Let $G$ be a finitely generated Fuchsian group.

(i.e. a discrete subgroup of $\mathrm{PSL}_2(\mathbb{R})$).

Is it true that $d(G) < 2\beta_{2}^1(G) + 1$ ?

Here, $\beta_{2}^1(G)$ stands for the first $L^2$-Betti number of $G$, and $d(G)$ is the smallest cardinality of a generating set of $G$.

  • 1
    $\begingroup$ Why would you think that? What is the background? $\endgroup$ – Igor Rivin Nov 18 '15 at 16:30
  • $\begingroup$ @IgorRivin Well, this holds for free nonabelian groups, and hopefully also for surface groups (right?). Maybe my claim can be deduced from these facts using the observation that Fuchsian groups are virtually surface groups. Furthermore, if this is true, I can give a quick proof of the well known fact that $G$ does not have a nontrivial finite normal subgroup. $\endgroup$ – Pablo Nov 18 '15 at 16:48
  • 1
    $\begingroup$ As @ThiKu's answer shows, the problem here is that the natural relationship is between $\beta^1_2(G)$ and the (orbifold) Euler characteristic, $\chi(G)$. In the setting of free or surface groups, $\chi(G)$ is closely related to $d(G)$, but that relationship breaks down in orbifold groups. $\endgroup$ – HJRW Nov 19 '15 at 10:21

A priori the inequality seems unlikely because for a group containing an index k surface subgroup the L^2-Betti numbers are those of the surface group divided by k, so the right hand side gets approximately divided by k, but one wouldn't expect the same for the left hand side.

To get an explicit counterexample look at the action of a Hurwitz group on a genus g surface, the Fuchsian group corresponding to the quotient orbifold being the (2,3,7)-triangle group.

The Hurwitz group has order 84(g-1) and the first L^2-Betti number of the closed surface is (2g-2), so for the (2,3,7)-triangle group it is 1/42. So the right hand side is 1+1/21. But the (2,3,7)-triangle group is certainly not cyclic, so $d(G)= 2$.

| cite | improve this answer | |

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.