4
$\begingroup$

I hear there is a semi-famous theorem from my advisor, but he didn't know the name and I was unable to find it online. Does anybody know of the following?

Let $S$ be a compact Riemann surface. Then for a given genus $g$, up to isomorphism (holomorphism in our case), there are only finitely many compact Riemann surfaces $Y$ of genus $g$ that are images of surjective holomorphisms $\varphi : S \twoheadrightarrow Y$

Has anybody heard of such a result? It may be true only for genus 2 and higher, I have no idea on this one.

It would just be nice to mention in a talk I am giving about the Hurwitz Theorem.

$\endgroup$
  • $\begingroup$ (+1) Though the statement doesn't sound surprising, I don't recall hearing of this theorem (and can't see how to get it from Hurwitz off the top of my head). Just remarking that it's not true for $g = 1$: A torus holomorphically covers infinitely many conformally-distinct tori, so the same is true of any connected Riemann surface mapping surjectively to some torus. $\endgroup$ – Andrew D. Hwang Oct 27 '15 at 19:58
  • $\begingroup$ Good catch - thanks! $\endgroup$ – John Samples Oct 28 '15 at 5:48
  • $\begingroup$ The proof I know uses Mumford' compactness theorem. One might be able to derive this also from Parshin's Finiteness theorem. None of these is elementary or reduces to Hurwitz theorem. If you are still interested I will write a detailed proof. $\endgroup$ – Moishe Kohan Nov 8 '15 at 16:39
  • $\begingroup$ I am very interested for sure! I would definitely appreciate it. $\endgroup$ – John Samples Nov 20 '15 at 5:49
2
$\begingroup$

This can be seen as an application of the Parshin-Arakelov finiteness theorem: There are only finitely many (up to a biholomorphic isomorphism) holomorphic families of Riemann surfaces of the fixed type $(g,n)$ (where $g$ is the genus and $n$ is the number of punctures) with fixed compact base $B$. (See here and here for some references.)

Consider now the special case when $B$ is your Riemann surface $S$. The finiteness theorem then translates to the statement that there are only finitely many nonconstant holomorphic maps $B\to M_{g,n}$ where the target is the moduli space of Riemann surfaces of type $(g,n)$ (I am simplifying a bit here, you should treat the target as an orbifold/stack). Note that $M_{g,1}$ is naturally isomorphic to the universal curve $C_g\to M_g$. (This is where you need $g\ge 2$.) Suppose now that there are infinitely many genus $g$ Riemann surfaces $Y_g$ with nonconstant holomorphic maps $B\to Y_g$. Identify $Y_g$'s with suitable fibers of $C_g$. Then you obtain infinitely many nonconstant holomorphic maps $B\to M_{g,1}$, contradicting the finiteness theorem. In your setting the genus of $Y$ does not exceed the genus of $S$ (this can be derived for instance from the Riemann-Hurwitz formula). qed

As an addendum, here is an easier proof of the easier result (due to de Franchis) that given two compact Riemann surfaces $X, Y$ of genus $\ge 2$, there are only finitely many nonconstant holomorphic maps $f: X\to Y$. This proof is a model of the proofs of the harder finiteness theorem above. By the Schwarz lemma, each map $f: X\to Y$ is 1-Lipschitz with respect to the uniformizing hyperbolic metrics on $X$ and $Y$. (This is one place where genus $\ge 2$ is used.) By Arzela-Ascoli theorem, any sequence of holomorphic maps $f_n: X\to Y$ has to subconverge to a nonconstant holomorphic map $f: X\to Y$. Therefore, after passing to a subsequence, for large $n$ the maps $f_n$ are homotopic to $f$. But then $f_n=f$ for large $n$. (Use the fact that a holomorphic 1-form is uniquely determined by its periods; here again we need the fact that the genus is $\ge 2$.)

Edit. An update. The finiteness theorem is due to Severi. See

Bounds on the number of holomorphic maps,

by M.Tanabe, Proc. of AMS, 2005, for the explicit bounds on the number of pairs $(Y,f)$, given the genus of $X$.

$\endgroup$

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.