Automorphisms of Riemann surface and mapping class - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T02:13:41Z http://mathoverflow.net/feeds/question/78275 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78275/automorphisms-of-riemann-surface-and-mapping-class Automorphisms of Riemann surface and mapping class Guangbo Xu 2011-10-16T18:40:26Z 2011-10-16T21:06:12Z <p>For a higher genus Riemann surface $\Sigma$, is it true that every nontrivial (holomorphic) automorphism is of nontrivial mapping class, i.e., not isotopic to the identity?</p> http://mathoverflow.net/questions/78275/automorphisms-of-riemann-surface-and-mapping-class/78276#78276 Answer by Francesco Polizzi for Automorphisms of Riemann surface and mapping class Francesco Polizzi 2011-10-16T18:54:11Z 2011-10-16T18:54:11Z <p>Yes, this is an old result due to Hurwitz, and it is often used in Teichmuller theory.</p> <p>It is cited, for instance, at p. 152 of <a href="http://www.math.jussieu.fr/~lochak/textes/gga.pdf" rel="nofollow">this paper</a> by P. Lochak. However, I do not know the original reference. </p> http://mathoverflow.net/questions/78275/automorphisms-of-riemann-surface-and-mapping-class/78286#78286 Answer by Tim Perutz for Automorphisms of Riemann surface and mapping class Tim Perutz 2011-10-16T20:39:00Z 2011-10-16T20:39:00Z <p>If a surface-diffeomorphism $h$ acts trivially on rational cohomology, the Lefschetz number of $h$ is equal to the Euler characteristic of the surface $\Sigma$. By the Lefschetz fixed-point formula, this number equals the intersection number, in $\Sigma \times \Sigma$, of the graph of the automorphism with the diagonal. </p> <p>In the case that $h$ is a non-trivial holomorphic automorphism, the intersections are isolated and the intersection multiplicities positive. This can occur only when the Euler characteristic is non-negative.</p> http://mathoverflow.net/questions/78275/automorphisms-of-riemann-surface-and-mapping-class/78289#78289 Answer by Tom Church for Automorphisms of Riemann surface and mapping class Tom Church 2011-10-16T20:58:20Z 2011-10-16T21:06:12Z <p>A much more general fact is true: <em>any</em> isometry of <em>any</em> closed negatively curved Riemannian manifold is not homotopic to the identity. There are many proofs of this; one (perhaps not the most natural) is as follows. Hartman proved that if two harmonic maps $f_0,f_1\colon M\to N$ are homotopic, where $M$ is compact and $N$ is nonpositively curved, then $f_0$ and $f_1$ are the boundary of an isometrically embedded product $f\colon M\times [0,1]\to N$. If $M=N$ is negatively curved, this is impossible (such a product region would have sectional curvature 0 in certain directions), so we conclude that no nontrivial isometry is homotopic to the identity.</p> <p>To apply this to your example, note that by uniformization the universal cover of the Riemann surface $C$ is the unit disk $\Delta$. Thus your curve $C$ is the quotient $\Delta/\Gamma$ by some group $\Gamma$ of biholomorphic automorphisms of $\Delta$, namely Aut($\Delta$)=PSL$_2(\mathbb{R})$. Now note that the action of PSL$_2(\mathbb{R})$ on $\Delta$ preserves the Poincaré metric (of constant curvature $-1$), which thus descends to a metric of constant negative curvature on $C$. The resulting metric is preserved by any automorphism of $C$ as a Riemann surface, so in particular your original map is an isometry in this metric.</p>