4
$\begingroup$

Let $\Gamma_g$ be a surface group of genus $g \geq 2$. That is, there is a presentation $$\Gamma_g = \langle x_1, y_1, \dots, x_g, y_g \vert \prod_{i = 1}^{g}[x_i,y_i] = 1\rangle$$

Is there a nontrivial, finitely generated $N \lhd \Gamma_g$ of infinite index?

More generally, Can there be a finitely generated $K \leq \Gamma_g$ of infinite index, containing a nontrivial $N \lhd \Gamma_g$?

$\endgroup$
7
  • $\begingroup$ Does not a comparison of "Euler characteristics indicate that $N$ must be free, and if $n$ is the number of generators of $N$ , then $1-n=q(2-2g)$ which shows that $q$ the index must be finite? $\endgroup$ Sep 18, 2014 at 8:46
  • 1
    $\begingroup$ You can only use the formula if you already know that the index is finite (note that you did not use the normality at all). Anyway, $N$ is indeed free. $\endgroup$
    – Pablo
    Sep 18, 2014 at 8:48
  • $\begingroup$ Pablo: Do you know about limit sets of Fuchsian groups? $\endgroup$
    – Misha
    Sep 18, 2014 at 10:24
  • $\begingroup$ @Misha: No I don't, but curious to learn about anything new. $\endgroup$
    – Pablo
    Sep 18, 2014 at 10:25
  • $\begingroup$ @Venkataramana: the same argument would suggest, incorrectly, that $F_2$ must have finite index in $F_3$... $\endgroup$ Sep 18, 2014 at 18:37

2 Answers 2

7
$\begingroup$

The answer to both questions is 'no'. This was proved by Greenberg for Fuchsian groups. One outline of the proof is as follows.

  1. Any finitely generated subgroup $H$ of a surface group $\Gamma$ is quasiconvex. So the map $H\to\Gamma$ induces an injection of Gromov boundaries $\partial H\to\partial\Gamma\cong S^1$.

  2. If $K\lhd \Gamma$ is non-trivial then the only closed, $K$-invariant subset of $\partial\Gamma$ is the whole of $\partial\Gamma$.

  3. Therefore $\partial H=\partial\Gamma$.

  4. Therefore $H$ is of finite index in $\Gamma$.

As you can see, this works for any quasiconvex subgroup of any hyperbolic group. This was proved by Gersten--Short.

$\endgroup$
6
  • $\begingroup$ How do you show the implication 4? $\endgroup$
    – Igor Rivin
    Sep 18, 2014 at 12:00
  • 1
    $\begingroup$ Igor, I guess to streamline things I should have said 'geometrically finite' instead of 'quasiconvex'. Anyway, geometrically finite subgroups act properly discontinuously and cocompactly on the convex hull of their limit set; in this case that's the whole of $\mathbb{H}^2$, so the quotient $\mathbb{H}^2/H$ is compact. $\endgroup$
    – HJRW
    Sep 18, 2014 at 13:21
  • $\begingroup$ @HJRW for Fuchsian groups, is there some really simple way to show that there are no finite normal subgroups? This is a special case of Greenberg's theorem. By "really simple" I mean that no geometry/topology should be involved - I want something that stems from the definitions. $\endgroup$
    – Pablo
    Sep 6, 2015 at 12:08
  • 1
    $\begingroup$ @Pablo: first things first, what's your favourite definition of a Fuchsian group? Mine is a discrete subgroup of $\mathrm{Isom}(\mathbb{H}^2)$, but perhaps you're thinking of defining them via presentations? $\endgroup$
    – HJRW
    Sep 7, 2015 at 9:41
  • $\begingroup$ @HJRW I would like to know of a definition using presentations. What I know is the definition given in en.wikipedia.org/wiki/Fuchsian_group which just says that it is a discrete subgroup of $\mathrm{PSL}(2,\mathbb{R})$. This is (seemingly) equivalent to your definition, but without the action on the hyperbolic plane in mind. So, I would prefer to have some simple argument using only matrices. $\endgroup$
    – Pablo
    Sep 7, 2015 at 10:48
3
$\begingroup$

If $N$ has infinite index in a surface group $S$ then it is free. (This can be seen topologically, or by invoking a theorem of Strebel on subgroups of Poincare duality groups.) Now $H^2(S;Z[S])\cong H^1(S/N;H^1(N;Z[S])$, by the LHS spectral sequence, since $H^p(N;Z[S])=0$ for $p\not=1$. If $N$ is finitely generated then this is in turn $H^1(S/N;Z[S/N)\otimes{H^1(N;Z[N])}$. But $H^2S;Z[S])$ is infinite cyclic, since $S$ is a surface group. Therefore $S/N$ and $N$ each have one end, i.e., are virtually infinite cyclic, and so $S$ is virtually $Z^2$. This contradicts genus $\geq2$.

$\endgroup$
1
  • $\begingroup$ Interesting argument! I think you mis-spoke slightly: having one end precisely does not imply being virtually infinite cyclic---infinite cyclic groups have two ends. $\endgroup$
    – HJRW
    Sep 22, 2014 at 7:09

Your Answer

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