1549055384-Symplectic_Geometry_and_Topology__Eliashberg_

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Then


H. HOFER, HOLOMORPHIC CURVES AND DYNAMICS

D81(uo)(h, k) T(h, k)

(hs, ks) + J( uo) (ht, kt) + [DJ( uo)(h, k )] · ( i, 0)


(hs + iht, ks+ ikt) + [D1J(uo) h] · (i, 0) +

+ [D2J(uo) k] · (i ,O)

(h s + i ht, ks+ i kt) + [D2J(uo) k] · (i, 0)

-. (hs + i ht+ ak, ks + i kt+ bk).


Here a,b E C^00 (D,.CIR(C)). We have used the fact that J(uo(z)) = J(z,0) = i and


(DiJ(u 0 ) · h)(z) = dd J(ht(z),0)1 = dd ii = 0


t t=O t t=O

where ti----; (ht, 0) is a smooth path in B with ho(z) = z and ftht(z) lt=o= h. D


Remark 3.8. Here is an important point which should be made. It is crucial for


the implicit function theorem that the linearisation Dfh ( uo) has t his particular


form, which says that the operator is upper triangular. In short


[

h ] [ Os + iOt a ] [ h ]

k -+ 0 os + iot + b. k.

This makes it possible to use particular features of linear Cauchy-Riemann type
operators in complex dimension 1.


Let us write L(z) = Tcz,o)F for z E oD. Then L is a loop of totally real
subspaces of (C^2 , i).
In the following proposition a formula for the index is given. This is a nontrivial
result and for a proof we refer to [l].


Proposition 3.9. The operator T is Fredholm and its index is given by


ind(T) = 4 + k.

Now we can prove one main theorem of this section: A criterion for surjectivity
of the operator T = D8 1 (u 0 ) in t erms of the number k. This result however does


not require the special boundary condition L( z ) = izffi. EB zkl^2 JR., so let 2 < p < oo


and assume L is a smooth arc of totally real planes in C^2 with Maslov index μ 2 ( L)
such that L(z ) ::) izffi. x {O} for all z E oD. Consider the linear operator T defined
by


T(h, k) = (h s + iht + ak, ks+ ikt +bk)

with domain Tu 0 B. Write L(z ) = Tcz,o)F , lzl = 1 and define Vf, by


Vf, = {f E W^1 ·P(D, C^2 )1f(z) E L(z) for all z E oD}.


The range is LP= LP(D, C^2 ). The following theorem holds:


Theorem 3.10. If k ;:=: -1 then the operator T is surjective and its kernel is

precisely k+4-dimensional. If moreover k = 0 then every element (h 1 , h 2 )) E ker(T)


with h2(z) f:. 0 for some z ED satisfies h2(z ) f:. 0 for every z ED.

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