1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you so much! $\endgroup$
    – Math1016
    Commented Feb 9, 2021 at 9:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.