Groups acting on Riemann Surfaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T23:55:54Z http://mathoverflow.net/feeds/question/56510 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56510/groups-acting-on-riemann-surfaces Groups acting on Riemann Surfaces Martin David 2011-02-24T10:00:57Z 2011-02-24T18:36:26Z <p>By Hurwitz theorem, order of a group $G$ of automorphisms (conformal homeomorphisms) of a compact Riemann surface of genus $g\geq 2$ is bounded above by $84(g-1)$.</p> <p><strong>1.</strong> Is there any example of a compact Riemann surface whose automorphism group is trivial?</p> <p><strong>2.</strong> Does $C_2$ act on every <strong>compact Riemann surface</strong> of genus $g\geq 2$ ? ($C_2$ acts on any <strong>compact surface</strong> of genus $g$).</p> <p><strong>3.</strong> If all Sylow-subgroups of a finite group act on the a compact Riemann surface, does it imply that the whole group acts on Riemann surface?</p> <p><strong>4.</strong> Can one suggest a survey article about groups acting on Riemann surfaces/ automorphisms of Riemann surfaces? </p> http://mathoverflow.net/questions/56510/groups-acting-on-riemann-surfaces/56511#56511 Answer by Charles Matthews for Groups acting on Riemann Surfaces Charles Matthews 2011-02-24T10:29:37Z 2011-02-24T11:12:05Z <p>Q1: A typical Riemann surface has no holomorphic automorphisms, and this implies a negative answer to Q2. I don't see that Q3 can work: the p-subgroups surely don't uniquely determine the group in general. </p> <p>The literature on these questions is quite large. <a href="http://www.jstor.org/pss/2160738" rel="nofollow">http://www.jstor.org/pss/2160738</a> is a paper on the issue of surfaces with no non-trivial automorphisms. It is a little hard to tell what you want, but some of the material on the inverse Galois theory problem (which does use curves) might help you.</p> http://mathoverflow.net/questions/56510/groups-acting-on-riemann-surfaces/56531#56531 Answer by JSE for Groups acting on Riemann Surfaces JSE 2011-02-24T16:07:15Z 2011-02-24T16:07:15Z <p>I agree with the answers above. <a href="http://www.rose-hulman.edu/~brought/" rel="nofollow">Allen Broughton</a> at Rose-Hulman is a guy who has written a lot about automorphisms of Riemann surfaces: his paper Classifying finite group actions on surfaces of low genus, J. of Pure &amp; Appl. Algebra 69 (1990), 233-270 will probably be of use.</p> http://mathoverflow.net/questions/56510/groups-acting-on-riemann-surfaces/56546#56546 Answer by Yuri Zarhin for Groups acting on Riemann Surfaces Yuri Zarhin 2011-02-24T18:36:26Z 2011-02-24T18:36:26Z <p>A counterexample to Q3 is provided by the genus 2 compact Riemann surface $X$ of $y^2=x^5-1$. Indeed, the order 10 cyclic group $C_{10}$ acts on $X$ (by changing sign of $y$ and multiplying $x$ by $5$th roots of unity). It is known that the jacobian of $X$ has endomorphism ring $Z[\zeta_5]$ - the $5$th cyclotomic ring of integers and any finite multiplicative subgroup of $Z[\zeta_5]$ is a subgroup of $\mu_{10}\cong C_{10}$. This implies that $Aut(X)=C_{10}$. On the other hand, the dihedral group $D_{10}$ of order $10$ has the same Sylow subgroups as $C_{10}$ but is not isomorphic to it. In other words, there is no faithful action of $D_{10}$ on $X$ while its Sylow subgroups $C_5$ and $C_2$ act faithfully on $X$.</p> <p>If $Y$ is a compact Riemann surface of genus $g$ and its jacobian $J$ has no nontrivial automorphisms (i.e., $End(J)$ is the ring of integers $Z$) then either $Y$ is non-hyperelliptic and $Aut(Y)={1}$ or $Y$ is hyperelliptic and $Aut(Y)=C_2$. For example, if $g>1$ and $Y_g$ is the hyperelliptic Riemann surface $y^2=x^{2g+1}-x-1$ then its jacobian $J_g$ has no nontrivial endomorphisms (Math. Research Letters 7 (2000), 123--132) and therefore $Aut(Y_g)=C_2$. If $p$ is an odd prime then for each integer $n \ge 5$ the automorphism group of the compact Riemann surface $y^p=x^n-x-1$ is the cyclic group $C_{p}$. Indeed, the endomorphism ring of the jacobian is the $p$th cyclotomic ring $Z[\zeta_p]$ (Math. Proc. Cambridge Philos. Soc. 136 (2004), 257--267) and one may easily check, using the differentials of the first kind that the curve is non-hyperelliptic.</p> <p>Using Del Pezzo surfaces of degree 2, one may construct non-hyperelliptic genus 3 curves $Y$, whose jacobian has no nontrivial endomorphisms (AMS Translations Series 2, vol. 218 (2006), 67--75; MR2279305, 2007k:14060) and therefore $Aut(Y)={1}$. For the genus 4 case see a paper of Anthony Várilly-Alvaradoa and David Zywina (LMS Journal of Computation and Mathematics (2009), 12: 144-165); their approach makes use of Del Pezzo surfaces of degree 1 (see also Math. Ann. 340 (2008), 407--435).</p>