# An analytic proof of the De Franchis theorem

The De Franchis theorem in its simplest form states that given two compact Riemann surfaces $\Sigma_{g_1},\Sigma_{g_2}$ where $g_1,g_2 > 1$, there are only finitely many non-constant holomorphic mappings from $\Sigma_{g_1}$ into $\Sigma_{g_2}$.

The complete proof can be found in P.Samuel's book Lectures on Old and New Results on Algebraic Curves. This proof uses quite a bit of Algebraic Geometry, in which I have very little background. Does anyone know how to prove this result using purely complex analytic ideas?

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Let me give a proof using the deformation theory of holomorphic maps developed by Horikawa [Journal Math. Soc. Japan 25]. This can be seen as a purely analytic proof in the spirit of Kodaira's "deformations of complex structures".

Set $X:=\Sigma_{g_1}$ and $Y:=\Sigma_{g_2}$, and fix a non constant holomorphic mapping $f \colon X \to Y$.

Let us denote by $\textrm{Mor}(X, Y)$ the space of holomorphic maps from $X$ to $Y$. It is an analytic space, whose tangent space at the point $[f]$ coincides with the space of first-order deformations of $f$, namely $H^0(X, f^*T_Y)$.

Since $Y$ is of genus $g \geq 2$, we have $\deg T_Y <0$, so $H^0(X, f^*T_Y)=0$.

This means that the morphism $f$ is rigid, in other words there are only finitely many first-order deformations of $f$ up to composition with automorphisms of $X$.

But it is well known that $|\textrm{Aut}(X)| \leq 84(g(X)-1)$, so there are only finitely many first - order deformations of $f$ at all.

This shows that every component of the space $\textrm{Mor}(X, Y)$ is a point. In general, it can happen that $\textrm{Mor}(X,Y)$ has countably many components; in this case, however, it has only finitely many of them, since the possible degrees of $f$ are bounded from above by the Riemann-Hurwitz formula. This implies that there are finitely many choices for $f$.

If you do not like deformation theory, there exists actually a purely analytic (and completely different) proof of a definitely strong result, the so called Kobayashi - Ochiai Theorem:

Theorem. Let $X$ be a Moishezon space and $Y$ a compact complex spece of general type. Then the set of meromorphic maps from $X$ to $Y$ is finite.

From the proof, that is a combination of techniques and uses in an essential way the Schwarz lemma, I refer you to the original Kobayashi-Ochiai paper [Meromorphic mappings onto compact complex spaces of general type, Inventiones Math. 31 (1975)]

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" But Mor(X,Y) is a quasi-projective variety, so it has only finitely many components": an analytic argument is given in my post below. –  Johannes Ebert Oct 25 '10 at 9:00
Right. Actually I wrote this better; the fact that there are finitely many components follows easily since \deg f must be bounded from above (Riemann-Hurwitz). –  Francesco Polizzi Oct 25 '10 at 9:09
Well, "easily" if one accepts standard facts on deformation theory. I actually would like to see a proof using only the standard theory of Riemann surfaces. Your argument about compactness seems nice, but it does not prove that every component is actually a point. Or I am missing something? Probably one can work out a simplified version of Kobayashi-Ochiai's proof for Riemann surfaces... –  Francesco Polizzi Oct 25 '10 at 10:05
@Francesco: this morning, I did not claim that my argument shows that the mapping space is dicrete. I found an argument on my wayhome from the department and updated my answer. –  Johannes Ebert Oct 25 '10 at 16:18

Here is an argument. It is a generalization of the argument that shows the finiteness of automorphism groups. I have not thought through the details, so there might be a bug.

First we show that the mapping space is compact. This is an analytic argument. You can pick, by uniformization, universal covers $E \to \Sigma_i$ ($E\subset C$ is the unit disc). Then assume that $f_n: \Sigma_1 \to \Sigma_2$ is a sequence of maps. The goal is to find a convergent subsequence. You can find lifts of these maps $g_n:E \to E$, such that $g_n(0)$ remains in a bounded ball in the Poincare metric. By Montel's theorem, you find a convergent subsequence of $g_n$; the limit is a function $g:E \to C$ (not yet to $E$). If $|g(z)|=1$ for some $z \in E$, by the maximum principle, $g$ is constant. You can exclude this case, because $g_n (0)$ was to stay in a ball, but the circle as $\infty$ away from $0$. So you can assume that $g_n$ converges to some map $g: E \to E$. $g$ will be invariant under the Deck transformation group of $\Sigma_1$ and so it descends to a map $f: \Sigma_1 \to \Sigma_2$. This shows that the space of holomorphic maps is compact. In particular, only finitely many homotopy classes can be realized by holomorphic maps.

Next we show that any homotopy class contains at most one holomorphic map (if it is not constant). I use basic facts from algebraic topology, but no algebraic geometry for that. Let $f: \Sigma_1 \to \Sigma_2$ be a holomorphic map and let $\Gamma \subset \Sigma_1 \times \Sigma_2$ be the graph; it is a complex submanifold. The normal bundle to $\Gamma$ can be identified with $f^{\ast} T\Sigma_2$. This bundle has negative degree if $f$ is not constant, namely $deg (f) \chi (\Sigma_2)$. The algebraic self-intersection number of (the homology class of) $\Gamma$ is thus negative.

If $g$ were another holomorphic map in the same homotopy class as $f$, then the graph $\Delta$ of $g$ has the same homology class as $\Gamma$. Thus the algebraic intersection number of the two graphs is, as argued above, negative. If $g$ and $f$ were different, then $\Gamma$ and $\Delta$ intersect in a finite number of point and the intersection index at each intersection point is positive (because $f$ and $g$ are both holomorphic), so the sum of the intersection indices is nonnegative. On the other hand, these intersection indices should add to the algebraic intersection number of $\Gamma$ with itself, which is negative. Contradiction.

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I like this. In fact, your argument on the normal bundle nicely agrees with the one I used to show that $f$ is rigid. The crucial point in both approaches is really that $f^∗TΣ_2$ has negative degree. –  Francesco Polizzi Oct 25 '10 at 16:38
Yes, I think this is the crucial point in most arguments about Riemann surfaces of genus larger than one. –  Johannes Ebert Oct 26 '10 at 18:05

We break the proof up into two steps: 1. The set of holomorphic maps is compact in the compact open topology. This is an easy consequence of the hyperbolicity of the Riemann surfaces

1. The space of holomorphic maps is discrete. Suppose that there is a one parameter family (f_t) of maps between the Riemann surfaces. The derviative (\frac{\partial f}{\partial t}\0 is then a holomorphic section of the pull-back of the tangent bundle of the range over the domain. By hyperbolicity there are no non-zero sections.

You can find arguments of this type in

M. Kalka, B. Shiffman and B. Wong, Finiteness and Rigidity Theorems for Holomorphic Mappings, Michigan Math. J. 28 (1981), 289-295.

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