7
$\begingroup$

Frucht's theorem is a theorem in algebraic graph theory conjectured by Dénes Kőnig in 1936 and proved by Robert Frucht in 1939. It states that every finite group is the group of symmetries of a finite undirected graph. More strongly, for any finite group $G$ there exist infinitely many non-isomorphic simple connected graphs such that the automorphism group of each of them is isomorphic to $G$,You can see here.

I am not familiar with Riemann geometry, but I think my question has some meaning.

Do we have such Frucht's type theorem for Riemann surface? Precisely, for a finite group $G$, is there Riemann surface $M$, which its fundamental group is isomorphic to $G$?

I am interested in the special case, where $G=D_{2n}$ is the dihedral group. Also, when $G=D_{\infty}$ is interesting for me.

Thanks in adavanced.

$\endgroup$
  • 1
    $\begingroup$ Unlikely if you stick to surfaces. The only nontrivial finite fundamental group of a surface is $\mathbb Z/2=\pi_1(\mathbb R P^2)$, but $\mathbb R P^2$ is strictly speaking not a Riemann surface as it admits no complex structure. On the other hand, every finitely presented group can occur as a fundamental group of a four-manifold. $\endgroup$ – Sebastian Goette Nov 20 '15 at 20:54
  • $\begingroup$ @ShahroozJanbaz: Is the question about Riemann surfaces, or about Riemannian manifolds? These are quite different things. $\endgroup$ – Peter Mueller Nov 20 '15 at 21:46
  • $\begingroup$ @PeterMueller: Actually, as you can see in my question, I am interested in the cases where the dihedral group appears as a kind of group related to the Riemann surface (Manifold). Since I know a little about the Spectrum of compact surface, I asked my question for Riemann surface. But, if there are something for groups corresponding to Riemann manifolds which dihedral groups appears there, I will be so thankful if you give me some references. $\endgroup$ – Shahrooz Janbaz Nov 20 '15 at 22:07
10
$\begingroup$

You seem to be asking about the group of isometries, not the fundamental group. If so, for every $n$ and every finite group $G$ there is a compact hyperbolic manifold of dimension $n$ whose isometry group is $G.$ See Belolipetsky and Lubotzky. They actually do $n\geq 4,$ but the paper has references to work in lower dimensions (by Greenberg and Kojima).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Dear Prof. Rivin, thanks for your helpful answer. Actually, I need such groups which we can correspond to Riemann surface and group of isometries is very interesting. Would you please introduce to me some more references for studying in this direction? Although, I accepted your answer. $\endgroup$ – Shahrooz Janbaz Nov 20 '15 at 21:02
  • $\begingroup$ I believe the reference I gave has very extensive references, and the Greenberg paper should be quite clear, though I admit I haven't looked at it for a while... $\endgroup$ – Igor Rivin Nov 20 '15 at 21:06
  • $\begingroup$ I can not find the paper which is written by Greenberg. Where can I find the paper? $\endgroup$ – Shahrooz Janbaz Nov 20 '15 at 22:09
  • $\begingroup$ The paper by Kojima is open access, and Kojima proves Greenberg's theorem. $\endgroup$ – Igor Rivin Nov 20 '15 at 22:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.