3
$\begingroup$

I was reading Wendl's notes on closed holomorphic curves, and could not seem to figure out the following assertion. I think I am missing some fact or the other about Riemann surfaces.

Let $\mathcal{M}_{g,m}$ be the space of pointed Riemann surfaces. To construct this, take the space of tuples $(\Sigma, j, \Theta)$ where $\Sigma$ is a topological closed surface of genus $g$, $j$ is a complex structure, and $\Theta$ is an ordered tuple of $m$ points in $\Sigma$. Then, quotient out by the equivalence relation $(\Sigma, j, \Theta) \sim (\Sigma', j', \Theta')$ if and only if there exists some biholomorphic diffeomorphism $\phi: \Sigma \to \Sigma'$ such that $\phi$ maps $\Theta$ to $\Theta'$ in an order-preserving manner.

We can define a related space as follows. Fix a surface $\Sigma$ and a tuple of $m$ points $\Phi$. Let $\text{Diff}_+(\Sigma, \Phi)$ be the set of all orientation-preserving diffeomorphisms on $\Sigma$ that restrict to the identity map on $\Phi$. Furthermore let $J(\Sigma)$ be the space of complex structures on $\Sigma$. Then one can take the quotient $J(\Sigma)/\text{Diff}_+(\Sigma, \Phi)$.

The claim is that $\mathcal{M}_{g,m}$ is homeomorphic to $J(\Sigma)/\text{Diff}_+(\Sigma, \Phi)$, likely by the map $[(\Sigma, j, \Theta)] \to [j]$ in one direction and $[j] \to [(\Sigma, j, \Phi)]$ in the opposite direction.

Right off the bat, to show that the forward map is injective, we must show that, for any complex structure $j'$ and points $\Theta$, the tuple $(\Sigma, j', \Theta)$ is equivalent to $(\Sigma, j', \Phi)$. This reduces to the following statement:

There exists a biholomorphic automorphism on $\Sigma$ that takes the set of points $\Theta$ to $\Phi$?

I am confused because this statement is certainly not true in general. For example, if we take $g = 0$ and $m = 4$, and $j$ the standard complex structure on $\Sigma = S^2$, the biholomorphic automorphisms are Mobius transformations. These are determined by the image of $\{0, 1, \infty\}$, so the action of the Mobius transformations is not transitive on quadruples of points.

$\endgroup$
1
$\begingroup$

Somewhat embarrassingly, this confusion was due to a failure on my part to absorb the necessary definitions.

There is always an orientation-preserving diffeomorphism from $\Sigma \setminus \Phi$ to $\Sigma \setminus \Theta$ for any set $\Theta$ of the same size, by virtue of the two surfaces being trivially homeomorphic.

This makes the fact that these two maps form a homeomorphism immediate. Any class in $\mathcal{M}_{g,m}$ admits a representative of the form $(\Sigma, j, \Phi)$, and in fact the composition of the two maps is the endomorphism on $\mathcal{M}_{g,m}$ that takes $[(\Sigma, j, \Phi)]$ to itself.

The other direction is clear by definition.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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