we conclude that any loop 0: E [0:] = a lifts to a loop based at p. But M is simply connected, and so is homotopic to a constant map to p, and its projection 0: is therefore homotopic to a constant map to p. Thus a = [0:] = 0 and so the homomorphism is 1-1. 0

a

a

1.6.2. Discrete group actions. Groups actions are ubiquitous in differential geometry and in mathematics generally. Felix Klein emphasized the role of group actions in the classical geometries (see [Klein]). More on classical geometries can be found in the online supplement [Lee, Jeff1- In this section, we discuss discrete group actions on smooth manifolds and show how they give rise to covering spaces. Definition 1.99. Let G be a group and M a set. A left group action on M is a map l : G x M -+ M such that

1) l(g2,l(gl,X)) = l(g2g1, x) for all gl,g2 E G and all x E M; 2) l(e,x) = x for all x E M, where e is the identity element of G. We often write 9 . x or just gx in place of the more pedantic notation l(g, x). Using this notation, we have g2(gIX) = (g2gdx and ex = x. Similarly, we define a right group action as a map l' : M x G -+ M with r(r(x, gl), g2) = r(x, glg2) for all gl, g2 E G and all x E M and r(x, e) = x for all x E M. In the case of right actions, we write r(x, g) as X· 9 or xg.

If l : G x M -+ M is a left action, then for every 9 E G we have a map 19 : M -+ M defined by 19(x) = l(g, x), and similarly a right action gives for every g, a map rg : M -+ M. For every result about left actions, there is an analogous result for right actions. However, mathematical conventions are such that while 9 f-t 19 is a group homomorphism from G to the group of

1.6. Coverings and Discrete Groups

41

permutations of M, the map 9 t-+ r 9 is a group anti-homomorphism which means that r91 0 r92 = r 9291 for all 91, 92 E G (notice the order reversal). Given a left action, the sets of the form Gx = {9X : 9 E G} are called orbits. The set Gx is called the orbit of x. Two points x and yare in the same orbit if and only if there is a group element 9 such that 9X = y. The orbits are equivalence classes and so they partition M. Let G\M be the set of orbits and let gJ: M -+ G\M be the projection taking each x to Gx. We give G\M the quotient topology. By definition, U is open in G\M if and only if gJ -1 (U) is open. This makes gJ continuous, but in this case, it is also an open map. To see this, let U be open. Then gJ-l (gJ(U)) is the union UgEc9U which is open and so gJ(U) is open by the definition of quotient topology. Definition 1.100. Suppose G acts on a set M by l : G x M -+ M. We say that G acts transitively if for any x, y E M there is a 9 such that 9X = y. Equivalently, the action is transitive if the action has only one orbit. We say that the action is effective provided that 19 = idM implies that 9 = e. If the action has the property that 9X = x for some x E M only when 9 = e, then we say that G acts freely (or that the action is free). In other words, an action is free provided that the only element of G that fixes any element of M is the identity element. Similar statements and definitions apply for right actions except that the orbits have the form xG. The quotient space (space of orbits) will then be denoted by MIG. Warning: The notational distinction between G\M and MIG is not universal. Example 1.101. Let gJ : M -+ M be a covering map. Fix a base point Po E M and a base point Po E gJ-1(PO). If a E 1fl(M,po), then for each x E gJ-l(PO), we define ra(x) := xa := a(1), where a is the lift of any loop a representing a. The reader may check that ra is a right action on the set p-l(PO). Example 1.102. Recall that if gJ : M -+ M is a universal C r covering map (so that M is simply connected), we have an isomorphism 1fl(M,po) -+ Deck(gJ), which we denote by a t-+ cPa. This means that l(a,x) = cPa(x) defines a left action of 1fl(M,po) on M. Let G be a group and endow G with the discrete topology so that, in particular, every point is an open set. In this case, we call G a discrete group. If M is a topological space, then we endow G x M with the product topology. What does it mean for a map a : G x M -+ M to be continuous? The topology of G x M is clearly generated by sets of the form S x U, where

42

1. Differentiable Manifolds

8 is an arbitrary subset of G and U is open in M. The map 0: : G x M -t M will be continuous if for any point (gO, xo) E G x M and any open set U c M containing o:(gO, xo) we can find an open set 8 x V containing (gO, xo) such that 0:(8 x V) c U. Since the topology of G is discrete, it is necessary and sufficient that there is an open V such that 0: (gO x V) cU. It is easy to see that a necessary and sufficient condition for 0: to be continuous on all of G x M is that the partial maps O:g := o:(g,.) are continuous for every g E G. Definition 1.103. Let G be a discrete group and M a manifold. A left discrete group action is a group action l : G x M -t M such that for every g E G the partial map 19(·) := l(g,·) is continuous. A right discrete group action is defined similarly.

It follows that if l : G x M -t M is a discrete action, then each partial map 19 is a homeomorphism with l;l = 19-I. Definition 1.104. A discrete group action is C r if M is a C r manifold and each 19 (resp. rg ) is a C r map. Example 1.105. Let ¢ : M -t M be a diffeomorphism and let ;:z act on M by n . x := ¢n(x) where

¢o := idM, ¢n := ¢

0 ... 0

¢ for n

> 0,

¢-n := (¢-l)n for n > 0. This gives a discrete action of;:Z on M. Definition 1.106. A discrete group action l : G x M -t M is said to be proper if for every two points x, y E M there are open neighborhoods Ux and Uy respectively such that the set {g E G : gUx n Uy i- 0} is finite.

There is a more general notion of proper action which we shall meet later. For free and proper discrete actions, we have the following useful characterization. Proposition 1.107. A discrete group action l : G x M -t M is proper and free if and only if the following two conditions hold:

(i) Each x

E M has an open neighborhood U such that gU

nU

=

0

for all g except the identity e. We shall call such open sets sel/avoiding.

(ii) If x, y

E M are not in the same orbit, then they have self-avoiding

neighborhoods Ux and Uy such that gUx

n Uy

=

0 for

all g E G.

Proof. Suppose that the action l is proper and free. Let x be given. We then know that there is an open V containing x such that gV n V = 0 except

1.6. Coverings and Discrete Groups

43

for a finite number of g, say, gl,' .. ,gk, which are distinct. One of these, say gl, must be e. Since the action is free, we know that for each fixed i > 1 we have giX E M\ {x}. By using continuity and then the fact that M is a regular topological space, we can replace V by a smaller open set (called V again) such that giV C M\{x} for all i = 2, ... ,k or, in other words, that x tJ. g2V U··· U gkV. Let U = V\(g2V U··· U 9kV), Notice that U is open and we have arranged that U contains x. We show that Un gU is empty unless 9 = e. So suppose 9 i= e = gl. Since UngU c VngV, we know that this is empty for sure in all cases except maybe where 9 = gi for i = 2, ... , k. If x E Un giU for such an i, then x E U and so x tJ. gi V by the definition of U. But we also have x E giU C gi V, which is a contradiction. We conclude that (i) holds. Now suppose that x, y E M are not in the same orbit. We know that there exist open sets Ux and Uy with x E Ux, Y E Uy and such that gUx n Uy is empty except possibly for some finite set of elements which we denote by gl, ... ,gk· Since the action is free, g1X, ... ,gkX are distinct. We also know that y is not equal to any of g1X, ... ,gkX, and so since M is a Hausdorff space, there exist pairwise disjoint open sets 01, ... ,Ok, Oy with giX E Oi and y E Oy. By continuity, we may shrink Ux so that giUx C Oi for all i = 1, ... , k, and then we also replace Uy with Oy n Uy (renaming this Uy again). As a result we now see that gUx n Uy = 0 for 9 = gl,"" gk and hence for all g. By shrinking the sets Ux and Uy further we may make them self-avoiding. Next we suppose that (i) and (ii) always hold for a given discrete action l. First we show that l is free. Suppose that x = gx. Then for every open neighborhood U of x the set gU n U is nonempty, which by (i) means that 9 = e. Thus the action is free. Next pick x, y E M. If x, yare not in the same orbit, then by (ii) we may pick Ux and Uy so that {g E G : gUx n Uy i= 0} is empty and so certainly a finite set. If x, yare in the same orbit, then y = gox for a unique go since we now know that the action is free. Choose an open neighborhood U of x so that gU n U = 0 for 9 i= e. Let Ux = U and Uy = goU. Then, gUxnUy = gUngoU. If gUngoU i= 0, then golgUnU i= 0 and so golg = e and 9 = go. Thus the only way that gUx n Uy is nonempty is if 9 = go and so the set {g E G : gUx n Uy i= 0} has cardinality one. In either case, we may choose Ux and Uy so that the set is finite, which is what we wanted to show. 0

It is easy to see that if U c M is self-avoiding, then any open subset V C U is also self-avoiding. Thus, if the discrete group G acts freely and properly on M, then the open sets of (i) and (ii) in the above proposition can be taken to be connected chart domains.

44

1. Differentiable Manifolds

Proposition 1.108. Let M be an n-manifold and let l : G x M -t M be a smooth discrete action which is free and proper. Then the quotient space G\M has a natural smooth structure such that the quotient map is a smooth covenng map. Proof. Giving G\M the quotient topology makes gJ : M -t G\M continuous. Using (ii) of Proposition 1.107, it is easy to show that the quotient topology on G\M is Hausdorff. By Proposition 1.107, we may cover M by charts whose domains are self-avoiding and connected. Let (U, x) be such a chart and consider the restriction gJlu. This restricted map is open since, as remarked above, gJ is an open map. If x, y E U and gJ(x) = gJ(Y), then x and yare in the same orbit and so Y = gx for some g. Therefore y E gU n U, which means that gU n U is not empty and so 9 = e since U is self-avoiding. Thus x = y and we conclude that gJlu is injective. Since gJlu is also surjective, we see that it is a bijection and hence a homeomorphism. I.e., since gJlu is also open, it has a continuous inverse and so it is a homeomorphism. Since U is connected, the connected components of gJ-l (gJ (U)) are exactly the sets gU for 9 E G. Since gJ 0 19 = gJ for all G, it is easy to see that gJ restricts to a homeomorphism on each connected component gU of gJ-l (gJ (U)). Thus, gJ(U) is evenly covered by gJ and so gJ is a covering map. For every such chart (U, x), we have a map x 0 (gJlu )-1 : gJ(U)

-t

x(U),

which is a chart on G\M. This map is clearly a homeomorphism. Given any other map constructed in this way, say yo (gJlv )-1, the domains gJ(U) and gJ(V) only meet if there is agE G such that gU meets V and 19 maps an open subset of U diffeomorphically onto a subset of V. In fact, by Exercise 1.109 below, the map (gJl v )-1 0 gJl U is defined on an open set each point of which has a neighborhood on which this map is a restriction of 19 for some g. Thus (gJlv )-1 0 gJlu is smooth and for the overlap map we have yo (gJlv) -1

0

(x 0 (gJlu )-1) -1

= yo (gJlv) -1

0

gJlu

0

x-I,

which is smooth. Thus we have an atlas on G\M and the topology induced by the atlas is the same as the quotient topology since we have already 0 established that the charts are homeomorphisms. Exercise 1.109. In the context of the proof above, show that (gJlv )-1 0 gJlu is defined on an open set 0 = (gJiu )-1 (gJ(U) n gJ(V)). Show that each x E 0 has a neighborhood on which the map (gJlv )-1 0 gJlu coincides with a restriction of a map x f--t gx for some fixed g. Conclude that (gJlv )-1 0 gJlu is a CT map. [Outline of solution: For x E 0, we must have (gJlv )-1 0 gJlu (x) = gx for some g. Now 0' = Un g-1 V is an open set that contains x. But also,

1.6. Coverings and Discrete Groups

45

r(g-l V) = r(V) so a' = Un g-l V c (rl U)-l (r(U) n r(V)). Let x' E Then since r(x') E r(U) n r(V), it follows that

a'.

(rlv)-lo plu (x') = x", where x" is the unique point in V such that r(x") = r(x'). But gx' E V and p (gx') = x" so gx' = x".] Example 1.110. We have seen the torus previously presented as T2 = Sl X Sl. Another presentation that uses a group action is given as follows: Let the group £:: x£::= £::2 act on ]R2 by

(m,n)· (x,y):= (x+m,y+n). It is easy to check that Proposition 1.108 applies to give a manifold ]R2 /£::2. This is actually the torus in another guise, and we have the diffeomorphism ¢: ]R2/£::2 ---+ Sl X Sl = T2 given by [(x,Y)]1--t (ei27rx,ei27rY). The following diagram commutes:

Exercise 1.111. Let p : M ---+ G\M be the covering arising from a free and proper discrete action of G on M and suppose that M is connected. Let fc := {lg E Diff(M) : 9 E G}; then G is isomorphic to fc by the obvious map 9 I--t 19 and furthermore fc = Deck(p).

Covering spaces r : M ---+ M that arise from a proper and free discrete group action are special in that if M is connected, then the covering is a normal covering, which means that the group Deck (p) acts transitively on each fiber p-1(p) (why?). Example 1.112. Recall the multiplicative abelian group £::2 = {I, -I} of two elements. Let £::2 act on the sphere sn C ]Rn+1 by (±1) . x := ±x. Thus the action is generated by letting -1 send a point on the sphere to its antipode. This action is also easily seen to be free and proper. The quotient space is the real projective space ]Rpn (See Example 1.41), ]Rpn = Sn /£::2.

If M is simply connected and we have a free and proper action by G as above, then we can define a map ¢ : G ---+ 7r1(G\M,bo) as follows: Fix a base point Xo E M with p(xo) = boo Given 9 E G, let, : [0,1] ---+ M be a path with ,(0) = Xo and ,(1) = gxo. Then

¢(g):= [po,]

E

7r1(G\M,bo).

1. Differentiable Manifolds

46

This is well-defined because M is simply connected. In fact, we already know from Theorem 1.98 that there is an isomorphism 1fl(G\M,bo) ~ Deck(gJ). But by Exercise 1.111, we know that G ~ Deck(gJ) by the map 9 I---t 19. Composing, we obtain an isomorphism 'ljJ : 1fl(G\M, bo) ---+ G. Recalling the definition of the isomorphism constructed in the proof of Theorem 1.98, we see that the map

The reader may wish to try to prove directly that

~

Z2.

1. 7. Regular Submanifolds A subset 5 of a smooth n-manifold M is called a regular submanifold of dimension k if every point p E 5 is in the domain of a chart (U, x) that has the following regular submanifold property with respect to 5:

x(U n 5) = x(U)

n (jRk

x {c}) for some c E

jRn-k.

Usually c is chosen to be 0, which can always be accomplished by composition with a translation of jRn. The terminology here does not seem to be quite standardized. If a subset 5 c M is covered by charts of M of the above type, then 5 itself is said to have the (regular) submanifold property. We will refer to such charts as being single-slice charts (adapted to 5). For every such single-slice chart (U, x), we obtain a chart (U n 5, xs) on 5, where Xs := pr 0 xl unS and pr : jRk X jRn-k ---+ jRk is projection onto the first k coordinates. In other words, if xl, ... ,xn are the coordinate functions of a single-slice chart, then the restrictions of xl, ... ,xk to un 5 are coordinate functions on 5. These charts provide an atlas for 5 (called a submanifold atlas) making it a smooth manifold in its own right. Indeed, one checks that the overlap maps for such charts are smooth. Exercise 1.115. Prove this last statement.

We will see more general types of sub manifolds in the sequel. An important aspect of regular submanifolds is that the topology induced by the smooth structure is the same as the relative topology. The integer n - k is called the codimension of 5 (in M), and we say that 5 is a regular sub manifold of codimension n - k.

47

1.7. Regular Submanifolds

y

Figure 1.7. Projection chart

Example 1.116. The unit sphere sn C ffi.n+1 is a regular submanifold of To see this, let Wi± := {(al, ... , an+1) E ffi.n+l : ±ai > a}. Then define 1/;; : Wi± -+ 1/;; (ffi.n+ I) by

ffi. n + l .

n'.±( I ... , an+l) -_ ( a, I ... , ai-I , aHI , ... , an+l , I a I - 1) . '{Ii a, These can easily be checked to give charts on ffi.n+l smoothly related to the standard chart. If p E S2 then p is in the domain of one of the charts 1/;;' On the other hand, identifying ffi.n+l with ffi.n x ffi., we have 1/lt(wi±nsn) = 1/;;(U)n(ffi.n x {o}) and so these charts have the submanifold property with respect to sn. Let pr : ffi.n+ I = ffi. n - l x ffi. -+ ffi.n-l. The resulting charts on sn given by pro 1/;;lw±nsn have the form (al, ... ,an+ l ) I-t h 'are proJec . t'IOns from sn ont 0 co ord'1( aI , ... , ai-I , aHI , ... , an+l) . Tese nate hyperplanes in ffi.n+1 and are easily checked to give the same smooth structure as the stereographic charts given earlier. Exercise 1.117. Show that a continuous map f : N -+ M that has its image contained in a regular submanifold S is differentiable with respect to the submanifold atlas if and only if it is differentiable as a map into M. Exercise 1.118. Show that the graph of any smooth map ffi.n -+ ffi.m is a regular submanifold of ffi.n x ffi. m . Exercise 1.119. Show that if M is a k-dimensional regular submanifold of ffi.n, then for every p EM, there exists at least one k-dimensional coordinate plane P such that the orthogonal projection ffi.n -+ P ~ ffi.k restricts to a coordinate chart for M defined on some neighborhood of p. [Hint: If (U, x) is a single-slice chart for M so that xl, ... , xk restrict to coordinates on M, then xk+l, ... , xn together give a map f : U C ffi.n = ffi.k x ffi.n-k -+ ffi.n-k

1. Differentiable Manifolds

48

such that MnU = j-l(O). Argue that if ul, ... , un are standard coordinates on JR. n , then after a suitable renumbering we must have

a( x k+l , .. . ,xn) .../.. 0 a( u k+l , .. . ,un ) r . Now use the implicit mapping theorem to show that M is locally the graph of a smooth function.]

1.8. Manifolds with Boundary For the general Stokes theorem, where the notion of flux has its natural setting, we will need to have the concept of a smooth manifold with boundary. We have already introduced the notion of a topological manifold with boundary, but now we want to see how to handle the issue of the smooth structures. Some basic two-dimensional examples to keep in mind are the upper half-plane JR.~2:o:= {(x,y) E JR.2 : y 2: O}, the closed unit disk D = {(x,y) E JR. 2 : x 2 +y2 ~ I}, and the closed hemisphere which is the set of all (x, y, z) E 8 2 with Z 2: o. Recall that in Section 1.2 we defined the closed n-dimensional Euclidean half-spaces JR.~>c := {a E JR.n : .\( a) 2: c}. Of course, JR.~-c so we are including -both JR.~kO. We also write JR.~=~ = {a E jRn : .\(a) = c}. Give JR.~>c the relative topology as a subset of JR. n . Since JR.~>c c JR. n , we already have-a notion of differentiability for a map U ---t JR. m, where U is a relatively open subset of JR.~>c. We just invoke Definition 1.58. We can extend definitions a bit more: -

Definition 1.120. Let U c JR.~ 1_ >c 1 and j : U ---t JR.:\2_ >c2 . We say that j is C r if it is C r as a map into JR.m. If both j : U ---t j(U) and j-l : j(U) ---t U are homeomorphisms of relatively open sets and C r in this sense, then j is called a C r diffeomorphism. For convenience, let us introduce for an open set U c JR.~>c (relatively open) the following notations: Let au denote JR.~=c n U and int(U) denote U \ au. In particular, aJR.~>c = JR.~=c. Notice that au is clearly an (n - 1)manifold. We have the following three facts: (1) First, let j : U C JR.n ---t JR.k be C r differentiable (with r 2: 1) and 9 another such map with the same domain. If j = 9 on JR.~>c n U, then D j(x) = Dg(x) for all x E JR.~2:c n U. (2) If j : U C JR.n ---t JR.~>c is C r differentiable (with r 2: 1) and j(x) E JR.~=c = aJR.~>c for all x E U, then Dj(x) must have its image in JR.~=O" -

1.8. Manifolds with Boundary

49

Figure 1.8. Manifold with boundary

(3) Let f : UI C ~~ 1_ >c 1 -+ U2 C ~r2_ >c 2 be a diffeomorphism (in our new extended sense). Assume that aUI and aU2 are not empty. Then f induces diffeomorphisms aUI -+ aU2 and int(U1 ) -+ int(U2). These three claims are not exactly obvious, but they are very intuitive. On the other hand, none of them is difficult to prove (see Problem 19). We can now form a definition of smooth manifold with boundary in a fashion completely analogous to the definition of a smooth manifold without boundary. A half-space chart x for a set M is a bijection of some subset U of M onto an open subset of some half-space ~~>c. A C r half-space atlas is a collection (Uo , xoJ of half-space charts such-that for any two, say (Uo , xo ) and (U{3, x{3), the map xoox~l is a C r diffeomorphism on its natural domain. Notice carefully that we allow the half-space to vary from chart to chart, but we will keep n fixed for a given M and refer to the charts as n-dimensional half-space charts.

Definition 1.121. An n-dimensional C r manifold with boundary is a pair (M, A) consisting of a set M together with a maximal atlas of ndimensional half-space charts A. The manifold topology is that generated by the domains of the charts in the maximal atlas. The boundary of M is denoted by aM and is the set of points whose image under any chart is contained in the boundary of the associated half-space. The three facts listed above show that the notion of a boundary is a welldefined concept and is a natural notion in the context of smooth manifolds; it is a "differentiable invariant" .

50

1. Differentiable Manifolds

Colloquially, one usually just refers to M as a manifold with boundary and forgoes the explicit reference to the atlas. Also we refer to an n-dimensional Coo manifold with boundary as an n-manifold with boundary. The interior of a manifold with boundary is M\oM. It is a manifold without boundary and is denoted int(M) or M. Exercise 1.122. Show that oM is a closed set in M. If no component of a manifold without boundary is compact, it is called an open manifold. For example, the interior int(M) of a connected manifold M with nonempty boundary is never compact and is an open manifold in the above sense if every component of M contains part of the boundary.

Remark 1.123. We avoid the phrase "closed manifold", which is sometimes taken to refer to a compact manifold without boundary.

Let M be an n-dimensional C r manifold with boundary and p E oM. Then by definition there is a chart (U, x) with x(p) E olR~>c' The image of the restriction xl un 8M is contained in olR~>c for some A and c depending on the chart. By composing this restriction WIth any fixed linear isomorphism olR~>c ~ IRn -l, we obtain a bijection, say X8M, of Un oM onto an open subs-;;t of IRn - 1 which provides a chart (Ua: noM, X8M) for oM. The family of charts obtained in this way is an atlas for oM. The overlaps are smooth and so we have the following: Proposition 1.124. If M is an n-manifold with boundary, then oM is an (n - 1) -manifold. Exercise 1.125. Show that the overlap maps for the atlas just constructed for oM are smooth. Exercise 1.126. The closed unit ball B(p, 1) in IR n is a smooth manifold with boundary oB(p, 1) = sn-l. Also, the closed hemisphere Sf- = {x E sn : x n+ 1 ~ O} is a smooth manifold with boundary. Exercise 1.127. Is the Cartesian product of two smooth manifolds with boundary necessarily a smooth manifold with boundary? Exercise 1.128. Show that the concept of smooth partition of unity makes sense for manifolds with boundary. Show that such exist.

51

Problems

Problems (1) Prove Proposition 1.32. The online supplement [Lee, Jeff] outlines the proof.

(2) Prove Lemma 1.11. (3) Check that the manifolds given as examples are indeed paracompact and Hausdorff. (4) Let M I , M2 and M3 be smooth manifolds. (a) Show that (MI XM2) x M3 is diffeomorphic to MI X(M2 x M 3 ) in a natural way. (b) Show that f : M ---+ MI XM2 is Coo if and only if the composite maps pri 0 f : M ---+ MI and pr2 0 f : M ---+ M2 are both Coo.

(5) Show that a CT manifold M is connected as a topological space if and only it is C T path connected in the sense that for any two points PI, P2 E M there is a CT map c: [0,1] ---+ M such that c(o) = PI and c(l) = P2.

(6) A k-frame in ffi.n is a linearly independent ordered set of vectors Show that the set of all k-frames in ffi.n can be given the structure of a smooth manifold. This kind of manifold is called a Stiefel manifold. (VI, ... , Vk)'

(7) For a product manifold M x N, we have the two projection maps pri : M x N ---+ M and pr2 : M x N ---+ N defined by (x, y) f------7 x and (x, y) f------7 y respectively. Show that if we have smooth maps h : P ---+ M and 12 : P ---+ N, then the map (J,g) : P ---+ M x N given by (J,g)(p) = (J (p), 9 (p)) is the unique smooth map such that pr I 0 (J, g) = f and pr 2 0 (J, g) = g. (8) Prove (i) and (ii) of Lemma 1.35.

(9) Show that the atlas obtained for a regular submanifold induces the relative topology inherited from the ambient manifold.

(10) The topology induced by a smooth structure is not necessarily Hausdorff: Let S be the subset of ffi.2 given by the union (ffi. x 0) U {(O, I)}. Let U be ffi. x and let V be the set obtained from U by replacing the point (0,0) by (0,1). Define a chart map x on U by x(x, 0) = x and a chart y on V by

°

y(x, 0)

= {

~

if x if x

# 0, = 0.

Show that these two charts provide a Coo atlas on S, but that the topology induced by the atlas is not Hausdorff.

52

1. Differentiable Manifolds

(11) As we have defined them, manifolds are not required to be second countable and so may have an uncountable number of connected components. Consider the set jR2 without its usual topology. For each a E jR, define a bijection cPa : jR X {a} --+ jR by cPa (x, a) = x. Show that the family of sets of the form U x {a} for U open in jR and a E jR provide a basis for a paracompact topology on jR2. Show that the maps cPa are charts and together provide an atlas for jR2 with this unusual topology. Show that the resulting smooth manifold has an uncountable number of connected components (and so is not second countable). (12) Show that every connected manifold has a countable atlas consisting of charts whose domains have compact closure and are simply connected. Hint: We are assuming that our manifolds are paracompact, so each connected component is second countable. (13) Show that every second countable manifold has a countable fundamental group (a solution can be found in [Lee, John] on page 10). (14) If C x C is identified with jR4 in the obvious way, then S3 is exactly the subset ofCxC given by {(ZI,Z2): IZ112+lz212 = I}. Letp,q be coprime integers and p > q ~ O. Let w be a primitive p-th root of unity so that Zp = {I, w, . .. , wp- l }. For (ZI' Z2) E S3, let w·(zI, Z2) := (WZl' wQz2) and extend this to an action of Zp on S3 so that wk. (ZI' Z2) = (WkZl' wQkZ2). Show that this action is free and proper. The quotient space Zp \S3 if called a lens space and is denoted by L(p; q). (15) Let SI be realized as the set of complex numbers of modulus one. Defin{ a map 0: SI xS l --+ SI xS l by O(z, w) = (-z, w) and note that 000 = id Let G be the group {id,O}. Show that M := (SI X SI) IG is a smootl 2-manifold. (16) Show that if S is a regular k-dimensional submanifold of an n-manifol( M, then we may cover S by special single-slice charts from the atlas c M which are of the form x : U --+ VI X V2 C jRk X jRn-k = jRn with

x(unS) = VI x {O} for some open sets VI C jRk, V2 C jRn-k. Show that we may arrange f( VI and V2 to both be Euclidean balls or cubes. (This problem should t easy. Experienced readers will likely see it as merely an observation.) (17) Show that

1rl (M

x N, (p, q)) is isomorphic to

1rl (M,

p) x

1rl (N,

q).

(18) Suppose that M = U U V, where U and V are simply connected ar open. Show that if UnV is path connected, then M is simply connecte (19) Prove the three properties about maps involving the model half-spac ajR~~o listed in Section 1.8.

Problems

53

Px(-l,l)

(]I)

Figure 1.9, Smoothly connecting manifolds

(20) Let !vi and N be smooth n-manifolds with boundaries aM and aN. Let P be a smooth manifold diffeomorphic to both aAf and aN via maps 0: and /3. Suppose that there are open neighborhoods U and V of aM and aN respectively and diffeomorphisms


--7

P x [0,1),


--7

P

X

(-1, 01

such that
= {u lR+ = {u

u i > 0 for i = 1,2, ... ,n},

E

lRn

E

lR n : ui ~ 0 for i = 1,2, ... ,n}.

:

A boundary point of lR+ is a point such that at least one of its coordinates u i is O. A corner point of lR+ is a point such that at least two of its coordinates ui,u j are O. We consider a set Af. An lR+-valued chart on M is a pair (U, x), where U c M and x : U --7 x (U) is a bijection onto an open subset x (U) of lR+, where the latter has the relative topology as a subset of lR n. A smooth atlas for M is a family {( Uu, xu) }uEA of lR+-valued charts whose domains cover M and such that whenever

1. Differentiable Manifolds

54

Uo:

n Uf3

is nonempty, the composite map xf3

0

x~l

:

xo:(Uo:

n Uf3)

-+ xf3(Uo:

n Uf3)

is smooth. A maximal atlas of lR+ -valued charts of this type gives M the structure of a smooth manifold with corners (of dimension n). We also use the terminology n-manifold with corners. (a) Suppose that p E M and xo:(p) is a corner point in lR+. Show that if p is in the domain of another chart (Uf3, xf3) in the atlas (as above), then xf3 (p) is also a corner point. Use this to define "corner points" on M. Do the same for "boundary points". Thus the boundary contains the set of corner points. Explain why a manifold with corners whose set of corner points is empty is a manifold with (possibly empty) boundary. (b) Define the notion of smooth functions on a manifold with corners and the notion of smooth maps between manifolds with corners. (c) Show that the boundary of a manifold with corners is not necessarily a manifold with corners. (22) Prove Theorem 1.113 and its corollary.

Chapter 2

The Tangent Structure

In this chapter we introduce the notions of tangent space and cotangent space of a smooth manifold. The union of the tangent spaces of a given manifold will be given a smooth structure making this union a manifold in its own right, called the tangent bundle. Similarly we introduce the cotangent bundle of a smooth manifold. We then discuss vector fields and their integral curves together with the associated dynamic notions of Lie derivative and Lie bracket. Finally, we define and discuss the notion of a l-form (or covector field), which is the notion dual to the notion of a vector field. One can integrate l-forms along curves. Such an integration is called a line integral. We explore the concept of exact l-forms and nonexact l-forms and their relation to the question of path independence of line integrals.

2.1. The Tangent Space If c : (-€, €) -+ ~N is a smooth curve, then it is common to visualize the "velocity vector" c(O) as being based at the point p = c(O). It is often desirable to explicitly form a separate N-dimensional vector space for each point p, whose elements are to be thought of as being based at p. One way to do this is to use {p} x ~N so that a tangent vector based at p is taken to be a pair (p, v) where v E ~N. The set {p} x ~N inherits a vector space structure from ~N in the obvious way. In this context, we provisionally denote {p} x ~N by Tp~N and refer to it as the tangent space at p. If we write c(t) = (xl(t), .. . , x N (t)), then the velocity vector of a curve c at time 1 t = 0 is (p, ddt1 (0), ... , dd,; (0)), which is based at p = c(O). Ambiguously, both 1 It is common to refer to the parameter t for a curve as "time" , although it may have nothing to do with physical time in a given situation.

-

55

2. The Tangent Structure

56

d:/

(d:

1 (p, (0), ... , d~; (0)) and t (0), ... , d~; (0)) are often denoted by c(O) or c'(O). A bit more generally, if V is a finite-dimensional vector space, then V is a smooth manifold and the tangent space at p E V can be provisionally taken to be the set {p} x V. We use the notation vp := (p, v). If vp := (p, v) is a tangent vector at p, then v is called the principal part of vp.

We have a natural isomorphism between ]RN and Tp]RN given by v r--+ (p, v), for any p. Of course we also have a natural isomorphism Tp]RN ~ Tq]RN for any pair of points given by (p,v) r--+ (q,v). This is sometimes referred to as distant parallelism. Here we see the reason that in the context of calculus on ]RN, the explicit construction of vectors based at a point is often deemed unnecessary. However, from the point of view of manifold theory, the tangent space at a point is a fundamental construction. We will define the notion of a tangent space at a point of a differentiable manifold, and it will be seen that there is, in general, no canonical way to identify tangent spaces at different points. Actually, we shall give several (ultimately equivalent) definitions of the tangent space. Let us start with the special case of a submanifold of ]RN. A tangent vector at p can be variously thought of as the velocity of a curve, as a direction for a directional derivative, and also as a geometric object which has components that depend in a special way on the coordinates used. Let us explore these aspects in the case of a submanifold of ]RN. If M is an n-dimensional regular submanifold of ]RN, then a smooth curve c: (-E, E) ---+ M is also a smooth curve into]RN and C(O) is normally thought of as a vector based at the point p = c( 0). This vector is tangent to M. The set of all vectors obtained in this way from curves into M is an n-dimensional subspace of the tangent space of]RN at p (described above). In this special case, this subspace could play the role of the tangent space of M at p. Let us tentatively accept this definition of the tangent space at p and denote it by TpM. Let vp := (p, v) E TpM. There are three things we should notice about vp. First, there are many different curves c : (-E, E) ---+ M with c(O) = p which all give the same tangent vector vp , and there is an obvious equivalence relation among these curves: two curves passing through p at t = 0 are equivalent if they have the same velocity vector. Already one can see that perhaps this could be turned around so that we can think of a tangent vector as an equivalence class of curves. Curves would be equivalent if they agree infinitesimally in some appropriate sense. The second thing that we wish to bring out is that a tangent vector can be used to construct a directional derivative operator. If vp = (p, v) is a tangent vector in Tp]RN, then we have a directional derivative operator at p which is a map coo(]RN) ---+ ]R given by f r--+ Df(p)v. Now if vp is tangent

2.1. The Tangent Space

57

to M, we would like a similar map COO(M) -t K If f is only defined on M, then we do not have D f to work with but we can just take our directional derivative to be the map given by

Dvp : f

t--t

(f

0

c)' (0),

where c : I -t M is any curve whose velocity at t = 0 is vp. Later we use the abstract properties of such a directional derivative to actually define the notion of a tangent vector. Finally, notice how vp relates to charts for the submanifold. If (U, y) is a chart on M with p E U, then by inverting we obtain a map y-1 : V -t M, which we may then think of as a map into the ambient space ]RN. The map y-1 parameterizes a portion of M. For convenience, let us suppose that y-1(0) = p. Then we have the "coordinate curves" yi t--t y-1(0, ... , yi, ... , 0) for i = 1, ... , n. The resulting tangent vectors Ei at p have principal parts given by the partial derivatives so that 8y-1 ) Ei:= ( p, 8yi (0) . It can be shown that (E 1 , . .. , En) is a basis for TpM. For another coordinate system y with y-1(0) = p, we similarly define a basis (E1, ... , En). If Vp = "n L."i=l aiEi = "n L."i=l a-iE-i, th en 1ett·lng a = (a 1, ... , an) an d a- = (-1 a , ... , a-n) , the chain rule can be used to show that

a = D(y 0

y-1) Iy(p) a,

which is classically written as n 8- i -i _ " Y j a-~8ja.

j=l

y

Both (a 1, ... , an) and (a 1, ... , an) represent the tangent vector vp , but with respect to different charts. This is a simple example of a transformation law. The various definitions for the notion of a tangent vector given below in the general setting will be based in turn on the following three ideas: (1) Equivalence classes of curves through a point. (2) Transformation laws for the components of a tangent vector with respect to various charts. (3) The idea of a "derivation" which is a kind of abstract directional derivative. Of course we will also have to show how to relate these various definitions to see that they are really equivalent.

2.1.1. Tangent space via curves. Let p be a point in a smooth nmanifold M. Suppose that we have smooth curves C1 and C2 mapping into M, each with open interval domains containing 0 E lR and with C1 (0) = C2 (0) = p. We say that C1 is tangent to C2 at p if for all smooth real-valued functions

2. The Tangent Structure

58

f defined on an open neighborhood of p, we have (f 0 cd (0) = (f 0 cd (0). This is an equivalence relation on the set of all such curves. The reader should check that this really is an equivalence relation and also do so when we introduce other simple equivalence relations later. Define a tangent vector at p to be an equivalence class under this relation. Notation 2.1. The equivalence class of c will be denoted by [c], but we also denote tangent vectors by notation such as vp or X p , etc. Eventually we will often denote tangent vectors simply as v, W, etc., but for the discussion to follow we reserve these letters without the subscript for elements of lRn for some n.

If vp = [cl then we will also write C(O) = vp. The tangent space TpM is defined to be the set of all tangent vectors at p EM. A simple cut-off function argument shows that Cl is equivalent to C2 if and only if (f 0 cd (0) = (f 0 C2)' (0) for all globally defined smooth functions f : M -+ Itt Lemma 2.2. CI is tangent to C2 at p if and only if (f 0 cd (0) = (f 0 for alllR k -valued functions f defined on an open neighborhood of p.

cd (0)

Proof. If f = (fl, ... , r), then (f 0 cd (0) = (f 0 C2)' (0) if and only if (t 0 Cl)' (0) = (Ii 0 C2)' (0) for i = 1, ... ,k. Thus (f 0 cd (0) = (f 0 cd (0) if Cl is tangent to C2 at p. Conversely, let 9 be a smooth real-valued function defined on an open neighborhood of p and consider the map f = (g, 0, ... ,0). Then the equality (f 0 cd (0) = (f 0 cd (0) implies that (g 0 cd (0) (g 0 cd (0). 0

The definition of tangent space just given is very geometric, but it has one disadvantage. Namely, it is not immediately obvious that TpM is a vector space in a natural way. The following principle is used to obtain a vector space structure: Proposition 2.3 (Consistent transfer of linear structure). Suppose that S is a set and {VaJaEA is a family of n-dimensional vector spaces. Suppose that for each a we have a bijection ba : Va -+ S. If for every a, (3 E A the map b~1 oba : Va -+ V,a is a linear isomorphism, then there is a unique vector space structure on the set S such that each ba is a linear isomorphism. Proof. Define addition in S by SI + S2 := ba (b;;I(SI) definition is independent of the choice of a. Indeed,

ba (b;;I(Sl)

+ b;;1(S2))

=

ba [b;;1

0

b,a 0 b~l(SI)

+ b;;1 0

b,a [b~l(Sl)

0

= b,a

(b~l(Sl) + b~1(S2)) .

0

b,a 0 b~1(s2)l

+ b~1(S2)]

= ba

b;;l

+ b;;I(S2))'

This

2.1. The Tangent Space

59

The definition of scalar multiplication is a· s := ba(ab~l(s)), and this is shown to be independent of a in a similar way. The axioms of a vector 0 space are satisfied precisely because they are satisfied by each Va. We will use the above proposition to show that there is a natural vector space structure on TpM. For every chart (xa, Ua) with p E U, we have a map ba : ]Rn -+ TpM given by v H bv]' where "tv : t H x~l(xa(P) + tv) for t in a sufficiently small but otherwise irrelevant interval containing o. Lemma 2.4. For each chart (Ua, xa ), the map ba : ]Rn -+ TpM is a bijection and b~l 0 ba = D (Xfj 0 x~l) (xa(P)). Proof. We have

(xa 0 "tv)' (0) = dd I Xa t t=O

= dd I t

0

x~l(xa(P) + tv)

(xa(P)

+ tv)

=

v.

t=o

Suppose that bv] = bw] for v, wE ]Rn. Then by Lemma 2.2 we have v

= (xa 0 "tv)' (0) = (xa 0 "tw)' (0) = w.

This means that ba is injective. Next we show that ba is surjective. Let [c] E TpM be represented by c : (-E, E) -+ M. Let v := (xa 0 c)' (0) E ]Rn. Then we have bQ(v) = bv]' where "tv : t H X~ 1 (Xa (p) + tv). But bv] = [c] since for any smooth f defined near p we have

I f 0 X~ 1 (Xa (p) + tv) = D (J 0 X~ 1 ) (Xa (p)) . v t t=O D (J 0 x~l) (xa(p))· (XQ 0 c)' (0) = (J 0 c)' (0).

(J 0 "tv)' (0) = dd =

Thus bQ is surjective. From Lemma 2.2 we see that the map [c] H (xQ 0 c)' (0) is well-defined, and from the above we see that this map is exactly b~l. Thus

b~l

0

bQ(v) = dd I xfj t t=O

0

x~l(xQ(p) + tv)) = D (Xfj x~l) (xa(P))v. 0

0

The above lemma and proposition combine to provide a vector space structure on the set of tangent vectors. Let us temporarily call the tangent space defined above, the kinematic tangent space and denote it by (TpM)kin. Thus, if Cp is the set of smooth curves c defined on some open interval containing 0 such that c(O) = p, then

(TpM)kin = Cp /"', where the equivalence is as described above.

60

2. The Tangent Structure

Exercise 2.5. Let CI and C2 be smooth curves mapping into a smooth manifold !v!, each with open interval domains containing 0 E lR. and with CI (0) = C2(0) = p. Show that

(J

0

cd (0) = (J

0

C2)' (0)

for all smooth f if and only if the curves x 0 CI and x 0 C2 have the same velocity vector in lR. n for some and hence any chart (U, x). 2.1.2. Tangent space via charts. Let A be the maximal atlas for an nmanifold M. For fixed p EM, consider the set r p of all triples (p, v, (U, x)) E {p} x lR. n x A such that p E U. Define an equivalence relation on r p by requiring that (p, v, (U, x)) '"" (p, w, (V, y)) if and only if

(2.1)

w =

D(y 0 x-I) Ix(p) . v.

In other words, the derivative at x(p) of the coordinate change yo x-I "identifies" v with w. The set r pi "-' of equivalence classes can be given a vector space structure as follows: For each chart (U, x) containing p, we have a map b(u,x) : lR. n -+ rpl"-' given by v H [P,v,(U,x)], where [P,v,(U,x)] denotes the equivalence class of (p, v, (U, x) ). To see that this map is a bijection, notice that if [p, v, (U, x)] = [p, w, (U, x)], then v = D(x 0

x-1)lx(p) .

v= w

by definition. By Proposition 2.3 we obtain a vector space structure on tangent vectors. This is another version of the tangent space at p, and we shall (temporarily) denote this by (TpM)phys. The subscript "phys" refers to the fact that this version of the tangent space is based on a "transformation law" and corresponds to a way of looking at things that has traditionally been popular among physicists. If vp = [p, v, (U, x)] E (TpM)phys' then we say that v E lR.n represents vp with respect to the chart (U, x).

r pi "-' whose elements are

This viewpoint takes on a more familiar appearance if we use a more classical notation. Let (U, x) and (V, y) be two charts containing p in their domains. If an n-tuple (vI, ... , v n ) represents a tangent vector at p from the point of view of (U, x), and if the n-tuple (wI, ... , w n ) represents the same vector from the point of view of (V, y), then (2.1) is expressed in the form (2.2)

i _

~

fJyi

w - L.t fJxj j=l

I

j

v ,

x(p)

where we write the change of coordinates as yi = yi(xl, ... ,xn) with 1 < i So n. Notation 2.6. It is sometimes convenient to index the maximal atlas: A = {(Uo, xoJ }OEA. Then we would consider triples of the form (p, v, ex) and let

61

2.1. The Tangent Space

the defining equivalence relation for (TpM)phys be (p, v, 0:) '" (p, w, j3) if and only if D(x(3 0 x~l) Ixa(p) . v = w. 2.1.3. Tangent space via derivations. We abstract the notion of directional derivative for our next approach to the tangent space. There are actually at least two common versions of this approach, and we explain both. Let M be a smooth manifold of dimension n. A tangent vector vp at p is a linear map vp : COO(M) ---+ ~ with the property that for f, 9 E COO(M),

vp(fg) = g(p)vp (f)

+ f(p)vp (g).

This is the Leibniz law. We may say that a tangent vector at p is a derivation of the algebra COO(M) with respect to the evaluation map evp at p defined by eVp(f) := f(p). Alternatively, we say that vp is a derivation at p. The set of such derivations at p is easily seen to be a vector space which is called the tangent space at p and is denoted by TpM. We temporarily distinguish this version of the tangent space from (TpM)kin and (TpM)phys defined previously by denoting it (TpM)alg and referring to it as the algebraic tangent space. We could also consider the vector space of derivations of C r (M) at a point for r < 00, but this would not give a finitedimensional vector space and so is not a good candidate for the definition of the tangent space (see Problem 18). Recall that if (U, x) is a chart on an n-manifold M, we have defined af /ax i by

af axi(p):= Di

(fox-I) (x(p))

(see Definition 1.57). Definition 2.7. Given (U, x) and p as above, define the operator 8~i Ip :

It is often helpful to use the easily verified fact that if ci is the curve defined for sufficiently small E by

Ci(t)

:=

x-l(x(p)

:

(-E, E) ---+ M

+ ei),

where ei is the i-th member of the standard basis of ~n, then

~I f = lim f(Ci(h)) - f(p). ax t p h--+O h From the usual product rule it follows that 8~i Ip is a derivation at p and so is an element of (TpM)alg. We will show that for the vector space (TpM)alg.

(-£r Ip , ... , 8~n Ip) is a basis

62

2. The Tangent Structure

Lemma 2.8. Let vp E (TpM)alg. Then

(i) if f, g E coo(M) are equal on some neighborhood of p, then vp (f) = vp (g); (ii) if hE coo(M) is constant on some neighborhood ofp, then vp (h) =

O. Proof. (i) Since vp is a linear map, it suffices to show that if f = 0 on a neighborhood U of p, then vp (f) = O. Of course vp(O) = O. Let;3 be a cut-off function with support in U and ;3(p) = 1. Then we have that ;3f is identically zero and so 0= vp(;3J) = f(p)vp (;3)

= vp (f)

+ ;3(p)vp (f)

(since ;3(p) = 1 and f(p) = 0).

(ii) From what we have just shown, it suffices to assume that h is equal to a constant c globally on M. In the special case c = 1, we have

Vp (1) = vp(I·I) = 1· vp (1)

+ 1· vp (1) =

2vp (1),

so that vp (1) = O. Finally we have vp (c) = vp (lc) = c (vp(I)) = O.

D

Notation 2.9. We shall often write vpf or vp' f in place of vp (f).

We must now deal with a technical issue. We anticipate that the action of a derivation is really a differentiation and so it seems that a derivation at p should be able to act on a function defined only in some neighborhood U of p. It is pretty easy to see how this would work for a~' Ip' But the domain of a derivation as defined is the ring coo(M) and not Coo(U). There is nothing in the definition that immediately allows an element of (TpM)alg to act on Coo(U) unless U = M. It turns out that we can in fact identify (TpU)alg with (TpM)alg' and the following discussion shows how this is done. Once we reach a fuller understanding of the tangent space, this identification will be natural and automatic. So, let p E U C M with U open. We construct a rather obvious map : (TpU)alg -+ (TpM)alg by using the restriction map coo(M) -+ Coo(U). For each wp E TpU, we define wp : coo(M) -+ lR by wp(f) := wpUlu). It is easy to show that wp is a derivation of the appropriate type and so wp E (TpM)alg' Thus we get a linear map : (TpU)alg -+ (TpM)alg' We want to show that this map is an isomorphism, but notice that we have not yet established the finite-dimensionality of either (TpU)alg or (TpM)alg' First we show that : wp f--t wp has trivial kernel. So suppose that wp = 0, i.e. wp(f) = 0 for all f E COO (M). Let h E Coo(U). Pick a cut-off function ;3 with support in U so that ;3h extends by zero to a smooth function f on all of M that agrees with h on a neighborhood of

2.1. The Tangent Space

63

p. Then by the above lemma, wp(h) = wp(flu) = wp(f) = O. Thus, since h was arbitrary, we see that wp = 0 and so

.

.

In eIther case the formula IS the same:

a Ip f

ax;

=

a(Jox- 1 ) au;

(x(p)).

Notice that agreeing on a neighborhood of a point is an important relation here and this provides motivation for employing the notion of a germ of a function (Definition 1.68). First we establish the basis theorem: Theorem 2.10. Let M be an n-manifold and (U, x) a chart with p E U.

Then the n-tuple of vectors (derivations) (a~llp"'" a~n Ip) is a basis for (TpM)alg' Furthermore, for each vp E (TpM)alg we have

Proof. From our discussion above we may assume that x(U) is a convex set such as a ball of radius E in ]Rn. By composing with a translation we assume that x(p) = O. This makes no difference for what we wish to prove since vp applied to a constant is O. For any smooth function 9 defined on the convex set x(U) let

gi(U):=

(log oui (tu) dt for all u

Jo

E

x(U).

The fundamental theorem of calculus can be used to show that 9

EgiUi. We see that gi(O) =

?url o'

= g(O) +

For a function f E COO(U), we let

9 := f 0 x-I. Using the above, we arrive at the expression f = f (p) + E fiXi , and applying aa; I we get Ji(p) = !!.La ax ; Ip . Now apply the derivation vp to x p

64

J = J(p)

2. The Tangent Structure

+ 2: Jix i to obtain vpJ = 0 + =

L vp(Jixi)

L vp(xi)Ji(p) + L OVpJi

" ' Vp(x i ) = " ~

This shows that vp =

2: vp(xi)

aJI p. axi

8~i Ip and thus we have a spanning set.

To see that (8~i Ip ' .. · , 8~i Ip) is a linearly independent set, let us assume that 2: a i 8~i Ip = 0 (the zero derivation). Applying this to x j gives 0 =

2: ai ~~~ Ip = 2: ai o1 = aj , and since j

was arbitrary, we get the result.

D

Remark 2.11. On the manifold JR. n , we have the identity map id : JR.n~ JR.n which gives the standard chart. As is often the case, the simplest situations have the most confusing notation because of the various identifications that may exist. On JR., there is one coordinate function, which we often denote by either u or t. This single function is just idjR. The basis vector at to E JR. associated to this coordinate is Ju Ito (or gt Ito)· If we think of the tangent space at to E JR. a..<; being {to} x JR., then Ju Ito is just (to, 1). It is also common to denote Ju Ito by "I" regardless of the point to· Above, we used the notion of a derivation as one way to define a tangent vector. There is a slight variation of this approach that allows us to worry a bit less about the relation between (TpU)alg and (TpM)alg. Let Fp = C~(M, JR.) be the algebra of germs of functions defined near p. Recall that if J is a representative for the equivalence class [f] E F p , then we can unambiguously define the value of [J] at p by [J](p) = J(p). Thus we have an evaluation map evp : Fp ~ JR..

Definition 2.12. A derivation (with respect to the evaluation map evp) of the algebra Fp is a map Dp : Fp ~ JR. such that Dp([J][g]) = J(p)Dp[g] + g(p)Dp[J] for all [f], [g] E Fp. The set of all these derivations on Fp is easily seen to be a real vector space and is sometimes denoted by Der(Fp).

Remark 2.13. The notational distinction between a function and its germ at a point is not always maintained; DpJ is taken to mean Dp[J]. Let M be a smooth manifold of dimension n. Consider the set of all germs of Coo functions Fp at p EM. The vector space Der(Fp) of derivations of Fp with respect to the evaluation map evp could also be taken as the definition of the tangent space at p. This would be a slight variation of what we have called the algebraic tangent space.

2.2. Interpretations

65

2.2. Interpretations We will now show how to move from one definition of tangent vector to the next. Let M be a (smooth) n-manifold. Consider a tangent vector vp as an equivalence class of curves represented by c : I --+ M with c(O) = p. We obtain a derivation by defining

vp ! := dd

t

!

I

0

c.

t=O

This gives a map (TpM)kin --+ (TpM)alg which can be shown to be an isomorphism. We also have a natural isomorphism (TpM)kin --+ (TpM)phYs, Given [c] E (TpM)kin' we obtain an element vp E (TpM)phys by letting vp be the equivalence class of the triple (p, v, (U, x)), where vi := It=o xi 0 c for a chart (U, x) with p E U.

1t

If vp is a derivation at p and (U, x) an admissible chart with domain containing p, then vp , as a tangent vector in the sense of Definition 2.1.2, is represented by the triple (p,v, (U,x)), where v = (vI, ... ,vn) is given by

vi

=

vpx i

('Up is acting as a derivation).

This gives us an isomorphism (TpM)alg --+ (TpM)phys· Next we exhibit the inverse isomorphism (TpA1)phys --+ (TpM)alg. Suppose that [(p, v, (U, x))] E (TpM)phys where v E ]Rn. We obtain a derivation by defining vp ! = D(f 0 x-I) Ix(p) . v. In other words,

v p! =

L vi 8x8 i I ! n

i=1

P

for v = (vI, ... , vn). It is an easy exercise that vp defined in this way is independent of the representative triple (p, v, (U, x)). We now adopt the explicitly flexible attitude of interpreting a tangent vector in any of the ways we have described above depending on the situation. Thus we effectively identify the spaces (TpM)kin' (TpM)phys and (TpM)alg. Henceforth we use the notation TpM for the tangent space of a manifold M at a point p. Definition 2.14. The dual space to a tangent space TpM is called the cotangent space and is denoted by M. An element of M is referred to as a covector.

T;

The basis for

T;

T; M that is dual to the coordinate basis (-£r I

p "

described above is denoted (dx1Ip" .. , dxnlp)' By definition dx i Ip 6j. (Note that 8}

= 1 if i =

j and 8}

= 0 if i i

.. ,

{)~n

I)

({)~j Ip) =

j. The symbols 8ij and

2. The Tangent Structure

66

8ij are defined similarly.) The reason for the differential notation dx i will be explained below. Sometimes one abbreviates 8~j Ip and dx i Ip to 8~j and

dx i respectively, but there is some risk of confusion since later 8~j and dx i will more properly denote not elements of the vector spaces TpM and T; M, but rather fields defined over a chart domain. More on this shortly.

2.2.1. Tangent space of a vector space. Our provisional definition of the tangent space at point p in a vector space V was the set {p} x V, but this set does not immediately fit any of the definitions of tangent space just given. This is remedied by finding a natural isomorphism {p} x V ~ Tp V. One may pick a version of the tangent space and then exhibit a natural isomorphism directly, but we take a slightly different approach. Namely, we first define a natural map JP : V -+ Tp V. We think in terms of equivalence classes of curves. For each v E V, let cp ,v : lR -+ V be the curve cp ,v (t) := p + tv. Then Jp( v) := [cp,v] E Tp V.

As a derivation, Jp( v) acts according to

J t---+

:t 10 J

(p

+ tv) .

On the other hand, we have the obvious projection pr2 : {p} x V -+ V. Then our natural isomorphism {p} x V ~ Tp V is just JP 0 pr2' The isomorphism between the vector spaces {p} x V and Tp V is so natural that they are often identified. Of course, since Tp V itself has various manifestations ((Tp V)alg' (Tp V)phys' and (Tp V)kin' we now have a multitude of spaces which are potentially being identified in the case of a vector space. Note: Because of the identification of {p} x V with Tp V, we shall often denote by pr2 the map Tp V -+ V. Furthermore, in certain contexts, Tp V is identified with V itself. The potential identifications introduced here are often referred to as "canonical" or "natural" and JP : V -+ Tp V is often called the canonical or natural isomorphism. The inverse map Tp V -+ V is also referred to as the canonical or natural isomorphism. Context will keep things straight.

2.3. The Tangent Map The first definition given below of the tangent map at p E M of a smooth map J : M -+ N will be considered our main definition, but the others are actually equivalent. Given J and p as above, we wish to define a linear map TpJ : TpM -+ Tf(p)N. Since we have several definitions of tangent space, we expect to see several equivalent definitions of the tangent map. For the first definition we think of TpM as (TpM)kin'

2.3. The Tangent Map

67

Figure 2.1. Tangent map

Definition 2.15 (Tangent map I). If we have a smooth function between manifolds

J:

M -+ N,

and we consider a point p E M and its image q = the tangent map at p,

J(p)

EN, then we define

TpJ : TpM -+ TqN, in the following way: Suppose that vp E TpM and we pick a curve c with c(O) = p so that vp = [c]; then by definition

TpJ . vp = [J 0 c] E TqN, where [J 0 c] E TqN is the vector represented by the curve J Notation 1.3.)

0

c. (Recall

Another popular way to denote the tangent map TpJ is Jp*, but a further abbreviation to J* is dangerous since it conflicts with a related meaning for f* introduced later. Exercise 2.16. Let J : M -+ N be a smooth map. Show that TpJ : TpM -+ TqN is a linear map and that if J is a diffeomorphism, then TpJ is a linear isomorphism. [Hint: Think about how the linear structure on (TpM)kin was defined.] We have the following version of the chain rule for tangent maps: Theorem 2.17. Let J : M -+ Nand 9 : N -+ P be smooth maps. For each p EM we have Tp(g 0 f) = (Tf(p)g) 0 TpJ.

2. The Tangent Structure

68

Proof. Let v E TpM be represented by the curve c so that v = [c]. Then J 0 c represents TpJ(v) and we have

Tp(g 0 f)(v) = [(g 0 f) =

0

c]

=

[g 0 (f 0 c)]

(Tf(p)g) (TpJ(v)) = ((Tf(p)g) o TpJ) (v).

D

For the next alternative definition of tangent map, we consider TpM as

(TpM)phys· Definition 2.18 (Tangent map II). Let J : M ---t N be a smooth map and consider a point p E M with image q = J(p) E N. Choose any chart (U, x) containing p and a chart (V, y) containing q = J(p) so that for vp E TpM we have the representative (p, v, (U, x)). Then the tangent map TpJ : TpM ---t Tf(p)N is defined by letting the representative of TpJ . vp in the chart (V, y) be given by (q, w, (V, y)), where w

= D(y 0 J 0 x-I) . v.

This uniquely determines TpJ . v, and the chain rule guarantees that this is well-defined (independent of the choice of charts). Another alternative definition of tangent map is given in terms of derivations:

Definition 2.19 (Tangent map III). Let M be a smooth n-manifold. Continuing our set up above, we define TpJ . vp as a derivation by

(TpJ . vp)g = vp(g 0 f) for each smooth function g. It is easy to check that this defines a derivation and so a tangent vector in TqM. This map is yet another version of the tangent map TpJ. In the above definition, one could take 9 to be the germ of a smooth function defined on a neighborhood of J(p) and then TpJ . vp would act as a derivation of such germs. One can check the chain rule for tangent maps using the above definition in terms of derivations as follows: If J : M ---t Nand 9 : N ---t P are smooth maps and vp E TpM, then for h E Coo (M) we have

(Tp(g

0

f). vp)h = vp(h 0 (g

0

f)) = vp((h 0 g)

= (TpJ . vp) (h 0 g)

=

0

f)

Tf(p)g . (TpJ . vp) h,

so since hand vp were arbitrary, we conclude again that Tp(gof) = (Tf(p)g)o TpJ. We have just proved the same thing using different interpretations of tangent space and tangent map, but the isomorphisms between the versions are so natural that we may use any version convenient for a given purpose and then draw conclusions about all versions. This could be formalized

2.3. The Tangent Map

69

using category theory arguments but we shall forgo the endeavor. Just as with the idea of the tangent space, we will think of the above versions of the tangent map as a single abstract thing with more than one interpretation depending on the interpretation of tangent space in play. Now we introduce the differential of a function.

Definition 2.20. Let M be a smooth manifold and let p E M. For f E Coo (M), we define the differential of f at p as the linear map df (p) TpM ---+ ~ given by

df(p) . vp = vpf for all vp E TpM. Thus df(p) E T* M. The notation dfp or dflp is also used in place of df(p). One may view dfp(vp) as an "infinitesimal" aspect of the composition f H f where ,'(0) = vp. It is easy to see that df(p) is just Tpf followed by the natural map Tf(p)~ ---+ R In a way, df(p) is just a version of the tangent map that takes advantage of the identification of Tf(p)~ with ~ (recall Remark 2.11).

0"

Let (U, x) be a denoted the basis now this notation (~I p have dxil pax]

chart with x = (xl, ... ,xn) and let p E U. We previously dual to (lxr(p)""'o~n(P)) by (dxllp, ... ,dxnlp), and is justified since we can check directly that we really do

)=

<5Ji .

Definition 2.21. Let I = (a, b) be an interval in IR. If c : I ---+ M is a smooth map (a curve), then the velocity at to E I is the vector c(to) E Tc(to)M defined by

c(to)

:=

Tto

c,:u Ito'

tu

where Ito is the coordinate basis vector at to E Ttol = Tto~ associated to the standard coordinate function on ~ (denoted here by u).

Note: We may also occasionally write c' for

c.

Thus if f is a smooth function defined in a neighborhood of c(to), then c(to) acts as a derivation as follows: . to) . f = dt dI c(

f

0

c;

t=to

c(to) is also denoted by ft Ito c. In this notation, to every tEl the tangent vector c( t) := Ttc·

c = ftc is a map that assigns

fu It'

and this is referred to as the velocity field along the curve c. Also note that we may view c(to) as the equivalence class of the curve t H c( t + to).

2. The Tangent Structure

70

The differential can be generalized: Definition 2.22. Let V be a vector space. For a smooth J : M --+ V with p E M as above, the differential dJ(p) : TpM --+ V is the composition of the tangent map TpJ and the canonical map Ty V --+ V where y = J(p),

The notational distinction between TpJ and dJp is not universal, and dJp is itself often used to denote Tpj. Exercise 2.23. Let J : M --+ N be a smooth map. Show that if TpJ = 0 for all p E M, then J is locally constant (constant on connected components of M).

We now consider the inclusion map L : U Y .Nf where U is open. For p E U, we get the tangent map TpL : TpU --+ TpM. Let us look at this map from several points of view corresponding to the various ways one can define the tangent space. First, consider tangent spaces from the derivation point of view. From this point of view the map TpL is defined for vp E TpU as acting on COO(M) as follows: TpL(Vp)J = vp (J 0 L) = vp (II U). We have seen this map before where we called it : (TpU)alg --+ (TpM)alg' and it was observed to be an isomorphism and we decided to identify (TpU)alg with (TpM)alg. From the point of view of equivalence classes of curves, the map TpL sends h'J to [L 0 ,),J. But while')' is a curve into U, the map L 0 ')' is simply the same curve, but thought of as mapping into M. We leave it to the reader to verify the expected fact that TpL is a linear isomorphism. Thus it makes sense to identify h'J with [L 0 ')'J and so again to identify TpU with TpM via this isomorphism. Next consider vp E TpU to be represented by a triple (p, v, (UCO xaJ) where (UC\:, xC\:) is a chart on the open manifold U. Since (UC\:' xC\:) is also a chart on M, the triple also represents an element of TpM which is none other than TpL' vp. The map TpL looks more natural and trivial than ever, and we once again see the motivation for identifying TpU and TpM. More generally, when S is a regular submanifold of M, then the tangent space TpS at pES C M is intuitively a subspace of TpM. Again, this is true as long as one is not bent on distinguishing a curve in S through p from the "same" curve thought of as a map into M. If one wants to be pedantic, then we have the inclusion map L : S y M, and if c : I --+ S is a curve into S, then L 0 C : I --+ M is a map into M. At the tangent level this means that C(O) E TpS while (L 0 c)'(O) E TpM. Thus TpL : TpS --+ TpL(TpS) c TpM. When convenient, we just identify TpS with TpL(TpS) and so think of TpS as a subspace of TpM and take TpL to be an inclusion.

2.3. The Tangent Map

71

Recall that a smooth pointed map f : (M,p) ---+ (N, q) is a smooth map f: M ---+ N such that f(p) = q. Taking the set of all pairs (M,p) as objects and pointed maps as morphisms we have an obvious category. Definition 2.24. The "pointed" version of the tangent functor T takes (M,p) to TpM and a map f : (M,p) ---+ (N, q) to the linear map Tpf :

TpM ---+ TqM. The following theorem is the inverse mapping theorem for manifolds and will be used repeatedly. Theorem 2.25. If f : M ---+ N is a smooth map such that Tpf : TpM ---+ TqN is an isomorphism, then there exists an open neighborhood 0 of p such that f(O) is open and flo: 0 ---+ f(O) is a diffeomorphism. If Tpf is an isomorphism for all p EM, then f : M ---+ N is a local diffeomorphism. Proof. The proof is a simple application of the inverse mapping theorem, Theorem C.l found in Appendix C. Let (U, x) be a chart centered at p and let (V, y) be a chart centered at q = f(p) with f(U) c V. From the fact that Tpf is an isomorphism we easily deduce that M and N have the same dimension, say n, and then D(y 0 f 0 x-I)(x(p)) : IR n ---+ IR n is an isomorphism. From Theorem C.l if follows that yo f 0 x-I restricts to a diffeomorphism on some neighborhood 0' of 0 E IRn. From this we obtain that flo: 0 ---+ f(O) is a diffeomorphism where 0 = x-I (0'). The second part follows from the first. 0 2.3.1. Tangent spaces on manifolds with boundary. Recall that a manifold with boundary is modeled on the half-spaces 1R~>c := {a E IR n : oX (a) 2: c}. If M is a manifold with boundary, then the tangent space TpM is defined as before. For instance, even if p E 8M, the fiber TpM may still be thought of as consisting of equivalence classes where (p, v, a) '" (p, w, (3) if and only if D(x{3 0 x~l) Ixa(P) . v = w. Notice that for a given chart (Uo:, xo:), the vectors v in (p, v, a) still run through all of IR n and so TpM still has dimension n even if p E 8M. On the other hand, if p E 8M, then for any half-space chart x : U ---+ 1R~>c with p in its domain, Tx- I (Tx(p)IR~=c) is a subspace of TpM. This is the subspace of vectors tangent to the boundary and is identified with Tp8M, the tangent space to 8M (also a manifold). Exercise 2.26. Show that this subspace does not depend on the choice of chart.

If one traces back through the definitions, it becomes clear that because of the way charts and differentiability are defined for manifolds with boundary, any smooth function defined on a neighborhood of a boundary point can be thought of as being the restriction of a smooth function defined

2. The Tangent Structure

72

Figure 2.2. Tangents at a boundary point

slightly "outside" M. More precisely, the representative function always has a smooth extension from a (relatively open) neighborhood in 1R~>c to a neighborhood in IRn. The derivatives of the extended function at points of alR~>c = 1R~=c are independent of the extension. These considerations can be -used to show that tangent vectors at boundary points of a smooth n-manifold with boundary can still be considered as derivations of germs of smooth functions. A closed interval [a, b] is a one-dimensional manifold with boundary, and with only minor modifications in our definitions we can also consider equivalence classes of curves to define the full tangent space at a boundary point. It also follows that if c is a smooth curve with domain [a, b], then we can make sense of the velocities c(a) and c(b), and this is true even if c(a) or c(b) is a boundary point. The major portion of the theory of manifolds extends in a natural way to manifolds with boundary.

2.4. Tangents of Products Suppose that f : MI x M2 ~ N is a smooth map. For fixed p E MI and fixed q E M2, consider the "insertion" maps &p : y H (p, y) and &q : x H (x, q). Then fo&q and fo&p are the maps sometimes denoted by f(-,q) and f(p, .).

Definition 2.27. Let f : MI x M2 tangent maps ad and a2f by

~

N be as above. Define the partial ~

(ad) (p, q)

:=

Tp (f

0

&q) : TpMI

(a2f) (p, q)

:=

Tq (f

0

&p) : TqM2 ~ Tf(p.q)N.

Tf(p,q)N,

Next we introduce another natural identification. It is obvious that a curve c: I ~ MI X M2 is equivalent to a pair of curves CI :

I

~

MI,

C2:

I

~

M2.

2.4. Tangents of Products

73

The infinitesimal version of this fact gives rise to a natural identification on the tangent level. If c(t) = (CI(t),C2(t)) and c(O) = (p,q), then the map T(P,q)prl x T(p,q)pr2 : T(p,q)(MI x M2) --t TpMI x TqM2 is given by [c] f---t ([CI], [C2]), which is quite natural. This map is an isomorphism. Indeed, consider the insertion maps ~p : q f---t (p, q) and ~q : p t--+ (p, q),

We have linear monomorphisms T~q(p) Ttp(q) : TqM2 --t T(p,q)(M I x M 2),

TpMI --t T(p,q)(MI x M 2) and

Then, we have the map

Ti q + Tip: TpMI

X

TqM2 --t T(p,q)(M I x M2),

which sends (v,w) E TpMI X TqM 2 to Tiq(V) + Tip(W). It can be checked that this map is the inverse of [c] f---t ([CI], [C2])' Thus we may identify T(p,q)(M I x M 2) with TpMI x TqM 2. Let us say a bit about the naturalness of this identification. In the smooth category, there is a direct product operation. The essential point is that for any two manifolds MI and M 2 , the manifold MI x M2 together with the two projection maps serves as the direct product in the technical sense that for any smooth maps f : N ~ MI and g: N ~ M2 we always have the unique map f x 9 : N ~ MI X M2 which makes the following diagram commute: N

~gJ~

MI ~ MI

X

M2 ---- M2

For a point x E N, write p = f(x) and q = g(x). On the tangent level we have

Tx(fxg)

Txg

which is a diagram in the vector space category. In the category of vector spaces, the product of TpMI and TpM2 is TpMI X TpM2 together with the

74

2. The Tangent Structure

projections onto the two factors. Corresponding to the maps Tpf and Tqg we have the map Tpf x Tqg. But

(T(P,q)prl x T(P,q)pr2)

0

Tx (f x g)

= (T(P,q)prl 0 Tx (f x g)) x (T(P,q)pr2 0 Tx (f x g)) =

Tx (prl

(f x g)) x Tx (pr2

=

Txf x Txg.

0

0

(f x g))

Thus under the identification introduced above the map Tx (f x g) corresponds to Txf x Txg. Now if v E TpMl and w E TqM2' then (v, w) represents an element of T(p,q) (Ml x M 2), and so it should act as a derivation. In fact, we can discover how this works by writing (v,w) = (v,O) + (O,w). If Cl is a curve that represents v and C2 is a curve that represents w, then (v,O) and (O,w) are represented by t f-7 (Cl(t),q) and t f-7 (p,C2(t)) respectively. Then for any smooth function f on Ml x M2 we have

(v,w)f = (v,O)f =

v[f 0

+ (O,w)f =

~ql

+ w[f 0

:tl

o f(Cl(t),q)

+

:tl

o f(p,C2(t))

~pl.

Lemma 2.28 (Partials lemma). For a map f : Ml x M2 -+ N, we have

T(p,q)f· (v, w) = (8d)(p,q) . v + (82f)(p,q) . w, where we have used the aforementioned identification T(p,q)(Ml x M 2) TpMl x Tq M 2 .

=

Proving this last lemma is much easier and more instructive than reading the proof so we leave it to the reader in good conscience.

2.5. Critical Points and Values Definition 2.29. Let f : M -+ N be a Cr-map and p E M. We say that p is a regular point for the map f if Tpf is a surjection. Otherwise, p is called a critical point or singular point. A point q in N is called a regular value of f if every point in the inverse image f-l{q} is a regular point for f. This includes the case where f-l{q} is empty. A point of N that is not a regular value is called a critical value. Most values of a smooth map are regular values. In order to make this precise, we will introduce the notion of measure zero on a second countable smooth manifold. It is actually no problem to define a Lebesgue measure on such a manifold, but for now the notion of measure zero is all we need. Definition 2.30. A subset A of ~n is said to be of measure zero if for any E > there is a sequence of cubes {Wi} such that A c UWi and L vol(Wi ) < E. Here, vol(Wi) denotes the volume of the cube.

°

2.5. Critical Points and Values

75

In the definition above, if the Wi are taken to be balls, then we arrive at the very same notion of measure zero. It is easy to show that a countable union of sets of measure zero is still of measure zero. Our definition is consistent with the usual definition of Lebesgue measure zero as defined in standard measure theory courses. Lemma 2.31. Let U c ]Rn be open and f : U has measure zero, then f(A) has measure zero.

-7

]Rn a C 1 map. If A

cU

Proof. Since A is certainly contained in the countable union of compact balls (all of which are translates of a ball at the origin), we may as well assume that U = B(O, r) and that A is contained in a slightly smaller ball B(O, r-8) c B(O, r). By the mean value theorem (see Appendix C), there is a constant C depending only on f and its domain such that for x, y E B(O, r) we have Ilf(y) - f(x)11 :S c Ilx - YII. Let E > be given. Since A has measure zero, there is a sequence of balls B(Xi' Ei) such that A c U B(Xi' Ei) and

°

'"' vol(B(xi' Ei)) < _E_. n n ~

Thus f(B(Xi, Ei)) also have vol

c

2 c B(j(Xi),2cEi), and while f(A) C

UB(j(Xi), 2CEi),

we

(U B(j(Xi), 2CEi)) :S L vol(B(j(xi), 2CEi)) :S

L vol(Bl) (2CEi)n :S 2ncn L vol(B(xi' Ei)) :S E,

where Bl = B(O, 1) is the ball of radius one centered at the origin. Since E was arbitrary, it follows that A has measure zero. D The previous lemma allows us to make the following definition: Definition 2.32. Let M be an n-manifold that is second countable. A subset A C M is said to be of measure zero if for every admissible chart (U, x) the set x( A n U) has measure zero in ]Rn. In order for this to be a reasonable definition, the manifold must be second countable so that every atlas has a countable subatlas. This way we may be assured that every set that we have defined to be measure zero is the countable union of sets that are measure zero as viewed in some chart. It is not hard to see that in this more general setting it is still true that a countable union of sets of measure zero has measure zero. Also, we still have that the image of a set of measure zero under a smooth map, has measure zero. Proposition 2.33. Let M be second countable as above and A = {(Uo:, xo:)} a fixed atlas for M. If xo:(A n Uo:) has measure zero for all a, then A has measure zero.

2. The Tangent Structure

76

Proof. The atlas A has a countable subatlas, so we may as well assume from the start that A is countable. We need to show that given any admissible chart (U, x) the set x(A n U) has measure zero. We have

x(A n U) =

Ux(A nUn Un) (a countable union).

Since xn(An U nUn) C xn(An Un), we see that xn(An U n Un) has measure zero for all cx. But x(A nUn Un) = X 0 x~l 0 xn(A nUn Un), and so by the lemma above, x(A nUn Un) also has measure zero. Thus x(A n U) has 0 measure zero since it is a countable union of sets of measure zero. We now state the famous and useful theorem of Arthur Sardo Theorem 2.34 (Sard). Let N be an n-manifold and M an m-manifold, both assumed second countable. For a smooth map f : N ---+ M, the set of critical values has measure zero. The somewhat technical proof may be found in the online supplement [Lee, Jeff] or in [Bro-Jan]. Corollary 2.35. If M and N are second countable manifolds, then the set of regular values of a smooth map f : M ---+ N is dense in N. 2.5.1. Morse lemma. If we consider a smooth function f : M ---+ JR, and assume that M is a compact manifold (without boundary), then f must achieve both a maximum at one or more points of M and a minimum at one or more points of M. Let Pe be one of these points. The usual argument shows that dfl pe = 0. (Recall that under the usual identification of JR with any of its tangent spaces we have dfl pe = TpJ.) Now let p be some point for which dflp = 0, i.e. p is a critical point for J. Does f achieve either a maximum or a minimum at p? How does the function behave in a neighborhood of p? As the reader may well be aware, these questions are easier to answer in case the second derivative of f at p is nondegenerate. But what is the second derivative in this case? Definition 2.36. The Hessian matrix of f at one of its critical points p and with respect to coordinates x = (x\ ... , x n ), is the matrix of second partials:

where Xo = x(p). The critical point p is called nondegenerate if H is nonsingular.

2.5. Critical Points and Values

77

Any such matrix H is symmetric, and by Sylvester's law of inertia, it is congruent to a diagonal matrix whose diagonal entries are either 0 or 1 or -1. The number of -1 's occurring in this diagonal matrix is called the index of the critical point. According to Problem 13 we may define the Hessian H f,p : TpM x TpM -t JR, which is a symmetric bilinear form at each critical point p of f, by letting Hf,p(v,w) = Xp(Yf) = Yp(Xf) for any vector fields X and Y which respectively take the values v and w at p. Thus, we may give a coordinate free definition of a nondegenerate point for f. Namely, p is a nondegenerate point for f if and only if Hf,p is a nondegenerate bilinear form. The form H f,p is nondegenerate if for each fixed nonzero v E TpM the map Hf,p(v,·) : TpM -t JR is a nonzero element of the dual space M.

T;

Exercise 2.37. Show that the nondegeneracy is well-defined by either of the two definitions given above and that the definitions agree. Exercise 2.38. Show that nondegenerate critical points are isolated. Show by example that this need not be true for general critical points. The structure of a function near one of its nondegenerate critical points is given by the following famous theorem of M. Morse: Theorem 2.39 (Morse lemma). Let f : M -t JR be a smooth function and let Xo be a nondegenerate critical point for f of index 1/. Then there is a local coordinate system (U, x) containing Xo such that the local representative fu := f 0 x-I for f has the form

fu(x I , . .. ,xn)

= f(xo) + L hijxix j i,j

and it may be arranged that the matrix h = (h ij ) is a diagonal matrix of the form diag( -1, ... , -1, 1, ... ,1) for some number (perhaps zero) of ones and minus ones. The number of minus ones is exactly the index 1/. Proof. This is clearly a local problem and so it suffices to assume that f : U -t JR for some open U C JRn and also that f(O) = O. Our task is to show that there exists a diffeomorphism ¢ : JRn -t JRn such that f 0 ¢( x) = xt hx for a matrix of the form described. The first step is to observe that if 9 : U c JRn -t JR is any function defined on a convex open set U and g(O) = 0, then

2. The Tangent Structure

78

Thus 9 is of the form 9 = 2:~=1 Uigi for certain smooth functions gi, 1 :::; i :::; n with the property that Oig(O) = gi(O). Now we apply this procedure first to f to get f = 2:~1 Udi where od(O) = fi(O) = and then apply the procedure to each fi and substitute back. The result is that

°

n

f(Ul, ... ,Un ) =

(2.3)

L uiujhij(ul"",un) i,j=l

for some functions hij with the property that hij is nonsingular at, and therefore near 0. Next we symmetrize the matrix h = (h ij ) by replacing hij with ~(hij + h ji ) if necessary. This leaves (2.3) untouched. The index of the matrix (hij(O)) is v, and this remains true in a neighborhood of 0. The trick is to find a matrix C(x) for each x in the neighborhood that effects the diagonalization guaranteed by Sylvester's theorem: D = C(x)h(x)C(X)-l. The remaining details, including the fact that the matrix C(x) may be D chosen to depend smoothly on x, are left to the reader.

2.6. Rank and Level Set Definition 2.40. The rank of a smooth map rank of Tpf.

f at

p is defined to be the

If f : M -t N is a smooth map that has the same rank at each point, then we say it has constant rank. Similarly, if f has the same rank for each p in a open subset U, then we say that f has constant rank on U.

Theorem 2.41 (Level submanifold theorem). Let f : M -t N be a smooth map and consider the level set f-l(qO) for qo E N. If f has constant rank k on an open neighborhood of each p E f-l(qo), then f-l(qO) is a closed regular submanifold of codimension k. Proof. Clearly f-l(qO) is a closed subset of M. Let Po E f-l(qo) and consider a chart (U, cp) centered at Po and a chart (V, 'lj;) centered at qo with f(U) c V. We may choose U small enough that f has rank k on U. By Theorem C.5, we may compose with diffeomorphisms to replace (U, cp) by a new chart (U' , x) also centered at Po and replace (V, 'lj;) by a chart (V', y) centered at qo such that J := yo f 0 x-I is given by (a l , ... , an) H (a l , ... ,ak,O, ... ,0), where n = dim(M). We show that U'

If p E U' or

n f-l(qo) = {p E U ' : Xl(p) = ... = xk(p) = a}.

n f-l(qo),

then yo f(p) =

°

and yo f 0 x-1(X1(p), ... , xn(p)) =

°

Xl(p) = ... = xk(p) = 0. On the other hand, suppose that p E U' and xl (p) = ... = xk (p) = 0. Then we can reverse the logic to obtain that yo f(p) = and hence f(p) = qo.

°

2.6. Rank and Level Set

79

Since Po was arbitrary, we have verified the existence of a cover of f-l(qo) by single-slice charts (see Section 1.7). D Proposition 2.42. Let M and N be smooth manifolds of dimension m and n respectively with n > m. Consider any smooth map f : M ---+ N. Then if q E N is a regular value, the inverse image set f- 1 (q) is a regular submanifold. Proof. It is clear that since f must have maximal rank in a neighborhood of f-l(q), it also has constant rank there. We may now apply Theorem 2.41. D Example 2.43 (The unit sphere). The set 5 n - 1 = {x E JRn : L (x i )2 = I} is a codimension 1 sub manifold of JRn. For this we apply the above proposition with the map f : JRn ---+ JR given by x J---7 L (xi) 2 and with the choice q = 1 E lR.. Example 2.44. The set of all square matrices Mnxn is a manifold by virtue of the obvious isomorphism Mnxn ~ JRn 2 • The set sym(n, JR) of all symmetric matrices is a smooth n(n + 1)/2-dimensional manifold by virtue of the obvious 1-1 correspondence sym(n, JR) ~ JRn(n+1)j2 given by using n(n+ 1)/2 entries in the upper triangle of the matrix as coordinates. It can be shown that the map f : Mnxn ---+ sym(n, JR) given by A J---7 At A has full rank on O(n, JR) = f-l(I) and so we can apply Proposition 2.42. Thus the set O(n, JR) of all n x n orthogonal matrices is a submanifold of Mnxn. We leave the details to the reader, but note that we shall prove a more general theorem later (Theorem 5.107).

The following proposition shows an example of the simultaneous use of Sard's theorem and Proposition 2.42. Proposition 2.45. Let 5 be a connected submanifold of JRn and let L be a codimension one linear subspace of JR n . Then there exist x E JRn such that (x + L) n 5 is a submanifold of 5. Proof. Start with a line l through the origin that is normal to L. Let pr: JRn ---+ 5 be orthogonal projection onto l . The restriction 7r := prls ---+ l is easily seen to be smooth. If 7r(5) were just a single point x, then 7r- 1 (x) = (x + L) n 5 would be all of 5, so let us assume that 7r (5) contains more than one point. Now, 7r(5) is a connected subset of l ~ JR, so it must contain an open interval. This implies that 7r(5) has positive measure. Thus by Sard's theorem there must be a point x E 7r(5) c l that is a regular value of 7r. Then Theorem 2.42 implies that 7r- 1 (x) is a submanifold of 5. But this is D the conclusion since 7r- 1 (x) = (x + L) n S.

We can generalize Theorem 2.42 using the concept of transversality.

2. The Tangent Structure

80

Definition 2.46. Let f : M --t N be a smooth map and SeN a submanifold of N. We say that f is transverse to S if for every p E f- 1 (S) we have Tf(p)N = Tf(p)S + Tpf(TpM). If f is transverse to S, we write f rh S. Theorem 2.47. Let f : M --t N be a smooth map and SeN a submanifold of N of codimension k and suppose that f rh Sand f- 1(S) i= 0. Then f- 1(S) is a submanifold of M with codimension k. Furthermore we have Tp(f-l(S)) = Tf-l(Tf (p)S) for all p E f-l(S). Proof. Let q = f(p) E S and choose a single-slice chart (V, x) centered at q E V so that x(S n V) = x(V) n (II~n-k x 0). Let U := f-l(V) so that p E U. If 7f : IR n - k x IRk --t IRk is the second factor projection, then the transversality condition on U implies that 0 is a regular value of 7f 0 x 0 flu. Thus (7f 0 x 0 flu )-1 (0) = f- 1 (S) n U is a submanifold of U of codimension k. Since this is true for all p E f- 1 (S), the result follows. D We can also define when a pair of maps are transverse to each other: Definition 2.48. If h : Ml --t Nand 12 : M2 say that hand 12 are transverse at q E N if

Tf(p)N = TP1h(TPIM)

+ Tp212(Tp2M)

--t

N are smooth maps, we

whenever h(pd = 12(P2) = q.

(N ote that h is transverse to 12 at any point not in the image of one of the maps h and h.) If hand 12 are transverse for all q E N, then we say that hand 12 are transverse and we write h rh 12. One can check that if f : M of N, then f and the inclusion according to Definition 2.47.

--t L :

N is a smooth map and S is a submanifold S y N are transverse if and only if f rh S

If h : Ml --t Nand 12 : M2 consider the set

--t

N are smooth maps, then we can

(h x 12)-1 (~) := ((pl,P2) E Ml x M2 : h(Pl) = 12(P2)}, which is the inverse image of the diagonal

~ :=

{(ql, q2)

E

N x N : ql = q2}.

Corollary 2.49 (Transverse pullbacks). If h : Ml --t Nand 12 : M2 --t N are transverse smooth maps, then (h x 12)-1 (~) is a submanifold of lvIt x M 2. If gl : P --t Ml and g2 : P --t M2 are any smooth maps with the property h o gl = 12 0 g2, then the map (gl,g2): P --t (fl X 12)-1 (~) given by (gl, g2) (x) = (gl (x), g2 (x)) is smooth and is the unique smooth map such that prl 0 (gl,g2) = gl and pr2 0 (g1,g2) = g2. Proof. We leave the proof as an exercise. Hint: h x 12 is transverse to ~ if and only if h rh 12. D

2.7. The Tangent and Cotangent Bundles

81

2.7. The Tangent and Cotangent Bundles We define the tangent bundle of a manifold M as the (disjoint) union of the tangent spaces; T M = UPEM TpM. We show in Proposition 2.55 below that T 111 is a smooth manifold, but first we introduce a couple of definitions.

Definition 2.50. Given a smooth map J : M -+ N as above, the tangent maps TpJ on the individual tangent spaces combine to give a map

TJ:TM-+TN on the tangent bundle which is linear on each fiber. This map is called the tangent map or sometimes the tangent lift of J. For smooth maps J : M -+ Nand 9 : N -+ M we have the following simple looking version of the chain rule:

T(goJ) = TgoTf.

If U is an open set in a finite-dimensional vector space V, then the tangent space at x E U can be viewed as {x} xV. For example, recall that an element vp = (p, v) corresponds to the derivation J r---+ vpJ := ftlt=o J(p+tv). Thus the tangent bundle of U can be viewed as the product U x V. Let Ul and U2 be open subsets of vector spaces V and W respectively and let J : U1 -+ U2 be smooth (or at least C 1). Then we have the tangent map TJ : TU1 -+ TU2. Viewing TU1 as U1 x V and similarly for TU2, the tangent map T J is given by (p, v) r---+ (J(p), D J(p) . v). Definition 2.51. If J : M -+ V, where V is a finite-dimensional vector space, then we have the differential dJ (p) : TpM -+ V for each p. These maps can be combined to give a single map dJ : T M -+ V (also called the differential) which is defined by dJ(v) = dJ(p)(v) when v E TpM.

If we identify TV with the product V x V, then dJ = pr2 0 TJ, where pr2 : TV = V x V -+ V is the projection onto the second factor. Remark 2.52 (Warning). The notation "df" is subject to interpretation. Besides the map dJ : T M -+ V described above it could also refer to the map dJ : p r---+ dJ(p) or to another map on vector fields which we describe later in this chapter. Definition 2.53. The map 1fTM : TM -+ M defined by 1fTM(V) = p if v E TpM is called the tangent bundle projection map. (The set T AI together with the map 1fTM : T AI -+ M is an example of a vector bundle which is defined later.)

82

2. The Tangent Structure

Whenever possible, we abbreviate JrTM to Jr. For every chart (U, x) on M, we obtain a chart (if, x) on T M by letting

if:= TU = and by defining

Jr- 1 (U)

c TM

x on U by the prescription

x(vp) = (xl(p), ... ,xn(p),v l , ... ,vn ), where vp E TpM, and where v!, ... , v n are the (unique) coefficients in the coordinate expres. a Ip' Thus ~-l( I ... , u n, v I , ... , v n) -_ " a Ix-l(u)" slOn v -_ "i...J v i axi xu, i...J v i axi Recall that if vp = L vi a~i Ip' then vi = dx i (vp). From this we see that x = (xl 0 Jr, ... , xn 0 Jr, dx l , ... , dxn). For any (U, x), we have the tangent lift Tx : TU -+ TV where V = x (U). Since V C IRn, we can identify Tx(p) V with {x (p)} x IRn. Let us invoke this identification. Now let vp E TpU and let, be a curve that represents vp so that ,'(0) = vp. Exercise 2.54. Under the identification of Tx(p) V with {x (p)} x IR n we have Tpx,vp = (x (p), (x 0 , ) ) . [Hint: Interpret both sides as derivations.]

ftlt=o

If vp = a~i Ip' then we can take ,(t) := x-l(x (p) i-th member of the standard basis of IRn. Thus

Tpx,

f)~i Ip =

(x (p) , :t It=O (x (p)

+ tei) )

+ tei), = (x

where ei is the

(p) , ed·

Now suppose that vp = Lvi a~i Ip' Then Tpx . vp

~ Tpx. =

(L {J~i IJ

(x (p),

vi

L viei)

=

(xl(p), ... , xn(p), vI, ... , vn).

x

From this we see that Tx is none other than defined above, and since if = TU, we see that an alternative and suggestive notation for (if , is

x)

(TU, Tx), and we adopt this notation below. This notation reminds one that the charts we have constructed are not just any charts on T M, but are each associated naturally with a chart on M and are essentially the tangent lifts of charts on M. They are called natural charts. Proposition 2.55. For any smooth n-manifold M, the set T M is a smooth 2n-manifold in a natural way and JrTM : TM -+ M is a smooth map. Furthermore, for a smooth map f : M -+ N, the tangent map T f is smooth and

2.7. The Tangent and Cotangent Bundles

83

the following diagram commutes:

TM~TN

~

f

M

~

>N

Proof. For every chart (U, x), let TU = 11"-1 (U) and let Tx be the map Tx: TU ---+ x(U) x jRn. The pair (TU, Tx) is a chart on TM. Suppose that (TU, Tx) and (TV, Ty) are two such charts constructed as above from two charts (U, x) and (V, y) and that U n V i= 0. Then TU n TV i= 0 and on the overlap we have the coordinate transitions Ty 0 Tx- 1 : (x, v) t--t (y, w) where y = yo X-I (x), w

= D(y 0

x- 1 )lx v.

Thus the overlap maps are smooth. It is easy to see that Tx(TU n TV) and Ty(TU n TV) are open. Thus we obtain a smooth atlas on T M from an atlas on M and this generates a topology. It follows from Proposition 1.32 that T M is Hausdorff and paracompact. To test for the smoothness of 11", we look at maps of the form x011"o(Tx) -1. We have X011"O

(Tx)-l (x,v)

=X011"

(vi aail ) =xox-1(x)=x, x x-1(x)

which is just a projection and so clearly smooth. The remainder is left for the exercise below. 0 In the above proof we observed that Tx(TU n TV) and Ty(TU n TV) are open. This must be checked because of (ii) in Definition 1.25 and is the kind of detail we may leave to the reader as we move forward. Exercise 2.56. For a smooth map

f : M ---+ N, the map

Tf: TM ---+ TN is itself a smooth map.

If p E U n V and x(p) = (xl(p), ... , xn(p)), then, as in the proof above, TyoTx- 1 sends (xl(p), ... ,xn(p),v 1, ... ,vn ) to (yl(p), ... ,yn(p), wi, ... , w n ), where

84

2. The Tangent Structure

If we abbreviate the i-th component of yo x-I(XI(p), .. . , xn(p)) to yi = yi (xl (p), ... , xn (p)), then we could express the tangent bundle overlap map by the relations ) " 8yi k Yi = Yi ( x I ( p, ... , x n (p)) an d w i = 'L..J 8x k v .

Since this is true for all p E x(U n V), we can write the very classical looking expressions Yi

= Yi( x I , ... , x n) and

wi

" 8yi k = 'L..J 8x k v ,

where we now can interpret (xl, ... , xn) as an n-tuple of numbers. Once again we note that local expression could either be interpreted as living on the manifold in the chart domain or equally, in Euclidean space on the image of the chart domain. This should not be upsetting since, after all, one could argue that the charts are there to identify chart domains in the manifold with open sets in Euclidean space.

Definition 2.57. The tangent functor is defined by assigning to a manifold M its tangent bundle T M and to any map f : M -t N the tangent map T f : T M -t TN. The chain rule shows that this is a covariant functor (see Appendix A). Recall that we also defined a "pointed" tangent functor. We have seen that if U is an open set in a vector space V, then the tangent bundle is often taken to be U x V. Suppose that for some smooth n-manifold M, there is a diffeomorphism F : TM -t M x V such that the restriction of F to each tangent space is a linear isomorphism TpM -t {p} X V and such that the following diagram commutes: TM--_F_~) MxV

~~ M Then for some purposes, we can identify TM with M x V.

Definition 2.58. A diffeomorphism F : T M -t M x V such that the map FITp M : TpM -t {p} X V is linear for each p and such that the above diagram commutes is called a (global) trivialization of TM. If a (global) trivialization exists, then we say that T M is trivial. For an open set U eM, a trivialization of TU is called a local trivialization of T Mover U. For most manifolds, there does not exist a global trivialization of the tangent bundle. On the other hand, every point p in a manifold M is contained in an open set U so that T M has a local trivialization over U. The

2.7. The Tangent and Cotangent Bundles

85

existence of these local trivializations is quickly deduced from the existence of the special charts which we constructed above for a tangent bundle. Next we introduce the cotangent bundle. Recall that for each P E M, the tangent space TpM has a dual space T; M called the cotangent space at

p. Definition 2.59. Define the cotangent bundle of a manifold M to be the set

T*M:=

U

T;M pEM M to be the obvious projection taking

and define the map 7rT" M : T* M -t elements in each space T; M to the corresponding point p.

Remark 2.60. We will denote both the tangent bundle projection and the cotangent bundle projection simply by 7r whenever no confusion is likely. Remark 2.61. Suppose that J : M -t N is a smooth map. It is important to notice that even though for each P E M, the map TpJ : TpM -t Tf(p)N has a dual map (TpJ)* : (Tf(p)N) * -t (TpM)*, these maps do not generally combine to give a map from T* N to T* M. In general, there is nothing like a "cotangent lift". To see this, just consider the case where J is a constant map. We now show that T* M is also a smooth manifold. Let A be an atlas on M. For each chart (U, x) E A, we obtain a chart (T*U, T*x) for T* M which we now describe. First, T*U = 7rr:M(U) = UPEu T;M. Secondly, T*x is a map which we now define directly and then show that, in some sense, it is dual to the map Tx. For convenience, consider the map Pi : {}p t---+ ~i which just peals off the coefficients in the expansion of any {}p E T; M in the basis

(dxll p "'" dxnl p):

Notice that we have

(}p(a~il) = L~j dxil p (a~il) = L~jol =~j = Pi ({}p) , and so

Pi(Op) = Op (

8~i

IJ .

With this definition of the Pi in hand, we can define

T*x =

(Xl 0 7r, ... ,

xn

0 7r,PI, ...

,Pn)

on T*U. We call (T*U, T*x) a natural chart. If x = (Xl, ... , x n ), then for the natural chart (T*U, T*x), we could use the abbreviation T*x =

2. The Tangent Structure

86

(xl, ... ,Xn,P1, ... ,Pn)' Another common notation is qi := xi notation is very popular in applications to mechanics.

07r.

This

We claim that if we take advantage of the identifications of TxjRn = jRn = (jRn)* = T*jRn where (jRn)* is the dual space of jRn, then T*x acts on each fiber T; M as the dual of the inverse of the map Tpx, i.e. the contragredient of Tpx: ((TpX)-l) * (fJp) . (v)

= fJp ((Tpx)-l . v) .

Let us unravel this. If fJp E T; M for some P E U, then we can write fJp = L~i dxil p

for some numbers We have

~i

depending on fJp which are what we have called Pi (fJp).

((Tpx)-l)* (fJp)' (v)

= fJp ((Tpx)-l. v) = L~i dxil p ' ((Tpx)-l. v)

IJ

~ Lei dxil (~>' a=' ~ Leivi p

Thus, under the identification ofjRn with its dual we see that

((T x)-l) * (fJ p

p)

is just (6, ... ,~n). But recall that T*x(fJp) = (x1(p), ... ,xn(p),6"'.'~n)' Thus for fJp E T; M we have T*x(fJp) = (x(p), ((Tpx)-l)* (fJp)).

Suppose that (T*U, T*x) and (T*V, T*y) are the coordinates constructed as above from two charts (U, x) and (V, y) respectively with unv i- 0. Then on the overlap T* U n T* V we have T*y

0

(T*x)-l : x(U n V) x jRn*

-t

y(U n V) x jRn*.

This last map will send something of the form (x,~) E U x jRn* to (x, () = (yox-1 (x), D(xoy-1)* .~), where D(xoy-1)* is the dual map to D(xoy-1), which is the contragredient of the map D(y 0 x-I). If we identify jRn* with jRn and write ~ = (6, ... , ~n) and ( = ((1, ... , &), then in the classical style we have: . 1 " ox k ft. = yt(x , ... , xn) and ~i = L...J ~k oyi . k

This should be compared to the expression (2.2). It is now clear that we have an atlas on T* M constructed from an atlas on M. The topology of T* M (induced by the above atlas) is easily seen to be paracompact and Hausdorff.

2.8. Vector Fields

87

In summary, both T M and T* M are smooth manifolds whose smooth structure is derived from the smooth structure on M in a natural way. In both cases, the charts are derived from charts on the base M and are given by the n coordinates of the base point together with the n components of the element of T M (or T* M) in the corresponding coordinate frame.

2.8. Vector Fields In this section we introduce vector fields. Roughly, a vector field is a smooth assignment of a tangent vector to each point of a manifold. Definition 2.62. If 1l' : M -+ N is a smooth map, then a (global) section of 1l' is a map (J : N -+ M such that 1l' 0 (J = id. If (J is defined only on an open subset U of Nand 1l' 0 (J = idu, then we call (J a local section. In case the section (J is a smooth (or cr) map, we call (J a smooth (or cr) section. Clearly, if 1l' : M -+ N has a (global) section, then it must be surjective. Definition 2.63. A smooth vector field on M is a smooth map X: M -+ T M such that X (p) E TpM for all p E M. In other words, a vector field on M is a smooth section of the tangent bundle 1l' : T M -+ M. We often write Xp = X(p). Convention: Obviously the notion of a section or field that is not smooth makes sense. Sometimes one is interested in merely continuous sections or measurable sections. In this book, by "vector field" or "section", we will always mean "smooth vector field" or "smooth section" unless otherwise indicated explicitly or by context. A local section of T M defined on an open set U is just the same thing as a vector field on the open manifold U. If (U, x) is a chart on a smooth n-manifold, then writing x = (Xl, .. . ,xn ), we have vector fields defined on Uby

:pi--t aa.j. aa. xt xt p The ordered set of fields (Ixr, ... , 8~n) is called a coordinate frame field (or also "holonomic frame field"). If X is a smooth vector field defined on some set including this chart domain U, then for some smooth functions Xi defined on U we have

X(p) = or in other words

L Xi(p) a~i jp,

2. The Tangent Structure

88

Notation 2.64. In this context, we will not usually bother to distinguish from its restrictions to chart domains and so we just write = L: a~i .

X

X

Xi

Lemma 2.65. If v E TpM then there exists a vector field X such that X(p) = v.

Ip·

Proof. Write v = L: vi a~i Define a field Xu by the formula L: vi a~i where the vi are taken as constant functions on U. Let (3 be a cut-off function with support in U and such that (3(p) = 1. Then let X := (3Xu on U and extended to zero outside of U. 0

Let us unravel what the smoothness condition means for a vector field. Let (TU, Tx) be one of the natural charts that we constructed for T M from a corresponding chart (U, x) on M. To test the smoothness of X, we look at the composition Tx 0 X 0 x-I. For x E x(U), we have Tx 0 X

0 X-I (x)

(I: Xi a~i) x- (x) = Tx(I: X (x- (x)) aa I

= Tx 0

1

0

1

i

= (x,

T

)

i

X

x- 1 (x)

X-l(X)X(I: X i (x- 1 (x)) aaX i Ix-1(x) ))

= (x, X I 0 x-I (x), ... , Xn 0 x-I (x)) . Our chart was arbitrary, and so we see that the smoothness of X is equivalent to the smoothness of the component functions in every chart of an atlas for the smooth structure.

Xi

Exercise 2.66. Show that if X : M -t T M is continuous and 7r 0 X = id, then X is smooth if and only if X f : p H Xpf is a smooth function for every locally defined smooth function f on M. Show that it is enough to consider globally defined smooth functions. Notation 2.67. The set of all smooth vector fields on M is denoted by X(M). Smooth vector fields may at times be defined only on some open set U c M so we also have the notation X(U) = XM(U) for these fields.

We define the addition of vector fields, say X and Y, by

(X

+ Y) (p)

:=

X(p)

+ Y(p),

and scaling by real numbers, by

(cX) (p) := cX(p).

89

2.8. Vector Fields

Then the set X(M) is a real vector space. If we define multiplication of a smooth vector field X by a smooth function f by

(f X) (p)

:=

f(p)X(p),

then the expected algebraic properties hold making X(M) a module over the ring COO(M) (see Appendix D). It should be clear how to define vector fields of class cr on M and the set of these is denoted xr(M) (a module over Cr(M)). The notion of a vector field along a map is often useful.

Definition 2.68. Let f : N ---+ M be a smooth map. A vector field along I is a smooth map X : N ---+ T M such that 1fTM 0 X = f. A vector field along a regular submanifold ScM is a vector field along the inclusion map S,-+ M. (Note that we include the case where S is an open submanifold.) We let Xf denote the space of vector fields along f. It is easy to check that for a smooth map COO(N)-module in a natural way.

f : N ---+

M, the set Xf is a

We have seen how individual tangent vectors in TpM can be identified as derivations at p. The derivation idea can be globalized. We explain how we may view vector fields as derivations.

Definition 2.69. Let M be a smooth manifold. A (global) derivation on COO(M) is a linear map V : COO(M) ---+ COO(M) such that

V(fg)

= V(f)g + fV(g).

We denote the set of all such derivations of COO(M) by Der(COO(M)). Notice the difference between a derivation in this sense and a derivation at a point.

Definition 2.70. To a vector field X on M, we associate the map LX COO(M) ---+ COO(M) defined by

(Lxf)(p)

ex

:=

Xpj.

is called the Lie derivative on functions.

It is important to notice that (Lxf)(p) = Xp' f = df(Xp) for any p and so Lxf = df 0 X. If X is a vector field on an open set U, and if I is a function on a domain V c U, then we take LX f to be the function defined on V by p f---t Xpf for all p E V. It is easy to see that we have CaX+bY = aLx + bLy for a, bE lR and X, Y E X(M).

Lemma 2.71. Let U c M be an open set and X E X(M). If Lxf all IE COO(U), then Xiu = o.

= 0 lor

2. The Tangent Structure

90

Proof. Let p E U be given. Working locally in a chart (V, x), let X = 'LX i8/8xi. We may assume p EVe U. Using a cut-off function we may find functions fi defined on U such that fi coincides with xi on a neighborhood of p. Then we have Xi(p) = Xpx i = Xpfi = (.cxf i ) (p) = O. Thus X (p) = 0 for an arbitrary p E U. 0

The next result is a very important characterization of smooth vector fields. In particular, it paves the way for the definition of the bracket of vector fields which plays a central role in differential geometry. Theorem 2.72. For X E X(M), we have.c x E Der(CC'O(M)), and if V E Der(COO(M)), then V = .cx for a uniquely determined X E X(M). Proof. That .cx is in Der(COO(M)) follows from the Leibniz law, in other words, from the fact that Xp is a derivation at p for each p. If we are given a derivation V, we define a derivation Xp at p (i.e. a tangent vector) by the rule Xpf := (V J) (p). We need to show that the assignment p H Xp is smooth. Recall that any locally defined function can be extended to a global one by using a cut-off function. Because of this, it suffices to show that p H Xpf is smooth for any f E COO(M). But this is clear since Xpf := (VJ) (p) and Vf E COO(M). Suppose now that V = .cXI = .cX2' Notice that .cXI - .cX2 = .cXI-X2 and so .cXI-X2 is the zero derivation on COO(M). By Lemma 2.71, we have Xl - X2 = O. 0

Because of this theorem, we can identify Der(COO(M)) with X(M) and we can and often will write X f in place of .c x f:

Xf:= .cxf. The derivation law (also called the Leibniz law) .cx(Jg) = g.cx f + f.cx 9 becomes simply X(Jg) = gXf + fXg. Another thing worth noting is that if we have a derivation of COO(M), then from our discussion above we know that it corresponds to a vector field. As such, it can be restricted to any open set U c M, and thus we get a derivation of COO(U). If f E COO(U) we write X f instead of the more pedantic Xlu J. While it makes sense to talk of vector fields on M of differentiability r where 0 < r < 00 and these do act as derivations on cr(M), it is only in the smooth case (r = 00) that we can say that vector fields account for all derivations of cr(M). Theorem 2.73. If VI, V2 E Der(COO(M)), then [VI, V2J E Der(COO(M)) where

91

2.8. Vector Fields

Proof. We compute

VI ('0 2 (fg)) = VI (V 2(f)g

+ f V 2(g))

= ('0 1'021) 9 + V2fV 1g + VdV 2g + fV 1V 2g. Writing out the similar expression for '0 2 (VI (f g)) and then subtracting we obtain, after a cancellation,

[VI, '0 2] (fg) = ('0 1'021) 9 + f V 1V 2g - ((V2V d) 9 + f V 2V 1g)

= ([VI, V 2]f) 9 + J[V1' V 2]g.

0

Corollary 2.74. If X, Y E X(M), then there is a unique vector field [X, Y] such that £[X,Yj = £x 0 £y - £y 0 £x.

Since £xf is also written Xf, we have [X, Y]f = X (Yf) - Y (Xf) or

[X,Y] = XY - Yx. Definition 2.75. The vector field [X, Y] from the previous corollary is called the Lie bracket of X and Y. Proposition 2.76. The map (X, Y) X, Y, Z E X(M) we have

f-7

[X, Y] is bilinear over JR, and for

(i) [X, Y] = -[Y, X];

+ [Z, [X, YlJ = 0 = fg[X, Y] + f (Xg) Y - 9 (Y f) X

(ii) [X, [Y, ZlJ+ [Y, [Z, X]]

(Jacobi Identity);

(iii) [f X, gY]

for all f, 9 E COO(M).

Proof. These results follow from direct calculation and the previously mentioned fact that £aX+bY = a£x + bey for a, bE JR and X, Y E X(M). 0

The map (X, Y) f-7 [X, Y] is bilinear over JR, but by (iii) above, it is not bilinear over COO(M). Also notice that in (ii) above, X, Y, Z are permuted cyclically. We ought to see what the local formula for the Lie derivative looks like in conventional "index" notation. Suppose we have X = LXi 8~' and Y = L yi 8~i. Then we have the local formula

Exercise 2.77. Verify this last formula.

The JR-vector space X(M) together with the JR-bilinear map (X, Y) f-7 [X, Y] is an example of an extremely important abstract algebraic structure:

92

2. The Tangent Structure

Definition 2.78 (Lie algebra). A vector space a (over a field IF) is called a Lie algebra if it is equipped with a bilinear map a x a -t a (a multiplication) denoted (v, w) t--t [v, w] such that

[v,w] = -[w,v] and such that we have the Jacobi identity

[x, [y, z]]

+ [y, [z, x]] + [z, [x, y]] = 0

for all x,y,z E a.

Definition 2.79. A Lie algebra a is called abelian (or commutative) if [v, w] = 0 for all v, w E a. A subspace ~ of a is called a Lie subalgebra if it is closed under the bracket operation, and it is called an ideal if [v, w] E ~ for any v E a and w E ~. (We indicate this by writing [a,~] C ~.) Notice that the Jacobi identity may be restated as [x, [y, z]] = [[x, y], z] + [y, [x, z]], which just says that for fixed x the map y t--t [x, y] is a derivation of the Lie algebra a. This is significant mathematically and also an easy way to remember the Jacobi identity. The Lie algebra X(M) is infinite-dimensional (unless M is zero-dimensional), but later we will be very interested in certain finite-dimensional Lie algebras which are subalgebras of X(M). Given a diffeomorphism ¢ : M -t N, we define the pull-back ¢*Y E X(M) for Y E X(N) and the push-forward ¢*X E X(N) of X E X(M) by ¢ by ¢*Y ¢*X

= T¢-l 0 Y 0 ¢ and = T ¢ 0 X 0 ¢ -1.

In other words, (¢*Y)(p) = T¢-l. Y<j>(P) and (¢*X)(p) = T¢·X<j>-l(p). Notice that ¢*Y and ¢*X are both smooth vector fields. Warning: Since many authors use the notation f* for the tangent map T f, the notation f*X might be interpreted to mean T foX, which is actually a vector field along the map f rather than an element of X(N). We shall not use f* as a notation for Tf.

To summarize a bit, if f : M -t N is a smooth map, then for each p we have the tangent map Tpf : TpM -t Tf(p)N, the tangent lift T f : T M -t TN (a "bundle map"), and if f is a diffeomorphism, we have the induced maps on the level of fields f* : X(M) -t X(N) and f* : X(N) -t X(M). Notice that if ¢ : M -t Nand 'ljJ : N -t Pare diffeomorphisms, then we have ('ljJ

0

¢)* = 'ljJ*

0

¢* : X(M)

('ljJ

0

¢)* = ¢*

0

'ljJ* : X(P) -t X(M).

-t

X(P),

We have right and left actions of the diffeomorphism group Diff(M) on the space of vector fields. The left action Diff(M) x X(M) -t X(M) is given by

2.8. Vector Fields

93

(¢, X) t-+ ¢*X, and the right action x(M) x Diff(M) -+ x(M) is given by (X, ¢) t-+ ¢* X.

On functions, the pull-back is defined by ¢*g := go ¢ for any smooth map, but if ¢ is a diffeomorphism, then we can also define a push-forward ¢* := (¢-1)*. With this notation we have the following proposition. Proposition 2.80. The Lie derivative on functions is natural with respect to pull-back and push-forward by diffeomorphisms. In other words, if ¢ : M -+ N is a diffeomorphism and f E COO(M), 9 E COO(N), X E x(M) and Y E x(N), then

and

Proof. We use Definition 2.19. For any p we have

(£<jJ*y¢*g) (p) = (¢*Y)p ¢*g = (T¢-l = (T¢-l . Y<jJ(p») [g

0

0

Y

0

¢)p [g

0

¢]

¢] = T¢ (T¢-lY<jJ(p») 9

= Y<jJ(p)g = (£yg) (¢(p)) = (¢* £yg) (p). The second statement follows from the first since ¢* = (¢ -1) *.

0

Even if f : M -+ N is not a diffeomorphism, it may still be that there is a vector field Y E x(N) such that

TfoX=Yof. In other words, it may happen that Tf· Xp = Yf(p) for all p in M. In this case, we say that Y is f -related to X and write X '" f Y. It is not hard to check that if Xi is f-related to Y; for i = 1,2, then aX l + bX1 is f-related to aYl + bYl .

Example 2.81. Let M and N be smooth manifolds and consider the projections pr1 : M x N -+ M and pr2 : M x N -+ N. Since T(p,q) (M x N) can be identified with TpM x TqN, we see that for X E x(M), Y E x(N) we obtain a vector field X x Y E x(M x N) defined by (X x Y) (p, q) = (X(p), Y(p)). Then one can check that X x Y and X are pr1-related

and

X x Y and Yare prTrelated. Exercise 2.82. Let M, N, X, Y and X x Y be as in the example above. Show that if ~q : M -+ M x N is the insertion map p t-+ (p, q), then X and X x Y are ~q-related if and only if Y(q) = O.

94

2. The Tangent Structure

Lemma 2.83. Suppose that f : M -+ N is a smooth map, X E X(M) and Y E X(N). Then X and Yare f-related if and only if X(g 0 f) = (Y g) 0 f for all 9 E COO(N). Proof. Let p E M and let 9 E Coo (N). Then

X(g

0

f)(p)

= Xp(g 0

f)

= (Tpf . Xp) 9

and (Y go f) (p)

so that X (g 0 f) = (Y g)

0

= Yf(p)g

f for all such 9 if and only if Tpf . Xp = Yf(p).

0

Proposition 2.84. If f : M -+ N is a smooth map and Xi is f -related to Yi for i = 1,2, then [Xl, X2] is f-related to [YI, Y2]. In particular, if ¢ is a diffeomorphism, then [¢*X I , ¢*X2] = ¢*[X I , X 2] for all Xl, X 2 E X(M). Proof. We use the previous lemma: Let 9 E COO(N). Then X I X2(g 0 f) = X I ((Y2g) 0 f) = (YIY2g) 0 f. In the same way, X 2X I (g 0 f) = (Y2Y I g) 0 f and subtracting we obtain

[XI,X2] (g

0

f) = XIX2(g 0 f) - X2 X I(g 0 f) = (YIY2g) 0 f - (Y2YIg) 0 f =

([YI , Y2]g)

0

f.

Using the lemma one more time, we have the result.

o

If S is a submanifold of M and X E X(M), then the restriction Xis E X(S) defined by Xis (p) = X(p) for all pES is i-related to X where i : S y M is the inclusion map. Thus for X, Y E X(M) we always have that [Xis, Yl s] is i-related to [X, Y]. This just means that [X, Y] (p) = [Xis, Yls] (p) for all p. We also have

Proposition 2.85. Let f : M -+ N be a smooth map and suppose that X "'f Y. Then we have £x (f*g) = j*£yg for any 9 E COO(N).

The proof is similar to what we did above and is left to the reader. 2.8.1. Integral curves and flows. Recall that if c : I -+ M is a smooth curve, then the velocity at "time" t is

:t c(t) = c(t) = Ttc· :u It'

Iu

where is the standard field on 1R given at a E 1R as the equivalence class a f = f' (a). of the curve t f-t a + t or by the derivation

Iu I

2.8. Vector Fields

95

Definition 2.86. Let X be a smooth vector field on M. A curve c: I -t M is called an integral curve for X if for all t E I, the velocity of c at time t is equal to X(c(t)), that is, if

c= X

0

c.

Thus if c is an integral curve for X and

f

is a smooth function, then

Xc(t)f = (f 0 c)' (t) for all t in the domain of c. If the image of an integral curve c lies in U for a chart (U, x), and if X = L: Xi 8~i' then c = X 0 c gives the local expressions

d . dt xt 0

.

C

= Xt

0

c for i

= 1, ... ,n,

which constitute a system of ordinary differential equations for the functions xi 0 c. These equations are classically written as d:1ti = Xi.

A (complete) flow is a map

X~(p) =

:t

10
E

TpM for p E M.

If one computes the velocity vector C(O) of the curve c : t H 0, an open set V with Xo EVe U, and a smooth map

2. The Tangent Structure

96

such that t H cx(t) := (t,x) is a curve satisfying t E (-a, a) and cx(O) = x.

c~(t) =

F(cx(t)) for all

Example 2.88. Consider the differential equation on the line given by

c'(t) = (c(t))2/3. There are two distinct solutions with initial condition c(O)

= o. Namely,

c(t) = 0 for all t and

1 c(t) = 27t3 for all t.

The reason uniqueness fails is the fact that the function F(x) = x 2/ 3 is not differentiable at x = O. Now let X E X(M) and consider a point p in the domain of a chart (U, x). The local expression for the integral curve equation c( t) = X (c( t)) is of the form treated in the last theorem, and so we see that there certainly exists an integral curve for X through p defined on at least some small interval (-E, E). We will now use this theorem to obtain similar but more global results on smooth manifolds. First of all, we can get a more global version of uniqueness: Lemma 2.89. If

Cl : (-El, Ed ---+ M and C2 : (-E2' (2) ---+ M are integral curves of a vector field X with Cl (0) = C2(0), then Cl = C2 on the intersection of their domains.

Proof. Let K = {t E (-El,Ed n (-E2,E2) : Cl(t) = C2(t)}. The set K is closed since M is Hausdorff. It follows from Theorem 2.87 that K contains a (small) open interval (-E, E). Let to be any point in K and consider the translated curves cia (t) = Cl (to + t) and c~a (t) = C2 (to + t). These are also integral curves of X and they agree at t = 0, and by Theorem 2.87 again we see that cia = c~a on some open neighborhood of o. But this means that Cl and C2 agree in this neighborhood, so in fact this neighborhood is contained in K implying that K is also open since to was an arbitrary point in K. Thus, since I = (-El, Ed n (-E2, (2) is connected, it must be that I = K and so Cl and C2 agree on I = (-El, EI) n (-E2, (2). 0

Let X be a Coo vector field on M. A flow box for X at a point p E M is a triple (U, a,
(1) U is an open set in M containing p. (2)
2.8. Vector Fields

97

(4) The map
x


0

x


x


x


0

x


whenever defined. This is the local group property, so called because if
tx

Notice that whereas is a complete vector field on ]R2 (using standard coordinates x,y), this vector field restricted to ]R2\{O} is not complete on the manifold ]R2\{O}. The reason is that integral curves starting at points on the x-axis will run up against the missing origin in finite time. This "running up to missing points" is not the only way a vector field can fail on R The integral to be complete. Consider the vector field (1 + x 2 ) curve that is at the origin at time zero is obtained by solving the initial value problem x' = 1 + x 2, x(O) = o. The unique solution is x(t) = tan t, and since limH±7r/2 tan t = ±oo, the solution cannot be extended beyond

!

±1f/2.

tx

t

Exercise 2.90. Show that on ]R2 the vector fields y2 and x 2 y are complete, but y2 + x 2 y is not complete. In particular, the set of complete vector fields is not generally a vector space (but this is true if M is compact).

tx

t

Theorem 2.91 (Flow box). Let X be a Coo vector field on an n-manifold M with r ~ 1. Then for every point Po E M there exists a flow box for X at Po. If (U I , aI,
2. The Tangent Structure

98

to be smaller so that the flow stays within the range of the chart map x. In this setting, a vector field can be taken to be a map U -+ ]Rn, so Theorem 2.87 provides us with the flow box data (V, a, 0 small enough that vt = (t, V) C U for all t E (-a, a). Now the flow box is transferred back to the manifold via x, U

rpx (t,p)

= x-I(V),

= x-I [(t, x(p))].

If we have two such flow boxes (UI' aI, rpr) and (U2, a2, rp?f), then by Lemma 2.89, we see that for any x E UI n U2 we must have rpr (t, x) = rp?f (t, x) for all t E (-aI, al) n (-a2, a2).

Finally, since rpt = rpx (t , .) and rp~t = rpx ( - t , .) are both smooth and inverses of each other, we see that rpt is a diffeomorphism onto its image Ut = x-l(vt). 0 Lemma 2.92. Suppose that Xl, ... , X k are smooth vector fields on M and let Po E M be given and 0 be an open set containing Po. If rpXl, ... , rpXk are the local flows corresponding to flow boxes whose domains UI, ... ,Uk all contain Po, then there is an open set U C UI n ... n Uk and an E > 0 such that the composition Xk Xl rptk 0'" 0 rptl is defined on U and maps U into 0 whenever it, ... , tk E (-E, E).

Proof. If the flow box corresponding to rpXi is (Ui , Ei, rpf), then by shrinking UI further we may arrange things so that rp;l maps UI into 0 for all t E (-EI' EI), and then inductively we arrange for rpXi to map Ui into Ui-l for all t E (-Ei' Ei). Now let E = min{ EI,"" Ek}. 0 Remark 2.93. When making compositions of local flows, we will not always make careful statements about domains, but the previous lemma will be invoked implicitly. If Cp(t) is an integral curve of X defined on some interval (a, b) containing

o and Cp(O) = p, then we may consider the limit lim Cp(t).

t~b-

If this limit exists as a point PI EM, then we may consider the integral curve cPl beginning at Pl. One may now use Lemma 2.89 to combine t I-t Cp(t) with t I-t Cpl (t-b) to produce an extended integral curve beginning at p. We may repeat this process as long as the limit exists. We may do a similar thing in the negative direction. This suggests that there is a maximal integral curve defined on a maximal interval := (Tp~x' Tp~x), where Tp~x might be -00 and Tp~x might be +00. We produce this maximal integral curve

J;

2.8. Vector Fields

99

as follows: Consider the collection Jp of all pairs (J, a), where J is an open interval containing 0 and a : J -t M is an integral curve of X with a(O) = p. Then let J; = U(J,a)EJpJ and define cmax(t) := a(t) whenever t E J for (J, a) E Jp. By existence and uniqueness, this definition is unambiguous. -t .M is the desired maximal integral curve and is The curve Cmax : easily seen to be unique.

J;

Definition 2.94. Let X be a Coo vector field on M. For any given p EM, let := (Tp~x' Tp~x) c ~ be the domain of the maximal integral curve c: -t M of X with c(O) = p. The maximal flow cpx is defined on the set (called the maximal flow domain)

J; J;

U J; x {p}

1)x =

pEM

by the prescription that t such that cpx (0, p) = p.

H

cpx (t,p) is the maximal integral curve of X

Thus by definition, X is a complete vector field if and only if 1)x IRxM. We will abbreviate (T;'x,Tp~x) to (Tp-, T p+).

Theorem 2.95. For X E X(M), the set 1)x is an open neighborhood of {O} x M in ~ x M and the map cpx : 1)x -t M is smooth. Furthermore,

(2.4)

cpx (t

+ s,p) = cpx (t, cpx (s,p))

whenever both sides are defined. If the right hand side is defined, then the left hand side is defined. Suppose that t, s ::::: 0 or t, s ::; O. Then if the left hand side is defined so is the right hand side. Proof. Let q = cpx (s,p). If the right hand side is defined, then s E (T;,T:) and t E (Tq-,T:). The curve 7jJ : T H cpx(s + T,p) is defined for T E (Tp- - s, T: - s), and this is the maximal domain for 7jJ. We have

ddT I 7jJ = dd cpx (s 7

+ T,p)

T

=

dd U

I

cpx (u,p)

U=S+7

= X(cpX (s + T,p)) = X(7jJ(T)).

17=0

We also have 7jJ(0) = cpx (T + s, p) = cpX (s, p) so 7jJ is an integral curve starting at q = cpX(s,p). Thus (Tp- - s,Tp+ - s) C (Tq-,Tq+) and 7jJ = !pX(.,q) on (Tp- - s,Tp+ - s). But the maximal domain for 7jJ is (Tp- - s, ~ - s) and so in fact (Tp- - s, T: - s) = (Tq-, T:) for otherwise cpx (., q) would be a proper extension. But then, since t E (Tq-, T:), we have that t E (T; - s, Tp+ - s) and so 7jJ(t) = cpx (t + s,p) is defined and

cpx (t

+ s, p)

=

cpx (t, cpx (s, p)).

2. The Tangent Structure

100

Now let us assume that t, s > 0 and that cpx (s+t,p) is defined (the case of t, s ~ 0 is similar). Then since s, t ~ 0, we have s, t, t + s E

(Tp-, T:) .

Let q = cpx(s,p) as before and let O(u) = cpx(s + u,p) be defined for u with 0 ~ u ~ t. But O(u) is an integral curve with 0(0) = q. Thus we have that cpx (u, q) must also be defined for u = t and O(t) = cpx (t, q). But cpX(t,q) cpx(t,cpX(s,p)), which is thereby defined, and we have

cpx (s + t,p) :_ O(t) _ cpx (t, cpx (s,p)). Now we will show that Vx is open in lR x M and that cpx : Vx --+ M is smooth. We carefully define a subset S C 'Dx by the condition that (t,p) E S exactly if there exists an interval J containing 0 and t and also an open set U C M such that the restriction of cpx to J x U is smooth. Notice that S is open by construction. We intend to show that S = 'Dx. Suppose not. Then let (tQ,po) E 'Dx n Sc. We will assume that to > 0 since the case to < 0 is proved in a similar way. Now let T := sup{t : (t,po) E S}. We know that (O,po) is contained in some flow box and so T > O. We also have T ~ to by the definition of to. Thus T E J:O and we define qo := cpx (T,PO). Now applying the local theory we know that qo is contained in an open set Uo such that cpx is defined and smooth on (-f, f) X Uo for some f > o. We will now show that cpx is actually defined and smooth on a set of the form (-8,r) x 0 where 0 is open, 8,r > 0, and (T,PO) E (-8,r) x O. Since this contradicts the definition of T, we will be done. We may choose tl > 0 so that T E (tl' tl +f) and so that cpX(tl'PO) E Uo. Note that (tl,PO) E S since tl < T. SO on some neighborhood (-8, tl +8) x U1 of (tl,PO) the flow cpx is smooth. By choosing Ul smaller if necessary we can arrange that cpx ({tl} x U1 ) c Uo. Now consider the equation

cpx (t, p) - cpX (t - tl, cpx (tl,p)). If It - tl < f and p E UI, then both sides are defined and the right hand side is smooth near such (t,p). But the right hand side is already known to be smooth on (-8, tl + 8) X UI. We now see that cpx is smooth on (-8, tl + f) X UI, which contains (T,PO) contradicting the definition of T. Thus S Vx. 0 Remark 2.96. In this text, cpx will either refer to the unique maximal flow defined on Vx or to its restriction to the domain of a flow box. In the latter case we call cpx a local flow. We could have introduced notation such as cp!ax, but prefer not to clutter up the notation to that extent unless necessary. We hope that the reader will be able to tell from context what we are referring to when we write cpx.

2.8. Vector Fields

101

If cpx is a flow of X, then we write CPt for the map p t---+ CPt (p). The (maximal) domain of this map is V~ - {p : t E (Tp~x' T:'-x )}' Note that, in general, the domain of CPt depends on t. Also, we have the tangent map Tocp; : ToJR ---t TpM and

:t It=o

cP; (t) = Tocp; :u 10 = X p,

where :'U 10 is the vector at 0 associated to the standard coordinate function on IR (denoted by u here). Exercise 2.97. Let sand t be real numbers. Show that the domain of is contained in V~+t and show that for each t, V~ is open. Show that


Definition 2.98. The support of a vector field X is the closure of the set {p: X(p) i= O} and is denoted supp(X). Lemma 2.99. Every vector field that has compact support is a complete vector field. In particular, if M is compact, then every vector field is complete.

J; J;

c;

Proof. Let be the maximal integral curve through p and = (T-, T+) its domain. If X(p) = 0, then the constant curve, c(t) = p for all p, is the unique integral curve through p and is defined for all t so = JR. Now (t) must always suppose X(p) i= O. If t E (T-, T+), then the image point lie in the support of X. Indeed, since X (p) is not zero, is not constant. If (t) were in M\ supp(X) , then (t) would be contained in some ball on which X vanishes and then by uniqueness this implies that is constantly (t) for all time a contradiction. But we show that if T+ < 00, equal to then given any compact set K c M, for example the support of X, there is an f > 0 such that for all t E (T+ - f, T+), the image (t) is outside K. If not, then we may take a sequence ti converging to T+ such that (ti) E K. But then going to a subsequence if necessary, we have Xi := (ti) ---t X E K. Now there must be a flow box (U, a, x), so that for large enough k, we have that tk is within distance a of T+ and Xk = (tk) is inside U. We are then guaranteed to have an integral curve ~ (t) of X that continues beyond T+ and thus can be used to extend which is a contradiction of the maximality of T+. Hence we must have T+ = 00. A similar argument gives the result that T- = -00. 0

c:

c: c: c:

c:

c;

c:

c:

c;

c;

c:

Exercise 2.100. Let a > 0 be any fixed positive real number. Show that iffor a given vector field X, the flow cpx is defined on (-a, a) x M, then in fact the (maximal) flow is defined on JR x M and so X is a complete vector field.

2. The Tangent Structure

102

@~ ~@ Figure 2.3. Isolated vanishing points

Figure 2.4. Straightening

If a vector field is zero at some point but nonzero elsewhere in a neighborhood of that point, then we say that the field has an isolated zero. The structure of a vector field near such an isolated zero can be quite complex and interesting. The qualitative structure of three such possibilities are show in Figure 2.3 for dimension two. Near zeros that are not isolated, the situation is also potentially complex. On the other hand, at non-vanishing points, all vector fields are the same up to a local diffeomorphism. This is the content of the following theorem which is sometimes called the straightening theorem. Theorem 2.101. Let X be a smooth vector field on M with X(p) some p E M. Then there is a chart (U, x) with p E U such that X

a

= axl

=1=

0 for

on U.

Proof. Since this is clearly a local problem, it will suffice to assume that M = ]Rn and p = O. Let (u l , ... , un) be standard coordinates on ]Rn. By a rotation and translation if necessary, we may assume that X(O) = The idea is that there must be a unique integral curve through each point of the hyperplane {u1 = O}. We wish to arrange for the new coordinates of q E ]Rn to be such that if an integral curve of X passes through (0, a 2 , ••• , an) at time zero and hits q at time t, then xi(p) = ai for i = 2, ... , n while xl(p) = t. Let r.p be a local flow for X near 0, and define X in some sufficiently

blo.

2.8. Vector Fields

103

small neighborhood of

°

by

x(a l , ••. , an) := <Pal (0, a2 , •. • , an). For a = (aI, a2 , ••• , an) in the domain of X, and 1 E COO(M), we have

TX' ~ 8u l 1a 1 = ~ 8u l 1a (f 0 X) = lim -hl [I(x(a l h-+O

+ h, a2 , •.. , an)) -

I(x(al, a2 , ••• , an))]

= lim-hl [I (<Pal +h(0,a2 , ... ,an )) -1(X(a l ,a 2 , ... ,an ))] h-+O

= lim -hI [J (
In particular, TX'

TX'

8~i

= (Xf)(X (a)).

Iur 10 = Iur 10' If i > 0, then at °we have

10

1=

8~t

10

lox

= lim -hl [J(X(O, '" h, .• .) - 1(0)] h-+O

= h-+O lim -hl [1(0, ... , h, ... ) -

1(0)]

= 88ut·1 0 f.

Thus ToX = id and so by the inverse mapping theorem (Theorem 2.25) we see that after restricting X to a smaller neighborhood of zero, the map x := X-I is a chart map. We have already seen that TX I = X 0 X. But then for 1 E Coo (M) we have

Iur

-; 1

8x

p

1=

81 1 1 0 8u x(P)

x-I =

8 1 1 lox 8u x(p)

=Tx 88 1 1 1= (Xox)(x(p))I=Xpl, u x(p) so that

Ixr = x.

o

2.8.2. Lie derivative. We now introduce the important concept of the Lie derivative of a vector field extending the previous definition. The Lie derivative will be extended further to tensor fields.

Definition 2.102. Given a vector field X, we define a map LX : X(M) -+ X(M) by

LXY:= [X, Y]. This map is called the Lie derivative (with respect to X).

2. The Tangent Structure

104

The Jacobi identity for the Lie bracket easily implies the following two identities for any X, Y, Z E X(M) -+ X(M):

..cx[Y, Z] = [..cxY,Z] + [Y,..cxZ], (Le . ..c[X,y] = ..cx o..cy - ..cy 0 ..cx). ..c[X,y] = [..cx,..c y ] We will see below that ..cx Y measures the rate of change of Y in the direction X. To be a bit more specific, (..cx Y) (p) measures how Y changes in comparison with the field obtained by "dragging" Yp along the flow of X. Recall our definition of the Lie derivative of a function (Definition 2.70). The following is an alternative characterization in terms of flows: For a smooth function f : M -+ R and a smooth vector field X E X(M), the Lie derivative ..cx of f with respect to X is given by

..cx f (p)

:t

=

10 f 0
Exercise 2.103. Explain why the above formula is compatible with Definition 2.70. We will also characterize the Lie derivative on vector fields in terms of flows. First, we need a technical lemma: Lemma 2.104. Let X E X(M) and f E Coo(U) with U open and p E U. There is an interval 10 := [-8,8] and an open set V containing p such that
f(
E

10

+ tg(t, q)

x V and such that g(O, q) = Xqf for all q E V.

Proof. The existence of the set 10 x V with '{)X (10 x V) C U follows from our study of flows. The function r(r,q) := f(
r

1 8r g(t,q):= Jo 8r(st,q)ds,

so that

tg(t, q) =

fal ~~ (st, q)t ds = fal :s r(st, q) ds = r(t, q).

Then f(
+ tg(t, q).

Also

g( O q) = lim ~r(t q) = lim f(
t-+O

t'

t-+O

t

q

0

105

2.8. Vector Fields

Proposition 2.105. Let X and Y be smooth vector fields on M. Let cP = cpx be the flow. The function t M Tcp-t . Y
dd I Tcp-t· Y
(2.5)

Proof. Let

I

E

COO(U) with p

E

U as in the previous lemma. We have

dI . Tcp-t . Yept(p) - ¥p . Ypi - (Tcpt . Yep tCP)) I -d TCP-t·Y
Ypi - (Tcpt . Yep t(p)) I

¥pI - Yep-t(p) (J 0 CPt)

t

t

_ ¥pI - Yep t(p) (J + tYt) t

where Y is as in the lemma and Yt(q) = y(t, q). Continuing, we have

Ypi - Y
Taking the limit as t side above becomes

t

---t

0 and recalling that Yo

= XI

on V, the right hand

lim (Y I)(p) - (Y J)(cp-t(p)) -lim Y t

t--+O

t--+O

ep_t(p)Yt.

ep-t (p)

Y t

= lim (Y 1)( CPt (p)) - (Y J) (p) - Yo X I t--+O t p

= X ,X I - YpX I = [X, Y]pf. All told we have

dd

I

t t=O

(Tcp-t . Y
for all f E Coo (U). If we let U be the domain of a chart (U, x), then letting I be each of the coordinate functions we see that each component of the TpMvalued function t M Tcp-t· Y
ft

We see from this characterization that in order for (£xY) (p) to make sense, Y only needs to be defined along the integral curve of X that passes through p. Discussion: Notice that if X is a complete vector field, then for each t E R the map cpf is a diffeomorphism M ---t M and

(2.6)

(cpf) * Y = (TCPf)-l 0 Y

0

cPf.

2. The Tangent Structure

106

One may write

cx Y -

:tl

o (
On the other hand, if X is not complete, then there exists no t such that O. It follows that for all 0 ~ t ~ € the map
(
It

Theorem 2.106. Let X, Y be vector fields on a smooth manifold M. Then d dt (
2.8. Vector Fields

107

Proof. Let cp := cpX. Suppressing the point p we have

:t

It cp;Y =

! 10

= cp; Exercise 2.107. Show that

=

CP;+sY

:s

! 10 cp;(cp~*Y) o

10 (cp;Y) = cp;£xY.

ftlo (cp~Y)(p) =

- (£xY) = -[X,Y].

Proposition 2.108. Let X E X(M) and Y E X(N) be f-related vector fields for a smooth map f : M ~ N. Then

f

cpf

0

= cpr 0 f

whenever both sides are defined. Suppose that f : M ~ N is a diffeomorphism and X E X(M). Then the flow of foX = (f-l)* X is f 0 cpf 0 f- 1 and the flow of f* X is f- 1 0 cpf 0 f. Proof. For any p E M, we have ft(f 0 cpf)(p) = Tf ·ftcpf(p) = Tf 0 X 0 cpf (P) = Yo f 0 cpf (p). But f 0 cp~ (p) = f(p) and so t t-+ f 0 cpf (p) is an integral curve of Y starting at f(p). By uniqueness we have f 0 cpf (p) = cpi (f(p)). The second part follows from the first. 0 Theorem 2.109. For X, Y E X(M), the following are equivalent:

(i) £xY

= [X,

Y] = O.

(ii) (cpf)*Y = Y whenever defined.

(iii) The flows of X and Y commute: cpf

0

cpr = cpr

0

cpf whenever defined.

Proof. (Sketch) The equivalence of (i) and (ii) follows easily from Proposition 2.105 and Theorem 2.106. Using Proposition 2.108, the equivalence of (ii) and (iii) can be seen by noticing that cpf 0 = 0 cpf is defined and true exactly when = Cp~t 0 0 cpf is defined and true, which happens exactly when

cpr

cpr cpr

cpr Y

CPs

(
= CPs

t

is defined and true. This happens, in turn, exactly when Y

= (cpf)*Y.

0

Example 2.110. On~.2 we have the flows given by ¢(t, (x, y)) = ¢t(x, y) := (x + ty, y) and 1/;(t, (x, y)) = 1/;t(x, y) := (x, y + t). We have 1/;1 0 ¢1(0, 0) = (1,1) while ¢l 01/;1(0,0) = (0,1). These noncommuting flows correspond to the noncommuting vector fields X = y8/8x and Y = 8/8y. Exercise 2.111. Find global noncommuting flows on S2.

108

2. The Tangent Structure

x y

x p Figure 2.6. Bracket measures lack of commutativity of flows

The Lie derivative and the Lie bracket are essentially the same object and are defined for local sections X E XM(U) as well as global sections. This is obvious anyway since open subsets are themselves manifolds. As is so often the case for operators in differential geometry, the Lie derivative is natural with respect to restriction, so we have the commutative diagram

X(U) r~ .} X(V)

.cxlu ---7

.c x

u

---7

X(U) .} r~ X(V)

where Xlu denotes the restriction of X E X(M) to the open set U and r~ is the map that restricts from U to V cU. If X and Yare smooth vector fields with flows 'Px and 'P Y , then starting at some p EM, if we flow with X for time 0, then flow with Y for time 0, and then flow backwards along X and then Y for time 0, we arrive at a point a(t) given by Y x Y x (t) . a

.= 'P -..;t 0 'P -..;t 0 'P..;t 0 'P..;t.

It turns out that a(t) is usually not p (see Figure 2.6). In fact, we have the

following theorem:

Theorem 2.112. With a(t) as above we have

dd

t

Proof. See Problem 8.

I

a(t) = [X, Y](p).

t=O

D

We know by direct calculation that if (xl, ... , xn) are the coordinate functions of a chart, then

2.8. Vector Fields

109

for all i,j = 1, ... , n. The converse is also locally true. This follows from the next theorem, which can be thought of as saying that we can simultaneously "straighten" commuting local fields.

Theorem 2.113. Let M be an n-manifold. Suppose that on a neighborhood V of a point p E M we have vector fields Xl, . .. ,Xk such that X 1(x), ... , Xk(X) are linearly independent for all x E V. If [Xi, X]] = 0 on V for all i, j, then there exist a possibly smaller neighborhood U of p and a chart x : U ---+ ]Rn such that 8 . -8. = Xi on U for z = 1, ... , k x~

and such that for the corresponding flows we have -1 ( 1 ... ,Ui , ... , Un) = (1 'Ptx, 0 XU, U , ... , Ui

+ t , ... , Un)

for i = 1, ... , k. Proof. Let p E V and choose a chart (0, y) centered at p with 0 c V. By rearranging the coordinate functions if necessary and using a simple linear independence argument, we may assume that the vectors

Xl (p), ... , Xk(p) ,

8x~+llp , ... , 8~n Ip

form a basis for TpM. Let 'P~l, ••• , 'P~" be local flows for Xl, .. . , Xk and let W be an open neighborhood of p contained in 0 such that the composition x" 0'" 0 'PtlXl 'Pt" is defined on Wand maps W into 0 for all t1, ... , tk E (-f, f) as in Lemma 2.92. Define

.- {( a k+1 , ... , an).. y -1(0 , ... , 0 ,ak+1 , ... , an)} S .and define the map 'I/; : (-f, f)k x S ---+ U by 0"( 1 k k+1 , ... , Un) -_ 'Pu" X" 0'" 0 'PUI Xl 'I" U , ... , U ,U

0

E

W,

Y-1(0 , ... , 0 ,Uk+1 , ... , Un) .

Since [Xi, X]] = 0 for 0 ~ i, j ~ k, we know that the flows in the composition above commute. Hence for any a E y (W) and smooth function f, we have for 1 ~ i ~ k,

TatP·

--

8~~la = 8~~lafO'l/;(u1, ... ,un) 8 i Ia f( 'Pu" X" Xl Y-1(0 , ... , 0 ,Uk+1 , ... , Un)) 8u

8 = 8u~

0'"

I f('P~' a

= Xi('I/;(a))f,

0

0 'PUI 0

'P~k"

0 .•• 0

--.

'P;"

0··· 0

'P~l

0

Y 1(0, ... ,0, uk+1, ... , Un))

2. The Tangent Structure

110

where the caret indicates omission. Thus we have Ta'ljJ·

a~i la =

Xi('ljJ(a)) for 1

~ i ~ k,

and in particular, To'ljJ·

For k + 1 ~ i

~

a~i 10 =

Xi(p) for 1

~ i ~ k.

n, we have

0"(0 - -1(0 , ... , 0 ,uk+1 , ... ,u, n) If' , ... , 0 ,uk+1 , ... ,un) -y

so To'ljJ·

a~i 10 = a~i Ip .

Thus To'ljJ maps a basis of TojRn to a basis of TpM. We can use the inverse mapping theorem (Theorem 2.25) to conclude that 'ljJ is a diffeomorphism from some open neighborhood of 0 E jRn onto an open neighborhood U of p in M. It is now straightforward to check that x = 'ljJ-1 is a chart map of D the desired type. For later reference, we note that if (U, x) is a chart of the type produced in the theorem above, then we can easily arrange that x (U) is of the form V x W C jRk X jRn-k, so that the fields are tangent to the submanifolds of the form S := {p E U: x k+1(p) = ak+l, ... , xk+1(p) = an}.

Ixr, ... ,b

The reader has probably surmised that the mathematics of vector fields and flows can be applied to fluid mechanics. This is quite true, but one needs to deal with time dependent vector fields. There is a trick that allows time dependent vector fields to be treated as ordinary vector fields on a manifold of one higher dimension, but it is not always best to think in those terms. The author has included a bit about time dependent vector fields in the online supplement [Lee, J eff1.

2.9. l-Forms Definition 2.114. A smooth (resp. or) section of the cotangent bundle is called a smooth (resp. or) l-form or also a smooth (resp. or) covector field. The set of all or l-forms is denoted by Xr*(M) and the set of smooth l-forms is denoted by X*(M). The set xr*(M) is a module over or(M). Later we will have reason to denote X*(M) also by 01(M). The analogue of Lemma 2.65 is true. That is, if O:p E T; M, then there is a l-form 0: such that o:(p) = p. The proof is again a simple cut-off function argument.

2. The Tangent Structure

112

which is interpreted to mean that at each p E Uo; we have

df(p) =

L ~:i Ip dxii p'

The covector fields dx i form what is called a coordinate co frame field or holonomic coframe field 2 over U. Note that the component functions X~ of a vector field with respect to the chart above are given by Xi = dx~(X), where by definition dx~(X) is the function p t-+ dx~ip (Xp). Thus ~. 8 Xlu = ~ dx~(X) 8xi '

Note. If a is a l-form on M and p EM, then one can always find many functions f such that df(p) = a(p), but there may not be a single function f such that this is true for all points in a neighborhood, let alone all points on M. If in fact df = a for some f E COO(M), then we say that a is exact. More on this later. Let us try to picture l-forms. As a warm up, let us recall how we might picture a tangent vector vp at a point p E ]Rn. Let "I be a curve with "I' (0) = vp. If we zoom in on the curve near p, then it appears to straighten out and so begins to look like the curve t t-+ P + tvp. So one might say that a tangent vector is the "infinitesimal representation" of a (parameterized) curve. At each point a l-form gives a linear functional in that tangent space, and as we know, the level sets of a linear functional are parallel affine subspaces or hyperplanes. We should imagine level sets as being labeled by the values of the function. A covector puts a ruling in the tangent space that measures tangent vectors stretching across this ruling. For example, the l-form df in ]R3 gives a ruling in each tangent space as suggested by Figure 2.7a. For a given p, the level sets of dfp are what we see if we zoom in on the level sets of f near p. The fact that the individual dfp's living in the tangent spaces somehow coalesce into the level sets of the global function f as shown in 2.7b, is due to the fact that the l-form df is the differential of the function

f· A more generall-form a is still pictured as straight parallel hyperplanes in each tangent space. Because these level sets live in the tangent space, we might call them infinitesimal level sets. These (value labeled) level sets may not coalesce into the level sets of any global smooth function on the manifold. There are various, increasingly severe ways coalescing may fail to happen. The least severe situation is when a is not the differential of a global function but is still locally a differential near each point. For example, 2The word holonomic comes from mechanics and just means that the frame field derives from a chart. A related fact is that [8~ '8~ J = O.

2.9. l-Forms

113

Figure 2.7. The form df follows level sets

Figure 2.8. Level sets of overlapping angle functions

if M = ]R2\{O}, then the familiar l-form a = (x 2 + y2)-1(_ydx + xdy) is locally equal to d() for some "angle function" () which measures the angle from some fixed ray such as the positive x axis. But there is no such single smooth angle function defined on all of ]R2\ {O}. Thus, globally speaking, a is not the differential of any function. In Figure 2.8, we see the coalesced result of "integrating" the infinitesimal level sets which live in the tangent spaces. While these suggest an angular function, we see that if we try to picture rising as we travel around the origin, we find that we do not return to the same level in one full circulation, but rather keep rising. Locally, however, we really do have level sets of smooth functions. Now the second more severe way that a l-form may fail to be the differential of a function is where there is not even a local function whose differential agrees with the l-form near a point. The infinitesimal level sets do not coalesce to the level sets of a smooth function, even in small neighborhoods. This is much harder to represent, but Figure 2.9 is meant to at least be suggestive. Nearby curves cross inconsistent numbers of level sets. AI:. an example consider the l-form {3 = -ydx + xdy.

The astute reader may object that surely radial rays do match up with the

2. The Tangent Structure

114

Figure 2.9. Suggestive representation of a form which is not closed

directions described by this 1-form. However, the point is that a covector in a tangent space is not completely described by the level sets as such, but rather the level sets are to be thought of as labeled according to the values they represent in the individual tangent spaces. Here we have a case where the infinitesimal level sets coalesce, but the values assigned to them do not; they are 1-dimensional submanifolds that fit the 1-form but they are not level sets of even a local smooth function whose differential agrees with the 1-form. This brings us to the most severe case which only happens in dimension 3 or greater. It can be the case that there is no nice family of (n - 1)-dimensional submanifolds that line up in any reasonable sense with the I-form (either globally or locally). This is the topic of the Frobenius integrability theory for tangent distributions that we study in Chapter 11, and we shall forgo any further discussion until then except to say that the reader should be ready to understand much of that chapter, including the Frobenius theorem, after finishing the next chapter.

Definition 2.116. If ¢ : M -t N is a Coo map, the pull-back of a 1-form a E X*(N) by ¢ is defined by (¢*a)p' v = a¢(p) (Tp¢ . v)

for v E TpM. This extends the notion of the pull-back of a function defined earlier. If we view a 1-form on M as a map T M -t JR, then the pull-back is given by ¢*a=aoT¢.

Exercise 2.117. The pull-back is contravariant in the sense that if ¢l : Ml -+ M2 and ¢2 : M2 -t N, then for a E X*(N) we have (¢2 0 ¢l)* =

¢i 0 ¢2' Next we describe the local expression for the pull-back of a 1-form. Let (U, x) be a chart on M and (V, y) a coordinate chart on N with ¢(U) c V. A

2.9. I-Forms

115

typical I-form has a local expression on V of the form a = Coo(V). The local expression for ¢*a on U is ¢*a = 2: (ai

2: aidyi for ai E 0

¢) d (yi

0

¢) =

. 3 L: (ai 0 ¢) a(y'o¢) ax) dx J • Thus we get a local pull-back formula convenient for computations:

(2.7) The pull-back of a function or I-form is defined whether ¢ : M --+ N happens to be a diffeomorphism or not. On the other hand, the pull-back of a vector field only works in special circumstances such as where ¢ is a diffeomorphism. Let ¢ : M --+ N be a Coo diffeomorphism with r ~ 1. Recall that the push-forward of a function f E Coo(M) is denoted ¢d and defined by ¢*f(P) := f(¢-l(p)). We can also define the push-forward of a I-form as ¢*a = a 0 T¢-l. Exercise 2.118. Find the local expression for ¢*f and ¢*a. Explain why we need ¢ to be a diffeomorphism. Lemma 2.119. The differential is natural with respect to pull-back. In other words, if ¢ : M --+ N is a Coo map and f : N --+ lR a Coo function, then d( ¢* f) = ¢* df. Consequently, the differential is also natural with respect to

restrictions. Proof. We wish to show that

(¢*df)p = d(¢* f)p for all p E M. Let v E TpM and write q = f(p). Then

(¢*df)lp v = dflq (Tp¢' v) = ((Tp¢' v) f) (q) =v(fo¢)(p) = d(¢*f)lpv. The second statement is obvious from local coordinate expressions, but also notice that if U is open in M and L : U <---+ M is the inclusion map (Le. the identity map idM restricted to U), then flu = l,* f and dflu = l,*df. So the statement about restrictions is just a special case of the first part. 0 The tangent and cotangent bundles T M and T* M are themselves manifolds and so have their own tangent and cotangent bundles. Among other things, this means that there exist I-forms and vector fields on these manifolds. Here we introduce the canonical I-form on T* M. We denote this form by Ocan and note that it is a section of T* (T* M). Let a E T* M and suppose that a is based at p so that a E T;M. Consider a vector U a E Ta (T* M). 3To ensure clarity we have not used the Einstein summation convention here.

2. The Tangent Structure

116

Notice that since 1r : T* M --+ M, we have Ta1r : Ta (T* M) --+ TpM. Thus Ta1r . U a E TpM. We define

Oean(u a) = a (Ta1r' u a). The definition makes sense because a E T; M and Ta1r . U a E TpM. Let (U, x) be a chart containing P and let (xl 0 1r, ... ,xn 0 1r, PI, ... ,Pn) = (ql, ... , qn,PI,'" ,Pn) be the associated natural coordinates for T* M.

Exercise 2.120. It is geometrically clear that Ta1r' a~.la = 0 since a~.la is tangent to the fiber T;(a)M along which 1r is constant. Deduce this directly from the definitions. Hint:

a~.la

can be represented by a curve in T;(a)M.

We wish to show that locally Oean =

Oean( a~.

L: Pi dqi.

U= Pi(a) and Oean( a~' U= 0 for all Oean(

since in fact Ta 1r'

8~i

a~' la = Oean(

I)

= a (Ta 1r '

It will suffice to show that

i. We have

8~i

IJ

= 0

0 by Exercise 2.120. Also, we have

8~i

I)

IJ

8~i ~ a ( /)~, ~ a' ~ p,(a), = a (Ta 1r '

I.)

a~.la = a~.lp' which follows from the Indeed, we know that Ta1r' a~.la = cf ~ Ip for some

where we have used the fact that Ta1r' definition qi constants

= xi 01r.

cf, but we have cf = dx k (Ta 1r ' 8~i = d ( xk

0

1r)

IJ

(8~i

= 1r*dxk

J

1

= dqk

(8~i

(8~i

I

IJ

J

=

8:'

This 1-form plays a basic role in symplectic geometry and classical mechanics. For more about symplectic geometry see the online supplement [Lee, J effj.

2.10. Line Integrals and Conservative Fields Just as in calculus on Euclidean space, we can consider line integrals on manifolds, and it is exactly the 1-forms that are the appropriate objects to integrate. First notice that all 1-forms on open sets in ]Rl must be of the

2.10. Line Integrals and Conservative Fields

117

form f dt for some smooth function f, where t is the coordinate function on RI. We begin by defining the line integral of a 1-form defined and smooth on an interval [a, bl C RI. If {3 = f dt is such a 1-form, then

r {3:= lb f(t) dt.

J[a,b)

a

Any smooth map '"'{ : [a, bl ~ M is the restriction of a smooth map on some larger open interval (a - c, b + c), and so there is no problem defining the pull-back '"'{*o.. If '"'{ : [a, bl ~ M is a smooth curve, then we define the line integral of a 1-form a along'"'{ to be

1 r 0.:=

J[a,b)

'Y

'"'{*o. =

lb

f(t) dt.

a

where '"'{*o. = f dt. Now if t = 4J(s) is a smooth increasing function, then we obtain a positive reparametrization 7 = '"'{ 0 4J : [e, d] ~ M. where 4J(e) = a and 4J(d) = b. With such a reparametrization we have

1

[e,d)

r 4J*'"'{*o. = r 4J*(f dt) = 1 f J[e,d) [e,d)

7*0. =

=

J[e,d)

ld

f(4J (s))4J' (s) ds

=

0

4J ~4J ds

lb

S

f(t) dt,

where the last line is the standard change of variable formula and where we have used 4J*(fdt) = d(~:t/»ds, which is a special case of the pull-back formula mentioned above. We see now that we get the same result as before. This is just as in ordinary multivariable calculus. We have just transferred the usual calculus ideas to the manifold setting. Definition 2.121. A continuous curve '"'{ : [a, bl ~ M into a smooth manifold is called piecewise smooth if there exists a partition a = to < tl < ... < tic = b such that '"'{ restricted to [ti' ti+11 is smooth for 0 ~ i ~ k - 1 (in the sense of Definition 1.58). It is convenient to extend the definitions a bit to include integration along piecewise smooth curves. Thus if'"'{ : [a, bl ~ M is such a curve, then we define for a 1-form a.

where ,",{, is the restriction of'"'{ to the interval [til ti+11.

2. The Tangent Structure

118

Just as in ordinary multivariable calculus we have the following: Proposition 2.122. Let, : [a, b] --+ M be a piecewise smooth curve with ,(a) = PI and ,(b) = P2. If a = df, then

~ a = ~ df = f(P2) -

f(p1).

In particular, I-y a is path independent in the sense that it is equal to Ie a for any other piecewise smooth path c that also begins at PI and ends at P2. Definition 2.123. If a is a I-form on a smooth manifold M such that Ie a = 0 for all closed piecewise smooth curves c, then we say that a is conservative.

We will need a lemma on differentiability. Lemma 2.124. Suppose f is a function defined on a smooth manifold M,

and let a be a smooth I-form on M. Suppose that for any P EM, vp E T M and smooth curve c with c(O) = vp, the derivative 1tlof(c(t)) exists and :t 10 f(c(t)) = a(v p). Then f is smooth and df = a. Proof. We work in a chart (U, x). If we take c(t) := x-1(x(p)+tei), then the hypotheses lead to the conclusion that all the first order partial derivatives of f 0 x-I exist and are continuous. Thus f is a1. But then also dfp . vp = 1tlof(c(t)) = ap(vp) for all vp, and it follows that df = a, and this also implies that f is actually smooth. D Proposition 2.125. If a is a I-form on a smooth manifold M, then a is

conservative if and only if it is exact. Proof. We know already that if a = df, then a is conservative. Now suppose a is conservative. Fix Po E M. Then we can define f(p) = I-ya, where, is any curve beginning at Po and ending at p. Given any vp E TpM, we pick a curve c : [-1, c) with c > 0 such that c( -1) = Po, c(O) = p and d (0) = vp. Then

-d f(C(T)) == - 1 id dT 0 dT 0 =

j

e[

1,7']

a

d~ 10 11[-1,0] a + ddT 10 1 [0,7'] a

= 0+

1

d~ 10 foT c*a - :1' 10 fo7' g(t) dt

= g(O),

2.10. Line Integrals and Conservative Fields

119

where c*o: = 9 dt. On the other hand,

O:(Vp ) = o:(c'(O)) =

= c*o: (:t

0:

IJ

(Toc.

:t IJ

= g(O) dtl o (:t

IJ

= g(O),

iT

where t is the standard coordinate on R Thus 10 f(C(T)) = o:(vp ) for any vp E TpM and any p E M. Now the result follows from the previous lemma. D

It is important to realize that when we say that a form is conservative in this context, we mean that it is globally conservative. It may also be the case that a form is locally conservative. This means that if we restrict the I-form to an open set which is diffeomorphic to a Euclidean ball, then the result is conservative on that ball. The following examples explore in simple terms these issues.

Example 2.126. Let 0: = (x 2 + y2(1 (-y dx + x dy). Consider the small circular path c given by (x, y) = (xo + £ cos t, Yo + £ sin t) with 0 ::; t ::; 27r and £ > O. If (xo, YO) = (0,0), we obtain

1

e 0:

{27r

= Jo

= Thus

0:

1 ( £2

1

27r

o

dx -y(t) dt

dY)

+ x(t) dt

1 2" ( - (£ sin t) (-£ sin t) £

dt

+ (£ cos t) (£ cos t)) dt = 27r.

is not conservative and hence not exact. On the other hand, if

(xo, Yo) f. (0,0), then we pick a ray Ro that does not pass through (xo, Yo) and a function O(x, y) which gives the angle of the ray R passing through (x, y) measured counterclockwise from Ro. This angle function is smooth

Y5, Ie

and defined on U = JR2\Ro. If £ < !Jx~ + then c has image inside the domain of 0 and we have that 0:1 U = dO. Thus 0: = O(c(O)) -O(c(27r)) = O. We see that 0: is locally conservative. Example 2.127. Consider (3 = y dx - x dy on JR 2. If it were the case that for some small open set U C JR2 we had (3lu = df, then for a closed path c with image in that set, we would expect that (3 = f(c(27r)) - f(c(O)) = O. However, if c is the curve going around a circle of radius £ centered at

Ie

120

2. The Tangent Structure

(xo, Yo), then we have

1~ 1(y( = =

t) ~~ - x( t) ~~) dt

1271' (( Xo + c sin t) (-c sin t) -

(Yo + c cos t) (c cos t)) dt

= -2c 2 7l', so we do not get zero no matter what the point (xo, Yo) and no matter how small c. We conclude that ~ is not even locally conservative. The distinction between (globally) conservative and locally conservative is often not made sufficiently clear in the physics and engineering literature. Example 2.128. In classical physics, the static electric field set up by a fixed point charge of magnitude q can be described, with an appropriate choice of units, by the 1-form q

q

q

gxdx+ gydy+ gzdz, p p p where we have imposed Cartesian coordinates centered at the point charge and where p = x 2 + y2 + z2. Notice that the domain of the form is the punctured space ]R3 \ {o}. In spherical coordinates (p, (), ¢), this same form is

J

q

p2dp= d

(-q) p ,

so we see that the form is exact and the field is conservative.

2.11. Moving Frames It is important to realize that it is possible to get a family of locally defined vector (resp. covector) fields that are linearly independent at each point in their mutual domain and yet are not necessarily of the form a~. (resp. dx i ) for any coordinate chart. In fact, this may be achieved by carefully choosing n 2 smooth functions f~ (resp. a~) and then letting Ek := 2:i f~ a~. (resp. ()k

:=

2:i a~dxi).

Definition 2.129. Let E 1, E 2, ... , En be smooth vector fields defined on some open subset U of a smooth n-manifold M. If E1 (P), E2 (P), ... , En (p) form a basis for TpM for each p E U, then we say that (E1' E2,"" En) is a (non-holonomic) moving frame or a frame field over U. If E1, E2, ... ,En is a moving frame over U C M and X is a vector field defined on U, then we may write

X = I::XiEi on U,

121

2.11. Moving Frames

for some functions Xi defined on U. If the moving frame (E1, ... , En) is not identical to some frame field (Ixr, ... , arising from a coordinate chart on U, then we say that the moving frame is non-holonomic. It is often possible to find such moving frame fields with domains that could never be the domain of any chart (consider a torus).

Ixn)

Definition 2.130. If E 1, E2, . .. ,En is a frame field with domain equal to the whole manifold M, then we call it a global frame field. Most manifolds do not have global frame fields. Taking the basis dual to (E1{p), ... , En{P)) in T;M for each P E U we get a moving coframe field (o1, ... , on). The Oi are I-forms defined on U. Any I-form 0: can be expanded in terms of these basic I-forms as 0: = L ad)i. Actually it is the restriction of 0: to U that is being expressed in terms of the 02 , but we shall not be so pedantic as to indicate this in the notation. In a manner similar to the case of a coordinate frame, we have that for a vector field X defined at least on U, the components with respect to (E1, .. . , En) are given by Oi{X):

Let us consider an important special situation. If M x N is a product manifold and (U, x) is a chart on M and (V, y) is a chart on N, then we have a chart (U x V, x x y) on M x N where the individual coordinate functions are xl 0 pr1, ... , xm 0 pr1, y1 0 pr2, ... , yn 0 pr2, which we temporarily denote bY -1 x , ... , x::::Tn , -1 y , ... , ::-:n y. Now we conSl·der wh a t·IS the reIa t·IOn b et ween the coordinate frame fields (Ixr, ... m ), (-/yr, ... -/yn) and the frame field

(Ixr, ... , a~n).

a:

The latter set of n + m vector fields is certainly a linearly independent set at each point (p, q) E U x V. The crucial relations are a~.f = a~i (f 0 prl) and a~' = a~' (f 0 p r 2)· Exercise 2.131. Show that Tpr2

a~' Ip

= Tprl ai?,I(p) and x ,q

l,Y I

q

a~Y I(p,q) .

Remark 2.132. In some circumstances, it is safe to abuse notation and denote Xi 0 pr1 by xi and yi 0 pr2 by yi. Of course we are denoting a~. by a~' and so on.

A warning (The second fundamental confusion of calculus 4 ): For a chart (U, x) with x = (xl, ... , x n ), we have defined for any appropri-

l!r

ately defined smooth (or C1) function

f. However, this notation can be

'In [Pen], Penrose a.ttributes this cute terminology to Nick Woodhouse.

122

2. The Tangent Structure

i!r

ambiguous. For example, the meaning of is not determined by the coordinate function xl alone, but implicitly depends on the rest of the coordinate functions. For example, in thermodynamics we see the following situation. We have three functions P, V and T which are not independent but may be interpreted as functions on some 2-dimensional manifold. Then it may be the case that any two of the three functions can serve as a coordinate depends on whether we are using the coordinate system. The meaning of functions (P, V) or alternatively (P, T). We must know not only which function we are allowing to vary, but also which other functions are held fixed. To get rid of the ambiguity, one can use the notations (Us) V and (U) T" In the first case, the coordinates are (P, V), and V is held fixed, while in the second case, we use coordinates (P, T), and T is held fixed. Another way to avoid ambiguity would be to use different names for the same functions depending on the chart of which they are considered coordinate functions. For example, consider the following change of coordinates:

Us

y2

yl = xl + x 2 , = xl _ x 2 + x 3 , y3 = x3.

Here y3 - x 3 as functions on the underlying manifold, but we use different symbols. Thus may not be the same as The chain rule shows that in fact = + This latter method of destroying ambiguity is not very helpful in our thermodynamic example since the letters P, V and T are chosen to stand for the physical quantities of pressure, volume and temperature. Giving these functions more than one name would only be confusing.

b

b

ib a?

a?

Problems (1) Show that if j : M -7 N is a diffeomorphism, then for each p E M the tangent map Tpj : TpM -7 Tf(p)N is a vector space isomorphism. (2) Let M and N be smooth manifolds, and j : M -7 N a Coo map. Suppose that M is compact and that N is connected. If j is injective and Tpj is an isomorphism for each p EM, then show that j is a diffeomorphism. (Use the inverse mapping theorem.) (3) Find the integral curves in ]R2 of the vector field X = edetermine if X is complete or not.

xtx + ~ and

(4) Which integral curves of the field X = x 2 fx + y /y are defined for all times t?

Problems

123

(5) Find a concrete description of the tangent bundle for each of the following manifolds: (a) Projective space IRpn. (b) The Grassmann manifold G (k, n).

(6) Recall that we have charts on IRP2 given by [x,y,z]1-t (Ul,U2) = (x/z,y/z) on U3 = {z i= O}, [x, y, z]1-t (VI, V2) = (x/y, z/y) on U2 = {y i= O}, [x, y, z] I-t (WI, W2) = (y/x, z/x) on Ul = {x i= O}. Show that there is a vector field on IRP2 which in the last coordinate chart above has the following coordinate expression:

a

a

WI--W2-· aWl aW2

What are the expressions for this vector field in the other two charts? (Caution: Your guess may be wrong!). (7) Show that the graph r(f) = {(p, f(p)) E M x N : p E M} of a smooth map f : M -+ N is a smooth manifold and that we have an isomorphism T(p,/(p)) (M x N) ~ T(p,j(p))r(f) EB Tf(p)N. (8) Prove Theorem 2.112. (9) Show that a manifold supports a frame field defined on the whole of M exactly when there is a trivialization of TM (see Definitions 2.130 and 2.58). (10) Prove Proposition 2.85. (11) Find natural coordinates for the double tangent bundle TTM. Show that there is a nice map s : TT M -+ TT M such that s 0 s = idTTM and such that T1f 0 S = T1fTM and T1fTM 0 s = T7r. Here 1f : TM -+ M and 1fTM : TTM -+ TM are the appropriate tangent bundle projection maps. (12) Let N be the subset of IR n+1 x IRn +l defined by N = {(x, y) : Ilxll = 1 and x . y = O} is a smooth manifold that is diffeomorphic to Tsn. (13) (Hessian) Suppose that f E COO(M) and that dfp = 0 for some p E M. Show that for any smooth vector fields X and Y on M we have that Yp(Xf) = Xp(Yf). Let Hf,p(v,w) := Xp(Yf) , where X and Y are such that Xp = V and Yp = w. Show that Hf,p(v, w) is independent of the extension vector fields X and Y and that the resulting map Hf,p : TpM x TpM -+ IR is bilinear. Hf,p is called the Hessian of f at p. Show that the assumption djp = 0 is needed. (14) Show that for a smooth map F : M -+ N, the (bundle) tangent map T F : T M -+ TN is smooth. Sometimes it is supposed that one can

124

2. The Tangent Structure

obtain a well-defined map F. : X (M) --t X (N) by thinking of vector fields as derivations on functions and then letting (F.X) f = X (f 0 F) for f E COO(N). Show why this is misguided. Recall that the proper definition of F. : X (M) --t X (N) would be F.X := T FoX 0 F-l and is defined in case F is a diffeomorphism. What if F is merely surjective?

(15) Show that if'l/J : M' --t M is a smooth covering map, then so also is T'l/J: TM' --t TM. (16) Define the map f : Mnxn(1R) --t sym(Mnxn(1R)) by f(A) := AT A, where Mnxn(1R) and sym(Mnxn(1R)) are the manifolds of n x n matrices and n x n symmetric matrices respectively. Identify TA (Mnxn(1R)) with Mnxn(lR) and Tf(A)sym(Mnxn(lR)) with sym(Mnxn(lR)) in the natural way for each A. Calculate TJ f : Mnxn(lR) --t sym(Mnxn(lR)) using these identifications.

(17) Let h, ... , fN be a set of smooth functions defined on an open subset of

T;

a smooth manifold. Show that if dh (p), ... , dfN (p) spans M for some p E U, then some ordered subset of {h, ... , fN} provides a coordinate system on some open subset V of U containing p.

(18) Let

~r

be the vector space of derivations on Cr(M) at p E M, where r ~ 00 is a positive integer or 00. Fill in the details in the following outline which studies ~r. It will be shown that ~r is not finitedimensional unless r = 00. (a) We may assume that M = lRn and p = 0 is the origin. Let mr := {f E cr(lRn) : f(O) = O} and let m~ be the subspace spanned by the functions of the form fg for f, 9 E mr . We form the quotient space mr/m~ and consider its vector space dual (mr/m~r. Show that if 8 E ~r, then 8 restricts to a linear functional on mr and is zero on all elements of m~. Conclude that 8 gives a linear functional on mr/m~. Thus we have a linear map ~r --t (mr/m~r. (b) Show that the map ~r --t (mr/m~r given above has an inverse. Hint: For a A E (mr/m~r, consider 8>.(f) := A([J - f(O)]), where f E cr(lRn) and hence [f - f(O)] E mr/m~. Conclude that by taking r - 00 we have ToRn = ~oo ~ (mr/m~)·. The case r < 00 is different as we see next. (c) Let r < 00. The goal from here on is to show that mr/m~ and hence (mr/m~r are infinite-dimensional. We start out with the case lRn = R First show that if f E mr, then f(x) = xg(x) for 9 E Cr-1(lR). Also if f E m~, then f(x) = x 2 g(x) for 9 E Cr-1(lR). (d) For each r E {I, 2, 3, ... } and each c E (0,1), define

o<

for x > 0, for x ~ O.

125

Problems

Then g; E mr , but g; ~ Cr +1(R). Show that for any fixed r E {1, 2, 3, ... }, the set of elements of the form [g;] := gr + m; for c E (0,1) is linearly independent in the quotient. Hint: Use induction on r. In the case of r = 1, it would suffice to show that if we are given 0 < cl < ... < Cl < 1 and if 2:i-l ajg~J Em;, then aj = 0 for all j. (Thanks to Lance Drager for donating this problem and its solution.) (19) Find the integral curves of the vector field on R2 given by X(x, y) := x

2

a

ax

a + xYay'

(20) Show that it is possible that a vector field defined on an open subset of a smooth manifold M may have no smooth extension to all of M. (21) Find the integral curves (and hence the flow) for the vector field on R2 given by X(x, y) := -yfx + x/y. (22) Let N be a point in the unit sphere 8 2. Find a vector field on 8 2 \ {N} that is not complete and one that is complete. (23) Using the usual spherical coordinates (cp, 8) on 8 2, calculate the bracket [¢,,81¢]. (24) Show that if X and Yare (time independent) vector fields that have flows r.pf and r.pf, then if [X, Y] = 0, the flow of X + Y is r.pf 0 r.pf, (25) Recall that the tangent bundle of the open set GL(n, R) in Mnxn(R) is identified with GL(n, R) x Mnxn(R). Consider the vector field on GL(n,R) given by X: 9 f--t (g,g2). Find the flow of X. (26) Let t f--t Qt =

(~~:: ~~~~t ~)

001 for t E R. Let ¢(t, P) := QtP, where P is a plane in R3. Show that this defines a flow on the Grassmann manifold G(3,2). Find the local expression in some coordinate system of the vector field XQ that gives this flow. Do the same thing for the flow

t

f--t

Rt =

(co;t

~ -~nt)

sin t 0

cos t

and find the vector field XR. Find the bracket [XR,XQ]. (27) Develop definitions for tangent bundle and cotangent bundle for manifolds with corners. (See Problem 21.) [Hint: A curve into an n-manifold with corners should be considered smooth only if, when viewed in a chart, it has an extension to a map into Rn. Similarly, a functions is smooth at a corner (or boundary) point only if its local representative

126

2. The Tangent Structure

in some chart containing the point can be extended to an open set in ]Rn.]

(28) Show that if p(x) for some mEN,

= p(Xl, .. . ,xn) is a homogeneous polynomial, so that

P(tXl, ... , tXn) = tmp(Xl, ... , x n ), then as long as c ifold of ]Rn.

=1=

0, the set p-l(c) is an (n - I)-dimensional subman-

(29) Suppose that 9 : M ~ N is transverse to a submanifold WeN. For another smooth map f : Y ~ M, show that f rh g-l(N) if and only if

(g 0 f) rh W. (30) Let M x N be a product manifold. Show that for each X E X(M) there is a vector field X E X(M x N) t~at is prl-related to X and prTrelated to the zero field on N. We call X the lift of X. Similarly, we may lift a field on N to M x N.

Chapter 3

Immersion and Submersion

Suppose we are given a smooth map J : M ---+ N. Near a point p E M, the tangent map TpJ : TpM ---+ TpN is a linear approximation of J. A very important invariant of a linear map is its rank, which is the dimension of its image. Recall that the rank of a smooth map J at p is defined to be the rank of TpJ. It turns out that under certain conditions on the rank of J at p, or near p, we can draw conclusions about the behavior of J near p. The basic idea is that J behaves very much like TpJ. If L : V ---+ W is a linear map of finite-dimensional vector spaces, then Ker Land L(V) are subspaces (and hence submanifolds). We study the extent to which something similar happens for smooth maps between manifolds. In this chapter we make heavy use of some basic theorems of multivariable calculus such as the implicit and inverse mapping theorems as well as the constant rank theorem. These can be found in Appendix C (see Theorems C.l, C.2 and C.5). More on calculus, including a proof of the constant rank theorem, can be found in the online supplement to this text [Lee, J effj. 3.1. Immersions Definition 3.1. A map J : M ---+ N is called an immersion at p E M if TpJ : TpM ---+ Tf(p)N is an injection. A map J : M ---+ N is called an immersion if it is an immersion at every p EM. Note that TpJ : TpM ---+ Tf(p)N is an injection if and only if its rank is equal to dim(M). Thus an immersion has constant rank equal to the dimension of its domain.

-

127

128

3. Immersion and Submersion

Immersions of open subsets of R2 into R3 appear as surfaces that may self-intersect, or periodically retrace themselves, or approach themselves in various limiting ways. The map R2 -+ R3 given by (u, v) H (cos u, sin u, v) is an immersion as is the map (u,v) H (cosusinv,sinusinv, (1- 2 cos2 v) cos v). The map S2 -+ R3 given by (x, y, z) H (x, y, z - 2z 3 ) is also an immersion. By contrast, the map f : 8 2 -+ R3 given by (x, y, z) H (x, y, 0) is not an immersion at any point on the equator 8 2 n {z = o}.

Example 3.2. We describe an immersion of the torus T2 := 8 1 x 8 1 into R3. We can represent points in T2 as pairs (eilh,ei02). It is easy to see that, for fixed a, b > 0, the following map is well-defined: (e l01 , ei(2 ) H (x(e l01 , ei(2 ), y(el01 , el(2 ), z(ei01 , ei(2 )), where

x(el01 , ei(2 ) = (a + bcos{h) cos 02, y(e~Ol,ei02) = (a z(e~Ol,ei02) =

+ bCOSOl) sin 02,

bsinOl.

Exercise 3.3. Show that the map of the above example is an immersion. Give conditions on a and b that guarantee that the map is a 1-1 immersion. Theorem 3.4. Let f : M -+ N be a smooth map that is an immersion at p. Then for any chart (x, U) centered at p, there is a chart (y, V) centered at f (p) such that f (U) c V and such that the corresponding coordinate expression for f is (xl, ... , xk) H (xl, ... , xk, 0, ... ,0) ERn. Here, n is the dimension of Nand k = dim(M) is the rank of Tpf. Proof. Follows easily from Corollary C.3.

o

Theorem 3.5. If f : M -+ N is an immersion (so an immersion at every point), and if f is a homeomorphism onto its image f(M) (using the relative topology on f(M)), then f(M) is a regular submanifold of N. Proof. Let k be the dimension of M and let n be the dimension of N. Clearly f is injective since it is a homeomorphism. Let f(p) E f(M) for a unique p. By the previous theorem, there are charts (U, x) with p E U and (V, y) with f (p) E V such that the corresponding coordinate expression for f is (xl, ... , xk) H (xl, ... , xk, 0, ... ,0) ERn. We arrange to have f(U) C V. But f(U) is open in the relative topology on f(M), so there is an open set o C V in M such that f(U) - f(M) n O. Now it is clear that (0, yi o ) is a chart with the regular submanifold property, and so since p was arbitrary, we conclude that f(M) is a regular submanifold. 0 If f : M -+ N is an immersion that is a homeomorphism onto its image (as in the theorem above), then we say that f is an embedding.

3.1. Immersions

129

Exercise 3.6. Show that every injective immersion of a compact manifold is an embedding. Exercise 3.7. Show that if I: M -+ N is an immersion and p E M, then there is an open U containing p such that 1Iu is an embedding. Exercise 3.S. Recall the definition of a vector field along a map (Definition 2.68). Let X be a vector field along 1 : N -+ M. Show that if 1 is an embedding, then there is an open neighborhood U of I(N) and a vector field X E X(U) such that X - X 0 I. Recall that a continuous map 1 is said to be proper if 1-1 (K) is compact whenever K is compact. Exercise 3.9. Show that a proper 1-1 immersion is an embedding. [Hint: This is mainly a topological argument. You may assume (without loss of generality) that the spaces involved are Hausdorff and second countable. The slightly more general case of paracompact Hausdorff spaces follows.] Definition 3.10. Let Sand M be smooth manifolds. A smooth map 1 : S -+ M will be called smoothly universal if for any smooth manifold N, a mapping 9 : N -+ S is smooth if and only if log is smooth.

Definition 3.11. A weak embedding is a 1-1 immersion which is smoothly universal. Let 1 : S -+ M be a weak embedding and let A be the maximal atlas that gives the differentiable structure on S. Suppose we consider a different differentiable structure on S given by a maximal atlas A2. Now suppose that f : S -+ M is also a weak embedding with respect to A2. Resorting to seldom used pedantic notation, we are supposing that both 1 : (S, A) -+ M and f : (S, A2) -+ M are weak embeddings. From this it is easy to show that the identity map gives smooth maps (S, A) -+ (S, A2) and (S, A 2) -+ (S, A). This means that in fact A = A2, so that the smooth structure of S is uniquely determined by the fact that 1 is a weak embedding. Exercise 3.12. Show that every embedding is a weak embedding.

3. Immersion and Submersion

130

Figure 3.1. Figure eight immersions

In terms of 1-1 immersions, we have the following inclusions: {proper embeddings}

c c

{embeddings} {weak embeddings}

c

{1-1 immersions} .

3.2. Immersed and Weakly Embedded Submanifolds We have already seen the definition of a regular submanifold. The more general notion of a submanifold is supposed to realize the "subobject" in the category of smooth manifolds and smooth maps. Submanifolds are to manifolds what subsets are to sets in general. However, what exactly should be the definition of a submanifold? The fact is that there is some disagreement on this point. From the category-theoretic point of view it seems natural that a sub manifold of M should be some kind of smooth map I : S ---7 M. This is not quite in line with our definition of regular submanifold, which is, after all, a type of subset of M. There is considerable motivation to define sub manifolds in general as certain subsets; perhaps the images of certain nice smooth maps. We shall follow this route. Definition 3.13. Let S be a subset of a smooth manifold M. If S is a smooth manifold such that the inclusion map L : S ---7 M is an injective immersion, then we call S an immersed submanifold. Notice that in the above definition, S certainly need not have the subspace topology! Its topology is that induced by its own smooth structure. The reader may rightfully wonder just how S could acquire such a smooth structure in the first place. If f : N ---7 M is an injective immersion, then S := f(N) can be given a smooth structure so that it is an immersed submanifold. Indeed, we can simply transfer the structure from N via the bijection f : N ---7 f(N). However, this may not be the only possible smooth structure on f(N) which makes it an immersed submanifold. Thus it is imperative to specify what smooth structure is being used. Simply looking at

3.2. Immersed and Weakly Embedded Submanifolds

131

v

Figure 3.2. Immersions can approach themselves

the set is not enough. For example, in Figure 3.1 we see the same figure eight shaped subset drawn twice, but with arrows suggesting that it is the image of two quite different immersions which provide two quite different smooth structures. Suppose that S is a k-dimensional immersed submanifold of a smooth n-manifold M, and let pES. Then using Theorem 3.4, we see that there is a chart (0, x) on S, and a chart (V, y) on M, with p E 0 C V, such that yo

La

x- 1

-

yo x- 1 : x (0) ---+ Y (V)

has the form (a\ ... , a k ) t--+ (a\ ... , ak , 0, ... ,0). This means that y (0) = yo x l(x (0)) is a relatively open subset of ]Rk x {O}. Thus there is an open subset W of y (V) c ]Rn such that y (0) = W n (]Rk x {O}). Letting Ul := y-l(W), we see that y (U1 nO)

= y(U1 ) n

(]Rk x

{O}).

Thus the chart (ylUl ,Ut) has the submanifold property with respect to 0 (but not necessarily with respect to S). The set 0 has a smooth structure as an open submanifold of S. But this is the same smooth structure 0 has as yk 0 a regular submanifold. To see this note that the restrictions combine to give an admissible chart on S. Indeed, using the functions we obtain a bijection of 0 with an open subset of ]Rk. We only need to show that this bijection is smoothly related to the chart (0, x), and this amounts to showing that yk 0 are smooth. But this follows immediately 1 from the fact that yo x- is smooth. Notice that unlike the case of a regular submanifold, it may be that no matter how small V,

y1lo ' ... , I

y1Io' ... ' I

y(V n 0) # y(V) n (]Rk

x {O}),

as indicated in Figure 3.2. So in summary, each point of an immersed submanifold has a neighborhood that is a regular submanifold. Proposition 3.14. Let ScM be an immersed submanifold of dimension k and let f : N ---+ S be a map. Suppose that L of: N ---+ M is smooth,

132

where /, : S y also smooth.

3. Immersion and Submersion

M is the inclusion. Then ij j : N -+ S is continuous, it is

Proof. We wish to show that j : N -+ S is smooth if j-1(0) is open for every set 0 C S that is open in the manifold topology on S. Let pEN and choose a chart (V, y) for M centered at /, 0 j (p), so that U

= {q E V

: yk+1 (q)

= ... = yn (q) = O}

is an open neighborhood of p in S and such that y1j u ' ... , yk jU are coordinates for S on U. By assumption j-1(U) is open. Thus (/, 0 f) (I-1(U)) C U. In other words, /, 0 j maps an open neighborhood of pinto U. To test for the smoothness of j, we consider the functions (y' j u) 0 j on the set j-1 (U). But (yijU) oj=yio/,oj, and these are clearly smooth by the assumption that /, 0 j is smooth. 0 Sometimes the previous result is stated differently (and somewhat imprecisely): Suppose that j : N -+ M is a smooth map with image inside an immersed submanifold S; then j is smooth as a map into S if it is continuous as a map into S. The lack of a notational distinction between j as a map into Sand j as a map into M is what makes this way of stating things less desirable. Let ScM and suppose that S has a smooth structure. To say that an inclusion S y M is an embedding is easily seen to be the same as saying that S is a regular submanifold, and so we also say that S is embedded in M.

Corollary 3.15. Suppose that ScM is a regular submanijold. Let j : N -+ S be a map such that /, 0 j : N -+ M is smooth. Then j : N -+ S is smooth. Proof. The map f., : S y M is certainly an immersion, and so by the previous theorem we need only check that j : N -+ S is continuous. Let 0 be open in S. Then since S has the relative topology, 0 = un S for some open set U in M. Then 1-1(0) = 1- 1(UnS) = 1-1(/,-1(U)) = (/, 0 f)-1 (U), which is open since /, 0 j is continuous. Thus j is continuous. 0 Definition 3.16. Let S be a subset of a smooth manifold M. If S is a smooth manifold such that the inclusion map /, : S -+ M is a weak embedding, then we say that S is a weakly embedded submanifold. From the properties of weak embeddings we know that for any given subset ScM there is at most one smooth structure on S that makes it a weakly embedded submanifold.

3.2. Immersed and Weakly Embedded Submanifolds

133

Corresponding to each type of injective immersion considered so far we have in their images different notions of submanifold: {proper submanifolds}

c c

{regular submanifolds} {weakly embedded submanifolds}

C {immersed submanifolds}.

We wish to further characterize the weakly embedded submanifolds. Definition 3.17. Let S be any subset of a smooth manifold M. For any XES, denote by Ox(S) the set of all points of S that can be connected to x by a smooth curve with image entirely inside S. It is important to be clear that Ox(S) is not necessarily the connected component of S with its relative topology since, for example, S could be the image of an injective nowhere differentiable curve. In the latter case, Cx(S) = {x} for all XES! Definition 3.18. We say that a subset S of an n-manifold M has property W(k) if for each So E S there exists a chart (U, x) centered at So such that x(Cso(U n S)) = x(U) n (jRk x {O}). Here jRn = jRk X jRn-k. Together, the next two propositions show that weakly embedded submanifolds are exactly those subsets that have property W(k) for some k. Our proof follows that of Michor [Michl, who refers to subsets with property W(k), for some k, as initial submanifolds. With Michor's terminology, the result will be that the initial submanifolds are the same as the weakly embedded submanifolds. Proposition 3.19. If an injective immersion I : S ~ M is smoothly universal, then the image I (S) has property W( k) where k = dim( S). In particular, if SCM is a weakly embedded submanifold of M, then it has property W(k) where k = dim(S). Proof. Let dim(S) - k and dim (M) = n. Choose So E S. Since I is an immersion, we may pick a coordinate chart (W, w) for S centered at So and a chart (V, v) for M centered at I(so) such that

volow I(y) = (y,O) = (yl, ... ,yk,O, ... ,O).

°

Choose an r > small enough that Bk(O, 2r) C w(W) and Bn(O, 2r) C v(V). Let U = v-I(Bn(O,r)) and WI = w-I(Bk(O,r)). Let x := vl u . We show that the coordinate chart (U, x) satisfies the conditions of Definition 3.18:

x-l(x(U) n (jRk x {a})) = x-I{(y,0) : Ilyll < r} =Iow Io(xolow 1)-I({(y,O): Ilyll
= low I({y: Ilyll < r}) = I(WI)'

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3. Immersion and Submersion

Clearly I(WI )

c 1(8), but we also have v 0 I(WI) c v 0 I 0 w-I(Bk(O, r)) = Bn(O, r)

n {]Rk

x {O}} C Bn(o, r),

so that I(Wd C v-I(Bn(o, r)) = U. Thus I(WI) C Unl(8). Since I(WI) is smoothly contractible to l(so), every point of I(WI) is connected to l(so) by a smooth curve completely contained in I(WI) C Unl(8). This implies that I(WI ) C C1(so)(U n 1(8)). Thus x-I(x(U) n (]Rk x {O})) C C1(so)(U n 1(8)) or x(U)n(]Rk x {a}) cx(C1 (so)(Unl(8))). Conversely, let z E C1(so)(U n 1(8)). By definition there must be a smooth curve c : [0,1] -+ M starting at l(so) and ending at z with c([0,1]) C Un 1(8). Since I : 8 -+ M is injective and smoothly universal, there is a unique smooth curve CI : [0, 1] -+ 8 with I 0 CI = c.

Claim: CI([O, 1]) C WI. Assume not. Then there is a number t E [0,1] with CI(t) E w I({r:::; Ilyll < 2r}). Therefore,

(VOI)(CI(t)) E (volow- I )({r:::; Ilyll < 2r}) = {(y,O): r:::; Ilyll < 2r} C {z E]Rn: r:::;

Ilzll < 2r}. (vo c)(t) E {z E ]Rn : r < Ilzll < 2r}, which

This implies that (vo I 0 CI)(t) = in turn implies the contradiction c(t) t/:. U. The claim is proven.

The fact that cI([0,1]) C WI implies cI(1) = l- I (z) E WI, and so z E I(WI). As a result we have C1(so)(U n 1(8)) = I(WI ) which together with the first half of the proof gives the result:

I(WI) = x-I(x(U)

n (]Rk

x {a})) C C1 (so)(U

n 1(8)) =

I(WI )

====>

x-I (x(U) n (]Rk x {O})) = C1(so) (U n 1(8))

====>

x(U) n (]Rk x {O}) = x(C1(so)(U n 1(8))).

0

Proposition 3.20. If 8 C M has property W(k), then there is a unique smooth structure on 8 which makes it a k-dimensional weakly embedded submanifold of M. Proof. (Sketch) We are given that for every s E 8, there exists a chart (Us,x s ) with xs(s) = and with x(Cs(Us n 8)) = x(Us) n (]Rk x {O}). The charts on 8 will be the restrictions of the charts (Us, xs ) to the sets Cs(Us n 8). The overlap maps are smooth because they are restrictions of overlap maps on M to subsets of the form V n (]Rk x {O}) for V open in ]Rn. If (USp X S1 ) and (US2 'X S2 ) are two such charts with corresponding sets 81 := C81 (US1 n 8) and 82 := CS2 (US2 n S), then we need to check that X S1 (81 n 8 1 ) is open in ]Rk x {a} (recall the definition of smooth atlas). For each p E US1 n US2 n 8 consider the set C(p) := Cp (US1 n US2 n 8). We

°

3.2. Immersed and Weakly Embedded Submanifolds

135

leave it to the reader to show that C(p) c 81 n 8 2 and that if p i- q then c(p) n C(q) = 0. Thus the sets C(P) form a partition of US! n US2 n 8. It is not hard to see that each C (p) maps onto a connected path component of XS ! (US1 n US2 n 8) and that every path component of this set is the image of some C(p). But this implies that X s ! (US1 n US2 n 8) open. Notice however that the topology induced by the smooth structure thus obtained on 8 is finer than the relative topology that 8 inherits from M. This is because the sets of the form Cs(U n 8) are not necessarily open in the relative topology. Since it is a finer topology, it is also Hausdorff. It is clear that with this smooth structure on 8, the inclusion ~ : 8 y M is an injective immersion. We now show that the inclusion map ~ : 8 y M is smoothly universal and hence a weak embedding. By the comments following Definition 3.11 the smooth structure on 8 is unique. Let 9: N ---+ 8 be a map and suppose that ~ 0 9 is smooth. Given x E M, choose a chart (Us, x s ) where s = 9(X). The set 9- 1 (Us ) is open since ~ 0 9 is continuous and (~O 9)-1 (Us) = 9- 1 0 ~-I(Us) = g-I(Us). We may choose a chart (V, y) centered at x with V c g-l(Us ) and we may arrange that y(V) is a ball centered at the origin. This means that ~og(V) is smoothly contractible in Ug(x) n 8 and hence 9(V) c Cg(x) (Ug(x) n 8). But then Xs

-1, IC.(u.nS) 0 goy -1 = Xs 0 (~ 0)goy

and so 9 is smooth because ~ 0 9 is smooth. To be completely finished, we need to show that with the topology induced by the atlas, each connected component of 8 is second countable. We can give a quick proof, but it depends on Riemannian metrics which we have yet to discuss. The idea is that on any paracompact smooth manifold, there are plenty of Riemannian metrics. A choice of Riemannian metric gives a notion of distance making every connected component a second countable metric space. If we put such a Riemannian metric on M, then it induces one on 8 (by restriction). This means that each component of 8 is also a separable metric space and hence a second countable Hausdorff topological space. 0 We say that two immersions II : Nl ---+ M and 12 : N2 ---+ M are equivalent if there exists a diffeomorphism
M

136

3. Immersion and Submersion

Figure 3.3. Tori converge to a point

If J : N -t M is a weak embedding (resp. embedding), then there is a unique smooth structure on S = J(N) such that S is a weakly embedded (resp. embedded) submanifold and J : N -t M is equivalent to the inclusion " : S y M in the above sense. We shall follow the convention that the word "submanifold", when used without a qualifier such as "immersed" or "weakly embedded", is to mean a regular submanifold unless otherwise indicated. What R. Sharpe [Shrp] calls a "submanifold", refers to something more restrictive than the weakly embedded sub manifolds , but still less restrictive that the regular submanifolds. Sharpe's definition of "submanifold" seems designed to exclude examples like that shown in Figure 3.3. Here the tori converge to a point on the plane (which is taken to be part of the manifold). Every neighborhood of that point will contain an infinite number of tori. Such a behavior is excluded by Sharpe's definition, but this is still a weakly embedded submanifold. One can also imagine the tori flattening while only the holes converge to a point. This example can easily be modified to be path connected and yet, it could never be the maximal integral manifold of a tangent distribution (see Chapter 11 for definitions). The celebrated Whitney embedding theorem states that any secondcountable n-manifold can be embedded in a Euclidean space of dimension 2n. We do not prove the full theorem, but we will settle for the following easier result. Theorem 3.21. Suppose that M is an n-manifold that has a finite atlas. Then there exists an injective immersion of Minto ]R2n+1. Consequently, every compact n-manifold can be embedded into ]R2n+1. Proof. Let M be a smooth manifold with a finite atlas. In particular, M is second countable. Initially, we will settle for an immersion into ]RD for

3.2. Immersed and Weakly Embedded Submanifolds

137

some possibly very large dimension D. Let {Oi, 'PihEI be an atlas with cardinality N < 00. By applying Lemma B.4 twice, the cover {Oil may be refined to two other covers {UihEI and {VihEI such that Ui C Vi C Vi C Oi. Also, we may find smooth functions fi : M --+ [0, IJ with supp(li) c Oi and such that fi(X) = 1 for all x E U, and f,(x) < 1 for x ¢ Vi. Next we write 'P, = (x;, ... , xi) so that Oi --+ ]R is the j-th coordinate function of the i-th chart, and then form the product

xi :

lij

:=

fixi,

which is defined and smooth on all of M after extension by zero. Now we put the functions fi together with the functions fij to get a map

f : M --+ ]Rn+Nn : f = (h,···, fn, fll, h2, ... , 121' ... ' fNn). Now we show that f is injective. Suppose that f(x) = f(y). Note that fk(x) must be 1 for some k since x E Uk for some k. But then Jk(y) = 1 also, and this means that y E Vk (why?). Since fk(X) = fk(Y) = 1, it follows that Jkj(x) = Jkj(Y) for all j. Remembering how things were defined, we see that x and y have the same image under 'Ilk : Ok --+]Rn and thus x = y. To show that Txf is injective for all x E M, we fix an arbitrary such x; then x E Uk for some k. But then near this x, the functions Jkl,!k2, ... , fkn, are equal to xl, ... , xk and so the rank of f must be at least n and in fact equal to n since dim TxM = n. So far we have an injective immersion of Minto ]RD where D = n + Nn. We show that there is a projection 7r : ]RD --+ L c ]RD, where L ~ ]R2n+l is a (2n + I)-dimensional subspace of ]RD such that 7r 0 f is an injective immersion. The proof of this will be inductive. So suppose that there is an injective immersion f of Minto ]Rd for some d with D ~ d > 2n + 1. We show that there is a projection 7rd : ]Rd --+ L d - 1 ~ ]Rd-l such that 7rd 0 f is still an injective immersion. To this end, define a map h : M x M x]R --+ ]Rd by h(x, y, t) := t(f(x) - f(y)). Since d > 2n + 1, Sard's theorem (Theorem 2.34) implies that there is a vector z E ]Rd which is neither in the image of the map h nor in the image of the map df : T M --+ ]Rd. This z cannot be 0 since a is certainly in the image of both of these maps. If pr.lz is projection onto the orthogonal complement of z, then pr.lz 0 f is injective; for if pr.lz 0 f(x) = pr.lz 0 f(y), then f(x) - f(y) = az for some a E ]R. But suppose x i- y. Since f is injective, we must have a i- O. This state of affairs is impossible since it results in the equation h(x, y, l/a) = z, which contradicts our choice of z. Thus pr.lz 0 f is injective. Next we examine Tx(pr.lz 0 f) for an arbitrary x E M. Suppose that Tx(pr.lz 0 f)v = o. Then d(pr.lz 0 f)lx v = 0, and since pr.lz is linear, this

138

3. Immersion and Submersion

amounts to pr..l.z 0 dflx v = 0, which gives dflx v = az for some number a E ~, and which cannot be 0 since f is assumed to be an immersion. But then dflx ~v = z, which also contradicts our choice of z. We conclude that pr..l.z 0 f is an injective immersion. Repeating this process inductively we finally get a composition of projections pr : ~D ~ ~2n+1 such that pr 0 f : M --t ~2n+1 is an injective immersion. The final statement for compact manifolds follows from Exercise 3.6. 0

3.3. Submersions Definition 3.22. A map f : M --t N is called a submersion at p E M if Tpf : TpM --t TJ(p)N is a surjection. f : M --t N is called a submersion if f is a submersion at every p E M. Example 3.23. The map of the punctured space ~3\ {O} onto the sphere S2 given by x t--t xllxl is a submersion. To see this, use any spherical coordinates (p, c/J, (1) on ~3\ {O} and the induced submanifold coordinates (c/J, (1) on S2. Expressed with respect to these coordinates, the map becomes (p, c/J, (1) t--t (c/J, (1) on the domain of the spherical coordinate chart. Here we ended up locally with a projection onto a second factor ~ x ~2--t ~2, but this is clearly good enough to prove the point. As in the last example, to show that a map is a submersion at some p it is enough to find charts containing p and f(p) so that the coordinate representative of the map is just a projection. Conversely, we have Theorem 3.24. Let M be an m-manifold and N a k-manifold and let f : M --t N be a smooth map that is a submersion at p. Then for any chart (V, y) centered at f(P) there is a chart (U, x) centered at p with f(U) c V such that yo f 0 x-I is given by (xl, ... , x k , ... , xm) t--t (xl, ... , xk) E ~k. Here k is both the dimension of N and the rank of Tpf. Proof. Follows directly from Theorem C.4 of Appendix C.

o

In certain contexts, submersions, especially surjective submersions, are referred to as projections. We often denote such a map by the letter 7r. Recall that if 7r : M --t N is a smooth map, then a smooth local section of 7r is a smooth map (J : V --t M defined on an open set V and such that 7r 0 (J = idv. Also, we adopt the terminology that subsets of M of the form 7r- 1 (q) are called fibers of the submersion. Proposition 3.25. If 7r : M --t N is a submersion, then it is an open map and every point p E M is in the image of a smooth local section.

139

3.3. Submersions

Proof. Let p E M be arbitrary. We choose a chart (U, x) centered at p and a chart (V, y) centered at 7I"(p) such that yo 71" 0 x-I is of the form (xl, ... , xk, xk+l, ... ,xm) r--+ (xl, . .. ,xk), where dim M = m and dim N = k. By shrinking the domains if necessary, we can arrange that x(U) has the form A x B C IRk X IRm - k and y(V) = B C IRI. Then we may transfer the section ib : a -+ (a, b) where b = x(p). More precisely, the desired local section is (j := x-I 0 ib 0 y on A. We now use the existence of local sections to show that 71" is an open map. Let 0 be any open set in M. To show that 71"(0) is open, we pick any q E 71"(0) and choose p E 0 with p E 7I"-I(q). Now we choose a chart (U, x) as above with p E U C O. Then q is in the domain of a section which is open and contained in 71"(0). 0 Proposition 3.26. Let 71" : M -+ N be a surjective submersion. P is any map, then I is smooth il and only il I 0 71" is smooth:

II I: N

-+

M

~!~~

N~P

Proof. One direction is trivial. For the other direction, assume that 1071" is smooth. We check for smoothness of I about an arbitrary point q E N. Pick p E 7I"-l(q). By the previous propositionp is in the image of a smooth section ( j : V -+ M. This means that I and (f 071") 0 (j agree on a neighborhood of q, and since the latter is smooth, we are done. 0 Next suppose that we have a surjective submersion 71" : M -+ Nand consider a smooth map g : M -+ P which is constant on fibers. That is, we assume that if PI,P2 E 7I"-l(q) for some q E N, then I(Pl) = I(P2). Clearly there is a unique induced map I: N -+ P so that g = 1 0 71". By the above proposition I must be smooth. This we record as a corollary:

Corollary 3.27. II g : M -+ P is a smooth map which is constant on the fibers 01 a surjective submersion 71" : M -+ N, then there is a unique smooth map I : N -+ P such that g = I 0 71" • The following technical lemma is needed later and represents one more situation where second count ability is needed. Lemma 3.28. Suppose that M is a second countable smooth manilold. II I: M -+ N is a smooth map with constant rank that is also surjective, then

it is a submersion. Proof. Let dim M = m, dim N = nand rank(f) = k and choose p E M. Suppose that I is not a submersion so that k < n. We can cover M

3. Immersion and Submersion

140

by a countable collection of charts (Uo , xo ) and cover N by charts (Vi, y,) such that for every a, there is an i = i(a) with f (Ua,) C Vi and Yt 0 f 0 x~l(xI, ... , xn) = (xl, ... , xk, 0, ... ,0). But this means that f (Uo,) has measure zero. However, f(M) = Uof (Uo ) and so f(M) is also of measure zero which contradicts the surjectivity of f. This contradiction means that f must be a submersion after all. 0

Problems (1) Let

°<

a

< b. Show that the subset of]R3 described by the equation

(J x 2 + y2 _ b) 2 + z2 =

a2

is a submanifold. Show that the resulting manifold is diffeomorphic to 8 1 x Sl. (2) Show that the map 8 2 ~ ]R3 given by (x, y, z) ~ (x, y, z - 2z3 ) is an immersion. Try to determine what the image of this map looks like.

(3) Show that if M is compact and N is connected, then a submersion f : M ~ N must be surjective. (4) Let f : M ~ N be an immersion. (a) Let (U, x) be a chart for M with p E U, and let (V, y) be a chart for N with f(p) E V such that yo fox 1(a 1,a2, ... ,an ) - (a 1,a2, ... ,an ,0, ... ,0).

Show that Tpf·

aax i Ip = aayi If(p)

for i - I , ... , n.

(b) Show that if f is as in part (a) and Y E X(N) is such that Y(p) E Tpf(TpM) for all p, then there is a unique X E X(M) such that X is f-related to Y. (5) Define a function s : ]Rn+1\{0} ~ ]Rpn by the rule that s(x) is the line through x and the origin. Show that s is a submersion. (6) Show that there is a continuous map f : ]R2 ~ JR2 such that f(B(O, 1)) C B(O, 1), f(]R2\B(0, 1)) C f(]R2\B(0, 1)) and faB(O,l) = idaB (o,l) and with the properties that f is Coo on B(O, 1) and on ]R2\B(0, 1), while f is not Coo on ]R2. (7) Construct an embedding of ]R x 8 n into ]Rn+1.

(8) Embed the Stiefel manifold of k-frames in]Rn into a Euclidean space JRN for some large N.

Problems

141

(9) Construct an embedding of G(n, k) into G(n, k + l) for each l ~ 1. (10) Show that the map f : 1~'p2 ---+ lR3 defined by f([x, y, z]) = (yz, xz, xy) is an immersion at all but six points p E lRP2. The image is called the Roman surface, and nice images can be found on the web. Show that it is a topological immersion (locally a topological embedding). Show that the map 9 : lRp 2 ---+ lR4 given by g([x, y, z]) = (yz, xz, xy, x 2 + 2y2 + 3z2) is a smooth embedding. (11) Let h : M ---+ lRn be smooth and let N c lRn be a regular submanifold. Prove that for each E > 0 there exists a v E lR n , with Ivl < E, such that the map p r-+ h(p) + v is transverse to N. (Think about the map M x N ---+ lRn given by (p, y) r-+ y - f(p).) (12) Define ¢ : SI ---+ lR by e'o r-+ () for 0 ~ () < 271". Define>. : lR ---+SI by () r-+ eiO • Show that >. is an immersion, that >. 0 ¢ is smooth, but that ¢ is not differentiable (it is not even continuous).

Chapter

4

Curves and Hypersurfaces in Euclidean Space

So far we have been studying manifold theory, which is foundational for modern differential geometry. In this chapter we change direction a bit to introduce some ideas from classical differential geometry. We do this for pedagogical reasons. The reader will see several ideas introduced here that will only be treated in generality later in the book. These ideas include parallelism, covariant derivative, metric and curvature. We concentrate on the geometry of one-dimensional sub manifolds ofIRn (geometric curves) and submanifolds of co dimension one in IR n , which we refer to as hypersurfaces. l Recall that a vector in TpIRn can be viewed as a pair (p, v) E {p} x IR n , and we sometimes write vp or (vI, ... , vn)p for (p, v). The element v is called the principal part of vp. Recall also that in IR n we have a natural notion of what it means for vectors in different tangent spaces to be parallel. By definition, vp E TpIRn is parallel to Wq E TqIR n if v = W E IRn. For any wp E TpIRn, there is a unique vector Wq E TqIR n which is parallel to wp. We call Wq the parallel translate of wp to the tangent space TqIRn. We often identify TpRn with IRn. If eI, ... ,en is the standard orthonormal basis for IRn, then el," . will denote the standard global frame field on IRn defined by e,(p):- 8~.lp for i = 1, ... , n, where xl, ... , xn are the standard coordinate functions on IRn. The reason for this separate notation is to emphasize the fiducial role of this frame field. If vp E TpIRn and Wq E TqIRn, then vp is

,en

1 Unless

otherwise stated, we will assume that a hypersurface is a manifold without boundary.

-

143

4. Curves and Hypersurfaces in Euclidean Space

144

parallel to

Wq

if vi - wi for all i, where

vp

L viei(p) and

=

Wq

=

L wiet(q).

Recall that if e : I -t M is a smooth curve into an n- manifold M, then a smooth vector field along e is a smooth map Y : I -t T M such that 71' 0 Y = e. The velocity of the map Y is an element of T (T M) rather than T M. On the other hand, if c : I -t ]Rn, then the special structure of ]Rn allows us to take a derivative of a vector field along e and end up with another vector field along c. If Y is such a vector field along e, then it can be written Y = ~ yi 0 e for some smooth functions yi : I -t R In other words, Y(t) = ~Yt(t)ei(e(t)) for tEl. For such a field, we define

ei

dY di(t)

=

Y'(t)

=

dyi ~ L Tt(t)ei(e(t)).

In notation that keeps track of base points, we can write Y so that Y(t) = (yl(t), ... , yn(t))c(t), and then

= (yl, ... , yn)c

dY (t) = (d d Y n (t) ) . -d -d Y 1 (t), ... , -d t t t c(t) Since ~r is obviously also a vector field along c, we can repeat the process to obtain higher derivatives ~r = y(k). In particular, we have the velocity e', acceleration e", and higher derivatives e(k). Except for the current emphasis on the base point of vectors, these definitions are the usual definitions from multivariable calculus. Another thing that is special about ]Rn is the availability of the natural inner product (the dot product) in every tangent space Tp]Rn. This inner product on Tp]Rn is denoted (-, ')p and is given by

(vp,wp) t--+ (vp,wp) = Lviw i , where vp = (p, v) and wp - (p, w) as explained above. This gives what is called a Riemannian metric on ]Rn, and it is obtained from the canonical identification of the tangent spaces with ]Rn itself. We study Riemannian metrics on general manifolds later in the book. For smooth vector fields on ]Rn, say X and Y, the function (X, Y) defined by p t--+ (Xp, Yp) is smooth. IT X and Yare fields along a curve e: I -t ]Rn, then (X, Y) is a function on I and we clearly have

d /dX) dt (X, Y) = \ dt' Y

dY) + /\ X, dt .

Remark 4.1. We shall sometimes leave out the subscript p in the notation vp if the base point is understood from the context. In fact, we often just identify vp with its principal part v E ]Rn.

4.1. Curves

145

An ordered basis VI,.'" vn is positively oriented if det(vI, ... , v n ) is positive. For v p , wp E T p]R3, we can obviously define a cross product vp x wp := (v x w)p, which results in a vector based at p which is then orthogonal to both vp and wp. If vp and wp are orthonormal, then (vp, w p , (v x w)p) is a positively oriented (right-handed) orthonormal basis for Tp]R3. The following lemma allows us to do something similar in higher dimensions.

Lemma 4.2. If VI, ... , Vn I E ]Rn, then there is a unique vector N(VI, ... , Vn-l) such that

(i) N( VI, ... , Vn-I) is orthogonal to each of VI, ... , Vn-I'

(ii) If {Vl, ... ,Vn -I} is an orthonormal set of vectors, then the list (VI, ... , Vn-I, N) is a positively oriented orthonormal basis.

(iii)

N(VI"'" Vn-I) depends smoothly on VI,.··, Vn -l.

Proof. Let L be the linear functional defined by L (V) := det (VI, ... , Vn-I, V). There is a unique N E ]Rn such that L(v) = (N, v). Now (i), (ii) and (iii) follow from the properties of the determinant. D

4.1. Curves If C is a one-dimensional submanifold of]Rn and pEe, then there is a chart (V, y) of C containing p such that y (V) is a connected open interval I c R The inverse map y-l : I ~ V c M is a local parametrization. Thus for local properties, we are reduced to studying curves into ]Rn which are embeddings of intervals. We can be even more general and study immersions. The idea is to extract information that is appropriately independent of the parametrization. If 'Y : I ~ ]Rn and c : J ~ ]Rn are curves with the same image, then we say that c is a positive reparametrization of 'Y if there is a smooth function h : J ~ I with h' > 0 such that c = 'Y 0 h. In this case, we say that 'Y and c have the same sense and provide the same orientation on the image. We assume that 'Y : I ~ ]Rn has IIT'II > 0, which is the case of interest. Such a curve is called regular, which just means that the curve is an immersion.

Definition 4.3. If 'Y : I ---t]Rn is a regular curve, then T(t) := 'Y'(t)/ 11T'(t)II defines the unit tangent field along 'Y. (Of course, we then have IITII = 1.) We have the familiar notion of the length of a curve defined on a closed interval'Y: [tI, t2J ~ ]Rn:

L=

l

t2

tl

!Ir' (t) II dt.

4. Curves and Hypersurfaces in Euclidean Space

146

One can define an arc length function for a curve 'Y : I --+ lRn by choosing to E I and, then defining s

= h(t)

:-I 1i'Y'(T)\\ t

dT.

to

Notice that s takes on negative values if t < to, so it does not always represent the length in the ordinary sense. If the curve is smooth and regular, then h' - 1i'Y'(T)II > 0, and so by the inverse function theorem, h has a smooth inverse. We then have the familiar fact that if e(s) = 'Y 0 h- 1 (s), then Ile'll (s) :- Ile'(s)11 = 1 for all s. Curves which are parametrized in terms of arc length are referred to as unit speed curves. For a unit speed curve, ~~(s) = T(s). Since parametrization by arc length eliminates any component of acceleration in the direction of the curve, the acceleration must be due only to the shape of the curve. Definition 4.4. Let c : I --+ lRn be a unit speed curve. The vector-valued function dT

K(S)

:=

ds (s)

is called the curvature vector. The function

K,

defined by

is called the curvature function. If K,(s) > 0, then we also define the principal normal dT 11-1 Ts(s), dT N(s):= Il Ts(s)

so that ~'!'

= K,N.

Let 'Y : I --+ lRn be a regular curve. An adapted orthonormal moving frame along 'Y is a list (El," . ,En) of smooth vector fields along 'Y such that El(t) = 'Y'(t)/Ii'Y'(t)11 and such that (E1(t), ... ,En(t)) is a basis of T'Y(t)lR n for each tEl. Identifying El (t), . .. , En(t) with elements of lRn written as column vectors, we say that the orthonormal moving frame is positively oriented if

Q(t) = [E1(t), ... , En(t)] is an orthogonal matrix of determinant one for each t. Definition 4.5. A moving frame El(t), ... , En(t) along a curve 'Y : 1--+ lRn is a Frenet frame for 'Y if 'Y(k) (t) is in the span of El (t), . .. , Ek(t) for all t and 1 ~ k ~ n. As we have defined them, Frenet frames are not unique. However, under certain circumstances we may single out special Frenet frames. For example,

147

4.1. Curves

if c: I -4 lR3 is a unit speed curve with /'i, > 0, then the principal normal N is defined. By letting

B=TxN we obtain a Frenet frame T, N, B. It is an easy exercise to show that we obtain dT ds dN

-

dB ds

=

/'i,N -/'i,T

ds

+

TB

-TN

for some function T called the torsion. This is the familiar form presented in many calculus texts. In matrix notation,

[ °~ -/'i,~ ~T° 1

:8[T,N,B] =[T,N,B]

•

Notice that for a regular curve, /'i, ~ 0, while T may assume any real value. Another special feature of the frame T, N, B is that it is positively oriented. If [ is injective, then we can think of /'i, and T as defined on the geometric image [(1). Thus, if p = ,(80), then /'i,(p) is defined to be equal to /'i,(80)' Exercise 4.6. Let c : I -4 lR3 be a unit speed regular curve with /'i, > 0. Show that (( c' (8) X e" (8)) , e"' (8)) c 11 I () T 8 = /'i, (8) 2 lor a 8 E . Exercise 4.7. Let c: I -4lR3 be a unit speed curve and 80 E I. Show that we have a Taylor expansion of the form

,(8) -,(80) = ((8 - 80) -

~(8 -

80)3/'i,2(8 0)) T(80) 80) 3 d/'i,)) d8 (80 N(80 )

+ ( 12 (8 -

80) 2 /'i,(80)

+ (~(8 -

80 )3/'i,(8 0 )T(80 )) B(8)

1 + 6(8

+ 0((8 -

80)3).

We wish to generalize the special properties of the Frenet frame T, N, B to higher dimensions thereby obtaining a notion of a distinguished Frenet frame. For maximum generality, we do not assume that the curve is unit speed. A curve, in lRn is called k-regular if {['(t), ,"(t), .. . ,[(k)(t)} is a linearly independent set for each t in the domain of the curve. For an (n - 1)regular curve, the existence of a special orthonormal moving frame can be easily proved. One applies the Gram-Schmidt process: If El (t), ... , Ek (t)

4. Curves and Hypersurfaces in Euclidean Space

148

are already defined for some k < n - 1, then

Ew(t)

:~ [-y(W) (t) - ~ U

k +1) (t),

Ck

E,(t)) E,(t)] ,

where Ck is a positive constant chosen so that IIEk+1 (t) II = 1. Inductively, this gives us El(t), ... , En-l(t), and it is clear that the Ek(t) are all smooth. Now we choose En(t) to complete our frame by letting it be of unit length and orthogonal to E1(t), ... ,En-1(t). By making one possible adjustment of sign on En(t) we obtain a moving frame that is positively oriented. In fact, En(t) is given by the construction of Lemma 4.2, from which it follows that En(t) is smooth in t. By construction we have a nice list of properties:

(1) For 1 ~ k ~ n, the vectors El (t), .. . , Ek(t) have the same linear span as -y'(t), ... ,-y(k)(t) so that there is an n x n upper triangular matrix function U (t) such that

b'(t), ... , -y(n) (t)]U(t) = [E1(t), ... , En(t)]. (2) For 1 ~ k ~ n - 1, the vectors E1(t), ... , Ek(t) have the same orientation as -y' (t), ... , -y(k) (t). Thus U (t) has diagonal elements which are all positive except possibly the last one.

(3) (E1(t), ... , En(t)) is positively oriented as a basis of T'Y(t)R n ~ JRn • Exercise 4.8. Show that the moving frame we have constructed is the unique one with these properties. We call a moving frame satisfying the above properties a distinguished Frenet frame along -y. For any orthonormal moving frame, the derivative of each Ei(t) is certainly expressible as a linear combination of the basis E1(t), ... , En(t), and so we may write

d

n

dtEj(t) = ~Wi3(t)Ei(t). ~=l

Of course, Wij(t) = (Ei(t),-9tEj(t)), but since (Ei(t), Ej(t)} = elij, we conclude that Wij(t) = -Wji(t), i.e., the matrix w(t) = [wtj(t)] is antisymmetric. However, for a distinguished Frenet frame, more is true. Indeed, if (E1(t), ... , En(t)) is such a distinguished Frenet frame, then for 1 ~ j < n we have Ej(t) - E{=l Ukj'Y(k) (t), where U(t) = [Uk3 (t)] is the upper triangular matrix mentioned above. Using the fact that U, -9tU, and U- 1 are all upper triangular, we have

149

4.1. Curves

But 'Y(k+1) (t) so that

= "k+l (U- 1) r,k+l E r (t) , and 'Y(k) (t) L.,.,r=l :tEj(t) =

=

t, (! tr (d j

Ukj ) ')Ik)(t)

dt Ukj

j

+

+

t,

= "k (U- 1) rk E r (t) L.,.,r=l

ukd k+1)(t)

) ?; (U-1)rkEr(t) k

k+1

L Uk) L (U- )r,k+1 Er(t). 1

k=l

r=l

From this we see that ;1tEj(t) is in the span of (Er(t)h~r~j+1' Thus w(t) = (Wij(t)) can have no nonzero entries below the sub diagonal. But w is antisymmetric, so we conclude that W has the form

o W(t) =

o

-Wn,n-l(t)

Wn,n-l(t)

0

We define the i-th generalized curvature function by

Wt+1,i(t) () Ki t := 1l'Y'(t) II . Thus if 'Y : I -t

]Rn

is a unit speed curve, we have

o W(s) =

o

-Kn-l(S)

Kn-l(S)

0

Note: Our matrix w is the transpose of the W presented in some other expositions. The source of the difference is that we write a basis as a formal row matrix of vectors. Lemma 4.9. IJ'Y : I -t ]Rn (n ~ 3) is (n - I)-regular, then Jor 1 :::; i :::; n-2, the generalized curvatures Ki are positive.

150

4. Curves and Hypersurfaces in Euclidean Space

Proof. By construction, for 1

~

i

~

=

L Uji(t)-y(J) (t),

n- 1

i

Ei(t)

J-1 i

'Y(i)(t) =

L (U)j/ (t)Ej(t), 3=1

with Un

> 0,

w.+1,.(t)

and hence

(U- 1 )ii > o.

Thus if 1 ~ i ~ n - 2, we have

~ (E,+J, ! E.) ~ ( E.+l, :t ~ Uj;(thu1 (t)) = / E~+l, \

t

J-1

'Y(J)(t) :t

UJ~(t)) + / E~+l, \

t UJ~(t)-y(j+1)(t)) J

1

= Un (E~+1(t)''Y(i+l)(t)) = Un (U- 1)i+1,i+1 > O. In passing from the second to the third line above, we have used the fact that E~+1 is orthogonal to all 'Y(J) for j ~ i since these are in the span of

{EJ}J=l, .... ~.

0

The last generalized curvature function II:n-l is sometimes called the torsion. It may take on negative values. Exercise 4.10. If 'Y : I -+ JRn is (n - l)-regular, show that E1 = T and E2 = N. If 'Y is parametrized by arc length, then fs 'Y - E1 and ~'Y = 1I:1E 2. Conclude that 11:1 II: (the curvature defined earlier). The orthogonal group O(JR n ) is the group of linear transformations A : JRn -+ JRn such that (Av, Aw) = (v, w) for all v, w E JR n . The group O(JRn) is identified with the group of orthogonal n x n matrices denoted O(n). The Euclidean group Euc(JR n ) is generated by translations and elements of O(JRn ). Every element ¢ E Euc(JR n ) can be represented by a pair (A, b), where A E O(JRn ) and bE JRn and where ¢(v) = Av + b. Note that in this case, D¢ - A (the derivative of ¢ is A). The elements of the Euclidean group are called Euclidean motions or isometries of JR n . If ¢ E Euc(JR n ), then for each p E JR n , the tangent map Tp¢ : TpJRn -+ T.p(p)JR n is a linear isometry. In other words, (Tp¢. vp, Tp¢· wp)/(p) - (vp, wp)p for all vp, wp E TpJR n . The group SO(JRn ) is the special linear group on JRn and consists of the elements of O(JR n ) which preserve orientation. The corresponding matrix group is SO(n) and is the subgroup of O(n) consisting of elements of determinant l. The subgroup SEuc(JR n ) c Euc(JR n ) is the group generated by translations and elements of SO(JRn). It is called the special Euclidean group.

4.1. Curves

151

Rn and '1 : I ~ R n be two (n - I)-regular curves with corresponding curvature functions Ki and K.i (1 < i ~ n - 1). If -y'(t) II I'1'(t) II and Ki(t) = Ki(t) for all t E I and 1 ~ i ~ n - 1, then there exists a unique isometry 4> E S Euc(Rn) such that

Theorem 4.11. Let"( : I

~

'1-4>0"(. Proof. Let (E1(t), ... , En(t)) and (E1(t), ... , En(t)) be the distinguished Frenet frames for "( and '1 respectively and let Wtj and i:hJ be the corresponding matrix elements as above. Fix to E I and consider the unique isometry 4> represented by (A, b) such that 4>b(to)) = '1(to) and such that

A(Ei(tO)) = Ei(tO) for 1 ~ i ~ n. Since 1I"f'(t) II = 11'1'(t)II and Ki(t) = K.i(t), we have that Wtj(t) = Wtj(t) for all i,j and t. Thus we have both

and

d

n

dt AEi(t) =

L Wji(t)AEJ (t). J

1

El and AEi satisfy the same linear differential equation, and since A(Ei(tO)) - Ei(to), we conclude that A(Ei(t)) = Ei(t) for all t and 1 ~ i < n. In particular, A-y'(t) = 1I"f'(t)II AEl(t) = 11'1'(t)II E1(t) = '1'(t). Thus Hence

4>b(t)) - 4> b(to)) =

it it

(4) 0 -y)' (T) dT =

to

=

to

A-y'(T) dT =

it

it

D4>· -y'(T) dT

to

'1'(T) dT = '1(t) - '1(to),

to

from which we conclude that 4>b(t)) - '1(t). For uniqueness, we argue as follows. Suppose that 'Ij; 0 -y = '1 for 'Ij; E Euc(Rn) and suppose that 'Ij; is represented by (B, c). The fact that D'Ij; must take the Frenet frame of "( to that of '1 means that A = D'IjJ = D4> = B. The fact that 'lj;b(to)) - '1(to) implies that b c and so 'Ij; 4>. 0 Conversely, we have Theorem 4.12. If Kl,' .. , Kn-l are smooth functions on a neighborhood of E R such that K t > 0 for i < n -1, then there exists an (n -1) -regular unit speed curve -y defined on some interval containing So such that Kl, ... ,Kn 1 are the curvature functions of -y.

So

152

4. Curves and Hypersurfaces in Euclidean Space

Proof. We merely sketch the proof: Let

A(s) :=

a

-KI

KI

a

a

a a

a a

a

a Kn-l

-Kn-l

a

and consider the matrix initial value problem

X'=XA, X(so) = I. This has a unique smooth solution X on some interval I which contains so. The skew-symmetry of A implies that X (s) is orthogonal for all s E I. If we let Xl be the first column of X, then

,(s):=

1 8

xI(t)dt

80

defines a unit speed (n - l)-regular curve with the required curvature func-

0

fuM.

Exercise 4.13. Fill in the details of the previous proof. If n > 3, then a regular curve, need not have a Frenet frame. However, a regular curve still has a curvature function, and if the curve is 2-regular, then we have a principal normal N which is defined so that T and N are the Gram-Schmidt orthogonalization of " and ,". We will sometimes denote this principal normal by E2 in order to avoid confusion with the normal to a hypersurface, which is denoted below by N.

4.2. Hypersurfaces Suppose that Y is vector field on an open set in IRn. For p in the domain of Y, and vp E TplRn, let c be a curve with c(O) = p and c(O) = vp. For any t near 0, we can look at the value of Y at c(t). We let V'vpY := (Y 0 c)' (0),

which is defined since Yo c is a vector field along c. Note that in this context (Y 0 c)' (0) is taken to be based at p = c(O). If X is a vector field, then a vector field V'xY is given by V'xY : pH V'xpY. In fact, it is easy to see that if X = E Xiei and Y = E Y~ei' then V'x Y =

It

L (Xyi)ei

since (Xyi) (p) = XpY~ = 10 yi 0 c. We have presented things as we have because we wish to prime the reader for the general concept of a covariant derivative that we will meet in later chapters. However, it must be confessed

153

4.2. Hypersurfaces

that under the canonical identification of IRn with each tangent space, \7 Xp Y is just the directional derivative of Y in the direction Xp. The map (X, Y) I---t \7 x Y is COO(IRn)-linear in X but not in Y. Rather, it is IR-linear in Y and we have a product rule:

\7xfY = (XI) Y

+ f\7xY.

Because of these properties, the operator \7 x : Y I---t \7 x Y, which is given for any X, is called a covariant derivative (or Koszul connection). In Chapter 12 we study covariant derivatives in a more general context. Since we shall soon consider covariant derivatives on submanifolds, let us refer to 'Vas the ambient covariant derivative. Notice that we have \7 x Y - \7y X -

[X,Yj. There is another property that our ambient covariant derivative \7 satisfies. Namely, it respects the metric:

X (Y, Z) = (\7xY, Z)

+ (Y, \7xZ).

Similarly, if vp E TpIRn, then vp (Y, Z) = (\7 vp Y, Zp) + (Yp, \7 vp Z). Consider a hypersurface M in IRn. By definition, M is a regular (n - 1)dimensional sub manifold of IRn. (If n = 3, then such a submanifold has dimension two and we also just refer to it as a surface in IR3.) A vector field along an open set 0 C M is a map X : 0 -+ TlRn such that the following diagram commutes: TIRn

Y1

o '---

IRn

Here the horizontal map is inclusion of 0 into IRn. If X (p) E TpM for all p EO, then X is nothing more than a tangent vector field on O. If N is a field along 0 such that (N(p), vp) = 0 for all vp E TpM and all p E 0, then we call N a (smooth) normal field. If N is a normal field such that (N(p) , N(p)) = 1 for all p E 0, then N is called a unit normal field. Note that because of examples such as embedded Mobius bands in IR3, it is not always the case that there exists a globally defined smooth unit normal field. Definition 4.14. A hypersurface Min IRn is called orientable if there exists a smooth global unit normal vector field N defined along M. We say that M is oriented by N. We will come to a more general and sophisticated notion of orientable manifold later. That definition will be consistent with the one above. Exercise 4.15. Show that for a connected orientable hypersurface, there are exactly two choices of unit normal vector field.

154

4. Curves and Hypersurfaces in Euclidean Space

In this chapter, we study mainly local geometry. We focus attention near a point p E M. We consider a chart (0, u) for ~n that is a single-slice chart centered at p and adapted to M. Thus if u = CuI, ... ,un), then the restrictions of the functions UI , ... ,un - I give coordinates for M on the set o = 0 n M = {un = o}. We denote these restrictions by ul, ... , un 1. Thus if we write u:= (u I , ..• ,un - I ), then (O,u) is a chart on M. We may further arrange that u( 0) is a cube centered at the origin in ~n and so, in particular, 0 is connected and orientable. Let us temporarily call such charts special. The coordinate vector fields a~. for i = 1, ... , n are defined on 0, while the vector fields a~. are defined on O. We have

r::J~' (p) =

-

uu t

r::J0. (P) for all p E 0 and i = 1, ... , n - 1. uu t

-

-

If X is a vector field on an open set 0, then its restriction to 0 - 0 Mis a vector field along 0, which certainly need not be tangent to M. If X is a vector field along 0, then there must be smooth functions Xi on 0 such that n

X(p) =

.

a

I: xt(P) ou (p) i

1

t

for all p EO. Then X is a tangent vector field on 0 precisely when the last component xn is identically zero on 0 so that n 1

a

X = '"' Xi_. L...J aut t

1

Now if X is a field along M, then it can be extended to a field X on 0 C ~n by considering the component functions Xi as functions on 0 which happen to be constant with respect to the last coordinate variable un. In other words, if 7r : 0 --+ 0 is the map (a I , ... , an) ~ (a I , .. . , an-I, 0), then

-

~

- a

X = L...J x taut' t

1

where it = xiou I 07rou• This last composition makes good sense because (0, u) is a special single-slice chart as described above. We will refer to X as an extension of X, but note that the extension is based on a particular special choice of single-slice chart. The constructions on the hypersurface that we consider below do not depend on the extension. Given a choice of unit normal N along a neighborhood of p EM, V'vpN is defined for any vp E TpM by virtue of the fact that N only needs to be defined along a curve with tangent vp. Alternatively, we can define V'vpN to be equal to V'vp IV for an extension IV of N in a special single-slice chart

155

4.2. Hypersurfaces

N

Figure 4.1. Shape operator

adapted to M and containing p. We note that ('VvpN,N(p)) = O. Indeed, since (N, N) 1, we have

0= vp (N, N) = 2 ('VvpN, N). Thus 'V vpN E TpM since TpM is exactly the set of vectors in TplRn perpendicular to Np = N(p). Exercise 4.16. Suppose we are merely given a unit normal vector at p. Show that we can extend it to a smooth normal field near p. For dimensional reasons, any two such extensions must agree on some neighborhood of p. Definition 4.17. Given a choice of unit normal Np at p E M, the map SNp : TpM ~ TpM defined by

SNp(Vp) := -'VvpN for any local unit normal field N with Np = N(p), is called the shape operator or Weingarten map at p. From the definitions it follows that if c is a curve with c(to) = p and = vp, then SNp(Vp) = - (N 0 c)' (to). If 0 c M is an open set oriented by a choice of unit normal field N along 0, then we obtain a map SN : TO ~ TO by SNITpM := SN p. This map is also called the shape operator (on 0). We have the inner product (" ')p on each tangent space TplRn and TpM C TplRn. For each p E M, the restriction of this inner product to each tangent space TpM is also denoted by (', ')p or just (', .). We also denote this inner product on TpM by gp so that gp("') = (-, ')p. The map p f--t gp is smooth in the sense that p f--t gp (X(p), Y(p)) is smooth whenever X and Yare smooth vector fields on M. Such a smooth assignment of inner product to the tangent spaces of M provides a Riemannian metric on M, a concept

c(to)

156

4. Curves and Hypersurfaces in Euclidean Space

studied in more generality in later chapters. We denote the function p r-+ gp (X(p), Y(p)) simply by 9 (X, Y) or (X, Y). In short, a Riemannian metric is a smooth assignment of an inner product to each tangent space. Definition 4.18. A diffeomorphism f : Ml ---+ M2 between hypersurfaces in JRn is called an isometry if Tpf : TpMl ---+ Tf(p) M2 is an isometry of inner product spaces for all p E MI. In this case, we say that MI is isometric to M 2·

Proposition 4.19. SNp : TpM ---+ TpM is self-adjoint with respect to (', ')p. Proof. Let (0, y) be a special chart centered at p and let 0 = 0 n M as above. Let Xp and Yp be elements of TpM and ex~end th.-:m to .:'ector fields X and Y on O. Then extend X and Y to fields X and Y on O. Similarly, extend Np to N and then fir. Note that Xp(fir, X) = Xp (N, X) = 0 and Yp(fir, Y) = O. Using this, we have

- (SNpXp, Yp)

+ (Xp, SNpYp)

= ('VxpN, Yp) - (Xp, 'VYpN) = ('V xfir, Y)p - (X, 'Vyfir)p

= Xp(fir, Y) - (fir, 'V xY)p - Yp(fir, X) + (fir, 'VyX)p = ('VyX - 'V xY, fir)p = ([Y, Xl, fir)p

= ([Y, Xl (p) , fir (p)) = ([Y, Xl (p) , N (p)) = 0 since [Y, Xl (p) sion map.

= [Y, Xl (p)

E

TpM by Proposition 2.84 applied to the inclu0

Definition 4.20. The symmetric bilinear form IIp on TpM defined by

IIp(vp,wp):= (Vp,SNpWp) = (SNpVp,Wp) is called the second fundamental form at p. If N is a unit normal on an open subset of M, then for smooth tangent vector fields X, Y, the function II(X, Y) defined by p H IIp(Xp, Yp) is smooth. The assignment (X, Y) H II(X, Y) defined on pairs of tangent vector fields is also called the second fundamental form. The form II is bilinear over COO(O), where 0 is the domain of N. As we shall see, the shape operator can be recovered from the second fundamental form II together with the metric 9 = (.,.) (first fundamental form). If the reader keeps the definition in mind, he or she will recognize that the second fundamental form appears implicitly in much that follows. We return to the second fundamental form again explicitly later. Exercise 4.21. Let Xp E TpJRn and let Y be any smooth vector field on JR n . Show that if f : JRn ---+ JRn is a Euclidean motion, then we have

157

4.2. Hypersurfaces

Tf· 'V xpY = 'VTj.xpf.Y. More generally, show that this is true if f is affine (i.e. if f is of the form f{x) - Ax + b for some linear map A and b E ]Rn).

Theorem 4.22. If Ml is a connected hypersurface in]Rn and if f : ]Rn ---t is a Euclidean motion, then M2 - f{M 1 ) is also a hypersurface and

]Rn

(i) the induced map flMl : Ml ---t M2 is an isometry; (ii) if Ml and M2 are oriented by unit normals Nl and N2 respectively, then, after replacing Nl or N2 by its negative if necessary, we have TfoSNl = SN2oTf.

Proof. That f(M I ) is a hypersurface is an easy exercise, which we leave to the reader. After noticing that flMl : Ml ---t M2 is smooth (why?), we argue as follows: Let ¢ := flMl and notice that T¢ . v = T f . v for all v tangent to Ml. Since T f preserves the inner products on T]Rn, we see that Tp¢ is an isometry for each p E MI. Since ¢ is clearly a bijection we conclude that (i) holds. Note that we must have Tpf· NI (p) = ±N2(p) and, since MI is connected, one possible change of sign on unit normals gives Tf

° NI ° f- I

-

N2.

Then by Exercise 4.21 Tf· SN1V

or Tf

0

SNl

= -Tf· 'VvNl = SN2

0

- -'VTj.vf.Nl

=

-'VTf·v N 2 = SN2 (Tf· v),

0

Tf.

Notice that an arbitrary isometry Ml ---t M2 need not be the restriction of an isometry of the ambient ]Rn and need not preserve shape operators. Definition 4.23. Let M be a hypersurface in ]Rn, take a point p E M, and let Np be a unit normal at p. The mean curvature H(p) at p in the direction Np is defined by 1 H(p) := --1 trace(SNp ) '

n-

Notice that changing Np to -Np changes H(p) to -H(P). If N is a unit normal field along an open set U c M, then the function p ~ H(p) is smooth. It is called a mean curvature function. If M is an orient able hypersurface, then it has a global mean curvature function for each unit normal field. Definition 4.24. Let M be a hypersurface in ]Rn. Let a unit normal Np be given at p. The Gauss curvature K(p) at p is defined by K(p) := det(SNp )'

4. Curves and Hypersurfaces in Euclidean Space

158

If N is a unit normal field along an open set U eM, then the function p H K(P) det(SN(p») is smooth. It is called a Gauss curvature function associated to the normal field, and always exists locally. If M is orient able, then it has a global Gauss curvature function for every choice of unit normal field. Notice that changing Np to -Np changes K(p) to (-lr- 1 K(p). It follows that if n - 1 is even, then there is a unique global Gauss curvature function regardless of whether M is orientable or not. In particular, this is the case for surfaces in IR3.

The shape operator S Np encodes the local geometry of the sub manifold at p and measures the way M bends and twists through the ambient Euclidean space. Since SNp is self-adjoint, there is a basis for TpM consisting of eigenvectors of SNp. An eigenvector for SNp is called a principal vector, and a unit principal vector is called a principal direction or a direction of curvature. The eigenvalues are called principal curvatures at p. If k1 , ... ,kn - 1 are the principal curvatures at p, then n-l

H(p)

_1_", k n 1 L...J 1 i

n-l

and

1

K(p) =

II k t

i•

1

If u E TpM is a tangent vector with (u,u) = 1, then k(u) = (SNpU,U) is the normal curvature in the direction u. Of course, if u is a unit length eigenvector (principal direction), then the corresponding principal curvature k is just the normal curvature in that direction. Notice that k( -u) - k(u). A vector v E TpM is called asymptotic if (SNpV,v) = O.

Proposition 4.25. Let M be a hypersurface and N a unit normal field. Let "( : I -+ M c ]Rn be a regular curve with image in the domain of N. Then (SNp "('(t), "('(t)) (N(-y(t)), "("(t)). Proof. Since (N (-y( t)), "(' (t)) - 0, differentiation gives

(N(-y(t)), "("(t))

+ (:t N(-y(t)), :t "((t)) = O.

Thus (SNp"(' (t), "(' (t))

= (-\} -y/(t)N, "(' (t)) = ( - :t N(-y(t)), :t "((t)) - (N(-y(t)) , "("(t)).

0

In particular, if u is a unit vector at p and u = C(O) for some unit speed curve c, then k(u) = (Np, d'(O)). This shows that all unit speed curves with a given velocity u have the same normal component given by the normal curvature in that direction. This curvature is forced by the shape of M and we see how normal curvatures measure the shape of M.

159

4.2. Hypersurfaces

Corollary 4.26. Let c : I --+ M c ]Rn be a unit speed curve. If I'\:(so) = 0 for some So E I, then k(c(so)) = O. If I'\: (s) > 0 and E2 is the principal normal defined near s, then k(c(s)) - ~(s) cosO(s), where O(s) is the angle between N(c(s)) and E2(S). Proof. Suppose ~(O) = O. Then k(d(O)) = (N(c(O)), c"(O)) 0, then E2 is defined for an interval containing s. We have

k(c'(s))

= O.

If ~ (s) >

(N(c(s)), c"(s)) = (N(c(s)), ~E2(C(S))) = ~(s) (N(c(s)), E2(S)) = ~(s) cos O(s).

=

o

If P is a 2-plane containing p E M and ~uch that Np is tan~ent to P, then for a small enough open neighborhood 0 of p, the set C = 0 n P n M is a regular one-dimensional submanifold of ]Rn. We parameterize C by a unit speed curve c with c(O) = p. This curve c is called a normal section at p. Notice that in this case E 2(O) - ±Np. Then k(c(O)) ±~l(O) where the" " sign is chosen in case Np - -E2(O). Thus we see that k(c(O)) is positive if the normal section c bends away from Np •

Definition 4.27 (Curve types). Let M be a hypersurface in I --+ M be a regular curve.

]Rn

and let

"y :

(i) , is called a geodesic if the acceleration ," (t) is normal to M for all t E I. (ii) , is called a principal curve (or line of curvature) if -y(t) is a principal vector for all tEl. (iii) , is called an asymptotic curve if -y(t) is an asymptotic vector for all tEl. Proposition 4.28. Let M be a surface in ]R3. Suppose that a regular curve "y : I --+ M is contained in the intersection of M and a plane P. If the angle between M and P is constant along " then, is a principal curve. Proof. The result is local, and so we assume that M is oriented by a unit normal N. Let v be a unit normal along P. Since P is a plane, v is constant. By assumption, (N, v) is constant along ,. Thus

0= :t (N

0"

v0

,)

= ("V"yN, v 0

,),

so 'V"yN is orthogonal to v along,. By the same token, "V"yN is orthogonal to N since (N, N) = 1. Thus "V"yN must be collinear with -y. In other words, SN-Y - "V"yN = >'-y for some scalar function >.. 0

160

4. Curves and Hypersurfaces in Euclidean Space

Example 4.29 (Surface of revolution). Let t t-+ (9(t), h(t)) be a regular curve in ]R2 defined on an open interval I. Assume that h > O. Call this curve the profile curve. Define x : I x ]R --+ ]R3 by

x(u, v) = (9(U), h(u) cosv, h(u) sin v). This is periodic in v and its image is a surface. The curves of constant u and curves of constant v are contained in planes of the form {x - c} and {z = my} (or {y = mz}) and so they are principal curves. For a surface of revolution, the circles generated by rotating a fixed point of the profile are called parallels. These are the constant u curves in the example above. The curves that are copies of the profile curve are called the meridians. In the example above, the meridians are the constant v curves. We know from standard linear algebra that principal directions at a point in a hypersurface corresponding to distinct principal curvatures are orthogonal. Generically, each eigenspace will be one-dimensional, but in general they may be of higher dimension. In fact, if SNp is a multiple of the identity operator, then there is only one eigenvalue and the eigenspace is all of TpM. In this case, every direction is a principal direction. It may even be the case that S Np = O. If c : I --+ M is a unit speed curve into a hypersurface, then rather than use the distinguished Frenet frame of c, we can use a frame which incorporates the unit normal to the hypersurface: Definition 4.30. Let c : I --+ M be a unit speed curve into a surface in R3 and let N be a unit normal defined at least on an open set containing the image of c. A frame field (Dl' D2, D3) (along c) such that Dl = T (= c), D3 = N 0 c and D2 D3 x Dl is called a Darboux frame. Exercise 4.31. Let c: I --+ M be a unit speed curve into a surface M in ]R3 and let D 1, D2, D3 be an associated Darboux frame. Show that there exist smooth functions 91, 92 and 93 such that 91 D 2 + 92D3, -91 D l + 93D 3, -92Dl 93D2. Show that 91 - 0 along c if and only if c is a geodesic. Show that 92 = 0 along c if and only if c is asymptotic, and 93 = 0 along c if and only if c is principal. dDl/ds dD 2 /ds = dD3/ ds =

The function 91 from the previous exercise is called the geodesic curvature function and is often denoted K- g • Let us obtain a formula for K- g • Define J : TpM --+ TpM by

4.2. Hypersurfaees

161

Figure 4.2. Curvature vectors

Notice that IIJvp 11 = Ilvpll and so J is an isometry of the 2-dimensional inner product space (TpM, (', ')p). But J also satisfies J2 = - id, and in dimension 2 this determines J up to sign since it must be a rotation by ±7l' /2. We now assume that N is globally defined so that the map J extends to a smooth map T M -+ T M. For a unit speed curve e : I -+ M, the geodesic curvature is given by (4.1)

Kg

/I dT JT ) = ( e, J e' ) - \/ ds'

.

Indeed, abbreviating No e to N we have from the first equation in Exercise 4.31 the following: dT

/I

,

,

ds = e = KgN x e + 92 N = KgJe + 92N. Taking inner products with Jd gives formula (4.1). Let e : I -+ M be a curve in a surface that is parametrized by arc length. The curvature vector K.(s) can be decomposed into a component Kg(S) tangent to M and a component K.n(s) normal to M:

K.(s) = K.g(s)

+ K.n(s).

The curve will be a geodesic if and only if K. g (s) = 0 for all s. The vector Kg(S) is the geodesic curvature vector at e(s). It is easy to show that Kg = ± II K.g(s) II. Figure 4.2 depicts a sphere of radius R with a conical "hat". The cone intersects the sphere in a curve of latitude. Since the cone is tangent to the sphere, the vectors K., K.g and K. n apply equally well to both surfaces at least along the curve. Using the Pythagorean theorem and similar triangles, it is possible to show that IIK.gll - 1/a, where a is the distance from the curve to the vertex of the cone. Now imagine cutting the cone along a generating line and unrolling it as shown in Figure 4.3. The result is a planar region with a circular arc of radius a and curvature whose

162

4. Curves and Hypersurfaces in Euclidean Space

magnitude is l/a. The reader might want to try and give a reason why this is to be expected. We will answer this later in this chapter.

Figure 4.3. Unrolling a cone

Definition 4.32. If SNp is a multiple of the identity operator, then p is called an umbilic point. If SNp = 0, then p is called a flat point. If every point of a hypersurface is umbilic, then we say that the hypersurface is totally umbilic. Example 4.33. If aIx l plane P in ]Rn, then

+ a2x2 + ... + anxn = 0 is the equation of a hypern

N = Laiei i-I

is a normal field when restricted to P. Since clearly SNp = 0 for all p E P, we see that every point of P is flat and that the Gauss and mean curvatures are identically zero. Example 4.34. Let sn-I be the unit sphere in ]Rn. Then the map p = (al, . .. ,an) f---t N(p) := ~~=I ai~ is a unit normal field along sn-I. We can calculate SN(V) = -\lvN. Let c be a curve in sn-I with C(O) = v. Then

-

dl

-\lvN = - dt

=-

N(c(t)) t=Q

d I L~ dt n

i=1

i

ei = -v.

t=Q

We are really just using the fact that up to a change in base point we have N (c( t)) = c( t). Thus S N = - id and we see that every point is umbilic and every principal curvature is unity. Also, K = 1 and the mean curvature H = -1 everywhere. Exercise 4.35. What is the shape operator on a sphere of radius r? Composition with a Euclidean motion preserves local geometry so if we want to study a hypersurface near a point p, then we may as well assume

163

4.2. Hypersurfaces

that p is the origin and that Np = en(O) = ~Io' Locally, M is then the graph of a function f : jRn-l ---t jR with ~(O) - 0 for i = 1, ... ,n -1. A normal field N which extends No = en(O) is given by n

1

.

(1 + L (8f /8x~)2)

-1/2

n-l

8t

(en - L 8xiei)' i-I

i=1

Let Z = en - E~:11 -£!'ei so that N = gZ, where 9 is the first factor in the formula above. For v = E~:ll v~ei tangent to M at the origin we have

V'v (gZ)

= (vg) Z + gV'vZ,

where vg means that v acts on 9 as a derivation. But vg vanishes at the origin since 9 takes on a maximum there. Since g(O) = 1, we have V'v (gZ) = V'vZ. Thus we may compute SNo using Z: SNoV

= -V'v Z .

We have n-l

SNoV

=

LV

(81 /8x i ) ~(O)

i=l

n-ln-l

.

81 I

= ~ ~ v 8xi8x j 0 e~(O). J

We conclude that the shape operator at the origin is represented by the (n - 1) x (n - 1) matrix

[D2 t] (0) =

[8:~XJ (0)] l~i,j~n-l

.

This is only valid at the origin. In other words, [D2 t] does not give us a representation of the shape operator except at the origin where M is tangent to jRn-l. We can arrange, by rotating further if necessary, that el, ... ,en 1 are directions of curvature. In this case, the above matrix is diagonal with the principal curvatures kI, . .. , k n - 1 down the diagonal. Example 4.36. Let M be the graph of the function f(x, y) = xy. We look at 0 E M. We have

[D21] = which diagonalizes to

[6 -.?d

[~ ~],

with corresponding eigenvectors ~

(el

+ e2)

and ~ (el - e2). Thus the Gauss curvature at the origin is -1 and the mean curvature is O. We can understand this graph geometrically. The normal sections created by intersecting the graph with the planes y = 0 and x = 0 are straight lines, and so the normal curvatures in those directions

4. Curves and Hypersurfaces in Euclidean Space

164

are zero. However, the normal section given by intersecting with the plane y = x is concave up, while that created by the plane y = -x is concave down. be a surface. If VI, V2 E TpM are linearly independent, then for a given unit normal Np we have

Proposition 4.37. Let M C

]R3

SNpVI x SNpV2 SNpVI x V2

+ VI

X

SNpV2

= K(p) (VI x V2), = 2H(p) (VI x V2) .

Proof. We prove the first equation and leave the second as an easy exercise. Let (s;) be the matrix of SNp with respect to the basis VI, v2. Then SNpVI x SNpV2

= (visi + v2s~) x (vls~ + v2s~) = sis~ - s~s~ = det(SNp)'

o

We remind the reader of the easily checked Lagrange identity:

(v x wax b) = '

I (v, a) (v, b) I (w,a)

(w,~

for any a, b, v, wE ]R3 (or in any Tp]R3). If N is a normal field defined over an open set in the surface M, and if X and Yare linearly independent vector fields over the same domain, then we can apply the Lagrange identity and the above proposition at each point to obtain

and also H=~

I(SNX,X) (Y,X)

(SNX, Y) (Y,Y)

I+ I (X,X)

(SNY,X)

(X, Y) (SNY,Y)

2~------~I-(~X-,X~)~(X-'-Y~)~I------~ (Y,X)

(Y,Y)

These formulas show clearly the smoothness of K and H over any region where a smooth unit normal is defined. Also, the formula k± =H± JH2_K

gives the two functions k+ and k- such that k+(p) and k-(p) are principal curvatures at p. These functions are clearly smooth on any region where k+ > k- and are continuous on all of the surface. Furthermore, if every point of an open set in M is an umbilic point, then k(p) := k+(p) = k-(p) defines a smooth function on this open set. This continuity is important for

165

4.3. The Levi-Civita Covariant Derivative

obtaining some global results on compact surfaces. Notice that for a surface in ]R3 the set of nonumbilic points is exactly the set where k+ > k-, and so this is an open set.

Definition 4.38. A frame field E I , ... , En I on an open region in a hypersurface M is called an orthonormal frame field if EI (p), ... , E n - l (p) is an orthonormal basis for TpM for each p in the region. An orthonormal frame field is called principal frame field if each Ei(P) is a principal vector at each point in the region.

Theorem 4.39. Let M be a surface in lR3 . If p E M is a point that is not umbilic, then there is a principal frame field defined on a neighborhood of p.

Proof. The set of nonumbilic points is open, and so we can start with any frame field (say a coordinate frame field) on a neighborhood of p. We may take this open set to be oriented by a unit normal N. We then apply the Gram-Schmidt orthogonalization process simultaneously over the open set to obtain a frame field FI, F2. Since p is not umbilic, we can multiply by an orthogonal matrix to assure that F I , F2 are not principal at p and hence not principal in a neighborhood of p. On this smaller neighborhood we have

SNFI

= aFI + bF2,

SNF2 = bFI + cF2 for functions a, b, c with b =1= O. Now define GI , G2 by

G1 = bFI + (k+ - a)F2' G2 = (k- - C)Fl

+ bF2

and check by direct computation that SNG I = k+G I and SNG2 = k+G2. We have used a standard linear algebra technique for changing to an eigenbasis. Since b =1= 0, we see that IIGtiland IIG211 are not zero. Finally, let EI = Gd IIGtil and

E2 =

G 2 / IIG211.

0

4.3. The Levi-Civita Covariant Derivative The Levi-Civita covariant derivative is studied here only in the special case of a hypersurface in lRn. The more general case is studied in Chapters 12 and 13. We derive some central equations, which include the Gauss formula, the Gauss curvature equation and the Codazzi-Mainardi equation. Let M be a hypersurface. For Xp E TpM, and Y a tangent vector field on M (or an open subset of M), define V xp Y by Vx"Y:= projT"MVX"Y,

where projTpM : TplRn -+ TpM is orthogonal projection onto TpM. For convenience, let us agree to denote the orthogonal projection of a vector

166

4. Curves and Hypersurfaces in Euclidean Space

v E TplRn onto TpM by v T and the orthogonal projection onto the normal direction by v.i. We call v T the tangent part of v and v.i the normal part. Then V'xl' Y = (V'xl'Y) T. Exercise 4.40. Show that for! a smooth function and Xp and Y as above, we have V'xl'!Y = (Xpl) Y(p) + !(p)V'xl'Y' If N is any unit normal field defined near p, then

V'xl'Y = V'xl'Y + (Np, V'xl'Yp) Np. Since

0= Xp (N, Y) = (-SNXp, Y) we obtain the Gauss formula:

+ (N, V'xl' Y) ,

(4.2) Notice that the right hand side of the above equation is unchanged if N is replaced by -N. It follows that if X and Yare smooth tangent vector fields, then p ~ V' xl' Y is smooth and we may then define the field V' x Y by (V'xY) (p) := V'xl'Y. By construction (V'xY) (p) = (V'zY) (p) if X(p) = Z(p). It is also straightforward to check that the map (X, Y) M V'xY is COO(M)-linear in X, but not in Y. Rather, like V', it is IR-linear in Y and we have the product rule V'x!Y

= (XI) Y + !V'xY,

which follows directly from Exercise 4.40. Thus V' is a covariant derivative on M and V' x Y is defined for X, Y E X(M). We remind the reader that X(M) is the space of tangent vector fields and not to be confused with vector fields along M which may not be tangent to M. Let Y and Z be smooth tangent vector fields on M and take Xp E TpM. We study the situation locally near p. Let (i5, \1) be a special chart ce~tered at p and let (~, u) be_the chart obtained by restrictio~ where 0 = 0 n M as before. If Y and Z are extensions of Y and Z to 0, then since S N IS self-adjoint, we have

(V'yZ - V' zY) (p) = (V'yZ - V' zY) (p)

= (V'yZ - V' zy) (p)

= [Y, Z]p = [Y, Z]p. Thus V' y Z - V' z Y = [Y, Z] for all Y, Z E X (M). This fact is expressed by saying that V' is torsion free. Also,

Xp (Y, Z) = Xp (y, Z) = (V'xl'y, Zp) = (V' xpY, Zp)

+ (¥p, V' xl'Z),

+ (Yp, V'xl'Z)

4.3. The Levi-Civita Covariant Derivative

167

so that if X E X(M), then X (Y, Z) = (V'xY, Z) + (Y, V'xZ). We express this latter fact by saying that V' is a metric covariant derivative on M. There is only one covariant derivative on M that satisfies these last two properties, and it is called the Levi-Civita covariant derivative (or Levi-Civita connection). We prove the uniqueness later in this chapter. With coordinates u\ ... ,un - 1 as above we have functions

If n-l

X

,8

= LX~-8i u

i=l

n-l

and Y

,8

= L Y3 -8" i=l

uJ

then (X, Y) =

L 9ijXi yj. i,j

The length of a curve in M is just the same as its length as a curve in the ambient Euclidean space. One may define a distance function on M by dist(p, q) := inf{L(c)}, where the infimum is over all curves connecting p and q. This gives M a metric space structure whose topology is the same as the underlying topology. This will be proved in more generality in Chapter 13. For now, the point is that metric aspects of M are determined by (" .) and locally by the 9'3 associated to each chart of an atlas for M. For example, the length of a curve c : [a, b] --+ 0 c M is given by

(4.3)

L(c) =

rL

J(1

b n-l

dci dd

9ij (c(t)) dtdt

dt,

a i,j=l

where ci(t) := ui

0 C.

Exercise 4.41. Deduce the above local formula for length from the formula for the length of the curve in the ambient Euclidean space. Returning to our covariant derivative V', we have

8 V'..L 8.. 1 = Bu' ·w

n-l

8

Lr~8U k

k=l

168

4. Curves and Hypersurfaces in Euclidean Space

for smooth functions rt known as the Christoffel symbols of V. For X and Y expressed as above, we have

and so we have

(4.4)

_~ ~ k j) t:i (~8Yk f-t X + /;::l rijX Y i

Vx Y -

Thus the functions

8 8u k '

i

8ui

rt determine V in the coordinate chart.

Also note that

o = [88., 88uJ ] = V...JL.. 8a - V...iL 88. = Lk (rfj - rji) 88u k . u~ au' u) au u~ 1

It follows that

rt =

(4.5)

rj~ for all i, j, k = 1, ... ,n - 1.

n-l

=

L

(r~igsj + r~jgsi) .

s=1

The matrix (gij) is invertible, and it is traditional to denote the components of the inverse by gi j , so that Er gjrgri = 6}. One may solve to obtain (4.6)

r~.~J = ~2 "" ks ~g

(8 9Si _ 8g~j 8ui 8us

+ 88u9 jsk )

.

II

Exercise 4.42. Prove formula (4.6) above. [Hint: First write the formula ~~i = Ell r~igsj + r~jgsi two more times, but cyclically permuting i, j, k to obtain three expressions. Subtract the second expression from the sum of the first and third. Use equality (4.5).] Proposition 4.43. The Levi-Civita connection on a hypersurface is determined uniquely by the properties of being a torsion free metric connection.

169

4.3. The Levi-Civita Covariant Derivative

Proof. In deriving the local formula for the Christoffel symbols, we only used the fact that V is a torsion free metric connection. We do this again in Chapter 13 in a more satisfying way. 0

If an object is determined completely by the metric on M, then we say that the object is intrinsic. Equivalently, if all local coordinate expressions for an object can be written in terms of the metric coefficients gij (and their derivatives, etc.), then that object is intrinsic. We have just seen that the connection V is intrinsic. It follows that if f : Ml ~ M2 is an isome1

2

try of hypersurfaces and V and V are the respective Levi-Civita covariant

derivatives, then 1

2

f*Vx Y = V/.xf*Y

for all vector fields X, Y E I(M). One way to see this is to examine the situation using a chart (V, u) on Ml and the chart (f(V), u 0 f 1) on M2. 2

A better way is to show that (X, Y) t--+ f*V /.xf*Y defines a torsion free metric connection on Ml and then use Proposition 4.43. (Exercise!)

If Y : I define

~

T M is a vector field along a curve c : I

~

M

c

]Rn,

then we

for any to E I and then define V a Y by at

( V ..!!. at

Y) (t) := V Yfor tEl. ..!!.I

at t

Suppose that c : I ~ M is such that c(J) and some chart (V, u). We can then write

Y(t) =

~ yi(t) a~i I i=l

c V for some subinterval J c I

for all t E J.

c(t)

Let us focus attention on a to such that c(to) =1= 0 and let J be an open interval with to E J. By restricting J if necessary ,!e may also assume that c is an embedding. Then there is a smooth field Y on a neighborhood of c(J) such that Yo c = Y. In fact, a simple partition of unity argument shows that we may arrange that Y be defined on all of V (in fact, on all of

170

4. Curves and Hypersurfaces in Euclidean Space

M). For t

E J we have

(V It Y) (t) = y' (t) T = ( (Y

0

c)' (t)) T

= (VC(t)Y) T = Vc(t)Y

n-l (n-l ~

= {;

=

,n-l + ij;l

ayk d aui (c(t)) d~

L (aY m + .L (rfj

n-l

k

n-l

k=l

0

c) (t)

d

rfj(c(t))

i

d~ (t)yj(c(t))

d~ (t)yj(t) auka I i

)

)

.

c(t)

l,j=1

We arrive at

We would like to argue that the above formula holds for general curves. If c(to) f:. 0, then there is an interval around to so that the formula holds as we have just seen. If c(to) = 0, we consider two cases. If there exists a sequence ti converging to to such that C(ti) = 0 for all i, then the formula holds for each ti and hence by continuity at to. Otherwise there must be an interval J containing to such that c is constant on J, say c( t) = p for all t E J, and then in this interval Y is just a map into the vector space TpM. In this case we have

(V:,Y) (t) = Y'W = =

(! ~>'(t) 8~.1J

ayk a I L m(t) auk

But this agrees with the formula since

c(t)

T

.

7t' = O.

Exercise 4.44. Show that for a vector field X along c : I --+ M and smooth function h E Coo (1) we have

I!.

Va/athX = hVa/atX + h'X.

Exercise 4.45. Show that for vector fields X, Y along c we have d dt (X, Y)

= (Va/at X , Y) + (X, Va/atY) = O.

The operator Va/at involves the curve c despite the fact that the latter is not indicated in the notation.

171

4.3. The Levi-Civita Covariant Derivative

We pause to consider again the question posed earlier about the circular arc on the unrolled cone in Figure 4.3. The key lies in the fact that the absolute geodesic curvature IKgl is intrinsic. We remarked earlier that the operator J is intrinsic up to sign (the latter being determined by orientation). On the other hand, in Problem 11 the reader is asked to derive the formula (Til, J,') K

But

g -

"'!"':--'--"';"":-

1Ir'11 3

IK 1- 1(Til, ±J,') 1_ I(V8/ 8 t'Y', ±J,') I g

-

1,'11 3

-

1Ir'11 3

'

and since both V and the pair ±J are intrinsic, we see that 1Kg 1 is intrinsic. A little thought should convince the reader that if a curve in one surface is carried to a curve in another surface by an isometry, then the curves will have equal absolute geodesic curvatures at corresponding points. The unrolling of the cone in Figure 4.3 can be thought of as inducing an isometry between the cone (minus a line segment) and a region in a planar surface in JR3. Thus we expect IKgl to be the same for both curves. But IKgl for a circular arc in a plane is just the reciprocal of the radius. Definition 4.46. A tangent vector field Y along a curve e : I -+ M is said to be parallel (in M) along e if V Jt Y = 0 for all tEl. Ilt

If e is self-parallel, Le. V8/8tC(t)

= 0 for all tEl, then it is easy to see

that e is a geodesic (in M) and in fact, this could serve as an alternative definition of geodesic curve, which will be the basis of later generalizations. Notice that if e is a curve in ]Rn, then Y can be considered as taking values in TJRn. However, Y being parallel in M is not the same as Y being parallel as a TJRn-valued vector field along e. In particular, a geodesic in M certainly need not be a straight line in ]Rn. For example, constant speed parametrizations of great circles on S2 C ]R3 are geodesics. Exercise 4.47. Given a smooth curve e : I -+ M, show that d' = 0 if and only if c is a geodesic in M such that (SNC(t), c(t)) = 0 for all t and choice of unit normal at e(t). Suppose that e" is never zero. Show that e is a geodesic in M if and only if e" is normal to M (Le. d'(t) ~ Tc(t)M for all tEl). The following simple result follows from the preceding exercise: Proposition 4.48. Let Ml and M2 be hypersurfaces in JRn. Suppose that c : I -+ ]Rn is such that e(t) E Ml n M2 for all t. If e" is not zero on any subinterval of I, then Tc(t)Ml = Tc(t)M2 for all tEl. Proof. By Exercise 4.47,

Tc(t)Ml

= d'(t).l = Tc(t)M2

for all t.

D

172

4. Curves and Hypersurfaces in Euclidean Space

~ M are parallel, then (X, Y) (t) := (X(t), Y(t)) is constant in t. In particular, a parallel vector field has constant length.

Proposition 4.49. If vector fields X, Y along a curve c : I

Proof.

ft (X, Y) = ('V 8/fJtX, Y) + (X, 'V 8/fJtY) = o.

o

Corollary 4.50. A geodesic has a velocity vector of constant length. Definition 4.51. Suppose Y E X(M) is a smooth vector field on Mj then Y is called a parallel vector field on M if 'V x Y = 0 at all points of M and for all smooth vector fields X E X(M). Obviously, Y is a parallel field if and only if Y all curves c.

0

c is parallel along c for

We now move on to prove two basic identities and introduce the curvat~r~ t~sor. It is easy to check by direct computation that for vector fields X, Y, Z on an open subset of ]Rn, we have

(4.8) If M is a hypersurface and X, Y, Z are tangent vector fields on a !le~h.E0r­ hood of an arbit~y p E M, then we may extend these to fields X, Y Z on a neighborhood 0 in ]Rn. Then we have

('Vx'VyZ - 'Vy'VxZ - 'V[X,YjZ) (p)

= ('V x 'VyZ - 'Vy'V xZ - 'V[X,y]Z) (p) = 0, and so (4.9) wherever the fields are all defined on M. Suppose that N is a unit normal field defined on the same domain in M. We apply the Gauss formula (4.2) to equation (4.9) above and then decompose it into tangent and normal parts:

0= 'Vx ('VyZ + (SNY, Z) N) - 'Vy ('VxZ + (SNX, Z) N) - 'V[X,YjZ = 'Vx'VyZ + (SNX, 'VyZ) N + X (SNY, Z) N - (SNY, Z) SNX - 'Vy'VxZ - (SNY, 'VxZ) N - Y (SNX, Z) N + (SNX, Z) SNY - 'V[X,YjZ - (SN[X, Y], Z) N. Equating the tangential parts of the above gives the Gauss curvature equation:

(4.10) 'V x'VyZ - 'Vy'V xZ - 'V[X,YjZ = (SNY, Z) SNX - (SNX, Z) SNY. The normal parts give

0= (SNX, 'VyZ) + X (SNY,Z) - (SNY, 'VxZ) - Y (SNX, Z) - (SN[X, Y], Z),

4.3. The Levi-Civita Covariant Derivative

173

or (VXSNY, Z) - (VySNX, Z) - (SN[X, YJ, Z) = 0 for all Z. From this we obtain the Codazzi-Mainardi equation:

(4.11) Let us give an application of the Codazzi-Mainardi equation and then return to the Gauss curvature equation.

Proposition 4.52. Let M be a connected hypersurface in ]Rn oriented by a unit normal field N. If every point of M is umbilic (i. e. M is totally umbilic), then the normal curvatures are all equal and constant on M. Furthermore, M is an open subset of a hyperplane or a sphere according to whether the normal curvatures are zero or nonzero. In particular, if M is a closed subset of]Rn, then it is a sphere or a hyperplane according to whether it is compact or not. Proof. There is a function k such that S N - kI, where I is the identity on each tangent space. This function is continuous since k = nIl trace S N. Let Xp E TpM and pick Yp E TpM so that Xp and Yp are linearly independent. Extend these to tangent fields X and Y on a neighborhood of p. By the Codazzi-Mainardi equation we have

0= VxkY - VykX - k[X, Yj = (Xk) Y + kVxY - ((Yk)X + kVyX) - k[X, Yj = (Xk) Y - (Yk) X, where we have used VxY - VyX = [X, Yj. In particular, at p we have (Xpk) Yp-(Ypk) Xp = O. Since Xp and Yp are linearly independent, Xpk = O. Since p and Xp were arbitrary and M is connected, we see that k is in fact constant. If the constant is k = 0, then SN = 0 on M, and this means that N = No is constant along M. This implies that M is in a hyperplane normal to No. If k =1= 0, then (changing N to -N if necessary) we may assume that k > O. Define a function f : M --t ]Rn by f(p) - p + tcN(p). We identify tangent spaces with subspaces of]Rn and calculate Df(p). Let v E TpM and choose a curve c: (-a, a) --t M with C(O) = v. Then we have

Df(p)·v= ddl f oc =C(0)+-k1 ddl t t=O t t 1

Noc 0

1

= v - "kSNV = V - "kkv - O. Thus D f (p) = 0 for all p EM, and since M is connected, f is constant (Exercise 2.23). Thus p + tcN(p) = q for some fixed q and all p. In other

174

4. Curves and Hypersurfaces in Euclidean Space

words, all p E M are at a distance 11k from q E ]Rn, and so M is contained 0 in that sphere of radius 11k. The left hand side of the Gauss curvature equation (4.10) is given its own notation

R(X, Y)Z := 'iJ x'iJyZ - 'iJy'iJ xZ - 'iJ[X,YjZ, and looking at the right side of (4.10) we see that (R(X, Y)Z) (p) depends only on the values of X, Y, Z at the point p, which means that we obtain 8 map Rp : TpM x TpM x TpM --t TpM defined by the formula

Rp(Xp, Yp)Zp := (R(X, Y)Z) (p). We say that R is a tensor since it is linear in each variable separately. We study tensors systematically in Chapter 7. The tensor R is called the Riemannian curvature tensor and the map p ~ Rp is smooth in the sense that if X, Y, and Z are smooth tangent vector fields, then p ~ Rp(Xp, Yp)Zp is a smooth vector field. We often omit the subscript p and just write R(Xp, Yp)Zp. (Notice that equation (4.8) just says that the curvature of lRn associated to the ambient covariant derivative 'iJ is identically zero.) Using (4.10) again, it is also easy to check that Rp is linear in each slot separately on TpM. In particular, for fixed X p, Yp E TpM we have a linear map R(Xp, Yp) : TpM --t TpM.

Theorem 4.53. If M is a surface in]R3 and (Xp, Yp) is an orthonormal basis for TpM, then

Proof. Using 4.10, and abbreviating SNp to S, we have

(R(Xp, Yp)Yp , Xp) = (SYp, Yp) (SXp, Xp) - (SXp, Yp) (SYp, Xp)

= det S = K (P).

0

We have shown that 'iJ is intrinsic and thus R is also intrinsic. Thus the previous theorem implies that the Gauss curvature K for a surface in lR3 is intrinsic. This is the content of Gauss's Theorema Egregium, which we prove (again) below using local parametric notation. If M is a k-dimensional submanifold in ]Rn, and (V, u) is a chart on M with U = u (V), then u- 1 : U --t M is a parametrization of a portion of M, and we denote this map by x: U --t M.

This notation is traditional in surface theory. Composing with the inclusion " : M y ]Rn, we obtain an immersion " 0 x : U --t ]Rn, but we normally identify" 0 x and x when possible. One may study immersions that are not

175

4.3. The Levi-Civita Covariant Derivative

necessarily one-to-one. A reason for this extension is that it might be the case that while x : U ---+ ]Rn is not one-to-one, its image is a submanifold M and so we can still study M via such a map. For example, consider the map x: ]R2 ---+ ]R3 given by

x(u, v) = ((a + b cos u) cos v, (a + bcos u) sin v, bsin u)

(4.12)

for 0 < b < a. This map is periodic and its image is an embedded torus. Notice that the restriction of x to sets of the form (uo - 7r /2, Uo + 7r /2) x (vo - 7r /2, Va + 7r /2) are parametrizations whose inverses are charts on the torus. Another example is the map x : ]R2 ---+ S2 C ]R3 given by x( 'P, 0)

= (cos 0 sin 'P, sin 0 sin 'P, cos 'P)'

The restriction of this map to (0,7r) set of measure zero.

X

(0,27r) parametrizes all of S2 but a

If x : U ---+ M is a parametrization of a hypersurface in ]Rn, then for UE U, the vectors (u) are tangent to M at p = x(u) and in fact form the

&:.

coordinate basis at p in somewhat different notation. In fact, if we abuse notation and write ui for ui 0 x-I, we obtain a chart (V, u) with x(U) = V. Then (u) is essentially just a~i ix(u)' We have = 9ij 0 x, but in the current context of viewing things in terms of the parametrization, we just change our notations slightly so that (::.':::,) = 9ij' These 9iJ are the components of the metric with respect to the parametrization. In these terms, some of the calculations look a bit different. For example, if 'Y : [a, b] ---+ M is a curve whose image is in the range of x, then the length of the curve can be computed in terms of the 9ij, which are now functions of the parameters u i . First, 'Y must be of the form t H x(u1(t), ... ,un-1(t)), where u : t H (u 1 (t), ... , Un-l (t)) is a smooth curve in U. Then we have

&:.

(&:., g;)

which is only notationally different from formula (4.3) due to our current parametric viewpoint. Exercise 4.54. Show that there always exists a local parametrization of a hypersurface M C ]Rn around each of its points such that 9~J (0) - 6~J and ~(O) = 0 for all i,j, k. [Hint: Argue that we may assume that M is written as a graph of a function f : ]Rn-l ---+ ]R with f(O) = 0 and Df(O) = O. Then let x: ]Rn-l ---+]Rn be defined by (u\ ... ,un - 1 ) H (U1 , ...

,un-l , f( u 1 , ... ,un-l)) .]

176

4. Curves and Hypersurfaces in Euclidean Space

Theorem 4.55 (Gauss's Theorema Egregium). Let M be a surface in]R3 and let p EM. There exists a parametrization x : U --t M with x(O, 0) = p such that 9ij = dij to first order at 0 and for which we have

K(p) = 82 912 (0) 8u8v

! 82 9222 (0) - ! 829112 (0). 2 8u

2 8v

Proof. In the coordinates of the exercise above, which give the parametrization (u, v) H (u, v, f(u, v)), where p is the origin of IR3 , we have

~f [ 911(U,V) 912(U,V)] _ [ 1 + (m)2 921(U, v) 922(U, v) ~~ 1 + (~)2

],

from which we find, after a bit of straightforward calculation, that

!

!

8 2912 (0) - 8 2922 (0) - 8 2911 (0) 8u8v 2 8u 2 2 8v 2

82 f 82 f

82 f

2

= 8u 2 8v 2 - 8u8v = det D f(O) = det S(p) = K(p).

0

Let us introduce some traditional notation. For x : U --t M c IR3, and denoting coordinates in U again by (u, v), and in IR3 by (x, y, z), we have

Xu=

( 8X 8y 8X) 8u'8u'8u x'

( 8X 8y 8X) 8v'8v'8v x' 82x 8 2y 8 2x) ( Xuv = 8u8v' 8u8v' 8u8v x' Xv

and so on. In this context, we always take the unit normal to be given, as a function of u and v, by

xuxXv x Xv I A careful look at the definitions gives (

)

N u,v =

(based at x(u,v)).

I XU

8N

8N

8u = SN(Xu) and 8v = SN(Xv).

Recall that the second fundamental form is defined by II(v,w) = (SNV,W), and this makes sense if the tangent vectors v, w are replaced by fields along x defined on the domain of N. The traditional notation we wish to introduce is E = (xu, xu) , 1=

(SNX u, xu) ,

F = (Xu,x v ) ,

G = (xv, xv) ,

m= (SNX u, xv) ,

n = (SNX v , xv) .

177

4.3. The Levi-Civita Covariant Derivative

Thus the matrix of the metric (.,.) (sometimes called the first fundamental form) with respect to Xu, Xv is

[:~~ :~:] = [~ ~], while that of the second fundamental form I I is

[! :]. The reader can check that Ilxu x Xv 112 = EG - F2. The formula for the length of a curve written as t r--t x(u(t), v(t)) on the interval [a, b] is

fa b

) a

E

(du)2 dt

dv (dV)2 + 2F du dt dt + F dt dt.

For this reason the classical notation for the metric or first fundamental form is

ds 2 = Edu 2 + 2Fdudv + Fdv 2, where ds is taken to be an "infinitesimal element of arc length" . Consider the map 9 : ]R3 -t (]R3)* given by v r--t (v, .). With respect to the standard basis and its dual basis, the matrix for this map is [~~]. Similarly, we can consider the second fundamental form as a map II: ]R3 -t (JR.3)* given by v r--t II (v, .), and the matrix for this transformation is [; ~]. Then since II (v, w) = (SNV, w), we have II = 9 0 SN and so SN = g-1

0

II.

We conclude that SN is represented by the matrix

This matrix may not be symmetric even though the shape operator is symmetric with respect to the inner products on the tangent spaces. Taking the determinant and half the trace of this matrix we arrive at the formulas nl-m2

K = EG-F2' H _ Gl + En - 2Fm 2 (EG - F2) .

Exercise 4.56. Show that 1 = (N, x uu ), m = (N, xuv) and n = (N, xvv). Exercise 4.57. Consider the surface of revolution given parametrically by

x(u, v) = (g(u), h(u) cosv, h(u) sin v)

178

4. Curves and Hypersurfaces in Euclidean Space

with h > O. Denote the principal curvature for the meridians through a point with parameters (u, v) by kll and that of the parallels by k7r • Show that these are functions of u only given by

k Il -

-

g' I g"

h' h"

I

-----;,------c~­

((g,)2 + (h,)2)3/2 '

g' k - ------..----=-----;;-----:7r h((g,)2 + (h,)2)1/2'

4.4. Area and Mean Curvature In this section we give a result that provides more geometric insight into the nature of mean curvature. The basic idea is that we wish to deform a surface and keep track of how the area of the surface changes. Let M C IR3 be a surface and let x : U --+ V C M be a parametrization of a portion V of M. We suppose that V has compact closure. The area of V is defined by A(V):=

L

Ilxu x

Xvii dudv.

The total area of M (if it is finite) can be obtained by breaking M up into pieces of this sort whose closures only overlap in sets of measure zero. However, our current study is local and it suffices to consider the areas of small pieces of M as above. Suppose x : U x (-c, c) --+ ]R3 is a smooth map such that for each fixed t E (-c, c), the partial map x(·, " t) : (u, v) t-+ x(u, v, t) is an embedding and such that is normal to Xu and Xv for all t. For each t, the image \It = x(U, t) is a surface. We have in mind the case where x(',', 0) is a parametrization of a portion of a given surface M so that Vo = V c M (see Figure 4.4). The normal N = Xu x Xvi Ilxu x Xv depends on t and at time t provides a unit normal to the surface \It. Thus \It is a one parameter family of surfaces.

tt:

Theorem 4.58. Let x : U x (-c, c) --+ ]R3 and let \It = x(U, t) be as above so that ~~ is normal to the surface \It. Let H (t) denote the mean curvature of the surface \It. Then

:t

A (\It) = -2

£I ~: I

Proof. Since ~ is parallel to N, we have ~~

axu at

=~ auat

H(t) dA.

=

I tt: II N.

Thus

ax = ~ S (x) +au ~ Ilaxll au (1laxll at N) = -II axil &t at N N

u

179

4.4. Area and Mean Curvature

Figure 4.4. Deformation of a patch

and so

Similarly,

(N,

Xu x

[}~v) = -II~~II (N, XU x SN (Xv)).

Now we calculate using the second formula of Proposition 4.37:

:t A = :t J Ilxu x xvii dudv = ! J (N,xu x xv) dudv = =

JI JI

d

\ dt N, XU

x Xv ) du dv +

[}xu

JI

[}x v )

\ N, Tt x Xv + Xu x [}t

= - J

= -2 J

[}xu \ N, Tt x Xv + Xu x [}Xv) [}t du dv

du dv

II~~II {II SN (xu) x Xv +xu X SN (Xv)II} dudv H(p)

I ~~llllxu x xvii dudv = -2 J I :11 H dA.

0

In particular, we can arrange that II ~ II = 1 at time t = 0, and then we have dd

I

t t=O

A = -2JHdA.

Thus, H is a measure of the rate of change in area under perturbations of the surface. Definition 4.59. A hypersurface in lRn for which H is identically zero is called a minimal hypersurface (or minimal surface if n = 3 so that dimM = 2).

4. Curves and Hypersurfaces in Euclidean Space

180

Example 4.60. For c> 0, the catenoid is parametrized by

x(u,v) = (u, c cosh(u/c)cosv, c cosh u sin v) and is a minimal surface. Indeed, a straightforward calculation gives [ EF FG]

[ c2 cosh2 (u/c)

[~

[~c I~C]'

It follows that H

0 ] cosh2(u/c)'

o

:] =

= O.

Example 4.61. For each b =1= 0, the map x(u,v) = (bv,ucosv,usinv) is a parametrization of a surface called a helicoid and this is also a minimal surface. The reader may enjoy plotting this and other surfaces using a computer algebra system such as Maple or Mathematica.

4.5. More on Gauss Curvature In this section we construct surfaces of revolution with prescribed Gauss curvature and also prove that an oriented compact surface of constant curvature must be a sphere. Let

x(u, v) = (g(u), h(u) cosv, h(u) sin v) be a parametrization of a surface of revolution. By a reparametrization of the profile curve (g(u), h(u)) we may assume that it is a unit speed curve so that (g,)2 + (h,)2 = 1. When this is done, we say that we have a canonical parametrization of the surface of revolution. A straightforward calculation using the results of Exercise 4.57 shows that for any surface of revolution given as above we have

g'

g' K=--

I g"

h ((g,)2

h' h"

I

+ (h,)2)

2'

If we assume that it is a canonical parametrization, then we have

K=

- (g,)2 h" + g' g" h' h

.

On the other hand, differentiation of (g,)2 + (h,)2 = 1 leads to g' g" = -h'h", and so we arrive at -h" K=h' which shows the expected result that K is constant on parallels (curves along which u is constant). Now suppose we are given a smooth function K defined on some interval I, which we may as well assume to contain O.

4.5. More on Gauss Curvature

181

We would like K to be our Gauss curvature and so we wish to solve the differential equation h" + Kh = 0 subject to h(O) > 0 and Ih'(O)1 < 1. The first condition simplifies the analysis, while the second condition allows us to obtain the canonical situation (g,)2 + (h,)2 = 1. In fact, we let

g(u)

=

fou V1 - (h'(t))2 dt,

and this will give the desired solution defined on the largest open subinterval J of I such that h > 0 and Ih'I < 1. With this solution, our surface of revolution will be defined, but only for u E J. Suppose we try to obtain a surface of revolution with constant positive Gauss curvature K = 1/c2 for some constant c. Then a solution of the equation for h will be h(u) = acos(u/c) for an appropriate a > O. Then

g(u)

=

fou

1- :: sin2 (t/c) dt,

and with the resulting profile curve, we obtain a surface of revolution with Gauss curvature 1/c2 at all points. There are three cases to consider. First, if a = c, then the interval J on which h > 0 and Ih'I < 1 is easily seen to be (-rrc/2, rrc/2) and we have

h(u)

=

ccos(u/c) and g(u)

= csin(u/c).

This gives a semicircle which revolves to make a sphere minus the two points on the axis of revolution. We already know that these two points can be added to give the sphere ofradius c = 1/.JK, which is a compact surface of constant positive Gauss curvature K. It turns out that the spheres are the only compact surfaces (without boundary) of constant positive Gauss curvature. Now consider the case 0 < a < c. The interval J is the same as before, but the surface extends in the x direction between Xl = limu -t-7rcj2 g( u) and X2 = limu -t7rc/2 g(u). It is easy to show that Xl < -a and X2 - -Xl> a. Since the maximum value of h is now smaller than c, the profile curve is shallower and wider than the semicircle of the a = c case above. Although liIIlu-t±7rc/2 h( u) = 0 as before, we now have the profile curve tangents at the endpoints given by lim

u-t±7rc/2

(g' (u), h' (u)) =

lim

u-t±7rc/2

(

1-

a:c sin (u/c), -~c Sin(u/C)) 2

( a2 a)

= 1 - c2 ' =t=~ • This shows that the revolved surface is pointed at its extremes and forms an American football shape. In this case, there is no way to add in the missing points on the axis of revolution to obtain a smooth surface. The two principal curvatures are no longer equal, but their product is still1/c2 •

182

4. Curves and Hypersurfaces in Euclidean Space

The third case a with boundary.

> 0 gives a surface that has an extension to a surface

Exercise 4.62. Analyze the case a

> O.

From above we see that there is an infinite family of surfaces with constant curvature (only one extending to a compact surface). The situation for constant negative curvature is similar, but we obtain no compact surfaces. One particular constant negative curvature surface is of special interest: Example 4.63 (Bugle surface). This surface has Gauss curvature and is given by

-lie?

x(u, v) = (u, h(u) cosv, h(u) sin v), where h is the solution of the differential equation h'-

- vc -h h 2-

2'

subject to initial condition limu-+o h( u) = c. The function h is defined on (0,00). Lemma 4.64. Let M be a surface in IR3. Let p E M be a nonumbilic point and EI, E2 a principal frame on a neighborhood of p (oriented by N) so that SNEI = k+ EI and SNE2 = k- E 2. If we define functions

then

Proof. Since (E2, E2) = I, we have ('VEl E2, E2) = 0 and so there is some function hI such that 'VEl E2 = hiEI. Similarly, 'V E2EI = h2E2 for some function h2. We find formulas for each of these functions. We have

and

0= EI (EI,E2) = ('VE 1 EI,E2)

+ (EI, 'VE 1 E2),

from which it follows that 'VEl EI = -hIE2' Similarly, 'V E2E2 = -h2El' We have

4.5. More on Gauss Curvature

183

We now apply the Codazzi-Mainardi equations (4.11):

0= 'VE1SNE2 - 'VE2SNEI - SN[EI,E2] = 'VEl k- E2 - 'V E2 k+EI - SN (hiEI - h2E2)

= (EIk- E2 + k-'VE1E2) - (E2k+ EI + k+'VE2EI) - (hIk+ EI - h2k- E2) = (Elk- E2

+ k-hIEI) -

(E2k+ EI

+ k+h2E2)

- (hlk+ EI - h2k- E2) = (k-h i

-

E2k+ - hIk+) EI

+ (Elk- + h2k- -

k+h2)E2.

Setting the coefficients of EI and E2 equal to zero we obtain

hI

-E2k+ k- and h2

= k+ _

E1kk- .

= k+ -

We compute as follows:

R(EI, E2)E2 = 'VEl 'V E2E2 - 'V E2 'V E1E2 - 'V[El,E2]E2 = 'VEl (-h2EI) - 'VE2 (hIEI) - 'V(hlEl h2E2)E2 = - (EIh2) EI - h2 'V El EI - (E2 hl) EI - hI 'V E2EI - hl'VE1E2 + h2'VE2 E2 = - (EIh2) EI - (E2hl) EI - h2 (-hIE2) - hI (h2E2) - hI (hIEI) + h2 (-h2EI) = - (Elh2) EI - (E2hl) EI - h~EI - h~EI'

Using the Gauss curvature equation (4.10) we arrive at

(4.13)

K

= (R(E I , E2)E2, EI) = -EIh2 -

E2hl - h~ - h~.

0

Corollary 4.65. If p is a nonumbilic point for which both k+ and k- are critical, then

Proof. Using equation (4.13), we have

K = -Elh2 - E2hl - h~ - h~

= -EI (k~~~- )

- E2

(k~E~~~ ) - (k~E~~~)

2-

(k~~~- )

2

At p we have E2k+ = E1k- = 0, and so (suppressing evaluations at p) we obtain -(k+ - k-)E?ko K(p) = (k+ _ k-)2

184

4. Curves and Hypersurfaces in Euclidean Space

Corollary 4.66 (Hilbert). Let M C ]R3 be a surface. Suppose that K is a positive constant on M. Then k+ cannot have a relative maximum at a nonumbilic point. Similarly, k- cannot have a relative minimum at a nonumbilic point. Proof. Since K = k+ k- > 0 is constant, k+ has a relative maximum exactly when k- has a relative minimum, so the second statement follows from the first. For the first statement, suppose that k+ has a relative maximum at p. Notice that X 2k+ ~ 0 and X 2k- ~ 0 for any tangent vector field defined near p. Near p, there is a principal frame as above, and if we use Corollary 4.65, we have E 2k+ - E 2kK (p) = 2 k+ _ k (p) < 0,

2

contradicting our assumption about K.

o

We are now able to use the above technical results to obtain a nice theorem. Theorem 4.67. If M C ]R3 is an oriented connected compact surface w~th constant positive Gauss curvature K, then M is a sphere of radius 1/.fK. Proof. The hypotheses give us the existence of a global unit normal field N. Note that k+ ~ ../K at every point, and since M is compact, k+ must have an absolute maximum at some point p. By Corollary 4.66, p is an umbilic point and so k+(p) = k-(p). But then (k+(p))2 = k+(p)k-(p) = K, and thus the maximum value of k+ is ../K. We have both k+ ~ ../K and k+ < ../K on all of M, and so k+ = k- = ../K everywhere (M is totally umbilic). By Proposition 4.52 we see that M is a sphere of radius 1/../K. 0

We already know from our surface of revolution examples that there are many noncompact surfaces of constant positive Gauss curvature, but now we see that the sphere is the only compact example. Actually, more is true: If a surface of constant positive Gauss curvature is a closed subset of ]R3, then it is a sphere. This follows from a theorem of Myers (see Theorem 13.143 in Chapter 13).

4.6. Gauss Curvature Heuristics The reader may be left wondering about the geometric meaning of the Gauss curvature. We will learn more about the Gauss curvature in Section 9.9 and much more about curvature in general in Chapter 13. For now, we will simply pursue an informal understanding of the Gauss curvature.

185

4.6. Gauss Curvature Heuristics

Negative Curvature

Zero Curvature

Positive Curvature

Figure 4.5. Curvature bends geodesics

On a plane, geodesics are straight lines parametrized with constant speed. If two insects start off in parallel directions and maintain a poliey of not turning either left or right, then they will travel on straight lines, and if their speeds are the same, then the distance between them remains constant. On a sphere or on any surface with positive curvature, the situation is different. In this case, geodesics tend to curve toward each other. Particles (or insects) moving along geodesics that start out near each other and roughly parallel will bend toward each other if they travel at the same speed. The second derivative of the distance between them will be negative. For example, two airplanes traveling due north from the equator at constant speed and altitude will be drawn closer and, if they continue, will eventually meet at the north pole. It is the curvature of the earth that "pulls" them together. On a surface of negative curvature, initially parallel motions along geodesics will bend away from each other. The second derivative of the distance between them will be positive. These three situations are depicted in Figure 4.5. For a more precise formulation of these ideas it is best to consider a parametrized family of curves, and we will do this in Chapter 13. It should be mentioned that according to Einstein's theory, it is the curvature of spacetime that accounts for those aspects of gravity (such as tidal forces) that cannot be nullified by a choice of frame (accelerating frames cause gravity-like effects even in a fiat spacetime). For example, an initially spherical cluster of particles in free fall near the earth will be deformed into an egg shape. For a wonderful popular account of gravity as curvature, see

[Wh]. Another way to see the effects of curvature is by considering triangles on surfaces whose sides are geodesic segments. These geodesic triangles are affected by the Gauss curvature. For instance, consider the geodesic triangle on the sphere in Figure 4.6. The angles shown are actually measured in the tangent spaces at each point based on the tangents of the curves at the endpoints of each segment. We have the following Gauss-Bonnet formula

186

4. Curves and Hypersurfaces in Euclidean Space

Figure 4.6. Curvature and triangle

involving the interior angles:

where D is the region interior to the geodesic triangle. This formula is true for geodesic triangles on any surface M. For a sphere of radius a, this becomes (31 + (32 + (33 = 7r + AIa 2 , where A is the area of the region interior to the geodesic triangle. The fact that the sum of the interior angles for a triangle in a plane is equal to 7r regardless of the area inside the triangle is exactly due to the fact that a plane has zero Gauss curvature. If one starts at p and moves counterclockwise around the triangle, then at the corners one must turn through angles Gl, G2, G3. In terms of these turning angles, the statement is 3

L i=l

Gi

= 27r

-1

K dB.

D

By an argument that involves triangulating a surface, it can be shown that if one integrates the Gauss curvature over a whole surface (without boundary) Me JR3, then something amazing occurs. We obtain

This result is called the Gauss-Bonnet theorem. Here X(M) is a topological invariant called the Euler characteristic of the surface and is equal to 2 - 2g, where g is the genus of the surface (see [Arm]). Thus while the left hand side of the above equation involves the shape dependent curvature K, the right hand side is a purely topological invariant! A presentation of a very general version of the Gauss-Bonnet theorem may be found in [Poor] (also see [Lee, J effj).

187

Problems

Problems (1) Let, : I -+ ]R2 be a regular plane curve. Let J : ]R2 -+ ]R2 be the rotation (x, y) t--+ (-y, x). Show that the signed curvature function K2 (t):=

,"(t) . J,'(t) 1Ir" (t) 11 3

determines, up to reparametrization and Euclidean motion.

+ g2 = 1. Let to E (a,b) and suppose that j(to) - cosfJo and g(to) = sinfJo for some ()o E R Show that there is a unique continuous function () : (a, b) -+ ]R such that ()(to) = fJo and

(2) Let j, 9 : (a, b) -+ ]R be differentiable functions with j2

f(t)

= cosfJ(t),

g(t)

= sinfJ(t)

for t E (a,b).

(3) Let, : (a, b) -+ ]R2 be a regular plane curve. (a) Given to E (a, b) and ()o with

" (to)

II,' (to) I

. = (cos fJo, sm fJo),

show that there is a unique continuous turning angle function ()"( : (a, b) -+ ]R such that ()(to) = fJo and

,'(t)

11r'(t) I

.

= (cos()"((t),smfJ"((t)) for

t E (a,b).

(b) Show that fJ~(t) = 1Ir'(t) I K2(t), where K2 is as in Problem 1. (4) Show that if K2 for a curve, : (a, b) -+ ]R2 is 1/r, then, parametrizes a portion of a circle of radius r. (5) Let,:]R -+]R3 be the elliptical helix given by t t--+ (a cos t, bsint, ct). (a) Calculate the torsion and curvature of ,. (b) Define a map F : ]R -+ SO(3) by letting F have columns given by the Frenet frame of ,. Show that F is a periodic parameterization of a closed curve in SO(3). (6) Calculate the curvature and torsion for the twisted cubic ,(t) := (t, t 2 , t 3 ). Examine the behavior of the curvature, torsion, and Frenet frame as t -+ ±oo. (7) Find a unit speed parametrization of the catenary curve given by c(t) := (a cosh(tla) , t). Revolve the resulting profile curve to obtain a canonical parametrization of a catenoid and find the Gauss curvature in these terms.

4. Curves and Hypersurfaces in Euclidean Space

188

(8) (Four vertex theorem) Show that the signed curvature function K2 for a simple, closed plane curve is either constant or has at least two local maxima and at least two local minima. (9) If 'Y : I ---+ ]R3 is a regular space curve (not necessarily unit speed), then _

show that B(t) -

-y'x-y" "'!'x"'!"II'

_ (-y'x"'!")·"'!'" _ I"'!'x"'!"11 I"'!' x",!" 112 and K 1"'1"113 •

T -

(10) With'Y as in Problem 1, show that 'Y"

=

(dt bill) T + b /11 2 KN.

(11) Let 'Y : 1---+ M C ]R3 be a curve which is not necessarily unit speed and 'Y" .J'Y') suppose M is oriented by a unit normal field N. Show that Kg =

W'

(12) Show that if a surface M c ]R3, either the image by an immersion of an open domain in ]R2 or is the zero set of a real-valued function f (such that df of. 0 on M) then it is orientable.

(13) Calculate the shape operator at a generic point on the cylinder {(x, y, z) : x 2 + y2 = r2}. (14) (Euler's formula) Let p be a point on a surface M in ]R3 and let Ul,U2 be principal directions with corresponding principal curvatures kl,k2 at p. If u = (cosO) Ul + (sinO) U2, show that k(u) = kl cos 2 0 + k2sin20. (15) Show that a point on a hypersurface in ]Rn is umbilic if and only if there is a constant ko such that k(u) = ko for all u E TpM with Ilull = 1. (16) Let Z be a nonvanishing (not necessarily unit) normal field on a surface M c ]R3. If X and Yare tangent fields such that X x Y = Z, then show that (Z,VxZ x VyZ) (Z, VxZ x Y +X x VyZ) 4 K = IIZI1 and H = 211Z11 3 (17) Find the Gauss curvature K at (x, y, z) on the ellipsoid x 2/a 2 + y2/b2+ y2/c 2 = 1. (18) Show that a surface in ]R3 is minimal if and only if there are orthogonal asymptotic vectors at each point. (19) Compute E, F, G, 1, m, n as well as Hand K for the following surfaces: (a) Paraboloid: (u, v) ~ (u, v, au 2 + bv 2) with a, b > O. (b) Monkey Saddle: (u,v) ~ (u,v,u 3 - 3uv 2). (c) Torus: (u,v) ~ ((a+ bcosv) casu, (a+ bcosv) sinu, bsinv) with a> b > O. (20) (Enneper's surface) Show that the following is a minimal but not 1-1 immersion: x(u,v):= (u-

~3 +uv2,

Chapter 5

Lie Groups

One approach to geometry is to view it as the study of invariance and symmetry. In our case, we are interested in studying symmetries of smooth manifolds, Riemannian manifolds, symplectic manifolds, etc. The usual way to deal with symmetry in mathematics is by the use of the notion of a transformation group. The wonderful thing for us is that the groups that arise in the study of geometric symmetries are often themselves smooth manifolds. Such "group manifolds" are called Lie groups. In physics, Lie groups play a big role in connection with physical symmetries and conservation laws (Noether's theorem). Within physics, perhaps the most celebrated role played by Lie groups is in particle physics and gauge theory. In mathematics, Lie groups play a prominent role in harmonic analysis (generalized Fourier theory), group representations, differential equations, and in virtually every branch of geometry including Riemannian geometry, Cartan geometry, algebraic geometry, Kahler geometry, and symplectic geometry.

5.1. Definitions and Examples Definition 5.1. A smooth manifold G is called a Lie group if it is a group (abstract group) such that the multiplication map J.I. : G x G ---t G and the inverse map inv : G ---t G, given respectively by J.I.(g, h) = gh and inv(g) = g-1, are Coo maps. If the group is abelian, we sometimes opt to use the additive notation 9 + h for the group operation. We will usually denote the identity element of any Lie group by the same letter e. Exceptions include the case of matrix or linear groups where we

-

189

190

5. Lie Groups

use the letter I or id. The map inv : G ~ G given by 9 H 9- 1 is called inversion and is easily seen to be a diffeomorphism.

Example 5.2. JR is a one-dimensional (abelian) Lie group, where the group multiplication is the usual addition +. Similarly, any real or complex vector space is a Lie group under vector addition. Example 5.3. The circle 8 1 = {z E C: IzI2 = 1} is a 1-dimensional (abelian) Lie group under complex multiplication. It is also traditional to denote this group by U(l). Example 5.4. Let JR* = JR\ {O}, C* = C\ {O} and IHI* = IHI\ {O} (here JH[ is the quaternion division ring discussed in detail later ). Then, using multipli. cation, JR*, C*, and IHI* are Lie groups. The Lie group IHI* is not abelian. The group of all invertible real n x n matrices is a Lie group denoted GL(n, JR). A global chart on GL(n, JR) is given by the n 2 functions x~, where if A E GL(n, JR) then xj(A) is the ij-th entry of A. We study this group and some of its subgroups below. Showing that GL(n, JR) is a Lie group is straightforward. Multiplication is clearly smooth. For the inversion map one appeals to the usual formula for the inverse of a matrix, A-I = adj(A)j det(A). Here adj(A) is the adjoint matrix (whose entries are the cofactors). This shows that A-I depends smoothly on the entries of A. Similarly, the group GL(n, C) of invertible n x n complex matrices is a Lie group.

Exercise 5.5. Let H be a subgroup of G and consider the cosets gH, g E G. Recall that G is the disjoint union of the cosets of H. Show that if H is open, then so are all the cosets. Conclude that the complement He is also open and hence H is closed. Theorem 5.6. If G is a connected Lie group and U is a neighborhood of the identity element e, then U generates the group. In other words, every element of g is a product of elements of U. Proof. First note that V = inv(U) n U is an open neighborhood of the identity with the property that inv(V) = V. We say that V is symmetric. We show that V generates G. For any open WI and W2 in G, the set WI W2 = {WI W2 : WI E WI and W2 E W2} is an open set being a union of the open sets U9EWlgW2. Thus, in particular, the inductively defined sets

V n = VV n- I , n

= 1,2,3, ... ,

are open. We have e EVe V2

c ... Vn

C ....

5.1. Dennitions and Examples

It is easy to check that each

191

vn is symmetric and so also is the union 00

V oo

:_

UV

n.

n=l

Moreover, V OO is not only closed under inversion, but also obviously closed under multiplication. Thus VOO is an open subgroup. From Exercise 5.5, yeo is also closed, and since G is connected, we obtain V OO = G. 0 In general, the connected component of a Lie group G that contains the identity is a Lie group denoted Go, and it is generated by any open neighborhood of the identity. We call Go the identity component of G. Definition 5.7. For a Lie group G and a fixed element 9 E G, the maps L9 : G --7 G and Rg : G --7 G are defined by

Lgx = gx for x E G, Rgx = xg for x E G, and are called left translation and right translation (by g) respectively. It is easy to see that Lg and Rg are diffeomorphisms with L;l = Lg 1 and Rg 1 = Rg 1. If G and H are Lie groups, then so is the product manifold G x H, where multiplication is (gl, hI) . (g2, h2) = (glg2, h1h2)' The Lie group G x H is called the product Lie group. For example, the product group 8 1 x 8 1 is called the 2-torus group. More generally, the higher torus groups are defined by Tn = 8 1 X ••. X 8 1 (n factors). Definition 5.S. Let H be an abstract subgroup of a Lie group G. If H is a Lie group such that the inclusion map H y G is an immersion, then we say that H is a Lie subgroup of G. Proposition 5.9. If H is an abstract subgroup of a Lie group G that is also a regular submanifold, then H is a closed Lie subgroup. Proof. The multiplication and inversion maps, H x H --7 Hand H --7 H, are the restrictions of the multiplication and inversion maps on G, and since H is a regular submanifold, we obtain the needed smoothness of these maps. The harder part is to show that H is closed. So let Xo E H be arbitrary. Let (U, x) be a single-slice chart adapted to H whose domain contains e. Let 8 : G x G --7 G be the map 8(gl' g2) = g1 1g2, and choose an open set Y such that e EVe V c U. By continuity of the map 8 we can find an open neighborhood 0 of the identity element such that 0 x 0 C 8- 1 (V). Now if {hi} is a sequence in H converging to Xo E H, then xo 1hi --7 e and Xo 1ht E 0 for all sufficiently large i. Since h-;lhi = (xo1h 3 f1 xo1h tl we

192

5. Lie Groups

have that hjl h~ E V for sufficiently large i, j. For any sufficiently large fixed j, we have Since U is the domain of a single-slice chart, un H is closed in U. Thus since each h j 1hi is in un H, we see that h-;lxo E Un H c H for all sufficiently large j. This shows that Xo E H, and since Xo was arbitrary, we are done. 0 By a closed Lie subgroup we shall always mean one that is a regular submanifold as in the previous theorem. It is a nontrivial fact that an abstract subgroup of a Lie group that is also a closed subset is automatically a closed Lie subgroup in this sense (see Theorem 5.81).

Example 5.10. 8 1 embedded as 8 1 x {I} in the torus 8 1 x 8 1 is a closed subgroup. Example 5.11. Let 8 1 be considered as the set of unit modulus complex numbers. The image in the torus T2 = 8 1 X 8 1 of the map JRl ---+ 8 1 X 81 given by t I--t (e~27rt, e~27rat) is a Lie subgroup. This map is a homomorphism. If a is a rational number, then the image is an embedded copy of 8 1 wrapped around the torus several times depending on a. If a is irrational, then the image is still a Lie subgroup but is now dense in T2. The last example is important since it shows that a Lie subgroup might actually be a dense subset of the containing Lie group.

5.2. Linear Lie Groups Let V be an n-dimensional vector space over IF, where IF = JR or C. The space L(V, V) of linear maps from V to V is a vector space and therefore a smooth manifold. A global chart for L(V, V) may be obtained by first choosing a basis for V and then defining n 2 functions {x~ h~~.j
5.2. Linear Lie Groups

193

the coordinate functions just introduced provide a global chart for GL(V). Let GL(n, IF) denote the group of invertible matrices with entries from IF. We obtain an isomorphism of GL(V) with the matrix group GL(n, IF) by choosing a basis and then simply sending each element of GL(V) to its matrix representative with respect to that basis. If we write mat(A) for the matrix that represents A E GL(V) with respect to a fixed basis, then A --t mat(A) is a group isomorphism and is clearly smooth. It follows that GL(V) is a Lie group, and for each choice of basis we have an isomorphism of Lie groups GL(V) ~ GL(n, IF). In practice it is common to work with the matrix group GL(n, IF) and its subgroups. The Lie group GL(V) is called the general linear group of V and is also denoted GL(V, IF) when we want to make the field apparent. The matrix group GL(n, IF) is also referred to as a general linear (matrix) group and is often identified with GL(IFn). More specifically, GL(n, lR) is called the real general linear group, and GL(n, q is called the complex general linear group. Lie groups that are subgroups of GL(V) for some vector space V are referred to as linear Lie groups and are often realized as matrix subgroups of GL(n, IF) for some n. Definition 5.12. Let V be an n-dimensional vector space over the field IF which we take to be either lR or C. Then the group SL(V) defined by SL(V) = SL(V,IF) := {A E GL(V) : det(A) = 1} is called the special linear group for V.

A bilinear form [3 : V x V ~ IF on an IF-vector space V is called nondegenerate if the maps V ~ V· given by [3 R : v ~ [3 (v , .) and fh : v ~ [3(" v), are both linear isomorphisms. If V is finite-dimensional, then f3 R is an isomorphism if and only if [3L is an isomorphism and then fJ is nondegenerate provided it has the property that if [3 (v, w) = 0 for all W E V, then v = O. Definition 5.13. A (real) scalar product on a (real) finite-dimensional vector space V is a nondegenerate symmetric bilinear form [3 : V x V ~ lR. A (real) scalar product space is a pair (V, (3) where V is a real vector space and f3 is a scalar product. (As usual we refer to V itself as the scalar product space when [3 is given) Definition 5.14. Let V be a complex vector space. An lR-bilinear map f3 : V x V ~ C that satisfies f3(av, w) - af3(v, w) and [3(v, aw) = a[3(v, w) for all a E C and v, wE V is called a sesquilinear form. If also f3(v, w) = [3(w, v), we call f3 a Hermitian form. If a Hermitian form is nondegenerate, we call it a Hermitian scalar product, and then (V, (3) a Hermitian scalar product space.

194

5. Lie Groups

The sesquilinear conditions imposed are described by saying that {1 is to be conjugate linear in the first slot and linear in the second slot. Nondegeneracy for a sesquilinear form is defined as for bilinear forms. Many authors define sesquilinear and Hermitian forms to be conjugate linear in the second slot and linear in the first. Obviously, f3( v, v) is always real for a Hermitian form. Definition 5.15. Let f3 be a (real) scalar product or Hermitian scalar product on a vector space V. Then

(i) f3 is positive (resp. negative) definite if f3(v,v) ~ 0) for all v E V and f3(v, v) = 0 ====} v = OJ

~

0 (resp. {1(v,v)

(ii) f3 is positive (resp. negative) semidefinite if f3(v,v) f3(v, v) ~ 0) for all v E V.

~

0 (resp.

What is called an inner product (a term we have already used) is a positive definite scalar product (or positive definite Hermitian scalar product in the complex case). In this book the term "inner product" always implies positive definiteness. Let us generalize the notion of orthonormal basis for an inner product space to include indefinite scalar product spaces. A basis (el, ... , en) for a scalar product space (V, f3) is called an orthonormal basis if f3(ei, ej) = 0 when i t- j, and f3(ei, ei) = ±1 for all i. An orthonormal basis always exists for a finite-dimensional scalar product space. Definition 5.16. Let V be an n-dimensional vector space over the field IF which we take to be either lR or C. If f3 is a bilinear form or sesquilinear form on V, then Aut(V, f3) is the subgroup of GL(V) defined by

Aut(V, f3) := {A E GL(V) : f3(Av, Aw) - f3(v, w) for all v, wE V}. If f3 is a scalar product, then the elements of Aut(V, f3) are called isometries of V. Theorem 5.17. SL(V) is a closed Lie subgroup of GL(V). If f3 is a bilinear or sesquilinear form as above, then Aut(V, f3) and SAut(V, f3) := Aut(V, f3) n SL(V) are closed Lie subgroups of GL(V). (The form f3 is most often taken to be nondegenerate.) Proof. It is easy to check that the sets in question are subgroups. They are clearly closed. For example, Aut(V, f3) = nv,wFv,w, where

Fv,w := {A E GL(V) : f3(Av, Aw) = f3(v, wH. The fact that they are Lie subgroups follows from Theorem 5.81 below. However, as we shall see, most of the specific cases arising from various choices of f3 can be proved to be Lie groups by other means. That they are Lie subgroups follows from Proposition 5.9 once we show that they are

195

5.2. Linear Lie Groups

regular submanifolds of the appropriate group GL(V, IF). We will return to this later, once we have introduced another powerful theorem that will allow us to verify this without the use of Theorem 5.B1. 0 Let dim V = n. After choosing a basis, SL(V) gives the matrix version SL(n,lF) := {A E Mnxn(lF) : detA - 1}. Notice that even when IF = C, it may be that fJ is only required to be lR-linear. Depending on whether IF = Cor lR and on the nature of fJ, the notation for the linear groups takes on special conventional forms introduced below. When choosing a basis in order to represent one of the groups associated to a form fJ in a matrix version, it is usually the case that one uses a basis under which the matrix that represents fJ takes on a canonical form. Example 5.18 (The (semi) orthogonal groups). Let (V, f3) be a real scalar product space. In this case we write Aut(V, f3) as O(V, f3) and refer to it as the semiorthogonal group associated to f3. With respect to an appropriately ordered orthonormal basis, fJ is represented by a diagonal matrix of the form

where there are p ones and q minus ones down the diagonal. The group of matrices arising from O(V, f3) with such a choice of basis is denoted O(p, q) and consists exactly of the real matrices Q satisfying Q'f/p,qQt - 'f/p,q' These groups are called the semiorthogonal matrix groups. With such an orthonormal choice of basis as above, the bilinear form (scalar product) is given as a canonical form on lRn where (p + q = n): p

(x,y):= Lxiyi i-I

n L

xiyi,

i=p+I

and we have the alternative description

O(p,q) = {Q E GL(n): (Qx,Qy) = (x,y) for all x,y E lRn}. If f3 is positive definite, we then have q = 0, and O(V, fJ) is referred to as a real orthogonal group. We write O(n,O) as O(n) and refer to it as the real orthogonal (matrix) group; Q E O(n) ¢:::::> QtQ = I.

196

5. Lie Groups

Example 5.19. There are also complex orthogonal groups (not to be confused with unitary groups). In matrix representation, we have O(n, C) := {Q E GL(n,C): QtQ = I}. Example 5.20. Let (V, (3) be a Hermitian scalar product space. In this case, we write Aut (V, t9, C) as U(V, t9) and refer to it as the semiunitary group associated to t9. If t9 is positive definite, then we call it a unitary group. Again we may choose a basis for V such that t9 is represented by the Hermitian form on cn given by p

(x,y) := Lxiyi 1-1

p+q=n

L

xiyi.

I-p+l

We then obtain the semiunitary matrix group U(p, q)

= {A

E GL(n, C) :

(Ax, Ay) = (x, y) for all x, y E lRn}.

We write U(n, 0) as U(n) and refer to it as the unitary (matrix) group. In particular, U(1) = 8 1 - {z E C: Izl = 1}. Definition 5.21. For (V, t9) a real scalar product space, we have the special orthogonal group of (V, t9) given by SO(V, t9) = O(V, (3)

n SL(V, lR).

For (V, t9) a Hermitian scalar product space, we have the special unitary group of (V, t9) given by SU(V, t9)

= U(V, (3) n SL(V, C).

Definition 5.22. The group of n x n complex matrices of determinant one is the complex special linear matrix group SL(n, C). We also have the similarly defined real special linear group SL(n, lR). The special orthogonal and special semiorthogonal matrix groups, SO(n) and SO(p, q), are the matrix groups defined by SO(n) = O(n) n SL(n, lR) and SO(p, q) = O(p, q) n SL(n, lR). The special unitary and special semiunitary matrix groups SU(n) and SU(p, q) are defined similarly. The group SO(3) is the familiar matrix representation of the proper rotation group of Euclidean space and plays a prominent role in classical physics. Here "proper" refers to the fact that SO(3) does not contain any reflections. In the problems we ask the reader to show that SO(3) is the connected component of the identity in 0(3). Exercise 5.23. Show that SU(2) is simply connected while SO(3) is not. Example 5.24 (Symplectic groups). We will describe both the real and the complex symplectic groups. Suppose that t9 is a nondegenerate skewsymmetric C-bilinear (resp. lR-bilinear) form on a 2n-dimensional complex

197

5.2. Linear Lie Groups

(resp. real) vector space V. The group Aut(V, {3) is called the complex (resp. real) symplectic group and is denoted by Sp(V, C) (resp. Sp(V,JR.)). There exists a basis {f~} for V such that {3 is represented in the canonical form by n

n

(v,w) = Lviw n+i

-

i=l

Lvn+jw j . 3=1

The symplectic matrix groups are given by Sp(2n, C) : ={A E M2nx2n(C) : (Av, Aw)

= (v, w)},

Sp(2n, JR.) := {A E M2nx2n(JR.) : (Av, Aw) = (v, w)}, where (v, w) is given as above. Exercise 5.25. For IF AtJA - J, where

= C or JR., show that A

E Sp(2n, IF) if and only if

Much of the above can be generalized somewhat more. Recall that the algebra of quaternions JHI is a copy of JR.4 endowed with a multiplication described as follows: First let a generic elements of JR.4 be denoted by x = (xO,x 1,x2,X3), y = (yO,y1,y2,y3), etc. Thus we are using {O,1,2,3} as our index set. Let the standard basis be denoted by 1, i,j, k. We define a multiplication by taking these basis elements as generators and insisting on the following relations:

= j2 = k 2 = -1, ij = -ji = k, jk = -kj = i, i2

ki = -ik =j.

Of course, JHI is a vector space over JR. since it is just JR.4 with some extra structure. As a ring, JHI is a division algebra which is very much like a field, lacking only the property of commutativity. In particular, we shall see that every nonzero element of JHI has a multiplicative inverse. Elements of the form a1 for a E JR. are identified with the corresponding real numbers, and such quaternions are called real quaternions. By analogy with complex numbers, quaternions of the form xli + x 2j + x 3 k are called imaginary quaternions. For a given quaternion x = x01+x 1 i+x 2j+x3 k, the quaternion xli + x 2j + x 3 k is called the imaginary part of x, and x01 =xo is called the real part of x. We also have a conjugation defined by

x

H

x

:=

x01 - xli - x 2j - x 3 k.

5. Lie Groups

198

Notice that xx = xx is real and equal to (X O)2 + (X 1)2 + (X 2)2 + (x 3)2. We denote the positive square root of this by Ixl so that xx = Ix1 2. Exercise 5.26. Verify the following for x, y E 1HI and a, bE R:

ax + by = ax + by,

(x) = x,

Ixyl = Ixllyl ,

Ixl = lxi, xy = yx.

Now we can write down the inverse of a nonzero x E 1HI: 1

-1

x

=

Ixl2x.

Notice the strong analogy with complex number arithmetic. Example 5.27. The set of unit quaternions is U(l,lHI) := {Ixl = I}. This set is closed under multiplication. As a manifold it is (diffeomorphic to) 8 3 . With quaternionic multiplication, 8 3 = U(l, 1HI) is a compact Lie group. Compare this to Example 5.3 where we saw that U(l, C) = 8 1 . For the future, we unify things by letting U(l, R) := Z2 = SO cR. In other words, we take the O-sphere to be the subset {-I, I} with its natural structure as a multiplicative group. U(l,lHI) = 8 3 , U(l, C) = 81, U(l,R) := Z2 = 8 0 . Exercise 5.28. Prove the assertions in the last example. We now consider the n-fold product lHIn , which, as a real vector space (and a smooth manifold), is R4n. However, let us think of elements of IHln as column vectors with quaternion entries. We want to treat lHIn as a vector space over 1HI with addition defined just as for Rn and en, but since 1HI is not commutative, we are not properly dealing with a vector space. In particular, we should decide whether scalars should multiply column vectors on the right or on the left. We choose to multiply on the right, and this could take some getting used to, but there is a good reason for our choice. This puts us into the category of right lHI-modules were elements of 1HI are the "scalars". The reader should have no trouble catching on, and so we do not make formal definitions at this time (but see Appendix D). For v, w E lHI n and a, bE 1Hl, we have

v(a + b) = va + vb (v+w)a=va+wa (va) b = v (ab) .

199

5.2. Linear Lie Groups

A map A : lHIn -+ lHIn is said to be lHI-linear if A(va) = A(v)a for all v E lHIn and a E 1HI. There is no problem with doing matrix algebra with matrices with quaternion entries, as long as one respects the noncommutativity of 1HI. For example, if A = (aj) and B = (b)) are matrices with quaternion entries, then writing C = AB we have

but we cannot expect that L aib; = L bjai. For any A = (a~), the map lHl!1. -+ lHIn defined by v f---t Av is lHI-linear since A (va) = (Av) a. Definition 5.29. The set of all m x n matrices with quaternion entries is denoted Mmxn(lHI). The subset GL(n, 1HI) is defined as the set of all Q E Mmxn(lHI) such that the map v f---t Qv is a bijection. We will now see that GL(n, 1HI) is a Lie group isomorphic to a subgroup of GL(2n, C). First we define a map t : C 2 -+ 1HI as follows: For (Zl' Z2) E C with Zl = x O+ xli and Z2 = x 2 + x 3 i, we let t(zl, z2) _ (x O+ xli) + (x 2 + x 3 i) j where on the right hand side we interpret i as a quaternion. Note that (x O+ xli) + (x 2 + x 3 i) j = x O + xli + x 2j + x 3 k . It is easily shown that this map is an lR-linear bijection, and we use this map to identify C2 with 1HI. Another way of looking at this is that we identify C with the span of 1 and i in 1HI and then every quaternion has a unique representation as zl + z2j for zl, z2 E C c 1HI. We extend this idea to square quaternionic matrices; we can write every Q E Mmxn(lHI) in the form A + Bj for A, BE Mmxn(C) in a unique way. This representation makes it clear that Mmxn(lHI) has a natural complex vector space structure, where the scalar multiplication is z(A + Bj) =zA + zBj. Direct computation shows that

(A + Bj) (C + Dj) = (AC - BD)

+ (AD + BC)j

for A + Bj EMmxn(lHI) and C + Dj EMnxk(lHI), where we have used the fact that for Q E Mmxn(C) we have Qj = jQ. From this it is not hard to show that the map 'I9 mxn : Mmxn(lHI) -+ M2mx2n(C) given by

'I9 mxn : A + Bj

f---+

(_AB

~)

is an injective lR-linear map which respects matrix multiplication and thus is an IR-algebra isomorphism onto its image. We may identify Mmxn(lHI) with the subspace of M2mx2n(C) consisting of all matrices of the form (_AB ~), where A, BE cmxn. In particular, if m = n, then we obtain an injective lRlinear algebra homomorphism 'I9 nxn : Mnxn(lHI) -+ M2nx2n(C), and thus the image of this map in M2nx2n(C) is another realization of the matrix algebra Mn n(lHI). If we specialize to the case of n = 1, we get a realization of 1HI as the set of all 2 x 2 complex matrices of the form ( _zw ~ ). This set of matrices

200

5. Lie Groups

is closed under multiplication and forms an algebra over the field R Let us denote this algebra of matrices by the symbol n.4 since it is diffeomorphic to lHI ~ R4. We now have an algebra isomorphism {} : lHI ---t n.4 under which the quaternions 1, i, j and k correspond to the matrices

(~ ~), (~ ~i)' ( ~1 ~)

(~ ~)

and

respectively. Since lHI is a division algebra, each of its nonzero elements has a multiplicative inverse. Thus n.4 must contain the matrix inverse of each of its nonzero elements. This can be seen directly: (

z w -w Z

)

1

-1

z-w ( )

Izl2 + Iwl 2

-

W

Z

•

Consider again the group of unit quaternions U(l, lHI). We have already seen that as a smooth manifold, U(l,lHI) is 8 3 • However, under the isomorphism lHI ---t n.4 C M 2x2 (C) just mentioned, U(l, lHI) manifests itself as SU(2). Thus we obtain a smooth map U(l, lHI) ---t SU(2) that is a group isomorphism. We record this as a proposition: Proposition 5.30. The map U(l, lHI) ---t SU(2) given by X

=

Z

. (z-w w)

+wJ

M

_,

Z

where x = xo + xli + x 2j + x 3 k, Z = xo + xli and w - x 2 + x 3i, is a group isomorphism. Thus 8 3 = U(l,lHI) ~ SU(2). Proof. The first equality has already been established. Notice that we then have

and so x E U(l, lHI) if and only if (-.!w ~) has determinant one. But such matrices account for all elements of SU(2) (verify this). We leave it to the reader to check that the map U(l, lHI) ---t SU(2) is a group isomorphism. 0 Exercise 5.31. Show that Q E GL(n, lHI) if and only if det({}nxn(Q))

=1=

o.

The set of all elements of GL(2n, C) which are of the form (_AB ~) is a subgroup of GL(2n, C) and in fact a Lie group. Using the last exercise, we see that we may identify GL(n, lHI) as a Lie group with this subgroup of GL(2n, C). We want to find a quaternionic analogue of U(n, C), and so we define b : lHIn x lHIn ---t lHI by

201

5.3. Lie Group Homomorphisms

Explicitly, if

then

b(v,w) =

[V' ...

vn

1[ }: ]

=

L .'w'.

Note that b is obviously JR.-bilinear. But if a E lHI, then we have b(va, w) = b(v, w)a and b(v, wa) = b(v, w)a. Notice that we consistently use right multiplication by quaternionic scalars. Thus b is the quaternionic analogue of an Hermitian scalar product. Definition 5.32. We define U(n, lHI): U(n,lHI):= {Q E GL(n,lHl): b(Qv,Qw) = b(v,w) for all v,w E JHr} U(n, lHl) is called the quaternionic unitary group. The group U(n, lHl) is sometimes called the symplectic group and is denoted Sp(n), but we will avoid this since we want no confusion with the symplectic groups we have already defined. The group U(n, lHl) is in fact a Lie group (Theorem 5.17 generalizes to the quaternionic setting). The image of U(n, lHl) in M2nx2n(C) under the map {}nxn is denoted USp(2n, C). Since it is easily established that {}nxnIU(n,lIlI) is a group homomorphism, the image USp(2n, C) is a subgroup of GL(2n, C). Exercise 5.33. Show that {}nxn(At) = ({}nxn(A)) t. Show that USp(2n, C) is a Lie subgroup of GL(2n, C). Exercise 5.34. Show that USp(2n, C) = U(2n) n Sp(2n, C). Hint: Show that {}nxn(Mnxn{IHr)) = {A E GL(2n, C) : JAJ- 1 = A}, where J = (-qd ig). Next show that if A E U(2n), then JAJ 1 = A if and only if AtJA = J.

5.3. Lie Group Homomorphisms Definition 5.35. Let G and H be Lie groups. A smooth map f : G -+ H that is a group homomorphism is called a Lie group homomorphism. A Lie group homomorphism is called a Lie group isomorphism in case it has an inverse that is also a Lie group homomorphism. A Lie group isomorphism G ~ G is called a Lie group automorphism of G.

5. Lie Groups

202

If f : G --+ H is a Lie group homomorphism, then by definition f(glg2) = f(gl)f(g2) for all gl, g2 E G, and it follows that f(e) = e and also that f(g-l) = f(g)-1 for all 9 E G.

Example 5.36. The inclusion SO(n, R) Y GL(n, IR) is a Lie group homomorphism. Example 5.31. The circle 8 1 C C is a Lie group under complex multiplication and the map z

= et(J

cos(O) sin(O) [ t---t - sin( 0) cos (0) o 0

is a Lie group homomorphism of 8 1 into SO (n). Example 5.38. The map U(l, lH[) --+ SU(2) of Proposition 5.30 is a Lie group isomorphism. Example 5.39. The conjugation map Cg : G --+ G given by x t---t gxg- 1 is a Lie group automorphism. Note that Cg = Lg 0 R g- l . Proposition 5.40. Let h : G --+ Hand 12 : G --+ H be Lie group homomorphisms that agree in a neighborhood of the identity. If G is connected, then h = h· Proof. By Theorem 5.6, any 9 EGis a product of elements in the set on which hand 12 agree, so the homomorphism property forces h = h. 0 Exercise 5.41. Show that the multiplication map J.L : G x G --+ G has tangent map at (e, e) E G x G given as T(e,e) J.L(v , w) = v + w. Recall that we identify T(e,e)(G x G) with TeG x TeG. Exercise 5.42. GL(n, IR) is an open subset of the vector space of all n x n matrices Mnxn(R). Using the natural identification of TeGL{n, IR) with Mnxn(IR), show that TeCg(x) = gxg-l, where 9 E GL(n,R) and x E Mnxn(R). Example 5.43. The map t t---t eit is a Lie group homomorphism from R to 8 1 C C. Definition 5.44. A Lie group homomorphism from the additive group IR into a Lie group is called a one-parameter subgroup. (Note that despite the use of the word "subgroup", a one-parameter subgroup is actually a map.)

203

5.3. Lie Group Homomorphisms

Example 5.45. We have seen that the torus 8 1 x 8 1 is a Lie group under multiplication given by (ei 7'J., et(h )(et7"2, et(2 ) _ (e t ('!"! +7"2), ei (9 1 +9 2 )). Every homomorphism of lR into 8 1 x 8 1 , that is, everyone-parameter subgroup of 8 1 x 8 1 , is of the form t I--t (e tai , etbi ) for some pair of real numbers a, b E JR. Example 5.46. The map R : lR

t I--t

~

80(3) given by

cost -sint ( sin t cos t o 0

is a one-parameter subgroup. Also, the map

t

I--t

cost -sint ( sin t cos t o 0

is a one-parameter subgroup of GL(3). Recall that an nxn complex matrix A is called Hermitian (resp. skewHermitian) if At - A (resp. At - -A). Let .5u(2) denote the vector space of skew-Hermitian matrices with zero trace. We will later identify .5u(2) as the "Lie algebra" of 8U(2). Example 5.47. Given 9 E 8U(2), we define the map Adg : .5u(2) ~ .5u(2) by Adg : x I--t gxg- 1 . The skew-Hermitian matrices of zero trace can be identified with lR3 by using the following matrices as a basis:

(01 -1) (-i0 0) ( -i0 -i) 0 0 ' i . I

These are just -i times the Pauli matrices 0'1,0'2,0'3, and so the correspondence .5u(2) ~ lR3 is given by -XiO'l - yi0'2 - iZ0'3 I--t (x, y, z). Under this correspondence, the inner product on lR3 becomes the inner product (A, B) = !trace(ABt) = -!trace(AB). But then

(AdgA, AdgB) = =

-21 trace(gAg- 1gBg- 1) 1

-2 trace(AB) -

(A, B).

So, Adg can be thought of as an element of 0(3). More is true; Adg acts as an element of 80(3), and the map 9 I--t Adg is then a homomorphism from 8U(2) to 80(.5u(2)) ~ 80(3). This is a special case of the adjoint map studied later. (This example is related to the notion of "spin". For more, see the online supplement.)

5. Lie Groups

204

Definition 5.48. If a Lle group homomorphism p : G -7 G is also a covering map then we say that G is a covering group and p is a covering homomorphism. If G is simply connected, then G (resp. p) is called the universal covering group (resp. universal covering homomorphism) of G. Exercise 5.49. Show that if p : M -7 G is a smooth covering map and G is a Lie group, then M can be given a unique Lie group structure ~ch that p becomes a covering homomorphism. (You may assume that M is paracompact. ) Example 5.50. The group Mob of Mobius transformations of the complex plane given by TA : z t-+ ~:t~ for A = (~~) E SL(2,C) can be given the structure of a Lie group. The map p : SL(2, C) -7 Mob given by p: A t-+ TA is onto but not injective. In fact, it is a (two fold) covering homomorphism. When do two elements of SL(2, q map to the same element of Mob?

5.4. Lie Algebras and Exponential Maps Definition 5.51. A vector field X E X(G) is called left invariant if and only if (Lg)*X = X for all 9 E G. A vector field X E X(G) is called right invariant if and only if (Rg)*X = X for all 9 E G. The set of left invariant (resp. right invariant) vector fields is denoted XL(G) (resp. XR(G)). Recall that by definition (Lg)*X = T Lg 0 X 0 L;1, and so left invariance means that TLgoXoL;l = X or that given any x E G we have TxLg·X(x) = X(gx) for all 9 E G. Thus X E X(G) is left invariant if and only if the following diagram commutes for every 9 E G: TL

TG~TG

x

IG

Lg

I 'G

x

There is a similar diagram for right invariance. Lemma 5.52. XL(G) is closed under the Lie bracket operation. Proof. Suppose that X, Y E XL(G). Then by Proposition 2.84 we have

(Lg)*[X, Y)

=

[Lg*X, Lg*Y) = [X, Y).

o

Given a vector v E TeG, we can define a smooth left (resp. right) invariant vector field LV (resp. RV) such that LV(e) = v (resp. RV(e) = v) by the simple prescription (resp. RV(g) = TRg · v).

5.4. Lie Algebras and Exponential Maps

205

A hit more precisely, LV(g) = Te (Lg) . v. The proof that this prescription gives smooth invariant vector fields is left to the reader (see Problem 7). Given a vector in TeG there are various notations for denoting the corresponding left (or right) invariant vector field, and we shall have occasion to use some different notation later on. We will also write L(v) for LV and R(v) for RV. The map v H L(v) (resp. v H R(v)) is a linear isomorphism from TeG onto XL(G) (resp. XR(G)): Exercise 5.53. Show that v H LV gives a linear isomorphism TeG ~ XL(G). Similarly, TeG ~ XR(G) by v H R(v). We now restrict attention to the left invariant fields but keep in mind that essentially all of what we say for this case has analogies in the right invariant case. We will discover a conduit (the adjoint map) between the two cases. The linear isomorphism TeG ~ XL(G) just discovered shows that XL(G) is, in fact, a vector space of finite dimension equal to the dimension of G. From this and Lemma 5.52 we immediately obtain the following: Proposition 5.54. If G is a Lie group of dimension n, then XL(G) is an n-dimensional Lie algebra under the bracket of vector fields (see Definition 2.75). Using the isomorphism TeG ~ XL(G), we can transfer the Lie algebra structure to TeG. This is the content of the following: Definition 5.55. For a Lie group G, define the bracket of any two elements TeG by

1J,w E

With this bracket, the vector space TeG becomes a Lie algebra (see Definition 2.78), and so we now have two Lie algebras, XL(G) and TeG, which are isomorphic by construction. The abstract Lie algebra isomorphic to either/both of them is often referred to as the Lie algebra of the Lie group G and denoted variously by .c(G) or g. Of course, we are implying that .c(H) is denoted ~ and .c(K) by t, etc. In some computations we will have to use a specific realization of g. Our default convention will be that g = .c(G) := TeG with the bracket defined above. Definition 5.56. Given a Lie algebra g, we can associate to every basis VI,"" vn for g, the structure constants ~j which are defined by [Vi, Vj]

=

2: C~jVk k

for 1 ~ i,j, k

< n.

206

5. Lie Groups

It follows from the skew symmetry of the Lie bracket and the Jacobi identity that the structure constants satisfy

i) (5.1) ii)

c:

l:k s4t + ~t4r + cfr4s -

O.

The structure constants characterize the Lie algebra, and structure constants are sometimes used to actually define a Lie algebra once a basis is chosen. We will meet the structure constants again later. Let a and b be Lie algebras. For (aI, bl ) and (a2' b2) elements of the vector space a x b, define

With this bracket, a x b is a Lie algebra called the Lie algebra product of a and b. Recall the definition of an ideal in a Lie algebra (Definition 2.79). The subspaces a x {O} and {O} x b are ideals in a x b that are clearly isomorphic to a and b respectively. We often identify a with a x {O} and 0 with {O} x o. Exercise 5.57. Show that if G and H are Lie groups, then the Lie algebra 9 x ~ is (up to identifications) the Lie algebra of G x H. Definition 5.58. Given two Lie algebras over a field IF, say (a, [, ]a) and (0, [, ]&), an IF-linear map (l is called a Lie algebra homomorphism if and only if

(l([V, w]a) = [(lV, (lwj& for all V, w E a. A Lie algebra isomorphism is defined in the obvious way. A Lie algebra isomorphism 9 ~ 9 is called an automorphism of g. It is not hard to show that the set of all automorphisms of g, denoted Aut(g), forms a Lie group (actually a Lie subgroup of GL(g)). Let V be a finite-dimensional real vector space. Then GL(V) is an open subset of the linear space L(V, V), and we identify the tangent bundle of GL(V) with GL(V) x L(V, V) (recall Definition 2.58). The tangent space at A E GL(V) is then {A} x L(V, V). Now the Lie algebra is T/GL(V), and it has a Lie algebra structure derived from the Lie algebra structure on XL(GL(V)). The natural isomorphism of nGL(V) with L(V, V) puts a Lie algebra structure on L(V, V). We now show that the resulting bracket on L(V, V) is just the commutator br~ket given by [A, B] := A 0 B - BoA. In the following discussion we let X denote the left invariant vector field corresponding to X E L(V, V) ~ T/(GL(V)). By definition we have [A, B] = [A, B]. We will need some simple results from the following easy exercises.

-----

5.4. Lie Algebras and Exponential Maps

207

Exercise 5.59. For a fixed A E GL(V), the map LA : GL(V) ---t GL(V) given by A ~ AoB has tangent map given by (A, X) ~ (AoB, AoX) where (A,X) E GL(V) x L(V, V) ~ T(GL(V)). Show that if X denotes the left invariant vector field corresponding to X E L(V, V), then X(A) = (A, AX). We are going to consider functions on G L(V) that are restrictions of linear functionals on the vector space L(V, V). We will not notationally distinguish the functional from its restriction. If I is such a linear function and X E L(V, V), let I,x be given by l,x(A) := I(A 0 X). It is easy to check that I,x is also a linear functional, and so for Y E L(V, V) we also have (f,x),y. It is clear that (f,x),y = I,Yox (notice the reversal of order). Exercise 5.60. Show that the map L(V, V) ---t (L(V, V))* given by X f,x is linear over JR.

~

Exercise 5.61. Let X be the left invariant field corresponding to X E L(V, V) as above. If I is the restriction to GL(V) of a linear functional on L(V, V) as above, then XI - I,x. Solution/Hint: (Xf)(A) = d/i A (XA) f(Ao X). Recall that if one picks a basis for V, then we obtain a global chart on

GL(V) which is given by coordinate functions x~ defined by letting x~(A) be the ij-th entry of the matrix of A with respect to the chosen basis. These coordinate functions are restrictions of linear functions on L (V, V) . Using this we see that if I,x = I,Y for all linear I, then I(X) = l,x(I) = l,y(I) f(Y) for all linear I and this in turn implies X = Y. Proposition 5.62. The Lie algebra bracket on L(V, V) induced by the isomorphisms XL(GL(V)) rv TJGL(V) ~ L(V, V) is the commutator bracket

[X, Y] := X

0

Y - Y a X.

Proof. Let X, Y E L(V, V) and let [X, Y] denote the bracket induced on L(V, V). Then for any linear I we have

f,[x,Y] Since

I

[X, Y]I = [X, Y]I = XYI - YXI = X (f,y) - Y (f,x) = (f,y),x - (f,x),y = l,xoY - I,Yox = 1,(xoY-YoX). =

was an arbitrary linear functional, we conclude that [X, Y] 0

XoY-YoX.

A choice of basis gives the algebra isomorphism L(V, V) ~ Mnxn(JR) where n = dim (V) . This isomorphism preserves brackets and restricts to a group isomorphism GL(V) ~ GL(n, JR). It is now easy to arrive at the following corollary.

5. Lie Groups

208

Corollary 5.63. The linear isomorphism of g[(n, IR) = TIGL(n, JR) with Mnxn(1R) induces a Lie algebra structure on Mnxn(IR) such that the bracket is given by [A, B] := AB - BA. From now on we follow the practice of identifying the Lie algebra g[(V) of GL(V) with the commutator Lie algebra L(V, V). Similarly for the matrix group, the Lie algebra of GL(n, IR) is taken to be Mnxn(IR) with commutator bracket. IT G c GL(n) is some matrix group, then TIG may be identified with a linear subspace of Mnxn. This linear subspace will be closed under the commutator bracket, and so we actually have an identification of Lie algebras: g is identified with a subspace of Mnxn. It is often the case that G is defined by some matrix equation or equations. By differentiating these equations we find the defining equations for g (as a subspace of Mnxn). We first prove a general result for Lie algebras of closed subgroups, and then we apply this to some matrix groups. Recall that if N is a submanifold of M and /. : N y M is the inclusion map, then we identify TpN with Tpt(TpN) and Tpt is an inclusion. IT H is a closed Lie subgroup of a Lie group G, and v E ~ = TeH, then v corresponds to a left invariant vector field on G which is obtained by using the left translation in G. But v also corresponds to a left invariant vector field on H obtained from left translations in H. The notation we have been using so far is not sensitive to this distinction, so let us introduce an alternative notation. Notation 5.64. For a Lie group G, we have the alternative notation v G for the left invariant vector field whose value at e is v. If H is a closed Lie subgroup of G and v E TeH, then v G E XL(G) while v H E XL(H). Proposition 5.65. Let H be a closed Lie subgroup of a Lie group G. Let

XL(H) := {X

E

XL(G) : X(e)

E

TeH}.

Then the restrictions of elements of XL(H) to the submanifold H are the elements of XL(H). This induces an isomorphism of Lie algebras of vector fields XL(H) ~ XL(H). For v, w E ~, we have [v, wh - [v H, wH]e = [v G, WG]e = [v, w]g I and so ~ is a Lie subalgebra of g; the bracket on inherited from g.

~

is the same as that

Proof. The Lie bracket on ~ = TeH is given by [v, w] := [v H, wH](e). Notice that if H is a closed Lie subgroup of a Lie group G, then for h E H we have left translation by h as a map G -t G and also as a map H -t H. The latter is the restriction of the former. To avoid notational clutter, let us denote

209

5.4. Lie Algebras and Exponential Maps

LhlH by lh. If t : H y G is the inclusion, then we have to lh = Lh a t, and so Tt a Tlh = TLh a Tt. If v H E XL(H), then we have Tht (v H(h))

Tht(Te1h(v)) = Tt a Tlh(V) = TLh a Tt(v) = TeLh (Tet(v)) = TeLh (v) = vG(h) = (v G a t) (h), =

so that v H and v G are t-related for any v E TeH C TeG. Thus for v, wE TeH we have [v,w]~ = [vH,wH]e = [vG,wG]e = [v,wjg, the formula we wanted. Next, notice that if we take Th" as an inclusion so that Tht (v H(h)) = v H (h) for all h, then we have really shown that if v E TeH, then v H is the restriction of vG to H. Also it is easy to see that

XL(H) = {v G : v

E

TeH},

and so the restrictions of elements of XL(H) are none other than the elements of XL(H). From what we have shown, the restriction map XL(H) --+ XL(H) is given by v G I-----t v H and is a surjective Lie algebra homomorphism. It also has kernel zero since if v H is the zero vector field, then v = 0, which implies that v G is the zero vector field. D Because [v, w]~ = [v, w]g, the inclusion ~ y 9 is a Lie algebra homomorphism. Examining the details of the previous proof we see that we have a commutative diagram

of Lie algebra homomorphisms where the top horizontal map is a restriction to H, the left diagonal map is v I-----t v G and the right diagonal map is v t----t v H • In practice, what this last proposition shows is that in order to find the Lie algebra of a closed subgroup H C G, we only need to find the subspace ~ = TeH, since the bracket on ~ is just the restriction of the bracket on g. The following is also easily seen to be a commutative diagram of Lie algebra homomorphisms:

XL(G)

('----~

where both vertical maps are v

I-----t

vG•

I 9

210

5. Lie Groups

The Lie algebra Lie group correspondence works in the other direction too: Theorem 5.66. Let G be a Lie group with Lie algebra g. If ~ is a Lie subalgebra of g, then there is a unique connected Lie subgroup H of G whose Lie algebra is ~.

The above theorem uses the Frobenius integrability theorem, so we defer the proof until we have that theorem in hand. Since the Lie algebra of GL( n) is the set of all n x n matrices with the commutator bracket, and since we have just shown that the bracket for the Lie algebra of a subgroup is just the restriction of the bracket on the containing group, we see that the bracket on the Lie algebra of matrix subgroups can also be taken to be the commutator bracket if that Lie algebra is represented by the appropriate space of matrices. We record this as a proposition: Proposition 5.67. IfG c GL(V) is a linear Lie group, then the Lie algebra of G may be identified with a subalgebra of End(V) with the commutator bracket. A similar statement holds for matrix Lie algebras. Example 5.68. Consider the orthogonal group O(n) C GL(n). Given a curve of orthogonal matrices Q(t) with Q(O) = I and -9t Q(O) = A, we compute by differentiating the defining equation 1= QtQ:

It=o

0=~1 QtQ dt t=o =

r

(! It-o Q

Q(O) + Qt(O)

(! It=o Q)

=At+A, so that the space of skew-symmetric matrices is contained in the tangent space TrO(n). But both TrO(n) and the space of skew-symmetric matrices have dimension n( n - 1) /2, so they are equal. This means that we can identify the Lie algebra o(n) = .c(O(n)) with the space of skew-symmetric matrices with the commutator bracket. One can easily check that the commutator bracket of two such matrices is skew-symmetric, as expected. We have considered matrix groups as subgroups of GL(n), but it is often more convenient to consider subgroups of GL(n, C). Since GL(n, q can be identified with a subgroup of GL(2n), this is only a slight change in viewpoint. The essential parts of our discussion go through for GL( n, q without any significant change. Example 5.69. Consider the unitary group U(n) c GL(n, C). Given a Q(O) = A, we curve of unitary matrices Q(t) with Q(O) = I and -9t

It=o

5.4. Lie Algebras and Exponential Maps

211

compute by differentiating the defining equation I = QtQ. We have

o=~1 clQ dt t-O =

(!It=o Q) tQ(O) + Qt(O) (!It=o Q)

=At+A. Examining dimensions as before, we see that we can identify u( n) with the space of skew-Hermitian matrices (At = -A) under the commutator bracket. Along the lines of the above examples, each of the familiar matrix Lie groups has a Lie algebra presented as a matrix Lie algebra with commutator bracket. This subalgebra is defined in terms of simple conditions as in the examples above, and the chart below lists some of the more common examples. Group

Lie algebra

SL(n,JR) O(n) SO(n) U(n) SU(n) Sp(2n,lF)

s[(n, JR) o(n) so(n) u(n)

su(n) sp(2n,lF)

Conditions defining the Lie algebra Trace(A) = 0 At=-A At=-A At=-A At = -A, Trace(A) JAtJ=A

=0

In the chart above the matrix J is given by J=

(~I ~).

Exercise 5.70. Find the defining conditions for the Lie algebra of the semiorthogonal group O(p, q). We would like to relate Lie group homomorphisms to Lie algebra homomorphisms. Proposition 5.71. Let f : G 1 --t G 2 be a Lie group homomorphism. The map Tel: gl --t g2 is a Lie algebra homomorphism called the Lie differential, which is denoted in this context by df : gl --t g2. Proof. For v E gl and x E G, we have

Txf· LV(x) = Txf· (TeLx . v)

= TeU 0 Lx) . v = Te(Lf(x) 0 f) . v = TeLf(x) (Tel . v) = TeLf(x) (Tel . v) = Ldf(v)U(x)),

5. Lie Groups

212

so LV "'/ Ld/(v). Thus by Proposition 2.84 we have that for any v,w E gl,

L[v,w] '" / [Ld/(v) I Ld/(w)]. In other words, [Ld/(v) I Ld/(w)]

0

f = T f 0 L[v,w], which at e gives

o

[df(v),df(w)] = [v,w].

Theorem 5.72. Invariant vector fields are complete. The integral curves through the identity element are the one-parameter subgroups. Proof. We prove the left invariant case since the right invariant case is similar. Let X be a left invariant vector field and c : (a, b) -7 G be an integral curve of X with c(O) = X(P). Let a < tl < t2 < b and choose an element 9 E G such that gC(tl) = C(t2)' Let t::..t = t2 - tl, and define c: (a + t::..t, b + t::..t) -7 G by c(t) = gc(t - t::..t). Then we have dd

I

t t=o

c(t)

= TLg . c(t = X

t::..t)

= TLg · X(c(t -

t::..t))

(gc(t - t::..t)) = X(c(t)),

and so c is also an integral curve of X. On the intersection (a + t::..t, b) of their domains, c and c are equal since they are both integral curves of the same field and since C(t2) = gC(tl) = C(t2)' Thus we can concatenate the curves to get a new integral curve defined on the larger domain (a, b + At). Since this extension can be done again for the same fixed At, we see that c can be extended to (a, 00 ). A similar argument shows that we can extend in the negative direction to get the needed extension of c to (-00,00). Next assume that c is the integral curve with c(O) = e. The proof that c(s + t) = c(s)c(t) proceeds by considering ,(t) = c(s)-lc(s + t). Then ,(D) = e and also

-yet)

TLc(s)-l . c(s + t) = TLc(s) l ' X(c(s + t)) = X(c(s)-lc(s + t)) = X(,(t)). =

By the uniqueness of integral curves we must have c(s)-lc(s + t) = c(t), which implies the result. Conversely, suppose c : R -7 G is a one-parameter subgroup, and let Xe - c(O). There is a left invariant vector field X such that X(e) = X e , namely X = LXe. We must show that the integral curve through e of the field X is exactly c. But for this we only need that c(t) - X(c(t)) for all t. We have c(t + s) = c(t)c(s) or c(t + 8) = Lc(t)c(s). Thus

c(t) =

:810 c(t + s) = (Tc(t)L) . c(O) = X(c(t)).

0

5.4. Lie Algebras and Exponential Maps

213

Proposition 5.73. Let v E 9 = TeG. We have the corresponding left invariant field LV and flow
(5.2) A similar statement holds with R V replacing LV.

Proof. Let u = st. We have that c1tlt=o
= sv 0

Theorem 5.74. Let G be a Lie group. For a smooth curve c: lR -7 G with c(O) = e and c(O) = v, the following are all equivalent:

(i) c is a one-parameter subgroup with c(O) = v. (ii) c(t) =
dd

t

It=O gc(t) =

dd

t

It=O Lg(c(t))

= TLgv = LV(g) for any g. In other words,

!I

Rc(t)g = LV (g)

t=O

for any g. But also, Rc(o)g = eg = 9 and
5. Lie Groups

214

Thus for s,t E R and v E 9 we have exp((s + t)v) = exp(sv) exp(tv), exp( -tv) = (exp(tv))-l . Note that we usually use the same symbol "exp" for the exponential map of any Lie group, but we may also write expG to indicate that the group is G. By Proposition 5.73 we have

exp(tv) = cptV(l) = cpV(t) =

cpf (e).

Thus by Theorem 5.74 we obtain the following: Proposition 5.76. For v E g, the map R ~ G given by t t--+ exp(tv) is the one-parameter subgroup that is the integral curve of LV. Lemma 5.77. The map exp: 9

~

G is smooth.

Proof. Consider the map R x G x 9 ~ G x 9 given by

(t, g, v) t--+ (g. exp(tv), v). This map is easily seen to be the flow on G x 9 of the vector field X : (g, v) t--+ (LV(g),O) and so is smooth. The restriction of this smooth flow to the submanifold {1} x {e} x 9 is (l,e,v) t--+ (exp(v) , v) and is also smooth. This clearly implies that exp is smooth also. 0 Note that exp(O) = e. In the following theorem, we use the canonical identification of the tangent space of TeG at the zero element (that is, To(TeG)) with TeG itself. Theorem 5.78. The tangent map of the exponential map exp : 9 ~ G is the identity at 0 E TeG = g, and exp is a diffeomorphism of some neighborhood of the origin onto its image in G, Teexp = id: TeG

~

TeG.

Proof. By Lemma 5.77, we know that exp: 9 ~ G is a smooth map. Also) 10 exp(tv) = v, which means that the tangent map is v t--+ V. If the reader thinks through the definitions carefully, he or she will discover that we have here used the identification of 9 with Tog. 0

it

From the definitions and Theorem 5.74 we have

cpf (p) = pexptv, cpf (p) = (exp tv) p for all v E g, all t E R and all pEG.

215

5.4. Lie Algebras and Exponential Maps

Proposition 5.79. For a (Lie group) homomorphism! : Gl following diagram commutes:

Proof. For v in the Lie algebra of Gl, the curve t a one-parameter subgroup. Also,

I-t

---t

G 2 , the

!(expGl (tv)) is clearly

and so by uniqueness of integral curves !(expGl(tV)) = expG2(td!(v)).

0

The Lie algebra of a Lie group and the group itself are closely related in many ways, and the exponential map is often the key to understanding the connection. One simple observation is that by Theorem 5.6, if G is a connected Lie group, then for any open neighborhood V C g of 0 the group generated by exp(V) is all of G. If H is a Lie subgroup of G, then the inclusion L : H y G is an injective homomorphism and Proposition 5.79 tells us that the exponential map on ~ egis the restriction of the exponential map on g. Thus, to understand the exponential map for linear Lie groups, we must understand the exponential map for the general linear group. Let V be a finite-dimensional vector space. It will be convenient to pick an inner product (.,.) on V and define the norm of v E V by Ilvll := J(v, v). In case V is a complex vector space, we use a Hermitian inner product. We put a norm on the set of linear transformations L{V, V) by IIAvl1 II All = sup -II-II . IIvll#o v We have IIA 0 BII ~ IIAIIIIBII, which implies that IIAkl1 ~ IIAllk, If we use the identification of gr(V) with L(V, V) (or equivalently the identification of g[(n, JR) with the vector space of n x n matrices M nxn ), then the exponential map is given by a power series

216

5. Lie Groups

which can be seen from the following argument: The sequence of partial 0 ~Ak is a Cauchy sequence in the normed space gl(V), sums SN :=

Ef

From this we see that

and so {SN} is a Cauchy sequence. Since gl(V) together with the given norm is known to be complete, we see that E:'o ~Ak converges. For a fixed A E gl(V), the function a : t H a(t) = exp(tA) is the unique solution of the initial value problem

a(O) = A.

a'(t) = Aa(t),

This can be seen by differentiating term by term: d dt exp(tA)

~1kk

= ~ k!t k 0

~

1

= ~ (k _1)!t

k

1 Ak

k=l

1 L tk k=l (k - 1)! 00

= A

A

1 A k- 1

= A exp(tA)

.

Under our identifications, this says that a is the integral curve corresponding to the left invariant vector field determined by A. Thus we have a concrete realization of the exponential map for gl(V) and, by restriction, each Lie subgroup of gl(V). Applying what we know about exponential maps in the abstract setting of a general Lie group we have in this concrete case exp((s + t) A) = exp(sA) exp(tA) and exp( -tA) = (exp(tA))-l. Let A, BE gl(V). Then

217

5.4. Lie Algebras and Exponential Maps

On the other hand, suppose that A

exp(A+B) =

L00

1

0

B = BoA. Then we have

m!tm(A+B)m

=

m-O

=

L00

1 m!

3+k=m J

m=O

~ ~ _1_ ti+ kAi Bk L..J L..J j!k!

i=Ok=O

(L0 0 ') %!AiBk

.

Thus in the case where A commutes with B, we have

exp(A + B) = exp(A) exp(B). Since most Lie groups of interest in practice are linear Lie groups, it will pay to understand the exponential map a bit better in this case. Let V be a finite-dimensional vector space equipped with an inner product as before, and take the induced norm on g[(V). By Problem 18 we can define a map log: U -+ g[(V), where

U = {B E GL(V) : IIBII

< 1},

by using the power series:

If we compute formally, then for A E g[(V), log (exp A)

= (A+

~!A2) - ~ (A+ ~!A2)2

+ ~ (A + ! 3

=A+

2!

A2) 3 + ...

(~!A2 - ~A2) + (;!A 3 _ ~A3 + ~A3) + ....

We will argue that the above makes sense if IIAII < 10g2 and that there must be cancellations in the last line so that log( exp A) = A. In fact, Ilexp A - III ~ e liAIl - 1, and so the double series on the first line for log(expA) must converge absolutely if e llAJJ - 1 < 1 or if IIAII < log 2. This means that we may freely rearrange terms and expect the same cancellations as we find for the analogous calculation of log(exp z) for complex % with Izl < log 2. But since log(expz) = z for such z, we have the desired conclusion. Similarly one may argue that exp(log B) = B if liB -

III < 1.

218

5. Lie Groups

Next, we prove a remarkable theorem that shows how an algebraic assumption can have implications in the differentiable category. First we need some notation. Notation 5.80. If S is any subset of a Lie group G, then we define

S-l = {s-l : s E S}, and for any x E G we define

xS = {xs: s E S}. Theorem 5.81. An abstract subgroup H of a Lie group G is a (regular) submanifold and hence a closed Lie subgroup if and only if H is a closed set in G. Proof. First suppose that H is a (regular) submanifold. Then H is locally closed. That is, every point x E H has an open neighborhood U such that un H is a relatively closed set in H. Let U be such a neighborhood of the identity element e. We seek to show that H is closed in G. Let y E Hand x E yU- 1 n H. Thus x E H and y E xU. This means that y E H n xU, and hence x- 1y E H n U = H n U. So y E H, and we have shown that His closed. Conversely, suppose that H is a closed abstract subgroup of G. Since we can always use left/right translation to translate any point to the identity, it suffices to find a single-slice chart at e. This will show that H is a regular submanifold. The strategy is to first find .c(H) = ~ and then to exponentiate a neighborhood of 0 E ~. First choose any inner product on T!}p so we may take norms of vectors in TeG. Choose a small ne.!ghborhood U of 0 E TeG = g on which exp is_a diffeomorphism, say exp : U ~ U, and denote the inverse by logu : U ~ U. Define the set Ii in fJ by Ii = logu (H n U). Claim. If h n is a sequence in H converging to zero and such that Un = hn / Ilhnll converges to v E g, then exp(tv) E H for all t E JR. Proof of the claim: Note that thn/ Ilhnll ~ tv while IIhnll converges to zero. But since IIhnll ~ 0, we must be able to find a sequence ken) E Z such that ken) Ilhnll ~ t. From this we have exp(k(n)hn) = exp(k(n) Ilhnllll~:II) ~ exp(tv). But by the properties of exp proved previously, we have that exp(k(n)hn ) = (exp(hn))k(n l . But also exp(hn ) E H n U CHand so (exp(hn))k(n l E H. Since H is closed, we have exp(tv) = lim (exp(hn))k(nl E H. n-+oo

Claim. If W is the set of all sv where s E JR and v can be obtained as a limit hn / Ilhnll ~ v with hn E Ii and h n ~ 0, then W is a vector space.

219

5.4. Lie Algebras and Exponential Maps

Proof of the claim: We just need to show that if hnf Ilhnll ~ v and h~J IIh~1I ~ w with h~, hn E ii, then there is a sequence of elements h~ from ii with h~ ~ 0 such that

"f I h"II n ~

hn

v +w Ilv +wll'

Using the previous claim, observe that

h(t) := logu(exp(tv) exp(tw)) = (logu op,) (exp(tv) , exp(tw)). Here p, is the group multiplication map. But, by the first claim, exp(tv) and exp(tw) are in H for all t, and so h(t) is in ii for small t. By Exercise 5.41 and the fact that Te log = id, we have that lim h(t)ft = h'(O) 40

= v + w.

Thus,

h(t)

--=

h(t)ft

~

v +w

.

Ilh(t)11 Ilh(t)ftll Ilv + wll 0 and let h~ := h(tn). Notice that by the first claim,

Now just let tn .!. exp(W) c H. Claim. Let W be the set from the last claim. Then exp(W) contains an open neighborhood of e in H. Proof of the claim: Let W 1. be the orthogonal complement of W with respect to the inner product chosen above. Then we have TeG = W1. EB W. It is not difficult to show that the map ~ : W EB W 1. ~ G defined by w + x ~ exp(w) exp(x)

is a diffeomorphism in a neighborhood of the origin in TeG. Denote this diffeomorphism by 1jJ. Now suppose that exp(W) does not contain an open neighborhood of e in H. Then we may choose a sequence hn ~ e such that hn is in H but not in exp(W). But this means that we can choose a corresponding sequence (wn, xn) E W EB W1. with (wn, xn) ~ 0 and exp(wn) exp(xn) E H and yet, Xn i= O. The space W1. is closed and the unit sphere in W1. is compact. After passing to a subsequence, we may assume that xnf Ilxnll ~ x E Wl., and of course IIxll = 1. Now exp(wn) E H since exp(W) CHand H is at least an algebraic subgroup, so we see that exp(wn ) exp(xn) E H. Thus it must be that exp(xn) E H also and so Xn E ii. But Xn ~ 0 and xnf Ilxnll ~ 0, and so, since we now know that Xn E ii, we have that x E W by definition. This contradicts the fact that Ixii = 1 and x E Wl.. Thus exp(W) must contain a neighborhood of e in

H. Finally, we let 0 C exp(W) be a neighborhood of e in H. The set 0 must be of the form 0 = H n V for some open V in TeG containing O. By

5. Lie Groups

220

shrinking V further we obtain a diffeomorphism 'I/Ilv' The inverse of this diffeomorphism,