Constant Mean Curvature Surfaces, Harmonic Maps and Integrable Systems
Birkhuser
Lectures in Mathematics ETH Zurich Department of Mathematics Research Institute of Mathematics Managing Editor: Michael Struwe
Constant Mean Curvature Surfaces, Harmonic Maps and Integrable Systems Notes taken by Roger Moser
Author's address:
CMLA Ecolc Normale Superiture do Cachan 61. avenue du President Wilson 94235 Cachan Cedex France
2000 Mathematical Subject Classification 53C42. 53C43; 53C28, 53035, 70006
A CIP catalogue record for this hook is available from the Library of Congress. Washington D.C.. USA
Deutsche Bibliothek Cataloging-in-Publication Data HElein, FridE ic: Constant mean curvature surfaces, harmonic maps integrable systems I Freddric Hdlein. Notes taken by Roger Moser. - Basel ; Boston ; Berlin : BirkhBuser, 2001
(Lectures in mathematics : ETH Zurich) ISBN 3-7643-6576-5
ISBN 3-7643-6576-5 Birkhauser Verlag, Basel - Boston - Berlin This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation. broadcasting, reproduction on microfilms or in other ways, and storage in data banks. For any kind of use permission of the copyright owner must he obtained. 0 2001 Birkhttuser Verlag. P.O. Box 133, CH-4010 Basel, Switzerland Member of the BertelsmannSpringer Publishing Group Printed on acid-free paper produced from chlorine-free pulp. TCF Printed in Germany ISBN 3-7643-6576-5 987654321
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Contents Preface
7
.
Introduction: Surfaces with prescribed mean curvature .
9
From minimal surfaces and CMC surfaces to harmonic maps Minimal surfaces 2.1 Constant mean irvature.surfaces 2.2 .
15 16 18
3
Variational point of view and Noether's theorem
22
4
Working with the Hopf differential 4.1 Appendix
34 39
5
The Gauss-Codazzi condition Appendix 5.1
41
1
2
6 7
8
9
50
Elementary twistor theory for harmonic maps . Appendix 6.1 Harmonic maps as an integrable system Maps into spheres 7.1 7.2 Generalizations A new setting: loop groups 7.3 7.4 Examples . . . . . . . . . .. . .
52
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60 63 63 68 71
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Construction of finite type solutions Preliminary: the Iwasawa decomposition (for Rc) 8.1 Application to loop Lie algebras 8.2 The algorithm 8.3 Some further properties of finite type solutions . 8.4
Constant mean curvature tori are of finite type . . . . . . . . . . . . . The result . . 9.1 Appendix . . . . . . . . . . . . . . . . . 9.2
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83 84 87 89 89 94
Contents
6 10
Wente tori CMC surfaces with planar curvature lines 10.1 A system of commuting ordinary equations 10.2 10.3 Recovering a finite type solution 10.4
11
Spectral curves
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107 107 109 110 .
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Weierstrass type representations Loop groups decompositions . 11.1 Solutions in terms of holomorphic data . 11.2 Meromorphic potentials 11.3 Generalizations . 11.4
Bibliography
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111
112 112 113 115 1119
117
Preface One of the most striking development of the last decades in the study of minimal surfaces, constant mean surfaces and harmonic maps is the discovery that many
classical problems in differential geometry - including these examples - are actually integrable systems. This theory grew up mainly after the important discovery of the properties of the Korteweg-de Vries equation in the sixties. After C. Gardner, J. Greene, M. Kruskal et R. Miura [44] showed that this equation could be solved using
the inverse scattering method and P. Lax [62] reinterpreted this method by his famous equation, many other deep observations have been made during the seventies, mainly by the Russian and the Japanese schools. In particular this theory was shown to be strongly connected with methods from algebraic geometry (S. Novikov, V.B. Matveev, I.M. Krichever... ), loop techniques (M. Adler, B. Kostant, W.W. Symes, M.J. Ablowitz...) and Grassmannian manifolds in Hilbert spaces (M. Sato... ). Approximatively during the same period, the twistor theory of R. Penrose, built independentely, was applied successfully by R. Penrose and R.S. Ward for constructing self-dual Yang-Mills connections and four-dimensional self-dual manifolds using complex geometry methods. Then in the eighties it became clear that all these methods share the same roots and that other instances of integrable systems should exist, in particular in differential geometry. This led K. Uhlenbeck [82] to describe harmonic maps on a two-sphere, with values in U(n) using families of curvature free connections depending on a complex "spectral" parameter. At the same period N. Hitchin [54] investigated
finite type tori into SU(2) starting from similar methods. Such formulations were already proposed in the seventies by K. Pohlmeyer [69], V.E. ZhakarovA.V. Mikhailov [91] and V.E. Zhakarov-A.B. Shabat [92]. A catalysor of these developments was the construction by H. Wente in 1984 of an immersed constant mean curvature torus in ]R3, which had the effect of removing an old inhibition due to the fact that people believed that such tori should not exist. Many results then followed quickly and we have now a very rich and fruitful theory for constructing constant mean curvature surfaces and harmonic maps of surfaces with values into symmetric manifolds using integrable systems methods. All that seems apparentely a new theory, but many features of the "completely integrable behaviour" of constant mean surfaces have been guessed by geometers of the nineteenth century. Namely the existence of associated families of such surfaces by O. Bonnet, the study of special surfaces with planar curvature lines by A. Enneper and his students and the various Backlund transformations discovered and studied by A.V. Biicklund, L. Bianchi, S. Lie, G. Darboux, E. Goursat and J. Clairin (see [71] for details and references).
This Monograph is intended to give an introduction to this old and new theory from the point of view of differential geometry. For that reason, it has
8
Preface
seemed more natural for me to introduce the existence of families of curvature free connections for harmonic maps starting from the associated family of constant mean surfaces. Note however that this is not the historical way the theory actually developed and contemporary people were initially more inspired by the example of the KdV equation and the twistor theory. We also presented here some basic exposition of the twistor theory for harmonic maps, which was initiated to my knowledge by E. Calabi. Indeed this theory was an important
stimulation for the integrable system theory, in particular in the work of K. Uhlenbeck, and shares some similarities with the integrable systems theory (for a complete exposition, see [231). 1 also made an effort to present the beautiful result of U. Pinkall and I. Sterling concerning constant mean curvature tori (Chapter 9) in the framework of loop groups theory, in order to show how this result connects with the rest of theory, which uses loop groups. The present text is just an introduction and is far from being complete. We may recommand as parallel lectures the books [46] and [43] plus of course ... reading the cited papers. These Notes come from a lecture that I gave at the Eidgenossische Technische Hochschule Zurich during Spring 1999. Most parts of the text were written and typed by R. Moser. I wish here to thank the ETH Zurich and more particularly
Prof. M. Struwe for his hospitality and to thank R. Moser for his very nice work.
I also want to thank Joseph Dorfineister and Pascal Romon for their valuable comments on this text.
1
Introduction: Surfaces with prescribed mean curvature
Curvature
For a curve I' in a plane, and for any point on this curve, there exists a circle (or a straight line) which is the best approximation at this point of the curve up to third order. The inverse of its radius k = R (or k = 0 if the best approximation is a straight line) is called the curvature of the curve at the given point.
For a surface E in R3 and a point m E E, we consider the 1-parameter family of affine planes which contain the straight line passing through m and being normal to E at m. Each of these planes locally intersects the surface along a planar curve containing in. We may label these planes by choosing one, say Po, and for 9 E R/27rZ naming by PO the image of Po by a rotation of angle 9 around the normal line to E at in. Then the curvature k(O) of PO fl E at m is of the form i2 + 2 2 cos(2(9 - Oo)), where kl and k2 are two numbers, called the principal curvatures of E at m. They are the extremal curvatures, achieved at two extremal positions of Po which are orthogonal (the principal directions). The average H = (k1 + k2) of kl and k2 is called the mean curvature of E at in, their product K2 = klk2 the Gauss curvature. If kl = k2, i. e. if the curvature k(9) is a constant, then the point in is called umbilic.
Experiments with soap
We are going to describe some experiments, carried out by J. Plateau, which may give some physical motivation for what is to follow. Imagine a piece of wire, bent in the shape of a closed curve r, that is dipped
into a solution of water and soap, such that it becomes spanned with a soap film. This film will take the form of a surface Eo with boundary 0EO = r, having least area among all surfaces with the same boundary. Computing the Euler-Lagrange equation of this variation problem, we see that Eo satisfies H = 0,
(1.1)
where H is the mean curvature of Eo as described above. Such a surface is called a minimal surface. Suppose now that there is some device that allows to have different pressure on either side of the soap film in question. Then instead of (1.1), the equation
H=C
(1.2)
10
1. Introduction: Surfaces with prescribed mean curvature
will hold for E0, where C is a constant different from 0 (depending on the difference of the pressures applied). In this case we call Eo a constant mean curvature (CMC) surface.
First and second fundamental form Let E be a piece of a surface, embedded in R3, which is diffeomorphic to the unit disk D2 = {z = x + iy E C : Izi < 1). Let X: D2 --' JR3 be a parametrization of E. Then we can define the following.
Definition 1.1 The first fundamental form of an embedding X: D2 - R3 is the quadratic form given by the matrix
ax aX
ax IZ dx
8x I
t3y
z
Ox ' ay >
1
ay
,
which depends on z E Dz.
= IdX(e)12. Now fix an orientation For = E R2, we compute of E by assuming that (e , ) is an oriented basis of Tx()E. Consider
u=
x Y x ax K
BX ' I
where x denotes the vector product in ]R3. The map u: D2 - S2 thus given is called the Gauss map of X. Definition 1.2 The second fundamental form of X is the quadratic form given by
a2x asay u
0)
Cuaway
+
>
8y
u
With the notions of the first and second fundamental form, we are able to define the principal curvatures of a surface in a more convenient way than before.
Definition 1.3 The principal curvatures k1, k2 are the eigenvalues of II with respect to I, i. e. the solutions of
det(II - Al) = 0.
1. Introduction: Surfaces with prescribed mean curvature
11
By a multiplication of II - AI with I-' from the right it is easily verified that the mean curvature and the Gauss curvature can be computed from I and II in the following way:
H = I tr(II I-' ),
K = det(II)/ det(I).
(1.3)
Conformal coordinates
To simplify things, we will use conformal coordinates. Choose E like above. The following is a well-known result.
Theorem 1.1 There exists an embedding map X : D2 --+ R3, such that
i) X(D2) = E, and ii) X is conformal, i. e.
12-IW12-
\8x'
lax
ay0.
(1.4)
Note that (1.4) means that the vectors az and have at each point the same length and are perpendicular to each other. This implies that there is a function w: D2 -+ R, such that
ex _
ax _
8x
+J
= ewe1,
= e e2,
and (e1, e2) establish an orthonormal basis of TX(Z)E for each z E D2. Thus we may write
I=e2m (1 1\
II=elm( h"
f
1
h22
h12
Then k1 and k2 are the solutions of det ( ( h11
\\ h
12
fi r 0
h21
11
))
`
h 22 )
0
L e. the eigenvalues of the matrix h11
h21
h12
h22
Consequently,
H = h11 + h22 2
h21 1
K = det h12
h22 J
'
1. Introduction: Surfaces with prescribed mean curvature
12
We see that, for conformal X,
2H = h11 + h22 = e-2-(AX, u), or
(OX, u) = 2e2i''H. Moreover, we have the following.
Lemma 1.1 If X is conformal, then
ax ay. ax oxl ax, This can easily be verified by differentiating (1.4). The details are left to the reader. It allows, however, the following conclusions.
Corollary 1.1 The map X satisfies
OX = (OX, u)u = 2He2i'u = 2H
ax
ax
ax x ay ,
as soon as it is conformal.
Apply this to the equation (1.2) in the form
H=C=Ho. We see that it is equivalent to
OX = 2Ho
Ox Ox ax x ay
In particular this implies that minimal surfaces are always images of harmonic and conformal parametrizations. This leads to the problem of finding maps X: D2
11 P3, such that
X is conformal, i. e.
/ax
o=laxl2-
- 2i ax , - e) I2
AX=0.
1. Introduction: Surfaces with prescribed mean curvature
13
Weierstrass representation Using the notation of complex differentiation, a
1
a
az = 2 (ax
a
a
a a a x + i ay)
1
- i ay) ' az - 2
a2
4 azaz'
we find that our problem can be expressed in the form
aX)2
= 0,
( az a (az) OX
= 0.
az
Write f = 2'9x: D2 -+ C3. We want to solve f 2 = 0,
02
= 0.
This can be done explicitly by the representation
f=
i(w2 - 1) ,(W2 + 1) iw
h,
where w, h: D2 - C are holomorphic functions. (The function w might be meromorphic.) Eventually, this gives us Weierstrass representation of a solution of our problem:
X(z) = Re
I
f
f(C)dC
.z
where zo is an arbitrarily chosen point in D2. Completely integrable systems The Weierstrass representation for minimal surfaces seems to be a miracle, since it describes all solutions of a nonlinear geometrical problem by a very simple algebraic construction. We shall see in these lectures more sophisticated miracles, occuring in various geometrical situations. People called them completely integrable systems. Classically, they are nonlinear equations on which mathematicians and physicists discovered unusual properties. Some of these properties are
existence of solitons. This terminology comes from evolution problems and the most famous example is the Korteweg-de Vries (or KdV) equation
14
1. Introduction: Surfaces with prescribed mean curvature at + 6u a2 + a = 0, modelling water waves in a flat shallow channel. The story started in 1834 with the experimental observation of a solitary wave travelling along such a channel along a very long distance by J.S. Russel [73). A model equation was derived in 1895 by D.J. Korteweg and G. de Vries [591. Solitons are solutions of nonlinear partial differential equations which are localised in space, whose profile is not dispersed after a long period of time and which resist to interactions with other solitons 1. Thus it is a smooth field, solution of some partial differential equation, which behaves like a particle. Backlund transformations. A baby example is the following: start from
a harmonic function f of two real variables x and y. We may write the Laplace equation for f as d (- 2L dx + a dy) = 0, which implies that, if we work on a simply connected domain, there exists a function g such that
- ay dx + a dy = dg. Then g is another harmonic function, namely the conjugate function off (i.e. f +ig is a holomorphic function of x+iy). Such transformations, producing a solution of some partial differential equation starting from another solution work also in nonlinear situations. by a nonlinear change of variable, the problem reduces to solving linear equations. Weierstrass representation does it obviously, reducing the minimal surface equation to the Cauchy-Riemann system. We shall meet the same situation, but involving a much more complicated "change of variable". a Hamiltonian structure. Solutions are spanned by the Hamiltonian flows of commuting functions on a symplectic (or Poisson) manifold. This is described by Liouville's theorem. Infinitely many symmetries. Infinite dimensional Lie groups acts on the set of solutions of these systems.
The theory of completely integrable systems is not a pragmatic point of view, in the sense that it does not really provide a method (like the analytic approach based on functional analysis) where, starting from some qualitative intuition, one works (often hardly) to construct the tools one needs for proving (or disproving) what one believe to be true (existence, regularity,...). Indeed working on completely integrable systems is rather based on a contemplation of some very exceptional equations which hide a Platonic structure: although these equations do not look trivial a priori, we shall discover that they are elementary, once we understand how they are encoded in the language of symplectic geometry, Lie groups and algebraic geometry. It will turn out that this contemplation is fruitful and leads to many results. 1for the KdV equation, the simplest solution involving the soliton behaviour is the onesoliton u(x, t) = 2a2sech2(a(x - 4a2t)), for a > 0
2
From minimal surfaces and CMC surfaces to harmonic maps
Let Il be an open subset of C and X: fZ - R3 a conformal parametrization of a surface E. We use all the notation form the previous section. In particular, we have for the mean curvature H =
h11 + h22
2
if we write for the second fundamental form hll
h21
h12
h22
II = e2"
It is clear that II is symmetric. So, for a fixed H, there exist functions a, b:11-- R
such that h11
h12
h2l
_
H + a
b
h22
-
b
H - a
(2.1)
For convenience, we denote coordinates on C alternatively (x, y) or (x1, x2), where x = xl, y x2. Because of the fact
\
\
which is a consequence of the orthogonality of u to II in yet another form:
II = - C/ au, aX
ax' axe
/
and a , we can write
i,i=1,2
(2.2)
Note furthermore that (u, 9'9 ) = 0 for i = 1, 2 (because of Jul = 1).
The map X provides a moving frame on the surface E, i. e. three vector fields that constitute a orthonormal and oriented basis of R3 at each point of E. It consists of the vector fields e1
Thus
au
_
_w
ax = e
ax _ "ay, ax u=el xe2. a -wax, e2=e
0'X ay
/ au aX \ _w / au el + e `ax ax ' ax J
e2,
Using (2.1) and (2.2), we finally obtain
au au ax' ay
__ _
aX aX
H+a
b
ax ' ay
b
H-a
i = 1, 2.
2. From minimal surfaces and CMC surfaces to harmonic maps
16
2.1
Minimal surfaces
Suppose now that E is a minimal surface. Then
- \a , ay/
(Ox,Oy/
\ b -a
Comparison with antiholomorphic functions
A function f : Q - C is defined to be antiholomorphic, if Of
= Of
Oz
Ox
iOf _ 0. Oy
In other terms, (ax
(b a)
ay)
,
if a denotes the real part and b the imaginary part of . This bears some resemblance to (2.4). We will see that this is no coincidence. As a matter of fact, u can be seen as an antiholomorphic function in a certain way. Another point of view on antiholomorphic functions comes form the following (equivalent) definition.
Definition 2.1 A map f : St - C is antiholomorphic if and only if i) it is weakly conformal, i. e. I af12
efl2-2i(af,of Oy
Ox
Ox Oy
0,
and ii) it reverses orientation, i. e.
detl Ox ,ay <0. The difference between condition i) and the definition for conformal maps is that weakly conformal maps are allowed to have vanishing derivatives.
Now note that (el, e2) is an oriented basis of
at each point in the sense
that it constitutes an oriented basis of R3 together with u. Thus (2.4) shows that u is both weakly conformal and orientation reversing. On the other hand,
2. Fi om minimal surfaces and CMC surfaces to harmonic maps
17
S2 can be mapped to C U {oo} by a conformal and orientation preserving map, the stereographic projection
(x', x2, x3) '-' x1 + ix2 1+x3
P: S2 --C U {oo},
which has the inverse map
P-'(z) =
1
+llzl2 (z + z, -i(z - z),1 - Iz12).
The composition P o u: S2 -+ C U {oo} is thus weakly conformal and orientation
reversing, hence antiholomorphic on the set it\u-1{(0,0,-1)}. This means
a Pou=0. Set v = -
= P o (-u). This function is clearly meromorphic. Furthermore, it allows the representation
v+v
U = -P-' o v = -1 +llvl2
-i(v - 2 ) 1 - Iv1Now
consider again (2.4). In complex notation it is
8 = -(a+ib),X. So by differentiation we obtain
ax
1
8
v+v
1
a+i6az (1+IvI2
az
8v
2i
B
1(vIv12)
:(v2-1) .
(a + ib)(1 + Iv)2)2 8z
2
+ 1)
,
iv
which corresponds to Weierstrass representation. Harmonic maps
We are going to exploit (2.4) some more. Take the vector product with u to get
ux Ux
OU
_
-(ac-68 )_-, a
ex
+b
OX)
OU
2. From minimal surfaces and CMC surfaces to harmonic maps
18
As a consequence of this we have
uxOu= /\
Ox
j+ lJ( uxay)
2
ax0y+5 yax-0'
which means that Du I. T S2. This property is worth a little consideration. Definition 2.2 A map u: S2
S2 is called harmonic if and only if it is a solution of Du 1 TuS2. More generally, let H be a submanifold of dimension n without boundary of an Euclidean space RN. Let S2 be an open subset of Rm. A map u: (1 --- H is called harmonic if and only if Du I. TuH everywhere.
For instance, if Il is an open interval (a, b), then a curve ry: (a, b) --- H is harmonic if and only if it satisfies ry(t) 1 T.y(t)H, which means that ry is a constant speed parametrization of a geodesic. Or, if H = R, then the harmonic maps f - R are just the solutions of the Laplace equation Au = 0.
2.2
Constant mean curvature surfaces
Now let's see what happens for CMC surfaces. In this case, instead of (2.4) we have to use (2.3). It implies
ex= + b-u x08xT = -(H + ax 8y 8x a)-
ux
On
-- - 2Hax 49U
8y
= (H -a) -- - bay = a+2H
8y '
.
Like before we compute
u x ou =
2 OfI u x xl + try OfI u x ay\I = ( ex axay + ayax
(u+2HX) = 0.
Hence also in this case u is harmonic. Thus we have proved half of the following, which is due to E. A. Ruh and J. Vilms, (72].
Theorem 2.1 Let X: St -- R3 be a conformal immersion of a surface, then its Gauss map u: St -' S2 is a harmonic map if and only if the image of X is a CMC surface.
Converse result
We will see a kind of converse: Any harmonic map u:11 S2 from a simply connected open set f2 in C into S2 is the Gauss map of two weakly conformal
CMC immersions, both having mean curvature H = 2. First we write the
2. From minimal surfaces and CMC surfaces to harmonic maps
19
harmonic map condition in a form that is more suitable for this purpose. Since U E S2, the condition Du 1 TUS2
is equivalent to
0=ux0u=Oxlux which implies that the 1-form u x
/+ay l uxTY j,
dx - u x ez dy is closed, i. e.
dlux OUdx-ux.Udy)
= 0.
But since 11 is simply connected, this 1-form is even exact. So we can find a map B: fl -+ R3
that satisfies
8B ex OB
y
Ux
Ou
ay ,
_ -ux
(2 . 5)
8u
ax.
This map will help us construct the CMC surfaces we are looking for. But first we have to check a few properties of B.
i) If B is an immersion (which is true if and only if u is so), then its Gauss curvature is 1.
To check this, note that the vectors and !LB- come from a= and by a rotation around u by an angle of 2 . It is clear therefore that OB OB au. ex x 8y = ex x 8y
8u 8y
Ou
Thus u is the Gauss map of B. Now we use a geometric characterization of the Gauss curvature of a surface: it is the ratio between the pull-back of the area form on S2 by the Gauss map and the pull-back of the area
2. From minimal surfaces and CMC surfaces to harmonic maps
20
form on the surface 2. So we have
`ux) =1.
K=
8B x
BOUB
U, T.-
Hence, if B is an immersion, its image has a constant Gauss curvature equal to 1. This means that the metric induced by this immersion makes B(ft) intrinsically isometric to a subset of S2, or in other words, that B(ft) is a (non-canonical) isometric immersion of the sphere. ii) Construction of X1 Also by differentiating (2.5), one can prove that OB = 2
au
au
ax x ay
-2
aB au &u OB OB ax x ay - ax x ay + ax x ay
OB
(2.6)
Hence B is a harmonic map onto its image. As mentioned earlier, we may see B as a harmonic map into a subset of S2, immersed isometrically in a non-canonical way. Actually, B is a kind of harmonic conjugate map to u into the sphere.
We want to compute 0(B + u) and 0(B - u). For u, we have 0 = 0(IuI2) = 2(u, Du) + 21du12,
where Idul2 = conclude that
Yx-
12 + p I2. Since we know that Du is parallel to u, we
Du = -ulduI2 au -T x(ux /- x(uxl OU
au
aB
OB
/
au
(2.7)
//
axx t?y+ax xt?y The equations (2.6) and (2.7) imply
A(B + u) =
a(B + u) x a(B + u) Ox ay
o(B - u) = a(B - u) ax
x
a(B - u) ay
2Indeed, since I = °dB.dB and II = -tdB.du, we deduce that K = det(II)/det(I) =
det(°dB.du)/ det(IdB.dB). But recall the following algebraic fact: if M = (MI, M2) and N = (NI, N2) are two 2 x 3 matrices (MI, M2, N1, N2 being column vectors in R3), then det(/M.N) = (MI x M2, NI x N2). Applying this here, it gives 8u 8u 88 8B 8u 8u
K=
2
xyl
`u'
-
X 85) -6i
2. From minimal surfaces and CMC surfaces to harmonic maps
Set X
21
B ± u. Then these maps satisfy ox}=8X±XOX±
ex
by
iii) The maps X± are weakly conformal. This can be checked by a mere calculation. Remember that -!M and a6 are obtained from as and gy- by a rotation by 2. iv) Conclusion We find that X+ and X- are both CMC immersions with mean curvature H = (cf. corollary 1.1). The images of X+ and X- are parallel to the image2 of B and pointwise at distance 1 from B. Of course they also have u as their Gauss map. This is a result of O. Bonnet [17] (see also [48]). The notion of harmonic map can be extended by replacing 92 by any Riemannian manifold (see the next Chapter), like for instance the n-dimensional sphere S". One may then ask whether one can associate to a harmonic map u:fCC S" a conformal immersion in a similar fashion as above. This is indeed the case: we can associate to u a conformal immersion X into the Lie algebra so(n+2), unique up to the addition of a constant in so(n+2), by a construction generalizing the above one. For n = 2, because of the decomposition so(4) ^_- so(3) ® so(3), this immersion eventually splits into two components X+
and X-. However the geometric interpretation of X in the general case is less clear: it is related to a kind of Wess-Zumino-Witten problem. For more details, see [48].
Exercise. Let u : St , S2 be a harmonic map and B : St -' R3 defined by (2.5). Consider the map P : fl - R3 defined by
P=uxB. Prove that
i) P(z) 1 u(z) everywhere. ii) P is a solution of OP + IduI2P = 0.
iii) For any smooth one-parameter family ut : R - S2 of harmonic maps, the map : Q --+ 1R3 is called a Jacobi field. Conclude that P is a Jacobi field.
3
Variational point of view and Noether's theorem
Harmonic maps Consider the set E consisting of all smooth maps u: ci -' H, where S2 is an open
subset of R' and H a smooth compact submanifold of RN without boundary of dimension n. For any u E E, we define its energy to be
E(u) =
f
Jd2 F2
dx,
n
fn
where Idu12 = El , TJ7 I2 and dx = dxl
dx"' We are interested in maps u that have finite energy (what we will assume subsequently). More generally, we could equip Il with a Riemannian metric gQp. The energy would then be defined similarly, except that instead of dx we'd have det(gap) dx and instead of IduI2 the expression I
m QEi
/axa 8u. Ou \ 5X#)
where (g°Q) is the inverse of the matrix (gQp). We will however not pursue this
any more, but restrict our attention to the flat case. We claim that harmonic maps in the sense of definition 2.2, i. e. solutions
uEEof
Du 1 are critical points of the energy functional E in E. To prove this, let V6N be a tubular neighbourhood ofH of width b. If b is sufficiently small, then for any y E V6N there exists a unique nearest point to y in N. Thus we can define the nearest point projection
rl:V6N-*N, which assigns that point to y. Moreover, this map is smooth. Now let u E E and 0 E C° ° (9, R'). The map ut = rl(u + t¢) belongs also to E, provided that Its is sufficiently small. We are now going to define what exactly is meant by a critical point of E. Definition 3.1 The map u E E is a critical point of E on E if and only if
E(ut) = E(u) + o(t) for all 0 E C' O (Q, RN).
3. Variational point of view and Noether's theorem
23
Our aim is to compute the Euler-Lagrange equation for this kind of variation. Consider first the derivative dHu(x) at a point u(x) E N. It is geometrically clear (and can be computed) that dIIu(x) is the orthonormal projection P(u(x)): RN -+ Tu(x)N onto the tangent space Tu(y)N. So we have ut = U + tP(u).4) + o(t),
and thus
8E (u)
dut dt
= 5E(u)(P(u).4))
r
O
M
(
a
fn a=1 \ axa ' axa
(P(u).o
/
dx
if (Au, P(u) 4)) dx
- J (P(u).Du, 0) dx. The Euler-Lagrange equation is therefore
P(u).iu = 0, which is just another way to express the condition
Du I TuN. In the case N = S2, we have seen that harmonic maps satisfy the equation
Du+ulduI2=0. A similar condition in the general case would be useful. To derive it, let P'(u) = idRN - P(u) be the orthonormal projection on (TuN)l. Differentiate the equation Pl (u)' axa = 0 to obtain '9U
F, aaQ (Pl(u)). a'a Q=1
+ Pl(u).Au = 0.
Hence
mN
Pl (u).Du =
au? au _: -A(u)(du, du). - a=1j=1 E oPla (u) axa axa
24
3. Variational point of view and Noether's theorem
The expression on the right hand side is a quadratic form in du. We call A the second fundamental form of the embedding N C RN. (As a matter of fact, it coincides with the second fundamental form as already defined for embedded surfaces.) If we apply this equation to a harmonic map, then it yields
Au = -A(u)(du,du). So we have indeed a generalization of the S2-case.
Noether's theorem: The effect of continuous symmetries Once again our aim is to generalize a fact observed for harmonic maps into S2. Recall that the harmonic map condition
Du 1 TUS2 for a map u: St , S2 on a domain Q C C can be written in the form
I+ JIuXOU)=o.
a CuX
That is, some divergence term vanishes. `Another observation to be made for
S2 is the fact that it is symmetric with regard to rotations. Moreover, this symmetry is compatible with the energy functional in the following way. If u: St -- S2 is a map in E and R E SO(3) is given, then Ru is still in E, and
E(Ru) = fn ld(2u)12 dx = fl, Idu12 dx = E(u). Therefore SO(3) is called a group of symmetries of E on {u: St _ S2}. We will see that these two observations are connected. Furthermore, this is merely a special case of a general principle, expressed in Noether's theorem.
Let us consider now a variational problem with several variables involving the first derivatives. In the following, fI is an open domain of R'n (m E N' ), we consider the class of maps
E={u:S2- 1R°}, and do not pay attention to the regularity of these maps (we could choose for instance these maps in C2). To any u E E we associate the action functional A[u] =
J in
L(x, u(x), du(x)) dx.
3. Variational point of view and Noether's theorem
25
Here, L is a function defined on S2 x R" x M(Rm, R'), with values in R, M(Rm, R") is the set of real matrices n x m, which we identify with the set of linear maps from Rm to R". du(x) is the Jacobian matrix of u, or differential of u at the point x:
ul
um
uln
um :
au'
au'
(W
&.n
1
du = n
)
=
49U
Any map u in E is a critical point of A if and only if, for any map v EC with a compact support in S2 and for s E R near 0, A[u + sv] = A[u] + o(s).
We may then show easily that u is a critical point of A, if and only if u is a solution of the Euler-Lagrange system:
M
aL
a
au - «- ax°
aL auQ
)
for i = 1, ... , n.
We want now to see what will happen when such a problem is invariant under the action of a one parameter symmetry group. Symmetries acting on IIt"
It is the most simple case. Assume that there exists a vector field U : Rn -+ IRn acting on the target space. We denote by 4, the flow of U, i.e. the solution of
4o(Y)=y,dyER", (y) ds
= U(' 8(y)),Vy E 1R",VS E R.
The family of diffeomorphisms t, induces a family of deformations of the set of maps from SZ to R". These deformations are described geometrically by the action of 4), on the graph of a map u: the graph of the deformed map u, is the image by (id, 4k,) of the graph of u. Actually it is clear that u, =4b,0U. So we assume the hypothesis that for any sub-domain w C 12, and for any map u from w to IRn, A,, [4a o u] = A. [u],
3. Variational point of view and Noether's theorem
26
where A, [u] = J L(x, u, du) dx. This condition implies in particular that for any w, A, [u + sU o u] = A,, [u] + o(s).
The last relation is true for all w if and only if the following holds: V(x, y, z) E Sl x Rn x M(Rm, Rn),
L(x, y + sU(y), z + sdU(y).z) = L(x, y, z) + o(s).
(3.1)
Theorem 3.1 Suppose that U is an infinitesimal symmetry of A, i. e. that (3.1) holds. Let u be a critical point of A, then the vector field J on 0 of components: n
r
U (u(x)) a (x, u(x), du(x))
J°(x) _ is divergence free, i. e.
divJ
= a=1 > °J = 0.
Remark Let P°(x, u(x), du(x)) = e o (x, u(x), du(x)) be the impulsion tensor, then J is constructed by contracting P° with U` o u, i. e.
J° (x) _
Proof.
U'(u(x))P°(x, u(x), du(x)).
The idea is to use the fact that the action functional A(u) of u is
stationary under the effect of modulation of U(u). We choose a function 0 E C' (S2, R) and we deduce from the critical point condition on u the relation A[u + sqU o u] = A[u] + o(s).
(3.3)
Let us develop A[u + sOU o u]. As we shall see it is not necessary to develop the full expression for taking into account the symmetry property of A: A[u + sOU o u]
=
L(x, u + sOU(u), du + s$dU(u) + sdOU(u)) dx t
f L(x, u + sgU(u), du + s$dU(u)) dx n n
m
+s
U' (u) i=1 0=1
a¢ aL ax° auQ
(x, u, du) dx + o(s).
3. Variational point of view and Noether's theorem
27
Now we use the symmetry relation and in particular the condition (3.1) by substituting so to s. We deduce Y,
,A[u + sqU o u]
in
= A[u] + s A[u] + s
Ui(u)
f
i=1 a=1 m
eO
axa
a' aL (x u., du) dx + o(s)
axa
Ja dx + o(s).
If now we compare this last expression with the relation (3.3), we obtain
a Ja dx = 0,
ax
E C(S, R).
J0t1
And this is precisely the weak formulation of our result.
O
Examples
i) Let us first consider the Dirichlet integral in one dimension. Let y: (a, b) 1R" be a curve and A[-y] =
fbm!2dt.
This corresponds to the problem of finding the trajectory of a free particle of mass m in classical mechanics. Note that A is invariant by translations in R'. Since in this case P, = mvy and U` is a constant, we conclude that the impulsion m-y is subject to a conservation law, provided that 'y is a critical point of A. ii) More generally, let L be chosen in a way such that the Legendre condition a-67 > 0 holds. Then we may use the Hamilton formalism. Suppose further that we have a symmetry vector field U E C' (R", R"). There are two ways to express this fact: L(t, ry + sU('y), %y + sdU(y).-y) = L(t, y, y) + o(s) (like before), or
where {., } denotes the Poisson bracket. These two conditions are equivalent. We see that:
Noether's theorem is immediate in the Hamiltonian setting, for the condition {E 1 U'P', H} = 0 implies the conservation law d
dt
E Ui(7(t))Pi((-Y(t)) i=1
= 0.
3. Variational point of view and Noether's theorem
28
It is clear that the converse is true. We have a generalization of that result: if F(x, p) is chosen such that IF, H) = 0, then F is conserved. iii) We return to the energy functional
A[u]=E(u)=
i2
Idurdx
for maps u: S2 -, S. The symmetries of S2 are given by rotations in SO(3), which is spanned by the vector fields a
a
Ukt: R3 - R3,
1 < k < l < 3.
Y
We compute furthermore Pa _
au' ax°,
and hence
jA'
au!
= uk ax;
auk - u1 ax°
are the components of the vector u x 3. Noether's theorem therefore gives
n
9
49U
0,
a fact that we know already for two-dimensional domains. This shows also how it is related to the symmetry of S2 via Noether's theorem.
Harmonic maps and conformal transformations We motivate our next step again first by considering harmonic maps u: St -, N, where ft is a domain in C. Define the complex valued function
f=l a
IZ-IayI2-2Z(a
'ay)'
We claim that f is holomorphic. It is convenient to write f in the form 2
=4()2,
f= Ca -Z ) (where y2 =
EN 1(y')2 for y E CN) for this implies immediately _ au 02u az - 8 ( az ' azaz) Of
-
2
au OZ ,
°")
,
3. Variations] point of view and Noether's theorem
29
where (yl, y2) = EN1 yiyy. The harmonic map condition now implies that = 0. Note that this fact is equivalent to the system a (IOU12 _ TX
16,U 12
+ a
ay
5 7X
IaiOU 12
12
TX
(2 \a ' Oi/)
ay
0,
ex ay
= 0.
Ia -
This means we have two conservation laws. We will see that this is again a consequence of certain symmetries and of (another version of) Noether's theorem. The symmetries involved in this case are given by conformal maps, for the energy functional is invariant by the action of the conformal group of C in the following sense. Let 0: f12 - fl, be an (oriented) conformal transformation (i. e. a = 0 and 310). Then for any map u: [11 --+ N, we have
Jn,
I dull dxl dyi =
J
I d(u o 0) 12 dx2 dy2.
The proof of this is based on the two following computations. First we compute Id(u o 0)12: Id(u0,0)12
a(uo0)I2+I8(uo0)I2 _ 4(8(u o0) 8(uo0)\
=
8x2
8y2
az2
822 12,
4
(az (0), az (0))
8u since 2"70±1 = 8z, (.0)8z2-M 2
+821, (0)
z2
a'
=
Idu(0)12 i dz2
and 8s = 0.
On the other hand, dx1 dy1 = det(dO(z2)) dx2 dye = I
.It 2
dx2 dye.
I
Putting these two facts together proves the claim. We conclude, that a map u is harmonic if and only if its composition with any conformal transformation is. In particular, harmonic maps on R.iemannian surfaces depend only on the conformal structure. Eventually, we can apply a version of Noether's theorem (that will follow in the next section) and see a connection of (3.4) with the invariance of the energy by conformal transformations. Symmetries acting on the domain St As mentioned before, there is a result similar to the version of Noether's theorem that we already know, dealing with the case where A is invariant by diffeomorphisms acting on ft. We will have to pay attention however to the fact that the
3. Variational point of view and Noether's theorem
30
domain might be deformed by such a transformation. Let X be a vector field on an open neighbourhood of S2, and W, the flow defined by it. This implies in particular that
',(x) = x + sX(x) + o(s).
(3.5)
The image of Q by %P, is of course in general different from Q.
Now consider a map u: St - R". It is transformed into a map u, by `Y, in a way such that the graph of u, is the image of the graph of u by ('P, id). This means that u, is defined on S2, = and satisfies u, o T, = u, Vs.
(3.6)
We say that A is invariant by X if and only if for any subset w C S2, Aq,. (W) [u8] = A. [u].
Changing variables x =
f
in the left integral, we can write this in the form
L (T. (p),
AW[ul.
W
On the other hand, a differentiation of (3.6) yields du(d), respectively
We insert this in the equation above and get
f
L
A,., [u].
(3.7)
W
An infinitesimal version of this relation, obtained by developing it up to first order, is then
A,, [u] = f L(x+sX(x),u(x),du(x).[Il-sdX(x)])det(Il+sdX(x))dx+o(s). W
Since this must hold for any w, we have necessarily V(x, y, z) E SZ x R" x M(lRm, IIt"),
L (x + sX(x), y, z.[Il - sdX(x)]) (1 + sdivX(x)) = L(x, y, z) + o(s).
(3.8)
3. Variational point of view and Noether's theorem
31
Theorem 3.2 Suppose X is an infinitesimal symmetry of A in the sense of (3.8). Let u be a critical point of A, then the vector field J on Sl of components J, (X)
X ,3(X)
_
axp a L (x, u(x), du(x)) - X«(x)L(x, u(x), du(x)) i=1 p=1
is divergence free.
Remark The vector field J« can be written in the form in
J«(x) = E XA(x)HR (x), Q=1
where HQ is the Hamiltonian tensor, defined by HQ (x)
(x, u(x), du(x)) - bQ L(x, u(x), du(x)).
Examples
i) Consider the case
A[y] =
J
-V(-Y)) dt,
6
where y: (a, b) -- Rn is a curve. This functional is invariant by translation with relation to t. Noether's theorem yields therefore
(mc + V (-Y)) = 0, which is known as the energy conservation law in mechanics. ii) Now consider the two-dimensional case A(u) =
-d22
dx dy
fn
for a map u: St - N, where ft C C. Again we have invariance by transla, tions (x, y) -4 (x, y) + F. The Hamiltonian tensor in this situation is N
8u' 8u= E - L 8x« 5x
HQ« _
IduI2 8p' 2
3. Variational point of view and Noether's theorem
32
hence HQ are the components of the matrix Iasi2eall ul2
2(8 eti ' ft 2
2
-112
We apply Noether's theorem and obtain OH 01
8H2
+
=0 forforj3 = 1, 2,
the two equations corresponding to the vector fields 8 and 0. But this is again (3.4).
Proof. We build a perturbation of u which is a modulation of the action of X by a function ¢ E C' (SZ, R). It gives us a map v, from St to R" such that v,(x + s$(x)X(x)) = u(x). Note that if s is sufficiently small, x 1 1 x + s4(x)X(x) is a diffeomorphism of Q. We may thus perform the change of variables x =1; + so(C)X(C) to compute the action of v,:
fL(xv(x)dv3(x))dx Jn
L (t; + sO(f )X
sd(O(C)X (C))]-1)
u(C),
det[Il + sd(O(t;)X (e))] dd
r L (l; + sq(t;)X (t;), u(t;), du(C).[Il Jn
(1;))J)
[1 + sdiv(¢(!;)X (l;))) d6 + o(s).
Let us develop this expression and then use hypothesis (3.8) (by replacing s by SOW): An[v,]
fn L (x + s¢(x)X (x), u(x), du(x).[Il - s¢(x)d(X (x))])
+s
J
[- E E 0,(3=1 i=1 m
+E
0=1
a
a a (x)
[1 + s¢(x)divX (x)] dx
(x)X s(x) OU, (x, u(x), du(x))
°
(x)Xc(x)L(x, u(x), du(x))] dx + o(s)
3. Variational point of view and Noether's theorem An [u] - s
E
J a,p=1 in
33
aX0 (x)X' (x)H; (x) dx + o(s).
Now we just use the fact that u is a critical point of A, and thus that An[v,] _ An[u] + o(s), to deduce
-
Jn
E a,A=1
8xa
(x)XA(x)H; (x) dx = 0.
This shows that J is divergence free.
0
Remark Noether's theorem plays also an important role in proving analytical results on harmonic maps. For more details see [48].
4
Working with the Hopf differential
Change of variables
Let 12 be an open subset of C and N C RN a Riemannian manifold as before. Let u: 12 - N be any map. Define the function (,,U I2
f=
_ i_ Iz=l Iz -
I
-2i(', )=4(5 )z.
Recall that f is holomor/phic if u is harmonic. Since this property of u is invariant by conformal changes of variables, one might be interested in the behaviour of f by such a transformation. So let's choose a conformal map ¢:122 -12,, where 121 is the domain of a map u: 12, -, N, and let's check how the corresponding function f transforms. We write z2 and z, = O(z2) for the coordinates of 122
and 12,, respectively. Set f,(z,) = 4(x)2 and f2(z2) = 4(a z, )2. Compute a(u o 0)
-
19Z2
au (46) ao au az, 09Z2 + az,
00 _ au
80
az,
az2
az2
Thus z
z
= 4 (azl (m(zz))) (8
f2(z2)
(zz)I lz
= f,(q5(z2))2 a0 (z2) J 1
f1 (z1) (00 (z2)J
=
We claim that although f is not invariant by
lz .
the so-called Hopf-differential
f (z)(dz)2 = f (z) dz ®dz E T'Q ®T'12 ® C
is. That is, c*(fi(zi)(dzi)2) = f2(z2)(dz2)2,
where ¢' denotes the pull-back by ¢. By a little abuse of notation we write fl(z1)(dzl)2 = f2(z2)(dz2)2. This property of the Hopf differential can easily be checked. We have 0*(dzi)2
=
(d(-O(z2)))2 = C
a lz/ I
(dz2)2,
O'(fi(zl)) = (fl 00)(Z2)Together with the earlier computation this proves indeed the claim.
4. Working with the Hopf differential
35
Applications
1) Harmonic maps from the sphere S2 Recall that there exists a map, called the stereographic projection, that maps the sphere S2 except one point conformaly to C. So if we write S2 = U1 U U2, where U1 = S2\{(0,0, -1)}, U2 = S2\{(0,0,1)},
and identify both U1 and U2 with C via the corresponding stereographic projections 01 and 02, then this gives a conformal structure on S2. Therefore we are so. say that a map u: S2 - N is harmonic if u o tij 1 and u o The stereographic projections 01 and ''P2 are given by X21
1
.,/,-1(z) =
z+z
21=
-i(z - z)
z+z
1
i(z - z)2
1 + IZ12
1 - IZ12
-1 + IZ12
It is easy to check that 7P2 1(1/z) = ip, 1(z) for z E C. Consider therefore the conformal map
0.
C` -+
z
C`'
H
1
z
and the two harmonic maps vi = u o ip 1 and v2 = u o ,p21, which satisfy
v2=v1o'onC*. Let
2
2
fl(zl) = 4 (C7z1) (zl),
f2(z2) = 4 (az2) (z2).
Then the Hopf differentials satisfy 2 fl(zl)(zl)2 = f2(z2)(dz2)2 = f2(1/z1)(dz1) 1
This means fl(zl) = f2(1/zl)z 1
for all z1 E C*. As z1 - oo, we get
lim fl (z1) = 0.
z,-+00
But recall that f1 is holomorphic. Liouville's theorem therefore implies that f = 0. Thus u is weakly conformal. Theorem 4.1 Any harmonic map u: S2 -+ N is weakly conformal. Using this we can prove the following.
4. Working with the Hopf differential
36
Corollary 4.1 Any harmonic map u: S2 -+S2 is either
i) constant, or ii) holomorphic, or iii) antiholomorphic. The proof of this Corollary uses the following lemma. We prove both results
in the appendix, at the end of the chapter. (The lemma is more or less wellknown, another proof can be found in [45].)
Lemma 4.1 Let U, V C C be two open subsets of C and u: U - V a weakly conformal function of class C2. If U is connected, then u is either a constant, or holomorphic, or antiholomorphic.
2) Hopf's theorem Theorem 4.2 Any CMC immersion of the sphere S2 in R3 is a round sphere. This was proved by H. Hopf in 1951 [55]. A. D. Aleksandrov [5] proved in 1956, that any embedded compact CMC surface in R3 without boundary is a round sphere. This led to the so-believed Hopf-conjecture (which was not formulated by Hopf) that there exist no immersed compact CMC surfaces besides the sphere at all in R3. However, H. Wente [86] found in 1984 an immersed CMC torus, and later N. Kapouleas [56] constructed in 1987 immersed CMC surfaces of even higher genus.
Proof of Hopf 's theorem. Let X : S2 - R3 be a conformal CMC immersion, and S2 be its Gauss map. In local conformal coordinates, we may write let u: S2 for the fundamental forms of X
I=e2"'(O 0
II=e2,(Hba
Hb
a
= -'du.dX,
as before. Recall that we have
e = -(H+a)ex -bay, I
_ -box -(H-aO"Y &Y
or, equivalently,
8
=-HEX
(4.1)
4. Working with the Hopf differential
37
Now compute
f 4H
(OX)2
X
. )2+2H(a-ib)
Since X is conformal, the expressions (
a , at )
)2 and (i)2 both vanish. Further-
more,
(OX
e2
laz
z z az,Fz
l2
So we have
f = 4H(a - ib)e2. But we know that u is harmonic, for X is a CMC immersion. Hence by Theorem 4.1
H(a-ib)-0, which implies either
H=Oandthus OX a
b
0, or 0, i. e. the surface is completely umbilic. In this case we have
II=e2"'H(1
\
?)
In the first case we apply the maximum principle to find that X is constant and hence no immersion. Thus this doesn't occur. In the second case we have
I-(u+HX)=0 because of (4.1). But u + HX is real. So d(u + HX) = 0, hence u + HX is constant, say
u+HX=_CER3. This means 2
X-H
lull
H2
1
H2-'
which proves that the surface parametrized by X is indeed a subset of a round sphere (of center C/H and radius 1/H). Lastly we can apply Corollary 4.1 to X and we conclude using the fact that X is an immersion. 0
4. Working with the Hopf differential
38
Remark We obtain the formula
f = 4H(a - ib)e2, for any conformal CMC immersion X: 0 -+ R3, where Sl is an open subset of C. Suppose H 3& 0. Then we have either
f - 0, i. e. a - b - 0, in which case the surface is totally umbilic and hence a sphere, or
f 0 0. Then f has (at most) isolated zeros. Hence the umbilic points are isolated. Non-umbilic case
Assume that X: Q -p R3 is a conformal CMC immersion without umbilics. Let
f (z) = 4
()2
(z)
be the component along (dz)2 of the Hopf differential of u. By hypothesis, f has no zeros. Assume H = z . Let t : fl -+ C be a holomorphic solution of
d (z) =
-f(z)-
We may consider Y = X o i/, instead of X, at least locally, for this is just another conformal parametrization of the surface. Its Gauss map is v = u o V)-'. Now, 4
(a((())2
4
( a (tP-'(())) (dd( 1(()) 2 2
-2
(F ,(Z))2 2 4
(z))
f(z)(-f(z))-1 = -1. Let us work now with these variables. We then obtain
2(a - ib)e2w = 4H(a - ib)e2' = -1, which means a = - e
-2W
and b = 0. We may thus write
II=e2"(Hba H-b a _- e ( sinhw 0
0 cosh es
}
Conclusion. The x-lines (images by Y of the straight lines y = const) are curvature lines for the smaller principal curvature, the y-lines are curvature lines for the larger principal curvature.
4. Working with the Hopf differential
4.1
39
Appendix
Here, we give the proofs of corollary 4.1 and lemma 4.1.
Lemma 4.1 Let U be an open connected domain in C and let u : U -+ C be a map of class C2. Assume that u is weakly conformal, i. e. I8xUI2 -
I8yU12
- 2i(8xu, 8yu) = 0.
(4.2)
Then, either u is constant, or holomorphic, or antiholomorphic. Proof. We decompose U = C U U- U U+, where
C:= {zEU : du.. = 0}, and U- _ {z E U : detduZ < 0}, U+ = {z E U : detduZ > 01. Since u is C2, C is clearly closed and U- and U+ are open. The fact that this is a partition of U follows from the fact that, by (4.2), for any z E U, du, is either of rank 0, or of rank 2. Moreover, simple algebra using (4.2) implies that u is holomorphic in U+ and is antiholomorphic in U-. We exclude the situation where C = U, which corresponds to the case where u is constant. Our task will be then to prove that C is composed of isolated points. Then, U- U U+ = U \C is connected and thus detdu has a constant sign on it, which means that either U \ C = U-, or U \ C = U+ and the conclusion follows easily.
Denote u(z) = u'(z) +iu2(z) and derivate (4.2) with respect to!. We obtain
2(
0 =
Caxayja ) +Z
2
e 2 > Dug ( Ox i=1
\
j
9
a
ay
y
2
)
Thus we deduce that Du = 0 on U \ C. Now let a -'dx + aul dy. We remark that da = Du'dx A dy, hence da = 0 on U \C.
(4.3)
C={zEU: aZ=0}.
(4.4)
But we also have
We claim that actually da vanishes at any point zo E U. suppose that zo is interior to C. Then by (4.4) there exists a ball B(zo, r) C C, on which a = 0 and thus da = 0 on B(zo, r).
4. Working with the Hopf differential
40
suppose that zo is not interior to C. Then there exists a sequence ((n)nEN of points in U \ C which converges to zo. By (4.3), dat = 0, and since u is C2, da is continuous and thus da., = 0.
Hence or is closed on U. Thus we can locally integrate a = dg, for some real valued function g. It turns out that ul + ig is a non-constant holomorphic function of z and thus its critical set, C, is composed of isolated points. Corollary 4.1 Let u : S2 - S2 be a weakly conformal map of class C2. Then either u is constant,
or u = Q.ovoPm Oru = Q.olloPm, where Pm : S2 -p cUfool and Q, : C U {oo} -' S2 are stereographic and inverse stereographic projections respectively and v : C U {oo} - C U {oo} is a (meromorphic) rational function of z. Proof. Using Sard's theorem, we can select a point s E S2 which is a regular
value of u, then we let P, : S2 -+ C U {oo} be a stereographic projection mapping s to oo. Similarly we choose some point m such that u(m) 0 s and we let Qm : C U {oo} -+ S2 be an inverse stereographic projection such that Qm(oo) = m. Then v := P, o u o Qm is weakly conformal from C \ {al, ... , ap}
to C, where (al, ... , ap) is the counter-image of s by u o Qm. Lemma 4.1 above proves that v is either constant, or holomorphic, or antiholomorphic. We exclude the constant case and without loss of generality, we shall assume that v is holomorphic. Then, since s is a regular value of u, each point ak is a simple
pole of v and v is a meromorphic function on C which tends to a constant P,(u(m)) at infinity. Let {b1, . . . , by} be the zeroes of v and denote dl E N* the multiplicity of each b1. Then, choosing a radius R > 0 such that a1,. .. , ap, b1,. .. , by are contained in B(0, R) and computing 0 = fOB(o,R) I using residue formula, we show that E l d, = Ek=i 1 = p. Lastly we apply Liouville's theorem to the function
z v -- f (z)
(z - a,)...(z - ap)
= (z - bl)d" ... (z - bq)d9 v(z) '
to deduce that f = c = P,(u(m)) everywhere. This terminates the proof with _ d _dy Qa=P, 1,Pm=Q;' andv(z)=c zb-ai...z-ar
.
5
The Gauss-Codazzi condition
We have seen at the end of the previous chapter, that given any conformal CMC immersion X: Il - 1R3 without umbilic points, with Gauss map u:11- S2, and with H 54 0, we can find a new parametrization Y = 2HX o 0-1, such that
I _- e2w (
1
0
1)
e"' 1
sink w 0 0 cosh w
Moreover, as we will see later, w satisfies the compatibility condition
Ow +sinhwcosh w = 0.
(5.1)
There is a converse result to that. Any solution of (5.1) on a simply connected domain gives rise to a "normalized" CMC immersion. We will prove that later, by considering a more general kind of inverse problem as follows. Given
I=e2"'(1
and II=e' (Hb a
01)
Hb a
where H is a constant and a, b, w: S2 --+ R are certain functions, is there a CMC immersion that has I and II as its fundamental forms?
Gauss-Codazzi condition as a zero-curvature condition
We want to find necessary and sufficient conditions such that the question above has an affirmative answer. Suppose first that X is a CMC immersion corresponding to I and II as given above. Introduce the moving frame (el, e2, e3): SZ - SO(3, R), where dX = e"'(eldx + e2dy), and u = e3 = el x e2 is the Gauss map of X. Write 3
det
=
E A; e,j, i=1,2,3, j=1 3
dX = E A44. j=1
That is, the 1-forms Ai and Ai are the components of de; and dX, respectively, with respect to the basis (el, e2, e3) of 1R3. Remarks
i) We know already that dX = e'°(eldx + e2dy), so Al
A2 A3
1=e"'
dx dy 0
5. The Gauss-Codazzi condition
42
ii) The fact that (ea, eb) = aab implies that the matrix with components AQ is skew symmetric, i. e. Ab + Ab = 0 for 1 < a, b < 3. iii) Consider the (4 x 4)-matrix
F-_
el
e2
e3 X
0
0
0
1
Note that the part R = (el, e2, e3) of this matrix is in SO(3,R). Define furthermore the matrix
A=
0
A2
Al
A2 Ai
0
A
A2
0
0
0 0
3
Al A2
A3 0
that has 1-forms as component. Then we can compute
dF = FA.
(5.2)
We may think of F as an element of SO(3, R) x R3, and analogously of A as being in the corresponding Lie algebra so(3, R) ® R3. Note that SO(3, R) oc R3 acts on R3 in the following way. For any t; E R3,
F
(1) =
(R+X)
Thus SO(3, R) x R3 may be seen as the group of affine isometries of R3.
Lemma 5.1 Assume that X is a CMC immersion such that
I=ez"(10 0)
b Il=e2' ( H+a b H-a
1
and
f := 4
'
(6,U)2 = 4H(a - ib)e2i''.
Then
A-
1
0 *dw
-*dw -Ai e"
Al
0 A2
0
0
-A2 e" dy d27 0
0
0
0
'
(5.3)
where
e-"
(5.4) A2 = e"H dy + e-" Re(i f dz), 4H and *: T`R2 -- T*R2 denotes the Hodge operator. In particular *dw = - Ty- dx+
Ai = e"H dx +
37 dy.
4H
R.e(f dz),
5. The Gauss-Codazzi condition
43
3 and A2 3 need to be computed, the rest Prof. Only the components A2 Al,
then follows from the definition of A. Since this can be done quite analogously for Ai and A2, we will confine ourselves to computing one of them. We have
Al2 =
(-ej' a ---
ax
2 / dx +
- -(,e2)dY ay
ax \ ae2
,/
el dx +
ay ' e2
dy.
\ ae l
But since
/
0=d2X =d(e"(eldx+e2dy))=e`" I
e2- +
Oft _j 7X
aw aeI -elf/ - ay
dandy'
we see that
Thus indeed
Al
=--dx+ y
-IW
xdy=*dw.
Now compute
Ai = (del, e3) _ - (du, el) _
-e_'
du,aXax
\J = e"((H + a) dx + 6 dy)
(recall (2.2)). This can be expressed in terms of f by
Ai = e`"H dx + 4H Re(f dz),
0
which concludes the proof.
Now, take the exterior derivative of (5.2). This yields
0 = d2F = d(FA) = FdA + dF A A = F(dA + A A A). (See Section 7.1 for more details on this notation.) Since F is clearly invertible, we obtain the necessary condition
dA+AAA=0
(5.5)
for X to be a CMC immersion. Remark Think of A as a connection form, then dA + A A A is the corresponding curvature 2-form. Thus (5.5) is a zero curvature condition.
5. The Gauss-Codazzi condition
44
Converse result
We have obtained (5.5) as a consequence of (5.2). The following lemma now states a kind of converse result. The proof will be given in the appendix of this chapter. Lemma 5.2 Let S2 be a simply connected open subset of R n and xo E 0. Let A E CI (S2, T`R n ® gl(n, R)), where gl(n, R) is the Lie algebra of the linear Lie group GL(n, R) (i. e. gl(n, R) consists of all (n x n)-matrices with real components). Write A = 1 Aa dx°. Assume that
dA+AAA= a
(A
09X01
09x# A,
+[A,,A#]) dxandxO=0.
Then for any Fo E GL(n, R) there exists a unique F E C2 (Q, GL(n, R)) such that
J F(xo) = F0,
dF = FA.
Remark The Lie group GL(n, R) and its Lie algebra gl(n, R) could be replaced by some other pair of Lie group and Lie algebra, and the lemma would still be true, e. g. respectively by SL(n, R) and sl(n, R), SO(n, R) and so(n, R), or in GL(n, C), U(n) and u(n). In particular, if we have 0
A=
-Ai -Ai Al
Al 2
0
A3
A2
0
0
ei,o 0
e2,0
-A23 A2 0 0
A3
E so(3, R) ® R3 R
0
and
Fo
-
0
e0° X1° ) E SO(3, R) a R3,
then the F obtained by lemma 5.2 takes values in SO(3, R) a R3 everywhere. Application
Lemma 5.3 Suppose that A has the form (5.3), (5.4), where H is a constant, w: S2 -' R some function, and f holomorphic on Q. Then
dA+AAA=(Ow+e2"'H2-e(4H)I2)
10
-1
0
0
0
0
00 0
0
00
dxAdy.
5. The Gauss-Codazzi condition
45
This can be proved by a mere computation, using the fact that d(Re(f dz)) _ d(Re(i f dz)) = 0, since f is holomorphic.
Now we put all facts together. On the one hand, if X is a CMC immersion, then (5.5) and lemma 5.3 show that
e(
Aw+e 2",HZ -
=0.
On the other hand, suppose that (5.6) is true for a constant H, a function R and a holomorphic function f. Then if A has the form (5.3), (5.4), it satisfies dA + A A A = 0. Furthermore, it takes values in so(3, R) ® R3. We can therefore find w: S2
e2
e3 X
0
0
1
with values in SO(3, R) v R3, such that dF = FA, according to lemma 5.2. It is now easy to see that X is a CMC immersion with mean curvature H. Thus (5.6) is a necessary and sufficient condition for w, H, and f to belong to a CMC immersion.
Note that in the non-umbilic case and in the "normalized" situation H = 21 If I = 1 (we know that we can even achieve f = -1), (5.6) takes the form Ow + sinh w cosh w = 0.
The corresponding second fundamental form is then
II=e2m
sinhw 0
1\
0 cosh w
Conjugate minimal surfaces and conjugate CMC surfaces Recall Weierstrass representation
X(z) = Re
J
JJJza
''-(v2 - 1)
h
2(v2 + 1)
(S)dK
iv
for a minimal surface. We have seen in Chapter 2 that if u is the Gauss map of X and P a stereographic projection, then we may choose v = P o (-u) and
h-
(a +ib)(1 + Iv12) 82*
5. The Gauss-Codazzi condition
46
For any A E S1 C C', we can modify this formula as follows. Set r:-(v2 - 1) X.\ (Z) = Re
J:a
.\'2h
2(v2 + 1)
(C) do
iv
This is equivalent to replacing a - ib by A-2(a - ib). We get thus a 1-parameter family of minimal surfaces, all with Gauss map u, called the conjugate family.
Now suppose that X is a CMC immersion. There is no simple formula similar to Weierstrass representation for this case. However, the following may be done instead. Replacing (a - ib) by A-2(a - ib) means replacing f by A-2f. Keeping H and w fixed, this gives rise to a deformation & of A. But... Key observation: we still have
dA,,+A,AA,=0, according to lemma 5.3, since we have not changed If 1. Hence also in this case we can find a conjugate family of CMC immersion, coming from this A,,. However,
the Gauss map needn't stay the same. This is a result due to 0. Bonnet [18]. We are going to look at this construction a little more closely. Write
Aa= (A,\ 0
where A,, is a (3 x 3)-matrix and OX a column vector. Since do dy
a = e'
0
this part of Aa doesn't bear much information. The important part will therefore be A,,. We compute
dAa+&A&
( dAa+A,,AA, do,,+A,,n¢a 0
The zero curvature condition
dA,,+& AA,=0 for Aa implies therefore
dA,,+AaAAa=0.
0
5. The Gauss-Codazzi condition
47
Furthermore, from lemma 5.1 and from the construction of AA, we see that
AA=
0
- * dw
*dw
0
_( dx+tRe(A-zjdz)) _(4dy+jRe(iA-2fdz)) )
dx+Re(A-2f dz)dy+-Re(iA'Zjds)
0
if we assume H = Z. Finally, (5.2) corresponds to dFA=FAAA, where F'A := (el,A, ez,A, e3,A)
and where ei,A, e2,A, e3,A are the deformations of el, e2, e3 which arise with the replacing of f by A-2 f as described above.
We want to modify (5.7) a little further. Choose 0 E R such that A = eie Consider the rotation cos 0 sin 0 0
RA =
-sin 0
0 0
cos 0 0
z
A-1-A z.
zi
2
0
0
A+A-t
=
A
1
0 0 1
and set GA := FARA,
and aA := Ra 1HARA.
Then (5.7) transforms into dGA = GAaA.
Moreover, the following can be computed:
aA =
0 0
A_i
-e" - e-"f
0 0
ie" - ie
e"+e-"f -ie"+ie-"f
4
+*dw A
+4
0
-1
0
1
0
0
0
0
0
0 0
0 0
e"+e-"f ie" - ie-" f
f
dz
0
-e" - e-" f -ie" + ie-" f
d2.
0
Setting A-lai, ao, and Aa' the first, second and third of the terms on the right hand side, respectively, this gives the splitting
ax = A-lai + ao + Aai,
5. The Gauss-Codazzi condition
48
where ai = a'. We observe that the matrix ao has non-vanishing entries according to the pattern 0
*
0
* 0
0 0
0 0
0 0
0 0
*
*
*
0
,
and a', ai according to *
In the particular case where f = -1, al takes the form 0 0
0 0
sinhw
-i cosh w
1= 2
- sinhw i cosh w 0
dz,
and al the conjugate of this. Interpretation. Let a = aaJa=1. Then we have
i) the splitting a = ao + al, where ao has non-zero entries only in the first two rows of the first two columns and al in the last row and in the last column,
ii) the further splitting al = ai +ai, where ai = al (a) dz and ai = ai, iii) the deformation
as = A-'a' +ao+Aai.
Thus i)-iii) give an algorithm that computes as from a without bothering with Aa, or even F,,. The following shows that we need just a harmonic map to start this construction. Theorem 5.1 Let G = (g1, 92, u): SZ - SO(3, R) and a := G-1dG. Suppose that as is constructed from a according to i)-iii). If u: (I - S2 is a harmonic map, then
daa+aanaa=0,
and vice versa.
To prove the one implication, recall that any harmonic map into S2 is the Gauss map of a CMC immersion. Thus as belongs to the corresponding conjugate family of CMC immersions, and satisfies therefore the zero curvature condition. The other implication will be proved later. Remark In the derivation of these equations using a complex parameters, we have followed a point of view close to [10).
5. The Gauss-Codazzi condition
49
Note that, beside CMC surfaces, there exist other examples of family of immersions which have the same first fundamental form and the same mean curvature function and which are geometrically different. They are solutions of the so-called Bonnet problem, see [13) for a review.
Recovering the CMC immersion from the S1-family of harmonic maps (Bobenko-Sym's formula) Once having constructed as as above, we can get the CMC immersions belonging to as back as follows.
Proposition 5.1 Let Ga: Q - SO(3, R) be a solution of
dGa = G,\ a,\, Ga(zo) = 13,
(5.8)
where as = \-'a' +ao+,\ai, and al = al
az
dz,
al = al.
Moreover, let ao, al E so(3, R) have non-vanishing entries in the two leftmost columns of the two uppermost rows, and in the last row and the last column, respectively. Let u: St S2 be the last column of Gl, and suppose u is harmonic. Write 0
Gi'
-B3
B3 0
-B1
B2
=o
-B2 B' 0
Then B ± u are weakly conformal CMC immersions and have u as their Gauss map.
Proof. Recall (see Chapter 2) that we can construct two weakly conformal CMC immersions B ± u belonging to u by setting
dB=ux
dx - ux a dy.
(5.9)
All we have to prove therefore is that these two constructions yield the same maps.
Write A = e`t, for real t (with small ItI). Then A = 1 + it + o(t), and
as = a + t(-iai + iai) + o(t), where a = ajIa=1. But it is easily computed that -iai + iai = *a1. So
as = a + t(*al) + o(t).
5. The Gauss-Codazzi condition
50
Now set b = ((AGe,,)jt=o)Gj 1, so that Ga = G + tbG + o(t).
(We abbreviate G = G1.) The condition (5.8) then implies
dG + t(dbG + bdG) = (G + tbG)(a + t(*a1)) + o(t), which is equivalent to the system
dG = Ga, dbG + bdG = bGa + G(*al ).
(5.10)
From system (5.10), using the first equation to eliminate dG in the second one, we conclude that db = *(Ga1G-1). (5.11) Now, write al
0
0
al3
0 a31
0
a3
a32
0
,
G = (el, e2,u).
Then (5.11) can be written in the form
db = *((al3e,
+ a2e2)EU - ut(agel + a382)), 3
whereas
du = age, + a2e2i according to (5.8). This means that db = *((du)tu - ut(du)).
Comparing this with (5.9), we see that b corresponds-up to a constant-to B, 0 and the proof is thus complete.
5.1
Appendix
We give here a proof of the following result.
Lemma 5.2 Let n be a simply connected open subset of 1R'", and let
A E C' (fl, T* R' 0 gl(n, R)) be a 1 form on f with coefficients in the set of matrices gl(n, ]R) = M(n, ]R) (the Lie algebra of GL(n, ]R)) such that
dA+AAA=0,
5. The Gauss-Codazzi condition
51
i.e., writing A = Fm j Aadxa,
iAs 1
(axa
+ JA, A$1 dxa A dx1 = 0. - aA,, ax0
Then, given any Fo E GL(n, R), xo E S2, 3!F E C2(fl, GL(n, R)) such that
F(xo) = Fo and dF = FA. Proof. Step 1: Local existence. We define the m following vector fields on f2 x GL(n,R),
Xa(x, M) := axa + (MA.(x))
aM'
where we denote
(MA. (x)) aM :_ E M,Ao(x)kaj . 1
k
Notice that
[(MAci(X)) J
[(MAti(x)) aM] = (MAa(x)Af(x)) am*
Then M, X01 (X, M) _
(M
(OOAOX0
- OA,, + [Aa, Ati])) a = 0.
Thus these vector fields commute and there exists an open subset U of R'n and a map : ft x GL(n, R) U t --- (O(t), F(t)), such that for all 1 < a < m, Wi(t) = XX(IP(t)). By a suitable redefinition of t,
one sees easily that :!,(t) = t = x and that i = FAa. Step 2: Local "uniqueness", up to a constant. Let F and F be two local solutions. On their overlapping domains, if dF = FA and dF = FA, then one compute easily that d(FF-1) = 0 and thus F = CF, for some constant C E CL(n, R). Step 3: Global result. Follows from the fact that Cl is simply connected. 0 We remark that the same result holds also with complex matrices (i.e. by replacing GL(n, R) by GL(n, Q. Also, if 0 is any Lie sub-group of GL(n, R),
with Lie algebra g, then it is easy to check that if F0 E 0 and A has its coefficients in g, then F takes its values in 0. Examples of such situations are SO(n) with Lie algebra so(n), SL(n) with Lie algebra sl(n), SU(n) with Lie algebra su(n), U(n) with Lie algebra u(n).
6
Elementary twistor theory for harmonic maps
Twistor theory as we are going to see it was first developed by E. Calabi [25] in 1967. Other contributions were made-among others-by J. Eells and J. Wood in 1982 [34], F. Burstall and J. H. Rawnsley in 1986 [23], and K. Uhlenbeck in
1989 (82]. We are going to consider harmonic maps u: S2 - S" C R"+1, i. e. maps satisfying Au + uJduI2 = 0,
which is equivalent to Du 11 u. In contrast to Chapter 4, where we considered the Hopf differential au 8u 8z 8z
(dz)
2
,
we will also use derivatives of u of higher order. Let us first introduce some notations. We write 8
8 8z'
8z'
If V, W E C"+1, then the complexification of the scalar product on R"+1 is n+1
(V, W) _ y2V'W`, i=1
and the standard Hermitian scalar product is denoted n+1
(V,W)=>V'W. i=1
Consider the expressions rla0 = (a°u, 0u) _ (0u, a0u),
a,#? 0,
in some local coordinate z E S2 on S. Note that these are no tensors. If we have a holomorphic change of variables 0: SZ2 -' f11, where f11, S22 are open subsets of C, then rhI (for instance) transforms as follows by 0: (82(u 0 0), 0(u ° 0)) = ((02u, Ou) 0
0)(.90)1
+ ((8u., au) ° 0)(020)(00)
So we have in general 0'(700) 36 (71-0 0
However, equality holds up to expressions containing 17Q,0, with a'+/3' < a+Q. We have seen this for X21; in general, it follows by induction. Furthermore, the following holds.
6. Elementary twistor theory for harmonic maps
53
Proposition 6.1 Let U: S2 -+ S" be harmonic. Then, in all holomorphic local coordinates z, and for all a, /3 > 0, such that a + Q > 1, we have 17.0 = 0. Definition 6.1 We say that u: S2
S" is isotropic, if and only if 17,,Q = 0 for all
a,,3 > 0 satisfying a +,3 >- 1. Proof of Proposition 6.1. We will apply induction on a +,3. The claim is clear for a + Q = 1, since 7701 = 7710 = (au, u) = 0.
Assume now that -y E N is given, and that dap = 0 for all a,,3 such that
1
ry-1
((ou 000), + E Pk (am, a2m, ... ,
Qy+1-k0)(aku
o (b), (au o O)(am
k=1 ry-1 m)(am)ry+1 +
= ((afu, au) o
Pk (aO, a2m,
... ,
ary+1-km)(71k1
o m)(am)
k=1
=
(')y1 0
0)(00)ry+1
for some polynomials Pk. This proves the claim. Step 2: t7y1(dz)ry+1 is holomorphic. We use the harmonic map equation
aau= 1Au=-1UjduI2. 4 4 It implies
a(ayu, au) _ (Gry-' (aau), au) + (ayu, aau) - 1 [(ay-1(ujduj2), au) + (ayu, ul dul2)].
The expression ay-1(uIduI2) gives a linear combination of u, au,..., ay-'u with coefficients that consist of the derivatives of IduI2. Hence a(ayu, au) = 0, and Step 2 is complete. Step 3: Use Liouville's theorem to conclude tyl = 0. Step 4: Completion of the proof. Note that for all a + Q = y, we have ,aa = 0 and hence aau) = (aa+1u, aQu) + (aau, av+1u) = j7a+1.a 0 = arlaa = a(OQu, + 11a,0+1
6. Elementary twistor theory for harmonic maps
54
So we can conclude from 1171 = 0, that 77-y-1,2 = 0,17-y-2,3 = 0, ... , i70,'Y+1 = 0,
0
which proves the proposition.
Isotropy is a strong property, but not easy to work with. However, with a little more preparations, we will finally be able to derive the following consequences. If u: S2 -' S" is so that u(S2) is not contained in a hyperplane in
R"+1 (such maps are called "full"), then i) n is even (see lemma 6.2), ii) there is some holomorphic data associated to u (Calabi [25], Eells-Wood [341),
iii) the energy of u is quantized (Calabi [25]).
We assume the following hypotheses: u: S2 -' S" is a full, isotropic map. We define, for any point z E S2, 0Q
= Span(Ou,... , 8°u), a E N,
1 91 = Span(8u,... , Yu),
I3 E N,
in the sense of complex vector spaces (i. e. O,,O are in the Grassmannian Gr(Cn+1)). Note that, for all a, i3, the conditions qµ, = 0, VA, v such that
µ
9'° 1911.
Here, 1 means perpendicular with respect to
Finally, we define
89=Span(8°Yu:00. Technical lemmas
We will state here a few properties of the spaces just defined. Some of the corresponding proofs will be given immediately, some in the appendix of this chapter.
Lemma 6.1 Assume that u: S2 - S" is harmonic. Then 89=919+9'9'+Cu for all q.
Lemma 6.2 Assume that u: St - S" is full, isotropic, and harmonic, and that S2 is connected. Let r := max dim 9Q. IEn ZO
Then 2r = n.
6. Elementary twistor theory for harmonic maps
55
A consequence of this result is that n is necessarily even. We shall use the notation n := 2r in the rest of this Chapter. Proof of 2r < n. By the definition of r, there exists a point z E Si, and there are ryl, ... ryr > 1, such that (a''''u, ... , Mru)(z) is of rank r. By isotropy, we have
=0, 1
(u, a", u) = (u, a',* u) = 0,
1 < i < r.
The spaces
Span(a"Iu, ... , afru)(z),
Span(a7` u, ... , a`u)(z),
and Cu
are thus mutually perpendicular. The direct sum
Span(a''u,...,f "u)(z) ® Span(&'u,...,8"'u)(z) ®Cu is a (2r + 1)-dimensional subspace of Cn+i Hence 2r < n.
0
Lemma 6.3 Let u: S2 - Sn be real analytic. Set
B:={zES2: dim0'r
Lemma 6.4 Let u: S2 - Sn be a full, isotropic, harmonic map on a connected domain 12. Let B be defined as in the previous lemma. Define
0Q(z) = 8 (z) E GrQ(Cn+1),
1 < a < r,
on S2\B. Then OQ has a unique real analytic extension on Si, and B is composed of isolated points.
Proof. Let
Then B is the set of all zeros of wr. We want to show that wQ behaves like a holomorphic function in the following sense: if w0(zo) = 0, then wa(t) = (z - zo)kwa(z) in a neighbourhood of zo, where k E N and WQ is real analytic with WQ(zo) 0 0. Then wQ has of course isolated zeros, and WQ provides locally an extension of ¢Q.
6. Elementary twistor theory for harmonic maps
56
For any z E Q, we have w°(z) E A° (ul ), since 8u, ... , 8°u I u. Consider the orthogonal projection P(z):C"+' -+ ul(z) and the unique linear mapping P ° (z): A°Cn+1 A°Cn+' such that f o r all v1, ... , v° E Cn+'
P°(z)(vl n ... A v°) = P(Z)VI A... A P(z)v°. Of course, we have (6.1)
Po(z)wc(z) = W.W. Now, compute
88uA02un...n8°u+NAO82un8un...AC7°u
8w°
+...+8uA...n8°-'un88°u _ -1[(uA82un...n&'°u)IduI2+(8uAUAO3uA...Aeu)BIdu12
+...+(8uA...
n8°uAu)8°-'IduI2).
Hence Po(z)8w°(z) = 0. Derivating (6.1) now yields
8w° = 8P°w° + P°8w° = 8P°.w°. Lemma 6.5 in the appendix will prove that this implies indeed the claim.
0
Conclusion
Suppose u: S2 -. S" is a full, isotropic, harmonic map on a connected domain. The lemmas of the previous section allow us to construct some holomorphic data that are associated to u. To this end, let ®9'.'-1)l.
1(z) = (B; ED CU
This is a 1-dimensional subspace of Cn+', we may therefore regard it as an element of CPn.
Claim The map 1: 1-+ CPn is holomorphic. Proof. Consider
ry=uAOun...AO''uAOuA...AOr-'u. Use the harmonic map equation to compute
97 =un8un...n8r-'un8'+'uAOuA...Aor-'.
6. Elementary twistor theory for harmonic maps
57
By the choice of r, we may write 8''+lu = V + µ O''u, where u E C and V E Span(Ou,... 0'-'u). So we get
8Y=117 Moreover, µ depends analytically on z E fl. There exists a vector i/l in the line 1 (depending on z), such that for all X E Cn+1, we have -yAX = (X,4G)E1 A...AEn+1, where E1," .. En+1 constitute the standard basis of Cn+1
Then we get
A...AEn+1
1L(X,tp)E1 A...AEn+1 =µyAX =C7(7AX) =
Hence O = p.m. Let zo E Q, and choose a simply connected open neighbourhood 1' of zo. Let A: St' - C* be a solution of 8a + µa = 0,
(6.2)
and set 0 = A . Obviously 0 is holomorphic. But 0 is a representation of 1, so also l is holomorphic at zo. Since zo can be chosen arbitrarily, l is holomorphic
0
in Q.
Remark To solve (6.2), first find a solution of 1
ON = 4 Ag = aµ. Then set P z) = J(Og - 11) d( - g(z). Note that this integral makes sense, since 8(8g-µ) = 0. The function f satisfies
Of = -µ. Hence if we set A = of, then 8A + µa = (a f + µ)ef = 0.
Example (for r = 1.) If u : ft - S2 is a full isotropic harmonic map, then ry = u A Ou and I = (Cu + COu)1 = COu. We know that we can write
u=fl+llvl2
v+v
-i(v-2) 1- Iv1
,
6. Elementary twistor theory for harmonic maps
58
where ± is + if u is holomorphic, and - if u is antiholomorphic. Then OV
c?u
(
l/1 + Ivl2)2 OZ-
i(v2 - 1)
v2 + 1 2iv
,
so that, in homogeneous coordinates, l(z) = [i(v2 - 1) : v2 + 1 : 2iv]. This is a holomorphic curve into the quadric {[z' : z2 : z3] E CP2 : (zl)2 + (z2)2 + (z3)2 = 0}
CP1.
Another point of view
Here we still denote n = 2r. Consider the complex submanifold
Zo={V EGr,.(C"+1): V 1V} of Grr(C"+1) We call a subspace V C C"+1 that satisfies V 1 V isotropic. The space Zo has the following properties.
Zo is fibered over S", i. e. there exists a projection II: Zo -+ S". It can be constructed as follows.
Let V E Zo. By isotropy, V ®V is a subspace of complex dimension n of Cn+1 Moreover, it is invariant by conjugation. The complex line (V ®V)1 is also invariant by conjugation. Thus we may choose a real vector y E R"+1 such that Jy) = 1 and (V ® V)1 = Cy. The vector y is obtained from any x E (V ®V )1 by setting either y = z+2 .tT or y = ` z z Taking into account that the decomposition Cn+1 = V ®V Cy gives an orientation of Cn+1 (or R"+1), we can make y uniquely determined by prescribing the orientation. More precisely, take an orthonormal basis
(vl,...,vr) of V. Set e2k-1 = *(vk + vk), e2k = ' 72'i(vk - Vk), for k = 1, ... , r. Choose y so that (el, ... , e2n, y) is an orthonormal, positively oriented basis of 1Rn+1. Now set
II(V) = y, and this gives the fibration of Zo. Zo ^_, SO(n + 1)/U(r), where U(r) ^_- {Hermitian bases of V}
for a given r-dimensional space V, and
SO(n + 1) - {oriented orthonormal bases of
Rn+11.
6. Elementary twistor theory for harmonic maps
59
Note that U(r) can be seen as a subgroup of SO(n) and hence of SO(n + 1) by the identification of Rn with Cr. The isomorphism Zo SO(n+1)/U(r) is given by the construction of an oriented basis of Rn+l as described above. Of course, this basis is only uniquely determined up to the choice of the basis (vl,... , v,.) Of V. The space Z = Zo U Zo is the sub-bundle of isotropic r-spaces in TSn+I ®C, for we can identify (TxSn+l ®C) U (T XSn+l ®C) with two copies of an element of the Grassmann bundle, for any x E S'+l We call Z the twistor bundle.
Now let u: S2 _ Sn be a full, isotropic, harmonic map. Consider the map V: Sl
Gr,.(Cn+l),
defined by V = Span(Ou,... , 8'u).
Then we have V = Span(au,... , a''u). This is perpendicular to V, hence V is a map into Zo. Moreover, we have aV C V. With the aid of the harmonic map equation, it is easy to see that OV C V ® Cu. Definition 6.2 A map V: St - Zo is called holomorphic, if and only if OV C V. It is horizontal, if and only if dV C V ® Cy, where y =1I(V).
So V: fl - Zo is both holomorphic and horizontal, if it is obtained from a full, isotropic, harmonic map as above. But the converse is also true. Proposition 6.2 Let V: SZ -- Zo be holomorphic and horizontal. Then u= fl(V) is harmonic and isotropic.
Proof. For all v E V, we have (v, u) = 0. Hence (av, u) + (v, au) = 0. But we have also
avEOVCVICu. So (v, &u) = 0, which means au 1 V. Since also au 1 u (recall Jul = 1), we have in fact su E V. Derivating with respect to z now gives
BauEOVCV+Cu. The same can be done with a and a exchanged. Then we get
aau E V + Cu. The orthogonality of V and V implies therefore that Du = 40au 11 u, and u is harmonic. From the fact 8u E V we conclude that 0u E V for all a, for V is holomorphic. Hence a°u E V, and the isotropy of u follows. 0
6. Elementary twistor theory for harmonic maps
60
Remark A slight generalization of this theory allows to associate holomorphic
datas to any harmonic maps S2 -+ CP' and conversely to construct all such harmonic maps starting from holomorphic curves in to CP" (the suitable twistor space here) (34). More precisely we can associate to each full holomorphic
0 : S2 - CP" a sequence of n + 1 harmonic maps into CP" named a harwhere in particular 00 = 0 is holomorphic, on is antiholomorphic and the other 4j's "interpolate" between 00 and On. Focusing on the relation between each element of a harmonic sequence leads to notions which may be further generalized to non isotropic harmonic maps and is connected with the Toda equation, as investigated by J. Bolton and L. Woodward monic sequence 00,
in [16).
6.1
Appendix
We complete here the proofs concerning the analysis of full isotropic harmonic maps from an open subset domain SZ C C into S". First we prove
Lemma 6.1 Let u : St _ S" be a harmonic map and 0' = COu + + CBau, 90" = Cau + + CBQu, and 9q = Span{&Yu : 0 < a + 0!5 q}. Then Bq = 8q + 9Q + Cu. Proof.
The non-obvious part is to prove the inclusion C. We prove it by
induction on q. The result is clear for q = 0. Assume that that the claim is true +1-a u, for some q E N. Any vector V in (),+I is a sum V = Vq + EQ±o ca8"a where Vq E 9q = 9Q + 9q + Cu. So we need only to prove the claim for each aaa+i-au. Exclude the easy case a = 0 or q + 1, then
8aa+1-au =
48a-,a-a(Du)
_ _4aa-la-a(ujduj2) E 9q-1,
by developing the last expression. This proves the Lemma.
0
Lemma 6.2 Assume that I is connected. Let u : SZ - S" be a full isotropic hannonic map and set
r:= max dim 9.. zEn,a>O
Then
n = 2r. Proof. Since we proved n _> 2r before, we show here that n < 2r.
Let us
assume the contrary: n > 2r. Then, by Lemma 6.1, choosing any point z, dim Span{Ba8Qu(z) : 0 < a, 3} < 2r + 1 < n + 1. So there exists some complex
6. Elementary twistor theory for harmonic maps
61
hyperplane HZ C C'+1, such that Span{8°3u(z)/O < a,#3} C H,t. Let fz : Cn+1 -+ C a linear form of kernel H2i we deduce that fZ on vanishes at infinite order at z and thus is identically zero by analycity, a contradiction.
Lemma 6.3 Let u : S2 - S" be a real analytic map and let
B = {z E c : dim 0'(z) < r}. Then S2\B # 0. In other words, there exists a point z E 0 such that dim 6;.(z) _ r.
Proof. For 1 < a < r, let us denote wa(z) = 8u(z) A ... A 8°u(z) E A° C"+1 and set B° = {z E 0: w°(z) = 0} and Bo = 0. Since all w°'s are continuous, all B°'s are closed subsets of Q. Let
ao:=sup{a: 0
and thus that ao < r. But on the nonempty open subset 0 \ B.O, w00+1 = 0 and w00 # 0, which implies that we may write 8°O+lu = Ep 1 cp85u, for some smooth functions co (these coefficients are smooth because they are unique on S2\B°p). Derivating this relation with respect to z an arbitrary number of times, we may easily prove by induction that, on 0 \ B00, 8ku E 9'O for all k E N*. To conclude, we consider maps fk = w0, A 8ku. They vanish on S2 \ B0O and are real analytic on 0, thus vanish on f2. It follows that r < ao, a contradiction. We now prove the following analytic result, crucial for the proof of Lemma 6.4. It basically says that a holomorphic section of a complex vector bundle with connection behaves locally like a holomorphic function, in the sense that either it is identically zero, or it has isolated zeroes with finite multiplicity.
Lemma 6.5 ("local behaviour") Let 0 be a connected open subset of C and A : 0 - GL(N, C) a smooth (C°°) map into the set of linear (matrices) operators acting on CN. Assume that w : S2 - CN is a real analytic map, solution of 8w(z) = A(z)w(z) on 0.
(6.3)
Set B = {z E St : w(z) = 0}. Then
either B = c or B has isolated points and for any zo E B, there exists some p E N' such that
w(z) = (z - zo)'W(z) around zo, when W is a real analytic and W(zo) 54 0.
6. Elementary twistor theory for harmonic maps
62
Proof. Let zo E B. Step 1: We prove the following claim: Let n E N such that for any 0 < k, l such
that k+l < n, C7k# w(zo)
= 0.
Then, for any 0 < k, l such that k + l < n, llka +lw(zo) = 0. This is a consequence of (6.3), for cr
0 +lw(zo) = 8ka (Aw)(zo) = 0,
by developing the last expression. Step 2: Now we suppose that, for all n E N, 8"w(zo) = 0. Then using the claim of Step 1, one easily deduce by induction that, Vk, I E N, 0 w(zo) = 0. And thus, since w is real analytic and S is connected, one concludes that to vanishes everywhere. Step 3: We assume that we are not in the case studied in Step 2. Hence there exists some p E N' such that d0 < k < p-1, 8kw(zo) = 0, but that 8pw(zo) 0 0. Then, arguing by induction from 0 to n = p -1 as in Step 2, by using the claim of Step 1, we show that, for any 0 < k, l such that k + l < p - 1,
ak8
+iw(zo)
= 0.
Hence a Taylor expansion at order p at zo gives:
w(z) = (z and the conclusion follows.
plo)piPw(zo) +O(Iz - zoID+l),
7
Harmonic maps as an integrable system
7.1
Maps into spheres
The sphere as a quotient of Lie groups
We consider once more maps into Sn. Note that Sn can be identified with the group SO(n + 1)/SO(n) as follows. Choose an orthonormal basis (f1, ... , En+1) of Rn+1 Then SO(n) can be looked at in the following way: SO(n)
{rg/E SO(n + 1) : g(En+1) = En+l }
A
_
R E SO(n) } C SO(n + 1).
0
{I
The subgroup A of SO(n + 1) is also one of the two components of
{g E SO(n + 1) : Adp(g): = PgP-1 = g},
/
where
P= l\
0 ).
In 0
1
An equivalence relation in SO(n + 1) is defined by glZg' q g-1g' E A and the set of equivalence classes {[g] = gA/g E SO(n + 1)} is denoted SO(n + 1)/SO(n). Now it can easily be seen that SO(n + 1) r/A
Sn
' - 9(En+1)
[9]
is a diffeomorphism.
Lifting a map into the sphere Consider a map u: S2 - Sn, where 1 is an open, simply connected subset of C. Then we can lift u to SO(n + 1), i. e. we can find a map
F: 1 - SO(n + 1), such that F(z)(En+l) = U(Z)-
Write F(z) = (el (z), ... en(z), u(z)), then el,... , en form an orthonormal basis of TuSn. We can thus write n
du=E¢ae., a=1
7. Harmonic maps as an integrable system
64
where 0. = (du, ea), and dea =
wQeb
- qau,
u = 1, ... n,
b=1
where wb = (deb, ea). The matrix 1
W21
wn Wn2
01
o2
a=
E T*Q 0 so(n + 1)
... 0 ... -on
WZ
_¢j2
on
0
satisfies
dF = Fa. We have seen in chapter 5, how the condition
da+aAa=0 can be derived from (7.1). Furthermore, lemma 5.2 showed that if SZ is simply connected (which we assume here), and if (7.2) is valid, then there exists an
F: St - SO(n + 1), such that
1 F(zo) = Fo,
dF = Fa,
where Fo E SO(n+1) may be chosen arbitrarily. The passing from F to a can be looked upon as a kind of linearization. The non-linear condition F E SO(n + 1) is replaced be the linear one a+ 'a = 0 (once we assume that Fo E SO(n + 1)).
We introduce the notation
[aA01=cc n/3+,0Aor = [,0Aa] for 1-forms a,,0 E T* Z ®gl(n + 1, C), where we write n
fa A /3i ...
aj' n Qn
aAJ3= i=1 for
an n Qi
...
an A
n
7. Harmonic maps as an integrable system
65
This can be generalized to general Lie algebras by setting [a A#] (C, n) = [a(C), 3(r!)] - [a(n), fl(C)I.
Thus (7.2) may be rewritten
da+ 1[aAa] =0. Decomposition of so(n + 1)
We are going to decompose so(n + 1) and hence the a defined above. To this end, consider the operator Adp:so(n + 1)
so(n + 1)
H PAP-1, which we have seen earlier. Since it is an involution, this operator acts linearly on so(n + 1) with eigenvalues f1. Moreover, it establishes a Lie algebra automorphism, i. e. we have
Adp(n)l =
nl)
Now, let so(n + 1)o so(n + 1)1
= =
{C E so(n + 1) : Adp(1;) = (-1)°!; = l;}, {C E so(n + 1) : Adp(t;) = (-1)11; = -Cl.
Recall that we considered earlier the subgroup .1 of SO(n + 1),
f=
J( R 0
R E SO(n) } ^_- SO(n).
It turns out that so(n + 1)o is the Lie algebra corresponding to A. Moreover, if 1; E so(n+ 1)Q, n E so(n+ 1)b then, using the fact that Adp is an automorphism, it can easily be checked that (t;, III E so(n+1)., where c a+b (mod 2). We now decompose a according to the decomposition so(n+1) = so(n+1)o®so(n+1)1:
a=ao+ati where
0
0
I=
ao = W.n_,
n
W,
0
0
...
n
0
0
0
0
i 11_
7. Harmonic maps as an integrable system
66
and ...
0
01
... -01 ...
0
0-
0
al =
0
-on
ET`1 ®so(n+1)1.
0
Then the integrability condition (7.2), written in the form
da+2[aAa]=0, and projected onto so(n + 1)o and so(n + 1)1, respectively, implies dao + 2lao A ao] + 2 (al A al] = 0 and
dal + (au A all = 0. (7.4) Remark Note that (7.4) is a linear condition on al. This will be an important fact for what is to follow.
The harmonic map equation We now assume that u is harmonic, that is Au + uIdu12 = 0,
which is equivalent to (7.5)
d(*du) 11 u.
Recall that du
= L Oaea, a=1 ra
dea
=
waebb
-
Oau,
a = 1,...,n.
b=1
We obtain therefore n
n
1
n
d(*du) = E d(*Ob) + EWn A (*Oa) I eb - E 0" A (*0a) U. a=1
b=1
a=1
Thus (7.5) means n
d(*cpb)+EWQA(*(pa)=0, a=1
b=1,...,n,
7. Harmonic maps as an integrable system
67
which may be written in the form d(*al) + [ao A *al] = 0.
Forgetting about u for the moment and thus focusing on a, we see that we have to study the system
dao+ 2[aoAao]+-[a, Aai]
dal + [ao A all d(*al) + [ao A (*al)]
0,
(7.6)
0,
(7.7)
= 0.
(7.8)
We think of (7.7), (7.8) as a nonlinear Cauchy-R.iemann system, in contrast to the linear one
d/3 = 0,
d(*,3)
=
0,
arising from a harmonic function f : 0 -i1R by setting 6 = df . The idea to treat (7.6)-(7.8) is to generalize the linear situation, in which case we are led to d/3a=0 for
Qa =
2
3+
where /3' = Q ( ) dz and with now, we use
ai
3(
(0) = .1-1,' + AO",
A E C',
) d2. In the situation that we are dealing
the fact that (7.7) and (7.8) are the same equation in al and *al, respectively, and linear with respect to al, the property
[(cos(h) al + sin(O) (*al)) A (cos(9) al + sin(g) (*ai ))] = [ai A all for all 0 E C, which is easy to verify.
We then observe that (7.6)-(7.8) is equivalent to
daa+2[cr,\ Aaa]=0, AEC*, where
aslai+ao+\ai = This proves the following.
+2
+ao+
2i(*al).
7. Harmonic maps as an integrable system
68
Theorem 7.1 Assume that fi is simply connected. Let u: ft -+ S" be a map having the lift F: 11 -+ SO(n + 1). Set a = F-1.dF, and decompose
a=ao+al according to the decomposition
so(n + 1) = so(n + 1)o ®so(n + 1)1.
Decompose further al = ai + ai , where ai = al (8) dz and a" Then u is harmonic if and only if the condition
= al
(a:)dz.
daa + 1 [aa A aa] = 0
holds for all A E C', where
a,\ = A-lai + ao + Aai. Conversely, each family of 1-forms as on 0 of the form as = a-'ai + ao + Aai, with coefficients satisfying the algebraic conditions above, and which is a solution of (7.9), gives rise to a S1-family of harmonic maps by the following algorithm. Integrate J F.\ (zo)
j
=
Fo
dFa = Faaa on S2
for allAES'. Set u,=[FA]=FA(e"+1). Then, for any A E S1, the map u,, is harmonic into S". We omit the details. Remark. In general, as has coefficients in so(n + 1) 0 C. However, if A E S' C
C', then as E T'Q ® so(n + 1). We thus recover the result of chapter 5 for harmonic maps into SS2, i. e. the existence of a conjugate family of harmonic maps.
7.2
Generalizations
SO(n) by a subgroup of 0. Suppose 0r is defined from an automorphism T:0 - t such We may replace SO(n + 1) by any compact Lie group 0 and . that T2 = 1m by
Or={gEG: T(.9)=g}
7. Harmonic maps as an integrable system
69
(r plays the role of Adp), and let (0,)o be the connected component of 0, containing the identity. If (0T)o c A c 6T, then the same results as before hold for harmonic maps u: fl - o/R. More precisely, we proceed as follows.
First lift u to 0, i. e. find a map F: fl -' 0 such that u is the composition of F with the projection 0 -, 0/A. Then set
=F-1dFET`fl®g, where g is the Lie algebra of 0. We have the Cartan decomposition
9:-- go ®gi =:1 ®gi, where
ga={l;Eg: drl(1;)=(-1)°£}, a=0,1. Note that t = go is the Lie algebra of A. We split a = ao + al according to this decomposition. We let a' = al ( ) dz and ai = al ( ) d2, and we set
as=A'1c +ao+Aai. Then u is harmonic if and only if
daa+2[aanaaJ=0
(7.10)
for all A E C. Moreover, if (7.10) holds, then we get an S'-family of harmonic maps by integrating J F.\ (zo)
= Fo,
dFa = F\ a.\,
and setting ua = [F\I. An example
The sphere S2 can also be represented by SU(2)/U(1), where
SU(2) _ {g E GL(2, C) : gtg = 1gg = 1, detg = 111 =
1(a -6 ) : a,bEC, b
a
)
and
U(1) = {g E SU(2) : Adp(g) = g}, where
Adp(g)=(0 1)g(0 °l)
111
7. Harmonic maps as an integrable system
70
Moreover, SU(2) acts on
R 3_V
{M E M(2,C) : trM = 0, 'M = Ml}
-zy)
:r.+iy
x,?I,zEIIY}
isometrically by M ; . gMg-'. Pick an M E V and a g E SU(2). Then (let M =
-(x2 + y2 + z2) (if M is represented by x, y, z as above). Since det(gMg-1) = det M, the mapping M '-- gMg-' may be looked upon as an element of SO(3). More precisely, the mapping SO(3)
4D: SU(2)
is a twofold covering (actually SU(2) is the spin group Spin(3)). Let now
a3=( 0
0
)E V,
then
V
SU(2)
i''
9
9as19
maps SU(2) into S2 C V. The fibers of this mapping - the Hopf fibration - are exactly the classes in SU(2)/U(1). This gives an identification of SU(2)/U(1) with S2. We are thus led to two different representations of a harmonic map u: St S2: . by
tta
= A-'
0
0
(01 ),
0
0
((h2)'
-(01)1
+A
-(02)1
0
0
0
0
1)l'
-1
0
+ w2
1
0
0
0 0 0
0 0
(i)l? W)n -(42)ll
E T'U ®so(3),
0
and by
A-' lYa
2
0
{
(0i +iO2)'
A
+ 2
(461
0 i02)11
+
(-01 + io2)' 0
)+
(-0' + i42)") 0
wj 2
{
-i
O)
0
E T*fJ ®su(2).
7. Harmonic maps as an integrable system
71
7.3 A new setting: loop groups We are going to take a different point of view by putting the parameter A in the target. Let 0, .fi, and r be as in the previous section. We may keep in mind the case 05 = SO(n + 1), .fi = SO(n), and r = Adp. However, this works actually for any Lie group. Let's first define the loop groups:
AO = =
AOC
{A H ga : A E S', ga E 0}, {A H ga : A E Sl, 9a E &C},
where OC is the complexification of 0, e. g. for 0 = SO(n + 1) or 6 = U(n), SO(n + 1)C
= {g E GL(n + 1,C): tgg =
detg = 1},
U(n)C = GL(n, G). We also need the twisted loop groups AOC
=
{A
g,\: A E S', 9a E OC, r(ga) = 9-a},
AO, = Aec n AO. We endow these sets with a topology, for instance by using the Fourier decomposition 9a = E 9kAk,
kEZ
and setting II9AIIH' = > I9ki2(1 + k2)'12 kEZ
for s > z . Then the point-wise product [A #.-* g,\].[,\ F-4 ha] = [A '-' 9aba]
is continuous and gives these sets some "Lie group" structure. The corresponding Lie algebras
AESl, aEgC}, Ag, A9;, etc have the Lie brackets [[A '-' G], [A '-' 17a]] = [A'-' [Ea, ria]]
Note that the twisting condition corresponds to
r(G) _ t-a
7. Harmonic maps as an integrable system
72
E AgC. But using Fourier's decomposition
for [.A
Ca =
E 4k,\k kEZ
once more, we see that this is equivalent to E E
go 9i
for all k E Z.
Link with harmonic maps
OIA is a harmonic map. We have seen how to construct from it a family of 1-forms Suppose u: S2
as = A-tai + ao + tai,
which we may regard as a 1-form with coefficients in Age. But furthermore, we have
as E T'12 ® Ag for A E S1 (reality condition), and
ai E T'S2®gi, ao E T*c®go. That is, as E T'Sl®Ag'. We have a simple characterization of harmonic maps in terms of loop groups.
We introduce further the loop groups
A+Or =
I [A H ga] E AOC : ga
go k } k>O
=
A-OT =
{ JA H ga] E AOC : ga admits a holomorphic extension inside the disk JAS < 1}, {[A
ga] E AOC: ga = EgkAk} k
_ A*OC
{[)1 -+ gA] E ACT : [a
9a-1J E A+OC}
_ {[A-'9a]EA 0c: goo=1},
and their corresponding Lie algebras A+gC, etc. (Note that the loops in A-OT 1}, so the condition can be extended holomorphically to (C U loo}) fl g,,. = 1 makes sense.) In particular, for any [A '- a] E AgC, the splitting
k
/
\k>O
7. Harmonic maps as an integrable system
73
shows that Agc = A.- gC, ED A+ 9c,.
the component of fa E Agc, in A-gC,, according to this We denote by decomposition. Then we have the following.
Proposition 7.1 Let FA: Q -' AO, be a map such that (7.11)
[Fa 1dFaJn.-e:
where /3 E T*12 ® gl satisfies
Q (8z) = 0.
(7.12)
Then, for any A, the map z'-- [Fa](z) is harmonic. Any map F,, as in Proposition 7.1 will be named an extended lift of a harmonic map. Proof. Consider the family of 1-forms
aa=Fa'dFa. The twisting condition Fa E AO, implies that as is also twisted, that is, as =
&k\1
k
kEZ
with &2k E go and &2k+1 E 9C. The condition (7.11) shows that &k = 0 if k < -1. By the reality condition Fa E AO we have also as E Ag and hence k>-1
E ak\k
k>-1
for .1 E S'. Hence a_k = &k. So we end up with as = a-1x1 + &o + a&1 = a-1Q + &o + Q,
with &o E go. Since we have furthermore Q () = 0 and hence Q () = 0, this is exactly the decomposition as required in Theorem 7.1. However, in this case the condition (7.9) holds trivially because of the definition of a.\. Therefore, Theorem 7.1 implies the claim. 0
7. Harmonic maps as an integrable system
74
Gauge transformations
The lift F of a harmonic map u: n QS/.f is of course arbitrary, since for any smooth map h: S2 --+ .g, the map Fh: f - 0 lifts the same u. Suppose that a lift F of u and a smooth map h: fl --+ .ft are given, and let F = Fh. Assume for simplicity that FA(zo) = 1 = h(zo). We want to compute the corresponding FA. To this end, first compute
& = F-' dF =h -' F-' (dFh + Fdh) = h- l ah + h-' dh = do + a1, where &o = h-' aoh + h-1 dh and a1 = h-' a1 h, for the usual splitting a = ao + a 1. Hence we have
A-'(h-'ash) + (h-'aoh + h-'dh) + A(h-'(.vi h), which means
Fa'dFA = &a = h-'aah + h-'dh = (FAh)d(Fah), and P.\(zo) = Fj(zo)h(zo) = 1. So we see that
Fa = F,\ h. Thus harmonic neaps u: S2
OIA correspond bijectively to claims of maps
[F.\] E A6,/A, such that (Fa 1dFa]A. mr = a-' fl and
(
) = 0.
Twistor interpretation Comparing the results with those of the previous chapter, we see some similarities. We may therefore give the following interpretation. Z = A0,./A is a kind of twistor space. The condition (F. ' dFAJA, mlr = A-1/3 is a horizontality condition. The condition (F; dFA]A er (e) = 0 means that FA is holomorphic in a certain way.
7.4
Examples
i) Let U: f2 -+ SO(3)/SO(2) is even holomorphic, i. e.
S2 be conformal. Assume for instance that u
ux c?u O Ux
-
a?/
Ou
_ --x F
,
7. Harmonic maps as an integrable system
75
or, equivalently,
*(u x du) = du.
(7.13)
Let F = (e I, e2, u) be a lift (or moving frame) belonging to u, and let
a = F-1dF =
0
W2
wl
0
02
-42
-q1
0
0
a3 a2
02
0
02
ai ai
=
We have by definition du = e1O' + e202,
and hence
uxdu =e2O'-e1O2. So (7.13) is equivalent to
=
*01 *02
Jl`
-01
and thus to
= -i(.01)',
/(42)'
{
02,
( 2)"
We find thus the representations
al =
0
0
0
0
-(¢1)' -(02)'
(¢1)' (0 2 )'
0
0
0
0
1
-i
-1 i
0
0
and
It =(O1)"
0
0
1
0
0
i 0
-1
-i
.
Set
Ao =
0
-1
0
1
0
0
0
0
0
A+ =
0
0
0 -1
0
-i
i
0
1
,
A_ =
Then a = (-01)'A+ + wi Ao + (01)"A-. We let 1
U=
1+IVI2
v+v -i(v-2)
1-IvI
,
0
0
0 -1
0
i
-i
0
1
7. Harmonic maps as an integrable system
76
where v: 9 - C U {oo} is meromorphic, and we choose F such that
i(v2 - 1) el +ie2 =
v2 + 1
1 +Ivl2
2iv
We then compute 8v
du
8z
dz = 1y Ivl2 (el - ie2) = (0')'(el - ie2), 8v
d"u
8z
Iy12 (e1
dz
+ ie2) =
(W')"(el
+ ie2),
and W2
= (del, e2) =
(vdv - vdv) _ - * d(log(1 + Ivl2)).
1 +lvl2
Hence we see that v2-02
1 - v2+ F
1+llvl2
-(v + v) r I
and
v+L
2
V2+02
V2 _ 2
i(v -.0)
-i(v -v) 1 - Iv12
2
+1Iv12 12 A++vA++1+1A- + 2A?
"'L-
dv 1W
Thus we may construct FA from F directly by the substitution of v by A- I v to obtain 22V2
FA =
l
+11v12
2-
A
+ \-1vA+ + 1 + AvA- + \2fA2l
Note that u is thereby deformed according to A-'+A 2 A
' -A
2i A -t+A
0
2i
2
0
0
0
1
hence the action of S' on u is just a rotation.
u,
7. Harmonic maps as an integrable system
77
ii) We are looking for "rotationally symmetric" harmonic maps u: C -+ S2 of the form sin ,3(x) cos(y/a) = u(x, y + 27ra) u(z) = u(x, y) = sin p(x) sin(y/a) cos p(x)
for some function p: R -i R. We then have
su
TX
- sin p(x) e2,
au ay
= p'(x)ei,
a
where
el =
cos p(x) cos(y/a) cos p(x) sin(y/a)
- sin(y/a)
- sin p(x)
0
cos(y/a)
We compute further
( / = plx)- sin p(x)a2cos N(x) \)e2
a ux au +8y a ux ay au ax ax
So u is harmonic if and only if p is a solution of the pendulum equation
p" - sin pa2cos p = 0,
(7.14)
in which case it has conserved energy
(p1)2 -
sin /3
= E0.
(7.15)
-
Note that the Hopf differential is then equal to Eo(dz)2. If Eo happens to
be positive, then we may-by a suitable rescaling z ' -, rz and a '-
assume that Eo = 1. Similarly, the case Eo < 0 can be reduced to Eo =r-1. Let F = (el, e2, u). Then we have F = e! A3 OA2 and dF = Fa, with
=
n0
0
a=
Cosdy -p'd x cos p
a
A3 dy
-e -
a
dy
p'dx einpdy °0
dy
sin p A l dy + p' A 2 dx, a
where 0
Al =
0
0
0 0 -1 0
1
0
,
A2=
0
0
1
0
0
0
-1
0
0
,
(Note that [A., Ab] = AC with c =- a + b (mod 3).)
A3=
0
-1
0
1
0
0
0
0
0
7. Harmonic maps as an integrable system
A solution of the pendulum equation can be obtained by integrating ordinary differential equations, or, geometrically speaking, by integrating vector fields. One way to do this is to eliminate /3 by letting
V-/3'
U=sin/3 2a '
a
21
so that
U V Z
d
Z-cos/3
'
-VZ -UZ
=
(7.16)
(7.17)
4UV
Exercise 1 Prove that (7.17) is equivalent to the assertion: There are a number a E R and a function /3: R -' R, such that (7.16) and the pendulum
equation (7.14) hold. (Hint: use 4U2 + Z2 = const.) Exercise 2 From the energy conservation (7.15) deduce that dx = -
dU
a + U2) \o
- 4U2)
Deduce the solution in terms of elliptic integrals and theta functions (see [32], [19[ ).
Now we are going to consider the extended lift Fa of u, defined by
dFa = F.,aa, and
=
as
A-1
c_+a
[r13Ai+cA2Jdz+____A3dY
r-iQ AI+ /32A2Jdz ' 1
2a
A-1(iUAI + VA2) dz - ZA3 dy + A(-iUAI + VA2) dz.
This is equivalent to replacing (U, V, Z) by (X ' U, A-1 V, Z). Note that still d
a-1U A-1V
Z
=
-(A-1U)Z
(2A-1U)-'V) + (A-'U)(A-1V)
7. Harmonic maps as an integrable system
79
for all A E S'. Thus this one-parameter family of vector fields is embedded
in the set of solutions U, V: R -, C, Z: R - R of d
dx
U V Z
-VZ -UZ
=
(7.18)
2(UV + UV)
One may check directly that any solution of (7.18) leads to some
as = A-'(iUAI +VA2)dz - ZA3dy+A(-iUAI +VA2)dz that solves
daA+1[aanaa]=0. Moreover, the following holds.
Proposition 7.2 Any solution U, V: R -* C, Z: R -, R of (7.18) is also a solution of drra = [rla, aa],
where
ria = A-' (UAI - iVA2) + ZA3 + A(UA1 + iVA2), and conversely.
Proof. We have
dria = A-' (dUA1 - idVA2) + dZA3 + A(dUA1 + idVA2) and
[r7,,, aa]
= A-' (-VZA1 + iUZA2) dx + 2(UV + VU)A3 dx + A(-VZA1 - iUZA2) dx.
So dna = [ria, aa] holds if and only if
dU = -VZdx,
dV = -UZdx, dZ = 2(UV + VU) dx,
Another key observation is that
as = i [.t_'ui i-ladz - Qa,
d 0
which concludes the proof.
dz - i
- iVA2) + 2 A3J
[A3 + A(UA1 + iVA2))
7. Harmonic maps as an integrable system
80
where
Ax = iZA3 dx + 2ia(UA1 + iVA2) dx. This decomposition irja dz = as + 13A
corresponds to Aso(3)c = Aso(3), ®
where
Ab so(3)c = {[A'-' a] E A+so(3), : o = o E b} and b = {itA3 : t E R} is the Lie algebra of the subgroup
'B =
cosh t
-isinht
isinht
cosh t 0
0 of AC
0 0
:
tER
1
SO(2)c. Hence, if r: Aso(3), ® Ab+so(3)c
Aso(3),
is the corresponding projection, then we have d r t a = [1j
,
r(iria dz)].
This example shows that some harmonic maps can be obtained by integrating vector fields. We will see more about this in the next chapter. All these observations have an interpretation in terms of some natural loop groups operations, this will be the subject of the next chapter. Remark The introduction of a complex parameter goes back to K. Pohlmeyer in 1976 [69]. The first extensive use of these ideas is due to K. Uhlenbeck in 1989 [82] and N. Hitchin in 1990 [54] and was followed by many other works [11], [12], [21], [22], [30], [41], [47], [49], [51]. The idea of loop groups, originated by works by M. Sato [74], was developped by G. Segal and G. Wilson in 1985 [75J. An algebraic theory of loop groups is constructed in the book of A. Pressley and G. Segal (1986) [70].
8
Construction of finite type solutions
At the end of the previous chapter, we have seen a kind of harmonic map that is constructed from a solution of the differential equation d
U V
=
Z
-VZ -UZ 2(UV + UV)
We want to find a similar way to construct a greater variety of harmonic maps To this end, we need a pair of appropriate vector fields X1, X2 on some finite dimensional vector space V, such that [X1, X2] = 0. This allows us to integrate them, i. e. to find a 17: R2 -+ V, such that 57x = X 1 o 77,
X2 0 i7.
Via some linear mapping II: V -+
for an appropriate Lie algebra g, we hope to find some solution as = f1(i?) of daa + 2 [ax A a,\] = 0,
and thus a family of harmonic maps. This is done by the so called AdlerKostant-Symes theory [4], [60], [77]. Our exposition can be compared with the one of F. Burstall and F. Pedit in [22].
8.1
Preliminary: the Iwasawa decomposition (for 0)
Let 0, A C 0 be Lie groups as before. Let AC be the complexification of A. For instance,
SO(n)c = {M E GL(n, C): 'MM = 1, det M = 1), U(n)c = GL(n, G),
SU(n)c = SL(n, Q. We want to find a subgroup B.q c Ac, such that for all g E .fic there exists a unique decomposition g = f b with f E A, b E Bit; BA is then called a Bonel subgroup of Ac. This is a generalization of matrix decompositions like M = RT for M E GL(n, R), R E SO(n), and a triangular matrix T.
8. Construction of finite type solutions
82
Example 1 Let .q = SO(2). Then we may choose Bso(2)
_
- sin it
cos it sin it
cos it
t E R C SO(2) c .
:
Since any g E SO(2)c may be written in the form
g=exp(0+it)1 0 we have indeed g = Rb with
R= exp (0 10
01
I
9,tER,
1
\\
b = exp (it
I E SO(2),
(0
1
0
I
I E Bso(2)
and the decomposition is unique. The Lie algebra of Bj is in this case bSO(2) = { t I
0
)
:
tE
R}
.
Example 2 Suppose A = SO(3), let 1
-i
X=
EC,
0
and define
Bso(3) = {b E SO (3)c : bX = vX for a v E (0, oo)}.
Now pick a g E SO(3)c, and let Y = gX. Since X is isotropic (i. e. X # 0 and (X, X) = 0), the same holds for Y. This means that Y can be written in the form Y = Y1- iY2 for real vectors Y1, Y2 0 0, where I Y1 I = I Y2I and (Yl, Y2) = 0.
Hence there exist a unique R E SO(3) and a unique number v E (0, oo), such
that
0
1
Y1=vR
0 0
J,
Y2=vR
1
.
0
Set b = R-1 g, then we have bX = vX, i. e. b E Bso(3) . So we have the unique decomposition g = Rb. The Lie algebra of Bso(3) is 0 b.(,(3) :=
it ix
-it -ix -x 0
8. Construction of finite type solutions
83
Note that here 93Sa(3) is not unique, a different choice of X (pick up any other isotropic nonvanishing vector of C3) would lead to another Borel subgroup.
Example 3 Let .fi = U(n), then it can be verified, that Ti
0
: T1,...,7n'E(0,oo), T; ECfori<j
ZU(n) = works.
In all cases we have furthermore that the mapping g x B-4
-4
(f,b)
.fc
fb
is a diffeomorphism, a property that we shall summarize by the relation AC = A.B.g. On the level of Lie algebras, this means
to C=tED bx, where t and b,a are the Lie algebras belonging to .fi and 93.R, respectively. We assume from now on that such a decomposition of Ac exists.
8.2
Application to loop Lie algebras
Consider again the Lie algebras
A9c = {[A
fa E E
Agc : fa extends holomorphically to the unit disk},
and the corresponding twisted Lie algebras. Define furthermore A R 9C = { [A I - a] E A+gc : o E b,a },
with the usual notation kEZ
The following decomposition
-kAk + (&b, +
kAk + kEZ
k<0
k>0
(i,k k>0
-
Ak
+
8. Construction of finite type solutions
84
where o = (to)t + (&b is the splitting according to t ® C = t ® bR, shows that we have the Lie algebra splitting AeC = Agr ®A+
C
(8.1)
Or
Let r: Agc7 -+ Agr
and
j: AgC 7 -' A R0C
be the two accompanying projections.
Remark Agr is the Lie algebra of Asr. One can also check easily that A+ gc, is a loop Lie algebra: we shall see in paragraph 8.4 the associated loop Lie group. This will turn to be crucial in the Chapter 11.
8.3
The algorithm
Now we are ready to construct harmonic maps. First choose an odd number d E N', and set d
Adgr =
JA H a) E Agr : Ca =
k=-d
k \k
This is no longer a Lie algebra. However, it is a finite dimensional vector space. Define the two vector fields XI, X2: Adgr - Ag by (8.2)
X2(G) = a,r(i)1d-lea)] _ Note that
f
r(.\d-ll;a) =
dt + (6,,cc-SlA
11;-d,
A-l -d + 06-d)t - Ail;-d.
We will use the notation tt a, (S,\) = r(.1d-lSA), a2(G) = r(iAd-lea),
/
tt
l3l(Sa)
t = j(Ad-ISA),
Lemma 8.1
i) The vector fields XI, X2 are tangent to Adgr. ii) Moreover, their flows preserve the quantity
2
,
dl(A)
is negative definite, the flows So, for 0 = SO(n + 1) or SU(n), since of Xl and X2 stay in spheres and thus exist for all times.
8. Construction of finite type solutions
85
Proof i) We first have to check that XI and X2 take values in AgT. But note that since d - 1 is even, r is well defined and we have r(ad-ISA) E AgT for CA E AgT. Thus the same holds for XI(CA), X2 (CA) by the Lie algebra property
of AgT. To show that XI and X2 actually take values in AdgT, we use the observation (8.3). Taking the Lie bracket of one of the expressions on the right hand side of (8.3) and CA yields only powers of A between -d and d. Another way to prove this is to take into account that 01 (CA), 02 (CA) E A'
gc
for CA E AdgT. Hence the Lie bracket of CA and /31(Ca) or f32(£,\) contains only
powers of A no less than -d. Since XI(CA) and X2(CA) are real, we have even XI (CA), X2 (CA) E AdgT.
ii) This property is easy to check by a direct computation.
O
Lemma 8.2 The vector fields XI and X2 commute, i. e. [XI, X2] = 0. Proof. Since a1 and a2 are linear, we have [XI, X2](CA) =
[XI ' A,
X1 '
-t [X2 ' CA, a1(. (G)] - [CA, X2
-
(bA/)],
I \CA)] CC
KA, a2(I* A, a1(SA)])]
[CA,al([CA,a2(CA)])]-
Now use the Jacobi identity for Lie algebras to conclude [XI, X2](CA) = [CA, KA],
where KA = a2([G,a1(SA)]) - a1([SA,a2(SA)]) + [a,(CA),a2( ,,)]
We substitute a1(CA) = \d-ICA - #I(CA) and a2(G) = iAd-ICA - Q2(CA) and obtain tt / ,,)] / K,\ = -a2([bA,Ql( i)1d/-I[CA,Q1(CA)]+Q2([CA,QI(CA)])+\d-I[CA,02(A)] _
-
Q2(eA)])
- \d-I[CA,Q2(CA)]
Ql`SA)]
+ [31(CA),132(G)]
= Q2([A,QI(A)]) - Q1([A,Q2(A)]) + WI(A),Q2(A)].
A - 02(CA)]
8. Construction of finite type solutions
86
Here we use the fact that AgT and A R gC are Lie algebras: since the ranges of al, a2 and of 31, /32 are AgT and A n gC, respectively, we see from these two representations of K,,, that Ka E AgT fl A A9C = {O}.
0
Thus [X1, X2] = 0.
Now, because of Lemma 8.2, we can simultaneously integrate Xl and X2 to construct a map rla: R2 - Adg" such that
8x
'rla
ay
= XI (77,0, = X2(77A) -
We denote as := r(ad-lr7adz) and /3a := j(Ad-1,%dz). Then we have the Lax equation dr1, = [rfa, r(ad-Ir1a dx)] = [rja, aa], as can easily be seen from the definition of rla. Moreover,
(8.4)
as = A-In-ddz + 01-ddz)t + An-ddz, so we have a representation of as according to our usual decomposition. But we have also the following. Lemma 8.3 The I -forms as and /3a satisfy
daa +2[a.,Aa.\]=0,
dfa-2[0aAQa]=0. Hence Lemma 5.2 and Proposition 7.1 may be applied to prove the existence
of an Fa: fl - 0, such that dFa = Faaa, and such that ua = [F,,] is harmonic. Proof. Recall /3a = j(Ad-Itrya dz). We know that
= 0, dta - [Qa, rla] = 0.
d7ja + [aa, rla]
We may write /3a = Ad-Ii,, dz - aa. Hence daa + 2 [aa A aa] + dQa -
2
[Qa n Qa] _ -Jid- I (drla + [aa, rla]) (az) dz A dz = 0.
8. Construction of finite type solutions
87
Since we have
daa + 2 [aa A ax] E Ag, and
doa - 2
A,3,\] EAR 9c
this means that both terms vanish because Ag, fl AKgcr = {0}.
We have thus indeed constructed harmonic maps by integrating a pair of vector fields on the finite dimensional vector spaces Adg,. Such maps are named finite type harmonic maps.
8.4 Some further properties of finite type solutions i) The algorithm developed above does not only create aa, but also a 1-form /3 E T'Q ® AgC, with the property
dpa-
1[QAApa]=0.
This is-up to the sign-the property requested in lemma 5.2. So in this case the same procedure provides a map BA: Q - Arc, such that dBA = 0.\B.\,
and it is unique up to a prescribed value at some point zo E Q. We may e. g. choose BA(0) = 16. Similarly, we may assume that FA(0) = 16. (Assume 0 E S2.)
In fact, we have even pa E T' St ®AR gc, that is, 00
QA=>QkAk, QOE11R. k=0
Hence, if we define the loop group A++ MA
OC r
= {[A'-' ga] E AOC : ga = >gkAk, 9o E `Bx}, k>O
where AC = A.s$jq , we can check easily that the Lie algebra of A+ OC T is
A+ 9C. Thus we obtain Ba E A++ OC
8. Construction of finite type solutions
88
ii) Consider once more the equation drla + [aa,lla] = 0, where as = FT 1 dFA. It implies that Fa-'d(Fa11AFa 1)Fa = drla + [aa, rla] = 0.
So 77,\: = Fa1l,, Fa ' is a constant in Ag,..
iii) Furthermore, we have Ad- l rla dz = a,\ + i3a,
a fact that can also be expressed in the form Ad-1 1)a dz
1 dz = F,\ (F;- 1dFa +dB,B_ 1)F_ 1 = = d(FABA).(FABA)-'.
But this means that FABA =
ead-1(8.5)
O
iv) Write 17a in the form 11a=
e-ad-'n"= 'la ead-I
= Ba 'F;-'. 17.\ F\B,, = Ba
We know that 17,\ E Adgr, i. e. d
11A _ E 11k '1k.
k=-d On the other hand, since Ba E A*+AQST, it has the representation
Ba =
BkAk. k>O
Hence in the expression B,1riABA, only powers AA; of A with k > -d appear.
But the reality condition 7IAE Ag, implies then that in fact 1IAE Adg,.
Constant mean curvature tori are of finite type
9
Recall that CMC surfaces in R3 correspond to harmonic maps into S2 by the Gauss map, which are neither holomorphic, nor antiholomorphic. We shall see now that for CMC tori, all such harmonic maps are of finite type, a result of U. Pinkall and I. Sterling. This result can be generalized to harmonic maps from torus into Lie groups [21] or more generally into symmetric spaces [21], [22].
9.1
The result
Theorem 9.1 (681 All CMC immersions T2 - R3 are of finite type. Sketch of the proof (details are in the Appendix) i) Any immersion of the torus T2 in R3 defines a unique conformal structure on T2 and so can be seen as a conformal immersion of R2/aZ + bZ, where (a, b) is some basis of R2. Such an immersion can be lifted into a conformal one X : R2 -- R3, periodic with period vectors a and b. Now the Hopf differential of the immersion has the form f(z)(dz)2, where f is holomorphic and periodic, so is constant by Liouville theorem. Would
f be equal to zero, then X would be a totally umbilic immersion, so its image would be a subset of a round sphere Sz in R3 (see the proof of Hopf's Theorem in Chapter 4). But this is not possible 3. So we are left with the situation where f is a constant different from 0. After suitable normalizations we see that all we need to do is to work with conformal CMC immersions X: R2 -+ R3, where X is periodic (with
respect to a lattice aZ + bZ), such that H = 2, and
f =la I
2-I
I2-2z\
=-1
for the Gauss map u: R2 - S2. ii) For any A E S1 C C*, construct a "normalized framing" e2,a
0
ua Xa) 0
1
J
: R2 -* SO(3) oc R3
3assume that it would be the case: endow R2/aZ + bZ with g, the pull-back of the metric on SX by X. Then on the one hand the Gauss curvature Kg of g is positive everywhere, but on the other hand the Gauss-Bonnet formula implies fTs K9dvol9 = 0, a contradiction
9. Constant mean curvature tori are of finite type
90
of X, such that
A,\ =Fa'dF,, =
2
0 0
sinhw
-icoshw
0
0
+*dw
A
+2
- sinhw e' icoshw -iew
0
0
0
-1
1
0
0
0
0
0
0 0 0 0
0 0
0 0
0 0 0 0
(9.1)
0
0
- sinhw
e'
0
0 i cosh w 0
-icoshw
iew
0 0
0 0
sinhw 0
-*dw
0 dw
0
- sinh w dx,, - cosh w dy,,
e' dx,, e' dy,,
0
0
0
0
sinhwdxa coshwdya 0
dz
0
'
where xA = Z(A-lz + Xi) and y,\ = 2i(A-lz - \z). (Cf. Chapter 5.) We have then the associated family of CMC immersions XA with mean curvature H = z and with f = -A. The function w: R2 -+ R satisfies Ow + cosh w sinh w = 0.
(9.2)
The map FA may be seen as an application with values in the loop group ASO(3) x 1R3. Moreover let
P:=
1
0
0
1
0 0
0
0
-1
0
0
0
-1
0
and
z:
M(4, IR)
M
0 0 I'
-+ M(4, R) '-- AdPM = PMP-1.
Then z is an automorphism of ASO(3) x R3. We also remark that -r(AA) = A_A,
so that AA has his coefficients in Aso(3) ®R3 :_ {A u-+ t;, E Aso(3) ®1R3/A E S', 7-(t
9. Constant mean curvature tori are of finite type
91
This is obviously a sub-Lie algebra of Aso(3) ® R3, and the Lie algebra of
G,, E ASO(3) a R3/A E S', rr(G,,) = G_,,}.
ASO(3) a R3
We deduce that if we choose F,,(0) E ASO(3) oc Rr (for instance the iden-
tity), then Fa takes its values in ASO(3) a R. We shall denote E the set ASO(3) v R such that (9.1) is satisfied, for some of maps F,, : 1R2 function w : R2 - IIt which is a solution of (9.2) (Fa isn't required to be periodic here).
iii) Study deformations of the given F,, E E that remain within E. Suppose that the surface Xa is thereby transformed into X,, + eO,ua + o(E).
(9.3)
What conditions are to be required of t/ia such that this is still a CMC surface with H = Z? Of course the surface given by (9.3) is no longer conformaly parametrized. This problem may however be avoided by considering instead the deformation X,, + E(tlel,,\ + tae2,A + 1/1),u),) + o(c)
for suitable ti,, t2. This amounts to deforming F,, into F,, + eF,,T,, + o(e) E E, where 0
ti A
TA =
t3
A
0
t2 a
0 t2 A
0
t3 ,, t3 A
t1
d
tA1
0
to 0 1
and t3 Note that T,, is completely determined by 0,\ up to some constant (see in the Appendix). One of the compatibility conditions on 0a that we are looking for turns out to be £
+bacosh 2w=0,
1
(9.4)
miraculously (!) the linearization of (9.3) (see Theorem 9.2 in the Appendix). iv) By this deformation, w is changed into W + EW + O(E).
One can compute that 1
9. Constant mean curvature tori are of finite type
92
where Xa is some function constructed starting from 0A by the following algorithm: we let 4a be a solution of a0a
I
'OX
ox
=
4Jiw:
=
sinh2w tG
X.\
:= 4
,
(9.5)
2
az
and we set
a,
(A2 a2)2
- aw: Oa
(09
(9 . 6)
The existence of solutions to system (9.5) is guaranteed because 4(iPaw.)Z +
sink 2w (ox
= 0,
2
consequence of (9.2) and (9.4). If Xa is constructed from tPa by means of (9.5) and (9.6) we shall write t'A X.\. An important property is that if t/ia X.\, then XA is also a solution of (9.4). Note that the solution XA of (9.5) and (9.6) is unique up to some constant (which may depend on a) times w..
v) There is now an easy way to construct an infinite sequence of formal complex solutions of (9.4). Namely: we start from tPa°) = 0 and then we build recursively a sequence of solutions of (9.4) such that
(t/
nEN
t/,jn+l), Vn E N. It is possible to choose these maps in the form IVin) _ \2n-2P(n) [w], where each P(n)[w] is a polynomial in wzf wZ;, w;t;, etc ... homogeneous in yoz- of degree 2n- 1. These functions can in principle be explicitely computed, the sequence begins by
,(n)
+G(°)
7Pa)
= a-2P(°)[w] = 0, = A°P(1)[w] = wz,
(3)
[w] =
lp(4)
A4P(3)[w]
2 ' ),
= 42a4(w:z::: -
10w:z=w2
- 10w2zw: +6:
etc.
vi) After some algebraic manipulations, we construct an infinite family of infinitesimal deformations T(n'b) E TF,,E, where n E N, b E C, characterized by n
n
t(n,b),3
=
bE k=1
ka-2kp(n-k+l)[w]
+
bEkA2kP(n-k+1)[w],
k=1
9. Constant mean curvature tori are of finite type
93
and the deformation rate of w induced by T("'b) is w(n,b)
(bp(n+l) [w]
=
4
+ bP("+1) [w])
.
In particular this expression does not depend on A. vii) This is the only analysis part in this process. Assume everything is periodic, i. e. we are working on a torus. Then the linear equation (9.4), which is an eigenvalue problem for the Laplace operator, has a finite dimensional space of solutions. In particular, there exists an integer N, such that
ranks{P(1)[w],...,P(N)[w]} = ranks{P(1)[w],...,p(N+1)[wJ} = N. viii) We now choose (b1, ... , bN+1) E CN+1 \ {O} (in particular bN+1
0) such
that b1 p(1)[w] + ...+ bN+1p(N+1)[w] = 0 on R2.
(9.8)
Then we let Tao,b,)
+ ... +TAN'bN+I) It is an infinitesimal deformation in TFAE which does not vanish. Indeed the T'A :=
family (A-2P(1)[w], ... , EN,
+Ta1,b2)
kA-2kP(N-k+1)[w])
is linearly independant
and so to 3 = q,* := q,(l,b2) + ... + T(N'bN+,) does not vanish.
ix) Think and interpret : Let
Fa+eFATT+o(e)Ee be the infinitesimal deformation of Fa induced given by T,*. The deformation rate of w induced by Ta is wa°'b1) + + t:J(N,bN+') (see (9.7)). But
because of (9.8) this is just zero: the surface is not deformed at all, but just moved around in R3. So FAT.\ = DAFA
(9.9)
for some constant D.% in the Lie algebra of ASO(3) a R3. x) We conclude that Ta = Fa 1 DAFa satisfies
dTa+[A,,TaJ=0. If we write
then we have
dTa + [Aa, Ta J = 0. But by construction, this includes only a finite number of powers of A (i. e.
Ta, Aa are polynomial in A). Hence we have a finite type representation of X, as claimed.
9. Constant mean curvature tori are of finite type
94
9.2
Appendix
In this Appendix we expose the technical details of the proof of Theorem 8.1. We shall use the notations Alto
{a '-- to E IR/t_A = to},
{A'--- to E R/t_A = -tA}.
AR1
Study of the deformation of FA Given some extended frame FA E E we want to study infinitesimal deformations FE,A := FA + EFATA + o(E) where TA is a smooth map from R2 to ASO(3) x IYT and such that FE,A still belongs to E. Thus TA may be viewed as an element of TF,,E, the set of tangent vectors WE at FA. We shall denote 0
TA =
t2 A t A
to
to
t2 A
t3 , A 0
0
0
0
t1 2 A
0
tl A 0
to
and sometimes drop the A when there is no ambiguity. The question is to characterize all such TA's. In view of (9.1) the answer is simple: one considers any one-parameter deformation of w, (wE)EE(_EO,EO) with wo = w such that for any e, wE is a solution of Ow,, + sinh wE cosh wE = 0.
Then
A E'A '
0
- *dwE
-sinhwEdxA
e"°dxA
*dwE
0
- cosh w4 dyA
el, d y,\
0
0
0
0
sinhw,dxj coshw(dyA 0
0
(9 10) .
is integrable, i.e. dAE,A + AE,A A A,,A = 0 and integration of the equation dFE,A = FE,AAE,A leads to the family of deformations. Notice that if one set wE := w + Ew + o(E), then cw is a solution of the linearized sinh-Cordon equation
Ow +w cosh 2w = 0.
A crucial point - which is remarked by U. Pinkall and I. Sterling in [681 - is that one may also construct a TA in TFAE knowing its component to (modulo some constant), a map into Alto.
9. Constant mean curvature tori are of finite type
95
Theorem 9.2 a) Let T,, E TFAE, then the component to of TA is a solution of the linearized sink-Gondon equation
/ to+tocosh2w=0. b) Let
a : R2
(9.11)
AR0. Then the overdetermined linear system
4aa a
a0A
=
doa
- . sink w
az
(9.12)
2
has a solution if and only if V)A is a solution of the linearized sink-Gordon equation (9.11).
c) (A partial converse to a)) Let
A
: R2 -+ AR0 be a solution of (9.11).
such that to = A if and only there exists a
Then there exists some TA E solution 0A of (9.12), such that z
T 1 A, A) := (a az - az A)
z
+ a- az z
- a- lawa - 2lA (9.13) 1
is a function which does not depend on A. The proof of Theorem 9.2 relies on the following Lemma. Lemma 9.1 Let TA E TFAE then
the component to of TA satisfies the linearized sink-Gordon equation (9.11)
there exists a map 0A : R2 - AR1 (where AR, := JA '--- t,, E R/t_A = -tA}), which is a solution of (9.12), with ipa =\ta, such that to + itA = e"' 12A
tia+it2A=em (\ a t21 'A
-- i
2
a20a az2
-
4A)z o
a - 0A 1
e-"' I
az2
w = n(IPA, OA)-
(9.14)
,
8 - 2a) az
,
(9.15)
(9.16)
(9.17)
The proof of Lemma 9.1 occupies the end of this section. We shall first prove Theorem 9.2 assuming that Lemma 9.1 is true.
9. Constant mean curvature Lori are of finite type
96
Proof of Theorem 9.2. The property in a) is a restatement of part of Lemma 9.1. To prove b), we need to write down the integrability condition of system (9.12), i. e. that
d 14Awz a dz -
asin22w Ad-
I=0
-
(Here we use the subscript w., for 0, etc... ). Using wZZ + s sinh 2w = 0, this simplifies into
/
Aw,z 1 4 z a 19
+ V i a cosh 2w'1 &Z- A dz = 0,
-z-
which is clearly equivalent to (9.11). Lastly for proving c) let us consider some i/ia, which is a solution of (9.11). Then Let us assume first that there exists a Ta E TFAe such that to
the infinitesimal deformation w of w induced by Ta needs to be independent of A. But because of (9.17), in Lemma 9.1, it means exactly that n(o.\, 0.\) is independent of A, for a suitable choice of 4a. Conversely if there exists some solution 0,\ of (9.12), such that I (th, 0,0 is independent of A, then formulas (9.14), (9.15) and (9.16) defines a map TA. A long computation (which basically uses the material of the proof of Lemma 9.1) 0 shows that actually Ta E TFE. Proof of Lemma 9.1 Step 1: Writing the relations satisfied by Ta One defines FE,a := Fa + FATA + o(e) E e and one writes that Ff.a dFE, a = AE,a,
9.18)
where A,,a has been defined in (9.10). The left hand side of (9.18) is
F-'dF(a = Aa +e(dT\ +
+o(e),
where dTa + [Aa, TA] = 0
dti _t2 sinhwdx,, +t3coshwdya dti-t2*dw+t1coshwdya 0
dt2 - t2 sinhwdxa + t 1 coshwdya 0
dtz+t1*dw+t2sinhwdxa 0
0
dt'-t2*dw-t3sinhwdx, -e"'t2dyA dt2+t'*dw-t3coshwdya-e"'tidxa dt3+t sinhwdxa+t2coshwdya-e"'(tidxa+tzdya)
0
0
dt3+t2*dw+t2coshwdya
dt3-t1*dw+tIsinhwdxa
9. Constant mean curvature tori are of finite type
97
Whereas the right hand side of (9.18) is
Aa +,E
0
- * dth
*dw
0
-wcoshwdxa we'dxa
-ii sinh wdya
c;,e"dye,
0 0
0 0
wcoshwdxa wsinhwdya 0
0
+ O(E).
Thus we are led to the following system of equations.
dt' -
- t3 sinh wdxa - e"t2dya = we"dxa, dt2 + t' * dw - t3 cosh wdya + e't2dxa = we"dya, t2 * dw
(9.19) (9.20)
dti - t2 *&A) - t2 coshwdya = wcosh wdxa,
(9.21)
dt2 + t * dw + t2 sinh wdxa = c;j sinh wdya,
(9.22)
1
dt2 - t2 sinhwdxa + ti coshwdya + *d = 0,
dt3 + t' sinh wdx), + t2 cosh wdya - e"(t1dxa + t2dya) = 0.
(9.23) (9.24)
Step 2: Useful relations Relation (9.24) allows to express ti and t2 in terms of t', t2 and t3: 3
e
e
a + t'sinhw a
(9.25)
3
+ t2
e
w
a
Furthermore, relations (9.19), (9.20) and (9.21) can be replaced by their
respective projections along 0 := a8 +)-' j and 0 := iA - i) ' &:
a
i a
+ t2 8y - t3 sinh w = we",
8t'
8w = t2e t2 8xa Oia 0t2
8xa 2
ay-,\
+ t'
- t' fla 8w
axa
(9.27)
-4e",
(9.28)
- t3 cosh w = we",
(9.29)
_ , cosh w,
(9.30)
' +t3
8xa
_
(9.26)
ava
9. Constant mean curvature tori are of finite type
98
3
aw t3 a= t2 cosh w, xa aya 1
at3e
- t;
axa
aya
(
-4sinhw.
.
(9.32)
= w si n h w,
aya + t1
9 31 )
(
9 33 ) .
Lastly, two further relations can be deduced from the four first ones of the previous series: substracting (9.26) - (9.29) gives 8(e-Wtl)
- 8(e "t2) _ -t se -2"
,
( 9. 34 )
and (9.27) + (9.28) gives
a(e-"t') aya
+
a(e-" t2) ax.\
= 0.
(9.35)
Step 3: We show that t3 is a solution of the linearized sink-Gordon equation Suming (9.30) + (9.33) gives we2w
= a(e"'t1) + a(e°'t2) axa
aya
which, using (9.25) to eliminate ti and t2, leads to
wee" = Ot3 +
a(t' sinh w) + a(t2 cosh w) aya axa
Now (9.26) + (9.29) gives CJe2w =
1
(O(ewt') + a(e"'t2) _ t3e2,,,) axa aya /I
These two relations implies by eliminating cwe2w
a(a t')
e1, =
Qt3 + 2
2 C
_ a(e-" t2)
a
Oya
/
And by using (9.34) one gets Ot3 + t3 cosh w = 0.
(9.36)
9. Constant mean curvature tori are of finite type
99
Step .¢: Equations on t 1 and t2 We sum/ - sinhw (9.26) + cosh w\(9.29) to obtain
- t3 sinh w I + cosh w I
- sinh w I a I + t2
+t1
and substracting e" (9.30) - e"(9.33) gives 3
ew
23
+t3l 52J1
'
- t3 cosh wJ
awl-ew(atl+t3aw,=-f.1.
2axA
/
A
Now we sum both equations and we use (9.25) in order to eliminate ti and t2. It leads to 2
a2
coshw - 2
W\
a1
sinh w - 2t2
ax,\ 82t3
-
+
(
cosh w + 2t1
A)2
Ow oxA
sinh w
+2---2= t3. aw 0t3
92t3
(OXA)2
c .1
C
gXA axA
v at3 A
A
A computation using (9.34) gives a(e-mtl)
ax)
a(e-"t2)
+
eyA
- 2w -e
a
a
at3)
-20
5X A \
OXA /
A
3
-2w
(e8Y,\
(9.37)
Similarly the combination - sinh w (9.27) - /cosh w (9.28) gives - sinh w I
-A
- t2 5x-,\
I - cosh w I
A
- tl
= t2,
and the sum ew (9.31) + ew (9.32) gives 3 -e" (at1 - t2
- r1\ 9x,\ - t1 aXA ) 3 2
OW
) W\ We sum these two relations and use (9.25) to obtain A
2
- aa -
_ -t2.
1
cosh w t3
axA O YA
v
+ aw
sinh w + t2 at3
+ aw
cosh w + t 1 at3
v
sinh w
=0.
a XA atA
a XA a XA
Then a computation using (9.35) gives
a(e wtl)
A-
a(e wt2) = e2e2 OXA
a 8XA
e -2w
0t3 OYA)
+ a OYA
e -2o
t3
(9.38)
a-A / J
9. Constant mean curvature tori are of finite type
100
We can conclude this step by the following equations which summarize relations (9.34), (9.35), (9.37) and (9.38).
8(e-"(tl + it2)) =
2
8zA
'
8(e-"(tl + it2))
where oia = w and We shall pose 2
3
8za
- 4 8w 8t3
82t3
8za
-
(9 39)
8ZA 8za'
(8ZA)2
e-2" 0 '
(9.40)
2
=A-la
- e-"(tl + it2)
. e-"(tl + it2) = 2
3
8za
- 0,\.
(9.41)
Then relations (9.39) and (9.40) allows to express 4a in terms of t3 (also using (9.36)): e46a 09Z,\
'90,\
=2
82t3 (8z,,)2 a2t3
=2
&. 8t3 - aza a (e-"(tl + it2)) = 4 8za aza
(9.42)
-
(9.43)
a (e-"(tl + it2)) = -sinhwt3.
8za 8za 8z-.\8za Step 5: Computation of cv Suming (9.26) + (9.29) we obtain e2"(2w + t3) =
2
8(e"(tl + it2)) + a(e"(tl + it2)) 8za
8za
And using identities like
8(e"(tl + it2)) = e2" 8(e-"(tl + it2)) 8za
8za
+ 2e 2" 8w (e-"(tl 8z,\
we get
W=-2+I +
ew
Ca
(e-"(tl +it2))
(e-"(tl - it2)) +
8tz
1
2
+ it2)),
it2))
1 8(e-"(tl - it2)) 2
8x
Now we use (9.41) in order to obtain a formula which relates w in function of t3 and 0a. Simplifications using (9.42) and (9.43) occur and give t3
02t3
- 2 + I\ ((OZa)2
We recognize the relation 9.2 by formula (9.13).
Ow
aZa
)
( (92t3
Ow
OX J
= fl(t3, 0a) where fl has been defined in Theorem
9. Constant mean curvature tori are of finite type
101
Step 6: Conclusion The map ¢a which was introduced through the relation (9.41) is actually completely determined from t/ia := to by relations (9.42) and (9.43), up to a constant. Then the relation (9.41) implies easily relation (9.14). And using (9.25) one deduces (9.15). Lastly one gets from (9.27) and (9.28) that 2O(e"'(t' e-2"' a(e"'t') _ 8(e"t2) \ I + it2)) 2 ( ayA = -e'IM t2 , aza axa s from which we deduce (9.16).
A Biicklund transformation of solutions of the linearized sinh-Gordon equation
In the previous subsection we have seen that for any ipa : R2 - ARo which is a solution of Ozpa + a cosh 2w = 0,
(9.44)
there exists a solution 0x : R2 -. ARI of a4
Ua oft
= 4X8 ft,\ stinh =
-
2
(9.45) 0x
Obviously 4x is unique up to a constant in ARI.
It will be useful to extend our setting to maps 0a : R2 -p ACo := {A '--+ to E C/t_a = t,\} which are solutions of (9.44). Then there exist solutions 0a : R2 --- AC1 := {A' -+ tX E C/t_a = -t,\} of (9.45). Definition 9.1 Let ika : R2 -- + ACo be a solution of (9.44) and Xx : R2 -. ACp. We write zpa
Xa if and only if there exists a solution 4
: R2 - ACI of
(9.45) such that Xa = 4 (X If zpa
2
(O z)2 a2 'A
A8Bzw4 1
XO, the set of maps XX such that zpa = Xa is exactly {X0 +CAwZ/C,, E
ACo } .
This definition is motivated by Lemma 9.1: indeed relation (9.17) can be restated as
2zGa,
for zpA = t3 and zp,,
X,, 4. Proposition 9.1 below leads to another charateri-
zation of Xa. 4 similarly relation (9.16) can be written as t2',.\
=4Xa-4
9. Constant mean curvature tori are of finite type
102
Lemma 9.2 If zya : R2 -p ACo is a solution of (9.44) and zlia = Xa, then Xa is also a solution of (9.44). Proof. It can be checked through a direct computation. Alternatively it follows from Lemma 9.1, since cw and to are both solutions of (9.44) and because of 0 (9.17). Lastly it is also a consequence of the Proposition below. Proposition 9.1 Let V)a : R2 -+ ACo is a solution of (9.44). Then the set of
maps Xa : R2 - AC0 for which there exists two maps aa, ba : it2 -+ AC such that
8
as
e
cosh w -i sinh w i sinh w cosh w i sinh w cosh w
8z
-isinhw
ba
i
ba
as
coshes
coincides with the set of Xa :
_ Ana
l
-i sinh 2w )
\
=
cosh 2w J i sinh 2w cosh 2w
2 -+ ACo such that z/ia
(9.46)
)
Xa.
Proof. We shall denote
( wa)
(
i sinh w
cosh w) (
(
(
i sinh w
cosh w
b,\
)
and
)
-
)(
ba
Step 1: Uniqueness of Xa modulo w,z
Since the system (9.46) is linear in 0a and Xa, it suffices to study the solutions of the system (9.46) for z/ia = 0. This reads 8
8 8z
(
= 0
V-\
0 cosh 2w
-i sinh 2w
wa i sinh 2w cosh 2w
v,\
) (wa)
_
- X-1 Xa
i sinh 2w cosh 2w
The first equation means that va and wa are holomorphic. The second equation gives
) (i sinh 2w) ('
(8zua
l az -
- Xa 1
2zvx
l cosh 2w
i sinh 2w cosh 2w
+
Oz
+ 2xwZwa
)(
cosh 2w
-i sinh 2w
9. Constant mean curvature tori are of finite type Since
U cosh 2w
,
-isinh2w
( i sinh 2w
t
cosh 2w
8w.\ 8x
)
103
l is a basis, this is equivalent to
- 2iwzva +
= 0
8Z- + 2iwzwa
=
0.
Assume that wa 36 0. Then since va and wj, are holomorphic, the second equation would imply that wZ is meromorphic. This is however impossible (excepted in the case where w - 0) because (w=) = - s sinh 2w 36 0. Thus we necessarily have wA - 0 and the second relation implies * = 0. Hence va is equal to a constant Ca E AC. Now the first equation gives Xa = 2iACAwZ. X Step 2: Existence of a solution to system (9.46) as soon as ?lia
We let 0a to be the solution of (9.45) such that X = 4 (\2 820
- a ¢a) and
we set W'\
4A
&Z
Then using (9.45) and (9.44) we obtain
8
Xa (
i sinh 2w cosh 2w
i sinh 2w cosh 2w
) ( wx )
va
(wa
=
We then find that
8
cosh 2w
8z
-i sinh 2w
- -4,\_1
A282OX
(
_
v',
)
(8z)2
i sinh 2w cosh 2w
0
and the conclusion follows.
Corollary 9.1 If /a : R2 -+ ACo and X : R2 (9.44), then 0a = X,, if and only if 3 T\.
ACo are two solutions of
Proof. The system (9.46) is invariant by complex conjugation provided one changes (ip.\ XA) into (-5F\, (and (aA, bA) (aa, ba)), so the result fol-
0
lows.
An algorithm to produce solutions to (9.44)
An obvious solution to (9.44) is wz. By Lemma 9.2 all maps X,, such that XA are also solutions of (9.44). A computation shows that such of the form
wZ
4,\2(W,
:: - 2wZ) + CAw2.
are
9. Constant mean curvature tori are of finite type
104
We may repeat this computation, finding that 2L,,3)
42A4(wzzzzz
- 10wzzzwZ - 10wZZw2 +6wZ),
etc... We can hence generate a unique sequence (Oan))tEN defined by ?
(O)
_ -2P(O)[w] := 0 P(1)[w] :_ Wz
Vn E N,
titan)
,P (n+1)
_'\2n-2P(n)[w] where P(n)[w] is a polynomial in Vn E N, O(n) C[wz, wz2, ...] homogeneous in of degree 2n - 1
The last assumption means that
aa'w Qj) ((OZ)ai
#'
/
...
8a,w 13,
((8z)aj)
,
where Cn (a l , ... , aj, Q1 i ... , Qj) = 0 if a1,01 + + aj,Oj 34 2n + 1. It can be deduced from an algebraic recursion formula which is proved in [68], Proposition
3.1, p. 149. An immediate consequence of Corollary 9.1 is that 'tan+1)
J1n), Vn E N.
Algebraic construction of a series of infinitesimal deformations in TFAE We wish to construct families of solutions zJ' : R2 - Alto of (9.44) such that there exist 0a : R2 -' AC solving (9.45) and such that St(t/ia, a) is independent of ) . 4 (23 - AWOa), this amounts to find pairs (tk.\, Xa) By denoting Xa ik,\ of solutions of (9.44) such that Xa and Xa + a Xa - ipa does not depend a z on X We look for maps t p,\ of the form n
E A-2kakP(n-k+1)[w] + [>2kQkP(n-k+1)[w]' kk=O
L
k=O
9. Constant mean curvature tori are of finite type
105
where ao, ..., an are complex constants to be determined. Note that this ansatz garantees that 1a is real valued. The maps Xa such that t/ Xa are n
n
X,\ = CAP(1)[w] + A2 E a-2kak p(n-k+2)[w) + X2 r` 2kak p(n-k)[wJ, k=0
l
where Ca E AC and so 1
1 (Xa + Xa
n-1
n
E A-2k+2 ak p(n-k+2) [w] + E X2k+2akP(n-k)[w] + CaP(1)(w]
4 n
k=O
Lk=O
A2k-2T-k+2)[w]
k=O
n-1 + > A-2k-2akP(n . k)[w] + CaP(1)[w] k=O
n X-2kakp(n-k+1) [wJ
-2 >2
n
- 2 r A2kak p(n-k+1) [w] k=O
k=O
- 2an)P(1)[,] +C,\P(1)[w] +,\-2n+2(an + an-2 - 2an-1)P(2)[w] L,\_2n(a,1_i
4
+a-2(a2 + ao - 2a1)P(n) [w1 + X-2a0P(n+2) [w) +(a1 - 2ao)P(n+1)[w] + (a1 - 2a-o) p(n+ 1) [W]
+.\2(a2 + a.0 - 2a1)P(n)[w] +.\2ap(n+2)[w]
+.12n-2(an-2 + an - n_1)P(2)[w] + A2n(an-1 - 2an)P(1)[w] +CaP(1)[w], aZ
.
It turns out that this quantity is independent of A if and only if we choose
ak=kb, VkENand Ca=(n+1)bA-2n+c for some constants b, c E C. We will choose c = 0. Then we are left with 4 (Xa + Xa - 2t(ia) = 4 (bp(n+1)[w] + hP(n+1)[w]) Sin genera) for c # 0,
(x + 4
-2i'A) = 4 (bP(n+1)[w)+6P(n+1)[wJ+cP(1)[w] +"cP(1)[w))
.
9. Constant mean curvature tori are of finite type
106
We conclude: for any it E N and b E C we define the neaps 41(n'b)
:= b> kA-2k p(n-k+1)[
n ]
+bEkA2kP(h+-k+l)[w],
k=1
k=1
Then there exists a unique snap an'b) such that (9.45) holds and such that IP(n.b),4,(n.b))
= 4 (bP(n+1)jw] +
and so in particular is independent of A. By Theorem 9.2 there exists a unique E T,.- E such that t(n.b).3 = *(n,b) and W(n.b) = fg p(n,b)' 4,(n,b)). T("") is defined by the formulas in Lemma 9.1. Examples a) For W(A'b) = 0, where b E C, we have (O'b) _ _ LA- I t
(O'b)
4
= lxdZ
and SZ(q,(D'b), 4 (O.b))
t(O, ) ' 1
b) For
q,al'b)
_
(bw, + & ) ,
= t (bu - boo .
= A-2bwz + A2bwZ,
where b E C, we have
A -:+2,\-I tx,,z -
.6)
2
and
w(I.b) _ t(t b),1 2.a
2bA-2w2
4 (1.b)) =
A31) cosh 2w,
+ 4b(ws,z-- - w3),
b(wzss - w3) + b(wzsa - wi)+
= ibA-2wZ + ib(wzzz - wz3) - ib(w-.:z - w3) - ib A2wz. 2
2
10 Wente tori Historically the first immersed constant mean curvature tori were constructed by H. Wente in 1984 [86] by analytical methods. A short time later U. Abresch simplified this construction. He remarked that Wente tori should possess planar curvature lines and thus studied all CMC surfaces with planar curvature lines. It leads to an overdetermined system of equations which can be solved by quadratures using elliptic integrals. And U. Abresch showed that some of the obtained immersed surfaces do close up, giving CMC tori [1] (see also [87]). A similar treatment was given in [84], [85] by R. Walter who remarked that each curvature line of these tori is either planar or spherical. Moreover, as pointed out in [66] special surfaces (including constant mean curvature ones) with planar or spherical curvature lines were already investigated by A. Enneper [37], [38], [39] and his students [63], [15], [83], (see also [27]), [28], [61], [3], in the last century. Explicit formulas were then produced.
10.1 CMC surfaces with planar curvature lines We again look at conformal CMC inmmersions X : R2 - R3 such that H = 1 /2 and such that their Gauss map u : R2 --0 S2 satisfies
Then the curvature lines are the image of the curves [x'-- X (x, y)] (small curvature lines) and [y a- X(x, y)] (large curvature lines). Let us focus at small curvature lines as an instance. Their Fhenet framing (tl, nl, 91) is defined by
=
nl
I-I8t1 11 'Ox
bl
ax
= E1 xn1.
The curvature and the torsion along these lines are defined by rcl := a
8x
,
rl :_ -e
ax ,
nl
/
108
10. Wente tori
Computing these datas, one finds that
tl
= et
(,,U, ) 2
e-"
fGl
ay
+ sinh2 w
Kln1
8w = e-" ( -a y e2+sinhwe3
K161
= e -" (-sinhwe2 -
gi2 rl
82w = e-3" Csinhw axay -
e3 coshw8axw 88yw)
The small curvature lines are planar if and only if the torsion rl vanishes identically. This is clearly equivalent to the condition 2
sinhw
wOW
- cosh
OW = 0.
(10.1)
Similarly the large curvature lines are planar if and only if cosh w& W
- sinhw
= 0.
(10.2)
U. Abresch showed that choosing one the two above conditions leads to an overdetermined system of equations which can be solved explicitely using elliptic integrals. This is based on the observation that for instance (10.1) can be restated as 8x (;17nhw Oy) 8y (si hw a 0' which implies that one may find two functions f and g of one real variable such
that 1
OW
I
Ow
sinhw 8x -f (x), sinhw 8y -g(y). Thus the geometrical hypothesis (10.1) leads actually to a separation of variables. Using these new variables together with the equation Ow + cosh w sinh w = 0,
one can then show by a tedious computation that f and g are solutions of
(f')2=f4+(1+a2-,02)f2+a2, (g )2 = g4 + (1 - a2 + /32)g2 + Q2, for some constants a and 6. Thus f and g can be given using elliptic integrals. A similar conclusion holds if one assumes (10.2) instead of (10.1). However only CMC immersions with planar small curvature lines can close up into immersed compact tori.
10. Wente tori
109
10.2 A system of commuting ordinary equations In a subsequent work [2], U. Abresch proved that the above solutions arise from commuting flows. Namely we consider R6 with the coordinates c
s
U:=
U1
U2 U3 U4
l
and we define on R6 the two vector fields sue
sui
I
XI
(U)
Cui
CU2
%F2
su3 - C8
2
CU4
X2(U) :_
V`
CU4
2
-8u3 - C8
-8ul - U2U4
8U2 + U1U4
-CU2 + U2U3
-CU1 - U1U3
One can check that XI and X3 commute by a direct computation. Moreover the quantities C2 - 82 and s2 + u + u2 + u3 + u4 are preserved by X 1 and X2 so their flows remain in a compactisubset of R6 and thus are defined for all time. So let us take any solution U : R2 R6 of
8U = Xi (U),
and
= X2(U)
(10.3)
with initial conditions (at (x, y) = 0) such that c2(0)
- 82(0) = 1.
(10.4)
Then since the quantity c2 - 82 is preserved by the flow, we can pose c = cosh w
and s = sinhw, where w is a function on R2. Equations (10.3) can then be rewritten as wa
8x =
1
i72
(u, - 2U2),
- i coshwu4), (a )2 = 4 (sinhwu3
iz = -coshw sinhw.
(10.5) (10.6) (10.7)
We see hence that w is a solution of the sinh-Gordon equation. So we can associate conformal CMC immersions to each solution of (10.3) with the initial condition (10.4). These immersions turn to enjoy two properties:
10. Wente tori
110
they contains all conformal CMC immersions with planar small curvature lines or planar large curvature lines (see [2]) and in particular "Wente" tori (actually (10.1) is equivalent in choosing a2 = 0 in (10.9) below) they are of finite type, as we will see in the next paragraph
10.3
Recovering a finite type solution
We analyze the conformal CMC immersions obtained by integrating (10.3) with the initial condition (10.4). We can first observe (as in [2]) that the following quantities are conserved along the flow of Xl and X2:
a1 := cu3 - 2 (ui - u2),
(10.8)
a2 := su4 - u1u2. (10.9) This is proved by a straightforward computation. We take a solution of (10.3)
with c2 - s2 = 1, so that we may pose c = coshw and s = sinhw. Then all higher derivatives of w can be expressed in terms of algebraic functions of U: the first relations are precisely (10.5), (10.6) and (10.7). The next one consists in computing 7"37'. Using first (10.6) and (10.3) (a
=8
)3
=
8
i9 x 1
-i
) (8Zl3 - icil4) .9y
(ul - iu2)(1 + 2cu3 - i2su4).
And now, by (10.8) and (10.9), "13W
(az)
=
3
1 W2
(ul - iu2)(1 + 2al - i2a2 + (ul - iu2)2)
1 + 2a - i2a2 w 4
z
+2
I
wz I 3
where we have used (10.5) in the last line. We thus conclude that 4.1
2
03W
(0z)3
_
raw\ 3 2 1\
8z
11
_ J2(1 + 2al - i2a2) 8w =
8Z
,
.12(1 + 2a1 - i2a2)w,t7 which is e., with the notations of Chapter 9, W'.' 0. In the light of the analysis of Chapter 9, it just means that w is a solution of finite type. In particular the relevant deformation T of the surface (such that property (9.9) holds) can be chosen so that i.
equivalent to w,,
ta'3
=
-2wz + '\2Gl1 .
10. Wente tori
10.4
111
Spectral curves
Any finite type solution is characterized by the existence of a map 77a : IR2 Adg, which is a stationary solution of (8.4): dpi + [aa, pa] = 0. Assuming that ,p, and as are matrices we can make sense of the characteristic determinant P(A,,u) := det(tA - Al).
Then equation (8.4) just implies that this polynomial does not depend on z. By compaetifying the complex algebraic curve of C2 of equation P(A, µ) = 0, we obtain a Riemann surface called the spectral curve of the finite type solution. Lastly it is possible to use tools from algebraic geometry (the Abel-Jacobi integrals and the Jacobi torus of the spectral curve, see [32], [19]) to produce expressions of the conformal CMC immersions in terms of theta functions. This has been carried out by A. Bobenko for surfaces in R3, S3 and H3 [11], [12] (see also [13]).
In particular Wente tori correspond to genus 2 spectral curves invariant by a complex involution [13].
Weierstrass type representations
11
The theory for finite type solutions developped in Chapter 8 can be generalized in order to represent all harmonic maps from a simply connected surface to symmetric spaces like the sphere S2. This has been developped by J. Dorfineister, F. Pedit and H.Y. Wu and leads to a Weierstrass type representation [30]. ead_`17AZ
The main idea is to try to generalize property (8.5) FaBa = which was found for all finite type solutions (see paragraph 8.4), by substituting the right hand side of (8.5) by a suitable holomorphic function of z with values in a complexified loop group. One then needs to find a way to split this holomorphic function into a product of the form FaBa, i.e. we need a non linear analog of the Lie algebra splitting (8.1). Such a splitting does exist and can actually be interpreted as an extension to loop groups of the Iwasawa decomposition described in paragraph 8.1. We will present here briefly this theory. We will be concerned with harmonic
maps u from a simply connected domain 12 C R2 into a symmetric manifold
s'3/A. We assume that there exists an automorphism r : 0 -- 0 which is an involution (r2 = 11) such that, if Or := {g E 0 : r(g) = g}, (fr)o C . C a,.. 11.1
Loop groups decompositions
We use here the notations AO, AOc concerning loop groups and twisted loop groups introduced in paragraph 7.2. We also will need the loop group A+ OC defined in paragraph 8.4. We assume that . is a compact semi-simple group, so that the following Iwasawa decomposition holds: .q xB.R
.
gb
(g,b)
is a diffeomorphism. We summarize by AC = A.BR this property. Then the first tool is an extension of this decomposition to the loop group AOT, namely Lemma 11.1 Assume that 0 is a compact semisimple Lie group. Then the mapping
AO, x ABA OC - ABC, (ga, ba)
'-'
gaba
is a diffeomorphism. We denote by AOC = AO,.A
,q
OC this property.
11. Weierstrass type representations
113
We also introduce the subgroups
A+Oc := {[A - a] E AOT extending holomorphically in the disk D2},
A OC := {[A'- a] E AOC extending holomorphically in C U {oo}\D2 and We then have the following.
Lemma 11.2 Assume that 0 is a semisimple Lie group. Then there exists a dense open subset C of the connected component of the identity of AOC, called the big cell, such that the mapping
A* OC x A+Oc -+
C
is a diffeomorphism. We denote by C = At 0c7 A+Oc 7 this property.
These two results are showed in [30]. The proofs are based on similar decomposition results for non twisted loop groups proved in [70].
11.2
Solutions in terms of holomorphic data
Extended lifts of harmonic maps are in correspondance with holomorphic datas as defined below. We first denote A-l,oogC
l;k'\k
E
k=-1
is the Definition 11.1 The set of holomorphic potentials, denoted set of holomorphic 1-forms on fl with values in A_1,,,,,gc. So any form IA., in has the expression 00 JIa = E 00 ftkAk = k k(z)Akdz,
k=-1
k=-1
where Vz E f2, rk -1 lk(z).\k E A-1,oogC Lemma 11.3 Let F,, : ft -+ Ae7 be the extended lift of a harmonic map u : fZ -- O/11 and assume that ft is contractible. Then
there exist a holomorphic map H,,
:
f2 - AOC and a map B,,
:
fl
A++ OC such that F,, = H,, B,, .
the Maurer-Cartan form µa := (Ha)-1dH,, is a holomorphic potential.
11. Weierstrass type representations
114
Proof. (see [301 for details) The existence of Ha and Ba relies on solving the equation 0 = a(FAB
1)
= Fa
(a,\ (') - BA
) (BA)-1,
1
which is equivalent to
/
1
= Ba(ao + Aa1 +'\2 a2) 1 YJ
,
with the constraint that Ba takes values in A+ Oc.\The existence of a solution is first obtained locally, then we can glue local solutions into a global one. This proves the first assertion. Now we write (Ha)-1dHa = B,\ (a,\ - Ba 1dBa)Ba 1,
and using the fact that Ba takes values in A++ OC and that z '- HA(z) is holomorphic, we deduce that µa := Ha 1dHa has the desired properties. Conversely any holomorphic potential in 11_1,,o(Q) produces harmonic maps as follows. po Theorem 11.1 Assume that 0 is compact semi-simple. Let µa E a point in 1 and Ha a constant in AOC. Then there exists a unique holomorphic map Ha : S2 -y AOC, such that dHa = H,\µ,\ and HA(po) = H.
we can apply the loop groups decomposition AOC = AO,.A A95C to HA(z)
for all value of z. It follows that there exist two maps Fa : SZ - As, and Ba : 11 -- A+
OC such that HA(z) = FA(z)BA(z),
`dz E Q.
Then Fa is an extended lift of a harmonic map SZ - 01A. Proof. Since µa = t:adz, with Pok = 0, it follows easily that dµ,\ + µa A µa = 0, hence the existence and the uniqueness of Ha. Using now Lemma 11.1 in order to perform the decomposition Ha = FABA, we obtain (11.1) (FA)-1dFa = Baµa(BA)-1 - dB.(BA)-1. Now using the fact that µa E 7{_1,.(ft) and Ba takes value in A+ Oc, it is easy to check that the right hand side of (11.1) has the form Ek _1 akAk. But (11.1) implies also that this quantity should be real, i.e. a 1-form with coefficients in + &0 + &1a and moreover AC,. Hence as := F-'dFA reduces to as Lastly a Taylor expansion in A of (11.1) proves that &o is real and &1 &_1 is a (1,0)-form, which ensures the result by Chapter 7.
11. Weierstrass type representations
11.3
115
Meromorphic potentials
The holomorphic potentials constructed in Lemma 11.3 are far from being unique. Moreover they involve in general infinitely many holomorphic maps. These defects can be mended, provided we allow meromorphic potentials and under some hypotheses on 0. We define A-19C := {[a -0, ] E A9c/Ca =
-is-11.
Definition 11.2 The set of meromorphic potentials, denoted M-1 (11), is the set of meromorphic 1-forms on SZ with coefficients in A_1gC,. So any form µa in M_1(1l) has the expression
µa = Fl-la-1 =
-1a-1(z)dz,
where t;-1(z)a-1 E A_l.oogc,.
Then using the same methods as in [30]; one can prove the following
Theorem 11.2 Assume that 0 is semi-simple. Let F., : 11 - AO, be the extended lift of a harmonic map into 01.tiv. Then there exists a finite subset {a1, ... , ap} of fl such that there exists a holomorphic map Fa : fl \ {a1 i ... , aP} -- A,-,Oc and a map Fa : SZ \ {a1i ... , ap} - A+Oc such that
F,,(z) = Fa (z)Fa (z),
dz E fZ \ {ai, ... , ap}
z I.-# Fa (z) extends to a meromorphic map on fl
the Maurer-Cartan form µa := (Fa)-1dF, of F;- is a meromorphic potential in M_1(fZ). Proof. (see [30] for details) The decomposition F,,(z) = Fa (z)F, (z) is possible as soon as we can prove that F,(z) belongs to the big cell C. Using Lemma 11.3
in the same way as in [30], one can show that this is true for all z, excepted maybe on a finite subset {ai, .... ap} CO. The second property is proved also in [30]. The last one follows easily by writing
µa=Fa
[aa-(Fa)-1dFa}(Fa)-1
which implies on one hand that u,\ is in
once one keep
in mind the fact that F' (z) E But on the other hand F; -(z) E A* OC and thus there is no nonnegative power of A in the Fourier expansion of µ,,. A+Oc.
This implies the conclusion.
0
11. Weierstrass type representations
116
11.4
Generalizations
This theory can be extended to other two-dimensional differential geometric situations. An instance is the Willmore surface problem [881: it consists in looking at surfaces E immersed in R3 which are critical points of the functional
W[EJ := fE H2dA, where H is the mean curvature along E and dA is the area element induced by the first fundamental form of the immersion. These critical points are named Willmore surfaces and they satisfy the order four Euler-Lagrange equation AEH + 2H(H2 - K) = 0. This problem was actually considered at the beginning of the century by G. Thomsen (se [81J, [9]). Willmore surfaces may be thought as an analog of the minimal surface problem in conformal geometry and turns to be completely integrable. A Weierstrass representation theory in the spirit of the above theory was constructed in 1491 (see also [501).
Another instance is the study of Hamiltonian stationary Lagrangian surfaces in a four-dimensional Kidder manifold. They are Lagrangian surfaces which are critical points of the area functional under Hamiltonian flow deformations. The Lagrangian constraint and the restriction on the testing deformation vector fields
lead to an order three Euler-Lagrange equation. In cases where the ambient space is a symmetric Ki hler manifold, it can he shown that this is a completely integrable system and a Weierstrass type representation can also be built [511, [521, [531.
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Frederic Helein Constant Mean Curvature Surfaces, Harmonic Maps and Integrable Systems
This book intends to give an introduction to harmonic maps between a surface and a symmetric manifold and constant mean curvature surfaces as completely integrable systems. The presentation is accessible to undergraduate and graduate students in mathematics but will also be useful to researchers. It is among the first textbooks about integrable systems, their interplay with harmonic maps and the use of loop groups, and it presents the theory, for the first time, from the point of view of a differential geometer The most important results are exposed with complete proofs (except for the last two chapters, which require a minimal knowledge from the reader). Some proofs have been completely rewritten with the objective, in particular, to clarify the relation between finite mean curvature tori, Wente tore and the loop group approach - an aspect largely neglected in the literature. The book helps the reader to access the ideas of the theory and to acquire a unified perspective of the subject.
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