# Is the automorphism group of free group of rank two relatively hyperbolic?

By Behrstock, Drutu and Mosher [BDM], we know that the (outer) automorphism groups $$\mathrm{Aut}(F_n)$$ and $$\mathrm{Out}(F_n)$$ of free group of rank $$n$$ are not relatively hyperbolic if $$n \geq 3$$ (Theorem 9.2 of [BDM]). If $$n = 2$$, then $$\mathrm{Out}(F_2)$$ is isomorphic to $$\mathrm{GL}(2,Z)$$ so that it is virtually free. I hope to know what happens to $$\mathrm{Aut}(F_2)$$.

Since the left transvection and right transvection commute, $$\mathrm{Aut}(F_2)$$ contains a free abelian subgroup of rank two which implies that it is not hyperbolic. Then, is it relatively hyperbolic? Or is it not relatively hyperbolic?

• Sorry to be pedantic, but I thought that groups had to be hyperbolic relative to a specified subgroup, so what does it mean to ask whether $G$ is relatively hyperbolic? Is the subgroup somehow understood? – Derek Holt Apr 20 at 8:26
• @DerekHolt: People sometimes talk about groups have a "proper" relatively hyperbolic structure, meaning that they are hyperbolic relative to some proper subgroup. (Of course, every group is hyperbolic relative to itself!) The question of whether $\mathrm{Aut}(F_2)$ has a proper relatively hyperbolic structure is perfectly meaningful. (As Sam Nead points out, the answer is "no".) – HJRW Apr 20 at 9:18
• There appears to be a serious typo in the question, however: Behrstock, Drutu and Mosher, showed that these groups are NOT (properly) relatively hyperbolic. – HJRW Apr 20 at 9:20
• Ah, good catch. I decided to be bold - I edited the the question to fix the typo and I added a ref to [BDM, Theorem 9.2]. – Sam Nead Apr 20 at 10:52

## 2 Answers

The group $$\mathrm{Aut}(F_2)$$ is not relatively hyperbolic. This is contained in (the proof of) Theorem 8.1 of Behrstock-Drutu-Mosher.

We first pass to $$\mathrm{Aut}^+(F_2)$$, the preimage of $$\mathrm{SL}(2, \mathbb{Z})$$. By Remark 7.2 of Behrstock-Drutu-Mosher it is enough to consider this index two subgroup.

Let $$S$$ be a copy of the two-torus. Let $$Z = \{x, y\} \subset S$$ be a pair of points. We now define two mapping class groups. There is $$\Gamma = \mathrm{MCG}(S, Z)$$ where $$Z$$ is fixed setwise. This is the usual mapping class group of (isotopy classes of) orientation preserving homeomorphisms. The pure mapping classes are those that fix the elements of $$Z$$ pointwise. We denote this index two subgroup by $$\Gamma' = \mathrm{PMCG}(S, Z)$$.

The point-erasing map (where we erase $$y$$, say) gives a homomorphism $$\beta : \Gamma' \to \mathrm{MCG}(S, \{x\}) \cong \mathrm{MCG}(S) \cong \mathrm{SL}(2, \mathbb{Z})$$ The kernel of $$\beta$$ is isomorphic to $$\pi_1(S - \{x\}, y) \cong F_2$$. This gives the Birman short exact sequence $$1 \to F_2 \to \Gamma' \to \mathrm{SL}(2, \mathbb{Z}) \to 1$$ It also (with a bit of work) gives an isomorphism between $$\Gamma'$$ and $$\mathrm{Aut}^+(F_2)$$.

We now appeal to (the proof of) Theorem 8.1 of Behrstock-Drutu-Mosher. They show that $$\Gamma'$$ is thick of order one, and thus not relatively hyperbolic. The proof uses the connectedness of the curve complex (Harvey) and the fact that axes of pseudo-Anosov maps are Morse (Behrstock's thesis).

• Thanks for your answer. This approach is based on BDM paper and I could directly understand your answer! – Sangrok Oh Apr 22 at 17:46

Here is more direct and elementary argument.

Lemma: $$\mathrm{Aut}(W_3)$$ and $$\mathrm{Aut}(\mathbb{F}_2)$$ are isomorphic, where $$W_3$$ denotes the free product $$\mathbb{Z}_2 \ast \mathbb{Z}_2 \ast \mathbb{Z}_2$$.

Sketch of proof. Let $$F \leq W_3$$ denote the kernel of the morphism $$W_3 \twoheadrightarrow \mathbb{Z}_2$$ sending all the generators $$a,b,c$$ to $$1$$. In other words, $$F$$ coincides with the elements that can be written as reduced words of even lengths. Because $$F$$ is a characteristic subgroup freely generated by $$\{ab,bc\}$$, we find a morphism $$\mathrm{Aut}(W_3) \to \mathrm{Aut}(\mathbb{F}_2)$$. It can be checked that it is an isomorphism. $$\square$$

I found this statement in Varghese's article The automorphism group of the universal Coxeter group. In the sequel, I work with $$\mathrm{Aut}(W_3)$$ instead of $$\mathrm{Aut}(\mathbb{F}_2)$$, but it is just because I am more confortable with it. As the isomorphism above is completely explicit, you can translate everything in $$\mathrm{Aut}(\mathbb{F}_2)$$ if you want.

Proposition: $$\mathrm{Aut}(W_3)$$ is not hyperbolic relative to proper subgroups.

Proof. Fix a basis $$\{a,b,c\}$$ of $$W_3$$. For every $$g \in W_3$$, let $$\iota_g$$ denote the inner automorphism given by the conjugation by $$g$$. Also, set $$\kappa_1 : \left\{ \begin{array}{ccc} a & \mapsto & a \\ b & \mapsto & b \\ c & \mapsto & c^{ab} \end{array} \right. \text{ and } \kappa_2 : \left\{ \begin{array}{ccc} a & \mapsto & a^{cb} \\ b & \mapsto & b \\ c & \mapsto & c \end{array} \right..$$ They are partial conjugations with infinite-order images in $$\mathrm{Out}(W_3)$$. (I use the notation $$g^h=hgh^{-1}$$.) Finally, set $$\varphi := \iota_{ab} \circ \kappa_1, \ \psi := \iota_b \circ \varphi \circ \iota_b, \text{ and } \xi := \iota_{cb} \circ \kappa_2.$$ The key observation is that $$\varphi$$ (resp. $$\psi$$, $$\xi$$) fixes $$ab$$ and $$bcb$$ (resp. $$ab$$ and $$c$$, $$bc$$ and $$bab$$). As a consequence, the subgroups $$\begin{array}{l} A:= \langle \iota_{ab}, \iota_{bcb}, \varphi \rangle \\ B:= \langle \iota_{ab}, \iota_c, \psi \rangle \\ C:=\langle \iota_{bc}, \iota_{bab}, \xi \rangle \end{array}$$ split as direct products between a(n infinite) virtually free group (namely, the part in the inner subgroup) and an infinite cyclic group. This implies that, if we assume that $$\mathrm{Aut}(W_3)$$ is hyperbolic relative to some collection of subgroups $$\mathcal{P}$$, then there exist $$I,J,K \in \mathcal{P}$$ such that $$A \subset I$$, $$B \subset J$$, and $$C \subset K$$. Now, we can conclude because $$\mathcal{P}$$ must be an almost malnormal collection of subgroups. Indeed:

• Because $$I \cap J \supset \langle \iota_{ab} \rangle$$, necessarily $$I=J$$.
• This implies that $$J \supset \langle \iota_{ab}, \iota_{bcb}, \iota_c \rangle \supset \langle \iota_{(bc)^2} \rangle$$, hence $$J \cap K \supset \langle \iota_{(bc)^2} \rangle$$, and finally $$J=K$$.
• Thus, $$I=J=K$$ contains $$\langle \iota_{ab} , \iota_{bcb}, \iota_{c}, \iota_{bc}, \iota_{bab} \rangle= \mathrm{Inn}(W_3)$$.
• Finally, from the fact that $$\mathrm{Inn}(W_3)$$ is a normal subgroup in $$\mathrm{Aut}(W_3)$$, we conclude that $$I= \mathrm{Aut}(W_3)$$.

Thus, we have proved that $$\mathcal{P}$$ must contain a subgroup that is not proper. $$\square$$

• Thanks for your answer! I haven't thought about the automorphism groups of coxeter groups and it would be a nice starting point to study various automorphism groups:) – Sangrok Oh Apr 22 at 17:43