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© P Baird 1983 First published 1983 AMS Subject Classifications: (main) 58E20, 53C40 (subsidiary) 58F05, 70E15
Library of Congress Cataloging in Publication Data Baird, P. (Paul) Harmonic maps with symmetry, harmonic morphisms and deformations of metrics. (Research notes in mathematics; 87) Includes bibliographical references and index. 1. Harmonic maps. 2. Submanifolds. I. Title. II. Series. QA614.73.B34 1983 514'.74 83-8186 ISBN 0-273-08603-0
British Library Cataloguing in Publication Data Baird, P. Harmonic maps with symmetry, harmonic morphisms and deformations of metrics.-(Research notes in mathematics; 87) 1. Harmonic maps 1. Title II. Series 514'.74 QA614.73 ISBN 0-273-08603-0
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Preface
We consider the basic problem of harmonic maps, that of finding a harmonic representative for a given homotopy class of maps between two Riemannian manifolds. Given that we are unable to solve this problem for fixed metrics, we can ask whether there exist metrics with respect to which there exists a harmonic representative this is known as the 'rendering problem'. In particular, we study these two problems for maps between spheres. We do this by considering maps where 'reduction occurs', that is, because the map possesses certain symmetries (equivariant), the problem of harmonicity reduces to solving a certain second-order nonlinear ordinary differential equation (the reduction equation). The origin of this method is the Thesis of R.T. Smith (3972), in which certain harmonic maps between Euclidean spheres are constructed. The symmetry which we make use of is that the map should preserve families of parallel hypersurfaces with constant mean curvature. It should be noted that this seems to be a more natural symmetry in the context of harmonic maps than the more commonly exploited symmetry of 'equivariance with respect to group actions'. We prove a reduction theorem for harmonic maps between space forms, and provide many examples of maps satisfying the conditions of the theorem. In some instances the reduction equation is easily soluble, on the other hand we find examples where we have little idea about appropriate solutions to the corresponding reduction equation.
Using the stress-energy tensor associated to harmonic maps we study harmonic morphisms, proving a theorem characterizing those harmonic morphisms with minimal fibres. We then consider harmonic morphisms defined by homogeneous polynomials, relating such maps to the construction of maps between spheres where reduction occurs. By allowing deformations of the metrics, first for harmonic morphisms and then for maps where reduction occurs, we are able to solve the rendering problem for all
classes of the homotopy groups 7r (Sn) = Z for all n. n Recently it has become apparent that harmonic maps have a significant role to play
in certain problems of theoretical physics. A particular example is the description of solitons in terms of harmonic maps from S2 into the complex projective space CPn . It is to be hoped that this work will lead to a greater understanding of solutions with symmetry of some of the variational problems of physics. I would like to express my most sincere thanks to Jim Eells who has provided continued support and encouragement during the preparation of this work. I would also like to thank L. Lemaire, J. Rawnsley, H. Sealey and J.C. Wood for many helpful conversations, the Science Research Council for their financial support, and the Institut des Hautes Etudes Scientifiques and the University of Bonn for their hospitality and support during 1981.
I am also especially indebted to my parents who have provided encouragement throughout.
Paul Baird
Contents
p.
INTRODUCTION
1
1.
FIRST CONSTRUCTIONS
7
1.1
The Laplacian on the sphere
1.2
1.4
Harmonic maps into spheres Joins of spheres and Smith's construction Outline of the solution of Smith's equation
1.5
Hyperbolic space
19
1.6
22
1.7
Polar coordinates on hyperbolic space and an analogous construction to Smith's Solving the equation for hyperbolic spaces
2.
ISOPARAMETRIC FUNCTIONS
27
2.1 2.2
Definition of isoparametric function Properties of isoparametric functions and Munzner's classification
2.3 2.4
theorem Examples of isoparametric functions Generalizing the notion of isoparametric families of hypersurfaces
3.
THE STRESS-ENERGY TENSOR
42
3.1
Derivation of the stress-energy tensor
42
1.3
3.2
7
9 11 16
24
27
29
34 39
46
3.3
The eigenvalue decomposition of the stress-energy tensor
4.
EQUIVARIANT THEORY
4.1
Maps which are equivariant with respect to isoparametric functions 49
47
49
4.2
Generalized equivariant maps between Riemannian manifolds
65
5.
EXAMPLES OF EQUIVARIANT MAPS
69
5.1 5.2 5.3
Maps from Euclidean space to the sphere Maps from hyperbolic space to the sphere Maps from sphere to sphere
69
6.
ON CERTAIN ORDINARY DIFFERENTIAL EQUATIONS OF THE PENDULUM TYPE
76
80
94
6.1 6.2 6.3
Existence of solutions Asymptotic estimates Smoothness of certain equivariant harmonic maps
7.
THE GENERAL THEORY OF HARMONIC MORPHISMS
122
7.1
General theory
122
7.2 7.3
Examples and non-examples of harmonic morphisms
125
94 111
116
Maps 0 : (M,g)-0(N,h) where 0 * h has two distinct non-zero eigenvalues
133
8.
HARMONIC MORPHISMS DEFINED BY HOMOGENEOUS POLYNOMIALS
8.1 8.2 8.3
Properties of harmonic polynomial morphisms 135 Some examples 139 Harmonic morphisms defined by homogeneous polynomials of degree bigger than two 142 Harmonic polynomial morphisms and equivariant maps between spheres
135
8.4
146
9.
DEFORMATIONS OF METRICS
153
9.1 9.2 9.3
Deformations of the metric for harmonic morphisms Examples Deformations of metrics for equivariant maps
153
160 162
9.4
Examples REFERENCES INDEX OF DEFINITIONS
172
177
180
Introduction
Let (M, g) denote the Riemannian manifold M together with its metric g.
Let 0 be
a map between Riemannian manifolds:
0: (M, g)
(N, h)
(all manifolds, metrics and maps will be assumed smooth unless otherwise stated). Then the derivative of 0, do , is a section of the bundle of 1-forms on M with values in the pull-back bundle Qf -1 TN: dO
c- W (T*M ® 0-1 TN).
The bundle T*M®O-1 TNis a Riemannian vector bundle and has a Levi-Civita connection V acting on sections:
V: W(T*M ®0-1 TN)-p_W(®2 T*M ®
0-1 TN).
Call v d 0 E e9 (02 T* M ® p1-1 TN) the 2nd fundamental form of the map 0 . Then we say that 0 is harmonic if trace V do = E vdp (X ,X .) is zero at each i 1 g point of M, where (Xi) I< i < dim M is a local orthonormal frame field for M. Given a map 0 as above; there are two basic problems of harmonic maps: (i) letting [0 ] denote the homotopy class of 0 , then does there exist a harmonic
representative 0 E [0 ] such that 0: (M,g)-w (N,h) is harmonic? ( ii) Do there exist metrics g and h on M and N respectively, such that there h) harmonic? exists a representative 0 E [0 ] with 0 : (M,g) The first of these problems has been considered throughout the history of elliptic analysis and differential geometry. The first existence theorem was proved by Eells and Sampson in [ 12 ]. The second problem is known as the rendering problem. This has been considered by Lemaire [28 ] in the case when the domain is a surface. One of the main objectives of this work is to study these two problems for maps 0: Sm-- PP.. Sn between spheres, where the homotopy classes are represented by the groups
Tr 1
Up until about 1970, the only harmonic maps 0: Sm -. Sn, m > n, that were known were a few homogeneous polynomial maps (defined by eigenfunctions of the Laplacian on Sm) :
(a) the identity maps In : Sn --- Sn for all n > 0 ; n : Stn-1 Sn , for n = 2, 4 and 8; (b) the Hopf maps H k, I where z e c , I z I2 = 1 and k is any : z -Z (c) the maps Gk : S1
-S
integer. In his Thesis in 1972, Smith [33 J gave a method whereby, given two such homogeneous polynomial maps, and provided certain "damping conditions" are satisfied, one can construct another harmonic map between spheres. Smith's method was to construct a 1-parameter family of maps all homotopic to the join of the two polynomial maps. This family is parametrized by a function a from the interval to itself. Smith then exploited the symmetry of the so constructed map OM to reduce the problem of whether 0a is harmonic or not to solving a 2nd order non-linear ordinary differential equation in a . This procedure of reducing the problem of harmonicity of a certain map to solving an ordinary differential equation we shall simply call reduction . The corresponding differential equation we shall call the reduction equation. Smith's equation has a simple physical interpretation - that of a pendulum moving under the influence of a variable gravity force with variable damping. The above damping conditions are sufficient conditions for this equation to have a solution. We place Smith's method into a more general setting by viewing his maps as sending "wavefronts to wavefronts"in the sense of geometric optics. That is, we have a commutative diagram of the form:
M
Oa
N
f
g If
a
g
where f, g are real valued functions on M,N respectively, with values in intervals I f9 19 2
respectively. , and a: I f - 19 is a smooth function. The appropriate class
of functions to which f and g must belong in order to provide a good theory turns out to be the class of isoparametric functions as defined by Cartan in 1938 [5 ]. These objects provide a rich and beautiful geometry on spheres, and it is indeed pleasing to find that they should be related to constructing harmonic maps between spheres. After Cartan's impressive study of isoparametric functions in papers dated 1938 1940 [5,6,7,8 ], the subject lay dormant until Nomizu's paper of about 1970 [32 ], surveying Cartan's work and giving some open problems. That revived interest in the subject, and Mummer proved an important classification theorem [30 ]. Also many new examples of isoparametric functions on spheres were found, see [32,34,40,33,17] In particular, examples of isoparametric functions on spheres whose level surfaces are non-homogeneous were found by Ozeki and Takeuchi [33 ] and Ferus, Karcher and Munzner [ 17 ]. These examples are very important from our point of view, since we make essential use of them in Theorem 5.3.8, Example 8.2.2 and in Section 8.4. This demonstrates that isoparametric functions are more natural for our purposes than families of homogeneous submanifolds of space forms (from which point of view one could conceivably derive our theory). R. Wood was the first person to make a connection between isoparametric functions and harmonic maps [43 ]. One of the essential features he observed was that such functions satisfy an eikonal equation of the form
Idf(x)I2
clxl2p-2
=
for some constant c, where p is an integer, p > 2. This remarkable fact makes the connection with geometric optics even more striking. Isoparametric functions as studied by Cartan are defined on space forms - that is, complete connected manifolds of constant curvature. In order to consider the construction of harmonic maps between more general Riemannian manifolds, we make what we consider to be a suitable definition of an isoparametric function on an arbitrary Riemannian manifold in Section 2.4. In Chapter 4 we develop the general framework, and define the necessary conditions on 0 in order that reduction occurs. Maps 0 satisfying these conditions we call equivariant. We prove our main theorem (Theorem 4.1.8) which is a reduction theorem for harmonic maps between space forms. 3
As a consequence of the above generalization, we can construct explicit harmonic maps from Euclidean spaces to spheres and from hyperbolic spaces to spheres - this is done in Sections 5.1 and 5.2. In Section 5.3 we find a very large number of classes n (Sn) where reduction occurs. However, many of the corresponding m
reduction equations have a qualitatively different nature to those of Smith, and we are
unable at present to solve many of them. One important class of equations does have similar properties to those of Smith, and the solutions yield some new and interesting harmonic maps between Euclidean spheres of the same dimension. This is proved in
Theorem 5.3.8. Two of the most significant examples in Smith's Thesis are
(i) the construction of harmonic representatives for all classes of Tn (Sn) for all n < 7, and which are parametrized by ( ii) the consideration of the classes of IT3(S2) their Hopf invariant (see [26 ]). The important point in example (i) is that the condition n < 7 is a consequence of the damping conditions. Thus Smith's method fails to give harmonic representatives for the classes of 11 n(Sn) with n > 7. In example (ii) the classes of TT 3(S2) are parametrised by their Hopf invariant d e Z. When d is the square of an integer, d = k2 for k e Z ; Smith gives a harmonic representative of this class. However, when d = k 1, k,1 c Z, k,1 0 and k 1; Smith provides a representative of the class where reduction occurs, but demonstrates that the corresponding reduction equation does not have a solution.
With the above two examples in mind we consider deformations of the metrics on space forms (Chapter 9) - this fits quite naturally into our general framework. We thus come to one of the main consequences of the reduction theorem. This is the solution of the rendering problem for all classes of nn(Sn) for all n (Theorem
9.4.5). The deformed spheres are familiar ellipsoids whose eccentricities depend only on a and the degree. Even allowing various deformations of metrics, we are still at present unable to solve the rendering problem for i 3(S2) . In Chapter 7 we undertake a general study of harmonic morphisms - maps between Riemannian manifolds which pull back germs of harmonic functions to germs of harmonic functions. We prove a theorem characterizing those harmonic morphisms for 4
which the fibres are minimal submanifolds. This generalizes a result of Eells and Sampson 1121 stating that every Riemannian submersion which is harmonic has minimal fibres. Harmonic morphisms were first considered in detail by Fuglede [ 19 ], and we make use of his ideas in Chapter 8 when we consider harmonic morphisms defined by homogeneous polynomials. Here we find an interesting connection with isoparametric functions, and we prove a theorem which associates to a certain class of harmonic polynomial morphisms an interesting harmonic Riemannian submersion. By using this theorem we can construct more maps between spheres where reduction occurs. One of the important tools used in Chapters 7, 8 and 9 is the stress-energy tensor a divergence free symmetric 2-tensor field on M. Such objects are well-known and important in relativity theory, where they in some sense model the matter distribution in a space-time model. For if (Sil ) 1 < i,j <3 are the space components of a symmetric 2-tensor, then Sij is the i-component of a force associated to the j-vector (the "stress "acting on a unit area orthogonal to j). Since force is a time rate of change of momentum, this represents the rate of flow of the i-component of momentum through a unit area orthogonal to j. We introduce the time components of S, to obtain the 4-tensor (S i7.) 0
div (S(X)) = 0
,
which corresponds to conservation of momentum if X is space like, and conservation of energy if X is timelike ( using the terminology of relativity theory, see [ 14 ]) . See [ 18 ] for more details. Hilbert was the first person to derive the stress-energy tensor from a variational principle [22 ], and it was following a suggestion of Taub ( 1963) that the stressenergy tensor was used to study harmonic maps in [2 ]. In chapter 3 we briefly derive and study basic properties of the stress-energy tensor for harmonic maps. For a much more detailed account see [2 ]. The emphasis with this work is on examples of harmonic maps. However, in chapter 4 we attempt to provide a general framework. This framework is rather unwieldy 5
because of (a) the large number of conditions on the map 0 , and (b) the necessary consideration of special cases. This detracts from the underlying simplicity, and often it is better to consider each example individually.
6
1 First constructions
1. 1
THE LAPLACIAN ON THE SPHERE
Let M and N be smooth Riemannian manifolds of dimension m and n respectively. Henceforth we shall write (M,g) to denote M together with its metric g. Let 0 : M - N be a smooth map, then the derivative dO is a section of the bundle of 0- 1 TN-valued 1-forms on M: d0
I t((T*M ® 0-'TN).
Let CAM denote the Levi-Civita connection on the bundle TM -- M, then v M extends to a connection on the bundle of tensors ®p TM ®® q T* M ~ M. The bundle
0-1 TN - M has a Riemannian structure induced from that of TN - N, allowing us to define the Levi-Civita connection V on the bundle T* M ® 0 -'TN-M. The section V (d0) (02 T*M ® 0-1 TN) is called the second fundamental form of the map 0 Lemma 1.1.1
If
X, Y E '' (TM), then
d0 (0 XY) + V d0 (X)
0 dO (X, Y)
Remark 1.1.2
d0 (Y)
This lemma is saying that contraction and covariant differentiation
commute.
We define the Laplacian A 0 of 0 , to be the trace with respect to g of the second fundamental form V d 0 :
A 0(x) =
a
V d0x(Xa(x),Xa(x))
at each point x of M, where Xa E '(TM), a = 1,2, ... , m, form a local orthonormal frame field about x. Then 0 is harmonic if A 0 = 0. Example 1.1.3
If M = R m with its standard Euclidean metric, and standard coordinates xl,x2, ... , xm, and 0: B m - R is a smooth function, then 7
a 0 2
'd 0 = E X
,
i
If 0 : Rm- Rn, 0 = 00=(001, ... , o0n
(01,
...
,
0 n),
.. , n, then
IR
We define the energy density Remark 1.1.4 e (0) : M - R given by
of the map 0 to be the function
e(0) (x) = 2 Id0(x) 12 1
E h(d0(X), x a do x(X a)) a
2
for each x e M, where (X a ) is an orthonormal basis for Tx M and h is the metric of N. Suppose that M is compact; using the canonical measure associated with g, we define the energy of 0 to be the number
E(0) = f
e(0) (x) dx
M
2
Then a C map 0.- M -- N is harmonic if and only if it is an extremal of the energy integral E [ 12 ]. Suppose 0 : M -- N and zb : N - P are two maps, then the second fundamental form of the composition is given by Vid(tb o 0) = dtb o V d0 + V
By taking traces we obtain the formula [ 12 ]
:
A(tJ 0 0) = dsb o AO + trace V dtb (dO, dO).
Let S M- I denote the standard unit sphere in R m , and let i : S m-1
Em be
the inclusion map.
Lemma 1.1.5 M-1 OS
8
If
(fo0
f: R m- R is a smooth function then m
Af
-
e2f
8r
(m-1) £`-fr )
oi
(1.1.4
Using the formula (1. 1.3) we have
proof
Sm-1
(f o i) = df (A i) + trace V df(di,di).
O
Also QR
2` f o i
m
(f) o i = trace V df (di, di) Sm-1
(foi) +
_
22
f
2r 2
+
2 r2
- df(
0 i
Sm-1
n.
0
Let x be a point of S m-1 , then at x Sm-1 p
i
= trace V di(x)
= E OR
m
dl x (X di x (X a )
a
a)
where (Xa) 1 < a < m forms an orthonormal basis for TxSm-1, and Xa = ya(0),
where ya(t) are geodesics in S m-1 with ya (0) = x. Thus S
M-1 i
=E a
ya (0) Sm-1 i
= - (m - 1)
Corollary 1.1.6 degree p, then Sm- i
0
F/2r. Thus
ya(0)
(foi)
If
`-
2r
f: R m - R is a harmonic homogeneous polynomial of
= - p(p +m - 2) f o i
so f o i is an eigenfunction of p
,
Sm-1 .
m-1
Remark 1. 1.7 All eigenfunctions of polynomials on R m in this way.
arise from harmonic homogeneous
1.2
Harmonic maps into spheres Let Sn-1 denote the standard sphere in Euclidean n-space, and j; Sn-1 --- I n the inclusion map. If 0: (M,g) _ Sn-1 is a map; let t denote the composition 9
j o o: M - En The map 0 is harmonic if and only if
Lemma 1.2. 1
p4 = - 2e (4b) fi
(1.2.1)
.
Proof From equation (1.1.3 ) p (j o 0) = dj (n 0) + trace V dj(d0, d0).
Since j is an isometric immersion V d j (X, Y) is perpendicular to d j (Z), for all TSn-1). in partiX,Y,Z (7 '(TSn-1) (so & 0 is the projection of A (4 ) onto cular, 0 is harmonic if and only if 6 (j o 0) is proportional to 4 Now, writing < , > for the Euclidean metric on JR n
< d(jo0), j o 0 > = 0. We differentiate this formula at x E M. Let (X a ) 1
OX a
n
d(j o 0) (Xa) d(j o 0) (Xa), j o 0 > +
(j o 0), 4, > + 2e (4')
.
0 : Sm-1 - Sn-1 is defined by harmonic homogeneous polynomials of common degree k, then 0 is harmonic and has constant energy density k(k + m - 2)/2. Corollary 1.2.2
Proof
If
Corollary 1.1.6 and Lemma 1.2.1 give the result.
E]
zk, z F C, k = 1,2, ... is defined Example 1.2.3 The map from C to ]; , z by harmonic homogeneous polynomials of degree k, and iestricts to a map G k S 1 -- S 1 harmonic of degree k.
Example 1.2.4 Let f : JR P x JR q - JR n be an orthogonal multiplication; i. e. f is bilinear and I f(x, y) I = I x I I y I , for all x c-7 1R p and all y c- )Et q. Then the Sn-1 which is a totally geodesic restriction of f produces a map tb : SP-1 x Sq-1 10
embedding in each variable separately, and hence is harmonic of constant energy density (p + q - 2) /2. If now p = q = n ; the Hopf map H n : s 2n-1
Hn(x,y) = ( Ix12 -
Iy1
2
,
2f(x,y))
-Sn
,
is a harmonic polynomial map with constant energy density 2n. In particular, if f is multiplication of complex, quaternionic or Cayley numbers, then we obtain the Hopf fibrations S3 examples see [33 ].
S2,
S 7- S4 or S 15 - S8 respectively. For other
Joins of spheres and Smith's construction Then Sp-1 * Sq-1 of the two Euclidean spheres Sp-1 and Sq-1 is the sphere Sm-1, m = p + q, obtained by writing each point z e Sm-1 as z = (cossx, sins y), where x Sp-1, y Sq-1 and s c- [0,TT/2 ]. We can think of the join as Sm-1 introducing "polar coordinates" on the sphere If g1 Sp-1- Sr-1 and g2: Sq-1_ Ss-1 are two maps; we can form the 1.3
join
gl*g2:
Sp+q-ly Sr+s-1 of g1 and
g2:
g1*g2(cossx, sinsy) = (toss g1(x), sins g2(y)) . Consider the case when g 1 and g2 are defined by harmonic polynomials with 2e(g1) = a 1 and 2 e (g 2) = a 2 say, where a 1 and a2 are constants. Smith's idea was to allow a "reparametrization "of s in equation (1.3.1) , by defining 0 : Sp+q-1-- Sr+s-1 as
0(coss x, sins y) = (cos a(s) g1(x), sina(s)g2(y)) ,
(1.3.2)
for some a: [ 0, TT/2 ] - [0,11/2] with a (0) = 0 and a (TT/2 )= TI/2. We will derive this map again later in Example 5.3. 1. The problem of 0 being harmonic then reduces to solving a second order non-linear ordinary differential equation in a . To see this we calculate A 0 , but in order to simplify later calculations we proceed differently, and by a slightly longer route to Smith [36 ), and for this we must introduce briefly the notion of harmonic morphism (much more will be said about these maps later in Chapter 7). A map 0: M --- N is a harmonic morphism if it pulls back germs of harmonic 11
functions to germs of harmonic functions, i.e. if f: V --- R is harmonic, V a domai in N, then f o 0 is harmonic on 0 -1 (V) in M. Consequently, if 0 is a non-constant harmonic morphism, then dim M > dim N, and if th : N -- P is a harmonic map, then so is zb o 0 : M - P. For x - M, let j x = ker dO x c T x M, and let J x be the perpendicular complement of '71 in T M with respect to g; say that 0 is horizontally conformal if x x
A
(x)2 g(X,Y) = 0 * h(X,Y) ,
for all X,Y s / X, where A : M -- R is some function (not necessarily smooth). Call A the dilation of 0 .
Theorem 1.3.1 [19,27 ]: A map 0: M - N is a harmonic morphism if and only if 0 is harmonic and horizontally conformal.
Corollary 1.3.2 If 0: M - N is a harmonic morphism and zb : N - P a harmonic map with energy density e 0b), then zb o 0 is harmonic and has energy density given by
e(ti o 0) (x) = A (x)2 e (zh) (0(x))
,
(1.3.3)
for all x E M. Proof The result follows from Theorem 1.3. 1 and the equation (1. 1.3) .
Consider the join Sm-1 = Sp-1 * Sq-1 , p + q = m. Regard s as a function
s: Sm-1-- R, by letting s(coss0x, sins 0y)) = s0, for all s0 C [0,11/2 ]. Let be the subset of S m-1 given by V, = S-10), then V 1 is isometric to SP-1. Similarly, let V2 = s (11/2) ; V2 is isometric to Sq-1. Define 1T1:Sm-1\V2 V
1
-- Vi
by rI 1 and
((cossx, sins y)) = x
T12 : Sm-1 \V1 - V2 by
172 ((cossx, sins y)) = y. Lemma 1.3.3
The maps T111T2 are harmonic morphisms, with dilations
A
1' A 2
respectively given by A1(z)2 = 1/cos2s, A2(z)2 = 1/sin2s for all z =(coss x, sins y). 12
proof For each x 0 E V1, the horizontal space with respect to 1l through (toss 0x 0, sins 0y 0) , s0 - [ 0, TT/2) , y0 r Sq-1 , consists of the tangents to all curves F (u) = (coss0 y(u), sins0y0), where y (u) is a curve in SP-1 withy (0) = x0.
Then R1 (coss0 y(u) , sins 0y0 )= y(u), so that, if I(u) =(coss0 y (u), sins 0y0) is another such horizontal curve, then TT
I
*h(r'(0), 1'' (0)) = h(y' (0),y' (0)) = g(I'(0),I''(0))/cos2s0
where h is the metric on SP-1 and g the metric on
Sm-1
Hence T11 is horizon-
2
tally conformal with X 1 = 1/cos 2 s.
To see that 171 is harmonic, we compute trace V d 1111 with respect to a particular orthonormal basis. Choose z 0 = (toss 0x 0, sins 0y0) C S m-1 \V2 and curves through z 0: T, (u)
= (coss0 yi (u), sins0y0)
T(u) = (coss0x0' sins0 ar (u)) where yi(u) is geodesic in SP-1 with yi (0) =x0, and cr(u) is geodesic in Sq-1
with cr(0) =y0. LetXi = F.' (0) and Yr = r' (0). Define v(s) =(cossx0,sins y0), so that f = v' (s 0)
Then 0
A rrl = trace V d 111
= EVdTTl(Xi,Xi) + ZVd111(Yr,Yr) +vdITl(E,E). Since yi(u) and c, (u) are both geodesics in Sp-1, Sq-1 respectively; both SmvX i X. and V Y i
1
r
Yr are proportional to t at z 0. Thus vSdT1 (Xi) d Tr1(X
V d TTI (Xi ,Xi) =
(1/cos2s0)V X -1 Xi i 0.
Also V dR1(Y,Y) = 0 since dTTI(Y) = 0, and V d TT1(t,t) = 0 since vt t = 0 and d TTl (t) = 0. Hence 111 is harmonic.
11 13
We now rewrite Smiths map 0:
Sm-1- Sn-1 of equation (1.3.2) as
(1.3.4)
0(z) = (cos a(s(z)) g1oTT1(z), sin a(s(z))g2o TT2(z))
for all z C Sm-1 \(V1 U V2); if z r V1 we define O(z) _ (g1(z),0), and if zE V2 we define 0(z) _ (0,g2(z)) Let t: Sn-1-- E be the function given by t((cost0v, sint0w)) = t0, where v C Sr-1 , w C Ss-1 and t0 C [0,1T/2 ]. Writing n is the inclusion, it is straightforward to compute fi = i o 0 , where i : Sn-1 and one obtains, using Lemma 1.3.3 and Lemma 1.2.1 on the maps g1 o A TT1 and g2oT12,
-R
4. (z)
- a'(s)2 ids I2 t(z) + (a"(s)Ids12 + a'(s),6s)VtO(z) -'-cos a(s)A(g1 o TT1)(z) + sina(s)A(g2 o TT2) (z) = - a,(s)2 Ids12 t (z) +(a"(s)IdsI2 + a'(s)AS)V
tO(z) a2(g20TT2)(z)
a1(g1o TT1)(z)
- cos a(s)
- sin a(s)
2
cos s
This lies in the plane spanned by V tO
(z)
and fi (z) ; hence 0 is harmonic if and
only if < i4 (z), V t6(z) > = 0, for all z e Sm-1 . TI
2
sin s
Now V
t9(z) _
cos a(s(z))g2 o TT2 (z)) , hence 0 is harmonic if and
only if a21
a"(s) Idsl2 + a'(s)a s + sin a(s) cos a(s)
-
1= 0, (1.3.5) sin s/ 2
with a(O) = 0 and a(11/2) = T1/2 .
Lemma 1.3.4: The Laplacian 0 s of s is given by
A s = (q - 1) cot s - (p - 1) tan s. Proof Let s0 C (0,11/2), then Ms = s-1 (s0) is a hypersurface of SM-1. 0 Define i s : M s0- S m-1 to be the inclusion map. Then 0
0 = 0(sois0 )=ds(dis ) +trace Vds(dis ,dis 0
14
0
0
= 2/2s is affine geodesic, hence °E f = 0, so that V ds ° ds() d s (E) = 0 (d s (f) is the unit tangent vector along E). That is trace °ds( di s , di s) _ As. Thus But
ds(V f)
0
0
o = ds(pis ) + As 0
But since A is is proportional to t , we conclude that 0
= - As t
Ais 0
Thus 6 s is - (m - 2) x (mean curvature of Ms
) .
0
Choose z r Ms0 ; z = (cos s0 x, sins0 y), x c- SP-1, y c-7 9q-1 . Let r(u) _ (coss0y(u), sins0y) be a curve in Ms0, r(0) = z , then x = r'(0) = (coss0y'(0),0). Now tr(u) = ( - sins0 y(u), coss0 Y). So d
°X = du r(u) I u=o _ (- sins0 -/, (0),0) tans0 X . Thus X is a principal curvature vector of Mso in Sm-1 , with principal curvature tan s0. Similarly, vectors tangent to curves 2; (u) = (coss0 x, sins0 c(u)) are principal curvature vectors with principal curvature - cot s0. The result follows by summing over the principal curvatures. Equation (1.3.5) now becomes
a"(s)IdsI2 + o'(s)((q- 1)cots-(p- 1) tans) + sin a(s) cos a(s)
0 .
(1.3.6)
The equation has singularities at s = 0, 1T/2 (that is the coefficients of as(s) and sin a (s) cos a(s) become infinite). We can remove these by reparametrizing (1.3.6) using the parameter u defined by e u = tan s. Then we obtain
15
a "(u) +
1
u
e +e +
u
-u
((q-2)e-u-(p-2)eu) a'(u)
-u
sin a (u) cos a(u) (a eu-a2 a-u) = 0 1
1
e +e
,
(1.3.7)
with a (- 00) = 0 and a(oo) = TI /2. This is the equation obtained by Smith in [36 1. 1.4
Outline of the solution of Smith's equation
Equation (1.3.7) is the equation of a pendulum with variable damping acted upon by a force of variable gravity. The position of the pendulum is described by the
angle 0 = 2a which is measured from the upward vertical. The idea is to find an exceptional trajectory of the pendulum, so it stands vertically upwards at time u = - co, and hangs straight down at time u = + cc
.
gravity
gravity
Smith shows that such a solution exists if either (i)
p = q and al = a2 , or
(ii) (p - 2)2 < 4a1 and (q - 2)2 < 4a2
We call (ii) the damping conditions. An outline of his construction is as follows. Fix u0 to be the time when gravity vanishes and manipulate the initial conditions aO = a(uO) and a, = a'(u0). For a given aO -(0, TI/2), throw the pendulum just hard enough (a' = a 0 ( a0)) so that a(u) -- 0 as u - - oo . Similarly, choose a0 +( a0) to get a(u) - TI/2 as u - + co. Then, since the coefficients in equation (1.3.7) are smooth; a0 and o., + are continuous in aO [9 j. Further, ,Y-0 -- 0 as aO - 0 and a U - 0 as aO - 17/2. The idea is to find an aO such that a+ and x- match. At opposite ends one requires c i+ bounded away from 0 for aO near 0, and a- be bounded away from 0 for aO near T1/2. This requires the 16
use of a comparison theorem for second order equations, together with the inequalities (ii) above. Smith gives a more sophisticated method of solving equation (1.3.7) in [36 ]. We give this method in Chapter 6. In [36 ] Smith gives many examples of harmonic maps constructed by the above methods. We give a few of these. SP-1 , and g2:S1 - Sl Example 1.4.1 Let g1 be the identity map Ip-1 Sp-1--be the map Gk of Example 1.2.3 of degree k. Then for p < 7 the damping conditions are satisfied, and so by Smith's construction one can construct a harmonic map
Sp-1
--
, p < 7, of degree k, k = 1, 2, ... (in general deg(g1* g2) deg(g1)deg(g2), g1: Sp-1_.. Sr-1, g2: Sq-1 Ss-1). By composing with an isometry of degree -1 we obtain harmonic maps of negative degrees. That is, we can represent Tin (S n) = Z harmonically for n < 7. 0:
Sp-1
Example 1.4.2 Let g1: S3 - S 2 be the Hopf map H2 of Example 1.2.4. Let g2 be the identity map I P-1 : S p-1 - S p-1. Then for p < 5 the damping conditions are satisfied, and we can represent the non-trivial class of TTn+1 (Sn) = 72 harmonically for n=3,4, ... , 8. Example 1.4.3 Let g1 : S 7 _ S 4 be the Hopf map H 4 of Example 1.2.4. Let g2 : S p-1 -- S p-1 be the identity map I p-1. Then g1 * g2 : S p+7 S P+4 represents the generator of 7 p; 7 (S p+4) = For p = 1, .. , 6 the damping conditions Z24. are satisfied, and we can represent that generator harmonically.
Example 1.4.4 Given an orthogonal multiplication r6 : Sp-1 Example 1.2.4 ; we can define a map 0 : S p+q-1 _ S n by
0 (coss x, sins y) = (cos a(s), sin a(s) rb (x,y))
x Sq-1 -
Sn-1
as in
,
where x r Sp-1, y -7Sq-1, s E [0,11/2 ], a(0) = 0 and a(J1/2) = TT . As in Section 1.3 we obtain an equation as a condition of harmonicity similar to equation 1.3.7, but with the gravity always having the same sign. The equation is of a pendulum whose position is described by i = 2 a, and one looks for a trajectory with the pendulum just completing a single rotation.
17
gravity
Similar considerations apply when th : S 1 x S 1 - S 1 is given by th (u,v) = uk vl, U. v r C,
lu 1 2 =
Iv 2
= 1. Then the relevant equation is (after reparametri-
zation)
a"(u) = (k2eu + 12e-u) e
u
+e
sin cv(u) cos cv(u)
(1.4.2)
-u
with a(-co) = 0 and a(uo) = TT. This is the equation of a pendulum with no damping
and variable gravity. If k = 1, then, due to symmetry considerations one can show
that equation 1.4.2 has a solution. However, if k # 1, then the gravity is either increasing or decreasing. If the gravity is increasing the pendulum will be unable to make a complete rotation. If the gravity is decreasing, the pendulum will go past the upward vertical after performing a complete rotation. Thus there is no solution of equation 1.4.2 with a (-oc) = 0 and a (co) = TT(this can be proved precisely using energy estimates for the pendulum). Now TT3 (S2) = Z can be parametrized by the integer k 1, which is the Hopf linking number - the Hopf invariant of each class. Thus we can represent harmonically the classes of TT3 (S2) with Hopf invariant k2 , but we cannot represent harmonically by the methods of Sections 1.3 and 1.4 the 2
classes of T7 3(S ) with Hopf invariant kl, k
In particular it is unknown whether the class of TT3(S2) with Hopf invariant 2 has a harmonic representative. An alternative method of finding harmonic representatives for the classes of T13(S2) is as follows. The Hopf map H2 : S3 S2 is a harmonic morphism, and hence by Corollary 1.3.2 , if f: S 2 S 2 is harmonic ; so is f o H 2: S 3- S 2. In particular, if f has degree k , then f o H2 has Hopf invariant k2 [26 ). Since we have harmonic maps f : S2 - - S2 of all degrees; we can find harmonic maps with Hopf
--
invariant k2 for all k. 18
1.
Similarly there are harmonic maps f: S4- S4 of degree k, for all k (from Example 1.4.1) ; by composing with the Hopf map H 4 : S 7 - S 4 , we can represent harmonically the classes of '17(S4) = Z ® $12 which have Hopf invariant k2 (the Hopf invariant parametrizes the Z factor of Z ® Z 12) . 1.5
Hyperbolic space
Let M in denote IR in equipped with the metric < < u, v >
M given by
u1vl + u2v2 + ... + um vm
M
for all u = (uI, ... , um) , v = (v1, ... , Vm) r IRm. The space Mm is called m-dimensional Minkowski space. Define (m-1)-dimensional hyperbolic space, or
the (m-1)-dimensional pseudo-sphere, to be the space H
m-1
={U
Mm
< u,u >M = -1)
;
,
together with the induced metric. HM-1, The pseudo-sphere is a "spacelike"hypersurface (if v is tangent to then < v, v >M > 0), whose unique normal vector field r is "timelike "(< n , n >M < 0) in the terminology of relativity theory. There are many analogies between the Euclidean sphere Sm-1 and the pseudo-sphere Hm-1, especially concerning harmonic maps. In particular, the usual stereographic projection sending the sphere less a
point to Euclidean space, can be modified to give the well-known isometry between M-1 m-1
and (B m-1, < , >) , where B m-1 is the open ball of radius 2 in R M-1 dx. ® dxi/ (1 - 4 £ x 2 2 , where (x.) are the and the metric < , > = E H
standard coordinates on Euclidean space. Let is H m-1
Lemma 1.5.1 H
M-1
where r/c^r
A M
m
M in be the inclusion map. If
f: Mm-- IR is a smooth function, then m Mm
(foi) _
.2 f+
c
+(m-1)
f2
2r
Ff
)oi
denotes differentiation in the normal direction to 2 `2
ex
2 +
8x2
+
,,.
+
(1.5.1)
,
C71
8
2
Hm-1
and
is the indefinite Laplacian.
rx m 19
Remark 1.5.2 The unit timelike normal to the pseudo-sphere at x is x itself. Thus, ^c (x) = x1Fx1() + x2
^c
.) r
(x) + ...
Fx2
+x
m m Fxm
(x)
where (x1. ... , xm) are standard coordinates on JR m Proof (of Lemma 1.5.1) : Recall equation 1.2.1 for the 2nd fundamental form of the composition :)f two maps:
vd(f o i) = df(Vdi) + Vdf(di,di)
-
this makes sense with M m having an indefinite metric (V is the Levi-Civita connection with respect to the indefinite metric on the bundle T * H m-l i- l T M m H m-l) . By taking the trace of this formula with respect to the Riemannian metric on H m- 1, we get A
H m-l
(foi) = df(A
H m-1
+ trace Vdf(di,di).
i)
f
But, (A
Mm
f) o i = trace V df(di,di) -` m
=-
2
1
2
22_
+
2
+
,,,
+
x2
(1.5.3,
F2
is the indefinite Lapla-
Fx2
m
f)oi = A HM-1 (foi) -
2 oi 8 2
`
Now
Hm-1 i
trace V d i m-1 Mm Vdi(X ) di (Xk) 1 k
k 20
oi
Equations (1.5.2) and (1.5.3) give
cian on M m. Mm
2
2 Fx
A
2
Fr2
where A M
(A
(1.5.2)
dfCHm
i
.
(1.5.4)
where we evaluate at a point x ,
Hm-1
with Xk = yk (0), yk (0) = x, yk(u) geo-
desic in Hk = 1, ... , m-1. Thus Thus
yk(0) = 8/aij
pH m-i
pH m-1
i=
11
yk (0). But yk (0) _
0
i =(m -1)
If f is a pseudo-harmonic polynomial (i.e. 6M Corollary 1.5.3 degree k on R m, then pHm-1
so foi
m
f = 0) of
(foi) = k(k+m-2) foi;
is an eigenfunction of
AHm-1
.
Let (M,g) be a Riemannian manifold and 0: M -be the inclusion map, and write 4 = i o 0 . Lemma 1.5.4
Hm-1
a map.
Leti:Hm-1y
Mm
The map 0 is harmonic if and only if
p = 2 e (fi) D
,
where a (1) = E M' (Xk) is an orthonormal basis for M. k Proof From equation 1.1.3
p(io0) = di(p0) + trace Vdi(d¢,d0). As in Lemma 1.2.1, V di is perpendicular to di with respect to the Minkowski metric. Thus 0 is harmonic if and only if A (4)) is proportional to t. That M -2e(4+) is as in Lemma 1.2. 1. Express each w = (x, y) C M2 as w = x + j y. Multiply w =x +j y and w' = x' + j y' using the rule w. w' = x x' + y y' + j (yx' + x y'). Then the map: M2 M2: w k is pseudo-harmonic, and induces a harmonic map 0 from H1 to H1 with constant energy density 1 k2 . If we express H1 as Example 1. 5.5
-w
H1 = {(coshs, sinhs); s C [0,-4} C
M2
then 0 is simply 0 ((coshs, sink s)).= (cosh ks, sinh ks). Example 1.5.6 Introduce polar coordinates on M2 by expressing each (x,y) E M2 2 .,2 as (x,y) = (p cosh s, o sinh s). The Laplacian AM2 = - ?2 + ° 2 goes over into ay ax 21
i2 p2
2
_
t2
c^2
2P2
_
1
P
P
FO
1
(illustrating Lemma 1.5.1) . Thus A H
is simply
2 P
This me an s that mapp ing s of the fo rm I, :
2
Pt
H1
S1 ,
r6 (cosh s , sinh s) =
(cos a (s), sin a (s)) are harmonic if and only if o is linear; or(s) = k s; such a map has constant energy density k2/ 2.
Polar coordinates on hyperbolic space and an analogous construction to Smith's Introduce "polar coordinates"on Hm-1 by expressing each point z (-- Hm-1 in the form z = (cosh sx, sinh s y ) , where x r Hp-1, y c Sq-1, p +q = m and 1.6
s c [0, oo)
Define a map 0: H p+q-1 y H r+s-1 by
0 ( (cosh s x, sinh s y) = (cosh a(s) g1(x), sinh a (s)g2(y)) , a(0) = 0 ,
where g1: HP-1 - H r-1 is harmonic with I dg 1 12 = a1 a constant, and g2: Sq-1 Ss-1 is harmonic with Idg2 12 = a2 a constant. By viewing s as a function, s : Hm-1 - B, s ((cosh sox, sinh s0y)) = s0, we can express 0 in the form
0(z) =(cosh a(s(z))g1oTY1(z), sinha(s(z))g20T12(z)),a(0)=0, Hm-1- Hp-1 is the harmonic morphism: (cosh s x, sinh s y) - x, and TT2: Hm-1 \ Hp-1 - Sq-1 is the harmonic morphism: for all z E Hm-1, where T11:
(cosh s x, sinh s y) -- y, s
0. The dilation A
1
of TT1 is given by A 12 = 1/cosh2s,
and the dilation A2 of TT2 is given by X22 = 1/sinh2 s (the proof of these facts is
similar to the proof of Lemma 1.3.3) . We compute the Laplacians: 6 (cosh a(s) g1 o T7 1) = g1 o TT1 (cosh a (s) a'(s) 2 Idsl2 + sinh a(s) A(a(s) )
+ cosh a(s) (a1/cosh2s)) A - (sinh a(s) g2 o TT2
,
= g20 T72(sinh a(s) a'(s)2 1 ds 12c o s h
- sinh a (s) (a2/sinh2 s))
.
Let t : Ha-1 -- B , n = r + s, be the function given by W (cosh cosh s0 v, sinh sow) = s0 Mn is the v F Hr-1, w r7 Ss-1. Then, writing cD = i o 0 where is Hn-1 22
for inclusion, we see that A 4 (z) lies in the plane spanned by 4 (z) and V t (z) m-1 . By Lemma 1.5.4, 0 is harmonic if and only if <6 if W, all z E H
Vt4, (z)>
M=O. Now vtO(z)=(sinha(s)g1on1(z),cosh a(s)g2o1T2(z)),
and remembering that I gl o T71 12 = -1, I g2 o TT2 12 = 1; the condition for harmonicity of 0 becomes
(,y (s)) - sinha(s)cosh a(s) (
al
2
=0
+
sinh2 s
cosh2s
or,
a"(s)Ids12 + a'(s)6 s - sinh a(s) cosha(s) (
a
2
+
`sinh2 s
a
1
cosh2 s
(1.6.1)
= 0, with a(0) = 0 Lemma 1.6.1
The Laplacian of s, 6 s, is given by
As = (p - 1) tanks + (q - 1) coth s. Proof
As for the spherical case, d s (z) is - (m - 2) x (mean curvature of
= s -1 (s0)) for each s0 a (0, Cr) , z C M s0 . A calculation similar to that of Lemma 1.3.4 shows that the principal curvatures of M s0 are - tanh s0 with multiplicity (p - 1), and - coth s0 with multiplicity (q - 1). Equation (1.6.1) now becomes M s0
a"(s) +((p-1) tanh s + (q - 1) coths) a' (s)
- sinh a(s) cosh a(s)
/
2
sinh2 s
with a(0) = 0 .
cosh1
+
2
s/
0
,
(1.6.2)
Equation (1.6.2) has a singularity at s = 0, we therefore reparametrize it using the substitution eu = sinh s. The equation then becomes
23
all(u)
1
+
e +e u
(peu
-u
_
+ (q-2)(eu + e-u)) a' (u)
sinh a(u)cosh a(u)
eu +e -u
1.7
a
a2 e
u
+
u1
e +e
u
a(-oo)
=0.
0
(1.6.3)
Solving the equation for hyperbolic spaces.
We provide an outline only of the solution of equation (1.6.3), giving a precise
derivation in Chapter 6.
Equation (1.6.3) can be thought of, in some sense, as the equation of a particle constrained to move on a hyperbola in M2 with a damping force, and with a variable "gravity" acting upon it.
gravity
The qualitative nature of the gravity and damping are illustrated by the following graphs: X2
gravity (q - 2) U = 00
u=-0
p u
Consider the situation when u is close to -cc, and so a(u) is close to 0. In this limit sinh 2 a (u) 2 a (u) - sin 2 a (u). Thus, for large negative time, the qualitative behaviour of the solution of equation (1.6.3) approximates the qualitative behaviour of the solution to Smith's equation. We choose a time u0 such that the gravity is greater than 0 at u0, and so is greater than 0 for all u < u0 . We choose a0 = a(u0) sufficiently close to 0 - note 24
that we have a 1-parameter choice for a0. The physical moden can now be approximated by a pendulum, and we pick a a (a0) = a' (u0) to be the velocity such that the pendulum just reaches 0 as u ---co in backward time (equations of the form p" = G (u, R, ( 3 ' ), where G, l G/i,/3, 2 G/8Q ' are continuous, have unique solutions through each point (u0, 001f0 ), [9 ]). In Chapter 6 we will demonstrate precisely the existence of non-trivial solutions of equation (1.6.3) which exist for all time. We have of course only established the existence of a non-trivial solution, and have not investigated the behaviour as u - oo . We remark that there is a 1-parameter family of solutions depending on the choice of a0 . Remark 1.7.1
We can write equation (1.3.5) for the spherical case in the form
A(a(s)) = g(s) sin2 a(s) ,
(1.7.1)
where s: Sm-1 - R and g(s) = a ((a2 /sin2 s) - (a1/cos 2 s) ). Similarly, we can write equation (1.6. 1) in the form
0(a(s)) = g(s) sinh 2 a(s) ,
(1.7.2)
where s: Hm-1 __ R and g(s) = z ((a2/sinh2 s)
(a1/cosh2 s) ). We remark that equation (1.7.1) bears a resemblance to the well-known sine -Gordon equation, and equation (1.7.2) resembles the sinh -Gordon equation [ 29]. +
Remark 1.7.2 The method of this chapter for constructing harmonic maps 0 between spheres and between hyperbolic spaces have a common feature, namely symmetry
with respect to certain functions s : M -E and t : N -- I R, where 0: M - N. That is the diagram 0
s
t
a
commutes, where s(M) = Is and t(N) = It are appropriate intervals. We might ask which functions are suitable as a symmetry of 0 , in order to reduce the problem 25
of harmonicity to solving a 2nd order ordinary differential equation. There is a
class of functions which adapts quite beautifully to our purpose. These shall be the subject of the next chapter.
2R
2 Isoparametric functions
2.1
Definition of isoparametric function
Let (M,g) be a space form (i.e. either the Euclidean sphere Sm, Euclidean space R m or hyperbolic space H m A smooth function f : M --- R is called isoparametric if
(2.1.1)
ri 1 (f(x))
Idf(x) 12
=
Af(x)
= th2(f(x))
(2.1.2)
,
for some smooth functions 0 1, 0 2 : R -R. Such functions were introduced by Cartan [5 ] in 1938. Their description on Euclidean space and hyperbolic space is relatively trivial, but on the sphere they are rich in geometry. More recently isoparametric functions have been studied in [ 17, 30, 32, 33, 39, 40 ].
Lemma 2.1.1 Let f: M - R be a function satisfying equation (2.1.1) , then the integral curves of V f are geodesics. " (Y"
= Vf/ Idf I , and let X c '(TM) be perpendicular to : g(X,f) = 0. Then X f = 0 so f (X f) = 0. Also t f = d f (V f) / Id f I = rV 1(f) is a function of f so Proof [32 ]
Let
X(ff) =0, hence
0 = [X,t] f (VXf - Vf X) f.
Thus g(VXf - VEX, f) = 0.
Since 1=g(f,t): g(Vf,f) = 0, so that g(VtX,i) = 0. Also g(X,f) = 0 implies g(VfX,f) = - g (X, V), so that Vf is proportional to f . But 1 = g (f ,f) implies g Q , Vf f) = 0. Thus V
= 0.
27
Let f: M - R satisfy equations (2.1.1) and (2.1.2) ; then if
Lemma 2.1.2
Mc = f-1(c) is a non-singular hypersurface of f; Mc has constant curvature. Proof Let i : M - M be the inclusion map. Then c
0 = A(foi) df(A i) + trace V df(di, di) I d f I (mean curvature of M
+ Af -
= Vf/ Idf I (df is non-zero on Mc since Mc is a hypersurface). Now, using Lemma 2.1.1 , where
vdf( ,0 = df(v
Vd (
df(i;) )
= (Vf) IdfI . Thus the mean curvature is equal to _ v f( Idf1)
- of
(2.1.3)
Idf I
which is a function of f.
Thus if f satisfies equations (2.1.1) and (2.1.2) ; the level hypersurfaces of f form a parallel family of hypersurfaces of constant mean curvature. Conversely, from equation (2.1.3) and by reversing the proof of Lemma 2.1.1, it is not hard to show that given such a family which are the level hypersurfaces of a function f: M -- E, then f satisfies equations (2. 1. 1) and 2.1.2). Furthermore, one can show
Proposition 2.1.3 [51 Given a parallel family of hypersurfaces of constant mean curvature on M, then the principal curvatures on each level hypersurface are constant on that hypersurface . We call a hypersurface of M with constant principal curvatures an isoparametric hypersurface. Example 2.1.4 Let M = R m and let f: R m - R be defined b y f (x1, ... ' ' m ) = x12 + ... + xp , p < m. Then Idf 12 = 4 f and Of = 2 p. The level hypersurface 28
f = s2 is given by x2 + ... Rm-P , s0 r- (O,ao ).
Example 2.1.5 Ix 12 -
xp2
= s2 and is isometric to the cylinder s0.Sp-1 x
Let M = Sand define F : R m --- R byF((x,y))
Iy12, where we write Em = Rp ® Rq, p+q=m, xERPand yERq
- R be the restriction FISm-1. Introduce polar coordinates on Sm-1 as in Sectionl.3 ; so that each z E Sm-1 is expressed as z = (toss x,sinsy), x r Sq-1, y E Sq-1. Then f(z) = costs - sings = costs, so that df = - 2 sing sds. Let f:
Sm-1
Thus Id f 12 = 4 sin2 2 s =4 (1-f2) .
Now Af =f" Ids 12
+
f' As, but from Lemma 1.3.4 As=(q-1) cots -(p-1)tans,
whence
Af = 4((p-2)cos2 s-(q-2)sin2 s) = 2((p-2)(l+f) - (q-2)(1-f))
,
and equations (2. 1. 1) and (2. 1.2) are both satisfied.
The level hypersurface M. = f-1 (cos2 s0), s0 E (0,1T/2) is isometric to a product of spheres coss0SP-1 x sins0 Sq-1 . Example 2.1.6 Let M = H m-1 . Introduce polar coordinates on H m-1 by expressing each point z E Hm-1 as z = (coshs x, sinhs y), x e HP-1 , y C Sq-1, m = p +q and s e [0,oo). Define f: Hni-1 R by f(z) =cosh2 s. Then df = 2sinh2s ds so that Idf 12 =4 (f2 - 1). Now A f = f"(s) Ids 12 + f '(s)n s, and from Lemma 1.6.1 6 s =(p-1) tanhs +(q-1) coths, whence
A f = 4((P-2) cosh2 s + (q-2) sinh2 S) = 2 ((P-2)(f-1)
+ (q-2) (f +1))
,
and f is isoparametric. The level hypersurface M. = f-1 (cosh2 s0) , s0 E(0,eo), is isometric to the pro0 duct cosh s0 HP-1 x sinhs Sq-1 . 0 2.2 Properties of isoparametric functions and Miinzner's classification theorem R. For Let Mc be a level hypersurface of the isoparametric function f: M
let A be the shape operator; A:X Tc M -- T M: X -- - VX x c
X E TxMc where t is the unit normal vector field to Mc. Then the 2nd fundamental form x E Mc,
.
,
29
,
h of Mc,
W
), is defined by g(AX,Y) = h(X,Y) for all X,Y ETXMc -X The eigenvalues of A are the principal curvatures , A W, ... , A p (x) say; let S1 (x) , ... , Sp (x) denote the corresponding eigenspaces. c-
T
c
The equation of Codazzi [381 is
g(R(X,Y) Z, ) = (VXh) (Y,Z) - (V h) (X,Z) ,
(2.2.1)
for all X, Y, Z ECt(T M c ), where R is the curvature tensor of M, and E ''(TM) is the unit normal vector field to M . Since M has constant curvature K say, c
R(X,Y) Z = K(g(Y,Z)X - g(X,Z) Y, is tangent to Mc, and equation (2.2.1, becomes
(Vxh)(Y,Z) = (VYh) (X,Z) ,
(2.2.2)
for all X,Y,Z E'(TMc). Let X C S. and Y,Z E S., then, since the A. are constant on )
]
1
M
c
(VXh)(Y,Z) = X(h(Y,Z)) - h(VXY,Z) - h(Y,V Z) X (Ajg(Y,Z)) - Ajg(V Y, Z) - Ajg(Y,V. Z) A. (V g) (Y,Z) l
= 0. Thus
0 = (VYh) (X, Z)
= Y(h(X,Z)) - h(VYX,Z) - h(X,vYZ) 0 - Ajg(VYX,Z) - Aig(X,VYZ) = (Aj - A i )
g(X, V Z) .
(2.2.3)
Thus, for all Y, Z E S .: IT (V Z) E Sj where II: TX M -- TX Mc is projection. Y
Proposition 2.2. 1
The distribution S. ,
j = 1, ... , p, is integrable,
and the
integral submanifolds are totally geodesic in M. Furthermore the leaves of S. are umbilical in M with mean curvature vector parallel to . So in the case Sm-1 . M = Sm-1 , the leaves are small spheres in >;
30
That the distribution is integrable and totally geodesic follows from equation
proof
(2.2.3) . Since, for Y,Z C S. J
VYZ = fl(VYZ) + ajg(Y,Z) E
;
the leaves of S. are umbilical in M with mean curvature parallel to
Suppose M is Sm-1 (the cases M = Em, Hm-1 are similar). Define y t : Mc -- M by
yt(x) = cost + sint E x , for all x r Mc For each i =1, 2 , ... , p, write X.(x) =cot 9. for some gi depending only on the level hyper-
surface Mc. Suppose XE S, i.e. X = p'(0), fl(O) = x, where p(s) is a curve in Mc. Then
d yei (X) = ds (yg o P(s)) I s = 0 1
coS01X + sing. ds ha(s) 1s=0
cos9iX + sin9i(Vf) X cos0.1 X + sing.1 (-cot 0.X) J Thus d yt (X) = 0 if and only if t = 0 . and X E Si (x), for some i = 1,2, Call y (M ) a focal variety of f . 8i
(2.2.4)
.
,
p.
c
Theorem 2.2.2 [321
Each focal variety is a minimal submanifold of
Sm-1
.
Remark 2.2.3 From equation (2.2.4) we see that dyt(X) is proportional to X for all X E Sj, j = 1, ... , p. Thus IE X E S. , where denotes Lie derivation on M, and since V = - A X ; we see that V X V t c- S , j = 1, ... , p, so the X X J t ) principal curvature distributions are parallel with respect to E. Proposition 2.2.4 [51
If M is R m or
If M is Sm-1 , there are at most two focal varieties HM-1
there is at most one.
Let V1,V2 denote the focal varieties (if there is only one, then let V2 = 0), and let TI: M \V2 V 1 be the projection map down the integral curves of t, then
-
31
Proposition 2.2.5 The map TI : M \V2 - V 1 is harmonic. If the number of distinct principal curvatures is less than or equal to two, then is a harmonic morphism (c.f. Lemma 1.3.3). TT
Proof Suppose dim M = m, and let x c M \V2. Choose a frame field X 1, ... , X m about x adapted to the S. spaces, in the sense that for each j, if dim S. = mj , a subset Xjl, ... , Xjm of X 1, ... , Xm forms an orthonormal basis for S. at each 1
point, and let Xm = . Furthermore, since the integral submanifolds of the S. distributions are totally geodesic in the level hypersurfaces of f, we can suppose for i = 1, ... , m. VX X i is proportional to 1 Now TI= yo for some j = 1, ... , m, so the horizontal space through x is 1
®
ij
S. (x), and the vertical space mar/ = S. (x) ® R t L. From equation (2.2.4) we see x
1
1
that 17 is horizontally conformal if and only if the number of distinct principal curva-
tures p is equal to 2. To see that TT is harmonic, we work out trace V d TI. If i = 1, ... , m - I then V d TT(Xi,Xi) = - d TT(VX X 1
i
)
+ VdTI(X.) d>I(Xi) 1
=0, Clearly v d TT( , ) = 0, from equation (2.2.4) and since VX Xi is proportional to i so that II is harmonic. On Rm , the number of distinct principal curvatures on each level hypersurface of f is 1, and on Hm-1 at most 2 [5 ]. Cartan classified all such families of isoparametric hypersurfaces. He also solved the classification problem for p = 3 on S m-1 [6 ] , finding that m could be only 2 , 5 , 8 ,14 or 26. Munzner [301 showed that every isoparametric hypersurface is algebraic, and only certain p are allowed; we state his remarkable theorem without proof.
Theorem 2.2.5 [30 ] Let M be Sm-1 and Mc an isoparametric hypersurface with p distinct principal curvatures al, ... , xp with multiplicities ml, ... mp respectively. Then (i) mi+2 = mi - i.e. there are at most 2 distinct multiplicities. (ii) Mc is the level surface of the restriction to Sm-1 of a homogeneous 32
polynomial F on F m satisfying IV F(x)12 = CF (x)
p2 lx
(2.2.5)
12p-2
dlxip-2
=
(2.2.6)
where d = p 2 (m2 - m1)/2 if p is even and 0 if p is odd. (iii) Conversely, any such F defines a family of isoparametric hypersurfaces on Sm-1 (iv) p can only be 1,2,3,4 or 6.
Remark 2.2.6 We will call p the degree of the isoparametric function, hypersurface or family of hypersurfaces. Remark 2.2.7 The function f = F ism_1 has image f(Sm-1) _ [-1,11, where F is as in Theorem 2.2.5 (ii). The sets f-1 (-1), f(1) correspond to the focal varieties.
Remark 2.2.8 If one puts f = cos p s, s E [0, it/p 1, then s represents the affine parameter such that E _ Vs. Lemma 2.2.9 The Laplacian 0 s(x) is minus the mean curvature of the level hypersurface M s0 = f -1 (cos p s0) , x e M s0 Proof
Let is0: M s0 -- M be the inclusion map. Then
0 = A(sois ) 0
= ds (&is0) + trace V ds(dis0, dis ). 0
= '/2s is affine geodesic, hence V = 0; so that V Vds() ds() = is the unit vector along R). That is But
trace Vds(dis0,dis0) = As.
ds(V E)
+
Thus
0 = ds(Aiso) + As. But since o iSO is proportional to
;
we conclude that A s.
= - A is0 . 33
Remark 2.2.10 With respect to the affine parameter s, whose level hypersurfaces we denote by Ms0 = s-1 (s0), the principal curvatures are X i = - cot(s0+ (i-1) 17/p), i = 1, ... , p, on Ms , and the corresponding integral small sphere of the distribu0 tion Si(x), x C Ms0 , has radius sin (s0 + (i-1)R/p). Hence we get Remark 2.2.11 The mean curvature varies from -o to + co on Sm-1 and is continuous, thus, for each family of hypersurfaces on Sm-1, there exists one hypersurface with zero mean curvature, i.e. minimal. In fact the minimal hypersurface
is MH/2p= f-1 (cos n/2) = f-1(0). 2.3
Examples of isoparametric functions
Example 2.3.1 The isoparametric families of hypersurfaces of degree 1 are all given by the restriction of linear functions: (i) On Rm; define f : R m - R by f(xl, ... , x m ) =X I' The level hypersurfaces are isometric to R m-1, and there is no focal variety.
(ii) On Sm-1; write z c- Sm-1 as z = (toss, sins x), x C Sm-2 and sE [0,11]. Define f: Sm-1 - R by f(z) = toss. The level hypersurfaces are (m-2) -dimensional small spheres. Each focal variety is a single point f -1(1) and f -1(-1).
(iii) on Hm-1 there are three distinct cases. m- 1 (a) For each z E H , write z = (coshs, sinhs y),y c S m-1 and s e [0 , oo) . Let f Hm-1 -- R be defined by f(z) = coshs. The level hypersurfaces are (m-1)dimensional spheres. There is one focal variety which is a point.
34
Let z E Hm-1 be written as z = (coshs x, sinhs), x e H m-2 , s E [0,_) . Define f: Hm-1 E by f(z) = sinhs. The level hypersurfaces are "equidistants" from the plane xm = 0 in Mm, and are (m-2)-dimensional hyperbolic spaces. There (b)
is no focal variety.
(c) Express H m-1 as the upper half plane in R m-1 with metric d s2
(dxi + ...
+
dxm_l2)/xm-12
=
, where x 1, ... xm_i are standard coordinates
- R by f(x 1, ...
xm_1) = xm-l' The level hypersurfaces are "horospheres 11 - that is, they are isometric to (m-2)- dimensional Euclidean spaces. There is no focal variety. on 1Rm-1 . Define f :
Hm-1
,
L Example 2.3.2
xm-1
The isoparametric families of hypersurfaces of degree 2 on R m,
Sm-1 and Hm-1 are all given in Examples 2.1.4, 2.1.5 and 2.1.6. In Example 2.1.4 there is one focal variety which is isometric to R m-p. In Example 2. 1.5 there are two focal varieties, one is isometric to SP-1 and the other is isometric to Sq-1. In Example 2.1.6 there is one focal variety isometric to Hp-1. Example 2.3.4 Define F: R3v+2- R, v= 1,2,3,4 or 8 by
F(u,v,X,Y,Z) = u3 - 3uv2 + 2 u(XX + YY- 2ZZ) + +
32 3
3Z3(XYZ+ZYX)
v(XX - YY)
(2.3.1)
where u, v c- I R, X, Y, Z C F, F is one of I R, C, quaternions or Cayley numbers, and is conjugation on F. Then f = F I 3 v + 1 is an isoparametric function of S
35
degree 3 (all such arise in this way [6 1). The focal varieties are antipodal, and are isometric to the projective plane P2 (F) , embedded as the standard Veronese minimal submanifold in the sphere. Example 2.3.5 [32 ] Identify (n+l )-dimensional complex space Cn+1 with R n+l ]R n+l, by writing a point z E Cn+1, z= (z1, ... , z n+1 ), z = x +Y ,)=1,...,n+1, j l l as z = x + iy = (x1, ... , xn+1) + i (y1, ... , yn+l) . We shall also write z as (x, y) whenever this is convenient. Let F : Cn+1 _ R be defined by
(2.3.2)
F(z) _ (Ix 12 - Iy12)2 + 4 <x,y >2 .
Then f = F I S2n+1 is isoparametric of degree 4 on S2n+1 Define is : S1 X s1,2 - S2n+1, where S n+1, 2 is the Stiefel manifold of orthonormal 2-frames in Rn+l, by i
s (e i0 , (x, y)) = e ig (toss x + i sins Y)
where s E [0,11/4 ]. Then is is an immersion which double covers the level hypersurface M S = f -1 (sin2 2 s). The hypersurface is obtained by the identification ( 0 , (x,y) ) - (0 + 11, (-x,-Y) ).
The principal curvatures of M. are -cots, -cot(s - 11/4) , -cot (s - 11/2), -cot(s - 311/4) with multiplicities n-1, 1, n-1,1 respectively. There are two focal varieties, one at s = 0, which corresponds to the set { e i0.x i = S1 XSn/SO, and one at s = 11/4, which corresponds to the set {ei0(x + iy)/22' It , i.e. the set{(x+iy)/22? _ Sn+122 . Let us study the Riemannian geometry of Ms in more detail. Define Sn+1 2 to be the analytic submanifold of Rn+le Rn+1 given by Ss
+1 2 =
{ (x,y) E Rn+le Rn+1; Ix 12 = cos2s, ly 12 = sings and <x,y> = 0} = {(toss x, sins y) C Ra+1® Rn+1 (x y) , E Sn+12 2 } . Let a1, ....en+1 be the orthonormal basis for Rn+l
such that ei = (0,... ,0,1,0,... ,0), with the 1 in the i'th place. Choose pESn+1,2 to be p = (cosse1,sinse2). Consider the following curves in Rn+1
Ai(u) =cos(u/sins) e2 + sin(u/sins) ei, i =3,...,n+1 36
,
(2.3.3)
and define the curves through p in Sts
1,2
yi(u) = (coss yi(u), sins e2) (u)
p(u)
= (cosset, sins Xi(u)), i = 3, ...
, n+1
= (coss(cos u e1 + sine e2) , sins (-sinuel + coss e2)) .
(2.3.4)
The tangent space to Sn+1,2 at p is given by
TpSn+1,2 = { (v,w)p; < v,w> = 0 = <w,y> and + <x,w> = 0 where p = (x,y))
,
and the vectors
y'(0) = (ei, 0)p x'i (0) = (0,ei)p
P'(0)
,
i = 3, ...
, n+1
= (coss e2, -sins el) ,
(2.3.5)
form an orthonormal basis at p for TpSn+1,2 Lemma 2.3.6
The curves (2.3.4) are all geodesic in S n+1,2 at p - that is,
E n+1 e )Rn+1 /.i'(0) )p is perpendicular to TpSn+1,2' where p (u) is one of (VQ,(0)
the curves (2.3.4). Proof A curve /3(u), (2(0) = p, in Sn+t,2 is geodesic at p if and only if p"(0) is perpendicular to TSn+1, p For example, p"(0) = (coss (-e 1 ), sins (- e 2) ). 2 Clearly the scalar product of p "(0) with the vectors of (2.3.5) is zero. Similarly for the other curves of (2.3.4). F-I We study M , the double cover of M s , which locally can be described as the set s of points ei6 (coss x + isins y), s E (O, 1T/4), (x, y) E Sn+1,2 and B E [0,211). Fix g , then S n+1, 2 is embedded in (rn+l as the manifold S n+1, 2 . Fix coss x + isins y, then as B varies, we trace out a great circle of S2n+1. How does this
circle intersect the Ssn+1, 2 37
Fix p = e
i00
(toss x0 + isins y0) E M s . Consider the two curves
y (u) = e
i9 (u)
6(u) = e 1
00
(toss x0 + isins y0)
(coss x(u) + isins y(u)) ,
(2.3.6)
both of which are contained in Ms, where x(0) = x0,y(0) = y0, 0(0) = 90 and (x(u), y(u)) E Sn+1 2 for all u. These are both curves through p, and without loss of generality 0 '(0) = 1. Then
y (0) = i9 =
(0) ai
00
(coss x0 + isins y0)
ei 00(-sins y0 + icoss x0)
and
6'(0) = ei00(coss x'(0) + isins y'(0))
.
The Riemannian scalar product
= i(cossx0 + isinsy0). (cossx'(0) + isinsy'(0)) = i(isins coss y0.x'(0) - icoss sins x 0' Y' (0)), since x0.x'(0) = y0.y'(0) = 0
= sins2sx0.y'(0), since x0.y'(0) + y0.x'(0) = 0 = cos(2s- TT/2)x0, y' (0) . Thus the angle of incidence of y and 6 is independent of 00, and is 2 s - TI/2. Furthermore, the plane of incidence at p is spanned by p'(0) and Y'(0), where p is as in (2.3.4). We therefore have The level hypersurfaces Ms, locally can be described as an "angular product"of S ' ! th Ssn+1,2 , where the angle of incidence is 2 s - TT/2, and the plane of incidence is spanned by the tangent vectors to the curve p(u) of (2.3.4) and the curve y (u) of (2.3.6).
Lemma 2.3.7
Example 2.3.8 [17) An n-tuple (P19 . . 38
.
, Pn ) of symmetric endomorphisms of
R21 is called a Clifford system if
P P, + P.P. = 2 S.. I
,
i,j = 1, ..., n.
(2.3.7)
Theorem 2.3.9 [ 17 1 Given a Clifford system (P1, ... , P) on R 21 ml = n and m2 = 1-n are both positive, then
such that
F(x) = IxI4 - 2E 2,xER2l i defines an isoparametric function f = F I S21_1 on of degree 4, with the multiplicities of the principal curvatures being (m1,m2). S21-1
Example 2.3.5 is a special case of this example. This family of examples also includes many where the level hypersurfaces of f are non-homogeneous (they are not the orbits of a subgroup of the orthogonal group) - including the non-homogeneous examples of Ozeki and Takeuchi [33 j (which we give explicitly in chapter 8). Furthermore these examples include ones where the focal varieties are also non-homogeneous.
2.4 Generalizing the notion of isoparametric families of hypersurfaces We wish to generalize the notion of isoparametric families of hypersurfaces to Riemannian manifolds whose curvature is not necessarily constant. There are various possible definitions, but we make one which is suitable for our purpose of constructing harmonic maps. Let (M,g) be a Riemannian manifold together with a family of hypersurfaces (Mc) cEI, where I is some indexing set, in the sense that there exists a closed nowhere dense subset K of M such that M\K is foliated by hypersurfaces Mc. Suppose that on M\K the following conditions are satisfied: (i) The locally defined unit normal vector field to the foliation t, satisfies 0 = 0. (ii) If M is a hypersurface in the foliation, then the principal curvatures c A l , ... , X are defined up to a sign. Suppose that for each i = 1, ... , p; A i is P constant on M c (iii) If Sk is the distribution on Mc corresponding to the principal curvature Ak, then projection down the integral curves of t , ps : Ms - Mc , preserves Sk in the sense that, if X E Sk (x), x E Ms, then ps * X E S k (ps (x) , provided that ps is a 39
diffeomorphism (i.e. we don't encounter a focal variety). Then we call the family of hypersurfaces a generalized family of isoparametric hypersurfaces. In certain circumstances, it may be that the family of hypersurfaces is no longer defined by a function.
Lemma 2.4.1 Let (Mc)ccI be a generalized family of isoparametric hypersurfaces on M which are the level sets of a function f : M -- E, then Idf 12 = rb IM
Af =
& 2(f)
for some functions r61 and ?b
2
.
Call such an f a generalized isoparametric function.
Proof The vector field 4 is defined, up to a sign, by 4 = Vf/Idf I. Let X E ker df, then, since g(X, 4) = 0 and o44 = 0; we conclude that g(Vf X, 4) = 0, and V4X C ker df. Also g(4,4) = 1 implies VI; E ker df, thus 0 = IX,4 If =X(i4f) 2) 4(Xf) = X(4f). But 4f = ( I d f l2)2 , df. Since the dimen1
sions are equal and finite ; we conclude that ker (d (Id f 12) 21 = ker d f , and that Idf 12 = r01(f) for some function tb1. Equation (2. 1.3) is still valid for non-constant curvature, giving Af = th2 (f) , for some function 02.
Remark 2.4.2 Since the curvature term in the Codazzi equation (2.2.1) is now no longer necessarily zero, it is not necessarily true that the Sk distribution on Mc is totally geodesic or integrable.
Example 2.4.3 All isoparametric functions f on Sm-1 define a generalized isoparametric family of hypersurfaces on R Pm-1. In the case when the degree of f is even; the family of isoparametric surfaces on R Pm-1 has two focal varieties. In the case when the degree of f is odd; there is just one focal variety.
Example 2.4.4 Express R 2m as R 2p ® R 2q where p + q = m, and identify R2p ® R2q with (Vp ®(rq in the obvious way. Consider F : R2my R, F(x,y) = Ix 12 - ly 12, x CR2p, y C R2p, yCR2q, whose restriction defines an isoparametric function on S 2m-1 with focal varieties S 2P-1 and S 2q-1. The function f factors through the action of Sl on S 2m-1, to give a generalized iso40
m-1 parametric function on (r P with focal varieties C Pp-1 and Q' Pq-1. The level hypersurfaces are diffeomorphic to S 2p-1 x S 2q-1 /Sl
Remark 2.4.5 The function f : S 2n+1 - R of Example 2.3.5 factors through the action of S1 on Stn+l to give a well-defined function g on C Pn, with focal varieties RPn and Sn+1 2 /S1. The level hypersurfaces are diffeomorphic to Sn+1 2/SO Then g is a generalized isoparametric function. Remark 2.4.6 Let 11 : Q P3 _ S4 be the Hopf map, and let f: S4 - R be the isoparametric function of degree 3 of Example 2.3.4. Define g: C P3 _ R by g(x) = f(11(x)), for all x E CP. Then g is a generalized isoparametric function. For example, if 11: M -- N is a harmonic Riemannian submersion, and Nc is a hypersurface of constant mean curvature in N, then Tf 1(N ) has constant mean curvature c
inM[15].
Example 2.4.7 Consider the tangent bundle TM of a Riemannian manifold (M, g). Let 11 be the projection map 11: TM - M. Define the metric G on TM to be the Sasaki metric [351 , and write D for the Levi-Civita connection on TM . The metric G and connection D have the properties that (i) the horizontal lift of any geodesic of M is a geodesic of TM .
(ii) Every straight line in the fibre of 11: TM - M is a geodesic of TM. (iii) Every fibre of H: TM -- M is totally geodesic in TM. Define f : TM - R by f (x, v) = Iv I2 , where x E M and v e Tx M. Then = of is a geodesic vector field in TM. Let Mc = f -1 (c) be a hypersurface of TM, c 0. Then for x s Mc ; let X E T xMc be horizontal with respect to 11 . Then D X = 0, so X is a principal curvature vector with principal curvature 0 of multiplicity equal to m = dim M. If X E T M is vertical with respect to 11 , then D X E = X/c, and X xc is a principal curvature vector with principal curvature -1/c of multiplicity m - 1. Clearly the principal curvature eigenspaces are preserved under projection down the integral curves of . Hence f : TM - R is a generalized isoparametric function. Since the horizontal distribution of TI: TM --- M is integrable if and only if M is flat; the eigenspace distribution corresponding to the principal curvature 0 will not in general be integrable. 41
3 The stress-energy tensor
3. 1
Derivation of the stress-energy tensor
Let (M,g) be a compact Riemannian manifold, and consider an action which, for simplicity, we suppose has the form r L(jl(0)) dx I (0) _ .'M
where 0 is a section of a Riemannian fibre bundle 1T: E --- M, and L is a function on the bundle of 1-jets of sections of TI: E - M ; L : J1(E) - I R, with L possibly depending on the metric of E.
Example 3.1.1 Let 0 : ( M,g) - (N,h) be a smooth map, E = M x N. Then 0 can be regarded as a section of E. Let L(j 1(0)) = Id O 12 be the square of the HilbertSchmidt norm of d 0 .
In general one looks for extremals of the action (3. 1. 1) with respect to variations of the section 0, and one finds that 0 is an extremal if and only if 0 satisfies the Euler-Lagrange equations
9eG(0) = 0
(3.1.2)
,
where F'(0) E'(0-1TE) (we writeg'ffor the Euler- Lagrange operator on sections). In Example 3. 1.1, the extremals are the harmonic maps and the EulerLagrange equation for 0 is (3.1.3)
L\ 0 = 0 .
Suppose now we vary the metric g. If g(u) is a smooth 1-parameter family of
metrics, g(0) = g, then 6g = 2g/2ulu= Proposition 3.1.2 dI (0) Iu=O du 42
0
lies in'(02 T*M).
For a fixed section 0 ,
-
fm
< S0 , bg> dx,
where S0 C' (02 T *M) , called the stress-energy tensor of 0 , is given by
-2
S L + 2 g L, and < , > is the metric induced on 02 T * M from g (by " L we g
0
g
mean the section of p T* M, which is given in components by (2g L) ab gabdxa dxb with respect gacgbd (summing over repeated indices), where g = cL 2gcd to a local coordinate system (xa ) on M). Proof Suppose g has a local representation in the form g = gabdxadxb (summing over repeated indices) with respect to local coordinates (xa) on M. Then
dI u 2L du u = 0-M 2gab gab dx + ./M a J
L
2(dx) 0gab
a gab
The volume element dx can be written in the form dx = (det g)Zdx1 A ... A dxm. Then
i (dx)
2(detg)2 (co-factor of gab) dxl A ... ndxm
2 gab
2(det g)-1 (co-factor of gab) (detg)2 dxl A ... A dxm g
Corollary 3.1.3
ab If
dx.
0 and L are as in Example 3. 1. 1, then
S0 = e(0) g - 0 * h Proof Let (x a, ya) be a local coordinate system on E = M x N. These induce coordinates (xa,yx,ya) on J1(E) with respect to which L(xa,y , ya ) gabyayb hap, where g = gabdxa dxb and h = ha#dyady are local representatives of the metrics g,h respectively (summing over repeated indices). Then
2gab But gij
gj k
=
(
ag1l
ab
) ya Yj#ha
= b k , so
0 = "gab
gjk
+ g t)
gjk
gab
43
Whence gll
_
gab
-
g
it
`glk
g
jk
gab
gil sa ab k
1
jk
g
gia gjb Thus 2L/P gab
- yaayfibhaR
Proposition 3.1.4 M
Or in coordinate free notation 2 L(0)
0 *h.
g
LJ
If X is a smooth vector field on M, then
dx - 2'r V* S0(X)dx = 0 M
where V * S0 denotes the divergence of S0 ; V * S0 (X) = E VX S0 (Xi ,X) , where (X. ) i is an orthonormaI frame field.
Proof Let to be the family of diffeomorphisms associated to the vector field X, then for each u I = '
1M
Ldx
fthu(M)Ldx
f ?u*(Ldx). M
Therefore
fM(Ldx - thu *(Ldx)) = 0, and so f 2X(Ldx) =0, M
where YX denotes Lie derivation with respect to X. Also
X(Ldx) _
dx + f <SO,.Xg> dx. M
Now, in coordinates, letting "; "denote covariant differentiation; (9 g) ab 44
= 2X (a; b)
(the bracket ( ,) means symmetrization of the indices) so that
f M(SOab 6 gab) dx = 2 f ((SOab x a ) Xa ) dx ,b - Sabb 0; M
= -2 .f MSab 0;b Xa dx. Corollary 3.1.5 If 0 is an extremal of I, and so satisfies the Euler-Lagrange equations (3.1.2) , then v * S0 = 0 The statement follows immediately from Proposition 3.1.4, since X was arbitrarily chosen. Proof
Proposition 3.1.6 Suppose (M,g) is any Riemannian manifold (i.e. not necessarily
compact), and 0: (M,g) - (N,h) a map. Let S0 C'(®2 T *M) be given by S0 = e(0)g- O*h, then for all X CC(TM) , = - v*S0(X). Proof In coordinates, as in Corollary 3.1.3, z(g110a0Qhix 9) gab
(S0)ab=
0aObhag .
Thus
(s)0 ab ;b
= (g1161 a0P j ;a
a
- 0aa, b 0g-0`a P;b) h oQ b b
Q; b
0a 0b
hap.
Corollary 3.1.7 If 0: (M,g) - (N,h) is harmonic, and S0 C((02T*M) is given by S0 = e(O)g- 0 *h, then
v*S0 = 0. Conversely, if 0 is a submersion almost everywhere, and v * S0 = 0, then 0 is harmonic.
45
3.2 Examples
,Example 3.2.1 Suppose 0: (M,g) --(N,h) is conformal with 0*h = pg for some smooth function p : M
S0 =
E. Then Corollary 3.1.3 gives (3.2.1)
2p(m-2)g
Thus if m = 2,S0 = 0; and if m > 2 and 0 is harmonic, Corollary 3.1.7 implies p is constant, i.e. 0 is homothetic. This generalizes a result of Hoffman and Osserman [ 24 ].
Conversely, if S0 = 0, then 0 = trace S0 = 2 (m - 2) e(0), whence, if 0 is nonconstant ; m = 2 and 0 * h = e (0) g, i.e. 0 is conformal. Thus Proposition 3.2.2 If 0 :(M,g)- (N,h), then S0 = 0 if and only if dim M = 2 and 0 is conformal.
Corollary 3.2.3 Suppose 0: (S2, g) -- (N,h) is a harmonic map of the 2-sphere into a Riemannian manifold N, then 0 is conformal. Proof Choose local isothermal coordinates z on S2, so that g = p(z) dz dz. Then a tensor T C W(E)2 T* M) has a type decomposition of the form T = T (2,0) + T (1,1) + T(0,2), where T(1,1) is spanned by dzldzl locally. A calculation shows that (0*h)(1, 1) = e(0) g. Thus S0 = S0(2,0) + S(0,2) , where SO(0,2) = That the divergence, v *S0 = 0, amounts to SO(2'0) being a quadratic holomorphic differential on S2. But it is well-known that there are no non-trivial quadratic holomorphic differentials on S2. Thus S0 = 0, and Proposition 3.2.2 shows that 0 is SS
.
conformal.
Example 3.2.4 Suppose 0: (M,g) - (N,h) is a Riemannian submersion, i.e. 0 is a submersion, and the differential is an isometry on the horizontal space. Then Proposition 3.2.5 (See [12 ] for a different proof) : the map 0 is harmonic if and only if the fibres are minimal. Proof Locally, about a point x e M, choose an orthonormal frame field for
TM, X1, .... Xn, Xn+1, ... , Xm, where the first n vectors are horizontal and the last n - m vectors are vertical. Choose indices i, j , ... to run from 1 to n ; r, s, .. . 46
to run from n + 1 to m and a, b, ... to run from 1 to m, and use the usual summation convention for r4eated indices.
Suppose 0 is harmonic, and y C-
(TM) is arbitrary, then
0 = (VX S0) (Y,Xa) a
- (VX 0 *h) (Y,Xa) a
- Xa((0*h)(Y,Xa)) - 0*h(VX Y,Xa) - 0*h(Y,VX Xa).
(3.2.2)
a
a Choose Y = X. to be horizontal, then l
o = Xig(X.,xi) g(V
XJ,X.) + g(xj,vX Xi)
xi
Xj'X ) +g(x i
Y'V j'
xi
x) i
0*h(VV1Xj,Xi) + 0*h(Xj,VXixi), where
denotes projection onto the horizontal space. So that
0 = 0*h(Y,VX Xr
r
and the fibre is minimal. Conversely, if the fibres are minimal, then V * S0 = 0. 3.3 The eigenvalue decomposition of the stress-energy tensor (N, h) be a smooth map, and let m = dim M, n = dim N. Suppose Let 0: (M, g)
0 *h has eigenvalues v1, ... , vm with respect to g, where v1 >. ... > vm > 0. Let X1, ... Xm be an orthonormal basis on a domain of M adapted to this eigenvalue decomposition, i.e. if T. is the eigenspace of v ., then a subset of X1, ... Xm l l forms an orthonormal basis of T,. The energy density of the map 0 is given by m
e(0) = E v /2 k=1
(3.3.1)
k
If P C ?(O2 T * M) , then let P.. = P(X, ,X,) denote the components of P with
respect to X1, ... , Xm. Then 47
Sij
= e($) gij - l0 *h)
ij
1(9)-v.)6.. Now
- vi = E q ( 1 22 6iq)
e(0)
vq
therefore
(1-26iq) S
it
=
E
q
v
2
q
6
(3.3.2)
i,
I
Example 3.3.1 In Example 3.2.4, when 0 is a Riemannian submersion; vl, ... , and vn+1, ... , vm = 0.
n=1
Example 3.3.2 If 0 ' is a harmonic morphism, as in Section 1.3, with dilation A:
E,
vi
,
V
A
n+1
.
vm = 0.
Example 3.3.3 In Proposition 2.2.5, the eigenspace of S corresponding to noni1 zero eigenvalues, correspond with the principal curvature spaces Si , i = 1, ...p, i j, where II= y9 . The eigenspace Tk of SR with vk = 0 corresponds to the
kernel of d II, i.e. the direct sum of S. with R field to the isoparametric family.
48
, where t is the unit normal vector
4 Equivariant theory
Maps which are equivariant with respect to isoparametric functions. Let M be a space form upon which is defined an isoparametric function f: M -
4. 1
R.
Let M * be the union of non-singular hypersurfaces of f with the induced topology,
and let TIf: M * - M*/f be the canonical projection. Then M*/f inherits a topology from M* with respect to which it becomes an open interval. If K E M*f; define a metric g at R as follows. If X C T- (M*/f) and x e iif 1 (R), then there exists a unique normal X to x at x such that d Tif(X) = X ; define g (X,Y) = g(X,Y). Lemma 4.1.1
The metric g defined above is well-defined.
Proof We must check that as defined above, is independent of the choice of 1 x E IIf (x), for all x s M*/f. Let x,y E of 1 (x), 6(u) a curve defining the unique
normal X to x at x, and 6(u) be a curve defining the unique normal X to x at y, such that d rtf (X) = d Tff (X) = X. Then d TTf (X) = (d/d u) (TTf o 6(u) ) I
u= 0 and d 11f(X) = (d/du) (T7f o 6(u)) I u=0. Since the level surfaces are parallel; we can choose 6 and 6 so that rlfo6(u) = TTf o 6(u), for all u E(-E, e) for some c > 0. Thus g(X,X) = g(X,X) and the result now follows. With respect to the metric g , TTf: (M*,g) -+ (M*/f,g) becomes a Riemannian submersion. We add in appropriate end points to the interval M*/f corresponding to the focal varieties of f, to obtain a closed, half-open or open interval If, depending on whether there are two, one or no focal varieties respectively. We extend Tif to a map TTf. M - If I. Similarly if g is an isoparametric function on a space form N, we can define Ti : N g
-
I g
Reparametrize f,g to become unit affine parameters s,t respectively (see Remark 2.2.8) ; so
= Vs and ;7 = Vt both have norm 1.
49
N is a smooth map such that
Suppose 0: M -
0
t
s
a
is commutative for some a : Is . It . If, in addition, dO (f)x =
,
(x)
(4.1.2)
where u : M - E is some function, then we shall call 0 wavefront preserving (WFP) with respect to the isoparametric functions s and t. Lemma 4.1.2
If 0: M
N is WFP with respect to the isoparametric functions
s : M - I s and - t: N- I,s then d0Q)x= a,(s(x))n0(x), for all xEM*, where
= V s, n = V t and a is as in Diagram (4.1.1) .
Proof From equation (4.1.2), d 0 ( )x = u(x)170 (x). Since t : N + It is a Riemannian submersion and d 0 (4) is horizontal with respect to t IdO(t)x12
=
Idtodo(t') x
12 .
But
d t o d O (4) = d a o d s Q), from diagram (4.1.1)
= d a(8/c^s)
a(s)
,
0
writing 8/8 s for the unit tangent vector to Is.
Suppose 0 : M - N is WFP with respect to s and t ; choose a particular hypersurface of M * for some fixed s0, and let Os0 = 0 I MS Ms ~Na (so) . Define Ms0
0
0
projections p : M* - Ms0 , a : N* + Na (so) to be the maps which project onto 50
Ms0. Na(s0) respectively along the integral curves of t , r respectively. Then os = P IM:Ms~MsO'Ot we can aefine a map 6
05,t(x) =
of
s
1o
INt:
t : Ms
Qs 0
Nt - Na(s0) are both diffeomorphisms. Thus Nt by
o os(x), for all x E Ms
.
(4.1.3)
The map fist is not necessarily independent of the choice of s0; however Lemma 4.1.3 Provided that O(M*) C N* (i.e. there is no s in the interior of Is with a (s) E a It - the end points of It) , then fist is well-defined and independent of the choice of s0. Proof Choose a hypersurface Ms1 of M* for some fixed sl, and define ps:Ms ~ Ms, to be the projection down the normal geodesics, and similarly for ot: Nt - Na(s ). Define
s, t :Ms
0s,t() =
vt
1
- Nt by -1
0 (dsl o ps(x), for all x E Ms
Let x E Ms ; then the unique normal geodesic y to Ms through x intersects Ms o at x0 say, and Ms, at x1. Suppose 0s0 (x0) =Y and 0s (x1) =Y V Since, for 0 1 all z e M*, u(z)rO(z); then ¢(y) = S, where O is the unique normal geodesic passing through y0 and y1. Thus, Ost(W ) =
Ot-1
Ct-1
ok1s1op(W) (YI)
= rt 1 (Y0)
rt
1 0 Os0oPs(x)
= 0s,t(x)
.
L_J
If 0 is WFP with respect to s and t and satisfies the conditions of Lemma 4.1.3, then we shall call 0 simply wavefront preserving (S-WFP) with respect to s and t. Lemma 4.1.4 Suppose that 0 is S-WFP with respect to the isoparametric functions 51
s and t, then
0I
Ms
Proof Let x e Ms 1
= 051'(S1) , for all s1 C int Is. 1
, then the unique normal geodesicy to Ms1 at x intersects Ms 0
at x0 say. Also y = 0(x) - N
a(sl), and there exists a unique normal geodesic S
to Na(s ) at 0(x) intersecting Na(s ) at y0 say (S is unique since we do not pass 0
1
through a focal variety).
Since for all z c- M*, d0 (4)z = u(z) r,0(z); 0(y) = b , and 0(x0) = y0. Thus
0s1,a(sl)(x) - °a(sl) -1 ° 0s0 ° Psl(x) = "cr(s1)-1 0 0s0 (xo) =
ca(s1)(y0)
= y. Thus 0 (x) = Osl,a(sl)(x) for all x c- Msl
.
Let the hypersurface Ms have distinct principal curvatures A1(s) , ... , (s), p and Nt have distinct principal curvatures µ1(t), ... pq(t). Let Sk(x) be the eigenspace of Ak(s) at x c- Ms, k = 1, ...,p, and Ti (y) the eigenspace of p. (t) at y C Nt, j = 1, ... , q. Denote the integral submanifolds of Sk(x) , T. (y) by'/ k(x) , yj (y) respectively. Let 0: M - N be S-WFP with respect to s and t. Suppose in addition there is a jk with (i) dOs0 (Sk (x)) C Tjk (0s0(x) ), for all x e and for all k. 'N
Ms0
Lemma 4.1.5 If 0: M - N is S-WFP with respect to s and t, and satisfies condition (i) above, then d0s t(Sk(x) C Tjk(0s t(x) ), for all x c Ms and for all
k, for some jk = 1. ..... q. Proof
Consider the map yu : Ms - M of Section 2.2; forall x c-- Ms
yu(x) = cosu x + sinu Ex 52
(we assume that M is a sphere - the other cases are similar). Then p
s
and ys p-s
are identical, and equation (2.2.4) shows that d ps (X) is proportional to X. The unit normal at yu(x), x e Ms, is given by
yu (x) = - sinu x + cosu tx , and it is now straightforward to compute Vdp (X) tP (x), whence whencewe weconclude conclude that that P.(X), s dps(Sk(W)) = Sk( PS(x) ). Similarly dot 1 (T .k (Os o ps(x))) _ 0
lk
't-10
0s0
o ps(x) ). Hence dOs t(Sk(x)) C Tj (0s t (x)). k
Corollary 4.1.6
For all x C M* and for all k ;
k(x) C.
0(
(0(x) k
for some jk = 1, ... , q.
If 0: M+ N is S-WFP with
respect to
s and t; let yk(s,t):Ms .
R be de-
fined by
yk(s,t)(x) = traces (x) h(d0s t' d0s t) k ' for all x E Ms, k = 1, ... , p, where h is the metric on N. Suppose (i) d 0s (S ( x ) ) C T . (0s (x)), 0 k k 0
(ii) 0s
t: Ms- Nt
for all x c- Ms
.
0
is harmonic for all s and t, and
(iii) for each k, yk(s,t) (x) depends only on s and t. Then call 0 S-equivariant with respect to the isoparametric functions s and t. P
Write y(s,t) = E yk (s,t) for such a 0. k=1
Example 4.1.7 Let 0: S m-1 y S n-1 be one of Smith's maps defined in Section 1.3. Clearly 0 is S-equivariant with respect to isoparametric functions s,t of degree 2 on Sm-1, Sn-1 respectively. Indeed, each level hypersurface M. is isometric to toss SP-1 x sins Sq-1 , and each level hypersurface Nt is isometric 53
to cost S r-1 x sint S s-1.
The map 0 t : Ms w Nt is defined by
os,t (toss x, sins y) = (cost g1(x), sint g2(y)) , where x C SP-1, y C Sq-1, gI : Sp-1 - Sr-1 is harmonic with Idg1 12 = al constant and g2: Sq-1 - Ss-1 is harmonic with Idg2 12 = a2 constant. Thus 0s t is harmonic, and y1(s,t) = cos2t a1/cos2 s, y2(s,t) = sin2t a2/ sin2 s depend only on s and t, and 0 is S-equivariant. Similarly the maps between hyperbolic spaces defined in Section 1.6 are S-equivariant with respect to isoparametric functions of degree two. Theorem 4.1.8 If 0: M - N is S-equivariantwith respect to isoparametric functions s and t, then 0 is harmonic if and only if p
a (s) + As a'(s) + E
k=1
yk = 0
µ
(4.1.4)
]k
for all s e int Is, if and only if
a"(s) + As a'(s) - zdty (s,a(s)) = 0 for all
s c-
(4.1.5)
int I s .
Remark 4.1.9 If the condition that a map 0 be harmonic reduces to solving a second order ordinary differential equation; we will say that reduction occurs. Remark 4.1.10 A reduction theorem for maps equivariant with respect to group actions is proved by Smith [36 ] (see [36 ] for the definition of such maps). We remark that when 0 is both equivariant with respect to the action of G on M and H on N, G and H are Lie groups, and harmonically r-equivariant in the sense that we have described, then the reduction equation of [36 ] and equation (4. 1.5) above both agree. Remark 4.1.11 We will give two proofs of Theorem 4.1.8, one direct proof and one using the stress-energy tensor and obtain equations (4.1.4) and (4. 1.5) respectively. Afterwards we will show more directly that equations (4.1.4) and (4.1.5) are in fact the same.
First proof of Theorem 4.1.8 We first of all show that there exists a smooth function 54
tg
: M * - II R, such that
06(x) = t6(x) r6(x) for all x E M*.
T
For each level hypersurface Ms0 of s, let is0: Ms0 - M denote the inclusion map. Then 0 o is0 is harmonic onto its image NON thus 0
0(6 o is )(x) = 01(x)n6(x) 0
,
for each x E Ms , where :L : Ms0 0
p(6 o is
= d6 (Dis
)
0
Since A- i5
R is some function. Now
1 )
+ trace Vd6(dis0, dis ).
0
is proportional to
0
; equation (4.1.2) implies
0
d O (e is0 ) (x) = tb2 (x )'16 (x) .
R. Therefore
and for some function tb : Ms
for all x c-- M5 0
2
0
3() T) trace V d6(dis0, dis ) (x) = thW 6(x), 0
for all x e Ms and for some function 0 3 : Ms 0
0
-R.
Furthermore, from Lemma 1.1. 1, Vd6
-dO (IM 0 + V 6(,;) pN
a (s)
a'(s)71
= LI4q ,
for some function 04: M* - R. Thus 55
A 0 = trace V d 0 =
_
trace Vd0(dis0 , dis0 ) + Vd0(4,U) ,
d)17
for some function th: M* _ R. We must now work out h (, 0,77). Let (Xa) 1 < a
e0 = trace VdO E V d9 (Xa,Xa) + V d0 ( a
)
,
(4.1.6)
,
and
Od0 (
,
)
d0()
_ - dO (V ,) + ON
Va,(s)n a' (s)77 a'(s)(d(a'(s))(r7),n)
=
= a'(s) .a"(s) . al(s)
77
= a"(s)ry .
(4.1.7)
Now
a
V d0 (Xa,Xa) = A(0 o is
0
- d0 (-A is) 0
But
0 = a(sois
)
0
= ds(c.i s0 ) + trace Vds(dis0,dis0 ) = ds(Ais
)
0
+ As,
since V d s Q, t) = 0 (s being a R iemannian submersion). Thus 56
L\ is 0
= - As
on Ms
(4.1.8)
,
0
and
-d9(ois0 = a'(s)Asr, , )
by Lemma 4.1.2. Finally, let (X a ) I< a < dim M-1 denote the orthonormal basis on Ms such that 0 dis0 (Xa) =X a for each a. For simplicity of notation write <X,Y> = h(X,Y) for
all X,Y EV(TN). Then, for each a, 0 = < d(0 o is0) (Xa) , 71 > ;
therefore, taking the covariant derivative with respect to d 0s
(X 0
a
), where we write
= 00 is
0s 0
p
0=
+ < dos (X a),'dNO 0
(X a < dOs0 (Xa)d0s0
>
E
(OX Xa,
<0 0s0,n> +
a
>
sp
(Xa )n>
+
a
< dOs0(Xa), - µjk d0s0 (Xa)> a
summing over the a, where we suppose aX C Sk for each a. Thus a
= 0
k
µ
]k yk
(4.1.9)
and from equation (4.1.6)
<00,'q> = a'' (s) + a,(s)0s + E p. yk Therefore the map 0 is harmonic if and only if equation (4. 1.4) holds. Second proof of Theorem 4. 1.8 Write the metric g on M * in the form g = d s2 + g
where gs is the induced metric on Ms. Similarly express h on N* as h = dt2 + ht' where ht is the induced metric on Nt . Then
57
s,
S0 = e(0) g - O*h = A (y(s,t) + lay(s) 12)(ds2 + gs) - 0 * (dt2 + ht )
(4.1.10)
,
where t = a(s). If (Xa ) 1 < a < dim M-1 is the orthonormal basis chosen as in the first proof, then for each b
V*S0(Xb) = V*(e(0)g)(Xb) -V*(0*h)(Xb) = V *(0 *ht) (Xb)
_ -Xa(0*ht(Xa,Xb)) + 0 *ht(VX Xa,Xb) + 0 *ht(Xa, V. a
- (0 * ht ( , Xb)) + 0 * ht ( V
Xb) a
, Xb) + 0 *ht ( , V x )
Xa(0 *ht(XaXb)) + 0 *ht(VX Xa,Xb) + 0 *ht(XaVX Xb) a
a
(4.1.11) summing over repeated indices. On the other hand, 0S0 is harmonic onto its image, so
v S08 = 0, 0
is the connection on M. .
where
Now
U
*
S0
s0
= i y (s0,t0)gs - 0s h, where t0 = a(s0) , 0
0
zy (s0,t0) gs - 0s* ht , since 0S* (dt2) = 0. 0
0
0
0
Thus
0 = - O * (Os* ht 0
(Xb
0
a - Xa(0s0 ht0(Xa,Xb)) + 0s0 ht0 (V-XXa,Xb) + OS0ht0 (Xa,v-XXb) , a
(4.1.12) 58
where (X a) 1
Xd) + vX Xd, for all c,d,
dis0 (Xc,Xd) _ - dis0 (V c
c
is an isometric immersion; V dis (X,Y) is
from Lemma 1. 1. 1. Since is 0
0
perpendicular to d i s0 (Z) for all X , Y, Z e'(T M s0) , so that
= for all c,d,e c
c
Substituting this into equation (4. 1. 12) shows that the right hand side of equation
(4.1.11) is zero, and v* S0 is zero on vectors tangent to the level hypersurfaces of s . It remains to evaluate V *S0 on f .
V *S0(f) = -4zdy(E) + a"(s)a'(s)-v*(O *dt2 + 0 *ht) (f)
= 2dy + 2dty. (dt/ds) + a" (s) as(s) - v*(0*dt2)(f) - V*(0*ht)(f)
(4.1.13)
.
We work out V*(O*dt2)(f ). Consider V*(ds2); for any X CW(TM);
V*(ds2)(X) = (Vx ds2)(Xa,X) + (VEds2)(f,X) a
= Xa(ds2(Xa,X)) - ds2(VX Xa,X) - ds2(Xa,VX X) a
a
+ (vds2)(f,X) ds(VX Xa). ds(X) + f(ds(f ). ds(X)) - ds(f).ds(vfX) a
On the other hand
As = v* (ds) = (vX ds)(Xa) + (vfds)(f) a
59
ds(VXaXa)
- ds(VX Xa) . a
Therefore V*(ds2)(4) =0s and V*(ds2)(Xa) =0 for each a, so that
V*(ds2) _ As ds. Now o * d t2 = a' (s) 2 d s2 ; therefore
V*Idal2ds2) = VX (a'(s)2ds2)(Xa) + V (0'(s)2ds2)(
)
a
= VX (a, (s)2).ds2(Xa) + V(a'(s)2).ds2(
)
a
+
al(s)2 As ds
= 2 a"(s)a'(s)ds + a'(s) 2 Asds. We now evaluate - v* (g *ht)(0:
- V*(0 *ht)(
)
= 0 *ht(Xa, VX k ) a
E Xk 0 *ht(Xa,Xa) a
a
_ -k X k
yk
Thus equation (4.1.13) becomes
2dsy + Zdsy. (dt/d s) - a" (s) a'(s) -
s(a'(s) )2 -
akyk.
(4.1.14)
k We therefore obtain the equation
a" (S) + A say(s) - z dt
21'(s) (dsy - 2 E akyk) = 0,
(4.1.15)
k
as a condition of harmonicity. Proposition 4.1. 12 below will show that the last term in equation 4.1. 15 is zero, and that we have indeed established that 0 is harmonic if and only if equation (4.1.5) holds. 11 60
Proposition 4.1. 12
If
0: M -
N is S-equivariant with respect to isoparametric
functions s and t, then (i) ds Yk - 2A k Yk =
0
(ii) dtyk + 2pj yk = 0, k
for each k = 1, ... , p . Proof From Corollary 4.1.6, we see that for all x EM*, 0 I
(0(x)
k(x)
fk(x)
Ik
has constant energy density. Without loss of generality assume that M is Sm-1 and
N is Sn-1, so that.'./ k and 9j are small spheres (the other cases are similar). Fix a level hypersurface Ms0 , and fix x E Ms0 , and consider 0 0
_ k (x0)
x 1.A (0
)
nj
1 j (0 (x0 )). Rescale the map to obtain a map zU :Sk
-
S k , where
k
mk is the multiplicity of ak' and n the multiplicity of p,. Then Id O 12 = ak = l
constant. From Remark 2.2. 10,.'fk(x) is a small sphere of radius sin (s - (k- 1) T7/p) for
x e Ms, and.9k(y) is a small sphere of radius sin (t - (jk- 1)Tf/q) for y E Nt. We therefore find that
yk(s,t) = ak. s i n
- (jk-1)Tl/q) /sin2(s -(k- 1) Tl/p).
Therefore
ds yk (s,t) = - 2 sin(s - (k- 1)iVp) cos(s-(k- 1)TVp) sin4(s - (k-1)Tl/p) .
sin2 (t - (jk - 1) Tl/q) ak
= - 2cot(s-(k-1)TI /p)yk(s,t) 2Akyk
'
and
61
2sin(t-Qk-1) TT/q)cos(t-(jk-1)ii/q. yk dt yk(s,t) =
sin2 (t-(jk-1) TT/q
= 2cot(t-(j k-1) TT/q) yk
= -2pj yk .
1-1
k
Remark 4.1. 13 In fact condition (iii) can be removed in the definition of S-equivariance on account of Proposition 4.1.14 below. However, we retain it in the statement of Theorem 4.1.8 in order to simplify the exposition.
Proposition 4.1.14 If 0: M - N is S-WFP, harmonic and satisfies conditions (i) and (ii), then µjk k(= dty It=a(s)) depends only on s. k
Proof By the conditions on 0 ,
= 0 . ,
Let (Xa) be a local orthonormal basis on Ms, adapted to the principal curvature spa-
spaces. Then, for each a, 0 = VX
a
,>
< dI(jto 0s't) (Xa) d(lto 0s,t) (Xa)
+
N
d(jt° 0s,t) (Xa)
a
where Xa C- Sk (a) say .
On the other hand 0 is harmonic, so , 0 = 0. But 0
62
00 = trace Vd0 (dis,dis) + V d0 (,t)
(4.1.16)
A(s,t)1), where A depends only on s and t (t= O (s)) (c.f.
and since V d 0
Lemma 4.1.2) , then
trace Vd0(dis, dis) = B(s,t) v, (t = a(s)) with B depending only on s and t.
Now
0(0 o i s ) = d0(, i s ) + trace V d0(di s , di s ) = C(s,t)17 , (t = (As))
where C depends only on s and t (since d0 (A is) = D(s,t) 71 (t = a(s)), with D depending only on s and t). Since 0 o is . jt 00 (t = a(s)) ;
s't
A(jt 00 s't ) = E (s,t)r, , (t = a(s) ), with E depending only on s and t , i.e.
d jt (AOs t) + trace V d jt (d 0s t' d 0s t) = E (s,t) 77 . But 0g t is harmonic, so that ,
trace djt (d 0s,t'd 0s,t)
E(s,t) 7
Whence, from equation (4.1.16) , we see that k pj yk depends only on s and t, k
with t = a(s). We now wish to remove the "S" condition on 0 ; i.e. we want to allow a with
a(s) E a It for s E int I covers N several times.
Furthermore, we want to allow the possibility that 0
Suppose N is Sn-1, and t: Sn-1 R is an isoparametric function of degree q. Sn-1 Let f: - [-1,1 ] be the restriction of the standard homogeneous polynomial F of Miinzner's theorem (Theorem 2.2.5) giving s. Call 0 :M + Sn-1 wavefront preserving (WFP) if 0
(i)
M
Sn-1
If
8
I Is
a
R -oos[-1,1 q
commutes, 63
and (ii) dO(Vs) x = u(x)vtO(x), for some u: M - R, for all x C M, where [-1,1 ] is given by cosq (t) = cosq t, for all t c- R. Then Lemma 4.1.2 cosq: R is still valid. If N is ]Rn or Hn-1 we remove the function cosq, and Diagram (4.1.1) applies. We may also wish to restrict a: I - R to have fixed boundary conditions, s and to have values in a certain domain D C R (for example, we may wish to find harmonic maps covering the sphere twice using isoparametric functions of degree 2, in which case we would suppose a : [0, 11/2 ] - [0, TI 1). Suppose in addition, that we can define a map 0s t: Ms - Nf-1 (cosqt) for all
s E int Is, and for all t E D\a (2 Is ), such that 0s ' a(s)
0
Ms, and the curves 6s(t)(x) = (6s t(x), for each s cc. int Is and for each x EMs, satisfy (d/dt) 6s(t)(x) = vs(t)7)O(x)
=
,
for some function vs: M* - R depending only on s. Suppose for all x E M*, (i) dO(Sk(x)) C Tjk (0(x) ), for all k, and for some
Jk'
-
Sn-1 is WFP with respect to s and t, and satisfies Lemma 4. 1.14 If 0: M condition (i) above, then dOs (Sk(x)) C t (x)) for all x e Ms and for all t
k=1, ..., p.
Proof From the proof of lemma 4. 1.5; it is clear that projection along the integral curves of ,ri preserves the eigenspaces Sk,T , respectively, provided that we do l not project through a focal variety. However, for all s E int Is, and for all t G D \a (8Is ), there exists a s c int Is with a well-defined projections p s:M - M~s along integral curves of l; , and a well-defined projection at: Nt - No(g) along integral curves of 17. This is because all s E int I s are allowed, and there exists at least one such s with close enough to Nt. Hence the map ¢s't preserves Na(s) the principal curvature eigenspaces. F-I If furthermore (ii) 0s ' t is harmonic for all s E Is and for all t E D \(a(2 Is )),
and (iii) yk(s,t) (x) = trace Sk(x) h(dO5 t , dOs t) depends only on s and t, for each k = 1, ..., p ; we call 0 equivariant with respect to s and t. Now locally, we can consider an equivariant map 0 as being S-equivariant, by considering t as varying in an interval of the form [1 TI/q, (1 + 1) II/q ] (in the case N 64
is a sphere), for some 1 = 0, ... , q - 1. Since the proofs of Theorem 4.1.8 are local in nature (one works in a sufficiently small domain about each point) ; it is clear that Theorem 4.1.8 is also true for equivariant maps. Thus Theorem 4.1.15 If 0 : M - N is equivariant with respect to isoparametric functions s and t, then 0 is harmonic if and only if I
p
a" (s) + A s a (s) + E
yk
=0
(4.1.4)
k=1
for all s
int Is, if and only if
a"(s) + es a'(s) - a dt.y(s,a(s)) = 0 for all s e int I
s
(4.1.5)
.
Remark 4.1.16 Equations (4.1.4) and (4.1.5) are valid for s e int I s ; it is conceivable (and often the case) that the equations become singular for s E a I , that is s limb I as and sim c I (E Frjk7k) are infinite. In this case, one looks for a res is s parametrization of the equations so as to remove the singularities. Remark 4.1.17 Given 0 : M* - N* which is S-equivariant with a satisfying equation (4.1.4) ; it is sometimes possible to extend 0 to a smooth map 9 : M - N. For example, this is the case for the Smith maps of Section 1.3. 4.2 Generalized equivariant maps between Riemannian manifolds. Let M,N be connected Riemannian manifolds each admitting a generalized family of isoparametric hypersurfaces (Mc)cEI' respectively. Let M* be the union of the hypersurfaces of M; M* = U M , then M* has a topology induced from CEI C that of M. Let 1T:M* - I be the projection, then I inherits a topology from M* with respect to which it becomes a set of open intervals. As before we can define a metric on I with respect to which TI becomes a Riemannian submersion (M* has the metric induced from M). We can add in appropriate end points to the intervals I corresponding to focal varieties, to obtain a closed, half-closed or open interval (or possibly a circle in the case that M is a torus or Klein bottle c.f. Section 5.1 (iii) ). Similarly we can define N* and the projection map a: N* J. (Nd)dr--J
65
Suppose that the generalized family of isoparametric hypersurfaces (Mc) c CI is defined by a function f on M, and furthermore, that f admits a reparametrization s = s(f) such that IVs 12 = 1, then we say that the generalized family of isopara-
metric hypersurfaces admits a unit speed reparametrization. Let 0: M - N be a map between Riemannian manifolds which admit generalized respectively, both of families of isoparametric hypersurfaces ) (Mc c (7-I , (Nd) dE
J
which admit unit speed reparametrizations. The notions of 0 being wavefront preserving and equivariant can be defined as in Section 4. 1 (by making the definitions locally). Lemmas 4.1.2, 4.1.3, 4.1.4 and 4.1.5 all remain valid. Similarly the proofs of Theorem 4.1.8 (replacing equation (4.1.5) with equation (4.1.15)) remain valid, and we obtain
Theorem 4.2.1 If 0: M - N is equivariant with respect to the generalized isoparametric functions f : M - R and g : N - E, where f and g admit unit speed reparametrizations s and t respectively, then 0 is harmonic if and only if
a"(s) + As as(s) +
E µ
k=1
lkyk = 0
(4.2.1)
for all s E int Is, if and only if
a"(s) + e s as(s) - A d t y(s,a(s)) P
- 2a1(s) (d s y- 2k1
yk) = 0
(4.2.2)
for all sEintIs Remark 4.2.2 Proposition 4.1.12 is no longer necessarily valid; however it can be proved directly that equations (4.2.1) and (4.2.2) are equivalent. Proposition 4.2.3 With the notations defined above and in Section 4. 1; if 0 is equivariant ;
a'(s) Nlkyk = Xkyk -
zdyk(s,t)(),
(4.2.3)
for each k = 1. ..... p. be a local orthonormal frame field on M, and suppose Proof Let (X a )1 < a< dim M-1 66
forms an orthonormal basis for Sk at each point (mk = dim SO that (Xki )1 <-I < mk
Then, for each k,
akyk = - E 0 *h(Xkl, vX
)
ki
< d0(Xki ), d0(OX
)>
ki i
=
E
as(s) ujk yk + E
VdO(Xki,
E
i
Since the functions are isoparametric, projection down the integral curves of preserves the Sk spaces. Thus, if d(u) is an integral curve of t, then we can choose our frame field (Xki) i m such that 6(u)*Xki is proportional to Xki. Thus k
-'2't Xki is proportional to Xki for each i; but OX E = akXki , therefore Of X. is ki
proportional to Xki.
But 1 = g(Xki,Xki) , thus V X. = 0. Thus we obtain
E
= idyk( ). Since d yk () = ds yk + dt yk (d t/d s) ; we see that equations (4.2.1) and (4.2.2 ) are indeed equivalent.
Remark 4.2.4 Remark 4.1.17 applies equally for maps 0 satisfying equation (4.2. 1). The theory of this chapter is best illustrated when applied to specific examples. With each example, the qualitative features of the above theory can be very different. We shall see how to adapt Theorem 4.1.8 in the next chapter. We shall leave the 67
application of Theorem 4.2.1 until Chapter 9, where we consider deformations of metrics.
68
5 Examples of equivariant maps
5.1
Maps from Euclidean space to the sphere
First of all we prove a lemma which will be useful throughout this chapter. Lemma 5. 1.1 Suppose f: M - R is a smooth function on a Riemannian manifold M with Idf I2 = t1(f) , Af = th2 (f) for some smooth functions zbI and zb2. Then, given an equation of the form
a"(f) Idf 12 + a'(f)Af = zb(f)
(5.1.1)
,
under reparametrization u = u(f), u'(f) 0, equation (5.1.1) remains invariant - that is, writing g (u) = a (f (u)) ; equation (5. 1. 1) become s for some smooth function :V
p"(u) Idu 12
;
+ p'(u)0u = zb(u)
,
where O(u) =rb(f(u)). Proof
We can write f as a function of u,f = f(u):
p'(u) = a'(f)f'(u) p"(u) = a"(f)f'(u)2 + a'(f)f"(u) Also
Idfl2 = f'(u)2 Idu12 Af
= f"(u) IduI2 + f'(u)Au
Therefore 1
a (f)IdfI2 + a'(f)Af =
f'(u)2
+
f
(p"(u) - p'(u) f-!(u)
)
fl(u) 2 IduI2
f (u)
(f"(u) IduI2 + f'(u) A u).
p"(u) IduI2 + p' (u) Au 69
Let 0: IRm _ Sm be defined by
Example 5.1.2
0 (s.x) = (cos a(s), sin a(s).x) , where x - Sm-1, s C- [0,- ) and a(0) = 0. The map 0 is equivariant with respect to the isoparametric functions s and t, where s (s0. x) = s0 , for all s0 C 10, co) and x E Sm-1, and t((cost0, sint0.x) = t0, for all t0E [0,11] and x c- Sm-1. The square of the norm of the derivative of s and the Laplacian of s are given by IdsI2 =
and y(s,t) =
1,
As = (m-1)/s
,
(5.1.2)
I dos t 12 is given by
y(s,t) = (m- 1). sing t/s2
(5.1.3)
From Theorem 4.1.8; 0 is harmonic if and only if aI I
(s) +
(M-1 3)'
as(s)
- (m 21), sin ce(s)cos a(s) = 0.
(5.1.4)
s
This equation is singular at s = 0, and we remove the singularity with a suitable substitution. Using Lemma 5.1.1 we make the substitution u = u(s), where u is given
by eu=s. Then Id u I2 =
1/e2u
and
o u = (m -2 )/e2u whence equation (5.1.4) becomes
(u) + (m-2 )a'(u) - 1(m-1). sin2 a(u) = 0
(5.1.5)
where u C(-OD,00), and as u - - ; a(u) - 0 . Equation 5, 1, 5 is a well-known equation - namely the equation of a pendulum with
constant gravity and constant damping. The variable u measures time, and a = 2 a is the angle the pendulum makes with the upward vertical.
70
gravity
We require an exceptional trajectory satisfying equation (5. 1.5), such that the pendulum is just standing vertically upwards at time u = - a . The solutions of the pendulum equation are well-known, see for example [23, p. 183 &p. 1961 . Consider first the exceptional case; m = 2. Then the damping is zero - a good treatmentof this case is to be found in [ 4, Section 6.3 ]. For each u0 c--(- oo, oo) , there is a unique solution to equation (5.1.5) with initial conditions
a(u0) = a0 and a'(u0) = a'0, for all a0 c [0,2 TI] and for all a' c R. Choose a0 = T7/2, then if
(i) a0 = 1; the solution a: F lim
a(u) = Ti and lim
R is strictly monotonic and has image (0, TT)with
a(u) = 0. In fact a(u) = sin- 1 tanh(u-u0) - this solution
u--u-00 is called critical. (ii) a < 1, then a: R - R is oscillatory with image contained in a closed subinterval of (0, TI).
(iii) a > 1, then the solution a is strictly monotonic increasing and surjective onto 1R.
In order to obtain a smooth harmonic map we choose the critical solution (i), giving a map 0 : R2 - S2, covering S2 except for the one point (-1, 0). Note in fact that we have a 1-parameter family of maps corresponding to the choice of a0. That the map 0 so constructed is smooth at 0 C R2 is demonstrated in Chapter 6. In the case when m , 2; the damping is non-zero. For each u0 c (_co, eo), and for each a(u0) = a0 ; there is again a critical solution with lim a(u) = 0. However,
u--00
in this case lim a(u) = T7/2, corresponding to the pendulum hanging straight down. U -00
In fact the pendulum does not reach the upward vertical on its first upward swing, and performs decreasing oscillations about the downward vertical position. As in the case when m = 2, the map 0 is smooth across the point 0 c R m . 71
Such solutions have been considered by J. C. Wood [ 42 ] , in connection with the
Dirichlet problem for the Euclidean disc, where he shows that there is no map from the Euclidean disc to the sphere of the above form with a0 > 11/2 + sin 1 tan h (Km ) where K m = (m - 1)z/(m - 2). Similar considerations apply to the map 0: Rm Sn given by
,
0(s.x) = (cos a(s), sin a(s).g(x)), a(0) = 0 Sn-1 is harmonic of constant for all x CSm-1, s r [0,co), and where g: Sm-1 energy density. The corresponding reduction equation has the same form as equation (5.1.5) but with a different value for the gravity. The qualitative behaviour of the solutions is the same as for the case when g is the identity map.
Example 5.1.3
Let 0 : Em - Sm be defined, for each integer k, by
0((x,sy)) _ (cosa(s).el , sina(s).y) ,
(5.1.6)
with a(0) = 0, where x C E, y CSm-2,m>3andsE[O,oo). Theno isequivariantwith respect to the isoparametric functions s and t, where s ((x, s0 y)) = s0, for all s0 C [0,oo), x c- R and y C Sm-2, and t((cos t0.u,sin t0.v))= t0, for all t0 e [0,11/2 ], u C S1 and v CSm- 2. The square of the norm of the derivative of s and the Laplacian of s are given by
Ids12 = 1
,
As = (m - 2)/s
(5.1.7)
,
and y (s,t) is given by
y(s,t) = k2cos2 t + (m - 2) sin2 t/s 2
(5.1.8)
.
From Theorem 4.1.8, 0 is harmonic if and only if a"(s)
+
(m
2)- k2). sin a(s)cos a(s) = 0. - 2) as(s) - ((m2 s
(5.1.9)
This equation is singular when s = 0, and so, using Lemma 5. 1. 1, we make the substitution u = u (s) where eu = s, to give the equation
a"(u) + (m - 3)a'(u) - '((m - 2) - k2.e2u).sin2 a(u) where u c(-oo,- ) and lim
72
a(u) = 0.
0
,
(5.1.10)
Equation (5. 1. 10) is that of a pendulum with constant damping and variable gravity.
As in Example 5.1.2, u measures time and a = 2cx is the angle the pendulum makes with the upward vertical.
gravity
gravity
For each u0 there exists a unique solution a(a0, a0) to equation (5.1.10 ) with prescribed initial conditions a(u0) = a0 and a'(u0) = a for all a0 C(0,1i/2) and for all 's R. We look for exceptional solutions such that the pendulum is just CIO standing vertically upwards at time u = - oD. We demonstrate the existence of such non-trivial solutions in Chapter 6. Note that we have a 1-parameter family of solutions depending on our choice of a0. Different choices of a0 lead to qualitatively different solutions. To see this we first of all state a comparison theorem for second order equations. Theorem 5.1.4 [ 9 ]
Let p, and ki be continuous on
[ a, b 1,
i = 1, 2, and let
0 < p2(u) < p1(u)
k2(u) > k1(u),
u e [a,b] .
Let L. be the operator L. v = (p. v') ' + k. v. If f. is a solution of L. f. = 0, let wi = tan-1 (fi/pifi' ). If w2 (a) > w1 (a) , then w (u) > wl (u) for all u s [a,b]. Moreover, if k2(u) > ki(u) on (a,b), then w2(u) > wi(u) for a
a"(u) +h(u) a'(u) + g(u) sin 2 a(u) = 0 We can write this in divergence form:
L a= (p a')' + ak a= 0, where 73
u
p(u) = exp (.u
(h (s) ds) 0
ka(u) = g(u) sin 2 a(u) p(u) /a(u) . Proposition 5.1.5 covers the sphere. Proof
If a0 is chosen close enough to 0 ; the solution a never
Let u1 be the time such that
(m-3)2
=
- (m-2)) -
4(k2.e2u1
A2
for some fixed A > 0 (such a u1 always exists), and let al = a(ul) . Then al depends continuously on a0 and a 0 (c. f. [ 9 ]) . As a0 - 0 ; a 0, 0 and al since in the limit a0 = 0, aU = 0 the solution is trivial. Thus there exists a0 such that al < 11/4. Consider the equation 6
(u) + b 6 '(u) + c 6(u) = 0 2
(5.1.11)
,
2u
with b = (m - 3 ) and c = k . e 1 - (m - 2 ). Given the initial conditions 6 (ul) = a(ul ) and 6'(u l) = a' (u1) , this has a periodic solution with period I'= 211/(4c - b2) = 211/A. Choose a0 such that there exists u2 with u2 > u1, I < u2 - ul and a2 = a (u2) < 11/2. Then over the interval [ ul , u2 1, we can use the comparison theorem (Theorem 5.1.4). Since the gravity of the equation (5. 1. 10) dominates that of equation (5.1.11) over the interval [u1,u2 ]; Theorem 5.1.4 shows that w1(u) > w2(u)
for u c [ul,u2 ] , where w1 = tan-1(a/p1a) and w2 = tan 1(6 /p26'), where p1 and p2 are defined as in Theorem 5.1.4. In particular wl passes through 11/2 if w2 does ; i.e. a' has become zero and the solution turns. Since the gravity is now increasing, the solution a will make decreasing oscillations, and lim a(u) = 0. U --00
In contrast to Example 5.1.2 we can show that
Proposition 5.1.6 If (m-3) 2 < 4(m-2), solutions covering the whole sphere.
i.e. m =3, ..., 7 ; thenthere exist
We take it as given that for each a0 c (0, 11/2) , there is an ao' with a(a0 , ad asymptotic to 0 as u - - co. This will be shown in Lemma 6.1.4 and lemma 6.1.5. Proof
Furthermore, the solution a is strictly decreasing as u decreases from u0. We 74
show that a0 is bounded away from 0 as a0 - 11/2. For in this case, the solution a(a0, a') will continue past 11/2 as u increases from u0 if a0 is chosen close enough to 11/2.
- 0.
Suppose there is a sequence a0n -+ 11/2 such that aoY n Since in the limit a0 = 11/2, a'0 = 0 the solution is trivial; we can assume u0 to be arbitrarily small,
i.e. for any u1, a(ao ,a0n)(u1) - 11/2 as a0 - 11/2 (the solution depends continuously on the initial conditions). Let a= a(a0n, a0'n ) be a solution of equation (5. 1.10), where n is chosen sufficiently large. Make two substitutions into equation (5.1.10) : firstly let 6 = Tl/2 - a, and secondly let v = - u. Equation (5.1.10) then becomes
6"(v) = (m-3)6'(v) - a ((m-2) - k2e-2v). sin2 6(v) Consider the equation of the harmonic oscillator:
Q"(v) _ (m-3) 3'(v) - M p(v)
,
where M is chosen such that ((m - 2) - k2 e2u ) sin 2 6 / 2 6 - M, 0 on a suitable interval [ v0, v1 ] with the period of p < v1 - v0 (v0 = u0) . Let R have the same initial data as 6 at v0 , and let w = 6 ',6 - 6 fl'. Then
w' = 60 [(m_3)(
!
w(v0) = 0 ,
-
)
+ (M- ((m-2) - e-2v) s26 6 )l
w'(v0) = 0
.
Provided 6'16 < Q'/p (i.e. w < 0) with 6./3 > 0, we have w' < 0. Also as long as w' < 0 we have w < 0. Now look at the first zero of p on [v0, v1 1. We must have 9'6 > 0 here since w < 0, which implies 6 has a zero on [v0,v1 1. This is impossible since 6 is strictly increasing and > 0. If we continue following the qualitative behaviour of the solution considered in the
proof of Proposition 5. 1.6, we see that since the gravity changes for u > u0 , the pendulum now moves under the influence of an upward gravity force past a = 11. The gravity now increases and continues to have the same sign. If a' is small enough, the pendulum will perform ever decreasing oscillations about a = 11.
75
gravity a0
Conceivably, there may be a certain a' such that the trajectory is exceptional and reaches 3 Tj/2 as u-co. If 00 is large enough the pendulum could make several rotations before tending to its equilibrium position. Similar conditions apply to maps 0 : R m - Sn of the form
9((x, s.y)) = (cos
a(s).eikx,
sina(s).g(x)), a(0) = 0 ,
where k is an integer, s s [O,ao), x s R, y C Sm-2 and g:
Sm-2 y Sn-2 is har-
monic of constant energy density. Again all the above constructed harmonic maps are smooth across the focal varie-
ties (c.f. Chapter 6). Example 5.1.7 [ 11 ] Consider the map 0 : R2
S2 given by
¢((x,s)) _ (cosa(s).e'tC, sina(s)),a(0) = 0 ,
(5.1.12)
where k is a non-zero integer, s C(-cc,co) and x C. R. Then 0 is equivariant with
IdsI2 = 1,
As = 0,
and
y (s,t) =
k2. cos
.
Equation (4.1.5) becomes
a"(s) = k2. sin a(s). cos a(s)
(5.1.13)
This equation has a periodic solution in terms of elliptic functions, and the map 0 factors to produce a harmonic map from the torus to the sphere.
5.2 Maps from hyperbolic space to the sphere Example 5.2.1 Let 0 : Hm Sm be defined by 76
0 ( (coshs, sinhs.x)) = (cosa(s), sina(s).x), a(0) = 0.
(5.2.1)
where x E Sm-1 and s E [ 0, -) . Then 0 is equivariant with respect to the isopara-
metric functions s and t, where s: Hm - R is given by s( (cosh s0, sinh s0. x)) =s0,
and t: Sm - R is given by t((cos t0, sint0.x)) = t0. From Lemma 1.6.1, we obtain
Ids12 = 1,
Os = (m-1).coths
and y (s, t) is given by
y(s,t)= (m-1). sin 2 t/sinh 2 s From Theorem 4.1.8, 0 is harmonic if and only if
ce"(s) + (m-1).coths.a'(s) - (m ), sina(s)cosa(s) = 0 . sink2s
(5.2.2)
This equation has a singularity at s = 0, and so we make the substitution u = u(s), where eU = sinh s. Then I du 12
= (eu + e-u)/eu
,
and
Au = (eu + (m-2)(e u + e-u))/eu
,
whence, from Lemma 5.1.1 the reduction equation becomes af 1
(u) +
(eu + (m-2)(eu + e +e -u u1
(m-1)
au(eu +e-u ) = 0
where u e(-oo,00) and lim
U ---00
,
sina(u)cos0(u)
(5.2.3)
a(u) = 0.
Equation (5.2.3) is that of a pendulum with variable gravity and variable damping. The graph of the damping has the form :
77
(m-1)
(m-2) u
CO
and the graph of the gravity has the form
(m- 1) u=-00
We look for an exceptional solution with the pendulum just standing up on end at
time u=- co. As for the Euclidean case; we have a 1-parameter family of such solutions with the resulting map smooth across the point s = 0. However, the asymptotic development of these solutions as u Co is not known.
Similar considerations apply to maps 0: Hm - Sn of the form
0 ((cosh s, sinh s.x)) = (cosa(s), sina(s).g(x)),a(0) = 0 where s E [0,0o), x ESm-1 and
g:Sm-1 y Sn-1
,
(5.2.4)
is harmonic of constant energy
density.
Let 0 : Hm - Sm be defined, for each integer k, by
Example 5.2.2
0((coshs.x, sinhs.y)) = (cosa(s).el , sin a (s). y)
(5.2.5)
a (0) = 0, where x E H1, Y E Sm-2 and s s [0, -o). Then 0 is equivariant with respect
to the isoparametric functions s and t, where s:Hm - R is given by s ( (cosh s0. x, sink s0. y)) = s0, for all s0 E [0, co) , x E H1, Y ES m-2 , and
t : Sm - E is given by t(cos t0 u, sin t0. v)) = to, for all t0 C [0, TT/2 ], u s S1 and v E Sm-2 . From Lemma 1.6. 1 we obtain
IdsI2 =
1 ,
As = tanhs + (m-1)coths ;
also y (s,t) is given by 78
y(s,t) = k2 cos2 t/cosh2 s + (m-2) sin2 t/sinh 2 s. From Theorem 4.1.8, 0 is harmonic if and only if
a?T(s) + (tanhs + (m-2)coths)a'(s) - sina(s)cosa(s'
2 m 21 - k
inh s
0
2
cosh s (5.2.6)
This equation has a singularity at s = 0, and so we make the substitution u = u(s) where eu = sinh s, to give the equation
a"(u) + u1 _u e+e
.
(2eu + (m-3)(eu+ a-u)) a, (u)
-sina(u)cosa(u) . (m-1 e
u
+e-u
e
u
k2
)
e +e -u u
=0
,
(5.2.7)
where u c(-co,ao) and lim a(u) = 0 U--00 Equation (5.2.7) is the equation of a pendulum with variable gravity and damping. The damping is given by the following graph:
(m-3)
U=-
u=
and the gravity by the graph
U=
79
As in Example 5.1.3, we have an interesting 1-parameter family of non-trivial solutions with the pendulum just standing up on end at time u = - co, and smooth across
the point s = 0. By similar arguments to those of Proposition 5.1.5 we can show that, provided k is large enough, there exist solutions which turn before a(u) = 1T/4 is reached. However, the qualitative aspect of the various solutions as u -. oo is not yet known.
5.3 Maps from sphere to sphere.
Example 5.3.1 The maps of Smith described in Section 1.3 are examples of equivariant maps between spheres. Then 0: Sm-1 y Sn-i is given by
0 ((coss.x , sins.y)) = (cosa(s).g1(x), since (s).g2 (y))
(5.3.1)
,
with a(0) = 0 and a(11/2) T1/2, where x ESP-1, y CSq-l, p +q = m, s E 0, r1/2 ] and g1: SP-1 - Sr- i , g2 : Sq-1 - Ss-1 are both harmonic with Id gi 12 = ai constant, i = 1, 2. Using Lemma 1.3.4 we obtain Ids I2 = 1
,
As = (q - 1)cots - (p- 1) tans .
Also 2 ,a y (s,t) = cost 2 1
+
sin2t . a sin2s 2 .
cos s
From Theorem 4.1.8 we conclude that 0 is harmonic if and only if equation (1.3.5) is satisfied. A more detailed consideration of this example is to be found in Sections
1.3 and 1.4.
y R be the isoparametric function of Example 2,3.5. Example 5.3.2 Let f: Recall that the level surfaces M s are parametrized by the sets leie (cos s.x + isins.y) E n+1 ; 0 E [0, 217], (x,y) a Sn+1,21 , for each s E [0, II/4 ]; in fact Ms is Stn+1
isometric to Sl X Sn+1 2 /S0, where Sn 1 2 is the analytic submanifold of Rn+1
®
]Rn+l
Sn+1 2
80
defined by the set
= {(coss.x, sins.y) E
Rn+1
®
Rn+l
;
(X,
S
.
Define 0: S
2n+1
-S
2
by
0:(eie (coss.x + isins.y)) = (cow(s), sina(s).e2ikO)
.
(5.3.2)
where k is some non-zero integer, a (0) = 0 and a (11/4) = U. Then 0 is S-equivariant with respect to the isoparametric functions s and t, where s: S2n+1 R is given by s(eie (coss0.x + isins0.y)) = s0, for all s0 E [0,11/4], 6 E [0,217] and (x,y) E Sn+1 2 , and t: S2 - 1 is given by t ((cos t0, sint0.u)) = to, for all t0 s [0,11] and u ES1. The principal curvatures of Ms are -cot s, -cot(s - 11/4), -cot(s - 11/2), -cot(s - 311/4) with multiplicities n- 1,1,n-1, 1 respectively, hence
- 1 +tans As = -(n-1)tans +(n-1)cots + 1-tans 1 +tans 1 -tans
(5.3.3)
Also
y (s,t) =
4k2 . sin2t
From Theorem 4.1.8 we conclude that 0 is harmonic if and only if
- 1 +tans ) a' (s) a"(s) + (-(n-1)tans + (n-1)cots + 11 +- tans 1 - tans tan s - 4k2 sina(s)cosa(s) = 0 ,
(5.3.4)
with a(0) = 0, and a(11/4) = U. This is singular at s = 0, 11/4, and so make the substitution u = u (s) , where eu = tan s/(1 - tans). Then I du l 2
= (1 +
'du= ((1+e')
e2u
2eu
+ 2e2u)/e2u
+ e u)
,
and
(5.3.5)
+ (n-1)
(1
- (n - 1)eu
1 + 2eu
C 2 (1 + eu) -
1 + eu
eu
+u u )
e
+
1
1 + 2e
u
- (1+2eu)
(5.3.6)
81
Using Lemma 5.1.1, equation (5.3.4) becomes
a" (u) +
+
[2(1+eU) eu (1+2e +2e 2) 1
1 +2e
- (1+2e
u
1
u )
I
J
e
(u
-leu_(n1)eu e`
e
1
a (u) -
4k2e2usina(u)cosa(u) (1 + 2e
u
+ 2e
)
=
2u 2
0
)
(5.3.7)
where u e(-oo,00), lim
a(u) = 0 and lim a(u) = TI.
u-00 Equation (5.3.7) is that of a pendulum with variable gravity and damping. u---00
Damping.
Gravity.
(n-2) U=- cc
As before a = 2 a measures the angle the pendulum makes with the upward vertical.
We look for an exceptional trajectory such that the pendulum starts at the top at time u=- oo; makes one circuit, and just reaches the top again at time u = Co. It is not known whether or not the equation has such a solution - we would certainly expect a solution in the case n = 2 due to the symmetric appearance of the gravity and damping. It has been pointed out by R. Wood that these maps are all homotopically trivial.
Example 5.3.3 Let f: S2n+1 -. R be the isoparametric function of Example 5.3.2. Consider the case when n = 3, and define a map 0 s: S 4 2 - S2 by 82
0s((x,Y))
sin2
=
gs
(-x
1y2
+ y2y1 - x3y4 + x4y3, -xly3 + x3y1 - x4y2+x2y4,
- x1Y4 + x4Y1 - x2y3 + x3y2)
,
(5.3.8)
where x= (x1, ...'x4), y=(y1, ... 'y4) cR4, Ix12=costs, ly12=sings
and
< x,y : = 0. Then
10112 = (4/sin22s)(Ix12 ly12 - <x,y>2) so that 01(S4
2
C S2 .
,
Furthermore, one can show that 01 is onto (see Example
8.2.1). Consider the point p = (coss(1,0,0,0), sins(0,1,0,0)) C S4 following curves through p in S4 2 (see Example 2.3.5): 01
(u)
2'
and consider the
_ (coss (cosu,0,0, sinu), sins(O,cosu,sinu,0))
p2(u)
_ (coss(cosu,O,sinu,O), sins(O, cosu,O,sinu))
33(u)
_ (coss(cosu,O,O,sinu), sins(O,cosu, -sinu,O))
$4(u)
_ (coss(cosu.O,sinu,O), sins(O,cosu,O,-sinu))
p (u)
= (coss(cosu,sinu,0,0), sins(-sinu,cosu,0,0)) ,
(5.3.9)
The tangent vectors to these curves at u = 0 span TS 2 and are all orthonormal. p 4 The acceleration vectors at u = 0 are all perpendicular to TPS4,2 in Tp(R4 ® R 4 2 p,(0) = 0 when p is one of the i. e. the curves are geodesic at p (that is V S4 )3'(0) ' curves (5.3.9)). Under 0s the curves map to
al(u) = 01 o p1(u) = (-cos2u, -sin2u,0)
a4(u)
01 o ft2(u)
= (-1,0,0)
01 0 fi3(u)
= (-1,0,0)
= Os o p4(u) = (-cos2u,0,sin2u) 01 o µ(u)
_ (-1,0,0)
.
(5.3.10)
From this we deduce that 01 is a harmonic Riemannian submersion, and in particular 83
Os
has constant energy density equal to 4. Define a map 0: S7
S3 by
¢(e iB (coss.x+isins.y)) = (cosa(s),sina(s).A (cossx, sinsy)) (5.3.11) with a (0) = 0 and a (Ti/4) = 11. Then 0 is S-equivariant with respect to the isoparametric functions s and t, where s: S7 - R is given by s(ej9 (cos s0.x + isin s0.y) ) = s0, for all s0 E [0,11/4 ], 0 - [0,2 TI ] and (x,y) E S42 , and t: S3- R is given by t( (cos t0, sint0.u)) = t0, for all t0 a [0,11]and u C S2.' The function y (s,t) is given by
y(s,t) = 8sin2t. From Theorem 4.1.8 we conclude that 0 is harmonic if and only if equation (4.1.5) is satisfied; this equation is precisely the same as equation (5.3.7) with n=3 and i k = (2) a . It is not known whether this equation has a solution or not. Neither do we know which homotopy class of TT7 (S3) = Z2, 0 would represent.
Similarly we can define submersions from S8 2 to S4 and from S 15 2 to S8, to give maps representing a class in i115(S5) _ Zs and T129(S9) respectively Example 5.3. 4 Let f: S2n+1 -- R be the isoparametric function of Example 5.3.2. Consider S 4 2 and the map Os: S 4 2 - R of Example 5.3.3. Define 0:S7 -. S4,
for each integer k. by O(e1B(coss.x
+ isins.y)) _ (cosa(s).e2i , sing(s). 0 (coss.x,sinsy))
(5.1.12) where a (0) = 0 and a (TI/4) = 17/2. Then 0 is S-equivariant with respect to the isoparametric functions s and t, where s: S7 - R is as in Example 5.3.3, and t: S4 -- R is given by t( cost0.u,sint0.v) = t0, for all t0 E [0,17/2 ], u ESl and vE
S2 .
ei0 (cosset + isinse2), where (ei)i=1, Let p c- S7 be the point p = 4 are the standard orthonormal vectors in R4. Consider the geodesic curve v in S7 V(U) = elu (cos set + isin se2 ). Then, in Example 2.3.5, we saw that v(u) intersects S 4 2 at an angle of 2 s - 11/2 and in the plane spanned by v'(0) and p '(0) ,
where p is defimd as in Example 2.3.5 (c.f. Lemma 2.3.7). By choosing an 84
orthonormal basis to Ms at p defined by tangent vectors to the curves (5.3.9); we find
y(s,t) =
cos2t.4k2
+
8
sin2t
.
(5.3.13)
cos22s
From Theorem 4.1.8; 0 is harmonic if and only if
- 1 +tans) a (s) a (s) + (-(n-1)tans +(n-1)cots + 1-tans 1 +tans 1 - tans I
- sin a(s)cosa(s). (8 -
\
4k2
(5.3.14)
cos22s
where n = 3, s E [0,11/4 ], a (0) = 0 and a (II/4) = 11/2. This equation has singularities at s = 0, 11/4, and we reparametrize it using the parameter u = u(s) where e u = tan s/ (1-tans) . Using Lemma 5.1.1 we obtain the equation It
a (u)
+
euu 2u (1+2e +2e )
C
2(l+e u) - 1+2eu u
- (n-1)eu + (n-1) 1+eu
e
1 +eu
eu
- (1+2eu) J a'(u)-sina(u)cosa(u)x
8e
2u
L1+2eu+ 2e where u E (-oo, oo) , lim
2u
U --.0
)2
4k2e2u
1
1+2eu)2
J
(5.3.15)
a (u) = 0 and lim a(u) = 11/2.
u-.0
Equation (5.3.15) is that of a pendulum with variable gravity and damping (a = 2a measures the angle the pendulum makes with the upward vertical). Damping
Gravity
u=oo u=-ao 85
The gravity has a negative component only if k = -1, 1. We would require an exceptional solution with the pendulum just standing up on end when u = - -0, and hanging straight down when u = co. We would expect such a solution for k = -1,1, since then the gravity has a balancing effect.
Example 5.3.5 (This example was suggested to me by H. Karcher) : Let f: Sm-1 -- [-1,11 be an isoparametric function on Sm-1 , defined as the restriction of a homogeneous polynomial F of degree p on ]Rm (c. f. Theorem 2.2.5). If the multiplicities of the distinct principal curvatures are equal; the polynomial is harmonic. Then we can normalize F such that IV F(x)12 = Ix I2p-2, for all x c Rm
,
(5.3.16,
In this case the map 0 = VF ism-1 defines a harmonic polynomial map from Sm-1 to Sm-1. That discovery was made by R. Wood [43 ]. Suppose now that the multiplicities of the distinct principal curvatures are not necessarily equal. Let 0 be defined as above, where F is now any polynomial Proposition 5.3.6 satisfying the conditions of Theorem 2.2.5, then 0 is equivariant with respect to the isoparametric functions f on the domain sphere and f on the range sphere.
Proof (due to H. Karcher)
We can express each point x E Em in the form
x = r.e , where r E [0, co) and e E Sm-1. Then F (r. e) = rpf (e) , and so
(1/p) VF (e) = f(e).e + (1/p) V f(e) .
(5.3.17)
Let MO = f 1(0) be the minimal hypersurface (any choice of hypersurfaces will do).
Let e(s) = e.cos s +t e sins, where e e MO, e is the unit normal to M0 at e, and s E [- 11/2p, II/2p ]. Then
f(e(s)) = sinps ,
(5.3.18)
and
Vf(e(s)) = pcosps.e'(s)
.
(5.3.19)
e. sins + e, coss
(5.3.20)
Also
e'(s) 86
so <
f,e' > = p.cosps
(5.3.21)
.
and
(1/p) V F(e(s)) = sinps.e(s) + cosps.e' (s)
sin(p-1)s.e + cos(p-1)s.
(5.3.22)
e
The result now follows from equation (5.3.22).
Remark 5.3.7
If we let s C [-11/2p, 11/2p I represent the affine geodesic parameter, so that f = cos p (s + 11/2p) . Then s = - 11/2p and s = 11/2p correspond to the focal varieties, and s = 0 corresponds to the minimal hypersurface. Now
O(e(s)) = sin(p-1)s.e + cos(p-1)s. 4e = cos(-(P- 1)s + TT/2).e + sin(-(p- 1)s+ 11/2. to
.
(5.3.23)
So that under the map 0 , the minimal hypersurface is mapped onto a focal variety and (a) if p is odd, both focal varieties are mapped onto a single focal variety, and (b) if p is even, both focal varieties are mapped to themselves. If we allow a re parametrization of the map as in Theorem 4.1.15, then the function a is such that a: [-T1/2p, TT/2p ] [ TI/2p, (2p- 1) 11/2p] with a (- 11/2p) = (2p- 1) 11/2p and a(11/2p) = 11/2p .
From Theorem 4.1.15, 0 is harmonic if and only if the corresponding reduction equation is satisfied. In the case when the multiplicities of the distinct principal curvatures are equal and F is harmonic, then 0 is harmonic and equation (4.1.5) is satisfied. In fact from equation (5.3.23) one sees that a (s) _- (p- 1)s + 11/2 is the solution. More generally, suppose p = 4. We will try to find an a: [-11/8, 11/8 ] [11/8,7 11/8 ] with a (-11/8) = 711/8 and a(11/8) = 11/8 satisfying equation (4.1.5). We remark that equation (4.1.5) is no longer that of a simple pendulum.
Recall from Remark 2.2.10 that the principal curvatures are - cot (s + 11/8) , -cot (s - 11/8) , -cot (s - 311/8) and -cot (s - 5 11/8) (remembering that s is now parametrized in the interval [-11/8, 11/8 ] instead of [0,11/4 ]) , with multiplicities mi,m2,
mi,m2 respectively.
Thus
87
y(s,t)
=
sin2 (t- Ti/8)
sin2 (t + TI/8 )
ml
+ m1
sin (s + 11/8)
+
m2
sin2 (s-11/8)
sin2 (t - 3 Ti/8 )
+
m
sin2 (t-511/8)
2
sin2 (s-3TI/8)
2
sin 2(s-5TI/8)
and equation (4.1.5) becomes
a" (S) + (-m1 tan(s+TI/8 + mlcot(s+ Ti/8) +
1-tan(s+17/8) - 1+tan(s+n/8 1-tan(s+11/8)) a (s) 1+tan(s+TI/8)
- m l sin(a(s) + TI/8)cos(a(s) + 11/8) m
2
sin(a(s)- II/8)cos(a(s)-11/8) =0
1
L
sin2 (s + Ti/8 )
cos2(s +TI/8 )
1
1
[51fl2511/8)
cos2(s - P/8) J
(5.3.25)
,
with s e [-TI/8, TI/8 ], a (-TI/8) = 7 II/8 and a (TI/8) = TI/8 . We now reparametrize this equation, using the parameter u defined by eu tan (s + 11/8) / (1- tan (s + TI/8)) , so u c-
A computation shows that
I du I2 = (1 + 2eu + 2e2u)2/e2u and,
Au = ((l+eu)2
+ e2u)
I
-m.
2(l+eu) - 1+2eu
e
u
e
u
+ m1
(1+e )+ eu
Using Lemma 5.1.1, equation (5.3.25) becomes,
88
m
1
2
1 +2eu
- m 2 (1+2eu) l
a"(u) +
e
r
u
2(1+eu) -
((1 +eu)2 +e2u)
m
(l+eu)
+
eu
u
u
e
+ ml
m
1+2e u
m2
-m 1
- m2 (1+2eu)
1 +2eu
sin(a(u) + if/8)cos (a(u) + 11/8)re-u e-u
1
2
+
LL
2
+
1
e +eu
2eu
sin(a(u) - I1/8)cos(a(u) - 11/8) e-u
+ 2+ 2eu
a'(u)
eu
(1+eu)2 J
2eu - 2eu L L
J
_
0.
(1 +2eu) 2
(5.3.26) At first sight this equation looks decidedly unpleasant, but if we look at the qualitative features, we find it has much in common with the Smith equation described in Section 1.3. Express the equation in the form (u)
= h(u)a'(u) + g1(u) sin(a(u) + fi/8) cos (a(u) + 1T/8)
+ g2(u)sin(a(u) - 11/8)cos(a(u) -
11/8) .
(5.3.27)
Then, for u near - cc ,
h(u) _ -(m1-1) - (m1-2)0(eu) +
(m1 +m2) 0 (e2u)
gl(u) = m1 - 0 (eu) g2 (u) = 0 (e3u) where by 0 (f (u) ) near u=-ac ", for some smooth non-zero function f, we mean that 0 (f (u) )/f (u) is bounded and positive for u near - -o . Also for u near + 40.
h(u) _
(m2-1)
+ (m2 - 2) 0 (e-u) - (m1 + m2) 0 (e -2u )
g1(u) = 0(e-3u)
g2(u) = m2 -
0(e-u)
89
Equation (5.3.27) can be thought of as that of a compound pendulum consisting of
two penduli fixed at right angles. Each pendulum is acted on by a different varying gravity force (g1 and g2) , and the system has a varying damping force acting upon it. The position of the system is described by the angle i = 2 a, located between the two penduli.
Label the two penduli 1 and 2 as in the above diagram. Then one wishes to find an exceptional trajectory of the system with a just starting off at time - at the angle 7 17/4, and just reaching the angle 17 /4 at time u = + oc .
Note that at time u= -co, the mass of the system lies entirely in pendulum 1. Then the mass transfers to pendulum 2 as time progresses, and then finally at time u=oo, the mass of the system lies entirely in pendulum 2. One can attempt to solve the equation in a way similar to the one Smith uses for equation (1.3.7) in [36 ] - this was described at the beginning of Section 1.4. One fixes the time u0 when g1 (u0) = g2 (u0) , and then the idea is to manipulate the initial conditions a0 = a (u0) and a = a' (u0) . However, the method is very long-winded 90
and involves several comparisons with other equations. The more elegant method used by Smith to solve equation (1.3.7) in [371 adapts to solve equation (5.3.26), and in Chapter 6 we show that equation (5.3.26) has a solution asymptotic to 7 P/8, 1T/8 as u tends to --,co respectively, if either (i1
m1 = m2 f
or
(ii) (m1 - 1)2 < 4m1 and (m2 - 1)2 <4m2, i.e. m1,m2 = 1, ... , 5.
(5.3.28)
As for Smith's equation we call (ii) the damping conditions. Note how conditions (i) and (ii) compare with the damping conditions described at the beginning of Section
1.4. Furthermore, in Chapter 6, we show that such a solution to equation (5.3.26 ) yields a smooth harmonic map between spheres. We therefore have
Theorem 5.3.8 Let F: Em - E be a homogeneous polynomial defining an isoparametric function of degree 4 on Sm-1, then 0 = VF I Sm-1 : Sm-1 Sm-1 is homotopic to a harmonic map, provided either of the above conditions (i) or (ii) are
-
satisfied .
Remark 5.3.9 There are several examples of inhomogeneous families of isoparametric hypersurfaces of degree 4 with m1 and m2 satisfying the damping conditions (see, for example , [16 1). Theorem 5.3.8 then gives us examples of harmonic maps equivariant with respect to inhomogeneous isoparametric functions. This illustrates that equivariance with respect to isoparametric functions is of some importance in the context of harmonic maps, i.e. we have generalized from the context of homogeneous hypersurfaces. Similar considerations apply when p = 6. Indeed one can again Remark 5.3.10 show that if ml and m2 satisfy (5.3.28), then the equivariant map described in Proposition 5.3.6 is homotopic to a harmonic map. However, we leave the details of this calculation to be described elsewhere.
We briefly consider an alternative way of describing the physical motion represented by equation (5.3.27). Transform the function a to a new function y which represents the centre of mass of the compound pendulum represented by equation
(5.3.27): 91
Referring to the diagram, we see that
e = tan (x/( 22-x)). Taking moments about pendulum 1 we obtain
(g1 + g2)x = 22
. g2
L e.
22 g X =
2
g1 +g2
So e = tan 1 (g2 / g1) , and
y = 2a + n/4 - tan-1 (g2 /gI )
(5.3.29)
Furthermore, a computation shows that
2glsin(2a+it/4)+2g2sin(2a-it/4) = 2(g1 +g2 )2sin'y g1 g2
g2 g1
2
g1
and a
z (y
+
2
)
+ g2
(g12 + g2) (g1g2 - g2g1) - 2 (g'1 + g2) (g1g2 - g2g,1) 2 (g12
+
822)2
Whence equation (5.3.27) becomes
y(u) = h(u)y'(u) + (g1(u)2 + g2(u) ) 2 siny(u) + p(u) 92
(5.3.30)
where, P =
2 (g1 + g) (g1g2 - g2g1) - (gi + g2) (g1g2 - g2g1 (g12
)
+ g22 ) 2
+
h (g1g2 - g2gi ) 2
g1
2
+ g2
This is the equation of a simple pendulum with varying damping, acted upon by a
variable gravity force, and with a force p of varying modulus acting tangentially to p(u) = 0. Furtherthe circular motion. One can check that lim p(u) = lim u--0o
u--00
more, if u is sufficiently large p(u) > 0 and p(u) = 0(e-3u). A smooth. harmonic map will be given by a solution of equation (5.3.30) which is asymptotic to 211, 0 as
u - - oo,co respectively.
93
6 On certain ordinary differential equations of the pendulum type 6.1
Existence of solutions
The equations that arise in the various examples of Chapter 1 and Chapter 5 behave in a similar way to equation (1.3.7) of Smith close to the asymptotic limit u = - Go. We adopt the methods used by Smith in [36 ] to investigate these equations. Equation (5.3.26) is dealt with separately using results of Hartman [211. These results were used by Smith to solve equation (1.3.7) in [37 ]. We use this method to solve both equation (1.3.7) and equation (5.3.26). First of all we list the equations to be considered, illustrating the various qualitative features of each one.
(i)
a"(u) +
((q-2)e-u
u
e
+e -u 1
4-
e
u
1
+e
-u
- (p-2)eu) a' (u)
sin a (u)cos a (u) (a1 eu - a2 a- u) = 0
where u E(-oo,oo), a(-oo) = 0 and a(-) = 1T/2. This equation arises with the Smith construction for harmonic maps from sphere to
sphere. The constants a1 and a2 are given by al = k (k + p - 2) and a2 = 1(1 +q -2), where k is the degree of a homogeneous polynomial map g1: SP-1 - Sr-1 and 1 is the degree of a homogeneous polynomial map g2 : Sq-1 -- Ss-1 (c.f. Corollary 1.1.6 and Section 1.3 ), However, in Chapter 9 we will consider deformations of metrics, and then the special relationship between the values of the gravity and damping need no longer hold. We will therefore suppose simply that a1 and a2 are positive numbers. Equation (6. 1. 1) is that of a pendulum with variable damping acted upon by a force of variable gravity as was described in Section 1.4. Write the equation in the form
a"(u) = h(u) a'(u) + g(u) sina(u) cos a(u) 94
,
(6.1.2)
where, near u = - oo ,
h(u) _ - (q-2) + 0 (e2u) and
g(u) = +
0(e2u) .
a2
By IT (f (u) ) near u = - co", for some smooth non-zero function f, we mean that 0(f(u)), f (u) is bounded and positive for u near - o . Also near u = +
h(u) = (p-2) -
0(e-2u)
g(u) = - a 1
0(e-2u)
and
a" (u)
+
e
u
+
.
(peu + (q-2) (eu + e-u))a' (u) -u +e 1
sinh a(u) cosh a(u) u -u e +e
al -u e +e u
a2
+
e
1
J=0
u
(6.1.3)
where u e (-oo, o) and a (- oo) = 0 .
This is the hyperbolic analogue of case (i) (c.f. Section 1.6). Here we assume that a1 > 0, and a2 > 0. Write this equation in the form
a"(u) = h(u) a'(u) + g(u) sinh a(u) cosh. a(u)
(6.1.4)
,
where, near u = - co,
h(u) = - (q-2) -
O(e2u )
and
g(u) =
a2
+
0(e2u)
.
a"(u) + (m-2) a' (u) - (m- 1) sina(u) cosa(u) = 0
(6.1.5)
,
a"(u) + (m-3) a' (u) - ((m-2)- k2e2u) sin a (u) cos a (u) = 0
,
(6.1.6)
where u c- (-so, o) and a (-so) = 0. Both these equations arise in the construction of 95
harmonic maps from Euclidean space to sphere. The first has been well studied (see, for example , [23 J), representing a pendulum with constant damping acted upon by a force of constant gravity. If we write the second of these equations in the form
a"(u) = h(u) a'(u) + g(u) since (u) cosa(u)
,
(6.1.7)
then, for u near - co ;
h(u) _ -(m-3) and
(rn-2) -
g(u) =
O(e2u)
.
(iv)
a"(u) + eu1+e -u
(eu + (m - 2) (eu + e-u )) a' (u)
(m-1)
-
eu ( e
u + e-u )
1
a"(u) + e
u
+ e -u
sin a(u) cosa(u) = 0
(2eu + (m-3) (eu + e-u )) a" (u)
m-1
sin a(u)cosa(u) e
u
(6.1.8)
e-u +
L
e
u
k
e
u
e-u J
+
f
(6.1.9)
where u E (-co, co) and a (-co) = 0. These equations arise in the construction of har-
monic maps from hyperbolic space to sphere. We can express them both in the form
ce"(u) = h(u)a'(u) + g(u)sina(u)cosa(u)
,
where, for u near - oo ;
h(u) = -b -
O(e2u )
and
g(u) = (m-1) - 0(e2u) . with b equal to (m - 2) in the first case, and (m - 3) in the second.
96
(6.1.10)
(v)
o (u)
+
u
leueu - m
u (1 +eu)e 2 + e2u
ml(e +e u)
+
[ 2(l+eu) eu +
m2 1 +2eu
- m 2 (1+2eu)
- m sin(a(u) + Tr/8)cos(a(u)+ TI/8) e-u +
1
m2
2
[e' -
+ 2eu
sin(a(u)- TT/8)cos(a(u)-TI/8) r 2e u
a-u + 2 + 2eu
a'(u)
J
eu (1 + eu )2
2e
-
u
(1 + 2e)
1
J
= 0. (6.1.11)
where u E (-co, oo) , a (-co) = 711/8 and a (co) = 11/8. This equation arises in Example
5.3.5 for harmonic maps from sphere to sphere. It is now no longer the equation of a simple pendulum, but a compound pendulum, consisting of two penduli fixed together, separated by an angle of TI/2 - the system having a variable damping and variable gravities acting on each pendulum distinctly.
Write this equation in the form
a"(u) = h(u) a'(u) + g1(u) sin (u(u) + TI/8)cos(a(u) + 11/8) + g2(u) sin(a(u) - iT/8)cos(a(u) -
11/8)
(6.1.12)
Then for u near - oo: 97
h (u) =-(m l- 1) - (ml- 2) 0(eu) + (m1 + m2) 0 (e2u) g1(u) =
m1 - 0 (eu )
and
g2(u) = 0(e3u )
and for u near -o ;
h(u) _ (m2-1) + (m2-2)0(e-u) - (ml+ m2)0(e-2u) g2(u) = m2 -
0(e-u )
and
g1(u) = 0(e-3u ) Note also that g1(u), g2(u) > 0, for all u r-(-co,co). Consider a general equation which includes (i) - (iv) as special cases:
a"(u) = h(u)a'(u) - g(u)f(a(u))f'(a(u))
,
(6.1.13)
where f is an analytic function with the property that f (x) - x = + 0 (x3) . Also h(u) and g(u) are smooth functions on (-co, o) and for u near
h(u) = -b+ 0(e 2u ) and
g(u) =
a + 0(e 2u
)
for some constants a and b. Now f '(x) = 1 + 0 (x2) , so that as x increases from 0, so does f(x); let x0 be the first x greater than 0 such that f (x0) f, (x0) = 0, or if no such x exists; let x0 be an arbitrary positive number. Choose a time u0 such that g(u) > 0 for all u< u0. By the uniqueness theorem for ordinary differential equations the above equation has a unique solution through (u0, a0, a0) which we denote by a(a0, a0) this exists for all time. The following parallels closely the methods of Smith [36 ]. Let A (ce ) = {ao s R ; a(a0,ao) decreases monotonically to zero in finite time as u decreases from u0 Then this set is bounded below by 0. Choose a, (a0) _ inf A(a0) 98
Lemma 6.1.1
a0 (x) is a well defined function on (0,x0).
Since A (a0) is bounded below by 0; it suffices to show that A (a0) is non-
proof
empty, i.e. given a0 (-- (0,x0); we wish to find an a such that Mao ,a0) decreases monotonically to 0 in finite time as u decreases from u0. As long as a '(v) > 0 on [u, u0 ] for some u < u0 , we see from equation (6.1.13 ) that, provided also a-, 0 ;
a"(v) < cI +c2a, (v) on [u, u0 ] for some positive constants c1 and c2. Integrating this inequality over [u,u0 ] yields
a'(u0) - a'(u) < c1(u0 - u) + c2(a0 - a(u))
< c1(u0-u), i.e. - a'(u) < - a0 + c1(u0-u) Choose a0 such that
- a0 + c1(u0 - u) <
-a0 < 0 . (u0-u)
Then, for all v, u .:S v < u0 :
- ac(v) < - a0 + cI(u0-v)< - a0 + cl(u0- u)
< - a0/(u0- u) . Integrating again over [u,u0 ] yields
- (a(u0) - a(u)) < - a0 . i.e. a(u) < 0, and a has decreased monotonically to 0 in finite time as required. Lemma 6.1.2
a'0 is strictly positive on (0, x0) .
Given a0 C (0,x0), suppose a0(a0) = 0, then at u0 ,
Proof aI I
(u0) = g(u0)f(a0)f' (a0)> 0 , 99
and so a"(u) > 0 for u sufficiently close to u0. Thus a(u) initially increases as u both increases and decreases from u0. Therefore there cannot be elements 0 in A(a0) arbitrarily close to zero, whence a' (a0) > 0. U af
Lemma 6.1.3
A(a0) is open for all a0 E (0,x0).
Proof Suppose a' E A(a0) and a(a0,a') arrives at 0 (as time travels backwards)
for the first time at time u. Then a'(u) > 0. For a' (u) > 0 and if a'(u) = 0 with a (u) 0, it follows from the uniqueness theorem for ordinary differential equations that a a 0. Hence a decreases through 0 with positive derivative. Also a'(v)> 0 for u < v < u0, since if a'(v) = 0, then a"(v) > 0 and a would not be monotonic. Again from the proof of Lemma 6.1.2 we see that a ' (u0) > 0. Hence c e' > 0 on [u - e , u0 ] for some e > 0, and so any function which is close enough to a in the C1 norm will decrease monotonically to 0 on this interval in backward time. Therefore points near a0 are in A (a0) and A (a0) is open.
For all a0 E(0,x0), a(a0,a0 (a0)) is strictly decreasing as u decreases from u0 and is asymptotic to 0 as u - - cc . Lemma 6.1.4
Proof We show that a' > 0 on (-eo,u0 ], and that 0 < a on (-co, u0 ]. For this will show that a decreases to some asymptotic value a , with 0 < a-00 < x0. But then as u - - eo, a (u) , a,t(u) -- 0, so equation (6.1. 13) -00 implies that the only possible value I
for a-00 is 0. Assume that one of these two conditions is violated. One must go wrong first, for if both are violated simultaneously, the uniqueness theorem implies that we have the trivial solution. On the other hand neither goes wrong at u0 from Lemma 6.1.2. Suppose a'(u) = 0 for u < u0 (for the first time), and 0 < a(u) < x0, then a"(u)> 0 and a increases as time decreases from u. But, by the definition of a0 , there are functions arbitrarily close to a on [u- e, u0 ] which are strictly decreasing or go past 0 on this interval - a contradiction.
Suppose now that a(u) = 0, and a' > 0 on [u,u0 ]. Then a (a0) E A(a0) - this is not possible since A(a0) is open and a' (a0) = inf A (a0) . We now demonstrate a local uniqueness result. 100
Lemma 6.1.5 Provided that u0 is chosen sufficiently close to - oo that h(u) < 0 for all u < u0, and a0 is chosen sufficiently close to 0 that if 0 < x < y < a0 , then f(x) f '(x) < f(y) f '(y) (i.e. ff ' is an increasing function), then a0 (a0) is the unique initial derivative for which we get a solution of the desired form in backward time. Specifically, if 0 < P < a (a0) then the associated solution a(a0, PO ) must eventually start to increase before reaching 0, and if p0 > ap (a0), then
Remark 6.1.6 The conditions of the above Lemma are always satisfied for any u0 and a0 in the case of equation (6.1.3), thus we have a global uniqueness result in that case. Proof (of Lemma 6.1.5)
Let
' < ao (a0) and write al = a(a0,a (a0)) and
a2 = a(a00). Then a1 (u0) = a2(u0) and al (u0) > a2(u0). Suppose that a'l(u0) - a2(u0) = e > 0. Define the function e(u) = (a1 - a2(u). Thene(u0) =E, and 0 = al (u0 ) e' (u)
- a 2 (u0
h(u0)(al (u0) - a2 (u0)) + g(u0)(f(al)f(al)- f(a2)f'(a2)) (u0) h(u0)(a,1 (u0) - a2 (u0)) < 0 Suppose there exists a u < u0, with e '(u) = 0
.
Then
h(u) al (u) + g(u) f (al(u)) f '(a1(u)) h(u)a2(u) + g(u)f(a2(u))f' (a2(u))
i.e. h(u)(a1(u) - a2(u)) = g(u)(f(a2 (u))f' (a2(u))-f(al(u))f'(a1(u)))
But al (u) > a 2(u) and al(u) < a2(u), so that the left hand side of this expression is less than 0, and the right hand side is greater than 0 as ff ' is an increasing function - a contradiction; so e '(u) < 0 for all u < u0, which implies that e(u) > e for all u < u0 . But lim a (u) = 0 - lira a' (u) > E E. Thus a'2 has become U- 00 u- _ 00 2 negative in finite time, and a2 starts to increase (for decreasing time) before reaching 0. 101
> a'0 (a 0 ) , then clearly a -=
If on the other hand p
0
g,
0
) cannot turn be-
fore reaching 0, since then the trajectories of al = a(a0,a (a0)) and a2 in phase space R2 (the curves u -- (a(u), a'(u)) C R2) would cross - an impossibility. Hence the only possibility is that QD C A(a0) or a2 is asymptotic to 0. But the above arguments show that if a2 is asymptotic to 0, then a1 must turn in finite time, i.e. a' (a0) C A(a0) - a contradiction. Hence RI C A (a0). Corollary 6.1.7
If the conditions of Lemma 6.1.5 are satisfied then a (a0) is a
continuous function of a0. Proof Suppose we have a sequence a0n -- a0 but a0 (a0 ) < a0 (a0) - e for some
e > 0. Let go = a0 (a0) - e/2. Then from Lemma 6.1.5 we see that (i)
a (an ,)3 ) decreases monotonically to 0 in finite time for all n (since
a0(a0)<00),and (ii) eventually begins to increase (as time decreases) before reaching 0. Letting an -- a0 this yields a contradiction. El 0
Corollary 6.1.8
a'(a0) --0
as
a0 - 0
In case (i) the problem is to find a solution asymptotic to 11/2 as u -- 00 . In his Thesis Smith's method was to derive the unique velocity a (a0) such that
a(a0, a (a0)) is asymptotic to 0 as u -- - oo as above, and similarly to derive the unique velocity a U + (a0) such that a (a0, a' + (a0)) is asymptotic to TT/2 as u -- ao. The damping conditions ensure (a0) and a'+ (a0) are bounded away from CIO
0 as a0 tends to TI/2, 0 respectively. Then Corollary 6.1.7 and Corollary 6.1.8 show there exists an a0 with a0 (a0) = a (a0) yielding the required solution. Later, in [34 ], Smith gave a more sophisticated method of solving case (i) by applying results of Hartman [21 ]. This method can be applied to solve equation (6.1.11) of case (v) and we use it now to solve both case (i) and case (v). Suppose we are given an equation of the form n
a = F(u,a,a') where F (u, a, a') is a continuous function on the set E (p,R) = { (u, a a') ; 102
(6.1.14)
0 < u < p, I a I < R, a' arbitrary] . Theorem 6.1.9 [ 211, [ 31 ] Suppose F is as above, and (i) F(u,R,O) > 0, F(u, -R, 0) < 0, for O< u< p, and (ii) IF I < '' ( Ian I) where q5(s) , 0 < s < p is a positive continuous function satisfying = oo. Let a , a be such that Ia , Ia < R. Then equation r°° s ds 0
0
ch(s)
p
0
I
pI
(6.1.14) has at least one solution in E(p,R) satisfying a(0) = a0, a(p) = a . Furp thermore, la I < C, where C is a constant depending only on ch, R and p. Write equation (6.1.1) in the form
a" = F(u,a,a)
,
then IF I< C0(1+ Ian I) for some constant C0 and F(u,0,0) = F(u,11/2,0) = 0. Adjusting the range of a (to [0,17/2 ] instead of [-R,R ]) and the time scale (to [-T, T ] instead of [ 0, p ]) , we can apply Theorem 6.1.9 to yield
Given T > 0 there is a solution aT of equation (6.1.1) satisfying
Lemma 6.1.10
aT (-T) = 0
,
aT(T) = P/2
,
0 < aT(u) < Ti/2 and IaT I < C on (-T, T) ,
(6.1.15)
with C depending only on C0 .
It suffices to show that 0 < aT(u) < IT/2 on (-T, T). We know from Theorem 6.1.9 that O< aT(u) < 17/2 on (-T, T). If u0 E (-T,T) has aT(u0) = 0, in order that aT remains in [0,11/2 ] (before or after u0); aT (u0) = 0. But equation 6. 1. 1 now implies aT (u0) = 0. The uniqueness theorem then implies that aT = 0 a contradiction. We now consider a sequence T = T(n) -- co. From standard equicontinuity arguconverges in C2 on compact ments (c.f. Hartman [21 ], Chapter 1, Section 2) aT(n) sets to a solution a0 of equation (6.1.1) with 0 < a0 < 11/2, and with Ia l bounded. Proof
Lemma 6.1. 11 Provided the damping conditions (1.4.1) hold, the solution a0 of equation (6. 1. 1) described above is non-trivial. Proof Suppose a0(u) = 0 for all u E R. Writing equation (6.1.1) in the form
103
a"(u) = h(u) a'(u) + g(u) sin a(u)cos a(u)
,
we see that there is a solution a = aT(n) of equation (6.1.1) satisfying conditions (6. 1. 15 ) which is very small out to a large value of u, say u1 < T(n). Then g(u1)
is close to -al. Choose a1 near a1 so that al + g sin acos a/a < 0 on a suitable interval [u0,u1 ] and also that the equation 01,=(p-2)fl'-a10,
has non-real roots and period less than u1 - u0. Let p have the same initial data as a at u0 , and let w = a 'fl - a P'. Then
w' = a/3 [h (u) a'/a - (p-2) f3'/p) + (al- + g(u) sin acos a/a) ]
w(u0) = 0
and
w'(u0) < 0
.
Provided a'/a < /3 '//3 (i.e. w < 0) with a,/3 > 0 we have w ' < 0. Also as long as w' < 0 we have w < 0. Now look at the first zero of p on [u0,u1 ]. We must have f3'a > 0 here since w < 0 - this is impossible if a> 0 on [u0,u1 ]. Hence a has a zero on this interval, contradicting (6.1.15). Therefore a0 = 0 is impossible. Similarly a0 = 11/2 is impossible. Lemma 6.1.12 If a is a non-trivial solution of equation (6.1.1) with 0 < a < 11/2 and I a' I < C, then a'(u) > 0 whenever lu I is sufficiently large, and a has the asymptotic limits 0, 13/2 as u -- -co, cc respectively. Proof Suppose u0 is close to + /3 and p-2 > 0, then we can assume h(u) > 0 and
g(u) < 0 for all u > u0. Then if a'(u0) < 0 a"(u0) < 0. Thus there exists a subsequent time u1 > u0 such that a '(uI) = - cc < 0. Then a" remains negative as long as a' remains negative, so that a' < - E until a has decreased to 0 and become negative - a contradiction. Hence a' (u0) > 0. Suppose p-2 = 0. Then if a'(u) < 0 (u large), a" must become negative before a' changes sign. Hence a' < 0 for all subsequent u. In fact, since a' is bounded, it follows that a" must become negative and remain so at least until a decreases to some small value. However, using the same comparison which showed that a was non-trivial leads to a contradiction. 104
Thus a'(u) > 0 for u sufficiently large, and a has an asymptotic limit < 11/2. The same argument used in the proof of Lemma 6.1.4 now shows the limit must in fact equal 11/2. The case when u is close to -co is similar. We now study case (v). Consider equation (6.1.12); express this in the form a = F(u,a,a') II
Then IF I < C0 (1 + I a' I) for some C0 and
F(u, 711/8,0) _ - g2(u)/2 < 0 F(u, 11/8,0) =
g1(u)/2 > 0
Applying Theorem 6.1.9, adjusting the ranges of a and time (remembering that 7 11/8 corresponds to -R and 11/8 corresponds to R), we have Lemma 6.1.13
Given T > 0, there is a solution aT of equation (6.1.11) with
aT(-T) = 7 P/8, aT(T) = 11/8
,
7 11/8 > aT(u) > 11/8 and I aT I < C on (-T,T)
(6.1.16)
with C depending only on CO. Proof It suffices to show that 7 7r/8 >a T(u) > 7r/8 on (- T , T) . We know from Theorem 6.1.9 that 77r/8>aT(u)>Tr/8 on(-T,T). Suppose u0E(-T,T)issuch that a T( u0 ) = 7 7r/8; in order
remains in [ 7 rr/8 , 7r/8 j ( before and after u0) we must have a'T(u0) = 0. Then that 0T equation (6. 1. 11) implies that a'T(u0) _ -g2(u0)/2 <0 , and a decreases past 7r/8 as u increases from u0 - a contradiction. Similarly we cannot have aT(u0) = 7r/8. As before we consider a sequence T = T (n) -. cc, and by equicontinuity arguments converges in C2 on compact sets to a solution a0 of equation (6.1.11) with 0T(n) 7 11/8 > a0 y 11/8 with a0 bounded. Clearly a0 cannot be constant since sin(2a -n'4) and sin (2a +n/4) cannot be 0 simultaneously.
Lemma 6.1.14 Let a be the non-trivial solution of equation (6.1.12) described above. Then provided the damping conditions (5.3.28) hold; as u -ao,a0(u) --11/8
and as u -. - cc, a0(u) -- 711/8. Proof Unlike case ( i), overdamping in this case could cause the solution to be asymptotic to 11/2. 105
Consider the case when u - co . We claim there are three possibilities: (a)
lim
a 0(u) = 5 11/8
(b)
lim
a (u) =
u -. u
-. W
17/8
0
(c) 2(a - 11/8) continually oscillates about the downward vertical TT. 0 Of course (b) and (c) could both happen. Case (c) could conceivably occur if lim h(u) = 0(m 2= 1). We will show that the damping conditions ensure that (b) u -00 is the only possible case.
Make the substitution y0 = 2 a0 - 11/4, then equation (6. 1.12) becomes
yo(u) =
+g1(u)cosy(u) + g2 siny0(u)
(6.1.17)
where 317/2 > y0 > 0. Suppose 3 17/2 > y0(u) > 17 for all u > u0, for some large u0. Then y'(u) > O for u > u0 , since if yo(u) < O ,together with the fact that sin y0 < 0 and co s -yo .5 0 ,
equation (6.1. 17) shows that y0 < 0 at least until y0 has passed through 11. But if y (u) > 0 ; lim y0 (u) exists and lies between 311/2 and Tl ; equation (6.1.7) shows U-00 this is impossible. Hence y0 must pass through IT and continue to do so every time y0 enters the range (17, 317/2 ).
Similarly if 17/2 > y0 (u) > 0 for all u > u0 for some large u0. Then y (u) < 0 for u > u0, since if yo (u) > 0 ; equation (6.1.17) shows that yo > 0 at least until y0 passes through 17/2. But if y0 (u) < 0 for u > u0; lim y0(u) exists and equaU -- OD tion (6.1.17) shows that this limit must equal 0. If on the other hand Tl > y0 (u) > TT/2 for all u > u0 for some large u0. Then since
g1(u) -. 0 as u - , the only possibility is that lim
y0(u) = 17.
For, if y > 0
u-°° for u > u0; y0 cannot become negative until y0 has become negative. And y0 must become increasingly close to 11 for this to occur. Inspection of equation (6.1. 17) then shows y0 must remain close to T1, or pass outside of [ TT/2, 111. The only other possibility is that y0 continually oscillates about the downward vertical 17, showing that one of (a), (b) or (c) hold. If we assume either (a) or (c) holds, then there is a solution y = yT of equation (6. 1. 17) which passes close to 17 for large u < T. The idea is to compare y with the solution of the equation with constant gravity and constant damping, to show that y 106
cannot possibly equal 0 at time T. Let us briefly consider the pendulum equation.
p"(u) = M(3'(u) - msin/3(u)
(6.1.18)
,
where M and m are constant, and we measure the angle p in a clockwise direction from the downward vertical. Suppose M < 0 and m > 0, then the damping resists the motion, and a generic solution performs decreasing oscillations about the downward vertical. Indeed, comparison with the equation of a harmonic oscillator: W" (u) = M W, (u) - m w(u)
,
for small p, shows that p must oscillate continually about the downward vertical if M2 - 4 m < 0. If on the other hand M > 0 and m > 0, then the damping increases the acceleration in the direction of the motion. By making the substitution u - u, equation (6. 1. 18) becomes
p"(u) = - Mp'(u) - msinp(u)
,
so that a generic solution p performs decreasing oscillations in backward time, i.e. p performs increasing oscillations as u increases. In this case p is either asymptotic to the position of unstable equilibrium I?, or p eventually moves around the circle in a single direction. Now consider the equation
p"(u) = Mp'(u) - msinp(u) - E
,
(6.1.19)
where M,m,E are all constant greater than 0, and we assume E to be small. This represents the motion of a pendulum with constant gravity and damping and with a
force of constant modulus tangent to the circular trajectory. A particular solution of equation (6.1.19) (which is approximately linear for small p) is p(u) = sin 1(-E/m). And a generic solution performs decreasing oscillations about the equilibrium position p = sin-1 (-E/m) in backward time. So as u progresses a generic solution performs sin-1 (-E/m), and either tends to a position of unstable increasing oscillations about equilibrium at 1T + sin 1(-E/m), or eventually moves around the circle in a single direction.
107
sin-1 (-E/m) For a detailed account of equation (6.1.19) see [34, Ch.6, Sec.2 ]. Perform the substitution v = y - TI + A 0 in equation (6. 1. 17), where 00 = sin 1(6/m), and the constants in equation (6.1.19) have yet to be chosen. Then equation (6.1.17 ) becomes
v'"(u) = h(u) v'(u) - g1(u)cos (v - 00) - g2(u) sin (v - 00) .
(6.1.20)
Similarly put tj = p + 60 into equation (6.1.19) , which then becomes
r"(u) = Mr'(u)-msin(tf-90)-E
(6.1.21)
.
First of all assume that m2 = 1 so that lim h(u) = 0. Assume also that v passes u -00 through 0 for large u, say u0. Let u1 be the last time this occurs (since v ( T) = - TT + g0 this must exist). Let v and n have the same initial conditions at u1. Define the Wronskian w = v'71 - vtf '. Then w
=
v
- v tl
1)
[(hv'
= vt)
w(u1) =0 Also, at u1 , 108
v
,
- M !i) + n
W, (u1) = 0
(- g2 sin(v- 00) + m sin(t)-90) v
.
tf
)
]
w
11
= 11'
(v'1
-
T11)
= n' [(h-M),1' - g1cos(-00) - g2 sin (-9) +msin(-00) +e] We now make careful choices of the constants M,m and E . Remembering that u0
can be chosen arbitrarily large, we choose M to be small so that h(u) - M < 0, m so that - g2 (u) + m < 0 and E = g1(u0) , where u0 is chosen sufficiently large that g1(u) < g1(u0), for all u > u0, Then, since ,i'(u1) < 0, we must have w"(u? 0. Therefore w' becomes < 0 immediately after u1, Also, since (v "- r, ") (u1) > 0 ; v' - r' > 0 and v - r > 0 immediately after ul, That is 0 > v > rf . Now the function x - sin x/x is increasing on [-11, 0 ] , so x - sin x /(x +a) is increasing on [- 11, 0 ], i.e. x - sin (x-a)/x is increasing on [- 11+ a, a ] Thus sin (v - 00) v > sin (rl - 001 /n > 0
immediately after u1. Whence
sin(v- 00) g2
v
m
sin (n- 00)
< 0.
71
Now if v' /v < iI'/Iq with v',17' <0 and 0 > v > 11, then v' v > n continues, and w' continues to be < 0 at least until 71 ' becomes zero for the first time. This must occur before rf reaches - 11 + 00, since n is bounded above by 17/2 + d on the previous swing, for some small constant d, and a and M can be made arbitrarily small.
Then w < 0 here, so that v'71 < 0, i.e. v'> 0 since rq < 0. So v' has become zero. But then equation (6.1. 19) gives v
" _ - g1 cos (v - 00) - g2 sin (v - 00 )
> - e cos(v-60) - msin(v-00) >
0
,
109
provided 0 > v > -TT/2 - 280. So that v' then becomes positive. Now v 1 cannot change
sign before v ", but v remains positive at least until v has passed through 0 again. This contradicts the definition of u1. Suppose now that m2 > 1, and (c) occurs. Let v0 = y0 - n+ 00, so v0 satisfies equation (6. 1.20). Then a comparison of v0 with Tl similar to the above shows that either lira v = - TT+ 80 1 i.e. case (b) holds, or v passes through T1/2 + 0 or u- o0
0
0
0
0
- TT+ 00 in finite time (since rT performs increasing oscillations) - contradiction. There remains the possibility that v remains close to 0, but always remains < 0 for large u. In this case, comparison with the equation Q
= M/ -mo ,
exactly as for Lemma 6. 1.11, shows that v must subsequently pass through 0. Thus (a) and (c) are impossible, and case (b) holds. Similarly, when u - - , a0(u) -711/8. Remark 6.1. 15 If the damping conditions do not hold, then it is conceivable that the above methods of solving equation (6. 1.2) and equation (6. 1. 12) lead in the first case to a trivial solution and in the second case to a solution asymptotic to 517/8. In the
first case the graph of aT for large T has the form 11/2
0
u=T
u=-T And in the second case the graph has the form 7 T1/8 5 TT/8
11/8
u=T
u=-T It is clear that if a0 = lira
n-
110
aT(n) , then a0 = 0 in the first case and
a0 (u) = 5 TI/8 in the second case.
lira u -- 00
6.2 Asymptotic estimates In this section we prove the following.
Theorem 6.2.1 Let a be a solution of any of the equations of cases (ii) - (iv) of Section 6. 1, which we express in the form
a"(u) = h(u)a'(u) + g(u)f(a(u))f'(a(u)) If a = lim g (u) and -b = lim u-. -oo
u-- Ch (u) , a, b > 0 ;
a = 1(1 +b), and assume that
1
.
let 1 be the number such that
> 1. Then for u close to - oo ,
(i) a(u) = O(elu
(ii) a(u) = 0(elu and
(iii)a'(u) - la(u) = O(e(1 +2)u)
We demonstrate the theorem by proving a series of lemmas, dealing with each case separately. Again we follow closely the methods used by Smith in (36 ] for case (i).
Lemma 6.2.2
If a is a solution of equation (6.1.3 ), then eventually (u close to -oo)
a' (u) > (1 - 0 (e2u ))sinh a (u)
,
where 1 is the positive number satisfying a2 = 1(1 +q - 2). Proof Choose u1 < u0 such that h(u 1) and g(ul) are close to their asymptotic va-
lues of - (q - 2) and a2 respectively. For u < ul, let 1(u) be the solution near 1 of the equation
(1(u)2 - 1(1 +q - 2))/1(u) = h(u)
Then 1(u) = 1-0(e 2u)
.
(q- 2) - 0(e2u)
We wish to show a'(u) > 1(u) sinha(u).
Let p be the solution of the equation Q
' (v) = 1(u) sinh p(v)
(3 (u)
= a(u)
Then R is asymptotic to 0 as v - - oo. For v < u ; 111
(v) = 1(u)2 sinh #(v) cosh p(v)
= h(u)cosh p(v) p'(v) + a2 sinh p(v)cosh p(v)
< h(v)p'(v)+a2 sinhp(v)cosh p(v) .
(6.2.8)
Now, suppose a'(u) < p'(u), then equation (6.2.8) shows that a"(u) > p"(u). Let , a' (u2) = (3'(u2) or a" (u2) u2 < u be the first time for which either a(u2) = /(u2),a' _ p"(u2). But if ca"> p"and a' < p' on (u2,u] with a(u) = p(u), then a'(u2) < /'(u2) and a(u2) > p(u2). Hence the only possibility is a"(u2) = 0 "(u2). However, a(u2) > /(u2) implies sinh a cosh a > sinh p cosh p at u2. Equation (6.2.8) shows this is impossible. Hence u2 = co. Thus p' - a' is non-decreasing on (- eo,u ] - a contradiction, since both a'(u) and p'(u) tend to 0 as u -- - co. Lemma 6.2.3
Eventually (u close to - co)
a'(u) < (1 +0 (e2u)) sinh a(u)cosh a(u) Proof Choose u1 as in the previous lemma. For u < u1, let 1(u) be the solution near 1 of the equation, (1 (u) 2 - g(u))/1(u) =- (q - 2)
Then 1(u) =1 + O(e2u). We wish to show that a'(u) < 1(u) sinha(u)cosh a(u) for u < ul.
Let fl (v) be the solution of the equation p'(v) = 1(u)sinh p(v)cosh p(v)
p (u) = a(u) . Then p is asymptotic to 0 as v - - uo .
For v
_
1(u)2 sinh p(v) cosh p(v)
(1(u)2 - g(u)) p'(v) +g(u) sinh p (v) cosh p (v) 1(u)
(q- 2) p' (v) + g(u) sinh p(v) cosh p(v)
> h(v) $'(v) + g(u) sinhp(v)cosh p(v) . 112
(6.2.9)
Now suppose a'(u) > p'(u), then equation (6.2.9) shows that a"(u) < yields a contradiction as in the previous proof.
There exist positive constants bl and b2 such that for u sufficiently
Lemma 6.2.4 close to - co
(u). This
.
b1elu < sinha(u) <
b2elu
Proof We know eventually that
a'(u) > 1-(u)sinha(u)
,
where 1 (u) = 1 - 0(e2u). Then for v < u
aI(v)
>
1- (v) sinh a(v)
>
1
(u) sinh a(v)
so a lies below the solution of Q
{
(v) = 1- (u) sinh f3(v)
(u) = a (u)
(6.2.10)
,
giving sinh a(v) < sinh p (v) for v < U. The explicit solution of equation (6.2. 10) is given by
cosech p (v) + coth f3(v) = (cosech a(u) + coth a (u) ) exp (1 (u) (u - v))
.
Now the left-hand side is equal to sinh P(v)/(1 + cosh G(v) ). We can assume u is sufficiently small that cosh a(u) < 2 say. Then
sinh j9(v) = (1 + cosh p(v)) cu exp (1 (u) v)
< 3cu exp (1(u) v) where c
u
,
is a constant depending on u
Hence
sinh a(v) < du exp (1 (u) v)
(6.1.11)
113
for some constant du depending on u . Let f(u) = sinh a(u).
Then
f ' (u) = sinh a(u) a' (u)
l+ (u) sinh a(u) cosh2 a(u) = (1 + sinh2 a(u)) l+f < (1 +sinha(u)) 1+ f 1+f + 1+ f2 = <
where 1+(u) = 1 + O(e2u), and we assume that u is sufficiently small that sinh a(u) < 1,
From equation (6.2.11) we have that sinh2a(u) < b0eµlu
where b0 and p are constants with p > 1. Hence f '(u) < 1+f +b eplu 0 We estimate a solution of the equation b0eµly
g'(v) = 1+g(v) + g (u)
= f(u)
,
so that g(v) < f(v), for v < u. The solution of the homogeneous part is given by v
9 (v) = c exp (.O 1+(s) ds) c exp Iv exp (rv 0 (e 2 s) ds ) .0 .. 1v
> ce
since the integral is uniformly bounded, where we assume c is a negative constant. A particular solution of the inhomogeneous equation is g1 (v) = gH(v)
.rvexp(0
f0s 1+(x) dx)exp(p is) ds
(Note that in order that gI(u) = f(u) > 0, we must choose c above to be negative).
114
Then
g1(v) > g1i(v) f v e(p- 1)isds 0
c e 1v Hence gI(v) > dely , for some constant d. So f(v) > bl elv , for v < u. For the other half of the inequality, we have that f ' ( u ) = cosh a(u) a' (u)
> cosh a(u) 1- (u) sinha(u) > 1-(u) f (u)
If g is a solution of
g'(v) = 1-(v)g(v) g (u) = f(u) , then
c exp fvl
(x)dx
0
cexp
fv(1-0(e
x)) dx
c elv exp(f v- 0(e2x) dx) 0
= deiv where c and d are constants. Hence f(v) < g(v) < b2ely, for v < u
.
The equations of cases ( iii) and (iv) are very similar. We prove the results for equation (6. 1.9) and similar considerations apply to equations (6. 1.5) , (6.1.6) and
(6.1.8). Lemma 6.2.5
If a is a solution of equation (6.1.9), then eventually (u close to -so)
.aI (u) < (1 - 0 (e
2u
))sin a (u)
,
where 1 is defined as for Theorem 6.2. 1.
Proof Choose ul < u0 such that h(ul) and g(ul) are close to their asymptotic values of -b and (m - 3) respectively. For u < ul, let 1(u) be the solution near 1 of the equation, 115
(1(u)2 - g(u))/1(u) = -b .
Then 1(u) =1- Me
2u
). We wish to show that a'(u) < 1(u) Sin a(u)
Let fl (v) be the solution of the equation
p'(v) = 1(u) sinp(v) p (u)
= a(u) .
Then (3 is asymptotic to 0 as v -- - co . For v < u, /3
(v) = 1(u)2 sing (v) cos /3(v)
- 1(u)2-g(u) . cosp(v)p'(v)+g(u)sinp(v)cosp(v) 1(u)
= -b cosp(v)p'(v) + g(u)sinp(v)cosp(v) > h(v) p'(v) + g(u) sinp(v)cosp(v) .
(6.1.12)
Now suppose a' (u) >/3'(u), then equation (6.1. 12) shows that a"(u) < p (u). The proof now concludes as for Lemma 6.2.2. C
Lemma 6.2.6
Eventually (u close to - eo)
a' (u) > (1 - 0 (e 2u )) sin a (u) cos a (u)
, .
Proof Choose u1 as before. For u < u1, let 1(u) be the solution near 1 of the equation
(1(u)2
- g(u))/1(u) = h(u)
Then 1(u) = 1 - O(e2u) (there being two contributions to - O(e2u ), one from g(u) and the otherfromh(u)). We wish to show that a'(u) > 1 (u) sin a (u) cos a (u). Let p be the solution of the equation
p'(v) = 1(u)cosp(v)sinp(v) p (u)
= a(u) .
Then p is asymptotic to 0 as v -- - co
116
.
For v < u,
1(u)2(cos2p(v) - sin2p(v))cosp(v)sinp(v) 1(u)2cosp(v) sin (v)
(1(u)2- g(u)) p (v) + g(u)cosp(v)sinp(v) 1(u)
h(u)/3'(v) + g(u) cos p(v) sin p(v)
< h(v) 6'(v) + g(u)cosp(v) sine(v) .
(6.1.13)
Now suppose a' (u) < p' (u) , then equation (6:1.13) shows that a" > p "(u) . The proof concludes as before.
Lemma 6.2.7 There exist positive constants bl and b2 such that for u suff iciently close to - co ;
bielu < sina(u) < b2elu
.
Proof The proof is similar to that for Lemma 6.2.4, using the estimates for a' obtained in Lemma 6.2.5 and Lemma 6.2.6. This concludes the proof of Theorem 6.2.1.
6.3. Smoothness of certain equivariant harmonic maps Recall that a solution to the reduction equation gives a smooth harmonic map from M * to N* . In the case when this map extends to a map from M to N, we need to know this extension is smooth at points of M\M
Theorem 6.3.1 The harmonic maps which arise from solutions of the equations of cases ( i) - (v) all extend to smooth harmonic maps across M\M*, where M is the appropriate domain depending on which case is being considered.
In cases (i) and (iv), this theorem can most readily be derived from the following. Suppose M and N are compact, and let i : N - V be an isometric immersion of N into a Euclidean space V. The completion of the space of smooth maps 41: M - V with norm
ICI = (r (Ib (X) 12
+
Ivfi(x)I2)dx)a
M
is a Banach space, denoted by Li (M, V) . 117
The subset L1 (M, N)
` L 1(M , V) ; 4 (M) C N) i s a smooth manifold. The energy
integral (c.f. Remark 1.1.4) defines a function E: L2 (M, N) - R. There are extrema of E which are not C0 [ 10 1. However, Hildebrandt has shown that any extremal
of E which is in L fl CO is smooth. Since the Laplacian is the Euler-Lagrange operator associated1 to the energy integral E, it is clear that the maps constructed in cases (i) and ( v) are extrema of E. Also these maps lie in L fl CO , and hence 1 they are smooth. In cases (ii), ( iii) and (iv) the domain is no longer compact, and in general the energy is no longer finite. However, we can now appeal to the regularity theorem of Eells and Sampson [ 12 1, which states that any C2 - map between smooth manifolds which
is harmonic is smooth. It suffices therefore to show the maps are C2 at points of M\M* . We demonstrate this by proving a series of lemmas , using the estimates of the last section. We remark that we are again adapting the programme of Smith [36 [ which he used for the maps of case ( i) (Hildebrandt's result was not known at that
time). Using the notations of Chapter 4, let V = M\M* be the focal variety of the isoparametric function s: M - R corresponding to s = 0(or u = - 00) , and let W C N be the focal variety or isoparametric hypersurface of N onto which V is mapped under 0. For a point x e V we pick suitable C2 charts about x and 0 (x). These charts can be thought of as generalized Fermi coordinates (see for example [ 1 1) , and they will be chosen in such a way as to induce an equivariant map between subsets of Euclidean
spaces. Since V is a smooth submanifold (Theorem 2.2.2), there is a smooth chart (131, B1) about x in V , with B1 -- Rp(p = dim V). We can form a tube TE (B1)about B1, R1: which through each point of B1 consists of the union of geodesics subtended by a certain small sphere. More precisely, recalling the notation of Chapter 4, let Ms = s-1(s0), so V = M0. Let TI: M* V be the projection down normal geodesics. 0
E
Then T E(B1) =
by defining a map
th1
from E(B1)to the set
We can choose coordinates for TE (B1) c-
B
s c- [0,E)? C Rm. Indeed Ms is a sphere bundle over V (s 0), and we choose B1 sufficiently small such that the induced bundle over B1 is a product bundle : 118
Ms = B1 x s Sq-1 . This defines the diffeomorphism 01: T (B1)
E1 , giving the
desired chart (t) 1,E1). Note that is equivariant with respect to isoparametric 01 functions. Similarly we can construct a chart (rh2 , E2) about OW, where E2 = { (u,tv) ; u E Rr , v E Ss-1, t C [0, 6)1, . With respect to these charts the equivariant map 0 induces an equivariant map 0 : E 1 -- E2 of the form
t,(x,sy) = (g2(x), a(s)g1(y)) , where s [ 0, e) and a (0) = 0. Also g2 is a smooth map and g1 is a homogeneous polynomial map of degree 1. We must check that the map 0 between Euclidean spaces is C2 when a satisfies the estimates of Section 6.2 . Rewrite th as
th(x,y) = (g2(x), a(log lyl)gI(y/Iyl)) where x C 1Rp, y C- Rq (m = p +q), and we have changed the parametrization from s to u where eu = s. Then it is sufficient to check that the map H defined by
H(y) = a(log ly1)g1(y/Iy1) extends over ly I = 0 to a C2 map.
Write H as H(y) = R2 (y)g1(Y) where R (y) = a(log ly 1) 2 / ly 121. Furthermore write
R=rop , where p(y) = ly12 = v say, and r(v) = a(2 logv)2/v Lemma 6.3.2 For v near 0 , the first derivative of r is uniformly bounded whilst d2r /d v2 is at worst of 0 (1/v). Proof The derivative
dr/dv = 'a( a' - 1 a) /v 1+1 From Theorem 6.2.1 we see that a' - 1 a = 0 (e (1+2)u ) and a= 0 (elu ), hence a (c e' - 1 a) = 0 (e2 (1+1)u ) = 0 (v1+1) as required. The second derivative d2r/dv2
=
la') +(a' - 2(1+1)a) (a' -
la))/2v1+2
119
Now a' - 2(1+1) a= 0(elu)and a' -la=0(e(1+2)u ), hence (a' - 2(1+1)a)(a-1a) (e2(1+1 )u
0
) = 0(v11), as required. For the first term on the right hand side we
note that a
-bay+l(1+b)a+0(e(1+2)u)
_
Hence It
0(e2(1+1)u)
- lay) = a(-ba'- 1(1+b)a) +
= Ml +b)(-a' + la) + 0(e2(1+1)u = 0(e 2(1+1)u) 0(v1+1
=
.
)
as required.
Lemma 6.3.3 (a) As y - 0, 8R/8yi derivatives remain bounded. (b) Similarly for R a .
-0 (y=(y1, ... ,yq)
whilst all second
Proof (a) R(y) =rop(y), where p(y) = ly 12 , hence
8R/8yi = 2yi 8r/8v , 2
2
8yi yJ
r
4yiyi + 2
e 2
Cr aij
The conclusion now follows from Lemma 6.3.2 . i
(b) It suffices to show that R2 is bounded away from 0 as y -- 0. But by
Theorem 6.2.1(i) Rz = a(u)/1y11 > b1elu/1y11=b1. Lemma 6.3.4
H extends to a C2 function i
Proof H(y) = R2(y) g1(y), so that 8H
8yi
=
8R2 8yi
g
+ 1
Rae g1 8yi
is continuous at y = 0 by Lemma 6.3.3 . Also
120
RZ1
82H
cyi 8y.
+
82
8yi 8yi
g1
8Rz 8yi
8g1 8yi
+
8R-
8g1
8yi
8y
i
+R2
is again continuous at y = 0 by Lemma 6.3.3 and by the fact that g1 is homogeneous and hence g1( 0) = 0. Indeed the first three terms tend to 0 as y - 0 .
121
7 The general theory of harmonic morphisms General theory We were briefly introduced to harmonic morphisms in Section 1.3 . Now we will give 7. 1
a much more detailed treatment. Harmonic morphisms have been extensively studied by Fuglede and Ishihara [19,27], and we will now describe some of their results. Let 0: (M,g) -. (N,h) be a map of Riemannian manifolds. If x c- M is such that d0 0, then TxM can be decomposed into rx = ker dox and Wx the orthogonal x complement with respect to g. The map 0 is called horizontally conformal if, for all x C- M where dox 0; d¢x I 4x - TO(x)N is conformal and surjective.
That is, for all X,Y C x h(d0 (X), do (Y)) = A2(x).g(X,Y), for some function A : M -. R, called the dilation of 0 .
Theorem 7.1.1 [19,271 A map 0: (M,g) -- (N,h) is a harmonic morphism if and only if 0 is harmonic and horizontally conformal.
Thus, for each x e M, dox either has maximal rank or dox = 0. Denote the critical set of 0 by C0 = { x c- M; dOx = 0lt . The function A2: M -- E is smooth and C0 _ (A2)-1(0). The critical set C0 is a polar set; see [19 ]. Roughly speaking this means that codim C > 2 in M. We remark also that 0 is an open mapping. 0
Remark 7.1.2 If C0 is non-empty; one can easily see from the condition that do = 0 on C0 , that each connected component of C0 is mapped under 0 to a single x point of N. For maps into En, where En has the standard metric < , > , the condition of harmonicity and horizontal conformality become the following.
If 0: (M, g) - (ER" , <, >), then 0 is a harmonic morphism if and only if the components y119 ... On of 0 are harmonic on M, and their gradients are mutually orthogonal and of equal length A (x) at each point x C- M: Theorem 7.1.3
122
[19 ]
(7.1.1) g(VOk,vO1) _ 26kl, for all k,1=1, ... n. Example 7.1.4 Any Riemannian submersion is horizontally conformal with A2 = 1. By a result of Eells and Sampson [12 ], a Riemannian submersion is harmonic if and only if the fibres are minimal submanifolds. Example 7.1.5 Define 0: Rn\{ 01 Sn-1 by fi(x) = x/ Ix I for all x ERnM 01, . Then 0 is a harmonic morphism with dilation X given by 2 (x) = 1 / Ix 12 Example 7.1.6
Let 0 be multiplication of real, complex, quaternionic or Cayley
numbers; 0: E2n -. En O(u,v) = uv, where n = 1,2,4 or 8 respectively. Then 0 is a harmonic morphism from Theorem 7.1.3 with dilation A given by X2(x) = Ix 12.
The critical set C0 consists of the point 0 in En Example 7.1.7 Let 01 : M -. N and 02: N - P be harmonic morphisms with dilations A and A2 respectively. Then, from equation (1. 1.3) , ¢2 o 01 :M P is a 1 harmonic morphism with dilation A given by A2 (x) = A 2 (¢1 (x)) . Ai (x) , for all x M. In view of Example 7.1.4 we might expect to find conditions when the fibres of a harmonic morphism are minimal submanifolds. This is what we consider in the next theorem.
Theorem 7.1.8 Let 0: (M,g) - (N,h) be a submersion which is a harmonic morphism. Then (setting r_ = dim N) (a) If n = 2, the fibres are minimal submanifolds (b) if n- 2, the following properties are equivalent: (i) the fibres are minimal submanifolds ; (ii) v A2 is vertical, where A is the dilation of 0 ; ( iii) the horizontal distribution has mean curvature V A2 /2 A2
Proof First of all we have e (0) = nx2/2, consequently the stress-energy tensor S0 is given by
S0 - Zn'A2 .g -
0
s
h.
(7.1.2)
Take a point x E M and an orthonormal frame field (Xa ) a=1,...,m near x with
X1, .... Xn horizontal and Xn+l , ... , Xm vertical. Use the following ranges of indices: 123
I < a,b, ... <m; 1 < i,j, ... < n; n+1 < r,s,... < m . The map 0 is harmonic so v * S0 = 0 ; therefore, summing over repeated indices,
0 = (vx S0) (Xb,Xa) a
2nXb(A2) - (Xa(0*h(Xb,Xa)) - 0*h(VX Xb,Xa a
-0h(Xb'VXXa).
(7.1.3)
a
Since the frame is orthonormal 0 = Xig(Xj,Xi)
g (v. Xj,X.) + g(Xj,vv1Xi) i
= g (%vXx.,Xi) + g (X.,
vX Xi) i
(1/A2)(0*h(VX X.,X.) + 0*h(Xj,vX Xi)) '.
(7.1.4)
where W'denotes horizontal projection. Choose Xb = Xj . then Xa (0 h (Xj , Xa)) = Xj (0), and using equation (7.1.4) ; equation (7.1.3) becomes
2(n-2)X.02) + 0*h(v XX ) + 0* h(X.,vXrxr j Xr r
0= =
=
'(n-2)X.(A2) + X2g(Xj,V
2
(n - 2 )Xj (A
Xr)
2) - A 2 (m - n) (mean curvature of fibre in the Xj direction)
(7.1.5)
Thus we have proved (a), and (i) if and only if (ii) in (b). Now choose Xb = Xr. Equation (7.1.3) becomes
0 = JnXr(A2) - 0 *h (VXiXr,X. ) = 21 nXr0,2)
_
JnXr(A2)
- A2g(yvXiXr,Xi +
A2g(Xl Xi.Xr)
(7.1.6)
a nXr (A2) - A2 n (mean curvature of horizontal distribution in Xr direction). 124
We now choose Xr to be in the direction of the vertical projection of vat, and we obtain ( ii) if and only if ( iii) in(b). To what extent can we regard Theorem 7. 1.8 as being valid for arbitrary harmonic morphisms (i.e. allowing C0 to be non-empty) ?In fact we can remove the "submersion "condition in the statement of Theorem 7.1.8 on account of the following recent result of Fuglede.
Theorem 7.1.9 [20 1 If VA2 is vertical, where A: M - R is the dilation of a harmonic morphism 0 : (M , g) -- (N, h) , then C0 is the empty set. 7.2
Examples and non-examples of harmonic morphisms
Example 7.2.1 The Hopf maps defined as follows are all harmonic morphisms. Let F denote either the real, complex, quaternionic or Cayley numbers, and let n = dim F. Define 0: R2n -- Rn+l by
0 (x,y) _ (Ix I2 - Iy 12, 2xy)
(7.2.1)
for all x,y E F. Then 0 is a harmonic morphism with dilation A given by A2((x,y)) _ I (x,y) 12, for all x,y e F. Also 10(z) 12 = Iz 14 , for z c R2n, so the fibres are all compact, and being in R2n cannot be minimal. Thus the condition (b) of Theorem
7.1.8 is indeed restrictive. Example 7.2.2 Isoparametric families of hypersurfaces with 2 distinct principal curvatures give rise to harmonic morphisms. Let M be Rm, and express z C Rm in the form z = (x, s. y) , where x c- Rp, y C Sq, p +q = m and s c [0,co). Then the level surfaces s = constant form an isoparametric family (c.f. Example 2.1.4). The map 0: Rm Rp ; 0(z) = x defines a harmonic Riemannian submersion.
Let M be Sm-l, and express z c Sm-1 as z=(coss.x, sins.y) where x C SP-1, y C Sq-1, p+q = m and s C [0,11/2 ]. The level surfaces s = constant form the isoparametric family of Example 2.1.5. Define 0: Sm \ S q -. S p by O (z) = x. Then 0 is a harmonic morphism with dilation A given by A2(z) = 1/cos2s. Here VA2 is vertical and the fibres are minimal. Let M be Hm-l , and express each z E Hm-1 as z = (coshs.x,sinhs.y), where
x e HP-1, y c Sq-1, p+q = m and s c [0,-). The level surfaces s = constant form 125
the isoparametric family of Example 2.1.6 . Define 0 : Hm-1 -- Hp-1 by 0(z) = x. Then 0 is a harmonic morphism with dilation given by A2(z) = 1/cosh2s. Also the Sq-1 defined by 0 (z) = y is a harmonic morphism with map 0 : Hm-I \ Hp-1 dilation p given byp2 (z) = 1/sinh2sS. All the above harmonic morphisms are applications of Proposition 2.2.5 Example 7.2.3 Isoparametric families of hypersurfaces on a space form M with 1 distinct principal curvature give rise to the following harmonic morphisms. Let Mc be a particular hypersurface in the family, and let V be the focal variety; then M\V is connected. Define 0: M\V - Mc to be projection down the normal geodesics. Then 0 is a harmonic morphism with geodesics as fibres. Conversely we have the following.
Example 7.2.4 Suppose 0: (M,g) - (N,h), dim M = dim N + 1, is a harmonic morphism, and V A2 is vertical, where A is the dilation of 0 . Then from Theorem 7.1.8, the integral curves of VA2 are geodesics, and so IVA2 12 is a function of A2 (one can see this by reversing the proof of Lemma 2. 1. 1). The horizontal distribution is integrable in this case, the integral submanifolds being the level surfaces of A2. Thus property (b) ( iii) of Theorem 7.1.8 implies the mean curvature of these level surfaces is a function of A2 - otherwise said, A2 is a generalized isoparametric function.
Example 7.2.5 Consider the tangent bundle (TM,G) of some Riemannian manifold (M,g), where G is the Sasaki metric of Example 2.4.7 . Let T1M be the unit sphere bundle:
T1M = {(x,v) C TM; Iv 12 = 11,
,
endowed with the metric induced from G. Then, as a consequence of Example 2.4.7
and Example 7.2.3, 0: TM\(zero section) -. T1M; 0(x,v) _ (x, v/ Iv I) is a harmonic morphism. The dilation A of 0 is given by A2(x,v) = 1/ Iv . Example 7.2.6 In Example 7.1.6 we saw that real, complex, quaternionic and Cayley multiplication are all examples of harmonic morphisms. Such multiplications are all orthogonal multiplications; i.e. a bilinear map 0: EP x Rq -. En such that I 0 (x, y) I = Ix 1. ly 1, for all x E EP, y C E q. Which orthogonal multiplications are 126
harmonic morphisms? It turns out that the multiplications of Example 7.1.6 are the only ones:
Theorem 7.2.7 If 0 : EP x Rq - Rn is an orthogonal multiplication, then 0 is a harmonic morphism if and only if p = q = n and n = 1, 2, 4 or 8 . Proof
Suppose 0 is a harmonic morphism . Since 0 is an orthogonal multiplication
10(x,y) I
Ix I. lyl
=
(7.2.2)
,
for all x E Rp and Y E Rq. Fix x0 E Rp and y0 a Rq, then 0y0 : Rp Rn given by 0y0(x) = 0(x,y0) and 0x0: Rq - Rn given by 0x0 (y) = 0(x0,y) are both linear. Therefore d(0
(u
)x
YO
0-
=
)
0
Y0
1
u
(7.2.3)
,
1
for all ul E EP, and has norm Id(0y0 )x0 (u1) I
=
lu I
1y01
I
(7.2.4)
Similarly Id(Ox0
I
=
lu21
Ix01
(7.2.5)
y0 (u2)
for all u2 E Rp . Now d 0(x0,y0)(v1,v2) = d(0 y0)x0(vI) + d(0 x0)y0(v2)
(7.2.6)
,
for all (vl , v2) E Rp x Rq. Thus ker dO(x
.Y) 0 0
= ker (dOy ) x 0
®
ker(d0x)
® { (vl,v2) E Rp x Rq ;
0y0
0
d(0
)
y0 x0
) = d(0
(v 1
From equations (7.2.4) and (7.2.5) we see that ker (df
)
x0 y0
)
(v
2
)?
.
(7.2.7)
=101, and ker(d0
YO x0
)
=
x0 y0
{0l, whenever x0 and y0 are non-zero, and so p,q> n. Henceforth assume x0.Y0 # 0. 127
From equation (7.2.7) ker d0 (x 0
_ {(v1,v2)C 1Rp x Rq; d(P ) x vi = -d(0x YO
,Y) 0
p
v2,
)
0
.
(7.2.8)
y0
Let L = d(0YO)x0 1Rp fl d(0x )y0 ]Rq . Since d0(xo,Y0) is surjective; dim L = 0
dim ker(dO)(
.Y 0
) = p +q-n. Let P: En -- L be the projection map, and define
= (ker(TTo 0
L YO
L
x0
))-L
.
y0
= (ker(T1o 0
x0
))1
= Lx- = {0; , then p = q and L has maximum dimension p. In which case, Ly = 1 and Lx = L. But then ) cannot be onto En unless p = d0(x If both Ly01 0
0'Y0
0
1
q = n, which is the required result. So without loss of generality, assume LY0 = fol. From equation (7.2.8), ker d 0(x0 9 YO) C Ly0 X LX Thus the horizontal subspace 0' (xO,yO) consists of
(i) the orthogonal complement of ker d 0
(ii) Ly
1
x
0
(x01Y0
) in Ly x Lx O and 0
0
1 0
1
Choose (u1,0) C Ly 1 x I. 0
, then
0
u1 dO(x0,y0 ) (u1,0) = d(0 Y0 ) x 0
and
I d(OY0 )x0 u1
12
=
u1
10
12
YO
lu112
and since I (u1, 0),12 = X2 (x0 , YO) =
lu1 12 ;
IY012
we conclude that the dilation A is given by
1y012
On the other hand, choose (ui,u2) as lying in that part of the horizontal space given by M. For u C- Ly0 , 0y0u c- L, so that 11 0 0yO (u) = 0yO (u), and 128
I ito 0
(u) I
y0
lul
=
Thus r 1o ¢ d((6
y0
)
y0 I
(u) I
10
=
yo
Iy01 .
is homothetic, and so
IL
y0
Lx0
=A
x0
A
(7.2.10)
XO
for some AXO E R, and some matrix
with AX * AX0 = identity. Similarly AX0
d (OX
)
L
0
for some A
tion (7.2.8)
y0
=A 0
y0
0
B , y0
e- R, and some matrix By0 with B * By0 = identity. From equay0
ker dO(XO0y0) = {(vl,v2) E Rp X Rq;
(vl) = - AyOByO(v2)1} AX0 Ax0
= j((-Ay0/Ax0) . AX0* BYO v2, v2 ) ; v2 E Ly0 }
Thus, if (ul,u2) E Ly0 X L XO, then (ul,u2) E (ker dO(XO'yO))
all v2ELy
1
.
if and only if, for
, 0
0 = <(-"YO /AX ). Ax * By0 v2, ul > + < v2,u2 > 0
0
_ (-Ay0Ax0 ). < v2, By0* Ax u1 > + 0
0
,
0
if and only if U2
(Ay0
/AX ). BYO * AX0 u1 . p
(7.2.11)
Thus, that part of the horizontal space given by (i) is spanned by { (AXOu 1
, Xy0 By0* AXOul) ; ul E
L[0
Now 129
I (Ax
u ,X 0
1
*
B YO
l
p
YO
Also, for all ul c Lx
2
Ax u ) 12 = (Ax 0
+ Ay2) lu112 0
,
0
d0(x0,Y0)(Ax0u1,Ay0By0*Ax0u1)
Ax 2 Ax 0 0
ul + AY 02 BY0 BY0 ; Ax0 ul
(Ax0 2 + AYO2) A x0 X0
1
and the square of the norm of this vector is equal to (Ax 2 + Ay 2) . Thus the dila0
tion A is given by (XX2
A2(xO,y0) _
+ Ay
2 3
0
.
0
0
But d(O 0
)x
Ax2 Ax
(Ax ul) = 0
0
0
0
ul
and equation (7.2.4) implies Ax4 lu112 = Ax2 lu112 0
ly012
0
that is Ax (y0)2 =
1y012
Similarly Ay (x0)2 = lx012
,
0
and so A2(xO,yO) = (x012 + Iy012
(7.2.12)
On account of equation (7.2.9) we see that 0 is horizontally conformal only when Lyo = Lxo = {0 j , i.e. only when p =q = n. It is well known [101 that in this case 0 must be one of the standard multiplications of Example 7.1.6.
Example 7.2.8 Example 7.1.5 is a special case of the following. Let Vn, k be the space of k-frames in ]Rn at 0, and 0n k the space of orthonormal k-frames in Rn at 0. The Gram-Schmidt orthonorialization process defines a natural deformation retract 0: Vn k -. 0n k. We might ask whether this retraction is a harmonic morphism. 130
Let Rn
denote Rn \O. Then Vn k can be regarded as an open subset of the Euclidean space RnX , , , x Rn#(k times), from where it inherits its metric. Similarly On k can be regarded as an open subset of S"-1X , , , x S n-1 (k times) with the induced metric. Let 1T: Sn-1 be the natural retraction TI(x) = x/ Ix I . The Gram-Schmidt orthonormalization process can be described as follows. Let (v1' ... 'vk) E Vn k ' then ul = TI(v1) C Sn-1. There exists a natural proRn#
jection Tiv
:S
1 \ { + Ti (vl )1
S u 2 which maps down normal geodesics to the
equator S n-2 subtended by the two poles + 11(vl) . Let u2 = ITv (TI (v2)) . Define 1
the projection 11 f2:
Su-2
\ {+ u2)
Su-3
1
which maps down normal geodesics to
2
the equator Su2 3 subtended by the poles + u2. Let u3 = TIv2 (TTv ( TI(v3))) , and so on. 1
In this way we obtain the desired orthonormal basis (u1, R
uk) C On,k n*
V2
S
The fibre over a point (u1, ... , uk ) C 0n k consists of products of various open half-spaces, and hence is minimal in Vn,k' The tangent space to Vn,k at each point can be identified with the Euclidean space Rn x ... x Rn (k times), and the horizontal space to the fibre over (ul, ... , uk) C On k consists of k-tuples (wl, ... 'wk)C , Rnx ... x Rn such that < u1,w1 > =
0
< u1,w2 >
= 0
= = ......... = = 0 .
(7.2.13) 131
We can identify the tangent space to 0n,k at (u1, ... , uk) with a subspace of Euclidean space, and the map 0* sends (w , ... , w ) ET V (v1,...,vk) n,k k (v1, ..., vk) 1
to
w2
1
(
1v1
,
1
w
Iv2 1
/
wk
vk.uk
I vk I
I Vk I
v2.u2
...
'
Iv2 1
1
w2
wk
v1. u1
V2.u2
vk.uk
'
Thus
1 0* (w1, ... , wk)
v)
(v
1'
.. ''
k
1
2
Iw1 12
= IV
1,
Iw
Iwk l2
12 1
Iv1.u1 I
2
Ivk. uk I 2
u1 1 2
and the dilation X of 0 is given by
X2 (vi, ...,vk) =
Iwk 12
Ivk. uk 12
/
(w12 +
... + wk)
(7.2.14) This is independent of the choice of (w1 , ... , wk) only in the following cases (i) k = 1; in which case we have 0: Rn Sn-1 as in Example 6.1.5 . (ii) k = n = 2; for then equation (7.2.13) implies that w2 = 0, and equation (7.2.14) shows that 0 is horizontally conformal. In case (ii); 0: GL(2) -- 0(2) . By choosing appropriate geodesics through (v1, ... , vk) one can see that 0 is harmonic, and so is a harmonic morphism (since dim 0(2) = 1 the map 0 is obviously horizontally conformal). Of course the above retraction may be a harmonic morphism with respect to some more natural metric on 0 n,k' such as the left invariant metric which arises from regarding On k as a homogeneous space.
Remark 7.2.9 Although the map 0: Vn,k - a,kof Example 7.2.8 is not in general a harmonic morphism, it is harmonic and has minimal fibres. In fact it is an example of a more general kind of map than a harmonic morphism, where 0 h(0:(M,g) (N,h)) has several distinct eigenvalues instead of just one (c.f. Example 3.3.2) (in Example 132
7.2.8 0*h has min (k,n - 1) distinct eigenvalues). In the next section we shall briefly consider maps with 0 7.3
*h
having more than one distinct eigenvalue.
Maps 0: (M,g) -- (N,h) where 0
*h
has two distinct non-zero eigenvalues.
Suppose 0: (M,g) - (N,h) is a submersion almost everywhere, with 0*h having at most two distinct eigenvalues. Denote these by A 1 and X2 , and let U be an open set in M on which Al and A2 are non-zero. Let Si denote the eigenspace of A. , with r. = dim S, , i = 1,2 . Choose a frame field X1, ... , Xm on U such that X1, .... Xp span S1,Xp+1, ... , Xn span S2 and Xn+1, ... , Xm are vertical. Choose the following ranges of indices:
1 < a,b, .. <m; 1 < i,j, .. < p; p+1 < r,s,..
h = (r1A1 + r2A2)/2
(7.3.1)
and the stress-energy tensor S0 is given by (7.3.2)
S0 = a(r1A1 + r2A2).g - 0*h
Suppose 0 is harmonic ; then calculations similar to those in the proof of Theorem 7.1.8 establish the following two equations (summing over repeated indices)
(rl - 2)dAl +r2dA2)(X+ (Al - A2)g(Xj,V Xr) + Alg(X.,V
Xa)= 0
,
(7.3.3) 2(r1dA1 +(r2 - 2)dA2)(Xs) +(A2 - A1)g(Xs,V X.) +A2g(Xs,V Xa) = 0 X. X i a (7.3.4)
We therefore have
Theorem 7.3.1 Let 0: (M,g) -- (N,h) be defined as above. If 0 is harmonic and the mean curvature of each eigenspace S i is vertical, i = 1,2 , then the fibre is minimal if and only if
((r1 - 2)dA1 + r2dA2) (Xi ) = 0
(7.3.5)
and
(rldA1 + (r2 - 2)dA2) (Xs) = 0
,
(7.3.6) 133
for each
i = 1, ...,p, s=p+1, ..., n. In particular if dim N=2 and
rl = r2 = 1, then equations (7.3.5) and (7.3.6) imply V(X1 - 112) is vertical. Example 7.3.2 Let M= Sm-1 , and f: Sm-l - 1R be an isoparametric function of degree 3, with focal varieties V1 and V2. Let IT: M\V2 -- V1 be defined as in Proposition 2.2.5, then from that proposition 1T is harmonic. Also O*h has two distinct eigenvalues (c.f. Example 3.3.3), where h is the induced metric on V1; the corresponding eigenspaces are precisely the principal curvature spaces. The mean curvature of these spaces is vertical, and so Theorem 7.3.1 applies. Since 111 and A2 are both vertical; equations (7.3.5) and (7.3.6) are satisfied and so the fibres are minimal.
134
8 Harmonic morphisms defined by homogeneous polynomials 8. 1
Properties of harmonic polynomial morphisms
Suppose 0: Rm -- Rn is a harmonic morphism defined by harmonic homogeneous
polynomials Ol, ... , 0n, each of degree p. Normalize 0 such that sup
10 (x) 12 = 1.
Ix 1=1
Following Fuglede [ 19 1, we define the set
1' _ {xCSm-1; 10(x)12 = 1? 10 (x)
Then Ix I2p -
12
.
> 0 in Em \{0l, with equality on the cone R+ F \{ 0 ?
.
Thus on
R+ r\{0 0 = a V( Ix I2p I2p-2
p Ix
1 0(x) 12)
(8.1.1)
x -k 1 0k(x) vOk(x) .
From Theorem 7.1.3 , this implies p2
Ix14p-2
=
A(x)2 10(x)12
(8.1.2)
where A : EM -- E is the dilation of 0. Thus on R+ F \{ 01 A2(x) = p2 Ix14p-2 / 10(x)12
= p2 Ix I2p-2
(8.1.3)
By the smoothness of A2, we see that A2 (0) = 0, and A2 (x) = p2 Consider the Laplacian A( Ix 12p
- 10(x)12) = p(2p - 2) Ix 12p-2 + mp Ix I2p-2 -
Ix
nX2 .
12p-2 on
R+r.
(8.1.4)
This is > 0 on R+I', thus p(2p - 2) + mp > np2 , that is
(m - 2) > (n - 2).p
.
Equality is obtained when A( Ix 12p - 10 (x) 12) = 0, that is when Ix I2p- 10(x)12==0
on Sm-1. We therefore have
135
Proposition 8.1.1 If 0: IRm Rn is a harmonic morphism defined by homogeneous polynomials of degree p , then
m - 2 > (n - 2).p
(8.1.5)
,
with equality if and only if
10(x)I2
= constant, for all x
ESm-1
.
Let 0: R4 - R3 be the Hopf map
Example 8.1.2
0((x,y)) _ (Ix I2 - IyI2, 2xy) where we regard x,y as being complex numbers. Then 0 is a harmonic morphism defined by homogeneous polynomials of degree 2. The map 0 I Sm-1 : Sm-1 ,--Sn-1
is a harmonic Riemannian fibration.
From now on we will assume that the dilation A : Rm -- R is given by A2(x) = p2 Ix 12P_2
(8.1.6)
,
for all x c- Rm. Let F: Em - R be defined by F (x) = 10(x) 12, for all xERm then F is homogeneous of degree 2p. Let f = F ISm_1
:
Sm-1 - R
If f is as above then
Lemma 8.1.3
Idf I2 = 01(f)
,
Of = 2(f)
for some smooth functions 01 and 02. In particular f is an isoparametric function
(c.f. Chapter 2). Proof First of all I df I2 =
IdF I2 - (8F/8r)2 n
n
4
(2pF)2
= 4X2 1012 - (2pF)2 4p2 (1 - F). F , using equation (8.1.6)
Write F as F = m ,&R
n k =l
0k2
F = 2Zi V 0.k*v ok 2n A2
136
' then
(8.1.7)
Also
iF = (1/r).2pF
Pr
,
and
e
2
F
_
-2pF
4p2F2
+
r2
c^r
where r2 = Ix I2
r2 .
Therefore from Lemma 1. 1.5
2? - (m - 2) 2pf .
AS m-1 f = 2n a2 - 4p
(8.1.8)
Theorem 8.1.4 If 0: Rm -- Rn is a harmonic morphism defined by homogeneous polynomials of degree p, with dilation 7, given by
A2
(x) = p2
Ix 12p-2
, then F is
a smooth submanifold of Sm-1 , and both R+ I' and the fibre over the origin in En are minimal cones through the origin in Em.
denote the fibre over the origin. Then F IE= 0, where F = 1012, and so r and En Sm-1 are both critical sets of f= F Ism_1 From Lemma 8.1.3 f is isoparametric, and hence from Theorem 2.2.2 I' and E n s m-1 are minimal submanifolds of S m-1. Proof
Let E = { x e Rm ; 0 (x) = 01,
Consider the map 0 r: r _ S n-1 , then by a result of Fuglede [ 19 1 this map is surjective. We prove the following theorem. I
Theorem 8.1.5
The map 0 I r: r
- Sn
defined above is a harmonic Riemannian
submersion. Proof Claim 1: The set r is precisely the set of x E Sm-1 such that Tx (fibre of 0 through x) C T {S m-1.
Proof of Claim I : Let y e Sn-1 be such that 0(x) = y, x E r, and write 0-1(y)
for the fibre over y. Let y(u) C 0(y) be a curve with y(0) = x. Then 0(y(u)) =y. Let
Then
S
0(µ(u)) = y/ Iy(u)Ip
,
by the homogeneity of 0 . Thus
137
d0(p'(0)) = (d/du) (y/ ly(u)Ip) u= 0 I
= -p .y;
(8.1.9)
so that -y(u) is tangent to S m-1 at u = 0 if and only if < y ' (0) , y (0) > = 0 if and only
if d0(p'(0) = 0. But we know that 10(x)12 = 1 is maximum on the sphere, i.e. 10 (p (u)) I2 < 1 and = 1 at s = 0. Therefore
0 = (d /du) 10(p(u))12 1u=0
(d/du) E
k=1
0k(p(u))2 lu=O
n
2E
0k(p(0))dOk(p (0))
k=1
2 < 0(p(0)), d0(p'(0)) > 2 y>
-2p
(8.1.10)
.
Thus y is tangent to Sm-1 at u = 0 and Claim 1 is proved. In particular PA2 is horizontal over r, and equation (7.1.5) of Theorem 7.1.8 implies that the mean curvature of the fibre is proportional to the horizontal projection of V A2 , i.e. to V A2 Sin-1 (in particular if the fibre were contained in then it would be minimal in S m-1) .
Claim 2: The set 0-1(y) fl r is minimal in T. Proof of Claim 2 : Suppose we are given a curve a (u) on Sm-1 such that
0(a(u)) = a(u).y
(8.1.11)
.
Then, since 0(p(u).a(u)) = p(u)p.a (u).y, if we let p(u) = a(u)-1/p whenever a(u) 0 ; the curve y(u) = a(u)/a(u)1/p is a curve in the fibre 0-1(y) whenever a(u) 0. Consider the particular case when a(u) is an integral curve of Vf through a point x C T. Then we find that (a) 0(a(0)) = y
(b) dO(Vf) = 2p20. (1 - f)
.
Both the conditions (a) and (b) imply that either (i) a (u) satisfies equation (8.1.11), 138
or (ii) O(a(u)) C
Sn-1 .
Consider case( 0. Since a(u) is a geodesic in S m-1; Va (0)" (0) is perpendicular to Sm-1, which implies that Vy,.( 0) y-'(0) is perpendicular to Sm-1
Consider case (ii). Then a(u) is horizontal, and so a'(u) is perpendicular to O-1 (y)
.
We can now perform the following construction.
Consider the set o-1(y) fl I'. Locally we can choose an orthonormal basis of
T(0-1(y);X1, ...,Xr, such that X1, ...,Xs span T((6-1(y)fl I') andXs+1,...,Xr arise as the tangent vectors to integral curves of Vf (this is always possible since 1' is the focal variety of an isoparametric function) as above, so that V X X. is perJ
pendicular to Sm-1, for j = s + 1, ... , r. We therefore find that the mean curvature vector of o -l(y) fl r is perpendicular to S m-1 , and in particular that (y) fl r is minimal in I'. Furthermore, since X2 I , = constant and V A2 (which is horizontal) is perpendicular to r, then the map 0 is a Riemannian submersion. The minimality of the fibres now imply that 0 I r is a harmonic Riemannian submersion.
8.2. Some examples
/Example 8.2.1 Let 0: )E28 - R5 be the Hopf map defined by 0 (x, Y)
(
1 x 1 2 - ly 12 , 2 (x1Y1 - x2y2 - x3y3 - x4y4) , 2 (x y +x2y1 +x3y4 - x4y3) , I 2
2 (x 1y3 +x3y1 +x4y2 - x2y4) , 2 (x
ly4
+x4y 1 +x2y3 - x3y2)) .
Then 0 is a harmonic morphism defined by homogeneous polynomials of degree 2.
Write 0 = (0,r, ... , 05). Then 61 = (x1,x2,x3,x4,-Y1,-y2,-y3,-y4)
f12
= (y1,-y2,-y3,-y4>x1,-x2,-x3,-x4)
139
v03 = 2(Y2,Y1,Y4,-y3,x2,x1,-x4,x3)
v04 = 2
(y3,-y 4,Y 1,Y 2,x 3,x 4,x 1,-x2)
v05 =2 (y4,Y3,-y2,Y1,x4,-x3,x2,x1)
from which vie see that X2(z)=4 Iz 12 for all z C R8 . We also note that there exists an A C 08 (R), such that 0i+1 = 0i o A ;
10001000 0001000-1 0 0-1 0 0 0 1 0 0 1 0 0 0-1 0 0 01000100 00100010 00010001 1 0 0 0-1 0 0 0
Let X : R 8 -- Rn , n < 5, be defined by a subset of { 01, ... , 05 ? . Let F = {x C S7 ; I x(x) 12 = 11, , F = I x 12 the associated isoparametric function and E the fibre over the origin in Rn. We consider the different cases in turn. Case 0
x (z) = 0, z = (x,y) E R8. Then I' is empty, 2; fl
Case 1
x(z)=01(z)= Ix 12- 1y12. Then F=(Ix12- Iy12)2,
X I I,:
I'
S0 is the obvious map, and z fl S7 =
S3
S7 =
S7 and F = 0. '= S3 U S3
,
x S3 /22 .
Case 2 x = (01, 02) . Let B (-7 08 (R) be the matrix
10000000 B =
0 1 0 0 0 0 0 0
00100000 00010000 0 0 0 0 1 0 0 0
0 0 0 0 0-1 0 0 0 0 0 0 0 0-1 0
0000000-1
then B preserves 01, while taking 02 into the function 2 < x,y > . Then F = (Ix 12 - ly 12 )2 + 4 < x,y >2, which is the isoparametric function of Example 2.3.5 The set r is given by r = S1 x S2/S0 = { e'' .x; x C S31 = {6os0.x,sin0.x); 140
x E S3
and X I1,: (cos®.x,sin0.x) - (cos20,sin 20), i.e. X Ir:(0,x) -- 20,
which is clearly a Riemannian submersion. Also z n S7 = S4/2 , the Stiefel manifold of orthonormal 2-frames in 4-space (defined in Example 5.3.3). the isoparametric function corLet sV Case 3 X = (01 10 = (02' 03' 04' 05) .Then is given by Fd) (x,y) = 4 Ix 12 ly 12 = (Ix 12 + ly 12 ) 2 -
2' 03) .
responding to ti, F
,
,
Thus the level sets of X coincide, (i.e. up to orthogonal transformation) with the level sets of X o A2, which coincide with the level sets of 10 12 - 02 , i.e. with the level sets of 1 - 0i - 02 . Using the matrix B, we can transform 02 into 2 <x,y> while preserving 01. Thus Ix 12 - ly 12 ) 2 = 1 - ( Ix 12 - ly 12 )2.
F(x,y) = 1 - (Ix l2 - 1y12 ) 2 - 4 <x,y >2
and r = S 1T/4 The map Xo
4,2
and 2; = S1 x S3/SO. We now perform a computation of X I r: r -- S2
X o B o A 2 is given by
BoA2(x,y) = 2(-x1y2+x2y1-x3y4+x4 y3,-x1y3+x3y1-x4y2+x2y4 - x 1y4 + x4y 1 - x2y3 + x3y2) .
Then I Xo B o A2(x,y)12 = 4( Ix
12
1y12_ <x,y >2), which is equal to 1 when Ix 12=
_ i and < x,y > = 0, i.e. on the focal variety S4 /2 = {(x+ iy)/2z ; (x,y) CS4,2 ly I 1 . We are thus precisely in the situation of Example 5.3.3, and the map X jr r-- S2 is given by the map 0 T7/4: S Tf/4 -. S2 of that example. 4,2 X = (011029 04) say, then X o A = (02'03'05)' The level sets of Similarly, if 2
I x o A 12 coincide with the level sets of Ith 12 - 042, which coincide with the level sets
of 1 - 0i - 04 O. If C E 08(R) is the matrix 1 0 0 0 0 0 0 0
01000000 00100000 00010000 00000010 00000001 00001000 L00000-100 J
,
then C takes the function 04 into the function 2 < x,y > , while preserving 01, i.e. we have the same case as above. 141
). Then x o A = +L , so that F= 1 - (lx l2 - ly 12 )2. 2' 0 3 10 4 i 2) = xy S3 S3 ; X 11,((x, y) /2 the set Fis isometric to S3 x S3/ 2 2 , and X r : multiplication of unit quarternions). The set Z fl S7 is isometric to S3 U S3. Case 4
X =(O
1
10
I
r-
1' -- S4 is Case 5 X =0. Then F = 1, 1'= S7, E fl S7 is the empty set and X the Hopf fibration. We remark on the duality between Case i and Case (5 - i), i = 1, ... , 5 - that I1,.
is, the set I of case i is identical with the set E n S7 of case (5 - 0. Example 8.2.2 Consider the example of an isoparametric function given by Ozeki and Takeuchi in [33 J, and defined as follows. Let H denote the space of quaternions, and write u = (uO, ul) , v = (vO, vl ), ui, vi C H; let u u denote the canonical involution. Define a function F : 1113 16 - 1 by
F((u,v)) = 4 ( tuv - < u,v >2) + ( 1u112 - Iv1 12 + 2 < u0,v0 >) 2 . Then f = F I S 15 is isoparametric.
Let 0 : IIi 16 - 1Ei4 be defined in the following way. Write u.1
i = 1, 2 , and let 0=(O
1
,
(u1, u 1
2, 3, u
i
i
u
4 i
... , 04) be given by
01 = lu1 12 - Iv1 12 + 2 < u0,v0 > 02
34 43 2 1 12 34 43 21 12 u0v0 - u0v0 - u0v0 + u0v0 + u1v1- u1v1- u1v1 +
u1v1
03
42 24 42 24 3 3 1 3 1 13 u0v0u0v0u0v0+u0 v0 +uivl - 1ulvlulvl+ulvl
04
14 23 32 41 23 32 41 14 u0v0 - u0v0 - u0v0 + u0v0 + ulvl - ulvl - u1 1 + ulvl .
Then the square of the norm of the gradients of the 0i , i = 1, ... , 4, are all equal and mutually perpendicular. Since each 0. is harmonic we see that 0 is a harmonic morphism. Furthermore F = 10 12 . Thus from Theorem 8.1.5 there exists an interesting harmonic Riemannian submersion from the set F onto S3. 8.3. Harmonic morphisms defined b6 homogeneous polynomials of degree bigger than two. We can easily construct harmonic morphisms defined by homogeneous polynomials 142
of degree bigger than 2, simply by composing two harmonic polynomial morphisms of
degree 2. Since we recall from Example 7.1.7 that the composition of two harmonic morphisms is again a harmonic morphism.
Example 8.3.1 Let 0: R8 - R4 be defined by O(x,y) = 2xy, where x and y are quaternions, and let t: R4 i by the Hopf map; O(u,v)_(Iu12- Iv12,2uv), where u and v are complex numbers. Then X = th o 0: R8 -- R3 is a harmonic morphism defined by homogeneous polynomials of degree 4. The critical set CX = 0-1(0) is a minimal cone in R8 . However, we ask whether there exist harmonic polynomial morphisms defined by homogeneous polynomials of degree bigger than 2, which do not arise in this way as the composition of two harmonic morphisms. In particular, do there exist harmonic morphisms defined by homogeneous polynomials of degree 3 ?
Recall the theorem of Munzner; Theorem 2.2.5. This states that an isoparametric function f: Sm-1 -- R arises as the restriction of a homogeneous polynomial F : Em -- R of degree p with IVF 12 = p2 Ix 12p-2
(i)
(ii)
AF
= c Ix Ip-2
where the constant c is zero when the multiplicities of the distinct principal curvatures,a.r-e equal.
Suppose we are given such a polynomial F: Em -- R with c = 0. Given any matrix
(E); define G: Rm -- R by G = F o A. Then, for all x E Em and for all vectors v E TxRm ,
AC0
m
< v, VG > = dGx(v) x = dFA(x)o dAx(v) = dFA(x)(Av)
=
VFA(x)>.
Thus VG
x
= A* VFA(x)
(8.3.1) 143
-
Em is an isometry; Therefore, from (i), IVG x 12 = p2 Ix 12p-2 . Since A: Em G is also harmonic, so that G satisfies condition (ii) also. In particular, if there exists an A E Om(]R) such that < VGX , V Fx > = 0, for all x E Em, then 0 = (F, G) gives a non-trivial harmonic morphism 0: Em _ E2, defined by homogeneous polynomials of degree p. This would give a method of constructing harmonic morphisms of various degrees.
Remark 8.3.2 All known harmonic morphisms defined by homogeneous polynomials of degree 2 arise from a single polynomial in this way. We now demonstrate that this procedure will now always work.
Let F,G be as above, and write f = F ISm_1 and g = G ISm-1
Let V M denote the focal varieties of f, and MO (f) the minimal
Lemma 8.3.3
hypersurface f-1(0). Then (a)
V(f) _
(b)
MO (f)
{xESm-1
x
ESm-1
VFXl
TSm-1)
; V Fx E
TXSm-1'
Proof The statement follows immediately from the equation
VFX = Vfx + pf.(8/ir)(x) ,
(8.3.2)
for all x E S Tn-1 , where r2 (x) = Ix I2
Proposition 8.3.4
If
< VF X, VG XS = 0
for all x r_ IRm, then
V(f) C M0(g) and
V (g) C M(f )
Proof From Lemma 8.3.3. (a) we see that x E V(f) and similarly 144
implies x C M0(g)
,
D
x E V(g)
implies
x E M0(f)
[J
.
R be the standard homogeneous polynomial of degree Theorem 8.3.5 Let F: R6 4 of Theorem 2.2.5, defining the family of isoparametric hypersurfaces of Example 2.3.5 with n = 2 (so the multiplicities of the principal curvatures are equal and F is harmonic). Then there exists no A E 06 (R) such that A (V (f)) c M0 (f), where f = F ISm-1 (hence from equation (8.3. 1) the above method of constructing a harmonic
morphism fails in this case). Proof The polynomial F is given by
F(x,y) = 2((Ix12 - 1y12)2 + 4 <x,y >2) - I (x,y) 14 where x,y E R3. As in Example 2.3.5 the level surfaces of f can be parametrized by the sets { eie(coss.x + isins.y); B E [0,211] and (x,y) E S3,21 for each s E [0,11/4 ] . Consider the focal variety V1 corresponding to s = 11/4. Then V1 = {(x+iy)/22; . The minimal hypersurface of f is given by (x,y) E 53,21,
M0(f)
S
{
Suppose the point z = (x + iy) /2 2 of V 1 is mapped under A to the point w = el th(cos ( 11/8) . u + i sin( TI/8). v) E M0(f) M. Then
w = (cos 4) cos(11/ 8).u - sin0 sin (11/8).v) + i(sin b) cos (11/8).u + cos 0 sin(17/8).v) .
Now, since 03 (R) acts transitively on S3,2 , there exists a B E 03 (R) with u = Bx and v = By. Thus the matrix A has the form A =
cos 0 cos (11/8) . B
- sin 0 sin M/8). B
sin 0 cos M/8). B
cos 0 sin (11/8) . B
We must check that this is indeed orthogonal.
145
We compute AA* and obtain
AA* =
Cos20Cos2 1 BB*+sin2rJ sin2 a BB* L sin2ticos
2 sin 2 O cos 4. BB
sin2 0Cos2
2
BB*+cos2th sin2 IBB*l
8 This can only be the identity matrix if sin2 b = 0, i.e. th = 0, T1/2, T1 or 3 TI/2. But in this case the non-zero diagonal entries are not all equal, and certainly do not
equal 1, e.g. if 0 = 0, then A.A* _ rcos2 (TT/8 ) 0
cos2 (TT/8
)
2
sin (TI/8) . sin2 (TT/8 )J
0
F-I
8.4 Harmonic polynomial morphisms and equivariant maps between spheres Recall Example 2.3.8 and Theorem 2.3.9 of apter 2. In that example we were given ann-tuple (P1, ..., P ) of symmetric endomorphisms of R21 with n
P.P. . = 2S P P. + P 1
1
j
if..I, i,j = 1, ..., n
,
called a Clifford system. To such a Clifford system, with ml = n and m2 = 1-n both positive, is associated the isoparametric function f = FIS21_1 , where F : R21 _ R is defined by
F(x) = 1 x 1 4 - 2 E < Px,x>2 i
1
We could equally well write F as
F(x) = E < P. x,x>2
1
(8.4.1)
i
and then, in view of Lemma 8.1.3 it is natural to ask whether such an isoparametric function arises from a harmonic polynomial morphism.
Theorem 8.4.1
f: 146
R21
Given a Clifford system (P 1 , P
..., Pn ) on R21 ;
define
i = 1, ..., n.
Then 0: R21 - Rn given by 0 = (f 1, ... , fn) , is a harmonic morphism defined by homogeneous polynomials of degree 2, with dilation A given by A2(x) = 4 Ix I2 , for each x r R21 Proof
Let (xi)i=1
m=21
xERm
be the standard coordinates on Rm. Then, for each
< Vfi(x), F/8x1> = dfi(x) (8/c^x1)
= (d/du)(fi(x +u(1,0, ...,0))lu=
0
= (d/du) lu
=0
= (d/du)( +
++0(u2 )) lu=0
+ Thus
m
Vfi(x) = Pi(x) + E
k=1
< Pi(8/Fxk), x > . 8/8xk
(8.4.2)
But Pi is symmetric, i.e. < Pi(x),y > = < Pi(y),x >
.
Thus, from equation (8.4.2),
Vfi(x) = 2Pi(x) ,
(8.4.3)
for each i= 1, ..., n. Then < Vf.(x), Vf.(x)> = 4 < P.(x), P (x) > 1 )
1
1
= 2 + 2<x,P.PW) ()> 1
1
= 4 <x,x> S..1) Thus 0 is horizontally conformal with dilation A given by A2 (x) = 4Ix 12, for all
xERm. Write P1 in matrix form as (Pb) 1 a,b-1,...,m , for each i = 1, ... , n.. Then ,
fi (x) = E a,b
P.ab
xaxb
Thus
Of.1 = Ea P.aa 1 147
Since P.2 = 1, the eigenvalues of P. are +1 and -1. If the eigenspaces of +1,-1 are E+(Pand E_(P.) respectively, then since P.P. + P.P.i = 0, i j, we see that i l i i l Thus dim E + (P.t ) P interchanges the eigenspaces E+(Pi ) and E- ( Pi of P.. P. i dim E (P.) = 1, whence
E Paa = 0 a
1
LI and f.1 is harmonic for each i . 1, ... , n . Examples of harmonic morphisms defined by Clifford systems, include as special
cases Examples 8.2.1. and 8.2.2. From Theorem 8.1.5, to each such example we can associate a harmonic Riemannian submersion onto a sphere. Amongst the family of isoparametric functions (8.4. 1), some are inhomogeneous ; furthermore in some cases the focal varieties are inhomogeneous [ 17 1. Thus we can associate harmonic Riemannian submersions from compact inhomogeneous spaces
onto spheres. We can modify the proof of Theorem 8.1.5 to obtain the following.
Theorem 8.4.2 Let 0: Rm -. Rn be a harmonic morphism defined by homogeneous polynomials of degree p, with dilation a given by X2 (x) = p2 Ix 12p-2. Let F: Em -- E be defined by F(x) I 0(x)12 , and write f = F ISm-1 , then 0 IM : Mc -- c2 Sn-1 is a harmonic homothetic submersion, where Mc = f-1(c) c and c # 0.
Proof Since c X 0, the projection p c : Mc -- I', which maps down normal geodesics,
is well-defined. Therefore 0 Mr -'c2. Sn-1 is onto, since it factors through IM : : c 2 S n-1 - Sn-1 ; i. e. the folpc , O I :F -- Sn-1 and the projection map c Jr 1
c2
lowing diagram commutes:
0
i 0 148
Ir
In order to prove the theorem, the crucial point to observe is that the proof of Theorem 8.1.5 will go through provided that the following claim is true.
For x c M c
Claim 8.4.3
(VA2 (x)) lies in the plane spanned by VX2 (x) and
is the unit normal vector field to M c For then the mean curvature of the fibre of 0 over O (x) would be perpendicular to M at x, and the rest of the proof goes through as before. x
where
c
Proof of Claim 8.4.3
Write 0 = (O1,
x c- Mc is spanned by f V Oi (X))
F=
... , n'
i=1
0n) , then the horizontal space through Now
n
2 E 0 k=l k
so
n
VF = 2 E
k=1
0k 00k
therefore
< VF,V0i > = 201 a2 Also X2 (x) = p2 Ix V X2
I2p-2
, therefore
= 2p2 (p- 1) Ix 12p-4
.
k
xk Vxk
Now k
< xk Vxk, V O > = p. Oi
by the homogeneity of 0i. Thus <
VA2, V Oi > =
2 p3(p - 1)0 .
Hence the horizontal projection of Vat is proportional to the horizontal . projectiop of VF. Since is proportional to the projection of VF onto the claim is proved. on Sm-1
TSm-1
Example 8.4.4 Consider Example 8.2.1 , Case (2). Here 0: R8 0 =(01,02) is given by 01(x,Y) =
Ix12
-
R2,
Iy12
02(X,Y) = 2 <x,Y >
,
149
where x,y are quaternions. The level hypersurfaces are parametrized by the sets Ms = { ei9 (toss. x + isins.y) ; 0 E [0,2 Tl], (x,y) E S3,21 , where s E [0, IT/4 ]. If z E M s then z can be expressed in the form
z = eiO(coss.x + isins.y) (cos0toss.x - sin6sins. y) + i(sin 0toss.x +cos0sins. y) Then
01(z) = Ix12- Iy"12 =
cost B costs +sin2B sin2s - sing 0 cos2s - cost 0 sin2s
= cos28.cos2s
,
and
02 (z) = 2 (cos B toss sin 0 toss - sin O sins cos g sins)
= sin 2 0 . cos 2 s . Thus 0 IM : Ms s
2s. S1 ; 0 (z) = cos 2s. a
2i6
.
Theorem 8.4.2 allows us to construct many equivariant maps x : Sm-1 -. Sn associated to a harmonic polynomial morphism 0: Rm - Rn with dilation A given by
X2
(x) = p2 Ix 12p-2 . We simply define X by
X (z) = (cos2a(s)(OIIIo p(z)), sin2 a(s))
,
where F = 10 12 is parametrized such that F = cosps, F= s-1(0), p is the projection down normal geodesics onto I' and a (0) = 0, a(TT/p) = TT/2 . In particular all the harmonic morphisms associated to a Clifford system give such equivariant maps, and hence we have examples of equivariant maps between spheres
with respect to isoparametric functions, where one of the isoparametric functions has non-homogeneous hypersurfaces. This then justifies our use of isoparametric hypersurfaces as opposed to homogeneous hypersurfaces. Furthermore, when the same isoparametric function Rives; rise to two distinct harmonic Riemannian submersions via Theorem 8.1.5, we expect tq be able to construct examples of equivariant maps similar to Example 5.3.4. Indeed the two Riemannian submersions of Example 8.2.1 given by Cases (2) and (3) give rise to 150
Example 5.3.4. Similarly Cases (1) and (4) of Example 8.2.1 generate the Hopf map from S7 to S4. In fact, suppose 0: Rm -- Rn is a harmonic polynomial morphism, satisfying
m - 2 = p(n-2) where p is the degree of the homogeneous polynomials defining position 8. 1. 1 we see that 10 (x) 12 =
0.
Then from Pro-
Ix 12p
for all x E Rm. Let 0 = (01, ..., 0n), and define harmonic polynomial morphisms p and a by
p = (01, ..., 0p) a = (0p+1, ... , 0n) Call such p and a complementary. Let Ms be a level hypersurface of the isoparametric function f defined by f(x) = Ip(x) 12 IS m_1 ,
1=
Ip12 +
IaI2
for each x E SThen since ,
must also be a level hypersurface of the isoparametric function g defined by g (x) = lc(x)12 I Sm-1 , for all x E S m-1. From Theorem 8.4.2, we obtain harmonic homothetic submersions p INl: Ms - a(s) SP-1, c IM : Ms -- b(s) Sq-1 M
s
s
s
p +q = n, for some functions a(s) and b(s). Since 1 = Ip 12 + 1x12 , we have a(s)2 + b(s)2 = 1, so we can choose a(s) = cos ps and b(s) = sin ps (by writing Ms = f 1(cos2ps)). Given two harmonic polynomial maps g 1: S p-1 - S r-1 and g2 : Sq-1 - S s-1, we can now define an equivariant map from S m-1 to S r+s-1 as follows. Let x E Sm-1, then( p (x) , a (x)) E SP - 1 * S q-1 ; we then compose with g1 * g2 , to obtain the point gl *
g2(p(x),(#(x)) E
Sr-1 * Ss-1 = Sr+s-1. The map so defined is clearly equi-
variant since the map of level hypersurfaces is harmonic of constant energy density
by Theorem 8.4.2. The Smith maps of Section 1.3 can be seen to arise in this way as follows. Consider 151
the harmonic polynomial morphism 0: Rm - Rm
given by
0(x1,...,xm) _ (x1, ...,xm) . Define the complementary maps o and a by
p (xl, ... ,xm) = (xl, ... , xp) and
c(x
,
1
...,xm
(x
p+1
, ...,x m ).
Then f (x) = Ip (x) 12 IS m_ 1 is isoparametric of degree 2. persurface given by f(x) = cos2 s. Then g(x) = Ir (x) 12
Let Ms be the level hy-
is given by g(x) = sin2s on M. For x ESm-1 ; (p (x), r(x) C SP-1 Sq-1, p + q = m. We now compose with harmonic polynomial maps g 1: S p-1 -. S r-1 and g2 : Sq-1 -. S 3-1 as above, to obtain the Smith map from Sm-1 to Sn-1 , r + s = n.
152
ISm_1
9 Deformations of metrics
9. 1
Deformations of the metric for harmonic morphisms
Let 0: (M , g) -- (N , h) be a horizontally conformal map between Riemannian manifolds with dilation X. Let U C M be an open set upon which d0 is non-zero. Let
x e U, then for horizontal vectors X, Y e`'x , we have 2g(X,Y) = 0*h(X,Y) . Thus we can decompose g as
g = (1/X).0 *h+k ,
(9.1.1)
over U, where (I/X2 ) 0 * h represents the horizontal part of the metric, and k the vertical part. The stress-energy tensor of 0 is given by
S0 = e(0)g- 0*h anA2.g - X2(g-k)
2'(n-2)X2.g+A2.k
(9.1.2)
,
where n = dim N. Thus
V S0 = '2 (n - 2).d(A2) + V *(A2k)
.
(9.1.3)
We therefore have
If 0 : (M,g) - (N.h) is horizontally conformal with dilation X and 0 is a submersion almost everywhere, then 0 is harmonic (and so a harmonic morphism) if and only if Lemma 9. 1. 1
21(2 - n). d X2 = V*(h2k)
,
(9.1.4)
where k represents the fibre metric (where defined) (assume both sides of (9.1.4) are zero when X = 0). 153
Define a new metric g on M by
g = (1/A2 c2).¢*h + (1/p2).k
(9.1.5)
,
where o2, p2 : M - R are smooth functions. The new metric g may not be welldefined everywhere if c2, p2 have zeros , and we remove such points to obtain the Riemannian manifold (ii , g ). Denote by V the associated Levi-Civita connection. The map 0: (M,g) - (N,h) induces a new map -0: (M,g) -- (N,h). If -0 is a sub-
mersion almost everywhere then, from Lemma 9.1.1,
9
is a harmonic morphism
if and only if
2'(2 - n).d(A2o2) _
*(A2(T 2.k/p2)
(9.1.6)
.
Our aim is to reformulate and solve equation (9.1.6), given that equation (9.1.4) is satisfied. The connection coefficients of p are described in terms of Proposition 9. 1.2 those of V by the following formulae. Use the following ranges of indices:
1 < i,j, ... < n = dim N; n+1 < r,s, ... < m = dim M; 1 < a,b, ... < m ; and let ( e a)1< a< m
=
(ei, e
r ) denote a local orthonormal basis with respect
g over a subset U of M where d0 # 0', and ei ,e r are horizontal, vertical, for each i , r respectively. Let 06a (e it er (cei , p e ) denote the corresponding r orthonormal basis with respect tog over U . Then to
g(et,V e es) = Pg(et,De e s ) + iP(g(et,dP(er)es) + g(esdp(et)er)
r
r
- g(er,dp(es)et ))
(9.1.7)
;
g(ei,V er es) = c2g(ei, a es) + (o2/2p)g(es,dp(ei)er) ;
(9.1.8)
r
g(ek, e e.) + i
)
o(g(ek,do(ei)e,) + g(e,,dc(ek)ei) I
)
- g(ei,d o(ej)ek)); Z(g(er,(a2+p2)
r a iej) - g(e,(a2
(9.1.9)
- p2) a ei) ]
+(o2/o)g(ej,da(er)ei)); 154
(9.1.10)
g(er, Veies) = p cg(er, V e s
+
s ag(e,dp(ei) e) r
(9.1.11)
i
g(el,Verei)= oag(ej, ere.) + za(1-(a2/P))g(er, [ei,e11) + (p/a)g(e,,da(er)ei));
g(e,V
r
(9.1.12)
-(p/u)g(ei,ve es)
(9.1.13)
r
Proof We use the fundamental formula for an orthonormal frame X, Y, Z E (see for example [ 14 1) :
g(Z,VXY) = a(g(Z, IX,Y1) + g(Y, IZ,X1) - g(X, IY,Z1))
.
TM
(9.1.14)
Similarly
g(Z,79-XY) = 2(g(Z, [X, Y1) + 9(X, [Z,X1) -g(X,
(9.1.15)
Let Y' denote Lie derivation. First of all we prove equation (9.1.7). From equation (9.1.15) , letting k, k denote the vertical part of the metric g, j respectively,
g (et, ve s) _ r
es) + k(esY er) - k (ergs et) )
z (k (et,
t
r
s
21(Pk(et,9e es) + k(et,dp(er)es) + ... ) r z(Pg(et, a es) + g(et,dP(er)es + ... ) r On the other hand
g(et, ve es) = k (e t, Ve es) r r _ (1/P)k(e,Vs) t r (1/p)g(et,ve es
r
Therefore
g(et,De es) = P2g(etV es) + iP(g(et,dP(er)es) + ...
r
)
.
r
The proof of equation (9.1.9 ) is similar. We prove one more, say equation (9.1.10) ; 155
the others use similar arguments. From equation (9.1.15), and writing H,H for the horizontal part of the metrics g,g respectively
gr,vele)) = z(k(er, ele)
ereil - H(ei,eer))
+ H(e)y
((1/P)k(er,-
e
ae.) +(o/a)H(e., a (a ei) r - H(ei, e. per)) l
21((0
2
/P)k(er, r e.) + PH(e., a e.) r
i
- p H(ei, a er) + (o/a) H(e,, do(er) ei)) . l
Now
H(e., 2' e - V er e r e.) i - H(ei,.e e r ) = g(ej, V er i e.
)
,
l
- g(ei, Ve,e r - v e.) er 7
l
= g(ej, a er) - g(ei, a er) = g(e , V e. + V e.).
Thus
g(er,
r
2
z (g(er, (a /p) (V
ei 1
-
ej
I
e ei
e ei )) + g(er,P (V V
+ V
j
g( ei . ,
d o(er )e.))
But
g (er, vale;) = k (er, vale; ) = (1/P) k(er, ve, e. ) = (1/P) g(er' V_ e. i
)
,
giving equation (9.1.10) We now compare V ; k with V * k
Lemma 9.1.3
156
Using the notations of Proposition 9.1.2
.
J
))
Vk(ej) = P2 V*k(ej) + AP(m-n)dp(ej)
(9.1.16)
for all j = 1, ... , n. Proof dices)
The divergence with respect tog is given by (summing over repeated in-
V k(e.) = eak(e a,e.) - k(V_eaaa,e,)) - MeaV- e.) ea )
)
- k(ea, Ve e
a)
(1/a) (k(ea, V-aej) - a k (ea, d a (ea) a ,) ) l
(p/a) k(ea, Ve ej ) a
(p/a) g(er, Ve ej)
r
(p2la2)(a2g(ej,Ve er) + (a2 /2p)g(er)dp(ej)er ))
r
,
from equations (9.1.11) and (9.1.13), =
P2g(ej , Ve e r + i P(m-n) dp(e.)
r
.
On the other hand V * k(e.)
k (e a,
g(er,
Ve
V
r
ej )
e.)
g(ej, erer
Lemma 9.1.4
Using the notations of Proposition 9. 1.2;
V *k(es) = P2dp(es) - zn(p/c) (da(es) + p2V*k(es)
(9.1.17)
for all s = n+l, ... , m. Proof
The divergence with respect tog is given by (summing over repeated indices) 157
V; k(e s ) = e ak06a ,e s ) - k(V_ eaaa ,e s ) - k(e a , V_ eaes
= per(ok(er,es)) - g(ve ei,es) - g(ve r, es ) r
i
- pg(e
ve es ) r
pdp(es) - p2g(es, V ei) - (P2/2 a) g(ei,do(es)ei)
ei
-p2 g(es, e er) - zPg(erdp(es)er) r
- P((1/P)g(er,ve e s ) - g(e r dp(e r )es))
r
where we have used equations (9.1.7) and (9. 1. 10) ;
pdp(es) - p 2 g(es, e ei) - n(P 2 /2 o)d a(es) - P 2 g(es, a er i r
a e s ) - pd P(e - z(m - n)pdp(e s - p2g(e, rp r + 4(m - n)Pdo(es) + p2dp(es) using equation (9.1.7) again ;
o2dp(es) - 2n(p2/o)da(es) - p2(g(es, a ei) 1
r e e s)) s e e r +g(e, + g(e,v
r
r
.
LI
Theorem 9.1.5 Assume equation (9.1.4) is satisfied, i.e. 0 is a harmonic (N,h) is a harmonic morphism if and only if, morphism. Then using the notations of Proposition 9. 1.2,
(2-n)d(a2)(e.)
=
(2 - n)d(X2a2)(es)
z(o /P2) (m-n)d(o2)(e.) =
2P2d(a2x2/P2)(es) +02A2 (2dp(es)
- 21 n(1/a2)d(a2)(es) - n(1/X2)dX2(es))
158
(9.1.18)
(9.1.19)
if and only if
(2 - n)d(log a2)(ej) = 2(m - n)d(logp2)(e.)
(9.1.20)
n)d(logX2o2)(es)
(2 -
=
2d(log(a a2/P2)(es) +2dp(es) - 2nd(log a2)(es) - nd(logX2) (es)
(9.1.21)
,
for all j = 1, ..., n and all s = n + 1, ..., m.
Remark 9.1.6 It may be more natural to write (9. 1.18) and (9.1.19) become
v2
=
a A2 , whence equations
(2 - n)(d(logµ2) - d(loga2))(e.) = (m - n)dp(ej)
(9.1.22 )
(2-n)d(logp2)(es) = 2d(log (p2/P2))(es) +2dp(es) - nd(log(u/X))(es) - nd(logX2)(es) for all j = 1 ,
,
(9.1.23 )
..., n and s = n + 1, ..., m.
Proof of Theorem 9.1.5 O*(A2k) =
First note that (summing over repeated indices)
dX2(e)k(e
V*((X2a2/P2)k)
=
+
X2 V*k
(9.1.24)
,
d(a2X2/P2)(er)k(er) +(a A2/P2) v*k
.
(9.1.25)
Let equation (9.1.6) act on e.; ) 22 22
2(2-n)d(A a )(ej) _ (a x /p2 )V k(ej) (a a2 /P2)(P2V*k(e.) + 20 (m - n)dp(e.))
from Lemma 9.1.3, a22(2 - n)d7X2(ej) +(o a2/P)12(m - n)dp(ej) from equation (9.1.4), giving equation (9.1.18) .
Now let equation (9.1.6) act ones, to give 2(2 - n)d(X2a2)(es) = d(a2A2/P2)(er)k(er,es) + (o2 A2/P2) V*k(es) =
p2d(o A2/P2)(es) +(a A2/P2) (P2dp(e8)
Zn(P2/a)da(es) +p2p*k (es)) 159
Now
V*k(es) _ (1/A2)V*(A2k)(es) - (1/A2)dX2(es)
'(2-00A 2)dA2(es) - (1/a2)dX2(es Zn(1/A2)dX2(es)
,
giving equation (9.1.19) .
Remark 9.1.7 If we put a2 = p2 = 1, then equations (9. 1. 18) and (9. 1.19) are satisfied - similarly, if a2 and p2 are both constant then the equations are also satisfied.
Remark 9.1.18 There is a striking analogy of the above methods with the classical notion of a Backlund transformation. There one has a hyperbolic surface M in IIR3, which is parametrized by a function a: M - R. This function is in fact the angle between the asymptotic coordinates, and the Codazzi equation for M is equivalent to a satisfying a certain second order equation called the Sine-Gordon equation. Conversely to each solution of the Sine-Gordon equation one can construct a hyperbolic surface. The idea of Backlund was to write down a first order equation in two variables a and a, such that if a is a solution of the Sine-Gordon equation, and a satisfies the first order equation, then a is also a solution of the Sine-Gordon equation. We view Backlund's idea in a more general context as the following fundamental principal:
(i) we are given a second order problem parametrized by a set of functions
(a,p,...), (ii) there is a set of first order equations in two sets of parameters; (a,Q,... ) and (a, p , ... ) , which associates to a solution (a , p , ...) of (i) another solution
(a, p., ...) of (i) In the context of harmonic morphisms, the parameter is the dilation A 9.2
Examples
Given a particular harmonic morphism 0: (M,g) -. (N,h), we attempt to find non-trivial solutions to equations (9.1.18) and (9.1.19) . We also consider instances when there are no non-trivial solutions. 160
Example 9.2.1 If n > 2 and p is constant, then equation (9. 1.18) implies that V a2 is vertical. Example 9.2.2 If p is constant and V A2 , Vat are both vertical, then equation (9.1.18) is satisfied. Equation (9.1.19) now becomes 2
d(X2a2) (e s ) = o x do(es) + a2dx2(es)
,
for all s = n+1, ... , m. This is satisfied if and only if 2 d a(e
s
= d o(e s
)
for all s = n + 1, ... , m ; if and only if
do = 0
,
i.e. the function a is constant. Example 9.2.3 If o = 1, n = 2, VU2 is horizontal and VA 2 is vertical, then equations (9. 1.18) and (9.1.19) are both satisfied. For example, let 0: R3\{0{ -. S2 be defined by 0 (x) = x/ Ix 1, for all x s R3 \ { 01, . Then if o : S2 -. R is any smooth function whidi does not take on the value 0 E R ; o : R3 \ { 0; -- I R, e(x) = -5(0(x)) has
the property that Dal is horizontal. We thus obtain a new harmonic morphism
fd: R3\{01 _ S2 .
Example 9.2.4 Let 0: S3 _ S be the Hopf fibration; so x2 = 1. Put p = 1, then equation (9.1.19) is satisfied if V c2 is horizontal. For example, if a : S2 _ R is smooth and does not take on the value zero; define a: S3 R by a(x) = a (0(x) ), for all x e S3 , then V a2 is horizontal. Example 9.2.5 Let 0: R4 - R3 be the Hopf map of Example 7.2. 1. Then 0 is a harmonic morphism with dilation A given by X2 (x) = 4 Ix 12 , for all x e R4. The VA2 is fibres of 0 consist of great circles of Euclidean spheres of R4, and hence perpendicular to the fibres and hence horizontal. Equation (9. 1.20) becomes
-d(log a2)(e.) =
d(log p2)(e.)
for all j = 1, ... n . This is solved if a-2 = constant x p .
If we choose p2 such
that V p2 is horizontal - for example p2 (x) = 0 ( Ix I2 ) for some function P. then 161
equation (9. 1.21) is also satisfied. More generally we have the following.
Let 0: R4 - R3 be the Hopf map of Example 9.2.5, and let Tip : Rp - R 4, p > 4, be the projection map. Define x p : Rp - R3 , by x p = 4 2 0 o TI4 . Then x p is a harmonic morphism with dilation A given by X (x) _ 4 I T14 (x )12 , for all x E Rp . Thus 0X2 is horizontal. Equation (9. 1.20) becomes Example 9.2.6
- d(log a2)(e.)
=
2'(p-3)d(Iogp2)(e.)
for all j = 1, ... , n. This is satisfied if a-2
= constant x p
(p-3)
If we choose a2 such that V a2 is horizontal, for instance a2 (x) is a function of Ti (x )12 , then equation (9. 1.21) is also satisfied. I
4
Remark 9.2.7 When V a2 is horizontal, then a2 16 _ 1(y) = constant, for all y E N. Thus a2 can be thought of as a function on N, and we can view the change from the map 0 to the map 0 as changing the metric h on N to h = h/a2 . That is, equation (9.1.5) can be seen as
g = (1/X2)0*h+(1/P2)k= 6*h(a2X2) +(1/P2)k. In this way, by suitable choices of the functions p2 and a , it may be possible, by removing and adding certain points, to change the topology of both M and N. 9.3. Deformations of metrics for equivariant maps Suppose 0: M -- N is equivariant with respect to the isoparametric functions
s: M - R and t: N -- E, where we suppose M and N are space forms. Using the notations of Chapter 4, we consider the map 0
S't
: Ms
Nt
-
between level hypersurfaces of s and t. Away from the focal varieties we can express
the metric g of M as
g = ds2 +
gs
'
where gs is the induced metric on Ms. Similarly on N, we can write h as
h = d t2 + ht 162
,
where ht is the induced metric on Nt There exist four reasonable kinds of deformations of the metrics g and h which we can consider. (i) Define the new metric g on M* by
(1/µ (s)) ds2 + gs
(9.3.1)
,
wherep(s) is smooth and positive and such that the resulting metric g extends smoothly across the focal varieties. (ii) Define the new metric h on N* by h
= (1/v(t) )dt2 + ht
(9.3.2)
where v(t) is smooth and positive and such that the resulting metric h extends smoothly across the focal varieties. (iii) If Ms has principal curvatures A1(s) , ..., Ap(s) with corresponding eigen-
spaces S1, ... , Sp, then we can express gs as
gs = 91(s) + g2(s)
+
...
+gp(s)
,
where, for each x c Ms, and for all X,Y C T Ms, then X
gi(s)x(X,Y) =
gs(X,Y), if X,Y C Si(x) 0
,
if either X C S. (x), j
or
Y C S. (x), j
i i
Define a new metric g on M*by g = d s2 + gs, where
gs = (1/a1(s)2 )g1(s) +
where al ,
.
+ (1/ap(s)2)gp (s)
,
(9.3.3)
...
, ap are smooth functions, which are chosen such that the resulting metric g extends smoothly across the focal varieties. Essentially, if M is a sphere, we are changing the sizes of the principal curvature small spheres of Ms by a factor depending on S.
(iv) We perform a similar construction to case (iii) for the level hypersurface Nt, by expressing ht in its principal curvature components as
ht = h1+ ...
+ hq 163
and defining
ht = (1/v1(t)2)h1 + ... + (1/v (t)2)hq
(9.3.4)
for some suitably chosen functions vI(t) , ... , vq(t) . (v) We can perform various combinations of (i) ... (iv) always in such a way that the resulting metric extends smoothly over all of M (N) We consider each case in turn. (i) We first of all work out the connection coefficients for the new metric. Let x E M* , and locally about x choose an orthonormal basis (X ) with respect to g, a' such that if y E Ms for some s, then (X (y)) is an orthonormal basis for T M Now g = (1/p (sZZ )) d s2 +g I and so (Xa, T) _ (X p ) is an orthonormal basis with s a9
respect tog .
Proposition 9.3. 1 VX
If
= p VX4
V is the Levi-Civita connection with respect to -g, then
(9.3.5)
,
for all X C TM, s for s E int Is Proof We can adapt Proposition 9.1.2 to apply to our present situation, and if we write a2 = 1 and p2 = p2 , then equation (9. 1. 10) tells us that
VXX) = 2 g(S, VXx) p therefore
-g(VX ,X) = g(U,V X)
-µ _ 164
(VX E,X )
-fig (VX
X)
In the new metric g , V s no longer has unit length. We therefore reparametrize s into 9(s) , such that -g (V s , 0 s) =1. Clearly the new map 0: (M,g) - (N,h) (we assume that g extends smoothly to M) is harmonically equivariant with respect to s and t. We must therefore compute the equations (4.1.4) and (4.1.5) in the new variables . Note we are using Theorem 4.2. 1, since s is now a generalized isoparametric function. X1(s0) , ... , X (s0) are the principal curvatures of M. with respect to Suppose P the isoparametric function s, for some s0 E int Is. Then from Proposition 9.3.1, the principal curvatures of Ms with respect to the generalized isoparametric func0 _ tion s are given by X 1(s0) , ... , )gyp (s0) , where Ak(s0) = µ(s0)Ak(s0)
(9.3.6)
for k = 1, ... , p. Therefore
As = P As
(9.3.7)
,
where 2i is the Laplacian with respect to g . Since p. (t) and yk (s, t) remain unchanged, equation (4.1.4) become s lk
as(s)
p
+
k=1 µl k yk
(9.3,8)
0;
this is the reduction equation in the new metric g . Alternatively this equation can be written in terms of the variable s (from Lemma 5.1.1) as Ids l2 + O s as (s) + g
p
(9.3.9)
k I p. k yk = 0 .
Now Ids 12 = µ(s)2. Whilst g
p s = s (s) Id 1 22 + s'(s)Zs
(9.3.10)
g
But without loss of generality assume inf Is = 0, then s is given by
s(s) = f(1/µ(u))du. 0
Therefore s' (s) =A (s) and s"( s) = MI( s )p (s) . comes
Thus equation (9.3.12) becomes
165
p(s)2 a,t(s)
+(p '(s)p(s) +p(s) Z s)a' (s) p +
(9.3.11)
0.
k 1 pikyk
We therefore have, substituting equation (9.3.7) into equation (9.3. 11) ;
0 : (M,g) - (N.h) is equivariant with respect to isoparametric functions s and t. Then 0 is harmonic with respect to the metric g of (i) if Theorem 9.3.2
If
and only if
O"(s) +(pp((s) + &s)a'(s)+(1/p(s)2) E
k=1
p yk=0
(9.3.12)
ik
for all s s int I s (ii) Such a change in the metric h still preserves the fact that 0 is equivariant, for we are simply reparametrizing the isoparametric function t. We therefore have
Theorem 9.3.3 If 0: (M,g) -. (N,h) is equivariant with respect to the isoparametric functions s and t. Then 0 is harmonic with respect to the metric h of (ii) on N, if and only if p
a (s) + As a (s) + v E k=1
"
p
y
lk k
=0.
(9.3.13)
for all s c- int I s Proof We have simply changed the principal curvatures p1, ... , pq on the level hypersurfaces Nt by an amount given by equation (9.3.9) , that is (t)
= v(t)p. (t)
Since t is a generalized isoparametric function on (N,h ), the result now follows from equation (4.2. 1). (iii) First of all assume p = 2. Let mi = dim Si, i = 1,2, and consider a local frame field adapted to the principal curvature spaces ; (X ,) =
0
a
a=1,...,m-1
(Xi'XrMi=1,... , m1, r=m 1 +1,... ,ml +m2 =m-1 Proposition 9.3.4 166
If
(Xa, ) is the frame field defined above, then
(X
a,
0=
(Xi , Xr , V
) =(a 1 Xi , o2Xr, ) is an orthonormal frame field with respect to
If
is the Levi-Civita connection for g then the following formulae hold: g(Xi, V
o2g(Xi,V
=0
(9.3.14)
g(Xr,V) _
a2g(Xr,V U) = 0
(9.3.15)
0
(9.3.16)
)
g(Xt, V . Xs = a2g(Xt,vX Xs)
r
g
(9.3.17)
r
Xj) = 1 g (Xk, vX Xi
(Xk' V
(9.3.18)
g(X., VX Xs) = 2g(Xi, V Xs)
(9.3.19)
g(Xr,V Xj)
=
(9.3.20)
g(Xr, V- Xs)
= al a2 g(Xr,OX Xs)
(9.3.21)
g(Xj,VX Xi) =U g(X.,VX Xi) 1U2
(9.3.22)
r
r
a2 (Xr,VxXI)
r
r
g(Xi'VX.t )
= g(Xi,VX U) - dal( )g(X.,Xi)/gal
j
(9.3.23)
j
g(Xr,OX
)
g(Xr, VX)
) - da(U)g(Xr,Xs)/2a2
= g(Xr,OX
s
(9.3.24)
s
=0
(9.-3.25)
1
etc., for all
i,j,k=1,...,m1 and r,s,t=m1+1,...,m1+m2'
Similar formulae hold for any p, and we now assume p is arbitrary, and the prin-
cipal curvatures are al, ... , ap with multiplicities ml, ... , mp respectively. Corollary 9.3.5 The integral curves of with respect to g .
are affinely parametrized geodesics
Proof This follows from equations (9.3.14), (9.3.15) and (9.3.16).
167
Corollary 9.3.6 If X is a principal curvature vector of Ms , with principal i Xi is still a principal curvature vector of Ms curvature , then in the metric Aki
with principal curvature
k = Ak + uk
(s)/2 vk
where X. E Sk
i Proof This follows from equations (9.3.23), (9.3.24) and (9.3.25).
Corollary 9.3.7
With respect to the metric
s is a generalized isoparametri c
function.
Corollary 9.3.8
If E is the Laplacian with respect to p
As = As - E k=1
then
,
Mk ak (s)/2 ok
Proof This follows from Corollary 9.3.6 and Lemma 2.2.9 Let S denote the stress-energy tensor of 0 in the metric
.
again, and use the ranges of indices 1 < i,j, ... < ml; m1 + I < a,b, ... < m1 + m2 = m- 1.
< r,s, ...<m I +m2,
6
Assume p = 2
We make the following further assumption on the equivariant map 0 . One of the conditions for equivariance is that d O(Sk) C Tjk , for each k = 1, ..., p and some
jk = 1, ... , q . If whenever k / 1, then jk j 1; call 0 p - equivariant. For example, the map 0: Sm-1 - Sn-1; 0(coss.x, sins.y)) = (cos a (s) g1 (x), yESq-1, p+q=m,g1: Sp-1 _ Sr-1 , g2: sina(s)g2(y)), xESp-l'
S q-1 -- Ss -1, harmonic of constant energy, is p- equivariant. On the other hand
the map ¢: S3 _ S2 ; 0 (coss.x, sins y)) _ (cos a(s), sina(s).xky1), x,y E S1, is not p- equivariant. Proposition 9.3.9
V SO(Xk) for all
=
In the above notation, and provided 0
,
0 is p- equivariant, then
V *S0(Xr) = 0
k = 1,...,m1 and all r=ml+1""'
m1+m2.
Proof The divergence is given by (summing over repeated indices) 168
V *S0(Xk)
V*(6*ht) (Xk)
-Xa(0*ht(Xa,Xk))
0*ht(OX +
a
a'Xk)+
0*ht(
a'VXXk) a
}
_ -Xi(0*ht(Xi,Xk)) - Xr(0 ht(Xr,Xk)) + 0*ht(vX Xi,Xk) + 0*ht(VX Xr,Xk)
r + 0 ht(Xi, OX Xk) + 0*ht(Xr,o- Xk) . i r i
*
Now the 2nd term vanishes since 0 preserves the orthogonality of the principal curvature spaces. The 4th term vanishes from equation (9.3. 19) and the fact that the mean curvature of the small sphere .,/ is proportional to t (Proposition 2.2.1) . 2 The 6th term is equal to a2 a1 0 ht(Xr, oX Xk) . Now
r
g(Xs,V Xk) X
r
g(VX Xs,Xk)
r
But OX Xs points in the
r
t;
direction, since by Proposition 2.2.1, `12 is totally
geodesic in Ms, and so g(X Xs,Xk) = 0. Thus X Xk is either in S1 space or proportional to . Both are perpendicular to S2 space, and since 0 preserves this orthogonality; the 6th term vanishes also. We have now established that
V* S (X k ) 0
=
*
) + 0*h(X,V a3 (-X(0*h i k i t(Xi ,Xk )) + 0 ht (VX.X,X t i X.Xk))
But this is zero since the right hand side of equation (4. 1. 11) is zero. Similarly
V S0(Xr) =0 Similarly for arbitrary p, and provided 0 is p-equivariant, one can show V * 0 is proportional to . Theorem 9.3.10 If 0 : M, g) -- (N,h) is p- equivariant with respect to isoparametric functions s and t. Then 0 is harmonic with respect to the metric g of ( iii) if and only if
169
aiT(s)
°I (s)
p
+(As
p
- k 1 mk 20 k(s) ) as(s) - zdt(k 1 °k2
'k
p
- (1/2 a') E
k=1 ak °k1 yk
0
,
for s E int I
s
Proof As before
v*S0() = z(dsy+dty.a'(s)a"(s)a'(s) - v*(0 *ht) (
- O s ( a ' ( s ) )2
)
P
where y = E y , with yk = °k2 yk' k=1 k
Then
(d/(i s)y = k=1 2 E°k ok ky+ k=1 °2d y k s k E Also
V*(0*ht)(T)
E A*ht(Xa,VX a
a
)
E E °k0*ht(X. ,-XkX. ), where (Xi )i span Sk k ik k k kk E
2;
6*ht(Xik,Xik
°k2
kk
Xk
°k2 A kyk + kE °k2 ak (s) yk/2 °k Thus
P
zd
s
V * (0 *ht)
°k2
E
k=1
(2 ds yk - Xkyk) + 2 k °k 1c yk
°k °k yk ' from Proposition 4.1.12 M
2
k
The result now follows from Corollary 9.3.8 .
Remark 9.3.11 We have defined the notion of 0 being p-equivariant in order to carry out the deformation. We can replace this condition by another condition.
If 0: (M,g) - (N,h) is equivariant (M,N are space forms ) and 0 170
I.`fk:-
f. k
Ik
is harmonic of constant energy density, for each k = 1, ... , p, then call 0 '/equivariant with respect to s and t. We remark that all the examples of equivariant maps which we have considered are also 'f - equivariant. If 0 is '/ - equivariant and ik: fk - M is the inclusion map, then d(0 o i k ) = dO(Aik) + trace Vd0(dik, dik)
is proportional to n , and since dik is proportional to proportional to
r7
;
trace VdO(dik,dik) is
.
The new metric g has simply changed the sizes of the small spheres .'/ k' and we conclude that 0 ,t: Ms _N is still harmonic and of constant energy density in the S, t. t metric 'j. Thus 0 is equivariant with respect to the generalized isoparametric functions s and t, and Theorem 9.3.10 is true with "p-equivariant" replaced by "jequivariant "as a consequence of Theorem 4.2. 1 (iv) As in case (i), the integral curves of n = Vt are geodesics in the metric h , and the function t: N -- R is a generalized isoparametric function. If we assume that 0: (M,g) -- (N,h) is.,/-harmonically equivariant, then in the metric h , 0s t: Ms _N is harmonic of constant energy density. Thus 0: (M,g) -t (N,h) is equivariant with respect to the generalized isoparametric functions s and t. Theorem 4.2.1 applies, and we have Theorem 9.3.12 If 0: (M,g) -- (N,h) is y-equivariant with respect to the isoparametric functions s and t. Then 0: (M,g) -. (N,h) is harmonic, where h is defined in GO, if and only if
a (s) +as cx (s) + Ek (µ.
+
lk
v.
lk
/2 v )y /v. 2 k lk lk
=0
for all s E int I s . Proof
For each k = 1, ..., p, we have
yk(s,t) = E 0 * h. (Xi lk .
k
(1/v (t)2 ) yk(s,t)
Xi k
k
lk
where h. = h /v. (t)2 ). In the metric h the principal curvatures become 1
7
l = µ + v.-12 P. 171
for each j = 1, ... , q. The result now follows from equation (4.2.1). 9.4
Examples
Example 9.4.1 Consider a deformation of the kind given by case (i) of Section 9.3. The problem here is to define u(s) in such a way that g is a smooth metric. For example, suppose µ (s) = K, where K is some constant not equal to 1. Consider the join S2 = S1* S0 . Then, with respect to the metric g , S2 still has constant positive curvature [ 38, vol. 2, chapter 7, addendum I 1 equal to K2 , and so has the appearance of a rugby ball [ 13 1.
The metric g is no longer C1. However, we can use bump functions as follows. Define
µ(s) = 1 +l(s), t > -1
,
where t : Is -- R is smooth, and supp t c int Is. Then (M, g) is a smooth manifold, since the question of smoothness only arises across the focal varieties of s, and g = g in a neighbourhood of these focal varieties. Suppose we are given one of Smith's maps between two spheres. Consider the reduction equation (equation (1.3.7)) - we use the reparametrized equation to avoid singularities. Use the same reparametrization for equation (9.3.12), so that time varies between - oo and + oo. Then the asymptotic form of the equation as time u -- + '0 will be the same in the deformed case as in the undeformed case. Theorem 6.1.9 and Lemma 6.1.11 will still apply to yield a non-trivial solution. Therefore for the maps of Example 5.3.1, the existence of solutions will be unaffected provided It 12 is small enough. For maps from Euclidean space to sphere and from hyperbolic space to sphere, the existence of solutions will again be unaffected. However, we would expect the asymptotic behaviour as time u - + oo to change substantially with such deformations of the metric. 172
Example 9.4.2 Consider the join of two harmonic polynomial maps as described in Section 1.3 and Example 5.3.1 . We thus consider a map 0 : S m- i _ S n-1 of the form
0(coss x, sins y) = (cos a(s)gI(x) , sin o(s)g2(y))
,
where x SP-1 , y C S q-1 , gl : SP-1 _ S r- i and g2 : S q-1 -- S s-1 are harmonic polynomial maps with I dg. 12 = ? . constant, p + q = m and r + s = n. Express the Euclidean metric on S m-1 in the form g
= d s2
+
cost s dx2 + sings. dy2
where dx2 is the Euclidean metric for SP-1 and dy2 that for Sq-1. We now perform a deformation of the metric incorporating deformations of types (i) and (iv) in order to make Sm-i ellipsoidal. The deformed metric has the form
g = (a 2sin 2s + b 2cos2s) ds 2 + a 2cos2s dx 2 + b2sin2 sdy 2 The sphere has now become ellipsoidal with one set of axes of length a having multiplicity (p - 1), one set of length b with multiplicity (q - 1), and the other axis retaining its original length of 1. Using the notations of equation (9.3. 1) and equation (9.3.3) we have µ(s)
2
2 2 2 2 = 1/(a sin s + b cos s)
2
,
al
= 1/a
2 ,
2 a2
= 1/b
2
(9.4.1)
The reduction equation before the deformation has the form
a "(s) + As a '(s)
+
2 E µ k=1 ik
where as =(q - 1) cots - (p- 1) tans, yl = A1cos2(s)/cos2s, y2 = µ.
= tan a(s) and p
j2
cot a(s).
X2sin2(s)/sin2s,
Using Theorem 9.3.2 and Theorem 9.3.10 ,
after the deformation the reduction equation becomes a 1 '(s)
+
W µ ( (s)
+ A s)
a'(s) + ( 1/u(s))2 k2lµj akyk
(9.4.2)
k
We reparametrize equation (9.4.2) as before, defining a new variable u by eu = tan s. Note that care must be taken since equation (9.4.2) no longer has the form of equatica (5.1.1) and so Lemma 5.1.1 no longer applies. 173
A computation shows that
p '(s) _ -(a 2- b )2 sins coss/(a 2sin 2s + b 2cos2s) 3/2 or
(a2- b2) u -u e +e
p' (s(u)) =
b2e-u)
(a2eu + u
e +e
3/2
-u
This tends to 0 as u tends to both - -a and + oo . Also u -u e + ep(s(u )) 2 2
a e
u +b2e-u
which tends to 1/b2, 1/a2 as u tends to becomes, in the variable u , 1
eu + e-u
-Go, + 00
respectively. Equation (9.4.2)
[µ +(q-2)e-u - (p-2)eu] a' (u) +
(12
X e )
eu +e-u
1
sin a(u)cos a(u) C
a
2
u
A 2e
b
-u
2
= 0. (9.4.3)
If we abbreviate this in the form
ait(u) + h(u)a'(u) +g(u)sina(u) cosa(u) = 0
then as u -- - 0 h(u) -- q - 2
g(u) -- - X2
and as u
h(u) - -(p-2) g(u) -- X1. The damping conditions become
174
,
Thus the damping conditions are as for the Smith maps, given by equations (1.4. 1). The methods of Section 6.1 (in particular Theorem 6.1.9 and Lemma 6. 1. 10) show that, provided the damping conditions are satisfied, equation (9.4.3 ) has a solution
yielding a smooth harmonic map 0: S m-1 - S n-1, where S m-1 has the ellipsoidal metric described above. We therefore have Theorem 9.4.3 The join of two harmonic polynomial maps between spheres for which the damping conditions are satisfied, can be deformed into a harmonic map from the ellipsoid described above into the sphere. The ellipsoid has distinct eccen-
tricities given by a, b and 1, for any a, b > 0. Theorem 9.4.4.
The join of two harmonic polynomial maps between spheres for
which the damping conditions are not satisfied, cannot be rendered harmonic by an ellipsoidal deformation of the above kind on the domain sphere.
In contrast, however, we see in the next example that such ellipsoidal deformations on the range sphere do yield harmonic maps.
Example 9.4.5 Again consider the join of two harmonic polynomial maps as in the last example. We can deform the range sphere into an ellipsoid, by using deformations of types (ii) and (iv). That is, we express the Euclidean metric on Sn-1 in the form
h = d t2 + cos2 t du 2
+
sin2t d v2
,
where d u2 , d v2 are the Euclidean metrics for S r-1. S s-1 respectively, then the deformed metric has the form h = (a2 sin2 t + b2 cos2 t) d t2 + a2 cos2 t d u2 + b2 sin2 t d v2
In the notations of equation (9.3.2) and the equation (9.3.4), we have v(t)2
=
1/(a2sin2t
+
b2cos2t), v1 = 1/a2, v2
=
1/b2
(9.4.5)
From Theorem 9.3.3 and Theorem 9.3.12, the reduction equation becomes 2
a (s) + As a '(s) + v(a(s)) E "
k=1
-yk
L.
lk
2
=0.
V.
)k
175
Using the parameter u defined by eu = tan s, we obtain
a" (u) +
eu+e -u 1
(eu
((q - 2)e-u - (p -2 )eu) a' (u)
sin a (u) cos a (u) +e-u) (a2sin2 a(u) +b2cos2a(u))2
(a2a eu - b 2 A a-u) = 0 . 1
2
(9.4.5) This equation fits into the category of equations described in Theorem 6.1.9. The arguments of that section apply, and we are assured of a solution provided the damping conditions are satisfied. These damping conditions are that
(q-2)2 < 4bX2 (p-2)2 < 4a X1.
(9.4.6)
Given the join of two harmonic polynomial maps, we can always find a and b such
that (9.4.6) are satisfied. We therefore have Theorem 9.4.6 The join of any two harmonic polynomial maps can always be rendered harmonic by a suitable deformation on the range sphere. In particular we can apply Theorem 9.4.6 to Example 1.4. ]. to yield
Theorem 9.4.7 For each n = 1,2, ... , there exists a smooth metric on the range sphere Sn (depending on n), such that each class of Tin (Sn) = Z contains a harmonic representative. The deformed spheres are familiar ellipsoids whose eccentricities depend only on n and the degree.
176
References
1
2
3
4 5
R. Abraham and J.E. Marsden Foundations of Mechanics (Second Edition), Benjamin/Cummings, (1978). P. Baird and J. Eells A conservation law for harmonic maps, Springer L.N.M. vol. 894 (1981), 1 - 25. F. Brauer and J.A. Nohel Ordinary Differential Equations, Benjamin, (1967). F. Brauer and J. A. Nohel Qualitative Theory of Ordinary Differential Equations, Benjamin, (1969). Families de surfaces isoparametriques dans les espaces E. Cartan a courbure constante, Ann. Mat. Pura Appl. , 17 (1938), 177 - 191.
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Sur des families remarquabies d'hypersurfaces isoparametriques dans les espaces spheriques, Math. Z. , 45 (1939) , 335 - 367.
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ii J. Eells and L. Lemaire
On the Construction of Harmonic and Holomorphic Maps between Surfaces, Math. Ann. , 252 (1980), 27 - 52.
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Cambridge Monographs on Math. Physics (1973). Riemannian submersions from complex projective space,
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(1924), 1- 32. Differential Equations, Dynamical Systems and Linear Algebra, McGraw-Hill, (1974). 24 D. Hoffman and R. Osserman The geometry of the generalized Gauss map, Memoir A,M.S. 236 (1980). 25 W.-Y. Hsiang and H.B. Lawson Minimal submanifolds of low cohomogeneity, J. Diff. Geom. , 5 (1971), 1 - 38. Fiber Bundles, McGraw-Hill, (1966). 26 D. Husembller 27 T. Ishihara A mapping of Riemannian manifolds which preserves harmonic functions, J. Math. Kyoto Univ. , 19 (1979), 215 - 229. 28 L. Lemaire On the existence of harmonic maps, Thesis, Warwick Univ., (1977). 29 1-I. P. McKean The Sine-Gordon and Sinh-Gordon Equations on the Circle, Comm. in Pure and Appl. Math., Vol. XXXIV. , p. 197 (1981). 30 H.F. Munzner Isoparametrische Hyperflachen in Sph'aren I, Math. Ann. 251 (1980). 57 - 71. 31 M. Nagumo Uber die Differentialgleichung y"= f (x,y,y'), Proc. Phys.-Math. Soc. Japan (3) 19 (1937). 32 K. Nomizu Elie Cartan's work on isoparametric families of hyper-
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179
Index of definitions
Clifford system Critical set of a map Damping conditions
Degree of an isoparametric function Dilation
Energy density
Equivariant with respect to isoparametric functions Focal variety Generalized family of isoparametric hypersurfaces Generalized isoparametric function Harmonic map Harmonic morphism Hopf invariant Hopf map
Horizontally conformal
Isoparametric function Isoparametric hypersurface Join of a map Join of two spheres Laplacian Orthogonal multiplication
p- equivariant with respect to isoparametric functions Reduction
Reduction equation
Rendering problem S-equivariant with respect to isoparametric functions S-wavefront preserving with respect to isoparametric functions 180
Second fundamental form of a map
1, 7
Stiefel manifold
36
Stress-energy tensor Unit speed reparametrization of a generalized family of isoparametric hypersurfaces Wavefront preserving with respect to isoparametric functions
43
66 50
181
