An analytic proof of the De Franchis theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T14:06:20Z http://mathoverflow.net/feeds/question/43484 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43484/an-analytic-proof-of-the-de-franchis-theorem An analytic proof of the De Franchis theorem Jaikrishnan 2010-10-25T05:43:09Z 2011-04-21T20:09:31Z <p>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 $\Sigma_{g_2}$.</p> <p>The complete proof can be found in P.Samuel's book <em>Lectures on Old and New Results on Algebraic Curves.</em> 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?</p> http://mathoverflow.net/questions/43484/an-analytic-proof-of-the-de-franchis-theorem/43494#43494 Answer by Francesco Polizzi for An analytic proof of the De Franchis theorem Francesco Polizzi 2010-10-25T08:46:45Z 2010-10-25T10:44:56Z <p>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". </p> <p>Set $X:=\Sigma_{g_1}$ and $Y:=\Sigma_{g_2}$, and fix a non constant holomorphic mapping $f \colon X \to Y$. </p> <p>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 $H^0(X, f^*T_Y)$.</p> <p>Since $Y$ is of genus $g \geq 2$, we have $\deg T_Y &lt;0$, so $H^0(X, f^*T_Y)=0$. </p> <p>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 $X$. </p> <p>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. </p> <p>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 $f$. </p> <p>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:</p> <p><strong>Theorem</strong>. 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.</p> <p>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)]</p> http://mathoverflow.net/questions/43484/an-analytic-proof-of-the-de-franchis-theorem/43496#43496 Answer by Johannes Ebert for An analytic proof of the De Franchis theorem Johannes Ebert 2010-10-25T08:57:43Z 2010-10-25T16:12:00Z <p>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.</p> <p>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.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/43484/an-analytic-proof-of-the-de-franchis-theorem/62568#62568 Answer by Morris Kalka for An analytic proof of the De Franchis theorem Morris Kalka 2011-04-21T20:09:31Z 2011-04-21T20:09:31Z <p>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</p> <ol> <li>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. </li> </ol> <p>You can find arguments of this type in</p> <p>M. Kalka, B. Shiffman and B. Wong, Finiteness and Rigidity Theorems for Holomorphic Mappings, Michigan Math. J. 28 (1981), 289-295.</p>