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Consider a one-sided ( say, internal) neighborhood $U$ of the unit circle $S$ ( $U$ contains $S$) on the plane with a choice of smooth complex stricture $\tau$ on $U$.

By smoothness of $\tau$ on $U$ we mean that corresponding operator of almost-complex structure $J$( defined in the interior of $U$) as a matrix with respect to the coordinate basis on the plane, has smooth extension to $S$.

Two such pairs $(U_1,\tau_1)$ and $(U_2,\tau_2)$ are equivalent if there are subneighborhoods $V_1\subset U_1$ and $V_2\subset U_2$ ( both containing $S$) and a diffeomorphism $f: V_1\rightarrow V_2$ which is biholomorphic in the interior of $V_1$ with respect to $\tau_1, \tau_2$.

What is the moduli space of such structures?

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I think that, even if you include the boundary $S$, they are all equivalent. In saying so, I'm assuming that the final 'it' in your first sentence refers to the plane, not to $U$ and that you are considering smooth complex structures.

The reason is this: Take a sufficiently small open neighborhood $W$ of the closure of the disk $\Delta$ that $S$ bounds that is also a disk. By the Riemann mapping theorem, there is a biholomorphism of $W$ (endowed with the given complex structure) with the unit disk $\Delta$ endowed with its standard complex structure.

The image $S'\subset \Delta$ of $S$ under that biholomorphism is a smoothly embedded curve that bounds a disk $D'\subset\Delta$. By the known boundary regularity in the Riemann mapping theorem, there is a map $(D',S')\to (\Delta,S)$ that is a biholomorphism in the interior and smooth up to and including the boundary (see the discussion at Riemann mapping theorem and smoothness on the boundary). Composing these two biholomorphisms gives the equivalence of the original $(U,S)$ with the given complex structure with a $(U',S)$ for which the complex structure is the standard one.

Now pass to germs, and you have the equivalence that you wanted.

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  • $\begingroup$ Robert, the final 'it' refers to $U$, not to the plane. Thank you for a comment, I will adjust the question now. $\endgroup$
    – Dmitri Scheglov
    Commented Jun 14, 2013 at 22:39
  • $\begingroup$ @Dmitri: I see, though, that you are assuming that '$J$ extends smoothly to $S$', which, as far as I know, means that it extends smoothly to at least some two-sided open neighborhood of $S$, in which case, my argument goes through essentially unchanged. I only need a complex structure on a small (two-sided) open neighborhood of $S$ to make the whole argument work. The conclusion is still that there is only one such up to your notion of equivalence. $\endgroup$
    – Robert Bryant
    Commented Jun 14, 2013 at 23:59
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I will assume you mean that $U$ is a "deleted" neighborhood, i.e. that it does not contain $S$, since that is the more general assumption. Then there are two equivalence classes: one represented by the standard complex structure on the open disc $D$ bounded by $S$ and the other, the ``cusp'', represented by a neighborhood of $\infty$ in $\mathbb{C}$. The proof uses that every complex structure on the annulus is biholomorphic to one of $\mathbb{C}-0$, $D-0$, or $D - \overline{r D}$ for some $r \in (0,1)$.

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  • $\begingroup$ Lee, I assume that $U$ does contain $S$, it is important. I improved formulation. $\endgroup$
    – Dmitri Scheglov
    Commented Jun 14, 2013 at 19:49
  • $\begingroup$ My answer is written with sufficient generality to cover that assumption: the cusp is ruled out, and only the other case remains. $\endgroup$
    – Lee Mosher
    Commented Jun 14, 2013 at 20:09
  • $\begingroup$ Here is what I do not quite understand. If you reduce everything to complex structures on the annuli, then how do you care about continuity near the boundary, which is not important in the annuli case? $\endgroup$
    – Dmitri Scheglov
    Commented Jun 14, 2013 at 20:18
  • $\begingroup$ @Lee, I think I now understand what you mean as it was not clear to me before. $\endgroup$
    – Dmitri Scheglov
    Commented Jun 15, 2013 at 1:57

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