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Let $S$ be a $2-$dim manifold and $q \in S$. Furthermore, let $j_{n}$ be a sequence of complex structures on $S$ converging in $C^{\infty}_{\text{loc}}$ to a complex structure $j$ on $S$ as $n\rightarrow \infty$. Furthermore, let $h_{n}, h$ be metrics of scalar curvature $-1$ compatible with $j_{n},\forall n$ and $j$ respectively, i.e. they are Poincare metrics with respect to the complex structures. Also assume that $h_{n}$ converges to $h$ in $C^{\infty}_{\text{loc}}$.

Question: Does there exists open neighbourhoods $U_{n},U\subset S$ of $q$ and biholomorphisms $f_{n}:(D,i)\rightarrow (U_{n},j_{n}),f:(D,i)\rightarrow (U,j),\forall n$ such that: 1) $f_{n}(0)=q,\forall n$ and $f(0)=q$; 2) $f_{n}$ converges to $f$ in $C^{\infty}_{\text{loc}}$ as $n \rightarrow \infty$?

I think the assumption on the metrics is not necessary, but I am not sure. Can this above statement be true? Do you know any reference on this? Hope for some answers.

Cheers, Tobi

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  • $\begingroup$ You might try to construct a solution of a boundary value problem on a disk around $q$, using the notion of Maslov index, near the point $q$, to build a convergent sequence of holomorphic functions, and then see if you can force them to be invertible somehow. $\endgroup$
    – Ben McKay
    Jan 22, 2015 at 19:48
  • $\begingroup$ What do you mean by maslov index? jow to construct the solution of boundary value problem using maslov index? $\endgroup$
    – Tobias
    Jan 23, 2015 at 4:44
  • $\begingroup$ @ Ben McKay: I do not understand your comment. Can you explain, please? $\endgroup$
    – Tobias
    Jan 23, 2015 at 5:51
  • $\begingroup$ If you specify boundary values around a disk for the real part of a holomorphic function, and specify the value of the imaginary part at a single point $q$, that will define a unique holomorphic function (if I remember correctly). Pick a local holomorphic coordinate function for $j$, and take its boundary values around a disk, and then, for each $j_n$, find the holomorphic function with those boundary values. The number of zeroes inside the disk (with multiplicities) is given by the Maslov index on the boundary. So it stays one, i.e. you still have a coordinate function for $j_n$. $\endgroup$
    – Ben McKay
    Jan 23, 2015 at 7:00
  • $\begingroup$ @ Ben McKey: For what is the maslov index needed ? $\endgroup$
    – Tobias
    Jan 23, 2015 at 7:15

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