# Deform a complex structure fixing marked points

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 $$K$$?

• 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 $0$. Commented Feb 5, 2021 at 3:19
• 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? Commented Feb 5, 2021 at 15:07
• 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. Commented Feb 5, 2021 at 15:14
• 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? Commented Feb 6, 2021 at 2:24

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.