Given a Riemann surface $X$, we have some fairly standard methods for identifying which groups $G$ admit an effective and holomorphic action $G \times X \to X$. For instance, some fairly elementary arguments show that the only finite groups acting effectively and holomorphically on $\mathbb{P}^1$ are finite cyclic groups, $D_4, A_4, S_4$, and $A_5$. (See e.g. Miranda, Ch. III p. 80 for these arguments.) Another result is Hurwitz' theorem, which says that given a Riemann surface $X$, the only such finite groups must have order $\leq 84(g(X) - 1)$.

Now let's instead fix a group $G$. (Maybe you want $G$ to be a Lie group, maybe finite.) What results tell us about the (non)existence of a Riemann surface $X$ on which $G$ acts effectively and holomorphically? I'd be surprised to hear that given $G$, we can always construct such an $X$.

Unlike the problem of finding groups given $X$, I have not heard anything about this kind of problem while learning about Riemann surfaces.

  • $\begingroup$ Presumably you want a connected Riemann surface? $\endgroup$
    – LSpice
    Sep 30 at 20:01
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    $\begingroup$ I like this question a lot. Your question also motivates if given a group Theres some kind of structure on the set of Riemann Surfaces which the group acts effectively and holomorphically upon. $\endgroup$ Sep 30 at 20:08
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    $\begingroup$ Every countable group acts effectively and holomorphically (and freely, properly discontinuously) on some Riemann surface. This is an elementary exercise in the theory of covering spaces. Of course, if you also consider actions of Lie groups of positive dimension, your options are very limited. $\endgroup$ Sep 30 at 21:12
  • $\begingroup$ @MoisheKohan I understand when you drop reference to complex structure, but I'd need to convince myself the argument still works when you do not. But I guess that just amounts to checking an explicit construction. And there aren't many to check. $\endgroup$ Sep 30 at 23:34
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    $\begingroup$ OK, I guess I will have to write an answer then. The thing is that every countable group is a quotient of a free group of countably infinite rank. The latter is obviously the fundamental group of a Riemann surface (complex plane minus the set of integers). $\endgroup$ Sep 30 at 23:55

2 Answers 2


In fact:

Theorem: Any finite group $G$ is the automorphism group of a compact Riemann surface, and more generally a smooth projective algebraic curve over any algebraically closed field.

The Riemann surface case is proved by L. Greenberg, Maximal groups and signatures, Ann. Math, Studies 1973, and second by M. Madan and M. Rosen, The automorphism group of a function field. Proc. Amer. Math. Soc(1992)


Donu's answer is correct but amounts to killing a fly with a gun shot: Greenberg proves a harder result than the one needed for the problem.

Theorem. Let $G$ be a countable group. Then there exists a Riemann surface $X$ such that $G$ is isomorphic to a subgroup of $Aut(X)$ acting properly discontinuously and freely on $X$.

Proof. Since $G$ is countable, it has (at most) countably many generators. Let $F_\infty$ denote the free group of countably infinite rank. Then there exists an epimorphism $\phi: F_\infty\to G$. There exists a Riemann surface $Y$ with the fundamental group $F_\infty$, namely, the complex plane ${\mathbb C}$ with a discrete closed countably infinite subset removed (say, the subset of integers). We obtain an epimorphism $\phi: \pi_1(Y)\to G$. Let $N$ denote the kernel of $\phi$ and $p: X\to Y$ the (connected) covering space corresponding to the subgroup $N< \pi_1(Y)$. The group $G$ acts as the Galois group of the covering $p$. Lifting the complex structure from $Y$ to $X$ via $p$, makes $X$ a Riemann surface and the group $G$ acts effectively and biholomorphically on $X$. qed

Incidentally, Greenberg's proof starts similarly, except he would use a compact Riemann surface $Y$, hence, $X$ will be also compact since $G$ is finite. The hard part of his proof is to observe (by a dimension count for suitable Teichmuller spaces) that if $Y$ is chosen generically, then $X$ has no conformal automorphisms other than those coming from $G$.

Lastly, if one allows, say, connected, nontrivial Lie groups $G$ in this problem then only few groups $G$ appear as subgroups of conformal automorphism groups: $S^1$, $S^1\times S^1$, ${\mathbb R}$, ${\mathbb C}, {\mathbb C}^*$, $PSL(2, {\mathbb R}), PSL(2, {\mathbb C})$, $SO(3)$, and few more solvable groups: The group of affine automorphisms of the real line, the group of complex affine automorphisms of the complex plane ${\mathbb C}$, and its subgroup ${\mathbb C}\rtimes S^1$.


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