LECTURE 4. THE WEINSTEIN CONJECTURE IN THE TIGHT CASE 89the following:Don Dv f:. 0,
DT n Dv = 0 for TE (-c:,O).
We also assume that the intersection points of Dv with Do are isolated, because
otherwise we would find by that the disks are identical. This is again some nontrivialfact. It is of course related to the fact that a holomorphic map having a cluster
point of zeros is the zero map near the cluster point. In our case, philosophically, if
we have two holomorphic curves which have a cluster point of intersection points,
they have a tangency at the cluster point. Then one curve may be viewed as the
graph of a section of the normal bundle of the other curve. This section and the
normal bundle can be taken in such a way that the section is holomorphic and
would therefore have a cluster point of zeros. Consequently the curves would bethe same near the cluster point. It is possible to give a rigorous proof along these
lines, see [1 J.
First of all we reduce our problem to the normal form previously derived. We
may assume therefore thatUo: D ~ C^2 ' uo(z) = (z,O) for all z E oD
and
F = aD x JR c c^2.The almost complex structure J , which we assume to be defined on the whole C^2satisfies J(z, 0) = i for all z E C^2 with lzl ~ 1. The totally real loop L defined by
L(z ) = T(z,o)F is just izJR EB R The differential equation we are going to study is
8Ju = 0 on fJ and u(oD) c F.
The solutions near Uo define a disk family (DT ). Moreover assume v : A ~ C^2is a solution of 8Jv = 0 on A, where A is a connected open subset of D. If
oA = oD n A is nonempty we assume that v(oA) c oD x R Moreover we assume
that v(A) n Do f:. 0 and v(A) n D 7 = 0 for all T E (-c:, 0). By our assumption
there are only a finite number of intersection points of v(A) with Do. Pick z E Awith v(z) E D 0. Assume first that v(z) tj. u 0 (8D) = oD x {O}. Then z does not
belong to the boundary of D and we have an interior intersection of the embedded
holomorphic curve u 0 with v. The local intersection number is positive, so that also
· v(A) would intersect neighbouring disks of Do which contradicts our assumption.
Hence we conclude that v(A) n Do c oD x {O}. Now comes the first crucial
assumption:
v(A \ oA) n F
(34)v(A \ oA) n (oD x JR)
0.Hence v(A\oA) n Do = 0 and therefore v(oA) n Do f:. 0. Since A c D we can
carry out a suitable rotation and may assume by replacing v by the corresponding
transformed map that there exists a point zo E oD such thatv(zo) = uo(zo) = (zo, 0).