12 Darboux tranforms
provided a mean to construct new Willmore surfaces by solving linear equations. Darboux transforms...
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12 Darboux tranforms
provided a mean to construct new Willmore surfaces by solving linear equations. Darboux transforms provide
Bdcklund transforms out of
a
given
one
another method for such construction, based on the solution of equation. For isothermic surfaces it is described in [6]. After
an
a
Riccati
introductory remark on Riccati equations, we first describe Dara special case of [6], namely for constant mean curvature R, because it displays a striking similarity with the Willmore
boux transforms for
surfaces in case
treated thereafter.
As with the Bdcklund transforms the is
again
theory
-
in the Willmore
only have a local existence of a solution to the Riccati problem, and moreover require this solution to be invertible in
algebra End(IV).
the
12.1 Riccati
equations
We consider Riccati type partial differential equations in below will be H or End(EP). Lemma 17. Let A be
manifold.
Let
a,#
E
A)
a
Then
for
any p E
R\ 10}, po dT
unitary algebra
associative
an
S?1 (M,
A
P
=
0
=
E M and
=
an
over
pTa T
-
0
=
=
=
_1'
the
reals, and
M
To
0,
d , PAa.
E
A the Riccati initial value problem
T(po)
=
To
a unique solution T on a connected neighborhood of po. Moreover, if S: M - A with
S2
algebra, which
with
da
has
-
local. We
initial value
a
case
Sa+aS=O,
and
F. E. Burstall et al.: LNM 1772, pp. 73 - 81, 2002 © Springer-Verlag Berlin Heidelberg 2002
dS=a-0,
(12.1)
12 Darboux tranforms
74
(T
-
S)2(po)
p-1
=
then
(T and T'S
everywhere,
S)2
-
=
P-1
(12.2)
S2T.
=
Proof. The integrability condition for (12.1) 0
is
aT +
=
pdT A
=
p(pTaT
=
-pp
pTdaT
0)
-
A aT +
A
T+
a
p-'dX
pTdaT
T is
X:= a
dT
pTa A
-
-
p(T
-
S)2
A
(pTaT
+ P
-
0)
-
dfl
do
and S
solution,
a
d#
pTa
-
pTdaT + pTa A 0
obviously satisfied. Now, if
Then X satisfies
-
_
as
above, then
we
define
1.
linear first order differential equation
=
(dT
=
(pTaT
=
Ta(pT2
=
TaX + XaT.
=
0
-
dS)(T -
-
a) (T
-
pTS
-
S)
+
(T
S)
+
1)
+
-
S)(dT
-
(T
dS)
S) (pTaT
-
(pT
-
2-
pST
-
-
a)
1)aT +
aS + Sa =0
X(po)
Hence
X
implies
=
T 2S_ ST 2
together
with
f
:
M
-+
mean
Im H be
a
"'tangential"'
into the K- and
S)2S
-
S(T
_
S)2
curvature surfaces in
R'
=
Ndf
=
N2
-dfN,
=
_1.
1-form: it anticommutes with N. We
ff-part
multiplication by
with respect to the
complex
structure
decompose it given by left
N to obtain
dN
Since that is also the traceless
_
conformal immersion:
a
*df dN is
(T
=
equation of the lemma follows from
(12.2).
12.2 Constant Let
0. The last
=
Hdf
+
w.
(12.3)
decomposition of the shape operator into "trace" and part,the function H : M --- R is the mean curvature, and *w
-Nw. Then
12.2 Constant
mean
1(dN + N (12.3)
resembles the formula dS
R3
75
dN).
*
2
Note that
curvature surfaces in
=
2
*
Q
-
2
*
A.
shape operator of f, and (12.3) gives its decomposition " the traceless part. Therefore H M -4 R is the mean and trace" the into Now -dN is the
curvature.
Differentiating (12.3)
get
we
O=dHAdf +dw. Hence H is constant if and only if dw We see that the theory of constant
parallels We
0, resembling d
=
mean
*
0.
Q
(=cmc)
curvature
surfaces in R3
that of Willmore surfaces in HP1.
now assume
0. Then
54
H to be constant
IN
g:=
H
satisfies 1
dg
=
df
dN
*dg
-
jy
=
=
dN
H(df
-
-Ndg
=
dg), dg N,
and
df Adg=O=dgAdf by type.
The map g is
an
immersion of constant
f, i.e. away from curvature surface of f.
from the umbilics of mean
For
simplicity
we
H P
,
case can
We put A
:=
parallel
constant
case
using the homothety f
-+
Pf, H
=
Therefore, for value problem
lemma 17.
dT
=
df dg, 0 0 O,pO E =
any jo
=
pTdgT
unique solution T
(locally)
has
in Im
because T satisfies the
A
curvature H away
-1.
=
be reduced to this
-H.) End(p, a
g -+ tig with [t
mean
0. It is called the
=
restrict ourselves to the H
(The general
w
,
-
df,
which
.
These match the
M and To E
T(po)
we assume
same
equation
fO
f +T.
We put =
=
assumptions of
ImH\10}
the initial
To to have
up to
a
no zeros.
minus
sign.
T
stays
76
12 Darboux tranforms
Then
*(df
*df
+
dT)
pT
=
-TNT-' (df
f0
This shows that is
+
*
dgT
dT)
=
-pTNdgT
=
=
-TNT-'pTdgT
-TNT-ldf 0.
is conformal with
M
Nf o
:=
=
-TNT-'. Moreover, f
immersion if and
only if g is an immersion. if g is immersive. Under what conditions does f 0 again have constant mean curvature? We
an
compute HO
T2
:=
=
Hf o, using
-IT12
TN+NT=TN+TN=-2,
,
and dN A
df
=
Hdf
A
df
-
Hdg
A
df
=
Hdf
A
df.
We find
HOdf 0
A
df 0
=
=
=
dNO
df
A
-d(TNT-1) A df 0 -(dTNT-1 + TdNT-1
=
-(dTN
=
(-(pTdgT
=
+ TdN -
-p(Tdg(TN 2 <
T,N
-
TNT-1dTT-1) TNT-1dT) A pdgT
df)N +
>
-
NT)
T,N
>
we
+
+
-jo-1
p-'Tdg))
df 0
A
pdgT
df
A
proved
Lemma 18.
HO
2 <
T,N
>
-p-I
=
IT12 Next
we
show
Lemma 19.
If HO
is
constant, then H
Proof. We differentiate
0
=
HOIT12
=
+ 2 <
A
TNT-' (pTdgT
pTdgT A pTdgT
IT12 Hence
Tdg
-p-I
IT12 2 <
-
-1.
T, N
>
-p-1:
pTdgT
-
df))
A
pdgT
12.2 Constant
0
=
HO
dT,T
<
> + <
=
HO(< pTdgT,T
=
HO(-IT12p T,df
<
-
-(HOIT12P
=
2p
=
-(HO P
+
2
+
2
+
2
1)
<
T,dg
T, df
+
1)
<
+
1)
<
df,T
<
>
pTdgT,N
<
77
>
-
<
pTdgT,N
> + <
T,df
> + <
T,dg
>
>
-(HO + 1) < T,df > + < pTdgT,N -(HO + 1) < T,df > + < pTdgT,N >,
T,dg >
>
>
>
-
(TdgTN + NTdgT))
T,df
(Tdg(TN + NT)
-(HO
=
+
><
<
HO
-
((TN + NT) (Tdg + dgT)
-(HO
=
T,N
1)
_
T,dg
T,dN
> + <
df,T >)+
<
-
T,dg >)
> + <
=
<
<
>
dT,N
curvature surfaces in
mean
T, df
+
(TN
NT) dgT
+
(TdgTN + NTdgT))
-
>
(TdgTN + TdgNT + TNdgT +NTdgT
(TdgTN + NTdgT))
-
% -V
=0 =
-(H#
If HO
Now
+
1)
1,
we are
df and
<
T, df
>.
done. Otherwise <
dT
=
dpN
dT
=
pTdgT
w
are
+
pdN -
djzN
-
>=
ydf
PP2 NdgN
df
and
tangential,
T, df
0, i.e. T
=
jLN, and
+ pw
-
df
=
ptt2W
-
d'
comparison of the above
two
equations
gives
dl-t
(I dl-t
=
_
tl)df
=
0,
(_tl
+
P,12)W,
0 and therefore
But then
f
has HO
-1.
As
=
a
g is the
parallel
consequence of the
Lemma 20.
HO
is
constant
preceeding
mean
curvature surface of
two results
we
f which
obtain
constant, if and only if
(T Proof. We know that HO equivalent with
-
N)2
=
P-1
ist constant, if and
(12.4) only
if it
equals -1, and this
is
>
12 Darboux tranforms
78
IT 12
2 <
-
T,N
>
+p-1
=
0.
But
(T
-
N)2
=
-IT
=
-(ITj2
point.
Definition 13. Let f mean
curvature H
:
M
-
2 <
p-1,
holds
R\101,
To
T,N
+p-1)
>
>
p-1
+
everywhere,
a
E
+1) -
I.
if it holds in
connected. This leads
Im H be
-4
2 <
-
and T is bounded with
simpli
is
T,N
-1, and immersed
=
p E
and
1
=
globally defined if M
be
can
IT- S12
-
-(IT12
=
(12.4)
Now recall from lemma 17 that Therefore
N12
-
no zeros. us
to the
a
single
Hence it
following
conformal immersion with constant f + N. Let parallel cmc surface g =
ImH\101,
M,
po E
assume
(To Let T be the unique solution
N(po))2
-
P-1
=
_
(12.5)
1.
the Riccati initial value
of
problem
T(po)=To.
dT=pTdgT-df, Then
fO:=f+T is called
a
Darboux
transform of f.
1. If H is constant
Remark 11. should be
0 0, -1,
then
(12.5)
in the above definition
replaced by H
(HTo
+
N(po ))2
1.
=
P
It turns out that 2. From
(12.5)
Hence there is
a
+ T has
again constant mean curvature H. a given p 54 0 there is an S2 of initial To 3-parameter family of Darboux transforms.
f
we see
that for
We summarize the previous results: Theorem 11. vature H in
The Darboux
transforms of surfaces
R' have constant
mean
curvature H.
with constant
mean cur-
12.3 Darboux transforms of Willmore surfaces
79
12.3 Darboux transforms of Willmore surfaces Let L C H
=
M
x
and dS
sphere S,
End(H2)-valued
EV be
=
maps
a
2(*A
Willmore surface in HP1 with
-
Since d
*Q).
F, G, locally
M
on
*
A
d
=
Q
=
mean
0
we
curvature can
define
by
G=F+S.
dF=2*A, Then
dG
dS
dG A dF
2
=
0
=
Hence the with
a
integrability conditions for dF are satisfied. dG, #
=
R\10},
unique (local) solution dT
we
=
dF, dF A dG.
the Riccati equation in A
End(fffl,
find for any
cmc case we
p E
which
-
=
As in the
a
Q,
*
dG
=
may
assume
To
GL(2, R,
E
po E M
T of the Riccati initial value
pTdG T
=
T (po)
dF,
-
=
to be invertible. As above let
(T0
_
S(P0))2
=
problem
To, us assume
that
(P
Then
(T everywhere by lemma 17,
and
-
S)2
we
Darboux
transform of
Our aim its
mean
now
Lemma 21.
:=
The
T-1L
L.
is to show that V is
curvature
(P-1
call
LO a
=
again Willmore. We
start
sphere.
mean
curvature
sphere of V
is
given by
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 tranforms
Proof. by
First note that the derivative
50
=
60
of LO is
0'(Hom(LO, HILO))
E
given
T-16T.
Therefore LO is immersed and
*50
T-1
=
6T
*
=
T-'SJT
yields *60
A similar computation we obviously have
=
60 SO
SOLO
T-'STT-16T
=
=
=
S060.
Due to the definition of
-
SO and LO
LO.
Moreover,
T-1ST
=
T-1ST2T-1
T-'T2ST--1
=
=
TST-1.
Now
dSO =,d(TST-1) =
dTST-1
+
TdST-1
dF)ST-1
=
(pTdGT
=
T((pdGT
=
T(pdG(TS +
=
=
=
-
TST-'dTT-1
+
T-'dF)S
TdST-1
+ dS
ST +
2(*QO
-
-
=
*AO),
-
decomposition of dSO T
similarly Finally,
*
(2
*
Q)T
=
into type:
-TS(2 * Q)T
C
=
-TST-'TdG T
for F.
QOILO AOH2
-
-
and
and
TST-l(pTdGT
dF)T-1 S(pdGT T-'dF))T-1 (T-1S + ST-1 I)dF)T-1 -
-
p-II) + -2 2jo-1T dF)T-1 T(pdGT T p-'T-'dF T-1 TpdG
which is the
*TdG T
-
-
LO,
0)
whence
dSOLO This proves that
=
SO
is the
mean
C
LO.
curvature
sphere of LO.
=
-SOTdGT,
12.3 Darboux transforms of Willmore surfaces
Theorem 12.
HP'
are
The Darboux
again Willmore
transforms.of surfaces.
an
immersed Willmore
surface
Proof.
-2p-ld * QO
=
d(TdGT)
=
(pTdGT
=
=
-
p(TdGT A
dT A dGT
dF) dG
A dGT
-
-
-
TdG A
TdG A dT
TdG A
TdGT)
(pTdGT =
0.
-
dF)
81
in
