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Let $\Sigma$ be a closed orientable surface of genus $g$ with $m$ marked points $x=\{x_1, \ldots, x_m\}$ and $j_0$ denote a complex structure on $\Sigma$. Take a neighborhood $U$ of the isomorphism class of $(\Sigma, x, j_0)$ in the moduli space $\mathcal{M}_{g,m}$ of genus $g$ Riemann surfaces with $m$ marked points.

My question is the following:

Can one take a compact subset $K$ of $\Sigma$ such that $K$ does not contain any marked point, and a representative of each element of a neighbourhood $U$ of $j_0$ can be obtained from $j_0$ by deforming it only on $K$?

Any comment will be helpful. Thank you in advance.

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  • $\begingroup$ Shouldn't taking $K$ to be $\Sigma$ minus a small disc around each marked point work? We deform $K$ by deforming $\Sigma$, then subtracting a small disc of the same radius. This works as long as the neighborhood $U$ is bounded in $\mathcal M_{g,m}$, so the distance between any two marked points does not go to $0$. $\endgroup$
    – Will Sawin
    Commented Feb 5, 2021 at 3:19
  • $\begingroup$ Thanks for the comment! No, I shouldn't take $K$ in such that way when $U$ is a large neighborhood of $j_0$. Then, what about the case $U$ is sufficiently small? $\endgroup$
    – Math1016
    Commented Feb 5, 2021 at 15:07
  • $\begingroup$ I don't understand your comment. I believe my argument shows that as long as $U$ is sufficiently small, the $K$ I described works. Does that answer your question, or would you like to put additional restrictions on $K$? If it does I can write it up as an answer. $\endgroup$
    – Will Sawin
    Commented Feb 5, 2021 at 15:14
  • $\begingroup$ Well, I believe that your argument works as long as $U$ is sufficiently small. In practice, I don't really see the reason such small deformations of $j_0$ form a small neighbourhood $U$ of $j_0$. Is this easy to see? $\endgroup$
    – Math1016
    Commented Feb 6, 2021 at 2:24

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This is true if $U$ satisfies an extra boundedness condition: There exists $\delta>0$ such that for all $(\Sigma',x', j_0') \in U$, the distance between any two marked points $x_i',x_j'$ is at least $\delta$.

This boundedness condition follows from the mild boundedness condition that the closure of $U$ is compact. (If so, then the closure of $U$ remains closed in $\overline{\mathcal M}_{g,n}$, hence does not intersect the boundary, but if the distance between two points goes to zero then in the limit those points collide and bubble off, forming a Riemann surface on the boundary.)

To check this, take $\epsilon < \delta /2$ and define $K$ to be $\Sigma$ minus the open ball of radius epsilon around each marked point $x_i$. Then $\Sigma \setminus K$ is a union of $n$ balls of radius $\epsilon$, with a marked point in the center.

Every other surface $\Sigma'$ in $U$ has the same form: We can write it as the union of $(\Sigma \setminus K)$ and a varying compact surface $K'$, that being $\Sigma'$ minus an open ball of radius $\epsilon$ around each part. Thus, it is obtained from $\Sigma$ by deforming $K$ and leaving $|Sigma \setminus K$ fixed.

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  • $\begingroup$ Thank you so much! $\endgroup$
    – Math1016
    Commented Feb 9, 2021 at 9:31

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