SI
10 Willmore Surfaces in
Let <
-,
> be
an
indefinite hermitian inner
product
on
EV. To be specific,
we
choo...
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SI
10 Willmore Surfaces in
Let <
-,
> be
an
indefinite hermitian inner
product
on
EV. To be specific,
we
choose < Vi
Then the set of
complementary
IT1W2 + IT2W1
1, 1 > 0 defines an S3 C RP1, while the hyperbolic 4-spaces, see Example 4. We have
isotropic lines 4-discs
>:--:::
W
are
<
=
(a db)
(10.1)
C
and the
same
holds for matrix representations with respect to
a
basis
(v, w)
such that
=O=<W'W>'
=l.
10.1 Surfaces in S3 an isotropic line bundle with mean curvature sphere S. We look >. Clearly S* adjoint map M -* Z7p -+ SP* with respect to < 1 L implies stabilizes Lj-, and L
Let L be at the
=
=
(dS*)L
=
S*L'
S*L
=
L'
=
L.
Similarly,
Moreover, if Qt belongs
to
S*,
(dS)*L
1
C L
then
Qt
(S*dS* 4 -
(dSS
-
-
*dS*)
*dS)*
I(SdS + *dS)*
4
-A*.
F. E. Burstall et al.: LNM 1772, pp. 61 - 66, 2002 © Springer-Verlag Berlin Heidelberg 2002
L.
10 Willmore Surfaces in
62
kerQt
Therefore We
proceed
=
=
(image(Qt)*)'
=
(imageA)-L
sphere,
curvature
mean
L'
D
to show that S and S* coincide
uniqueness of the S*
S3
on
L and
HIL. By
the
Theorem 2, it then follows that
see
S.
Let
0
E
1'(L),
and write
SO
=
OA,
S*O
0,60
>=
Op
and <
Note that <
0, 0
>= 0
0, JO yields
> makes sense, because of <
+C0
W
From 0 =<
0, So
>
0 =<
we
=
0, L
>= 0.
Differentiation of
0.
obtain
JO, SO>+< 0,(dS)O %-
>+ <
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
We conclude A Now
assume
i.e.
S I HIL
=
AOA
SIL =
=
Ip
0 =< =<
+
pwA
S* IL
and
(p
-
)CDA.
-
S* I HIL
50' so
JO'O
I u. Then
=
> + <
S*JO' 0
=& <
=
S*0,50
>
0, SJO
>
> + < > +
> P
&6j + WP.
But
*(.,) =<
=<
0, JSO
>=<
S*O' 90
>=
0, S60 XW.
>= WP
10.2
Comparison
with
(10.2)
shows p
0 It follows
a
A
=
=
p, i.e.
=
& W
A, and
=
WA
+
SIHIL
we
(
=
S*IHIL.
=
Hyperbolic
2-Planes
63
get
&)WA.
-
This
completes the assumptions
of Theorem 2, and S* = S by uniqueness. OA, then Conversely, if S* = S and So =
A < Now S2
0,0
=
-I
SO,O
>=<
implies A2
0,SO
>=<
>=<
1, and therefore
Proposition 18. An immersed holomorphic S*. i.e. a surface in S', if and only if S
>
A
get
<
0,,0 we
curve
=
A <
0, V)
L in
0,0
>.
> = 0.
HP1
is
isotropic,
=
2-Planes
Hyperbolic
10.2 In the
half-space
Poincar6 model of the
hyperbolic space, geodesics are orthogonally intersect the boundary. We consider the models of hyperbolic 4-space in HP', and want to identify their totally geodesic hyperbolic 2-planes, i.e. those 2-spheres in RP' that orthogonally intersect the separating isotropic S3. Using the affine coordinates, from ExS3 This ample 4, we consider the reflexion H -+ H, x -+ -.t at Im H preserves either of the metrics given in the examples of section 3.2. In particular, it induces an isometry of the standard Riemannian metric of RP' which fixes S3. Given a 2-sphere S E End(EV), S2 -I, that intersects S3 in a point 1, we use affine coordinates, as in Example 4, with 1 =:F vH and w or
euclidean circles that
=
.
=
such that < V,
v
>=< W,
W
>=
0,
< V,
W
Then N -H
S with N 2
=
R2
1, NH
=
=
(0
-R
HR, and S'
C Ifff is the locus of
Nx+xR=H. If S' is invariant under the reflexion at
iR
=
H
S3,
then it also is the locus of -Nd
-
or
Rx + xN
=
T1.
According to section 3.4, the triple (H, N, R)
is
unique
up to
sign. This implies
either
(H, N, R)
=
(ft, R, N)
or
(H, N, R)
-R, -N).
By (10.1) either S* S, and the 2-sphere lies within the 3-sphere, intersects orthogonally, and S* -S. We summarize: =
=
or
it
10 Willmore Surfaces in
64
S3
Proposition 19. A 2-sphere S E Z intersects the hyperbolic 4-spaces determined by an indefinite inner product in hyperbolic 2-planes if and only if S*
=
-S.
10.3 Willmore Surfaces in S' and Minimal Surfaces. in
Hyperbolic 4-Space Let L be set of
a
an
connected Willmore surface in S3 C
indefinite hermitian form
HP',
where S3 is the
H. Then its
on
mean
isotropic sphere
curvature
satisfies
S* Let
us assume
that A
$ 0,
and let
S.
L
ker A and
L
L.
image Q
be the
2-step
B.icklund transforms of L. Lemma 13.
Proof.
First
we
have I
Q*=
4
(SdS
-
*dS)*
4
(dSS
-
*dS)
I
4
Now
imap Q
Therefore < L,
L L
is
-
*dS)
=
-A.
(10-3)
S and So S-stable, and S* OA imply and dense on a 0, open subset of M =
>=
=
(-SdS
L'
=
(image Q)'
=
=
kerQ*
=
<
0, 0
> = 0.
ker A
Lemma 14.
-S
for
the
mean
curvature
sphere 9 of L.
L is obviously (-S)-stable. It is trivially invariant under Proof. First L and Q and, therefore, under d(-S) 2(*A *Q). Finally, the Q of (-S) =
=
-
A is
I
4
and this vanishes
sphere by
on
these three
((-S)d(-S)
-
*d(-S))
=
A,
L. The unique characterization -S. properties implies 9 =
of the
mean
curvature
10.3 Willmore Surfaces in
We
now
turn to the
S3 and Minimal Surfaces
in
1-step Bdcklund transform of
d(F+F*)=2*A+2*A*
Hyperbolic 4-Space L. If dF
=
2
*
A,
65
then
2*A-2*Q=-dS.
=
(10.3) Because S*
We
=
now use
S,
we can
choose suitable initial conditions for F such that
F+F*
-S.
(10.4)
affine coordinates with L
[,I.
Then the lower left entry g of
I
F is
a
Bdcklund transform of
f,
(7.9)
and g +
9
and
H.
=
We want to compute the mean curvature Bdcklund transforms we know
(9.3), (9.9). Likewise, 1 if
NO
we
obtain
0)
(0 1) ( R) (1-f)=(lf (R (1 (1 f) (1 Hg) ( -j ) (1 -Hg) (1 (1 f) (9 HgfI (1 -f) H
1
0
0 1
0 1
9 -fi
1
0
properties of
(10.5)
From Lemma 14
-R..
=
From the
sphere Sg.
Hg=f-f,
Nq=-R, see
(10.4) imply
-
ft
-f
0
0
1
0
0
-f 1
-
01
This
implies H
In
particular f
on
that set. It follows that
E Im
-
=
(I 1) ( 0
=
g
=
HH,
-
-N +
=
since H
0
=
-R
f
-I
whence
(f
-
on an
f)H. open set would
0 -N +
N and H E R for
f
(f
M
:
-
-4
-g)
I)H) O Im H
=
mean w
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*
H,
=
g
Sg and, because, R
N-
and -N
-Rg
1
0
-
01
-
0
1
0
1
f
-S9-
f
f
N
01
H
0
0
N
1
0
0
1
-
1
0
10 Willmore Surfaces in
66
We have
now
mean curvature spheres of g intersect S' orhyperbolic planes. We know that, using affine Euclidean metric, the mean curvature spheres are tangent
shown that the
and therefore
thogonally,
S3
coordinates and
a
to g and have the
under conformal
are
same mean
curvature vector
as
of the ambient metric.
changes
g. This
Therefore,
property remains in the
hyperbolic
g has mean curvature 0, and hence is minimal. If A =- 0, then w = 0, and the "Micklund tr'ansform" is constant, which may be considered as a degenerate minimal surface. In general g will be singular in the (isolated) zeros
metric,
of
dg
=
1w, 2
but minimal elsewhere.
We show the in
converse:
hyperbolic 4-space,
Let L be
i.e. with S*
=
4
(SdS + *dS)*
4
holomorphic
curve, minimal
-S. Then I
1
1
A*
immersed
an =
(dSS
-
*dS)
4
(SdS + *dS)
-A,
and therefore also
(d * A)* From
Proposition 15
we
=
-d
*
A.
have I
(f
A
d
dw
dw
4
-f dw f) -dw f
Therefore
jw-,
dw
f dw
=
dwf,
and hence
dw(f But
f
is not in
sition 12
S1,
and therefore
f)
dw=O,
=
0.
i.e L is Willmore.
=
-A
implies
w
=
=
(W
-77D. Rom S*
-S
1w
the backward Bdcklund transform h with dh
conditions is in Im H To
Similarly, Propo-
yields *A
and A*
+
=
we -
know TI
=
-H,
and
dH and suitable initial
R.
summarize
Theorem 9
(Richter [11]).
Let <
.,.
> be
an
indefinite
hermitian
product
IV. Then the isotropic lines form an S' C HP', while the two complemen-. tary discs inherit complete hyperbolic metrics. Let L be a Willmore surface in S' C HP'. Then a suitable forward Bdcklund transform of L is hyperon
bolic minimal.
Conversely, an immersed holomorphic curve that is hyperbolic Willmore, and a suitable backward Bdcklund transformation is a Willmore surface in S'. (In both cases the Bdcklund transforms may have
minimal is
singularities.)