Darboux transforms

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


-

NT)

T,N

>

we

+

+

-jo-1

p-'Tdg))

df 0

A

pdgT

df

A

proved

Lemma 18.

HO

2


-p-I

=

IT12 Next

we

show

Lemma 19.

If HO

is

constant, then H

Proof. We differentiate

0

=

HOIT12

=

+ 2


12 Darboux tranforms

78

IT 12

2


+p-1

=

0.

But

(T

-

N)2

=

-IT

=

-(ITj2

point.

Definition 13. Let f mean

curvature H

:

M

-

2


>

p-1

+

everywhere,

a

E

+1) -

I.

if it holds in

connected. This leads

Im H be

-4

2