1549055384-Symplectic_Geometry_and_Topology__Eliashberg_

LECTURE 3. THE WEINSTEIN CONJECTURE IN THE OVERTWISTED CASE 75

Proof. Consider first elements of the form (h, 0) in ker(T). These correspond

precisely (via hf---' (h, 0)) to the solutions of the problem h 8 + iht = 0 on D with

boundary conditions h(z) E izlR. One can compute all these solutions. Set

(X)

h(z) = 2::>jZj

j=O

and take z = eit E 8D. Then

h(z ) = ~ ~ ajei(j-I)t

iz i L......,,

j=O

must be real because of the boundary condition. This is exactly the case if a 0 = -ii 2 ,

a 1 = -ii 1 and aj = 0 for j 2: 3. So h is given by

h(z ) =a+isz-az^2

where a E C and s E JR. Let N C ker(T) be the three-dimensional subspace with

elements of the form (h, 0) as just discussed. Let M be an algebraic complement

of Nin ker(T). Denote by f2(z) a smooth loop of I-dimensional real subspaces of

C^2 complementing £ 1 (z ) = izJR x {O} in L(z). Let K be the loop of real lines in C

obtained by projecting £ 2 (z) onto the second component of C.

Consider the operator To with domain Vf< and range LP(D, q defined by

T 0 (k) = h s + iht + bh. The index of this operator is k + l. The reader should

try to verify it for the case k = 0 assuming that b = 0. The general case is nontriv-
ial. A proof of the index formula can be found in [1].

It can be shown that To is surjective iff k 2: -1 and injective iff k :::; -1, see [l].

Coming back to our geometric model situation we conclude that the linearisation of

the Cauchy Riemann operator at an embedded disk is surjective provided k 2: -1.

Now bringing all facts together the claim of the proposition follows.

D

By the previous discussion the situation near an embedded holomorphic disk
is the following. The embedded disk is given by Uo: D-+ C^2 ,uo(z) = (z,O). The
totally real surface F is

{(z, azkl^2 ) E C^2 I z E 8D ; a E JR}.

Let the associated loop

8D 3 z f---' Tuo(z)F = izJR EB zkl^2 R

of totally real subspaces in C^2 be denoted by L. We assume that k = 0, i.e.

F = 8D x JR. This implies by the previous discussion that the linearisation of the
Cauchy Riemann operator is surjective and has index four. Moreover the kernel has
a three-dimensional subspace N associated to the symmetries by the action of the
biholomorphic disk maps (we will explain this soon). Any algebraic complement
of N in the kernel however is spanned by an element (h, k) with k(z) '/- 0 for all
z E D. This latter observation will be crucial in the following.
Recall the definition of the Banach manifold B. A map u belongs to B iff

u E W^1 ·P(D, C^2 ) with u(D) CU and u(8D) CF. Here U is an open neighbourhood

of D x {O}, and 2 < p < oo. The tangent space of Bat uo is given by

Tu 0 B = {f E W^1 •P(D, C^2 )lf(z) E T(z,o)F = izJR EB JR for z E 8D}.