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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?

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  • $\begingroup$ 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? $\endgroup$ – Derek Holt Apr 20 at 8:26
  • $\begingroup$ @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".) $\endgroup$ – HJRW Apr 20 at 9:18
  • $\begingroup$ There appears to be a serious typo in the question, however: Behrstock, Drutu and Mosher, showed that these groups are NOT (properly) relatively hyperbolic. $\endgroup$ – HJRW Apr 20 at 9:20
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    $\begingroup$ 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]. $\endgroup$ – Sam Nead Apr 20 at 10:52
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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).

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

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  • $\begingroup$ 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:) $\endgroup$ – Sangrok Oh Apr 22 at 17:43

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