3 Berlin Heidelberg New York Barcelona Hong Kong London Milan Paris Tokyo
RE. Burstall
D. Ferus
K. Leschke R Pedit
Conformal
U. Pinkall
Geometry
of Surfaces in S4 and Quaternions
4
1011. Springer I.,
Authors Francis E. Burstall
Franz Pedit
Dept. of Mathematical Sciences University of Bath
Dept.
Claverton Down
University of Massachusetts
of Mathematics
and Statistics
Bath BA2 7AY, U.K.
1542, Lederle
E-mail. fie. burs tall@maths. bath. ac. uk
Amherst, MA 01003, U.S.A. E-mail: franz@gang. umass. edu
Dirk Ferus Katrin Leschke
Technical
Ulrich Pinkall
University of Berlin
Technical
University of Berlin
MA 8-3
MA 8-3
Strasse des 17.
Strasse des 17.
10623
10623
Juni 136 Berlin, Germany
E-mail. ferus@math. tu-berlin.de
Juni 136 Berlin, Germany
E-mail.
[email protected]
E-mail:
[email protected]
Cover
figure from D. Ferus, R Pedit:Sl-equivariant Minimal Tori in S'
in S3. Math. Z. 204,269-282
and
Sl-equivariant Willmore Tori
(199o)
CatalogIng-in-Publication Data applied for. Die Deutsche Bibliothek
-
CIP-Einheitsaufnahme
Conformal geometry of surfaces in S4 and quaternions / E E. Burstall Berlin; Heidelberg; New York; Barcelona; Hong Kong; London; Milan; Paris ; ....
Tokyo
Springer, 2002 (Lecture notes in mathematics; 1772) ISBN 3-540-43Oo8-3
Mathematics
Subject Classification (2000): 53C42,
53A30
ISSN 0075-8434 ISBN 3-540-43008-3
Springer-Verlag Berlin Heidelberg New York
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Preface
This is the first
comprehensive introduction to the authors' recent attempts understanding, of the global concepts behind spinor representations of surfaces in 3-space. The important new aspect is a quaternionicvalued function theory, whose "meromorphic functions" are conformal maps into ff- which extends the classical complex function theory on Riemann surfaces. The first results along these lines were presented at the ICM 98 in Berlin [10), and a detailed exposition will appear in [4]. Basic constructions of complex Riemann surface theory, such as holomoiphic line bundles, holomorphic curves in projective space, Kodaira embedding, and Riemann-Roch, carry over to the quaternionic setting. Additionally, an important new invariant of the quaternionic holomorphic theory is the Willmore energy. For quaternionic holomorphic curves in HP' this energy is the classical Willmore toward
a
better
energy of conformal surfaces. The present lecture note is based
given by Dirk Ferus at the September, 1999, [3). It centers on Willmore surfaces in the conformal 4-sphere HPI. The first three sections introduce linear algebra over the quaternions and the quaternionic projective line as a model for the conformal 4-sphere. Conformal surfaces f : M -+ HPI are identified with the pull-back of the tautological bundle. They are treated as quaternionic line subbundles of the trivial bundle M x H. A central object, explained in section 5, is the mean curvature sphere (or conformal Gauss map) of such a surface, which is a complex structure on M x IV. It leads to the definition of the Willmore energy, the critical points of which are called Willmore surfaces. In section 7 we identify the new notions of our quaternionic theory with notions in classical submanifold theory. The rest of the paper is devoted to applications: We classify super-conformal immersions as twistor projections from (Cp3 in the sense of Penrose, we construct Bdcklund transformations for Willmore surfaces in HPI, we set Up a duality between Willmore surfaces in S3 and certain minimal surfaces in hyperbolic 3-space, and we give a new proof of the classification of Willmore 2-spheres in the 4-sphere, see Ejiri [2], Musso [9] and Montiel [8]. Finally we explain a close similarity between the theory of constant mean curvature spheres in Summer School
on
Differential
on a course
Geometry
R3 and that of Willmore surfaces in
at Coimbra in
EEP1,
and
use
it to construct Darboux
transforms for the latter.
Bath/Berlin, August
2001
Francis
Burstall,
Dirk
Ferus, Katrin Leschke, Pedit, Ulrich Pinkall
Franz
'Table of Contents
I
Quaternions The Quaternions 1.1 The Group S3 .......................................... 1.2
3
Linear
5
..............................................
.......................................
2
2.1
2.2 3
4
.......................................
7
.........................................
9
..................
9
.........................................
11
Metrics
Moebius Transformations
3.4
Two-Spheres
HPI
Vector Bundles
4.3
7
5
3.3
on
HP .
........................
13
......................................
13
...........................................
15
in
S4
on
Quaternionic Vector Bundles Complex Quaternionic Bundles Holomorphic Quaternionic Bundles
The Mean Curvature Sphere 5.1
S-Theory
5.2
The Mean Curvature
5.3
Hopf Fields
5.4
The Conformal Gauss
.............................
15
...........................
18
.......................
20
.............................
23
...............................................
Sphere
23
.............................
24
............................................
27
Willmore Surfaces
Map
..............................
29
........................................
31
6.1
The
6.2
The Willmore Functional
Energy
1
.........
3.2
4.2
6
..............
Projective Spaces 3.1 Projective Spaces and Affine Coordinates
4.1
5
Algebra over the Quaternions Maps, Complex Quaternionic Vector Spaces Conformal Maps
Linear
1
Functional
..................................
................................
Metric and Affine Conformal
Geometry
31
35
..................
39
..............................
39
7.1
Surfaces in Euclidean
7.2
The Mean Curvature
7.3
The Willmore Condition in Affine Coordinates
Space Sphere
in Affine Coordinates
..........
42
.............
44
Table of Contents
VIII 8
Twistor 81 8.2
9
10
Project"ions
......................................
Twistor
Projections Super-Conformal Immersions
.....................................
47
.............................
50
Bhcklund Transforms of Willmore Surfaces 9.1
Bdcklund Transforms
9.2
Two-Step Bdcklund
Willmore Surfaces in S3
..........................
53 57
..................................
61 61
10.2
63
Hyperbolic
2-Planes
....................................
S3 and Minimal Surfaces in Hyperbolic
...............................................
Spherical Willmore Surfaces in HPI 11.1 Complex Line Bundles: Degree and Holomorphicity 11.2 Spherical Will1nore Surfaces Darboux tranforms 12.1 Riccati
equations
12.2 Constant
mean
Appendix 13.1 The bundle L 13.2 Holomorphicity ......
67
.........
67
.............................
71
.......................................
73
.......................................
curvature surfaces in
R3
...................
..................
73 74
79
...........................................
83
..........................................
83
and the
Ejiri/Montiel
theorem
.............
84
.................
87
..........................................................
89
References Index
64
......................
12.3 Darboux transforms of Willmore surfaces 13
53
10.1 Surfaces in S ...........................................
4-Space
12
...............
...................................
Transforms
10.3 Willmore Surfaces in
11
47
...................................
Quaternions
I
Quaternions
The
1.1
symbols i, j,
H
quaternions
The Hamiltonian
k with
i2 ii
non-zero
-ii
--
multiplication
The
=
ik
k,
has
division
a
=
multiplicative algebra over
k2
-kj
=
ao +
=
generated
by
the
_1' ki
i)
ali
a2i
showed in 1877 that
Frobenius
the reals.
+
and each commutative, and a a skew-field,
not
R-algebras
finite-dimensional For the
-ik
=
We have
inverse:
a
=
=
obviously
but
R, C and H are in fact the only and have no zero-divisors. ciative
we
j2
=
associative
is
element
4-dimensional
R-algebra
the unitary
are
the relations
that
are
asso-
element
a3k,
+
al
(1.1)
R,
C-
define a:=
ao
ali
-
a2i
-
-
a3k,
Rea:= ao, Ima:= Note
that,
and that
with
in contrast
the
ali
+
a2i
+
a3k.
complex numbers,
Im
a
is not
Wb_
=
the real vector space H in the obvious identify with R3: of purely imaginary quaternions subspace
W
=
The reals
are
identified
canonical.
imaginary
unit,
number,
with
RI.
embedding equally i,-j,k
The
The
and in
fact
F. E. Burstall et al.: LNM 1772, pp. 1 - 4, 2002 © Springer-Verlag Berlin Heidelberg 2002
way with
R,
and
IMH.
quaternions imaginary any.purely would do the job. Rom now on, however, we shall C C ffff generated by 1, i. C is less
real
b a.
Weshall the
a
obeys
conjugation
complex numbers for the complex qualify of square -1 quaternion usually use the subfield
of the
Quaternions
1
2
Occasionally be written
we
shall
Euclidean
need the
product
inner
on
R4 which
can
as
< a, b >R=
Re(ab)
=
Re(ab)
2
(ab
+
ba).
Wedefine a
>R
a
a,
=
vfa-
d.
Then
jabj A closer
study of
the
=
jal Ibl.
(1.2)
multiplication
quaternionic
displays
nice
geometric
as-
pects. Wefirst usual
finds
one
mention
that
and scalar
vector
for
the quaternion
products
a, b E Im Eff
=
on
multiplication using
a
consequence
R-
we
ba
=
In
(1-3)
have
if and only if Im a and Im b the reals are the only particular,
ab
(1.1)
state
For a, b G H
Lemma1. 1.
we
both the
the representation
R'
ab=axbAs
incorporates
V. In fact,
are
linearly
dependent that
quaternions
the reals.
over
commute with
all
others. 2.
a' a
-1
=
if and only if Jal two-sphere
=
1 and
S2
Proof.
=
Im
a.
Note that
the set
of
all
such
Write
a
=
a',
b
=
aobo
+
=
aobo
+
ao +
C
V
=
+
Y,
aob'
+
a'bo
+
aob'
+
a'bo
+ a'
for
the
=
bo
IMH.
where the prime
denotes
the
imaginary
Then
part.
ab
All
a
usual
the
is
these
follows. vanishes obtain
products, Romthe if
(2).
and
except same
only
if
a
formula is real
x
Y-
<
cross-product,
with or
a'b'
a
purely
=
b
we
a', Y
are
-
and (1) commutative, Im a 2 2aoa'. This Together with (1.2) we
obtain
imaginary.
>R
=
The
1.2
quaternions
The set of unit
S3 in H =
3-sphere
the
i.e.
it
interpret
RI,
Ip
:=
forms
the group of linear
as
E
HI IM12
maps
51b.
< a, b >:=
group is called We now consider
By (1.2)
this
action
product. complement R'
It
scalar
=
the
action
the
obviously
'53
norm on
a
of
compute the differential
H=
map, in fact
SO(3),
-+
y -+
y
R' and, hence, the Euclidean its orthogonal and, therefore, a representation,
P-1 II.H-
...
For p E
7r.
by
pap-1.
1-4
R C H
stabilizes
Im H. We get :
(/-t, a)
H,
H __
X
preserves
,X
us
group Sp(l). of S3 on H given
symplectic
the
S3
Let
We can also the hermitian
px of H preserving
-+
x
multiplication.
under
group
a
1}
=
product
inner
This
3
S'
The Group
1.2
Group S3
S3 and
E
v
Tt'S3
we
get
dl -,ir(v)(a)
=
-
Now y-1v
vaIL-1
commutes with
pap-1vp-1
-
all
a
E ImH
=
if
p(p-'va only if
and
ap-'v)/-t-1.
-
v
=
rp
for
some
real
of S3 diffeomorphism 0, p. manifold orthogonal preserving onto the 3-dimensional SO(3) of orientation this is of R3. Since S3 is compact and SO(3) is connected, transformations all a E ImH if and only if p E R, i.e. if And since pap-I a for a covering. 2:1. It is obvious that antipodal points of and only if M 1, this covering'is and therefore we S3 are mapped onto the same orthogonal transformation, r.
But then
v
=
because
Hence
I
v
ir
is
a
local
=
=
see
that
Rp3
SO(3)- S31111 Wehave of
SO(3).
now
displayed
This
group
the group of unit the is also called
S3 If
we
identify
H = C G)
=
C?,
SP(j)
quaternions spin group: =
we can
S3
-
as
the universal
covering
Spin(3). add yet
SU(2).
another
isomorphism:
4
In
j (a has
I
fact,
Quaternions let
p
=
(a iP) the following +
=
Ito -
+
1-iij
i,8)j.
C-
Therefore
matrix
po, pi
the
representation
AA1 A,.j Because of yopo + /-tlpj.
S' with
yo +
=
+
=-/-tl =
1,
we
(
jLjj
E C.
C-linear with
map
poj =1(-pi)
have
Po tL1
f4l)
Po
E
SU(2).
+
+
AIL
:
a,
C2
to the
respect
Imo
=
Then for
ifil
jpo.
-
C- R we
-4
basis
C2
,
x
have
1-4
fix
1, j of 0:
Algebra
2 Linear
Maps, Complex Quaternionic
Linear
2.1 Since
we
consider for
options
two
spaces to from the
Quaternions
the
over
be
vector
right
spaces V
vector
the skew-field
over
multiplication
the
i.e.
Spaces
quaternions,
there
We choose quaternion
by scalars.
spaces,
of
Vector
vectors
are
multiplied
are
vector
by quaternions
right: V
x
H -+
V, (v, A)
F-+
vA.
dimension, subspace, and linear map work as in the usual algebra. The same is true for the matrix representation of linear maps in finite dimensions. However, there is no reasonable definition The linear for the elementary symmetric functions like trace and determinant: (i) when using 1 as basis for H, but map A : H -+ IH x -+ ix, has matrix matrix (-i) when using the basis j. If A E End(V) is an endomorphism, v E V, and \ E H such that The notions
commutative
of basis, linear
Av then
for
any p E
H\ f 01 A(vp)
If
A is real
then
we
=
vA,
find
=
(Av)p
=
vAp
=
(vp)(p-lAp).
the
eigenspace is a quaternionic subspace. Otherwise it is vector subspace, and we obtain a whole quaternionic 2-sphere of "associated eigenvalues" (see Section 1.2). This is related to the fact that multiplication from the right) is not by a quaternion (necessarily of the In of V. between Iff-linear an H-linear endomorphism fact, space maps vector vector quaternionic spaces is not a quaternionic space itse f. vector Any quaternionic space V is of course a complex vector space, but this structure on depends choosing an imaginary unit, as mentioned in instead 1.1. We shall section have an additional Complex (quite regularly) from the left, and hence commuting with the quaterstructure on V, acting In other words, we consider nionic J E End(V) such that structure. a fixed a
J'
real
=
-
-1.
but
not
a
-
Then
(X
+
iy)v
:=
F. E. Burstall et al.: LNM 1772, pp. 5 - 8, 2002 © Springer-Verlag Berlin Heidelberg 2002
VX
+
(Jv)y.
Algebra
2 Linear
6
In this
case
(W, J)
and
split
as
a
we
(V, J)
call
(AJ
-JA)
=
Hom(V, W)
Hom+(V, W)
=
Hom(V, W)
and
If.
space.
(V, J)
maps from V to W complex linear (AJ = JA)
linear
homomorphisms.
Hom(V, W) fact, plication
(bi-)vector
complex quaternionic
a
such spaces, then the quaternionic direct vector sum of the real spaces of are
and anti-linear
In
Quaternions
the
over
ED Hom- (V,
complex
are
W)
vector
space with
multi-
given by
(x
+
iy)
Av
:
(Av)
=
x
+
(JAv)
y.
vector example of a quaternionic space is H" An example vector complex quaternionic, space is HI with J(a, b) := (-b, a). is simply left-multiplication On V H, any complex structure by some -1. lemma describes N E IHI with N2 The following that a situation for produces such an N, and that will become a standard situation naturally us. But, before stating that lemma, let us make a simple observation:
The standard
of
.
a
=
=
Remark 1. J E
for
End(U)
any
x
.0
On
a
2-dimensional
real
induces
vector
0. Wethen
compatible
J
call
space
the
following,
three
2
=
-
I
NU= : U U The pair 2.
(N, R)
is
=
and this
is
a
fx
E
HI
Definition normal
1.
as
real
unit
1V'
=
subspaces
U in the geometric
oriented
N, R (-= H
exist
=
2
(2.1)
R
(2.2)
UR, =
x}.
(2-3) there
is
only
one
normal -1
=
R,
vector
Ul of
of U
in Im H =
R3.
R2, the sets
:=IxEHINxR=-x}
dimension
2.
by (2) of the lemma, N and R are called a left of U, though in general they are not at all orthogonal
Motivated
vector
=
If U is oriented, up to sign. compatible with the orientation. above, and U C Im H, then
U:=jxEHjNxR=x},
right
structure
N is
Euclidean
orthogonal
Then there
unique
Given N, R E H with
are
2.
NxR
N=
3.
complex positively
properties: N
such pair such that If U, N and R are
is
0.
with
Lemma2 (Fundamental lemma). 1. Let U C H be a real subspace of dimension with
U each
(x, Jx)
0 such that
orientation
an
sense.
and to
Conformal
2.2
(of
the
to 1, then
a2
Proof
x
U\10}
E
(1).
I E U and if
If
(N, R)
1. Hence
-
easily from put U := x-'U.
follows
sign,
up to
lemma). =
then
1
NI
E U is
a
(a, -a)
=
a
works for
unit
U,
Clearly,
7
orthogonal
vector
and the
E U and Na E U. If
uniqueness, and arbitrary,
U is
U. Moreover,
1 E
Maps
(N, R)
0.
works for
Nx, R) basis of U, then R, and u, v is an orthonormal Use the geometric the requirements: N R u x v uv satifies properties of the cross product. of (3). The above argument shows that u(x) := NxR has I-eigenspaces 2. Since a is orthogonal, real dimension so are its eigenspaces.
only
U if and
(2).
If
(x
if
Example
=
=
Let
1.
mension 1. Then v
works for
U C Im H
=
=
-
and w, and
(V, J), (W, J) be complex quaternionic Hom+(V, W) is of real dimension 2. To
Then N2= -1 =
=7R'.
JF
FJv
-#=
But the set of all same
complex
such
result
vector
A linear
=
:
all
x, y
orthogonal
E V.
basis
"normalized"
Jw
=
Hom(V, W)
this,
of di-
choose bases
wN.
given by F(v)
is
=
wa, and
JFv
J(wa)
(Jw)a
=
dimension
Hom- (V,
and therefore
=
wNa 4=*
2, by the last
aR
part
=
Na.
of the lemma.
Hom(V, W) are W). As stated earlier, isomorphic with C. (non-canonically)
Maps
V
-+
Wbetween Euclidean
vector
spaces is called
conformal
A such that
positive
Fx,Fy
<
for
vR,
is of real
for
spaces,
map F exists a
if there
a
holds
Conformal
2.2
=
Now F E
4==> waR =
The
spaces
assume
Jv
FJ
vector see
This
equivalent
is
of V into
a
>=
A < x,y the
to
normalized
fact
>
that
orthogonal
F maps a normalized of F(V) C W. Here
basis
all vectors have the same length, possibly 54 1. C, and J : C -+ C denotes multiplication by the For x E C, IxI 4- 0, the vectors imaginary unit, then J is orthogonal. (x, Jx) form a normalized orthogonal basis. The map F is conformal if and only if On the other hand (Fx, JFx) is (Fx, FJx) is again normalized orthogonal. normalized if and only if Hence F is conformal, orthogonal. If
V
=
means
that
W= R'
=
FJ where the
sign depends
on
=
JF,
the orientation
behaviour
of F.
Algebra
2 Linear
involves
complex
the
for
is fundamental
If F
R'
:
condition
this
Note that
C
=
subspace
2-dimensional
of this
and injective,
H is R-linear
=
product,
scalar
the
generalization presented here. J. A
theory
R4
-+
involve
does not
structure
the
Quaternions
the
over
fact
to
U
then
but only quaternions
=
F(W)
is
a
H, oriented
of
by J. Let N, R E H be its left and Then NU U UR, and N induces an orthogonal right normal vectors. endomorphism of U compatible with the Euclidean scalar product of V. The real
=
map F
R'
:
U is conformal
-+
if and
conformal
exist
FJ
*F:=
This
leads
*df
df
:=
o
surface, TM-+
a
a
exist
N2= -1
*df
If f
is
an
Remark 2.
-
right
and
=
(2.4)
then
immersion
the left
called
functions
sense
f
(2.5)
C
:
conformal
-+
RI,
=
is
an
C,
maps into
i.e.
=
i.e.
a
:
C
-+
H is
such that
analog =
2-dimensional
-1. TM, j2 N, R : M -+ =
manifold
A map f : M H such that with
2
(2.4)
-dfR.
(2.5)
from (2.5), Of f.
vector
*df for
-I
R
=
Ndf
follows
normal
Equation
NF. Hence F
=
fundamental Riemann
Let M be
2'.
endowed with
R4 is
following
=
NF = -FR.
=
J: complex structure called conformal, if there J,
Definition H
the
to
only if FJ N, R E H, NI
if and
if there
only
=
and N and R
are
unique,
of
idf
In this Cauchy-Riemann equations. of generalization complex holomorphic
of the
H are
a
maps. -
-
real subspace. f is an immersion, then df (TpM) C H is a 2-dimensional there Lemma exist to a N, Hence, according 2, R, inducing complex strucThe definition that coincides with J ture J on TM1--- df (TpM). requires the complex structure already given on TpM. For an immersion f the existence of N : M -4 H such that *df Ndf for already implies that the immersion f : M-+ H is conformal. Similarly If
=
R. -
If
f
:
M-+ Im H
normal not
vector
orthogonal
of to
RI is
an
But for
df (TM).
then
immersion
general
f
:
M --+
N = R is
H,
the
"the
vectors
classical"
unit
N and R
are
3
Projective
Spaces
theory the Riemann sphere CP1 is more convenient as for than the complex plane. holomorphic functions target space Similarly, the natural for conformal immersions is HP1, rather than H. We target space therefore of the quaternionic give a description projective space. In
complex function
a
Projective
3.1
Spaces and Affine
The
quaternionic complex cousins,
tinuous)
projective as
canonical 7r
The manifold
defined,
is
the set of quaternionic
lines
in
7r(X)
=
similar
to
its
real
and
H1+1. We have the (con-
projection :
Ep+1 \f 01
__+
ff-lpn,
X
of Rpn is defined
structure
For any linear
HP'
space
Coordinates
form
P
(Hn+')
E
U:
+
as
[X]
=
XH.
follows:
P 54 0,
7r(x)
-+
<
X
P,
X
>_1
well-defined and maps the open set 17r (x) I < P, x > 54 0} onto the affine to Hn. Coordinates of this type are hyperplane P 1, which is isomorphic called affine coordinates for Hpn. They define a (real-analytic) atlas for ffffpn. often use this in the following We shall We choose for a basis setting: H1+' such that # is the last coordinate function. Then we get is
=
XJX-1 n+
X1
Xnxn+l
Xn
(Xnx*n+l) XlXn+1
or
I
i
Lxn+lj The set
17r(x) is called
the
hyperplane
at
I
<
0,
infinity.
F. E. Burstall et al.: LNM 1772, pp. 9 - 14, 2002 © Springer-Verlag Berlin Heidelberg 2002
X
>=
01
3
10
Spaces
Projective
Example 2. In the special point: HP' is the one-point
case
n
1, the hyperplane
=
compactification
R4,
of
at
infinity
hence "the"
is a single 4-sphere:
Elp, that
however, 4-sphere, but Note
not
the on
unless
-
antipodal
of the
notion
HP'
we
map is
natural
additional
introduce
on
usual
the
like
structure,
a
metric. For space
our
it
purposes
Tilapn
for
For that
coordinates: h
=
#
If u o 7r
IT+'
:
\101
(Hn+')*
E
good description
a
purpose,
ffp+l
7:
in affine
to have
important
is
1 E ffffpn.
is
we
above,
ffP+l,
\10}
the
of the tangent projection
Hpn
_-,
as
consider
x
-+
then x
<
0,
x
satisfies
d h(v)=v<#,x>-'-x-'-
.
Therefore ker dx h
dx,xh(vA) for
H\10},
A E
and the
same
holds
for
ker d,,
=
=
dx7r induces
an
dx7r
7r
of real
vector
dependence,
but it,
spaces, we
note
that
E
x
1\101
is
a
'-
In other
this
words,
Hn+'/l
as
follows:
we
that
identify x
1
_+
T1 Rffpn,
dx to 7r,
F
-+
of
x
this
E 1. To eliminate
dx -7r (F (X)),
isomorphism:,
well-defined
maps
(3.2)
Tllffpn,
.
Hom(l, RP+l /1) xH to
(3-1)
depends on the choice by (3.2) the map'
Hom(l, ffP+1 /1) with
xH,
dx7r(v).
isomorphism
EP+1 /1
:
dxh(v)
7r:
dx,\7r(vA) By (3.1),
xH,
ir
(v) (v)
with :=
v
c_
the
(3.3)
T Hpn.
homomorphism practical
mod 1. For
from 1 use,
=
we
7r
(x)
=
rephrase
Metrics
3.2
Proposition M,
Let p E
f (p),
dpf is
1:
Let
.1.
1:=
M-4 H'+1
E
v,
TpM.
differential
the
We denote
this
in
16f (v) The tangent
is identified
the
mod
f (p).
dpl(v)
3.2
Metrics
Given
a
Hn+l,
we
on
Hom(f (p), ffV+1 If (p))
follows:
For
a
(possibly
E
x
<
(dp f (v) A).
7rl
by Jf:
interpretation
(f)
df (v)
=
(3.4)
di(p)
=
is well-defined
7r
< x,
(v), dx 7r (w)
since,
for
0
54
f (p)
:
(p)
arbitrary
v,
>
=
w
> 54 0
3.
IRpn.
For < v, (In the
Riemannian
and v,
H,
we
maps
j(p)
w
>=
1:
The corresponding conformal considerations. following
w
E
<
(x4
we
>
.,.
metric
on
on
Hpn
define
>=<
>
have
d ,7r(v),
WX <
x,
dx7r(w)
W>
>
< V'W >< X'X
>
< x,
>
x
>
>
< X'X
complex
product
inner
Pseudo-)
-
< X'X
on
(p), that,
by
Re
Example
Hpn Hn If
--+
hermitian
A E
to
dx 7r (v), dx 7r (w)
Tf
1
dxX7r(wA)
<
E
Re < v,w > =< X'x >
d ,.Xir(vA),
extends
F
x
<
It
(dp 1(v))
quaternionic degenerate
Hn+1 with
d,
ir
Hpn
non-degenerate define
=
homomorphism
with
into
ric
M_+ Elpn.
vector
dp f (v)
This
=
(f(p) A)
dp f (v)
as
7rl:
=
11
Then
TpM ---> Tf (p)Hpn
:
f
and
HP'
by
given
Proof.
\10}
on
< V,X ><
>2
IX'W
>
(3-5)
.
ITkWk we obtain the standard Riemannian metthis is the so-called Fubini-Study metric.)
case,
structure
is
in
the
background
of all
of the
12
induces
Riemannian
standard
Wetake this it
Spaces
Projective
3
R4 via the affine
on
h
:
H2,
H -+
DIP'
t-+
1XI
S' and ask which metric
=
parameter
HP',x
and let
"=-"
(v)
(v0) (x)
:
(x, 1),
-+
x
on
H -+
h
Let
metric
_
equality
denote
(X)
mod
ff.
1
Then
6xh(v)((x)) The latter metric
d,,h(v)
1
is
vector
on
0
<
1
>-orthogonal
.,.
VV)
'7-
xv
-
1 + xx -
(x, 1),
to
-
-,t
I I + xx-,
the
and therefore
induced
given by
H is
1
h*
>x
< v,w
stereographic
But
=_
=
(1
+
X.:t)3
(I
+
X;,-)
V
Re <
Re(;vw)
2
(1
of S4 induces
projection
V)
)
W
-.t
< V,W >R
Xj )2
+
>
W
the metric
4 < V5
(I+ Xj )2 on
R4. Hence the standard
Example 4. If
we
consider of
above construction but
these
points
metric
a
an
metric
< V,
lines
Isotropic
by
x
The
points -
is
E ImH
point
at
=
for
the
metric
on
lines
isotropic
case
n
=
4.
curvature
Hn+', (< 1,
then
the
1 >=
0),
1, and the hermitian
in
the
induced factor
metric -
affine
(X) (X) I
H is
0
S3
coordinates
h* < v,w
constant
(1)
infinity
3-sphere
a
in affine
ITIW2 + IT2W1
I
,
h
coordinates
x
:
-+
(xl)
by
,
+ X,
>= j
R3.
As in the previous
for
Hpn fails
W >=
characterized
are
0 =<
or
hermitian
indefinite on
constant
product
inner
i.e.
HP' is of
We consider
scarce.
are
on
W >R
C
S',
and its
two open
-
example,
(2 on
isotropic,
we
Re X)2
the
the standard
Therefore,
too.
complement half-spaces.
the set of
consists
isotropic
of two open discs
find Re ( v-
half-spaces hyperbolic
2 Re metric
:A
ReX)2
< ViW >R
0 of H. on
these
This
is
-
half-spaces.
up to
a
Two-Spheres
3.4
Moebius
3.3
Transformations
The group Gl(2, i.e. the set of all
4
acts
How is this
action
hermitian
metric
nite
IdG(d,,-x(vA))I'
=
=
compatible with the metric of V? Using (3-5) we find
G(O), G(vA)
"
Re
=
I
IA12
Re
Gv, Gv
<
we see
length-preserving
But the
same
an
arbitrary
call
these
that
for
up to
obviously
is
G,
a
><
-+
factor,
defi-
Gx > <
Gx, G(vA)
><
Gx, Gv
>
Gx >2
Gx,
i.e.
the metric
for
is
a
conformal
induced
=
this
Two-Spheres
with
the
S4,
see
+
d)
I
+
is the full
of
b) (cx
+
=
cX
diffeomorphisms
complex
group of all
case.
orientation
preserving
confor-
[7].
S4
in
the set Z
-
Rj.
positive
given by
are
is known that
For S E Z
G(vA),
action,
p E
isomorphism. under by the pull-back 1 S4. We on RP GL(2, R acts conformally Moebius transformations In affine on Rp'.
constant
the
emphasises the analogy
We consider
a
I
d,,7r(vA)
+
3.4
<
-
A
=
mal
-
by
of this
jjol
0 the map
T,( ,)EIP',
and therefore
they
54
v
true
c
It
>
is
Gx,Gx >2 < Gv, Gx Gx, Gx >
(a db) [x] [ax db11 [(ax This
v,
induced
Gx, Gx
<
transformations
coordinates
><
all
12
1,\12 IdG(d,,7r(v))
H -+ is
GvH. The kernel
:=
Gv E vH for
<
=
Taking G
such that
13
HP1
on
RP' by G(vH)
on
G1(2, P
GE
S4
in
we
=
IS
E
End(EO) I S2
define
S'
:=
fl
E
RP1 I Sl
=
11.
,
We want to show 2. 1. S' is a 2-sphere in KPI, i.e corresponds Proposition to a plane in H R4 under a suitable affine coordinate. 2. Each 2-sphere in this way by an S E Z, unique can be obtained
real
2-
=
sign.
-
up to
>
Spaces
Projective
14
3
Proof
Weconsider
i.
H'
Then S is Clinear
as a
(right)
and has
complex
(complex)
a
S(vH) We choose
a
basis
and Sw = -vH
v,
w
affine
=
-1
E
=
h
parametrization
[vx +,w]
S'
:
R2,
IMP',
a
real-linear
equation
vNx
w)
+
Nx
this
thus the
Lemma2 this
is of real
-
-R
vN for
some
-
(vx
+
we
get:
w)-y
+
wR
w]
=
vx-y
+ w7
H = x-y -y
H.
for x, with
dimension
=
vH
-
Nx + xR
By
Sv
i.e.
[vx
x
Nx + xR is
S',
NH= HR.
3 S(vx
37
vH E
implies
-1
=
H -+
3.y
This
=
vN H = vH.
of EV such that
wR. Then S'
-
N2 For the
eigenvalue
imaginary unit vN, then
Sv
N. If
0.
Hence S'
N,
=
space with
vector
associated
=
homogeneuos equation
0.
2, and
any real
2-plane
can
be realized
way.
S and -S define Obviously, of the orientation an fixing lemma can be paraphrased Z is the
set
the
same
2-sphere. But S determines (N, R), 2-plane and thereby of S'. Hence
above real as
follows:
of oriented
2-spheres
in
S4
=
fflpl.
Bundles
4 Vector
We shall troduce
need vector
bundles
Quaternionic
4.1
the quaternions,
over
action
of H
vector
spaces.
Example the first
V from
on
The
5.
factor called
product
bundle
and the obvious the trivial
such
right
the
7r
that
a
-+
on
bundle
with _p
Zi
=
:=
1 (1, V)
:
Z
1. More
and vector
f*V:=
is
just
the obvious
the fibre
projection of V
f x}
on
x
Hn
E lffpn
X
Ep+1 I
_+
space structure
over
I(x,v)
IV
and vector
are
the
1-dimensional
precisely
E
Vf(,,)}
bundle
f (x).
F. E. Burstall et al.: LNM 1772, pp. 15 - 22, 2002 © Springer-Verlag Berlin Heidelberg 2002
V
C
1},
1. are
vector Example 7. If V -+ Mis a quaternionic 1 1 is a map, then the "pull-back" f *V 4Mis
with
projection
Hpn
___,
Rpn, (1, V)
7rz
The differentiable
the
each fibre
bundle.
7r_,
line
fibre-preserving
become quaternionic
M with
space structure
vector
smooth manifold
a
smooth
fibres
the
M x Hn
:
n over
Example 6. The points of the projective space Elpn bundle subspaces of Hn+'. The tautological
is the
in-
Bundles
Vector
A quaternionic bundle 7r : V -+ Mof rank vector with Mis a real vector bundle of rank 4n together
is also
briefly
and therefore
them.
the
bundle defined
obvious over
ones.
M, and f
:
M
by
CMX V structure.
The fibre
over
x
E M
Bundles
4 Vector
16
maps f the associate
be concerned
We shall
projective space. x is f (x) c ffP+l product bundle
f
To
with
we
JxJ
=
M -4, RP'
:
bundle
L
H1+1. The bundle
x
L is
an
Maps
All tions
natural
bundle
vector
H induces
quaternionic
two
all
x
the
Over lffpn
Example
8.
inside
the
vector
spaces
bundles.
For
A section It
is
maps
of isomorphism
notion
it,
V2 ).
4ilv,.
restriction
obvious
!P E
fibre-wise, a
to
subbundle
F(Hom(Vi, V2)) V,
:
V1,; homomorphically
for
vector
M
operaL of a
--
into
is called
a vec-
V2 such that for V2x. There is an
bundles.
product
bundle
H
=
HP'
x
H1+1 and,
Z. Then
subbundle
Hom(Z, HIZ),
Let L be a (and Definition). -+ E HIL -P(Hom(H, HIL)) 7rL .P(L) C F(H) is a particular map dO(X) E Hp Hn+', and 9
:
subbundle
line
H
be the M
-+
of H
projection. Hn+1
.
If
=
(do (X))
WL
E
(HIL)p
=
X E
EP+1 ILp.
A: M-+ R Then 7rL
(d(0A) (X))
7rL
=
(dO(X)I\
+
Odl\(X))
:--:
Wesee that
0 is tensorial
in
0,
i.e.
we
J(X)
-+
7rL
(4 (X))
:;--:
6 (X)
(0)
obtain :--
JL (X)
G
Hom(Lp, (HIL)p).
7rL
Mx Hn+'.
A section
=
Let
:
(3.3).
Example
Let
f
Mx Hn+1
=
extend, example,
smooth map !P
have the
we
tautological
a
THP' see
map
a
bundle HIL with fibres Given quotient H,;IL_-. V1, V2 the real vector bundle Hom(V1, V2)
a
homomorphism.
bundle
over
of the
subbundle
bundles
vector
Hom(Vi.,
has the fibres tor
for
of vector
category
the
subbundles
Line L c H
constructions
the
in
line
into
identification
M_+ ffffpn
:
f Z,
a
surface
whose fibre
L of H over Mdetermines
HP'
f
a
*
EP+l.
H:= Mx
Conversely, every line subbundle by f (x) := L,,. We obtain
:
from
=
(dO(X))I\-
TpM,
0
c
then
Quaternionic
4.1
Of
is R-linear
this
course
values
Mwith
in X
us
0
F(L)
E
Given
repeat:
p E
0(p)
such that
=
I Jp(X),Oo similarity
Note the
=
E
submanifold
of
a
to
TM and H1+1
M, X E TpM, and 00 Oo. Then
7rL(dv0(X))
clearly comparison
Euclidan
the
change
=
L
with
of
there
is
a
section
m
form
.
In the
case
at
hand,
bundle.
normal
subbundle
a
as
a
map
f
M -+ HP'.
:
L in
a
corresponds the general
L
This
is
connected)
(covariantly this
Even if
the second fundamental
to do with
is
an
form of
immersion, f. Instead,
I shows that
Proposition
Hom(L, HIL)
6: TM-+
corresponds of L
on
1
(dY (X))
the
's to
Lp,
E
dpV)(X)
=
space.
A correspond
nothing
has
1-form
H.
We can view 6
as a
(4.1)
the second fundamental
to
Min
measure
bundle
vector
be viewed
17
f2l (Hom(L, HI
a(X, Y)
method to
6 should
so
Bundles
Hom(L, HIL):
in
IJ Let
well,
as
Vector
to the derivative
of
f
and
,
we
shall
therefore
call
it the
derivative
.
Example
10.
The dual
space V is,
vector
quaternionic structure:
in
vector
For
w
E
a
V`
:=
natural
spaces
fw
-4
left be right
way, to
V
:
vector
extends
rank
1, then L*
to
space. spaces,
of
quaternionic
a
But since
the
we use
we
choose
opposite
V* and A E H we define
w.
wA := This
Hjw H-linearl
H-vector
a
quaternionic
vector
is another
quaterionic
A quaternionic vector product bundle Mx H,
bundle
bundles. line
is called
E.g., if L is a line bundle, i.e. of bundle, usually denoted by'L-1. trivial
if it
with the isomorphic M-+ V : global 0,, that form a basis of the fibre line everywhere. Note that for a quaterniQnic bundle over a surface the total space V has real dimension 2+4 6, and hence 0 : M -+ V has codimension 4. It follows from transversality any section deformed so that it will not hit the theory that any section can be slightly Therefore there exists nowhere vanishing 0-section. section: a global Any line bundle over a Riemann surface is (topologically) trivial. quaternionic i.e.
if there
exist
sections
is
,
=
4 Vector
18
Bundles
Complex Quaternionic
4.2
complex quaternionic
A
nionic
bundle
vector
bundle
vector
V and
Bundles is
a
J E
section
a
pair (V, J) consisting with 1'(End(V))
of
a
quater-
j2 section
see
2.1.
Example L
=
Given
11.
Mx H has
a
Example 12. For
f : M -+ H, *df Ndf the quaternionic complex structure given by =
a
given S
E
S'
Jv:=
Nv.
End(H2)
with
=
111
Sl
=
line
,
1}
S2 C
=
-1,
we
bundle
identified
HP,
in RP1, see. section 3.4. Wenow compute J, or rather the image line bundle L. In other words, we compute the corresponding tangent space of S' C HP'. Note that, because of SL C L, S induces a complex structure on L, and it also induces one (again denoted by S) on HIL such that irLS S7rL. Now for V) E r(L), we have as a
of
6,
2-sphere for
the
=
6SO This
=
7rLd(SO)
real
But the
S' is
vector
an
For
our
Lemma3.
image 6
a
example
=
S60.
U is Note:
-R rather
C
is
an
quaternionic
we
real
End(V),
then
2,
see
Example 1,
and
equality:
Hom(LI, (HIL)j)
V, W'be 1-dimensional
oriented,
the
has rank
Lemma2.
2-dimensional J E
Hom+(L, HIL).
generalize
next
U
with
S7rLdo
the inclusion
=
vector
TjHP1.
spaces,
and
Hom(V, W)
subspace. Then there exists a pair of complex End(W), unique up to sign, such that
vector
j
E
ju
If
=
Hom+(L, HIL)
surface,
UC
structures
C
Hom+(LI, (HIL)I)
=
Let
=
bundle
embedded
TIS'
be
7rLSdV)
shows
TS'
since
=
=
fF
there
E
is
=
U=
Ui,
Hom(V, W) I jFJ only
one
=
such pair
-Fj such that
J is
compatible
orientation.
Here
than
we
R.
choose
the sign
of
J in
such
a
way that
it"corresponds
to
Complex Quaternionic
4.2
Proof.
Choose
Hom(V, W) 1
The
EIPI
basis vectors v E V, w E W. Then elements endomorphisms of V or of Ware represented by quaternionic
and therefore
following
Proposition with
is
now
HIL,
M x W be
=
Definition
with
A line
3.
the
J
on
=
HPI, is
a
if
i.e.
real
subbundle
1'(End(H/L))
f
Let
13.
normal
E.F(L), J(
:
we
define
=
by J: T ,M
M x Hn-- '
over
in
curve
-+
a
6(TxM).
Riemann
Rpn, if there
sur-
exists
a
6j.
L is
2,
immersed
an
such there
then
also
is
holomorphic
J(TM)
that
a
C
complex
curve
in
Hom(L, HIL) structure
J E
(f) I
J E
(4.2)
jj
=
if and only into HP1 is a holomorphic curve with the are compatible given by. the proposition
on
M-
vector
Min the H be
R, and
a
let
sense
of
(4.2).
conformally L be the
M-*
immersed line
bundle
Riemann surface
corresponding
HPI.
and
R)
=
=
If
=
If
:
Then
E M
immersed
complex structures given complex structure the
Example right
x
such that
A Riemann surface
with
induced
injective,
of rank
*6
if
all
ji,
L C H
we see:
addition
in
for
L such that
proposition 6 is
surface in unique complex
oriented exist
6(Txm)j,
holomorphic
a
*6 From the
=
orientation
or
immersed
such that
6(T,:M)
subbundle
conformal
face M is called complex structure
of Lemma2.
that
to
Then there
by J, j,
J6
compatible
an
(Hom(L, HIL)). denoted
h(TXM)
and J is
reduces
in
evident:
5 E J?I
L and
on
the assertion
Let L C H
3.
derivative
structures
19
non-zero
and
1-matrices,
x
Bundles
7rLd((fl) d '") 7rL
End(L) by
0
J
R) =
7rL((dof) (*Of)d
R+
-7rL
R then
(fl)
dR)
(fl)
to
Bundles
4 Vector
20
jj
(L, J)
hence
is
=
1
right
normal
4.3
Holomorphic
Let
(V, J)
I
some
Quaternionic
complex quaternionic
a
R
is
holomorphic
a
is conformal
E and f
M-+
:
(L, J)
if
with
R.
vector
be
R for
-
*J,
Conversely,
curve.
M M
J
curve,then
holomorphic
a
=
Bundles bundle
vector
the Riemann surface
over
M. We decompose
HomR(TM, V)
KV (D
kV,
where
jw:
KV:=
jw:
KV:=
Definition
A
4.
TM
V
TM-+ V
holomorphic
structure
I
I
* w
* w
on
=
Jwj,
-Jwl.
=
(V, J)
is
linear
quaternionic
a
map
D
for
such that
0
all
1'(V)
E
0 E.P(V)
is
=
(DO)A
called
valued
-1 (dA
1.
For
+ i
(dA
*
a
dA).
+ i
*
In
only
(D,O)A 2.
+
=
if Do
=
*
dA).
(4.3)
0, and
we
put
ker D C F (V).
part
of
this,
k-part)
(the
note
for
that
of dA is
given
complexby OA
fact, =
structure
is
way to make
OOA".
1(OdA + JO
2
understanding
dA) (JX)
A holomorphic natural
better
anti-C4inear
A the
+
holomorphic
HO(V) Remark 3.
P (EV)
-4
and A: M-+ H
D(OA) A section
P (V)
:
*dA(X) a
-
i
dA(X)
=
-i(dA
generalized 0-operator. of a product rule
sense
+ i
*
dA) (X).
Equation of the form
(4.3) is "D(,OA)
the
in ELI"', does this mean L carries curve a natural holomorphic structure? This is not holomorphic yet clear, but we shall come back to See also Theorem I below. this question.
If
L is
a
Holomorphic
4.3
Quaternionic
-1, Example 14. Any given J E End(H), j2 Then F(H) bundle. vector complex quaternionic
turns
=
1(do
D,O is
holomorphic
a
+ J
2
H
Bundles
=
21
M x H'
H},
M-4
into
a
and
do)
*
structure.
line bundle and 0 E F(L) has no complex quaternionic D on (L, J) such exactly one holomorphic structure In fact, any 0 E r(L) that 0 becomes holomorphic. can be written as'O Op with p : M-* H, and our only chance is
Example
If L is
15.
then
zeros,
there
a
exists
=
1
Do This,
indeed,
Example that
2
(Odp of
the definition
+
(4.4)
JO * dy).
holomorphic
a
structure.
f : M-+ H is a conformal surface with left normal vector N, Mx H, and there exists a unique D such for L a complex structure A is 0. section 0 holomorphic if and only if dp + N * dp lp 0, If
16.
then N is D1
satisfies
:=
=
=
=
=
i.e.
*dl-t holomorphic
The left
normal
sections
f
N as
are
In this
.
case
=
Ndp.
therefore
the
dim HO(L)
! 2,
since
maps with the same I and f are independent
HO(L).
in
Theorem 1.
by
Jw
:=
wJ.
characterized H induces all
If
a
the
The
by
the
section
M x H'+1
with complex curve holomorphic structure a complex defined D pair (L-', J) has a canonical holomorphic structure linear form W : Hn+1 -4 following fact: Any quaternionic to the fibres by restriction WL E V(L-') of L. Then for
L C H
then
J,
structure
=
dual
L-1
bundle
is
a
inherits
w
DWL::::::
Proof. a
conformal
The vector
total
has
bundle
space of real
DWL
15
yields
0. Now any a E F(L-') by (4.4), for any section
<
Da,O 2
I 2
1 2
1 >=
2
<
L-L
<
wA,O
(d
<
a,O
> +
> +
*
fw
0
w
E
(EP+I)*
there
exists
I WJL ,
=
W such
0}
that
has WL
D such that holomorphic structure of the form a WLIX for some /\: M-+ H.
is
> + <
(d
=
unique
a
=
E
r(L)
WLdA+JWL *dA,o
(< wdA,o
X
4n + 2. Therefore
=
Then,
fibre
with
dimension
Example
no zero.
Lj-
0-
*dl\,Jo
*d < d < a,
we
>
>)
WA, J'O >)
JO >)
have
1 -
2
I -
2
<
<
wA, do
wA, do
+
+
*d(JO)
*d(JO)
>
>
4 Vector
22
*J
Note that wA by
<
This
a
=
in the
Da, 0
Bundles
6J implies last
contains
with
(Hn+l)*.
E
Remark 4.
ffffP1
carries
tive
spaces
prominent
2
no
WL has
w
-(d
>=
such that
As a
this role
term
< a,
reference no zero.
we
shall
is
+
*d(JV))
E
F(L),
and this
allows
us
to replace
well:
> +
to
*
d < a,
w, hence
But the last
JO >)
D is
equality
-
2
< a,
independent shows Da
do
+
of the =
0
for
*dJO
>
choice any
a
of =
W
WL
the next section, see in L in a holomorphic curve In higher dimensional structure. holomorphic projecL-1 rather than L plays a the case. Therefore no longer higher codimension.
natural in
do as
Sphere
5 The Mean Curvature
S-Theory
5.1
Mbe a'Riemann
Let
surface.
Let H:= Mx
denote to
product
the
S2
with
-1 be
=
a
bundle
M, and let S: M-+ End(EV) E F(End(H)) the differential on H. Wesplit according
over
complex
19
structure
type:
do where cP and d'
=
denote the Glinear
*d'
do
d"O,
and anti-linear
Sd,
=
+
*d"
=
components,
respectively:
-Sd".
Explicitly,
do So d" is
(do
-
d"O
do),
S*
2
(do
+ S
while d' is structure on (H, S), holomorphic of (H, -S). a holomorphic structure general d(SO) i4 Sdo, and we decompose further:
structure, In
2
a
*
dio).
an
anti-holomorphic
i.e.
d=,9+A,
d"=a+Q,
where
a(SO) AS For
example,
we
explicitly
S90'
5(so)
-SA,
QS
I
on
0 defines
H, while
a
Sao,
-SQ.
have
2
Then
=
=
(d"O
holomorphic structure Q are tensorial:
Sd'(So)).
-
and 0
A and
F. E. Burstall et al.: LNM 1772, pp. 23 - 30, 2002 © Springer-Verlag Berlin Heidelberg 2002
an
anti-holomorphic
structure
Sphere
5 The Mean Curvature
24
QEI(kEnd-(H)).
AEr(KEnd-(H)), For
M-+
(dS)O
IV
E F
(H)
have, by definition
we
=
d(SO)
=
(,g
=
ASO+
QSo
=
-2S(Q
+
=
2(*Q
*A)O.
+
dS,
+
A)O
SdO
-
A)SO + (5
-
of
(5.1)
Q)SO
+
SAO
-
S(o9
-
-
S(5
+
Q)0
SQO
-
A)O
Hence
dS
=
2(*Q
*A),
-
*dS
2(A
=
-
Q).
(5.2)
*dS).
(5-3)
Then
SdS
Q Since
Remark 5.
Q 0. If dS S decompose H =
deviation
5.2
+
A),
(SdS
4
*dS),
-
(M
=
x
C)
"complex
L and
an
E
on
HIL
(SdS
+
ED
(M
x
C).
Therefore
A and
immersed
the
measure
Sphere
holomorphic
HIL)).
HP1 with deriva-
L C H in
curve
Then there
exist
complex
structures
J
such that
We want to extend
J and
j
=
to
S E
a
jj
complex
structure
F(End(H))
such that
SL this
Q
case".
*J
Note that
=
=
S?'(Hom(L,
JL
4
0 if and only if A 0 Q are of different type, dS of the complex endomorphism 0, then the i-eigenspaces
The Mean Curvature
J
A
A and =
from the
Wenow consider tive
2(Q
conversely
whence
and
=
implies
=
L,
SIL
=
Ji
7rS
=
j7r-
of
H,
i.e.
find
an
on
7rdS(O)
7r(d(SO)
=
SdO)
-
Sphere
The Mean Curvature
5.2
JJO
=
ho
-
=
25
0,
and therefore
(5.4)
dSL c L. of S is'clear:
The existence bundle
L'
Since
L'
Identify not unique,
L. is
H
=
0,
R can be
Q:
interpreted
I((S
4
+
1(SdS 4 Q+ V)
E
F(L),
SIL
complementary Ji SIP := j.
:=
S + R is
kerR,
and
We compute
If
some
if and
RS + SR Note that
for
and define
7r,
RH c L c whence R2
L E) L'
=
It is easy to see that S is not unique. if M-+ End(H) satisfies R: only
such extension
another
Write
HIL using
with
1
4
+
*dS)
+
-
element
as an
R)d(S
(SdR
0.
=
R) I
*d(S
-
(SdR
4
of
Hom(H/L, L).
+
R))
+ RdS + RdR
+ RdS + RdR
-
-
Then RV
*dR)
*dR).
then 0
=
d(RO)
=
RdRO
=
dRO + Rdo,
-R 2do
=
0
and, by (5.4), R
We can therefore
=
0
continue I
00=QO+ (SdRO 4 QO+
dSO
I
4
-
(-SRJO
*dRO) + R*
=
JO)
QO+ =
1
4
(-SRdo
Q0 +
+
(-SRJO
*Rdo) +
Rj =RS=-SR
Hence, for 0
E
F(L),
00
=
Q0
-
-SRJO. 2
JO).
R.
5 The Mean Curvature
26
with
Now we start
54
-2SQ(X)6(X)-'7r:
0. First
note
of X
In
fact,
F-+
c
=
sinO
some
X
54 0. cosO, s
=
Q(cX
+
X
SJX) (J(CX
that
this
H
definition
SJX)) -')
is
Q(X) (CI
=
in view
+
of
(5.5),
define
(5.6)
H
-+
positive-homegeneous
R is
+
(J, J) and,
S of
any extension
R= for
Sphere
of the choice independent of degree 0, and with
SS) (J(X) (cl
+
ss))
-1
Q(X)6(X)-'. Next
RS
=
=
=
By
-2SQ(X)6X-'7rS -2SQ(X)SJj1-7r
=
=
-SR.
(5.6)
definition
kerR,
L c
and from
-2SQ(X)6x-lj7r 2S2Q(X)6 X1jr
(5.3)
and
(5.4)
we
L D
get
I(SdS
*dS)L
-
4
=
QL,
whence RH c L. Wehave
(5.5),
we
find
now
for
00
shown that
0 =
=
This
S + R is another
extension.
Finally,
using
F(L)
Q0 QV)
-
-
-1SRdo 2
=
QJ-lirdo
Q0
I -
2
S(-2SQ6-'7r)do
0.
=
shows
Theorem 2.
HP'.
E
Let L C H
Then there
exists
a
=
M x EV be
complex
unique SL *6
a
=
=
L, 6
holomorphic structure
dSL C 0
S
QJL
=
=
0-
S
0
S
curve on
immersed
into
H such that
L,
(5.7)
6,
(5-8) (5.9)
Hopf
5.3
S is
SpLp
a
terms. 2-spheres, a sphere congruence in classical sphere Sp goes through Lp E HP', while dSL JS S6) implies it is tangent to L in p, see examples
In
an
affine
coordinate vector
Definition
f
as
the
mean
Equations 0 + Q
Remark 6.
whence d"
=
holomorphic particular,
a
Example 17.
holomorphic S E
Let
2-sphere
KPI.
in
11
S2 E
complex sphere congruence of L by definition, S: Wehave SL and Q 0. !(SdS *dS) 4 structure
the
In the
shall
we
definition
of the w
can or
be
for
Hence
C
0
F(L),
E
immersed
an
and,
in
HP1
corresponding
line
bundle
from the immersion.
and endow
Then the
mean
map S' -+ Z of value
constant
dSL
implies
A
encounter differential frequently wedge product of 1-forms
O(X, Y)
generalized verbatim to Vi, provided there
bundles
End(V)
composition
F(L)
E
10}
=
C L
Fields
following
the usual
The
L.
=
-
Hppf
11
=
and the constancy
=
5.3
sphere (congruence) of the Hopf fields of L.
simply the
L is
the
motivates
Then
-I.
HP1 I Sl
inherited
curvature
=
=
Let L denote
the
same mean
subbundle of (H, S, d") a holomorphic bundle itself. vector quaternionic
End(H2), S'=
S' with
9 and 12.
called
are
(5-7), (5.8) imply do + S * do 12 (d + S * d) leaves L invariant.
HP is
in
curve
is
=
Remark 9. This
see
curvature
forms A, Q E S?'(End(H))
differential
a
H at p,
=
sphere Sp has the
L the
=
1
M-4 R4
:
S is called
5.
I,]
system
(or,
C L
=
curvature
is
Because
the
equivalently,
27
of
family
Lp
=
Fields
x
w(X)O(Y)
=
a
End(V)
-+
product
(Vi)
with
V,
x
or
the
End(V)
Note that
w(Y)O(X)
-
forms wi E 01 is
forms.
V2
-+
values V.
in vector
Examples between
pairing
spaces are
the
the
dual
V* and V.
On
a
use
the latter.
form As
be written
any 2-form
o-(X, JX)
an
example,
A
O(X, JX)
w
will
M,
Riemann surface
by the quadratic
=
=:
u(X),
a
E
and
w(X)O(JX)
-
S22 is completely we
shall,
for
determined
simplicity,
w(JX)O(X)
as
(5.10)
wAO=w*0-*wO. We now collect curvature
often
sphere
some
information
congruence
S
:
about
M-+ Z.
the
Hopf fields
and the
mean
Sphere
5 The Mean Curvature
28
Lemma4.
d(A+Q) Proof.
(5.2)
from
Recall
2(QA Q +AAA).
=
SdS
using AS
Therefore,
d(A
-SA, QS
=
Q)
+
=
2(A
=
+
Q).
(dS
A
-SQ,
=
Id(SdS) 2
1 =
2
dS)
Q) =2(AAA+AAQ+QAA+QAQ). =
But A A
Q
0
=
by
2S(A
following
the
QA A
Similarly stabilizing
Notice
that
QxO 0 QzO independent if
=
=
Proof. which
=
A*
=
ff",
Q
*AQ
-
0, because A
Let L C H be
Lemma5.
7r
A
S(A
+
Using
and
that
k))
Q "left
have
AAQ
H
Q)
type argument:
"right
A is we
+
L such that
an
A(-SQ)
=
is left
K and
Q is right
surface
immersed
dSL C L.
Then
(-AS)Q
-
QJL
and S =
0 is
=
0.
(5-11)
K.
a complex equivalent
structure
on
to AH c L.
images of the 1-forms A and Q are well-defined: also QjXO -SQxO 0, and thus for any Z E TM. In other words, the kernels of Q and A are of X E TM. The same remark holds for the respective images. the kernels
0 for
Wefirst stabilize
(dw (X, Y) 0)
and
X E TMthen
some
need
formula
a
=7r
for the derivative
wL C L. If
L, i.e.,
(d(wo) (X, Y)
=7r(x
-
=
(W(Y)O)
7r
+ -
=
w
Y
*7rL,
A -
then
=
of 1-forms for
0
E
w
E
Q'(End(H))
F(L)
dV) (X, Y))
(W(X)O)
-
W([X, YDO %vo
.11
EF(L) +
w
(X) do (Y)
=6(X)w(Y)O =6(X)w(Y),O =(J A w + -7rw wedge
-
w
(Y) do (X))
J(Y)w(X)O
7rw(X)dO(Y) J(Y)w(X)O +,7rw(X)JO(Y) A J)(X, Y)O, -
-
+
-
-
7rw(Y)dO(X) 7rw(Y)JO(X)
Note that the composition 7rWJ makes composition. L, and L is annihilated by 7r. We apply this to A and Q. Since AL C L, QL C L we have on L, by lemma 4,
where sense,
we
because
w
over
(L)
C
The Conformal
5.4
0
I7r(QA
=
similar
type argument
a
to
(5-11),
7rA A 6
similarly
and
the
for
remaining
=
7rA
=
-2S7rA6,
Since
AL C L and 0
5.4
=
QJL
L
:
For
6.
End(V)
-+
quaternionic
a
Example 18.
of
is taken
indefinite
an
for
0
=
-7rQ
Further,
A 6.
--xSA6
=
SJQIL
or
-
-
X
54
0 is
an
isomorphism,
we
get
n
and
Gauss Map vector
A >:
<
We obtain
HIL
JQIL
=
space
bundle
or
V
of rank
define
we
where the trace
=
A 6.
AJ
7r *
-
29
0-
=
The Conformal
Definition
A E
J(X)
5
*
AA
Weobtain
term.
-7rA6
irA
Q + 7rQ
get J
we
Map
7r(dA+dQ)
=
6 A A + 7rA A 6 + 6 A
=
By
Q +AAA)
2
Gauss
For A
=
Tn
traceR A,
endomorphism A. In particular product < A, B >:=< AB >.
the real
scalar
(a)
=
with
<
a
=
ao +
ial
14ao
A >=
4
ja2
+
=
+
ka3
E H
we
< I
>= 1.
have
ao,
and <
Proposition face L satisfies
The
4.
<
i.e.
S
:
M-+ Z is
Because of this
Bryant
[1].
AA >= Rea2 mean
dS, dS
=
curvature
>=<
a20
_
a21
_a
sphere S of
*dS, *dS >,
<
2 2
_a23
an
immersed Riemann
dS, *dS
>=
sur-
0,
conformal.
proposition,
S is also
called
the
conformal
Gauss map,
see
30
5 The Mean Curvature
Proof.
Wehave
QA
=
Sphere
0, and therefore <
Q,A
>=<
A, Q >=
(5.12)
0.
Then, from (5.2), <
dS, dS
>=4
< -S(Q
+
Q
A, Q
=4 <
-
A), -S(Q
+
A >=<
-
A)
>= 4 <
*dS, *dS
Q + A, Q + A
>
>
Similarly, <
dS, *dS >=4
<
=4(<
-S(Q
+
SQQ>
A), -
A
-
< S
Q>
QA >
+ <
SAQ >
=0
But,
by
a
of the real
property
<
trace,
SAQ >
=<
QSA>=< -SQA
SQQ>
=<
QSQ>=< -SQQ >= 0,
SAA >
=<
ASA >=< -SAA
>=
0,
>= 0.
-
<
SAA >).
Surfaces
6 Willmore
Throughout
The
6.1
this
Mdenotes
section
Energy
a
compact surface.
Functional
The set
IS
E
RP1 is
a
Z
of oriented
2-spheres
in
TsZ
=
-Z
=
S
Here
we use
the
in
Section
Definition
surface
A,
B >:=<
Critical
defined
points
Proposition d
*
from
5.
of
End(ffV)
with
product 1
AB >=
-
8
traceR(AB)
functional
of
a
map S
:
M --*
Z
Of
a
Riemann
by
S of this
maps
_jj
5.3.
E(S)
harmonic
submanifold
=
E
inner
The energy
7.
M is
End(V) I S2
Fnd(EV) I XS -SX}, I YS SY}. JY E End(fff)
IX
(indefinite) <
defined
=
fm
:=
functional
< dS A *dS >.
with
respect
to
variations
of S
are
called
M to Z.
S is harmonic
if
and
only if the Z-tangential
component of
dS vanishes:
(d This
condition
is
equivalent
to
*
any
dS)T of
the
F. E. Burstall et al.: LNM 1772, pp. 31 - 38, 2002 © Springer-Verlag Berlin Heidelberg 2002
=
0.
following:
(6-1)
Surfaces
6 Willmore
32
d(S
In
dS)
*
d
*
A
d
*
Q
A
=
=
o,
(6.2)
=
0,
(6-3)
=
0.
(6.4)
fact,
d(S Proof.
St be
Let
a
Q
*
=
4d
*
of S in
variation
fm
d
d
Wt- E(S) the
4d
=
S(d
dS)T
*
Z with
=
(Sd
variational
*
dS)T.
(6-5)
field
vector
Y.
-YS and
Then SY
Using
dS)
*
Wtwedge
<
fm
dS A *dS >=
(5.10)
formula
< dS A *dY > =<
< dY A
*dS > + < dS A *dY
traceR(AB)
and
dS(-dY)
*dS
-
traceR(BA),
=
we
dY >=< dY A *dS >
*
>.
get .
Thus d
Wt- E(S)
2
=
Therefore
f'
< dY A
*dS
>= -2
JM
S is harmonic
For the other 0
only
if and
first
equivalences, =
d
=
(d
d(S')
* *
with
*Q
-
8d
*
*
dS >= -2
*
fm
<
Yd*dS >.
dS is normal.
note
+ S
2(*dS)2
2dS A *dS +
Now, together
if d
Yd
dS)
*
*dS A dS + dS A *dS + Sd * dS
-
-
<
d(*dSS
=
dS)S
-2(dS)2
fm
(d
*
(d
+
dS)S
+ Sd
WSand A
*A
2
Q
8d
*
A
=
dS)S
*
2d(S
dS
dS.
*
1 4
=
+ Sd
(SdS
+
*dS),
this
implies
dS)
*
2dS A *dS + 2Sd
*
dS
-(d
dS)S
+ Sd
*
dS
S(d
dS +
S(d
*
dS)S).
=2(d*dS)T We now consider immersed
holomorphic
the
where S is the
case
curve.
mean
We decompose dS into
curvature
the
Hopf
sphere of
an
fields.
Lemma6.
dS A *dS >=
4(<
A A *A > + <
Q A *Q >),
(6-6)
dS A SdS >=
4(<
A A *A >
Q A *Q >).
(6.7)
-
<
Proof.
Recall
dS
from section
='2(*Q
Functional
33
5.1
*A),
-
Energy
The
6.1
*dS
2(A
=
Q),
-
SdS
2(Q
=
+
A).
Further
*QAA=O, by type.
*AAQ=O
Therefore < dS A *dS >
4 <
=
(*Q
*A)
-
A
(A
Q)
-
>
=-4<*QAQ> -4< *AAA> 4 <
=
similarly
and
for
Lemma7.
S,
B E
>
+4 < A A *A >,
< dS A SdS >.
Let V be
End(V)
Q A *Q
a
quaternionic
vector
space,
L CV
a
quaternionic
line,
such that
S2
_j
SB
T,
-BS,
=
image
B C L.
Then 2 traceR B
with
if and only if BIL
equality
Proof. SL
=
=
We may assume B 54 0. L. Let 0 E L\101, and
So Then 'X2 =-1,
0-
OA,
p is
imaginary,
traceR B2 This
can
be
applied
to
A
=
or
OIL.
-PA.
B20
follows
It
too.
BO
-BS implies
implies
AP Therefore
BV, and SB
Then L
=
and BS = -SB
0,
<
traceR B2 IL
Q instead
of
=
_JpJ2 0,
=
-41p 12.
B,
since
and
their
rank
is
< 1.
We
obtain Lemma8.
For
an
immersed
<
holomorphic
A A *A >=
I 2
<
curve
AIL
A
L
*AIL
we
>i
have
(6.8)
and
>O, In
particular
E(S)
> 0.
>O.
(6.9)
Surfaces
6 Willmore
34
Proof, *A >
< AA
traceR(-A'
8
(*A)'
-
4
2 traceR A
=-ASSA=A2
Because dim L
2
<
see
AIL
*AIL
A
Because AH c
5.3.
section
similarly
dim H we
L,
have
>=
we
2 traceR A
proves
(6.8).
Proposition
6.
This
The
=
is
closed.
If
S
:
Z, and dS
M-*
S*w In
=
=
2-form
-
w
E
S?2 (Z)
defined
by
forSEZ,X,YETsZ,
*A)
usual,
as
2 < A A *A > -2
see
Q A *Q
<
5.1
section
(5.3),
then
>.
particular, 1
degS:=
7r
is
2(*Q
IL7
from Lemma7.
ws(X,Y)=<X,SY>,
2.
12
traceR A 12L
(alternating!)
The
1.
2
have
follows
positivity
A
traceR
-
a
topological
f
- A M
of S.
invariant
S maps the surface Minto the 8-dimensional Z, deg S cermapping degree of S. But for immersed holomorphic curves of two mapping degrees deg S it is the difference deg N deg R, where N, R : M-4 S2 are the left and right normal vector in affine coordinates, see chapter 7. Remark 7.
tainly
Since
is not the
=
Proof.
(i).
the 2-form
Weconsider
COS(X, Y) Then
dsCo (X, Y, Z) is
a
1
:=
linear
-(< X, 2
X, Y, Z
E
TsZ,
S E Z,
End(EP) SY
>
combination <
But if
on
we
Y'XZ get
-
defined
<
by
Y, Sx >).
of terms >
-
of the form
< y )XZ > =<
dw
(ii).
S2yXZ
Therefore,
hence < Y, XZ > = 0.
S2yXZ
<
if
=
t
dt*Cv
-+
End(ffR2)
t*dCo
=
< y IXZ
_
35
SyXZS >
>=<
>=
Z
:
Fanctional
The Willmore
6.2
>'
inclusion,
is the
0.
=
Wehave
S*w(X, Y)
dS(X), SdS(Y)
=<
1 2
1 2
I =
2
>
(< dS(X)SdS(Y)
>
(< dS(X)SdS(Y)
>
< dS A
-
-
<
SdS(X)dS(Y)
>)
<
dS(Y)SdS(X)
>)
(X, Y),
SdS >
yields the formula. under deformations invarlance topological 3 deforms So: M-4 If 9 : Mx [0, 1]
and Lemma6 The
theorem:
fmx[0
0
f fm
6.2
that
integral
=
4
=
8
fm fm
g*W
g*"' J MX0
S*1W-
<
A A *A > + <
<
A A *A > +
for
fm so*
W.
Q A *Q
4fm (< Q
>
A
*Q
>
problems the
variational
energy
-
A A *A
<
>)
invariant
functional
can
be
replaced
by
of < A A *A >.
The Willmore
Definition
Hopf field
Stokes
1]
topological
the
from
then
From
E(S)
we see
S1,
dg*w ,
MAx 1
Remark 8.
of S follows 3 into
8.
A.
Let L be
Functional a
The Willmore
compact immersed holomorphic
functional
W(L)
:=
-I 7r
f
A M
defined
of
L is
<
A A *A >
as
curve
in
HP' with
36
Surfaces
6 Willmore
If we vary holomorphic
the
J
structure
Willmore not
On the
on
Msuch that
3.
Critical
Proposition
this
treat
HP, it will in general not hand, any immersion induces
M --+
:
other
with
Wwith
called
Willmore
Proof. Lt
if
(Ejiri and
[2],
curve,
of
fixing surfaces,
L
Willmore
the
R4
in
conformal
but
we
As usual,
we
7r, o
its
a
2
(H
K
-
have
we
Kj-) ldf I 2
-
An immersed
holomorphic
sphere S
is harmonic.
curvature
mean
variational
abbreviate
=
7r(SO)*
dt
-
irS
d
structure
d
E(St)
dt
t=0
=
general
fm
on
t=o f <
< A A *B
d,
d
0),
dt
curve
Y E
F(Hom(L, HIL))
Ot
F(Lt).
E
a
dot
YSO
-
Note that
.
57r
=
for
(YS
of the energy, which is L. By contrast, in the
Mwas
dSt
L is
for too.
given by
t=0
by =
<
field
vector
t=o
Wenow compute the variation functional as long as we vary the conformal
shall
A Am
[12]).
Rigoli
only if
has
see
called
are
13.
Y,O:= 7r(
In
holomorphic
a
such variations
a
complex
Let Lt be a variation, and St its mean curvature sphere. Note that i.e. the operator stay conformal the complex structure, *, varies,
to
The variation
dt
f
47r
7.3, Proposition
Theorem 3
remain a
here.
case
W(L) section
is to
variations
constrained
Example 19. For immersed surfaces
see
it
respect
only
consider
we are
to this
respect
of
points
surfaces. If of M they
structure
L
immersion
curve.
fixed,
A
and
*tdSt
no
L
E F
-
(L)
SY)O.
(6.10)
good as the Willmore 5 proof of Proposition
as
was
involved.
>
A M
A *dS > +
>=< B A *A
fm >,
<
dS A dS > +
because
Hence
fm
traceR(AB)
< dS A
=
*d
>.
traceR(BA).
(6-11) Next
we
claim II
=
0.
(6-12)
The Willmore
6.2
B,
On TMlet <
dS A dS
>
w(X)
i.e.
(X, JX)
conformal,
see
dS(X) dS(JX)
=<
dS(X)dS(BJX)
-
with <
for
all X, Now,
Y. we
dS(X)dS(JY) with
Using this
Y
d
> = 0 for
all
X.
BX
get (6.12).
we
>= 0
I.
,d*dS
<
>
< ,Sd*Q>
(, Sd
traceR
*
Q).
A M
lemma that
following
consider
we can
-
dS(Y)dS(JX)
=
imaged Therefore
-
A M
2
show in the
dS(JX)MS(X) > > < dS(JX)dS(BX) < dS(BX)dS(JX) >
> + <
fm 4f f
(6.5)
0, and
yields
integral
compute the I
We shall
(JX)
X
to
=
<
-
>
have BJ + JB
37
4, therefore
Proposition
respect
>
we
dS(X)dS(JBX)
<
< dS (X) dS
Differentiation
Then
=<
=
But S is
w(BX).
=:
Functional
QCL
*
Q as
c ker d
Q.
2-form
d
*
*
Q Ep2 (Hom(H/L,
a
*
L),
and continue 1
fm (, Sd Q H) fm (7r Sd Q HIL HIL) i fm traceR(7r JLSd Q HIL HIL) 1f traceR((YS-SY)(Sd*Q):H1L-4H1L) fm traceR(Yd Q) I traceR(SYSd 1
traceR
2 1
*
:
traceR
2
*
H -+
*
2
-+
:
--+
:
,
2
(6.1o)
A M
I
*
2
Now,d
Q is tangential
by (6.5),
-
2
*
Q).
M
and hence anti-commutes
with
S. Thus
>.
Surfaces
6 Willmore
38
1=:
-
fm traceR(Yd Q) fm traceR(Yd Q) fm Yd*Q *
2
-
-8
<
E(St)=-8f
t=O
Y (-= Q2 (Hom(H/L,
L),
this
. A M
vanishes
for
variational
all
fields
vector
if
d*Q In the
QS)
>
d
only
traceR(SYd A M
*
Tt
if and
f
2
showed
Wetherefore
Since
+
proof
we
made
of the
use
0.
=
following
Lemma9.
imaged
Proof.
Foro 0
QJL
because
This
=
E
shows the 7r
(d
*
C kerd
Q.
*
F(L)
d(*Qo) =
QCL
*
0. But
right
Q) (X, JX)
(d
-
*Q
Q)o
*
is
right
(d
*
-
Q)o
*Q
K, and 6
*Q
=
A
(d
=
is left
60
=
Q)o
*
K.
-
*Q
A
do,
Hence, by type,
0.
Also,
hand inclusion.
(d
do
A
A) (X, JX)
*
=
7r
=
7r(X
=
J(X)
=
-J(X)A(X)
(*A(JX))
-
-
JX
-
(*A(X))
-
*A([X,
JX]))
L-valued
=0.
-
*
A(JX)
-
-
J.(JX)
*
A(X)
6(X)SSA(X)
Y
7 Metric
the metric
Weconsider
associated
quantities
extrinsic
brevity
write
Surfaces
7.1 Let
we
<
> instead
.,.
and
right
*df
M
HPI.
of <
>R-
II(X, Proof.
Y)
We know from
the tangent
space
as
normal
=
Ndf
Lemma 2 that
fixed
its
point
(-I)-eigenspace
is the normal
II But differentiation
in. relation
(X, Y) of
dN(X)df (Y)R
(7.2)
1 =
-
H,
M
i.e.
R.
dN(X)
Y)
*
N(x)vR(x)
=
(X
df
is
an
-
-
df (Y)
-
NX
-
space,
so we
(X df (Y)
df (Y)R
df (Y)R
of
(7.1) involution
with
(7.2)
-
need to compute
NX
-
df (Y) R).
yields
+ NX
(Y))--L
df (Y)).
df (Y)
-
2
:
form II(X,
-+
v
f
+
Ndf (Y)dR(X)
=
X
-
df (Y),
or
X
to the
set:
Ndf (Y)R Its
-df
=
(*df (Y)dR(X)
2
of
vector
fundamental
The second
7.
M-+ R
:
Space
Euclidean
in
N, R denote the left
Proposition f is given by
f
of
geometry
to
L:=
For
Geometry
Conformal
and Affine
=
dN(X)df (Y)R
=
-dN(X)
F. E. Burstall et al.: LNM 1772, pp. 39 - 46, 2002 © Springer-Verlag Berlin Heidelberg 2002
*
+
df (Y)
Ndf (Y)dR(X) +
*df (Y)dR(X).
40
Proposition
The
8.
'Rdf Proof.
mean
(*dR
2
By definition
+
'H
vector
curvature
dfR
RdR),
2
II
trace
2
is
given
by
(7-3)
NdN).
(*dN
+
dR +
*dNdf
(7.4)
NdN)df,
(7.5)
of the trace,
jdfJ2
4'H
Geometry
Conformal
and Affine
7 Metric
=
*dfdR
=
-df (*dR
dN *
-
df
RdR)
+
df
-
+
*
(*dN
+
but
(*dN
If follows
+
NdN)df
=
*dNdf
=
-df
A
-
dN *
dR
=
df
-dN A
=
-df (*dR
df
=
-d(Ndf)
RdR).
+
that
27ildfI2
=
-df (*dR
RdR),
+
and
2 ldfTf Similarly
for
Proposition and let
K'
dR +
(*dR
=
RdR)Tf
+
N.
Let K denote
9.
denote
the normal
Kjwhere X E
dRR)Tf
TpM,
and
E-Lp 1
KIdf 12
=
ldf 12
=
K
1
:=<
2
the
Gaussian
curvature
of f defined
Rj- (X, JX) ,
M are unit
(< *dR,
-1 (< *dR, 2
of (M, f
curvature
N
vectors.
Therefore
=<
II(X,
<
>R)
by
>R) Then
RdR > + <
*dN,
NdN >)
(7.6)
RdR >
*dN,
NdN >)
(7.7)
-
<
Proof.
Kldfl4(X)
*
X), I.T(jX,
jX)
>
_III(X,
jX)12.
Surfaces
7.1
4KIdf 14
=<
*df dR
*df
<
-
(df
dR +
< N
=
<
<
N(df
<
df
df
dR +
*dNdf,
df
dfdR, dfR
<
dNdf, df
+ <
df
+ <
*dNdf,
+
dfRdR
ldf 12
I dyl2(<
21dfI2 (<
Kjafter
find,
we
4K
1
jdfJ2
=< N
similar
a
=< *dR
As
we use
a
Proposition
to
we
10.
this
The
(R
=
N)
df
dN,
< *
<
N*
dNdf
>
N * dN >
dR, NdNdf *dN,
>
NdN >
dN,N* dN
>
*
and the Ricci
X)
II(JX,
-
dN
>).
equation
JX)
>,
-
< *dN
>
NdN, NdN >
-
-
<
(*dN
-
NdN)df, dfRdR
(7.7).
obtain
pull-back
for compact 1
3-space
dfdR,
-ldf 12 + <
NdNdf
> >
have
T7r In
dR,NdNdf
*dNdf, <
dNdf
N
dR > + < dN,N
RdR), NdNdf
-
R*dA
Integrating
df
*
*dN, NdN >)
JX), II(X,
II(X,
>
-
Using (7.1)
K.
RdR, RdR >
-
(7.5)
corollary
*
dNdf,
computation,
df (*dR
+ <
On this
for
the formula
proves
dR,R
>
>
<
> + <
<
-
dNdf
>
+jdf 12
dR,R*dR
*dR, RdR >
<
This
Ndf
,
dR >
>
NdNdf
+
> >
dfdR, N dNdf
> + <
dR >
dNdf
<
> + <
*dR, RdR >
<
*dNdf
<
-
RdR >
dNdf, Ndf
+ <
-
R * dR >
dR,
<
dfRdR
dR >
*
dR,dfRdR
*
ldf 12
-
*dNdf,
<
dR + N *
*
>
dNdf)
+
>
*dNdf)
dR +
*
N(-dfdR
dNdf, dfR
+
-df
dR +
41
>
*dNdf
dR +
*
*dNdf),
dNdf, N(-df
dR +
*dNdf
df, -df dR + dNdf
*
dNdf ), -df
dR +
*
dfdR
+ <
*dN
-
dR +
*
dR +
*
<
=-
dR
*
df, -df
dN *
-
Space
Euclidean
in
this
f
of =<
2-sphere
area
under R is given
*dR, RdR >
M yields
Kjdf 12 A M
is
the
a
version
2
(deg
ofthe
R+
deg N).
Gauss-Bonnet
theorem.
by
>
Proposition
We obtain
11.
(J-H12 particular, integrand
if f given
In
is
:
K
-
-
-
Kj-) ldf12
Ij
12
K)Idfl2
-
4
(7.7)
(7.6),
11
K-L)Idfl2
4
11
4
*
dR +
*
dR12
*
dR
describe
characteristic
the
S relative
to the frame
equivalent
C L is
S=
of
vectors,
dR
*
-
<
section,
we
RdRI
2
to
Coordinates
affine
of S in i.e.
we
coordinates.
S
write
=
GMG-1,
We where
f)
01
EV
S
--
H2 having
(1f) (' -R) =
the
following
matrix
1f)
0
0
-H
01
(7.9)
_I 2 ,
RH
vanishing
(7.10)
HN.
,
L, has the nowhere
*dR, RdR >
<
2
N and R turn symbols is deliberate: related of f while H is closely to
compute
(7.8)
*dR, RdR >
4
X The bundle
Willmore
RdRJ2.
1IRdRI2
+
1
N2=-l=R
The choice
1
RdRj2_
-
0
where N, R, H : M-+ H. From S2
normal
0, and the classical
=
properties
G
First, SL representation:
RdRJ2
-
Sphere in Affine
The Mean Curvature
We now discuss
dR
*
give
1
41 7.2
1
4
then Kj-
M-+ Im H = R'
(7.3),
Equations
(I,HI2
K
-
by
(I Proof.
Geometry
Conformal
and Affine
7 Metric
42
section
out
its
to be the mean
(fl)
left
curvature
E V
and
right
vector
(L). Using
this
(f)
(*df) ird(S (f 7rd( (f (f) Ir((Ndf)
(f)
irSd
=
1
*6
Therefore
1
+
0
S6
=
we
7r((-df R) (f ) (f) (-Hdf)) (Ndf
R))
have identified
=
=
Ir
(-dR))
1
-df R,
=
N and R.
of the Hopf fields, is we need dS. This computation but lengthy forward somewhat the simplified computation, by G-dG. We skip the details and give the result: GdG dG
dS
SdS
Romthis =
SdS
=
G
=
straightfact
that
-dfH
G
-dfR
+ dN
-dH
-NdfH
-
-dR +
Ndf) P7
Hdf
+ NdN
+ RdH
'
dR)
0
HdN Hdf R + R
-
G-1.
obtain
*dS
-
NdN *dH +
HdfH
-
*dN
0
+ RdH
HdN 2HdfR + RdR +
-
*dR)
G-1
SdS + *dS
.
far-,
(
NdN + *dN -
dH +
*
QIL
The condition so
(
( (HdfH
=
G
=
we
G
used
a
=
=
4A
-7r
0
For the
Q
43
to
Ndf
=
+
0
1
JS is equivalent
=
*df and
Coordinates
0
=
sj
Affine
in
,
=
is
Sphere
The Mean Curvature
7.2
=
have the
HdfH
-
2dfH with
2Hdf
=
dR
2dfH
=
dN
(7.3)
equations -
-
H
0
HdN RdR
-
R N
dR
=
dN
dR dN we
L,
G-1.
which
we
have not
(7.11)
N * dN.
(7.12)
find
-R(*dR
=
*dR)
dR,
R*
-
-
-
AH C
0, and the corresponding equivalents: following
2Hdf
Together
2Ndf
+ RdH
+
-N(*dN
-2RTtdf, 2NdfR NdN)
RdR) +
=
=
=
-2dfRR,
and therefore H=
-RN
=
-RR.
(7.13)
-df 0
R)
Remark 9.
the
have,
Given
vector
vature
is
mean
f
of
at
(7.12)
Proposition
simplify
write
(01), f
Using (7.12)
and
=
( (w
-
0
dfH
dH + H *
+ R * dH
IH
*
2
of the
1H(NdN
(7.14) (7.15) H * dN.
-
*dN).
-
2
of
the reformulation
(dN
1H*
N*
-
dN)
-
previous
2
Affine
in
Proposition
But
w.
H * dN
(dN+N*dN)
Condition
The Willmore
0) G-1, dR) G-',
w
dR + R *
dH + R * dH +
=
H * dN
We use the notations
-2dH +
0
2
H(NdN
-
v,
=
12, and
in addition
dR+R*dR.
Note that V
Proposition
For
f
:
13.
=
The Willmore
A A *A >
M-4
-dR +. *dRR
R, this
1 =
16
JRdR
is the
-dR
--
integrand -
*dR12
classical
< A A *A >=
-
is
1 =
4
R * dR
-
integrand 1
4
(Ih 12
-
=
-v.
by
given
(IHI2
*dN).
Coordinates
viate
<
Hopf
the
rewrite
-
7.3
for
dN + N * dN 0
have to consider
Weonly
dfH
w
we can
w
H*
expressions
12.
4*A= G
Proof.
coordinate
the
follows
as
4*Q=G
where G
see
curvature
same mean
we now
mean cur-
by Sx. On the other hand, S" S,' and f Example 17. Therefore the name mean at x, justifying vector
Sx,
of
sphere
(7.11),
Equations fields, which
the
L
curve
M is determined
E
x
Geometry
holomorphic
immersed
an
curvature
fact, the sphere.
in
curvature
Conformal
and Affine
7 Metric
44
K)Idfl2.
K
-
K-L)JdfJ2.
abbre-
Condition
The Willmore
7.3
Affine
in
Coordinates
45
Proof. < A A *A >
traceR(-A'
8
1
4
4 Now see
Re(
Proposition
11
(*A))
-
1V)
2
4
16
4
IV12
and
A
*
=
16
4d
*
=
A
Most entries
of this
matrix
Proposition
14.
We have
=
G
we
(
A
jRdR
0 for
=
A
w
=
df
df
A
v
dv +
w
A
w
dw
A dH +
A dH
df
Aw=O
df
Av=O
df
dG)G-1,
df)
G-1.
(7.16) (7.17)
-
Idf
A R * dH +
*df
2
(7.18)
=0.
A
IdfH
2
dfH
H(NdN
A
2 1
A *dH+
io
A
(NdN
(NdN
-
*dN)
-
-
*dN)
*dN),
but
*(NdN
-
Willmore
vanish:
=df AdH+dfRA*dH+ df
*dR12.
then
We have
df
-
find
easily
df
A
GMG-1,
dv+wAdf =-(2dH-W)Adf Proof.
*
dG A M+ dM + MA G-1
dG
=
A
*
16
(7.8). d
equation 4
dR12
R*
and, for the second equality,
G(G-1
again using G-'dG
JdR +
=
We now express the Euler-Lagrange If we write coordinates. surfaces in affine 4d
traceR(A2)
=
*dfH
=
df (-R)H
*dN)
=
(N
*
dN
-df 2
-
N
HN
dN)
=
-N(NdN
Hence, by type, the second term vanishes as well, and A similar, but simpler, shows (7.17) computation Next, using (7.11), we consider
we
-
*dN).
get (7.16).
dv+wAdf =d(dR+R*dR)+wAdf =
d(-2Hdf)
=
(-2dH
=
(-dH
+ +
w)
w
A
A
df
df
+ R * dH +
1H(NdN
2
"o
-
*dN))
A
df.
46
and Affine
7 Metric
Again we Clearly
show
*a
*(NdN
-
aN
ON. Then .(7.18)
aN,
*dN)
-
flN.
showing *0 *a
=
Geometry
Conformal
will
(NdN
N * dN + NdNN=
by type.
follow
*dN)N,
-
Further
*
dH
*
dH
RdH + dHN
-
d(RH)
-
+
R(*dH)N
-
(dR)
d(HN)
H+
-
HdN
-
R(d(HN)
=RH
+R2* dH +
(dR)H (dR
-
=
As
a
HdN
-
R*
2HdfH
corollary
=
+ RH*
N*
-
+ RdH
-
HdN)
dN)
dN)
H(2dfH)
-
get:
we
15. 1
d*A=
w
H(dN
-
((dR)H
R*
-
(dR)H
R*
-
dR)H
HdN
-
4
G
dH + R * dH +
Therefore
f
is
(dw 0) 0
0
!H(NdN 2
Willmore
if
4
-fdwf) -dw
*dN).
-
only if
and
(dw
fdw
G-1
Let -y Example 20 (Willmore Cylinder). R2 the H defined -+ : f cylinder by
dw :
0.
RIm H be
a
unit-speed
curve,
and
f (S' t) with
after
the conformal some
structure
computation,
Ir.3+
is
exactly
the
J-L as
f
that
2
This
HdN)
0.
Proposition
with
(dR)H
-
=RH
condition
K11
=
is
-r.7,
that
=
-Y(S)
-2-.
+ t
Then
at
15, we obtain, using Proposition if and only if Willmore,
(non-compact) 2
=
-y be
(r,2-ol
0, a
free
elastic
=
0.
curve.
Projections
8 Twistor
Projections
Twistor
8.1
E C H
Let
subbundle from
M x H'
:=
a
over
Mx
=
C' be
Riemann surface
multiplication by i on IV. JE E f2l (Hom(E, HIE))
right
by
We define
JEO:= 7rEdo, where 7rE
:
H -+
Definition
9.
HIE
E is
is the
called
a
is
equivalent
to the
fact
0
E
F(E),
projection.
holomorphic *6E
This
line complex (not a quaternionic) JE induced complex structure
a
M with
that
d"O
=
:-.,:
the
curve
in
CP1, if
6E JE
holomorphic
1(do
2
+ i
*
structure
(8.1)
do)
and hence induces a holomorphic structure F(E) into itself, complex line bundle E. line bundle A complex line bundle E C H induces a quaternionic
of H maps the
on
L=EH=EE)Ej CH. to the structure of a JE admits a unique extension complex structure bundle (L, J), namely right-multiplication complex quaternionic by (-i) on line bundle (L, J) C H induces a a complex quaternionic Ej. Conversely, line bundle complex
The
E:=
Definition
lift
10.
We call
(L, J)
JOE L1JO=Oi}. the twistor
of (L, J).
F. E. Burstall et al.: LNM 1772, pp. 47 - 52, 2002 © Springer-Verlag Berlin Heidelberg 2002
projection
of E, and
E the twistor
48
Projections
8 Twistor
quaternionic case, any map f : M -+ CP' induces E, where the fibre over p, is f (p), and vice versa. above correspond to holomorphic curves in as defined curves of complex analysis. The correspondence between E and (L, J) is Cp3 -+ ffffpl. by the Penrose twistor projection
Remark 10.
'As
in
complex Holomorphic
line
bundle
a
the
sense
mediated
Theorem 4.
surface 1.
the
E C H be
Let
M, and (L, J)
Then
(L,J)
is
a
its
complex projection.
a
line
a
twistor
holomorphic
over
a
Riemann
i.e.
curve,
*6L
subbundle
---:
(8.2)
JLJ)
if and only if -
2
In
this
(JE
+
case
*JEJE)
we
have
(1, 0) -part
is
If (L, J)
is
a
C
S?'(Hom(E, HIE)).
operator
DO:=
S?'(L),o
-+
T(do
+
2
*d(JO))
by
given
AL 2.
differential
a
F(L) Its
S?'(Hom(E, LIE))
E
1 :=
2
(D
holomorphic
JDJ)
+
curve
1(Je
2
+
E
F(K
(L)).
End-
(8.3)
then
*JEJe)
=
7r-eALIE-
Moreover,
1(6E
2
In other
exactly 3.
Let
words: the
L be
congruence
The twistor
holomorphic an
S
+
0 4==
AL
1
0-
(SdS
curves
in
Cp3
are
=
holomorphic
4
=
of holomorphic projections in HP' with AL 0-
F(End-(H)), A
=
curves
immersed E
*JEJE)
curve
and J
=
*dS)
E
+
=
mean
SIL.
Then
1'(k
End-
satisfies
AIL
with
AL-
(H))
curvature
sphere
(i).
Proof,
(L, J)
If
is
a
holomorphic I
-7rL(dO
*d(Jo))
+
2
for
then,
curve
=
Projections
Twistor
8.1
49
F(L),
any
0.
But then
1(do
+
*d(J'O))
(E).
It
follows
2 a
fortiori
for
c F
all
2
17rE (do 2
Conversely,
+
(do
7rE
+
*d(JEO))
1(do
2
*d(jEo))
S?1 (L)
E
01 (LIE).
01 (LIE)
E
+
E
*d(JEO))
0
for
E
F(E) implies
E
S?1 (L),
and therefore
*6LIE Again
for
I(d(oJ)
0 +
2
E
=
6LJIE-
F(E)
*d(Joj))
2
((do)j
*d(JO)j))
+
2
(do
+
*d(JEO))j
N
E
S?1 (L).
V
ES-23-(L) This
shows
*6L
By the preceding,
=
01
maps into
1(-b
-
2
but
for
0
E
proves
(ii).
For
0
E
Its
(1, 0)-part
is
J
2
(*do
-
d(Jo))
=
-No.
(8-3). 0
E
F(L)
we
ALO But for
(L).
F(L)
*AO This
6LJ-
F(E)
we
have
=
I(do
4
have
+
J(dJo
*d(JO) -
*do)
+
=
J(dJO J(do
-
+
(8.4)
*dV))) *doi)i,
and hence
Projections
8 Twistor
50
4((do
ALO
!(do 2
By assumption Ej-component, ticular,
namely
7rE(ALO)
and
(iii).
0 if
=
17rE(do 2
linearity,
For
E
has values
1(do
IrE
and
only
*d(oi)) F(L) 4
(SdS
if 0
=
+
ALO =*
1(S(d(So)
-
I(S(d(So)
-
4
(8.4)
with
of lemma 8 this
In view
AIL
shows
Given circle an
a
ellipse
so-called is
surface
under in
quadratic
the
normal
Since
*d(SO)
*dO)
+
*d(SO)
Of
AL 0
JIL.
its
is
In par-
*JEJ)O)
+
ALIE
AL by
determines
*Sdo)
-
+
do).
AL.
following
the
curves of holomorphic Willmore vanishing
in
with
immersed second
space,
ellipse.
curvature
and
Cp3
are
ex-
functional.
Immersions
conformally
the
E G Ej,
0-
+
=
implies
Super-Conformal
0. =
(6E
Sdo)
1. The twistor Corollary projections the in HP1 curves actly holomorphic
8.2
=
=
(-i)-eigenspace
2
AL
*d(JO))i).
*dS),o
4
Comparison
+
L
in
in the
*d(Jo))
+
2
+
J(dO
+
the component
=
A0
*d(JO))
*d(Jo))
+
irEALO
+
R4,
into
fundamental
centered
The surface
the
at
is called
the
image of
form
(a
is
mean
a
double
curvature
super-conformal
tangential cover
vector, if this
of) the
ellipse
circle.
a
If
N and R
Proposition
7
are
we
II(X, and therefore
the left
and
right
normal
vector
of
dN(X)
*
f,
then
have
Y)
2
(*df (Y)dR(X)
-
df (Y)),
according
to
Super-Conformal
8.2
H(cos
OJX,
OX + sin
2
(*df(cosOX+sinOJX)dR(cosOX+sinOJX) dN(cos
-
I(df
(cos
2
2
OJX
+
+
H(cos
=
.1
(1
OX + sin
20),
OJX,
cos
OX + sin
(X)dR(X)
=-(*df 4
2
cos
-
sin
=!(I2
0
dN(JX)df (JX)).
-
cos
-
df (X)dR(X)
-
20)
we
+
4
+
4
dN(X)
df (X)
*
*df (JX)d ?(JX)
+
+
4
20(df (JX)dR.(X)
sin
20(df (JX)dR(JX)
dN(X)df (JX)
-
+
I
This
is
df (X)dR(X)
-
20(df (X)(*dR(X) 20N(a
circle
a
length. This only possibilities:
same
the
*
df (JX))
dN(X)df (,Y).
+
d1V(JX)df (X))
-
-
dN(JX)df
(JX))
12,
cos
sin
4
dN('JX)
df (X)dR(JX)
RdR(X))
-
-
(*dN(X)
-
+
NdN(X))df
(X))
=:b
=:a
+
-
=211(3X,JX)
cos
y (X)
=7 Ild
get
OJX)
=211(X,X) I
OX))
sin
(X))
dN(JX)df
-
dN(X)df (X)
+
2
OJX)
dN(X)df (JX))
-
0(df (JX)dR(JX)
0 sin
-
OJX))
sin
OX + sin
df (cos 0 JX
sin20(df (X)dR(JX)
cos
df (cos OX +
*
OX)dR(cos
sin
-
(COS2 0(df (JX)dR(X) -
Using COS2 0
OJX)
OX + sin
dN(cos OX + sin 0 JX)
-
1
51
OJX)
OX + sin
cos
Immersions
b).
only if clearly the
b and N(a +
if and
a
is
case
-
Assume that
N(a
if
a
there
+
b)
=
aR,
=
0
=
-
are
orthogonal
and have
0, but these are -1 with H, p2
=
P E
exists
P(a
b) b
or
in fact
=
b),
(8.5)
and note that Na
(8-5)
We multiply
-(a respectively. hence
a
=
by +
b)
=
0
or
=
0,
NP(a
(PN
Therefore b
the left
N from
or
a-
b
-
-
or
b),
NP) (a
=
Nb
=
bR. R from the
by
-(a -
b)
0. But then
+
=
b)
=
right
PN(a
to
-
obtain
b)
0, which implies P a+b 0, whence
also
=
=
a
J-N, and =
b
=
0.
52
8 Twistor
follows
It
Projections the immersion
that
*dR(X)
is
=
super-conformal
0,
or
*
dN(X)
if and
-
only if
NdN(X)
=
0.
choice of X, but preceding argument, this holds for a particular obviously follows for all X. Wemention that f -+ I exchanges N and R, hence f is super-conformal, and only if *dR RdR 0 for f or for 1. In view of proposition 12, this and to equivalent 0, AIL by Theorem 4 we obtain:
By
then if
RdR(X)
-
is
the
it
=
Theorem 5.
A
super-conformal twistor
=
-
projection
conformally if
and
of
a
immersed Riemann
only if
M 1
holomorphic
:
M -+ ELO, curve
in
surface or
Cp3.
If I I
f :
:
M-+ H =
M -+ HP'
RI is is
the
In this
section
we
describe
shall
a
Surfaces
of Willmore
Transforms
9 Micklund
method to'construct
new
surfaces
Willmore
depends on the choice of a point 00, and given one. The construction On therefore generously offers a 4-parameter family of such transformations. and therefore the other hand, the necessary computations are not invariant, ought to be done in affine coordinates. be hidden This fact will local: The transformation theory is essentially immersions. We shall also that the transforms in the assumption are again ignore period problems. from
a
9.1
Micklund
Let
f
:
Transforms be
M-+ ffff
w
surface
Willmore
a
=
with'N,
R, H, and
dH+H*dfH+R*dH-H*dN.
Then
dw and hence
integrate
we can
(Note
that
integral
the
of
of surfaces
f
we
to the
We denote to
prove
Using
.
construct
classical the
dw, (9.2),
=
assumption
Proposition
this
16.
=
g
problem.) is again
name,
new
we
g
M-+ H is
an
immersion
1W.
periods, a
so
transforms
general
in
Willmore
refer
with
(9.1)
surface
to the fact
examples from old
Biicklund
:
2
may have
this
show that
We want to
transform similar
w/2
of
of M. Weignore
covering
0',
Assume that
it.
dg
a
=
of
that
g is
defined
called in
a
a
only
on
Bdcklund
given category
by solving an ODE(9.1), see Tenenblat K-surfaces, [13]. a by subscript (-)g, and want ones
to g symbols associated of wg can be done under the weaker The computation which holds in the case above, see Proposition 14.
0.
Let
f
,
g
:
M-+ H be immersions
F. E. Burstall et al.: LNM 1772, pp. 53 - 59, 2002 © Springer-Verlag Berlin Heidelberg 2002
such that
Transforms
9 Micklund
54
of Willmore
df Then
f and
g induce
the
*
which
*dg
implies
wg)
using the conformal 0
For the next
=
df
=
A
-Rdg.
dg
structure
(9.3)
df
*
dg
(9.4)
-wdf.
=
-
HN=
2dfH=dN-N*dN,
Then
df (-R)dg,
Hence g is conformal, recall the equations
computations
by f.
induced
structure
=
M, and
on
-R,
=
-
(9.2)
Q.
=
conformal
same
dg(2dHg Define
dg
A
Ng Proof.
Surfaces
RH,
Ng
and
too,
(7.10),
and
=
-R.
(7.11),
(7.12):
-
2Hdf =dR-R*dR,
w=dH+H*dfH+R*dH-H*dN. Then Rw =RdH + RH
dfH
=RdH + HN
dfH
=RdH =RdH With
HdfH
-
dHN
-
*dH
*
dN
dN)
-
-
HdN
(9.5)
*dH.
-
dRH + RdH = dHN + HdN this Rw
HN dN
-
H(N
-
HdfH
RH dN
-
*dH
-
*dH
-
HdN +
-
*dH
-
becomes -
(9.6)
HdfH.
dRH +
Next
2dgHg
dNg
-
Ng
*
dNg
=
-dR
R * dR.
-
Therefore
-dg
A
dHg
2
d(-dR
-
R*
dR)
2
d(dR
R*
-
dR)
=
dH A
df,
or
dg(*dHg We now
use
(9.5)
+
(9.7)
and
RgdHg)
to
-(dHN
=
-
compute
wg) wg) dgRg (2dHg + =dg(-2RgdHg RgdHg HgdNg + HgdgHg dg (Rg dHg + *dHg) + dgHg (dgHg dNg) =(dHN *dH)df + dgHg(dgHg dNg)
Ngdg(2dHg
(9.7)
*dH)df.
-
-
-
-
-
*dHg)
-
-
-
-
-
=(dHN =(dHN
-
-
-
*dH)df *dH)df
I +
4 1
-
4
(dNg (dR
-
Ng * dNg) ((dNg
+ R*
dR)(dR
-
R*
-
Ng * dNg)
dR).
-
2dNg)
Bkklund
9.1
Similarly,
=
Rwdf
=
(dHN
=
(dHN
=
(dHN
f
dg(2df
+
-
-
(dR
-
Hdf)Hdf
-
*dH)df
-1(2dR 4
*dH)df
-1(dR 4
dR)(dR
dR + R *
-
dR)(dR
+ R*
by (9.1),
and g is defined
Willmore,
is
*dH)df
-
R*
-
-
R*
dR)
dR).
(9.4).
Comparison yields If
55
(9.6),
using
-Ngwdf
Transforms
2dHg
wg)
-
=
2dgdf
+
then
dg(2dHg
wg)
-
(2dg
=
-
w)df
=
0.
Hence
wg and g is Willmore, too. Now assume that h
g
=
2dh A
Proposition
16
df
We find
ward,
Wh
and h
reference
a
to 9
Note that is also
=
a
=
applied
-whdh
H is
again
2dH)
A
-
2df,
(2dg
(h, f )
to
=
df (2dH
-
whence h is
backward
Bdcklund
d(g
by integrating f is a forward
backward
-
Theorem,6.
(EV)*,
E
a
w)
Let
L be
a
<
d < g, h
:
M-+ H C HP'
(9-8)
Hg),
Then, by Proposi-
immersion.
an
a
-
H)
are
then
,
df (2dH
1w
=
-
df
>= 0
called
=
surface. h
0.
-2df A.
=
We call
g
a
for-
without
be obtained
can
of h because
!wg 2
-
depends in
df
dH.
this
surface
>= 0.
=
A
says
2dg)
-
2
clarifies
immersions
again Willmore surfaces, of L. The free choice
(f g)
of
transform
dg=2<0,*Aa>, transforms
2dH)
-
again a Willmore transform of f.
Willmore
0,
(w
=
of g because transformations
P, *Aa
are
=
Bdcklund
transform
IV such that
df
instead
The concept of Bdeklund theorem coordinates. The following
they
+
14,
tion
If
2d(f
dHg, on
!Wh, 2 (9.8). =
the choice
situation.
RP'.
df see
and
of affine
,
Choose
non-zero
0
E
Then =
d <
0, *Qa
that
satisfy
>
.
dh=2<0,*Qa>, forward
of 0 implies
respectively
that
choices pairs of Bdcklund transforms. (Different forms g -+ gA, or h -+ hA, for a constant A.)
there
of
a
is
result
backward Micklund a
whole
S' of such
in Moebius
trans-
56
9 Bhcklund
'hansforms
Proof.
Choose b E
H2,
2 <
Proposition
see
We can To do so,
(IV)*
such that
I
P, *Aa
>=
2
Surfaces
#,*Qa
2 <
W'.
fl
a, b and a,
are
1w
>=
-
2
dual bases.
Then
dH,
12.
proceed from integrate yields
g with
now
must
we
this
constant,
E
a
of Willmore
'Fw
-9
2
f
f
Hg). But,
+
+
Bdcklund
forward
another
d(f
up to
a
transform. translational
(9.9)
Hg.
We now observe Lemma10.
fil) Proof.
Note that
ker A
=
ker *A.
By Proposition
( )( ( if) (0 I
4*A
f
w
01
w
(0 1) f
the twofold
But this ker A
or
means
of
L
=
that
image Q.
Weobtain
a
chain
In
V
=-dN,+Ng*dN. 0
2dgHg
-
Bdcklund
surface
D
particular
h
the
Of course, it
the
may close
chain up.
may break
=
0.
f
transform
of A
zeros
HP1
satisfies
there
can
are
Q the 2-step Backlund simply as L periods arising. or
be described
no
transforms
-4f
-+
L
or
)
image Q.
L in
of Bdcklund
-+j-+
1
wHg+ dR+R*dR
away from
Willmore
a
1
0
f )H transforms
+
0
1
f
backward
have
0
0 1
Similarly
we
) ( _f) (f Hg) dR+R*dR) Hg) dR + R * dR
( ) ( 2dgHg 1
12
0
0
0 1
1
E ker A.
down if
g
-+f
-+
we
arrive
L
-+
at non-immersed
surfaces,
Two-Step
9.2
L C H
Let
a
57
Transforms
Micklund
M x E? be
=
'kansforms
Two-Step Bdcklund
9.2
surface,
Willmore
and
=fi
A
assume
0
each
on
Backlund
the two-step component of M. Wewant to describe directly form L -+ L, and compute its associated quantities (mean
trans-
sphere,
curvature
Hopf fields). We state see
fact
a
singularities
about
Proposition
Let L be
17.
proved
be
of
open dense subset
exists
appendix,
the
in
Weshall
and A
L
bundle
$
0
on
each
C H such that
on
andH=L(DL.
for image Q. L
that
assume
EIP',
in
line
M we have:
holds
assertion
unique
a
L=kerA, A similar
surface
Willmore
a
component of M. Then there an
will
that
13.
section
immersed,
is
again that
and want to prove
L
is
Willmore.
Q Hence
Let
again
is
S, 6, Q,
a
=
L
of
L
we
have
(9.10)
A.
surface.
Willmore denote
etc.
transform
Bdcklund
For the 2-step
Theorem 7.
associated
the operators
with
L.
Lemma11.
All
Proof.
Since
subset
of Mthen
For
0
E
=
F(L). 0
0
we
A(X) we
:
AE
interpret
HIL
-+
H is
f?'(Hom(H/!,, injective
H)).
for
any X
*A
*
On
54
a
dense open
to
-S rather
0.
get
=d(*A)o
=
d(*Ao)
+ *A A
do
=
do
+
Ado,
=0
AS * The
injectivity
(of
Proof than
to
the
of A then
theorem).
60
+
Ao
proves
Motivated
=
-AS(*S
SS)O.
the lemma.
by
the
lemma,
S. We put -S + B.
Then
+
we
relate
9
9 Biicklund
58
T ansforms
Q
=
=
=
=
The -
like
proof A
will
SdS
*dS
-
Bd9
-
Bd9
-
(Sdg
+
*dg)
(SdB
+
*dB)
Bd9
4A +
(SdB
-
*dB). lemma which take
values
in
shows that
L
image Recall
L
that
is S-stable.
immersive, FrBdo
-kBdo
for
+
0 for
0
=
*Sdo
=
-*so+
F(H)
X E
-R(*dB
=
kerB,
(9-13)
0
(9.14)
C
C L.
9-stable,
also
course
and therefore
L.
(9.15) =
follow
*SO
=
S-Rdo
+
9-kdo
=
SSO+ 9SO
=0.
have
we
SdB)X
(9.12)
F(L).
E
irgdo
+
15,
c
image S HIL Thus (9. 11) will But, using Lemma 11,
and therefore
show
SdB)
is of
It
B!,
=
-k(*d(BX)
+
Sd(BX)
-
B
*
dX
-
SBdX)
L-valued
This
proves
=
(*J
=
0.
+
SJ)Bx (Lemma 11)
(9.12).
On the other
Fr(*dB
hand, for 0 -
SdB)o
F(L),
E
=
ir
=
ir
(*dS
SdS)o +Fr(*dg
-
-
Sdg)o
=-4QO=O
(*dg
+
gdg)o
--k
(Bdg)o EF(L)
=
Together
Q
(9.11)
B C +
L c
Next,
*dS)
+
We have
image(*dB
we can
+
(SdS
completed with the following in L, while the "B-terms"
be
image
Now L is
+
has values
-
Lemma12.
Proof.
Surfaces
of Willmore
with
the previous
0.
equation
we
obtainTrdB
IL
=
0, and, for 0
E
F(L),
if
Two-Step
9.2
SBO -k(d(BO)) =
L
But
is
an
9d90
-
9d9o
-
9(dgo =
=
7r(d(SO)
But is
S
under
stable
59
0.
=
(9.13).
*d9O +
9
9do
+
*do)
+
*d b)
do +
(9 *) (d(90) -(9 *)(d(So) -
-
*do)
-
*dBO
=
0, proving
=
T ansforms
F(L),
C-
4Q50
Bdo)
-
BO
and therefore
immersion, for
Finally,
*((dB),O
=
Micklund
=
(6S
=
B
-
-
do
*
-
(-do
=
S. Therefore
0.
+
*(*do
9
+
*d9O) 9dO + d9o) do
*
+
using (9.13).
*do)
-
*6)0
-
So
QL
d(SO)
MO-C 1-(L),
-
Since
c L.
QL
=
and this
0, this
proves
(9.14). Taking L.
the two-step
Hence L
to the
corollary
=
L.
backward
of
(9.10)
backward transform
We remark
two-step and its
that
the
Bdcklund.
of
results
we
of this
transformation
analog A
Q we obtain
Theorem 8. L
L,
=
L
=
L.
get image Q section
L
=
image A
similarly apply image Q. As a
10 Willmore
Let
<
> be
-,
an
Surfaces
indefinite
SI
in
hermitian
product
inner
on
EV. To be specific,
an
S3
we
choose < Vi
Then the
set
complementary
of isotropic 4-discs
1, 1 hyperbolic
lines are
IT1W2 + IT2W1
W >:--:::
<
> = 0 defines
4-spaces,
see
C
Example
RP1,
while
(a db)
(10.1)
C
and the
same
for
holds
matrix
the
4. Wehave
with
representations
respect
to
a
basis
(v, w)
such that
=O=<W'W>'
Surfaces
10.1 Let at
L be the
an
adjoint
stabilizes
Lj-,
S3
in
bundle
line
isotropic
map M -*
and L
=l.
=
L
Z7p 1
with
SP*
-+
mean
with
to
implies S*L
S*L'
=
=
L'
=
L.
Similarly,
(dS*)L Moreover,
if
Qt belongs
to
S*,
1
(dS)*L
=
C L
then
Qt
(S*dS* 4 -
(dSS
I(SdS
4
-
-
*dS*)
*dS)* +
-A*.
F. E. Burstall et al.: LNM 1772, pp. 61 - 66, 2002 © Springer-Verlag Berlin Heidelberg 2002
sphere
curvature
respect
*dS)*
L.
<
>.
S. We look
Clearly
S*
10 Willmore
62
kerQt
Therefore
We proceed
S*
=
S3
in
(image(Qt)*)'
=
sphere,
curvature
mean
(imageA)-L
=
S and S*
show that
to
of the
uniqueness
Surfaces
L'
D
coincide
on
L and
HIL.
By
Theorem 2, it then follows
see
the that
S.
Let
0
E
1'(L),
and write
SO
=
OA,
S*O
0,60
>=
Op
and <
Note that
0, 0
0, JO yields
<
>= 0
because of <
> makes sense,
W+C0
From 0 =<
0, So
>
we
=
L >= 0. Differentiation
of
0.
obtain
JO, SO>+< 0,(dS)O
0 =<
0,
%-
0,SJO
>+ <
>
I.,
=0
=<
JO, SO >
=<
JO,O >A+ P
=
Now we
apply
*
+ <
S*O'JO
>
O,JO
>
<
CDA + pw.
using *W
=<
0, JSO >= WA,
(10.2)
and obtain 0
Weconclude
A
Now assume S I
i.e.
HIL
=
AOA+ pwA
SIL =
=
I p
0 =< =<
S*
and
IL
(p
-
)CDA.
-
S* I HIL
50' so
JO'O
I
=
> + <
S*JO' 0
=& <
=
Then
S*0,50
> + < > +
u.
>
0, SJO >
> P
&6j + WP.
But *(.,)
=<
=<
0, JSO >=< 0, S60 S*O' 90 >= XW.
>= WP
10.2
Comparison
(10.2)
with
shows p 0
It
follows
a
A
=
& W + WA
=
SIHIL
i.e.
p,
=
A, and
=
63
&)WA.
-
S*IHIL.
=
2-Planes
get
we
(
=
Hyperbolic
completes
This
the
assumptions
of Theorem 2, and S* = S by uniqueness. if S* = S and So OA, then Conversely, =
0,0
A < Now S2
>=<
Proposition i.e. a surface
S',
if
Hyperbolic
10.2
0,,0
>=<
1, and therefore
we
An immersed
18. in
0,SO
>=<
A2
implies
-I
=
SO,O
and
holomorphic only if S S*.
>
get
A <
=
A <
0, V)
>.
> = 0.
HP1
L in
curve
0,0
is
isotropic,
=
2-Planes
model of the hyperbolic are space, geodesics the the boundary. intersect We consider orthogonally models of hyperbolic their 4-space in HP', and want to identify totally those 2-spheres i.e. in RP' that 2-planes, geodesic hyperbolic orthogonally S3. Using the affine coordinates, the separating from Exintersect isotropic H -+ H, x -+ -.t S3 This at Im H ample 4, we consider the reflexion of the metrics either given in the examples of section 3.2. In parpreserves of the standard it induces Riemannian metric of RP' an isometry ticular, S3 which fixes S3. Given a 2-sphere S E End(EV), S2 -I, that intersects in a point affine in and with 1 as w 1, we use coordinates, Example 4, =:F vH In
half-space
the
euclidean
circles
Poincar6
or
that
=
.
=
such that < V,
>=< W,
v
W>=
0,
< V,
W
Then
S N2
with
=
R2
1, NH
=
(0
N -H
=
-R
HR, and S'
is the
C Ifff
locus
of
Nx+xR=H. If S' is invariant
iR
=
H
under the reflexion
at
S3,
is the locus
also
of -Nd
-
or
Rx + xN
According
then it
to section
=
T1.
(H, N, R)
3.4, the triple
unique
is
up to
sign. This implies
either
(H, N, R) By (10.1) intersects
either
S*
orthogonally,
=
=
(ft, R, N) S,
and the
and S*
=
(H, N, R)
or
2-sphere
-S.
lies
-R, -N). within
We summarize:
the
3-sphere,
or
it
64
Proposition termined by S*
=
in
S3
A 2-sphere S E Z inner product indefinite
the
intersects
19. an
hyperbolic 2-planes
hyperbolic
in
de-
4-spaces
if
and
only if
-S.
10.3
Willmore
Hyperbolic Let L be set
Surfaces
10 Willmore
of
Surfaces
Surfaces.
in
4-Space Willmore
connected
a
indefinite
an
S' and Minimal
in
surface
hermitian
form
S3
in
C
HP',
where S3 is the
Then its
H.
on
mean
curvature
isotropic sphere
satisfies
S* Let
us
assume
B.icklund
A
that
transforms
$ 0,
and let
S.
L
ker A and
L
L.
image Q be
the
2-step
of L.
Lemma13.
Proof.
First
have
we
I
Q*=
4 I
4
(SdS
(-SdS
imap Q is S-stable,
Now
Therefore
<
L, L
L
and
0,
>=
L'
=
=
-
*dS)*
*dS)
-
and S* on a
4
=
=
(dSS
-
-A.
S and
(10-3) So
=
dense open subset
(image Q)'
=
*dS)
kerQ*
OA imply
<
0, 0
> = 0.
of M
ker A
=
Lemma 14.
-S
for
the
Proof. and
mean
First
sphere 9 of L.
curvature
L
=
L
is
Q and, therefore,
under I
4
and this
vanishes
sphere by
these
on
three
(-S)-stable. d(-S) 2(*A
obviously
((-S)d(-S)
=
-
It -
*d(-S))
trivially
is
=
L. The unique characterization -S. properties implies 9 =
invariant
*Q). Finally,
the
under
Q of (-S)
A, of the
mean
curvature
A is
10.3
Surfaces
Willmore
We now turn
in
to the
S3 and Minimal Bdcklund
1-step
Surfaces
transform
d(F+F*)=2*A+2*A* Because S*
We now F is
a
=
use
S,
we can
affine
Bdcklund
initial
F+F*
-S.
with
[,I.
transform
f,
of
L
9
g +
(9.9).
1
Likewise,
for
(10.4)
sphere Sg.
0 1
0
left
From the
0 1
9 -fi
1
g of
entry
properties
of
(10.5)
From Lemma 14
1
then
imply
we
obtain
0)
0
A,
F such that
Then the lower
NO
H
*
(10.4)
(0 1) ( R) (1-f)=(lf (R (1 (1 f) (1 Hg) ( -j ) (1 -Hg) (1 (1 f) (9 HgfI (1 -f) if
2
Hg=f-f,
-R..
=
=
65
H.
=
Nq=-R,
(9.3),
conditions
and
We want to compute the mean curvature Bdcklund transforms we know
see
of L. If dF
I
(7.9)
and
4-Space
Hyperbolic
2*A-2*Q=-dS.
=
(10.3)
choose suitable
coordinates
in
-
ft
-f
0
0
0
1
0
-f 1
-
01
This
implies
H
and -N
-Rg In
particular
on
that
set.
f It
E Im
-
follows
Sg and, because, R
=
=
g
=
H,
N=
=
since
HH,
-
-N +
H
0
=
whence
(f
on
1
-
an
f)H. would
open set
that
(I 1) (
-R
g
0
f
-I
-g)
0 -N +
N and H E R for
f
(f
-
I)H)
O
M-4 Im H =
:
1
R3,
(I H) (-N ; (f-f)H g), (Ig-H) (N+(I-f)H 0) (1H-g) (I g) ( 1-f N+(I-f)H ) (i -g) g
S*
0
-
01
-
0
1
0
1
f
-S9-
f
f
N
01
H
0
0
N
1
0
0
1
-
1
mean w
0
Surfaces
10 Willmore
66
We have
the
shown that
now
thogonally,
and therefore
coordinates
and
are
Euclidean
a
S3
in
mean
g and have the same mean curvature under conformal changes of the ambient
g has
mean
and the
"Micklund
generate
minimal
1w, 2
of
dg
in
hyperbolic
=
but
1 =
4
(SdS
and therefore
+
Let L be
S*
with
i.e.
1
*dS)*
4
This
hyperbolic
in the
If A
is minimal.
remains
property =-
0, then
which may be considered in the (isolated) be singular
g will
w
=
as a
0,
de-
zeros
-S.
(dSS
holomorphic
immersed
an =
I
*dS)
-
minimal
curve,
Then
4
(SdS
+
*dS)
-A,
also
(d From
g.
Therefore,
elsewhere.
converse:
4-space,
as
metric.
constant,
general
In
minimal
Weshow the
A*
'ansform"
surface.
is
or-
that, using affine spheres are tangent
curvature
vector
0, and hence
curvature tr
S'
of g intersect
Weknow
planes.
to
metric,
spheres
curvature
mean
hyperbolic metric, the
Proposition
15
we
*
A)*
=
-d
A.
*
have
(f
I
A
d
4
dw
dw
-f
dw
-dw
f)
f
Therefore
jw-,
dw
f
dw
=
dwf,
and hence
dw(f f
But sition
is not in
S1,
dw=O,
and therefore
*A
=
-A
the backward
conditions To
f)
=
i.e
0. L is Willmore.
Similarly,
Propo-
yields
12
and A*
+
implies Bdcklund
w
=
-77D.
transform
=
(W
Rom S*
h with
dh
-S
1w
we -
know TI
=
-H,
dH and suitable
and
initial
is in Im H = R.
summarize
hermitian > be an indefinite product [11]). Let < lines form an S' C HP', while the two complemen-. isotropic Let L be a Willmore metrics. complete hyperbolic surface tary discs inherit Then a suitable in S' C HP'. forward Bdcklund transform of L is hyperbolic minimal. that is hyperbolic an immersed holomorphic curve Conversely, minimal is Willmore, and a suitable backward Bdcklund transformation is a Willmore surface in S'. (In both cases the Bdcklund transforms may have
Theorem 9 on
IV.
(Richter
Then the
singularities.)
.,.
Spherical
11
In
this
chapter
[8],
Montiel
spheres
suitable
we
which
projection
a
an
following result
in
HP'
affine
[2],
[8]).
Montiel
differs
from
and therefore
what
we
global
[1]
Bryant
A Willmore
have treated
methods
Complex Line Bundles:
of
Degree
of
theorem of
of a holomorphic or anti-holomorphic coordinates, corresponds to a minimal
requires from complex function theory.
11.1
of the earlier
proof
generalizes
(Ejiri
The material
global,
sketch
Surfaces
S3. See also Musso [9].
in
Theorem 10 twistor
Willmore
sphere
so
proof.
and
far:
in
and
Willmore
EEP1 is
Cp3'
in
a
or,
in
The theorem
is
curve
surface
[2]
Ejiri for
in
These
R1.
are
imported
Holomorphicity
complex vector bun le. We keep the symbol J E End(H) for the with the imaginary unit i. endomorphism given by multiplication We denote by R the bundle where J is replaced > is a by -J. If < hermitian metric on E, then Let
E be
a
.,.
R -+ E*
=
E-1, 0
-+<
0,.
>
isomorphism of complex vector bundles. Also note that for complex line E1, E2 the bundle Hom(Ei, E2) is again a complex line bundle. There is a powerful for complex line bundles E over a integer invariant Riemann surface: the It classifies these bundles up to isocompact degree. Here two definitions for are the morphism. equivalent degree. is
an
bundles
-
Choose
a
hermitian
Then < R(X,
R(X, Y)
metric
<
.,.
>
and
a
compatible
Y) 0, 0 > 0 for the curvature tensor (X, Y) J with a real 2-form w E fl2 (M). =
deg(E)
1 :=
27r
F. E. Burstall et al.: LNM 1772, pp. 67 - 72, 2002 © Springer-Verlag Berlin Heidelberg 2002
fm
W.
connection
R of V. Define
V
on
E.
Therefore
68
-
Spherical
11
Choose
Willmore
0
section
a
E V
Surfaces
(E)
with
deg(E)
isolated
0
ord
:=
HP1
in
Then
zeros.
E
:=
indp 0.
O(P)=O of
The index
is defined
0
p of
local
section non-vanishing b z (0) holomorphic parameter z '0 (z) A (z) p. Then 0 (z) for some complex function A : C C U -+ C with isolated zero at 0, and
and
a zero
using
a
where -y is
circle
small
a
fundamental
We state
deg(B)
0.
of the
properties =
=
dA
2-7ri
around
=
ZY A(z)'
1
indp
degree.
deg E-1
=
deg Hom(Ej, E2) More
a
for Mwith
We have
deg E,
-
deg El
+
deg E2.
generally,
deg(El
0
E2)
deg El
deg E2.
+
Example 21. Let M be a compact Riemann surface of genus g, and E its tangent bundle, viewed as a complex line bundle. We compute its degree The curvature of a surface with Riemannian tensor using the first definition. metric < > is given by R(X,Y) K(< Y,. > X- < X,. > Y), where =
.,.
Gaussian
K is the
compatible
with
W(X' Y)
curvature.
=
1traceR 2
(< Y,
_
<
=
=
K we
:=
therefore
this
=
Z ><
a
(local)
unit
field
vector
and <
(<
X'Z Y, Z
J
X,
X,
<
X,
JZ > + <
X,
Z >
X,
JZ >< <
K dA(X,
using
For the
E-1
Kdet
Y)
JZ ><
Y,
K(< Y,
deg(E).
R(X,
2
K
27r
Z be
J. Then
_
We integrate
Let
Z >
> < > <
<
-
X'JZ Y, jZ
X,
JZ >< Z ><
JZ ><
Y,
Y, Y,
Z >
JZ Z
>)
>)
>
Y).
Gauss-Bonnet,
canonical
-
and find
21rX(M)
=
27r(2
-
bundle
Hom(TM, C) ='fw
E
HomR(TM, C) I w(JX)
find
deg(K)
=
2g
-
2.
=
iw(X)}
2g)
.,
>
Complex
11.1
Definition
for
Let E be
11.
a
complex linear complex anti-linear
E is
valued
complex
map
a
Bundles:
Line
bundle.
A
vector
0 from the RE
Holomorphicity
holomorphic of E
sections
69
structure into
the E-
-+.V(KE)
r(E)
:
and
map
a
1-forms a
Degree
satisfying +
6A
Here
if (90
=
:=
!(dA+i*dA). 2
0.
We denote
(Local)
E is
If
vector
then
holomorphic deg E < 0,
maps preserve
then
any
proof and holornorphicity
are called holomorphic, sections of holomorphic
space
and of
In
orientation.
and
structure, index
positive
particular,
V)
E
because
if
M is compact in E vanishes identically.
and
concepts.
If (L, J)
12.
is
a
&
complex line
:=
10
E L
bundle,
then
I JO =,Oil
We define
bundle.
If LI, L2
line
complex quaternionic
deg Lemma15.
holomorphic
isolated
are
global holomorphic section Ejiri theorem we shall apply the concepts of degree to several complex bundles obtained from quaternionic
these
Werelate
Definition
0
of
zeros
of the
In the
ones.
the
with
bundle
line
complex
a
HO(E)\10},
a
the
U.
over
is
0 EF(Eju)
sections
by HO(Eju)
deg EL.
L
line
complex quaternionic
are
bundles,
and Ej
:=
ELj,
then
Homc(El,
Hom+(L1, L2)
is
an
of complex
isomor Phism
vector
bundles.
deg Hom+(L1, L2) The
proof
Example
is
22.
straightforward. Weconsider
immersed
In
deg L,
We now discuss an
E2)
BjEj
B
particular +
one
holomorphic
deg L2 example
in
detail.
curve
LCH=MxEV in
HPI with
vector
bundle,
S. For B E r
mean
the
(K
curvature
complex
End-
(H))
sphere S. The bundle K End- (H) is a complex with being given by post-composition
structure we
define
11
70
Spherical
Surfaces
Willmore
(,9xB)(Y)V5
c9x(B(Y),O)
=
HP1
in
B(OxY)o
-
-
B(Y)o9x0,
where
Oxy:=
0,0 Direct
that
=
I(d+S*d),O,
on on
V, Y1
YD'
J1jx'
+
i90=_I(d-S*d)0f6r'0E.V(H). 2
shows that
computation induced
and 0
2
2
is in fact
this
holomorphic
a
namely
structure,
on
K End-
by 6
-
(H)
K Hom+(TI,
=
H)
TM, and the above (quaternionic) ft.
=
K Homc(ft,
holomorphic
H) 0
structures
on
H
Lemma 16.
(d Proof.
Let
mark
X be
12, and- 0
(d
*
a
E
*
=
holomorphic
local
F(H).
A) (X, JX)o
A) (X, JX)
-
2 (Ox
A) (X).
field,
vector
[X, JX]
i.e.
=
0,
see
Then
(-X
=
A(X)
-
(JX)
-
-
SA(X)
-
A([X, JX])o 1--le-I =0
-(d(A(X),O) + A (X)
(do
+
+
do (X)
*d(SA(X)O))(X)
+ SA(X)
*d(So))
(X)
+
*
d b (X)
A(X) (do
-
S
*
do) (X).
Now
do
+
*d(SO)
=
=
=
=
=
(c9
0
+
Q)o
+ A+
+
*(,9
0
+
+ A+
Q)So
(0+O+A+Q)o+(S0-S5+SA-SQ)So (,9+O+A+Q)o+(-c9+O+A-Q)o 2(6 + A)o 20(A(X)0)
+
2AA(X)O.
Similarly
(o9+5+A+Q)0-S*(o9+O+A+Q)0
do-S*do= =
(a
+
0
+ A+
=
(0
+
0
+ A+
=
2(0
+
A)O.
Q)o Q),0
-
-
S(SO (-,9
SO + SA
-
+
6
-
A+
-
Q)o
SQ)0
Re-
Surfaces
Willmore
Spherical
11.2
71
Therefore
(d
A) (X, JX),o
*
-20x (A(X)O)
=
-2(0x(A(X)0) _2(6xA)(X)06A
=
A E Ho (K EndAs
a
consequence,
exists
and there zeros
bundle
line
a
of A. For local
(H))
0
E
F(L)
6A (Y)o
L
A
0,
_=
or
O(A(Y)O)
=
under
A E Ho (K
Spherical to the
We turn
Proof
(of
A
0.
This
of A
are
=
implies
the
zeros
isolated,
=
the
A(Y)&O.
-
49, like L is structure holomorphic above, get A defines and a K Hom+(TIlL, holomorphic L)
11.2
*
ker A away from C H such that Y we have and holomorphic E HI(TM)
a
we
2A(X)2,0
=0
invariant
is
mark 6. As
d
L
L
=0
Therefore
+
Ho (K Hom+(TI, H)).
=
Lemma23, either
see
2A(X)L9xO
and therefore
L is
that
assume
+
A(X)c9xO)
-
Willmore, 0, and A is holomorphic:
Now
2A(X)2,0
-
Hom+(RIL,
under
invariant on
the
see
Re-
bundle:
of this
section
0,
complex line bundle
L)).
Surfaces
Willmore
.
Theorem
10).
If A
0
=-
or
Q E 0,
then
L is
a
by
projection
twistor
Theorem 5.
Otherwise coincides
we
with
Proposition holomorphic
image of Q almost
the
We have
20.
line
A E
bundle
have the line
We proved
holomorphic
following
the
L)),
JLEHO(KHom+(L,H/L)), if AQ the
=
a
line
bundle
sections
about
in the
then
Q E Ho (K Hom+(HIL,
L
that
of complex
yields
LA
AQEHO(K Hom+(HIL, L)) 2
J1
0 then
statement
appendix. The degree formula
others
similarly everywhere. and
bundles:
Ho (K Hom+
and
L,
E
Ho (K Hom+(L, HIL))
A. We give
the
(similar)
proofs
of the
72
Spherical
11
Willmore
ord
Surfaces
JL
ord(AQ)
=
deg
=
2
deg
K
3
deg
K
6(g
S2,
For M=
i.e.
g
=
0,
K
-
IFffP1
in
deg
-
-
+
deg
L
6L
ord
-
get ord(AQ)
we
deg HIL
deg HIL
-
1)
L +
ordJL. <
0, whence AQ
0. Then
and ordA
=
deg
ordQ
=
degK
=
deg
ord
Addition
H
follows =
mean
-
K+
degH/L degH/L deg L
-
+ -
degL
degL
deg
HIL.
yields
ordJ.L
It
6.L
K+
that
Mx
IBF.
curvature
coordinates
with
corresponds
to
+
ord
ordQ
51
<
+ ordA
0,
i.e.
=
3
degK
=
4
deg
K
-
-
deg HIL ord
JL
+
=
-8
0, and L is d-stable,
6i
A
degL -
ord
JL.
hence constant
in
L. Therefore all spheres of L pass through the fixed point Choosing affine L oo, all mean cur vature spheres are affine planes, and L
From AS
=
-SA
0
=
a
minimal
surface
in
W.
we
conclude
tranforms
12 Darboux
transforms
Bdcklund of
out
a
given
one
equation. After
in
again
is
the
local.
displays
it
Riccati
Let
Let
differential
partial
type
Riccati
we
describe
first
constant
mean
with
Dar-
curvature
the
Willmore
in
-
Willmore
the
case
-
A be
a,#
any p E
equations
in
algebra
over
an
algebra,
which
reals,
and M
End(EP). E
an
R\ 10},
T solution a unique Moreover, if S: M-
S2
=
unitary
associative
S?1 (M,
A)
A
P
=
=
0 0
=
pTa
0,
on
_1'
a
T
-
connected
d,
=
To
=
the
with
po E M and
dT has
theory
the
transforms
a
for
for
similarity
striking
a
da
Then
[6], namely
of
a
[6].
equations,
Riccati
on
case
of
equations
be H or
Lemma17.
manifold.
in
solution
of a solution to the Riccati existence We only have a local in to be invertible problem, and moreover require this solution
Riccati
We consider
a
is described
the
End(IV).
algebra
below will
special
Bdcklund
the
value
12.1
on
provide
transforms
thereafter.
As with
initial
a
R, because
treated
case
for
based
surfaces
Willmore
new
Darboux
equations.
it
remark
introductory
an
boux transforms
surfaces
surfaces
For isothermic
construct
to
a mean
linear
construction,
such
for
method
another
provided by solving
PAa.
E
A the Riccati
T(po)
=
neighborhood
initial
and
F. E. Burstall et al.: LNM 1772, pp. 73 - 81, 2002 © Springer-Verlag Berlin Heidelberg 2002
problem
(12.1)
To
of
A with
Sa+aS=O,
value
dS=a-0,
po.
tranforms
12 Darboux
74
(T
S)2(po)
-
p-1
=
then
(T and T'S
everywhere,
0
is
condition
=
pdT
pTdaT
=
p(pTaT
=
-pp
A aT +
A
Now, if
p-'dX
linear
a
dS)(T
=
(pTaT
=
Ta(pT2
=
TaX + XaT.
=
0
-
pTa
S)
-
-
pTS
0
A
d#
pTa
+ P
S)
+
1)
+
(T
-
-
dfl
above, then
as
we
define
equation -
dS)
S) (pTaT
2-
(pT
0)
-
1.
_
S)(dT
-
(pTaT
do
-
differential
(T
A
and S
S)2
-
+
-
-
-
solution,
a
p(T
a) (T
-
+
order
(dT
A dT
pTdaT
T is
first
=
-
pTa
-
T+
a
X:=
Then X satisfies
(12.2)
(12.1)
for
pTdaT
A aT +
satisfied.
obviously
0)
-
P-1
=
S2T.
=
integrability
The
Proof.
S)2
-
pST
-
a)
-
1)aT
+ aS + Sa =0
X(po)
Hence
implies
X
0. The last
=
T 2S_ ST2
together
f
:
dN is
M-+ Im H be
a
"'tangential"'
the K- and
by
multiplication
conformal
a
Since
that
-Nw.
is also
part,the Then
-
S(T
_
S)2
Ndf
=
1-form:
ff-part
surfaces
with
in
R'
immersion: =
it
N2
-dfN,
anticommutes to
the
Hdf
+
respect
=
_1.
with
complex
N.
We decompose it
structure
given by left
N to obtain
dN
traceless
S)2S
curvature
mean
*df
into
_
from
(12.2).
Constant
12.2 Let
with
(T
=
of the lemma follows
equation
the
=
decomposition
function
H
:
M ---
of the R is
(12.3)
w.
shape operator the
mean
into
"trace"
curvature,
and
and *w
Constant
12.2
1(dN
(12.3)
Note that the
into
shape operator
Therefore
part. get
(12.3)
Differentiating
curvature.
=
of
and the traceless
trace"
"
2 * Q dS f, and (12.3)
the formula
resembles
Now -dN is the
R3
in
75
dN).
+ N*
2
surfaces
curvature
mean
2
-
*
A.
gives
decomposition
its
M -4 R is the
H
mean
we
O=dHAdf +dw.
parallels
of Willmore
that
Wenow
theory
the
Wesee that
only
if and
Hence H is constant
assume
if dw
=
of constant
mean
surfaces
HP1.
in
54
H to be constant
d
0, resembling
*
Q
0.
(=cmc)
curvature
in R3
surfaces
0. Then
IN
g:=
H
satisfies
dg
=
1
df
dN
*dg
-
=
=
dN
jy
H(df
dg),
-
-Ndg
=
dg N,
and
df Adg=O=dgAdf of constant
The map g is an immersion of f, i.e. away from curvature surface of f.
by type.
from the umbilics mean
For
simplicity
0. It
=
to the
ourselves
restrict
we
w
mean
is called
H away
curvature
parallel
the
constant
case
H = -1.
(The general H P
,
g -+ tig
We put
case
with
A
[t :=
can
be reduced
-H.) End(p,
using the homothety
f
Pf,
-+
H
=
Therefore, value problem
lemma 17.
for
dT
(locally)
has
in Im
because T satisfies
A
to this
unique
a
=
any jo
=
solution
df dg, 0 0 O,pO E =
pTdgT T
the
,
-
These match the
.
M and To E
T(po)
df,
which
we assume
same
equation
fO
f +T.
We put =
=
assumptions
ImH\10}
the
of
initial
To to
up to
have a
no zeros.
minus
sign.
T
stays
76
Darboux
12
tranforms
Then
*(df
*df
+
dT)
-TNT-'
(df
0 is
This
shows that
f
is
immersion
if and
an
=
T2
=
Hf
-IT12
o,
+
dgT
*
dT)
=
conformal
only
Under what conditions
compute HO :=
pT
-pTNdgT
=
M:=
-TNT-'pTdgT
0.
-TNT-ldf
with
=
Nf
o
=
-TNT-'.
Moreover,
if g is an immersion. if g is immersive. does f 0 again have constant mean curvature?
f We
using
,
TN+NT=TN+TN=-2,
and dN A
df
=
Hdf
A
df
df
0
Hdg A df
-
=
Hdf
A
df.
Wefind
HOdf 0
A
df
0
=
=
=
dNO A df
-d(TNT-1) -(dTNT-1
=
-(dTN
=
(-(pTdgT
=
-p(Tdg(TN 2 <
A
TdNT-1
+
+ TdN
df)N
-
T,N
-
+ >
TNT-1dTT-1) TNT-1dT) A pdgT -
NT)
-p-I
T,N
>
-jo-1
we
0 A
A
A
pdgT
pTdgT
df
proved
Lemma 18.
HO Next
we
T,N
>
-p-I
IT12
show
Lemma 19.
Proof.
2 < =
If HO is constant,
We differentiate
0
=
then H
HOIT12
+ 2 <
=
T,
A
TNT-' (pTdgT
+
p-'Tdg))
+
df
IT12 Hence
Tdg
pTdgT
IT12 2 <
-
-1.
N>
-p-1:
pTdgT
-
df))
A
pdgT
Constant
12.2
0
=
HO < dT,T
=
HO(< pTdgT,T
=
HO(-IT12p <
-
<
T,df
-(HOIT12P
=
2p
=
-(HO +
=
P
2
T,N +
2
+
2
_
><
<
+
1)
1)
+
1)
<
T, df
<
(TdgTN
in
77
>
pTdgT,N
>
-
<
pTdgT,N
> + <
T,df
> + <
T,dg
>
-(HO + 1) < T,df > + < pTdgT,N -(HO + 1) < T,df > + < pTdgT,N >, >
>
>
>
NT) (Tdg
<
<
surfaces
curvature
>
T,dg
T,dg
>)+
HO < df,T
-
T,dg
T,dN
> + <
df,T
<
-
T,dg >)
+
dgT)
(TdgTN
-
NTdgT))
+
T,df
(Tdg(TN
-(HO +
1)
((TN
-(HO +
=
<
>
> + <
=
dT,N
> + <
mean
NT)
+
T, df
(TN
+
NT) dgT
+
(TdgTN
-
NTdgT))
>
TdgNT + TNdgT +NTdgT
+
+
%
-
(TdgTN
+
NTdgT))
-V
=0 =
-(H#
If HO
Now
+
1)
1,
we are
df and
<
T, df
>.
Otherwise
done.
T, df
<
dT
=
dpN + pdN
djzN
dT
=
pTdgT
PP2 NdgN
w are
df
-
tangential,
and
ydf
-
T
0, i.e.
>=
=
jLN, and
+ pw -
df
=
ptt2W
of the
comparison
-
d'
above two
equations
gives
dl-t
(I dl-t
=
_
tl)df
0,
=
(_tl
+
P,12)W,
0 and therefore
But then
f
g is the
parallel
constant
mean
curvature
surface
has HO = -1.
As
a
consequence
Lemma20.
of the
preceeding
HO is constant,
we
f which
obtain
if and only if
(T Proof. Weknow that with equivalent
two results
of
-
N)2
HO ist constant,
=
P-1
if and
(12.4) only
if it
equals
-1,
and this
is
>
tranforms
Darboux
12
78
IT 12
2 <
-
T,N
>
+p-1
0.
=
But
(T
can
defined
globally
Definition mean
=
-(ITj2
N12
-
13.
f
Let
(12.4)
p-1,
-
T,N holds
simpli
To
>
2 <
T,N
+p-1)
+
E
>
p-1
everywhere,
+1) -
I.
if it holds
This leads
connected.
no zeros. us
to the
conformal immersion with cmc surface f parallel g
a
and immersed
R\101,
-
and T is bounded with
M-4 Im H be
:
-1,
H=
curvature
1
=
-(IT12
=
2 <
-
if Mis
p E
and
-IT
IT- S12
Therefore be
=
from lemma 17 that
Now recall
point.
N)2
-
=
ImH\101,
in
a
single
Hence it
following constant + N.
Let
M,
po E
assume
(To Let T be the
of
solution
unique
N(po))2
-
=
P-1
_
initial
the Riccati
(12.5)
1.
value
problem
T(po)=To.
dT=pTdgT-df, Then
fO:=f+T called
is
Remark 11. should
1.
be
of f.
transform
Darboux
a
0 0, -1,
If H is constant
replaced
(HTo It 2.
turns
From
out
(12.5)
Hence there
that
We summarize Theorem 11. vature
H in R'
a
(12.5)
in the
above definition
+
N(po ))2
H =
1.
P
+ T has
f
again constant mean curvature H. a given p 54 0 there is an S2 of initial of Darboux transforms. 3-parameter family
we see
is
then
by
the
that
for
previous
To
results:
The Darboux
transforms
have constant
mean
of surfaces
curvature
H.
with
constant
mean cur-
Darboux transforms
12.3
Darboux
12.3 Let
L C H
sphere
S,
=
transforms
Mx EV be
and dS
End(H2)-valued
a
2(*A
=
maps
of Willmore
Willmore
-
of Willmore
*Q).
surface
Since
F, G, locally
d
*
79
surfaces HP1 with
in
A
surfaces
Q
d
=
mean
0
=
curvature
we can
define
Mby
on
dF=2*A,
G=F+S.
Then
dG = 2
dS
dG A dF Hence the with
a
=
integrability dF dG, #
As in the
cmc case
p E a
(local)
unique
we
assume
find
for
for
R\10},
solution
may
0
Q, dF,
-
=
dF A dG.
the Riccati
equation
in
A
End(fffl,
satisfied.
we
=
any
To
T of the
dT
which
=
conditions are
=
*
dG
=
E
GL(2, R,
Riccati
pTdG T
_
value
T (po)
=
As above let
to be invertible.
(T 0
initial
dF,
-
po E M
S( P0))2
=
problem
To, us
assume
that
(P
Then
(T everywhere
by lemma 17,
and
-
S)2
we
Darboux
transform
Our aim its
mean
now
The
:=
T-1L
of L.
is to show that
V is again Willmore.
Westart
sphere of V
by
sphere.
curvature
Lemma21.
(P-1
call
LO a
=
mean
curvature
is
given
S0:=T-1ST=TST-1, and the
corresponding
Hopf fields
are
2*AO:=P-'T-'dFT-1,
2*QO:=pTdGT.
by computing
80
12 Darboux
Proof. by
First
tranforms
that
note
60
the derivative
50
0'(Hom(LO, HILO))
of LO is
given
T-16T.
=
LO is immersed and
Therefore
*50 A similar we
E
=
T-1
*
T-'SJT
=
*60
yields
computation
obviously
6T
=
=
60 SO
T-'STT-16T
=
S060.
Due to the definition
-
of
SO and LO
have
SOLO
=
LO.
Moreover,
T-1ST
=
T-1ST2T-1
=
T-'T2ST--1
=
TST-1.
Now
dSO =,d(TST-1) dTST-1
=
(pTdGT
=
T((pdGT
=
T(pdG(TS
-
+
T-'dF)S
TdST-1
+ dS
2(*QO
is the
=
similarly Finally,
TST-l(pTdGT
S(pdGT
-
+ ST +
+
-
T-'dF))T-1 I)dF)T-1
ST-1
-
decomposition T
*
for
(2
*
Q)T
of =
dSO
-TS(2
into *
type:
Q)T
=
-TST-'TdG
F.
=
0)
AOH2 C LO, whence
dSOLO C LO. This
dF)T-1
-
*AO),
-
QOILO and
-
-
=
and
-
TST-'dTT-1
-
p-II) + (T-1S -2 2jo-1T dF)T-1 T(pdGT T p-'T-'dF T-1 TpdG
=
*TdG T
TdST-1
dF)ST-1
=
=
which
+
proves
that
SO is the
mean
curvature
sphere of LO.
T
=
-SOTdGT,
12.3
Theorem 12.
The Darboux
HP'
Willmore
are
again
Darboux
transforms
transforms.of surfaces.
an
of Willmore
immersed
surfaces
Willmore
surface
Proof.
-2p-ld
*
QO
=
d(TdGT)
=
(pTdGT
=
p(TdGT
=
-
dT A dGT
dF)
A dG
A dGT -
-
-
TdG A
TdG A dT
TdG A
TdGT)
(pTdGT =
0.
-
dF)
81
in
Appendix
13
The bundle
13.1
If L
Lemma22.
is
is
immersed
an
AIL Proof,
AIL
Let
0
*Q
Note that
AIL 0
=
0,
d(*Qo)
=
do
A
(d
=
=
obtain
we now
d(*Ao)
=
=
*
A)O
(d
*
Q),O
*
AA
we
*Q
-
*Q A JO similarly -
0,
=
A
0 4=*
-=
QJL
0. Since
=
holoMorphic
=
A
find
=
by type,
0
do
0
(d
=
*
E
F(L)
Q)O
(d
=
(5.2)
Q
where
AA
then
0.
for
dO
HP1,
in
curve
&0
H
:
-+
JO
A*
A) V).
*
-
HIL.
Since
AJO
=
-2AJO.
=0
But L is
AIL
Therefore
immersion.
an
0
AJ
=
A
implies
0. The
=
converse
is obvious.
are
section T E HO(Hom(V, W)), a holomorphic complex vector bundles, there exist holomorphic
Given
Lemma23.
holomorphic
VO
Vo
such that
Proof. an
Let
r
T7
=
V, fV
image
may
of the
T away
from
a
subset.
discrete
maxf rank Tp I p E MI and G := fp I rank Tp r}. This is of M. Let po be a boundary point of G, an let 01, be , on sections of V on a neighborhood U of po. By a change of indices =
.
assume
that
To,
A
bundle
...
dimc
M = 1. We assume that
k E a
E
a
holomorphic
HO(ArWlu)
A
To, $0. But this
AIWIU,
exist section
W
C W
:=
holomorphic N,
V,
subbundles
open subset
holomorphic we
ker T and
=
C
where
po is
its
coordinate
only z
zero
centered
such that
To,
A
...
is
a
holomorphic
and hence has isolated
A
To,
F. E. Burstall et al.: LNM 1772, pp. 83 - 86, 2002 © Springer-Verlag Berlin Heidelberg 2002
=
zku.
within at po,
U.
zeros,
.
.
section
because
Moreover, there a holomorphic
and
84
Appendix
13
Off po the section
a
decomposable,
is
dle of
Ar(W), it defines a section WJUextending imageTIU\p..
easily
using the fact
closed
Proposition 0.
of
subset
exists
Assume
now
Then L
=
and
L
kerA
a
bundle
=
is
line
as
this
5yY
=
0
=
field
all
p in
constant
vector
W*
follows
HP',
in on
V*.
-+
and A
I 2
to
a
0
by
Fields).
U C M.
set
This
Lemma 22.
+
a
in
curve
The tangent carries
([X, Y]
By set.
is
of
bundle
holomorphic
J[JX'
UP'. Riemann
a
structure:
Y]).
vanishing of the Nijenhuis tensor implies holomorphic if OY 0. This is equivalent either of these conditions simply says
in X. The
but
field
=
=
in C is therefore
Riemann surface
can
always
Proposition holomorphic
22.
0.
holomorphic, be extended
a Willmore surface of complex holomorphic
Let L be
sections
Ho (K Hom+(RIL,
A
L)),
Ho (K Hom+(L, HIL)),
JL
and
proof
if AQ we
=
need
0 then
a
theorem
Y is called
OjyY,
vector
=
22.
discrete
a
non-empty
open
holomorphic
bundle
=
by Example ker A off
Ejiri/Montiel
immersed
Vector
an
=
isolated.
are
and the an
L
to
and a
a given tangent holomorphic vector
field.
For the
;t
open dense
an
L.
section
Alu
But then
of A
zeros
complex line
is tensorial
0. A vector
for
=
(Holomorphic
viewed
holomorphic L such that
a
[Y' JY] Any
surface
and H = L (D
bundle
H,, =,4 Lp E) Lp 0 on U. AIL
L denotes
section
Note that
with
T*
unique
oxy
=
the kernel
image
:
is
subbun-
about
lVillmore
because the
Remark 12
5J
of
L C H such that
Holomorphicity
surface
The statement
r-dimensional
line
exists
that
L,
contradiction,
In this
an
(connected)
a
F(KEnd-(H))
Lemma23 there
13.2
i.e.
M we have:
A E
Proof.
a
G,(W),
ker T is the annihilator
Let L be
21.
Then there
that
G,(W)
the Grassmannian
and since of
in
in
HP'.
line
Wehave the
Q (=- Ho (K Hom+(HIL, AQ E
JL
H'(K
2
Hom+(HIL, 7;
E
following
bundles: 7
L)), L)),
Ho (K Hom+(L, HIL)). -
~
Holomorphicity
13.2
The curvature
Lemma24.
R'9+'9 and
for
a
holomorphic
In
general,
-(A
=
field
vector
R'9+'5(Z, Proof.
of the
tensor
if V and
=
we
theorem
on
H is
85
given
by
Q A Q),
have
2S(Ozaz
V+
a +0
connection
AA+
Z
JZ)
Ejiri/MOntiel
and the
(13.2)
connections,
two
w are
OZOZ).
-
then
RV =RV+d7w+wAw. We apply
this
to
0 + 0
Ra+'9
R
d
=
d(A
-
-2(A -(A Equation
(13.2)
follows
Ro+'9 (Z, JZ)
=
=
=
becauseOZ2
=
0
=
d
-
+
AA+ AA+
(A
+
Q)
Q)
+
(A
QA
Q)
and
Q)
+ +
(A
Lemma4:
use
A
(A
+
AA+
Q)
Q A Q)
Q A Q).
from
5z) (aiz + Oiz) S 09Z + 6z) (az 6z) 2S(-o9z6z + Ozaz), (az
+
-
-
-
(Oiz
+
S (az
-
Oiz) (az + 6z) 6z) (az + 6z)
.
a2Z*
(Proof
of the proposition). of A was shown in example 22, and that of Q can be holomorphicity, shown in complete analogy. vector (H, S) is a holomorphic complex quaternionic bundle, and L is Remark 6. L Therefore and a holomorphic see subbundle, HIL are hololine and the morphic complex quaternionic bundles, complex line bundle K Hom+(L, EIL) inherits structure. a holomorphic Then, for local holomorphic sections 0 in L and Z in TM, Proof
The
(49Z6L)(Z)'O
By (13.1)
and
(13.2)
=
Mh(Z)O) (Z)
=
aZ
=
7rD9Z(dO(Z))
we
JL(aZZ)O
6L(Z)(09ZO)
19Z (7rL (10 (Z)) =
7TL49Z(19ZV))-
have
OzOzO
=
azOzO --IS R9+9(Z, JZ)O, 2 ,
=0
hence
-
I.-
EL
86
Appendix
13
(OZ JL) (Z)
=
0
-
Then also
(,9jzjL)(Z) 06L
and therefore
:--
=
0,
0.
holomorphicity
the
To prove
S(,9zjL)(Z)
=
AQ E r(K 2Hom(H/L, L)),
of
we
first
note
that 2
K
Hom(H/L, L)
natural
carries
a
phicity
of
A, Q,
Finally holomorphic find, for 0
holomorphic structure. product rule.
6.E
structure E
The rest
follows
from
L))) the
holomor-
and the
interpret
we
Homc(TM, Homr (TM, Hom+(HIL,
=
on
1'(L), 0
as
section
a
(L,
K HoM+
in
Note that
TI is given by a. From the holomorphicity
(aA)O
=
a(AO)
=
+
the
of A
we
under
A
MO.
=0
This
shows that
and,
as
that
1,
Z and
a
is invariant a
L
is a-invariant.
of
consequence
local
under
AQ
R9+5,
holomorphic
and that
0
section
-
azazO
Moreover, it is obviously invariant 0, also under Q. RomLemma 24
=
of
-
=
Oz azO + =0
for
local
a
L,
I -
2,
holomorphic
-
SRO+'9 (Z, JZ),O. V
EL
Then
(OZ h) (Z) 0
=
19 (h
(Z) 0)
-
09Z (JI (Z)O) =
7rLaZOZO
h (OZ Z) 0 h (Z) az 0 az(ir_LdO(Z)) -7rIaz(dO(Z)) -
=
=
0.
=
it
follows
vector
field
References
1. 2.
3.
4.
Geom. J. Differential Bryant, Robert. A duality theorem for Willmore surfaces. 20, 23-53 (1984) Lond. Math. Proc. in SN (1). Norio. Willmore Surfaces with a Duality Ejiri, 383-416 III Ser. (1988) 57, No.2, Soc., In: SumFerus, Dirk. Conformal Geometry of Surfaces in S4 and Quaternions. on Differential mer School Geometry, Coimbra 3/7 September 1999, Proceedings, Ed.: A.M. d'Azevedo Breda et al. Phicker Ulrich. Quaternionic Pedit, Franz; Pinkall, Ferus, Dirk; Leschke, Katrin; MathematEstimates. To appear in Inventiones Formula and Dirac Eigenvalue icae
5.
Friedrich,
Thomas.
On
Superminimal
Surfaces.
Archivum
math.
33,
41-56
(1997) 6.
Franz. Remarks on Darboux Udo; Pedit, Hertrich-Jeromin, Isothermic Surfaces. Doc. Math. J. 2, 313 333 (1997).
Transforms
of
-
7. 8.
(www.mathematik.uni-bielefeld.de/documenta/vol-02/vol-02.html) Ulrich Conformal Geometry. (Eds.). Ravi; Pinkall,
Kulkarni, schweig Montiel,
Vieweg,
Braun-
1988
Sebasti6n.
Spherial
Willmore
Surfaces
in
the
Fo,4r-Sphere.
Preprint
1998 9.
10.
Ann. Global Anal. Geom. Willmore surfaces in the four-sphere. Musso, Emilio. 13, 21-41 (1995) and Ulrich. on Riemann surfaces Quaternionic analysis Pedit, Franz; Pinkall, geometry. Doc. Math. J. DMV, Extra Volume ICM 1998, Vol. II, differential 389-400.
(www.mathematik.uni-bielefeld.de/documenta/xvol-icm/05/05.html) 11.
12. 13.
Richter, J8rg. Conformal Maps of a Riemannian Surface into the Space of Berlin 1997 Dissertation, Quaternions. Rigoli, Marco. The conformal Gauss map of Submanifolds of the Moebius Space. Ann. Global Anal. Geom 5, No.2, 97-116 (1987) Keti. to Differential of Manifolds and Applications Tenenblatt, Transformations Chapman& Hall/CRC Press 1998 Equations.
Index
affine
coordinates
Bdcklund
line
9
transform
53,'55-57
complex quaternionic conformal 7, 8
bundle
conformal
curve
conformal
Gauss map
constant
mean
Darboux
degree
curvature
transform of L
differential
Jf
bundle
energy
19 29
surface
mean
curvature
sphere
mean
curvature
vector
minimal
surface
normal
curvature
normal
vector
projective
78, 81
27,44 40,42
66 40 6
9
space
quaternionic
17
Riccati
21
functional
31
curvature
harmonic
map
15
73
suPer-conformal
40
31
63
equation
second fundamental
19,50 holomorphic curve bundles holomorphic quaternionic 84 holomorphic section 69 holomorphic structure 27 Hopf fields 64 hyperbolic 2-planes minimal 66 hyperbolic curve
bundle
vector
11
Gaussian
isotropic
75
16
67
derivative dual
18
subbundle
20
form
tangent space of projective bundle 15 tautological twistor projection 47, 50 13 two-spheres in S'
Willmore
cylinder
Willmore
functional
Willmore
surface
constrained
space
28
type argument
-
39
50
46 35
36, 55, 36
81
10
