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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 $84(g-1)$.

1. Is there any example of a compact Riemann surface whose automorphism group is trivial?

2. Does $C_2$ act on every compact Riemann surface of genus $g\geq 2$ ? ($C_2$ acts on any compact surface of genus $g$).

3. 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?

4. Can one suggest a survey article about groups acting on Riemann surfaces/ automorphisms of Riemann surfaces?

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    $\begingroup$ I would have thought that "almost all" compact Riemann surfaces of genus >2 have trivial automorphism group. Any hyperelliptic curve (in particular any Riemann surface of genus two) has a hyperelliptic involution (i.e. action by $C_2$), but these surfaces are quite special. $\endgroup$
    – Daniel Loughran
    Commented Feb 24, 2011 at 10:27
  • $\begingroup$ @Daniel: Can you explain first statement in your comments? $\endgroup$
    – Martin David
    Commented Feb 24, 2011 at 10:37
  • $\begingroup$ @Martin: A high-brow (and perhaps slightly vague) way to think about it is as follows. Let $M_g$ be the moduli space of curves (=Riemann surfaces) of genus $g>2$ (i.e. the space which parametrises all algebraic curves of genus $g$). Then the collection of curves with non-trivial automorphism group forms a closed subset, as do the collection of hyperelliptic curves. When you construct this moduli space you need to take these automorphisms into account, and the automorphism groups of your objects better be finite or things get funny. These ideas are prevalent in Geometric invariant theory. $\endgroup$
    – Daniel Loughran
    Commented Feb 24, 2011 at 11:18
  • $\begingroup$ Sorry I should have also made clear that these closed subsets are also not open, i.e. not the empty set or the whole space! $\endgroup$
    – Daniel Loughran
    Commented Feb 24, 2011 at 11:20
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    $\begingroup$ If a complex curve of genus $g>1$ has nontrivial automorphism group, one can show that the (complex) dimension of the deformation space is $3g-3$, and use another theorem of Hurwitz to compare that with the deformation space of the quotient curve equipped with the finite collection of branch points. If $g>2$, you find that there are directions you can deform the curve such that any symmetry is destroyed - that is, the locus of curves with nontrivial automorphisms has strictly smaller dimension than the full moduli space. $\endgroup$
    – S. Carnahan ♦
    Commented Feb 24, 2011 at 15:45

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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 $X$.

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.

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).

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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.

The literature on these questions is quite large. http://www.jstor.org/pss/2160738 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.

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I agree with the answers above. Allen Broughton 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 & Appl. Algebra 69 (1990), 233-270 will probably be of use.

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