Homogeneous hyperbolic groups

October 18, 2010

Homogeneous hyperbolic groups

Filed under: Geometric Group Theory,Logic — Henry Wilton @ 11:15 pm

I just got back from an extremely enjoyable meeting in Montreal, where I learned some nice new results about homogeneity, a natural logical property of groups. Now, I realise that logic may seem like a distant and irrelevant subject to many topologists, but I hope you’ll bear with me, as in this case I think there’s a very interesting relationship between logic and geometric group theory. If you need to be further convinced, perhaps it would help if I told you that this circle of ideas is intimately connected to Sela’s solution to the homeomorphism problem for hyperbolic manifolds. (Perhaps I’ll write some more about that on another occasion.)

If you’re interested in the logic of a group $\Gamma$ , then it makes sense to talk about the type of an element $\gamma$ . This is just the logical characterization of that element: to be precise, it’s the set of all formulae in first-order logic that the element satisfies. To take a simple example, the formula

$\phi(x)=(\forall y,~ xyx^{-1}y^{-1}=1)$

is in the type of an element $\gamma$ if and only if $\gamma$ is in the centre of $\Gamma$ . It’s obvious that elements of the same $\mathrm{Aut}(\Gamma)$ -orbit will have the same type, so it’s natural to ask whether the converse holds.

Definition: A group $\Gamma$ is 1-homogeneous if, whenever $\gamma,\delta\in\Gamma$ have the same type, there is an automorphism $\alpha$ of $\Gamma$ with $\alpha(\gamma)=\delta$ . You can make the same definition for $k$ -tuples $\underline{\gamma}=(\gamma_1,\ldots,\gamma_k),\underline{\delta}=(\delta_1,\ldots,\delta_k)$ , in which case $\Gamma$ is called $k$ -homogeneous. If it’s $k$ -homogeneous for all $k$ then I will simply call $\Gamma$ homogeneous.

(Note that this only coincides with the standard definition when $\Gamma$ is countable.)

As a result of the epic projects of Sela and, independently, Kharlampovich and Miasnikov, the logic of finitely generated non-abelian free groups is fairly well understood, so it makes sense to ask whether free groups are homogenous. Free groups have enormous automorphism groups, so you might guess that they are indeed homogeneous, and this has been confirmed by Perin–Sklinos and Ould Houcine.

Theorem (Perin–Sklinos, Ould Houcine): Finitely generated, non-abelian free groups are homogeneous.

Having dealt with that case, the obvious next thing to do is to look at groups of the form $\pi_1(\Sigma)$ , where $\Sigma$ is a closed, orientable, hyperbolic surface. These also have a very large number of automorphisms, so one might hope that surface groups are also homogeneous. But it turns out that this is not the case, and it’s easy to see this if one uses some known facts. A subgroup $H\subseteq\Gamma$ is called elementary if the inclusion map preserves types. The following is a very special case of a theorem of Sela and Perin (which actually classifies all elementary subgroups of all torsion-free hyperbolic groups).

Theorem (Sela, Perin): The elementary subgroups of $\pi_1(\Sigma)$ are precisely the non-abelian free factors of subgroups of the form $\pi_1(\Sigma_0)$ , where $\Sigma_0\subseteq\Sigma$ is a retract of $\Sigma$ and $\Sigma_0$ is not a three-punctured sphere.

This statement may seem a little complicated, but the reason for excluding the three-punctured sphere is fundamentally topological: its mapping class group is too small.

Corollary: If the genus of $\Sigma$ is greater than two then $\pi_1(\Sigma)$ is not homogeneous.
Proof: There is a retract $\Sigma_0$ of $\Sigma$ homeomorphic to a twice-punctured torus. Let $\gamma$ be a non-separating simple closed curve on $\Sigma_0$ , and let $\delta\in\pi_1(\Sigma_0)$ be an element of a basis but not represented by a simple closed curve. Then $\gamma$ and $\delta$ are both elements of a basis, and hence are related by an automorphism of $\pi_1(\Sigma_0)$ ; in particular, they have the same type in $\pi_1(\Sigma_0)$ , and hence in $\pi_1(\Sigma)$ . But automorphisms of $\pi_1(\Sigma)$ send conjugacy classes of simple closed curves to conjugacy classes of simple closed curves, so $\gamma$ and $\delta$ are not represented by an automorphism of $\pi_1(\Sigma)$ . QED

In summary, the problem is that the logic can’t distinguish simple closed curves from non-simple closed curves on $\Sigma$ .

At this point, it’s quite natural to think that this must be the end of the story for homogeneity in hyperbolic groups. Take, for example, $M$ to be a closed hyperbolic manifold of dimension greater than two. Mostow Rigidity implies that any automorphism of $\pi_1(M)$ is induced by an isometry of $M$ , and there are at most finitely many of these: that is, $\mathrm{Out}(\pi_1(M))$ is finite. So the heuristic I mentioned above, that to be homogeneous you need a large automorphism group, suggests that there’s no hope. But a very nice observation of Ould Houcine shows that heuristic to be completely wrong. Let $\Gamma$ be a torsion-free hyperbolic group, and we will define $\Gamma$ to be rigid if $\Gamma$ does not split as $\Gamma=A*_CB$ or $\Gamma=A*_C$ , where $C$ is trivial or cyclic. You can keep $\pi_1(M)$ , for $M$ a hyperbolic manifold of dimension greater than two, in mind as an example.

Lemma: If $\Gamma$ is rigid then there is a finite set $\underline{\sigma}\subseteq\Gamma$ with the property that any map $\Gamma\to\Gamma$ either kills some $\sigma_i\in \underline{\sigma}$ or is an automorphism.
Sketch proof: First, we note that torsion-free rigid hyperbolic groups are co-Hopfian (Sela). This means that an monomorphism $\Gamma\to\Gamma$ is an automorphism, so it is enough to show that the map $\Gamma\to\Gamma$ is injective.

Fix a generating set $\underline{a}$ for $\Gamma$ . Now, if the lemma fails then there’s a sequence of homomorphisms $f_n:\Gamma\to\Gamma$ that are injective on larger and larger balls in the Cayley graph of $\Gamma$ relative to $\underline{a}$ . Using a standard limiting argument, we can take a limit in the pointed Gromov–Hausdorff topology, and extract a nice, faithful action of $\Gamma$ on a real tree $T$ . We can then apply the Rips Machine to contradict the hypothesis that $\Gamma$ is rigid. QED

Ould Houcine uses this result to deduce homogeneity, as follows.

Theorem (Ould Houcine): Rigid, torsion-free, hyperbolic groups are homogeneous.
Proof:
Let $\underline{a}$ be a generating set for $\Gamma=\langle \underline{a}\mid \underline{r}\rangle$ and let $\underline{\gamma}$ , $\underline{\delta}$ be tuples with the same type. Then $\underline{\gamma}$ satisfies the formula

$\exists \underline{a},~(\underline{x}=\underline{\gamma}(\underline{a}))\wedge (\underline{\sigma}(\underline{a})\neq 1)\wedge (\underline{r}(\underline{a})= 1)$ .

The fact that $\underline{\delta}$ also satisfies this formula shows that

$\exists \underline{b},~(\underline{\delta}=\underline{\gamma}(\underline{b}))\wedge (\underline{\sigma}(\underline{b})\neq 1)\wedge (\underline{r}(\underline{b})= 1)$ .

The assignment $\underline{a}\mapsto\underline{b}$ now defines a homomorphism $\Gamma\to\Gamma$ that sends $\underline{\gamma}$ to $\underline{\delta}$ and that does not kill any elements of $\underline{\sigma}$ . By the lemma, this map is an automorphism. QED

In summary, the class $\mathcal{H}$ of homogeneous, torsion-free, hyperbolic groups is a naturally occurring class of hyperbolic groups that includes free groups and hyperbolic $n$ -manifold groups for $n>2$ , but not surface groups. I think that’s pretty interesting! The obvious next question to ask is:

Question 1: Is a free product of homogeneous hyperbolic groups homogeneous?

In fact, Perin–Sklinos’s methods can be used to give many further nice examples of non-homogeneous hyperbolic groups. For instance, let $F$ be a non-abelian free group and let $w\in [F,F]$ be such that $F$ is rigid relative to $w$ (meaning that $F$ has no free or cyclic splittings $A*B$ , $A*_CB$ or $A*_C$ , as above, with $w$ conjugate into $A$ or $B$ ). Then we can build a nice, one-ended, hyperbolic group $\Gamma$ as follows.

Example: Take $\Sigma$ to be a compact, orientable surface with one boundary component, of genus equal to the commutator length of $w$ . Let $\Gamma=\pi_1(\Sigma)*_{\partial\Sigma=w}F$ . Note that $F$ is a retract of $\Gamma$ . By another instance of the theorem of Sela and Perin, it turns out that $F$ is an elementary subgroup of $\Gamma$ . But every automorphism of $\Gamma$ restricts to an inner automorphism of $F$ , so just as before we see that any pair of non-conjugate basis elements of $F$ will serve to prove that $\Gamma$ is not homogeneous.

More generally, I would guess that the Perin–Sklinos argument can be adapted to show quite quickly that most non-free elementarily free groups (ie group with the same first-order logic as a free group) are not homogeneous, but perhaps I’m missing something.

This still leaves an enormous number of hyperbolic groups that remain mysterious.

Question 2: Let $w\in F$ . Is the double $\Gamma=F*_{\langle w\rangle} F$ homogeneous?

Of course, in some cases you get a free group or a surface group, but the majority of cases will look very different to either of these. On the one hand, typically the outer automorphism group is quite small ( $\mathbb{Z}$ , in fact, generated by the Dehn twist in $w$ ). On the other hand, if I have absorbed the Perin–Sela theorem correctly, these groups have no elementary subgroups, so there’s no hope of using an argument like the one above to show that they’re not homogeneous.

Comments (3)

3 Comments »

Maybe the heuristics of “a homogeneous model has lots automorphisms” can be refined as “a homogeneous model either has lots of automorphisms, or you can say a lot about its elements in first order”. A good example is that of $\mathbb{R}$ in the language of fields: it has no automorphisms, but each element is the only one of its type (it’s uniquely defined by which rationals are smaller), so it is homogeneous.

Comment by Chloé Perin — October 21, 2010 @ 7:11 am | Reply
“More generally, I would guess that the Perin–Sklinos argument can be adapted to show quite quickly that most non-free elementarily free groups (ie group with the same first-order logic as a free group) are not homogeneous, but perhaps I’m missing something.”

Actually, it follows from general model theory that every group $G$ whatsoever (or, more general, structure in the sense of first-order logic) can be embedded into a group $G^*$ which is $k$ -homogeneous for every $k$ (in the sense of your definition) and such that $G$ is an elementary subgroup of $G^*$ (in particular, has the same first-order theory). (Keyword: saturated models.)

Comment by Matthias Aschenbrenner — October 21, 2010 @ 3:10 pm | Reply
- Thanks for pointing this out, Matthias. I missed out a crucial qualification, viz: I really mean finitely generated elementarily free groups (in my universe, all groups are finitely generated!).
  
  In the case of a finitely generated elementarily free group $G$ , one expects to find an elementary subgroup $H$ with an automorphism that doesn’t extend to $G$ .
  
  Comment by Henry Wilton — October 21, 2010 @ 6:28 pm | Reply

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