JSJ VII: Tits alternatives

This talk was never given at the reading group due to time constraints but the original plan was to end with this topic.

 In this talk we reap some of the rewards of our previous work (more accurately we watch the magic show that Elia has put on for us).

Recall that a class of groups satisfies the Tits alternative if any finitely generated subgroup is either virtually soluble or contains a non-abelian free subgroup. The first result in this vein was proved by Tits for the class of linear groups; see a discussion here. There are multiple versions of the Tits alternative for different classes, and we will prove that $Out(G)$ satisfies what we will call the Tits* alternative: any finitely generated subgroup is either virtually polycyclic or contains a non-abelian free subgroup.

Boundary amenability is also known as coarse amenability, exactness, or Yu's property A. If $U$ is boundary amenable, then it satisfies Novikov's conjecture on higher signatures. If in addition $U$ is finitely generated, then $U$ is also known to satisfy the coarse Baum--Connes conjecture and the strong Novikov conjecture. Unfortunately, it remains unknown whether $\mathrm{Out}(G)$ is finitely generated for all special groups $G$.

The cohomological dimension ${\mathrm cd}(U)$ is the supremum of degrees in which $U$ has nontrivial group cohomology, over all coefficient modules. If $U$ has torsion, then ${\mathrm cd}(U)$ is infinite. If $U$ is torsion-free, we have ${\mathrm cd}(U)={\mathrm cd}(U')$ for all finite-index subgroups $U'\leq U$. Thus, if $U$ is virtually torsion-free, one can define the \emph{virtual cohomological dimension} (vcd for short) ${\mathrm vcd}(U)$ as the cohomological dimension of any finite-index torsion-free subgroup of $U$.

Today we will prove the following:

Theorem 1: For any virtually compact special group $G$:

• $\mathrm{Out}(G)$ is boundary amenable;
• $\mathrm{Out}(G)$ is virtually torsion-free with finite virtual cohomological dimension;
•  $\mathrm{Out}(G)$ satisfies the Tits* alternative.

Remark 2:  Let $U$ be a countable, discrete group. Let ${P}$ mean any of the following four properties: boundary amenability, the Tits$^*$ alternative, virtual torsion-freeness, finiteness of vcd. The property ${P}$ is stable under the following basic constructions.

1. Arbitrary subgroups. If $U$ satisfies ${P}$ and $V\leq U$ is any subgroup, then $V$ satisfies ${P}$. 
2. Finite-index overgroups. If $U\leq V$ has finite-index and $U$ satisfies ${P}$, then $V$ satisfies ${P}$.
3. Finite direct products. If $U_1,\dots,U_k$ satisfy ${P}$, then the product $U_1\times\dots\times U_k$ satisfies ${P}$. 
4. Extensions. Suppose that we have a short exact sequence $1\rightarrow N\rightarrow U\rightarrow Q\rightarrow 1$. If the groups $N$ and $Q$ are boundary amenable, then so is $U$. If $N$ and $Q$ satisfy the Tits$^*$ (resp.\ Tits) alternative, then so does $U$; indeed, if $N$ and $Q$ are virtually polycyclic (resp.\ virtually solvable), then so is $U$. Say that we have $1->A->B->C->1$ with $A$ and $C$ virtually polycyclic. Since polycyclic subgroups are finitely generated, $A$ is finitely generated, and so it has a finite-index polycyclic subgroup $A_0$ that is characteristic in $A$. In particular, $A_0$ is normal in $B$, and we get $1->A/A_0->B/A_0->C->1$. Since $A_0$ is polycyclic, it suffices to show that $B/A_0$ is virtually polycyclic. However, $B/A_0$ is finite-by-polycyclic by the above discussion, and finite-by-polycyclic groups are polycyclic-by-finite (prove it by induction on the length of the polycyclic group).
   If $N$ and $Q$ have finite vcd and $U$ is virtually torsion-free, then ${\mathrm vcd}(U)\leq{\mathrm vcd}(N)+{\mathrm vcd}(Q)<+\infty$. Finally, if $N$ is torsion-free and $Q$ is virtually torsion-free, then $U$ is virtually torsion-free. 
5. Quotients. Suppose that $N\lhd U$. If $N$ is amenable and $U$ is boundary amenable, then $U/N$ is boundary amenable. If $N$ is finite and $U$ satisfies the Tits$^*$ alternative, then $U/N$ satisfies the Tits$^*$ alternative (finite-by-polycyclic groups are polycyclic-by-finite).

The main ingredient will be the complexity-reduction homomorphism we construct below.

First of all, we need to introduce the correct notion of ``complexity'', as the proof of theorem 1 will be by induction on this quantity. For this --- and through most of the coming discussion --- we work with an actual special group $G$.

Definition:  The ambient rank $\mathrm{ar}(G)$ is the smallest integer $r$ such that $G$ can be embedded as a convex-cocompact subgroup of a right-angled Artin group $A_{\Gamma}$ where the graph $\Gamma$ has $r$ vertices.

If $G$ does not virtually split as a direct product, then $\mathrm{ar}(S)<\mathrm{ar}(G)$ for all $S\in\mathcal{S}(G)$; indeed, $S$ virtually splits as a product, and so it must be contained in a reducible parabolic subgroup of $A_{\Gamma}$. 

Before we continue, we need to make the following observation.

Remark 3: Let $G$ be special, let $H\leq G$ be convex-cocompact, and let $\mathcal{O}\leq\mathrm{Out}(G)$ be a subgroup preserving the $G$--conjugacy class of $H$. Let $C^G_H\leq\mathrm{Out}(H)$ be the finite subgroup determined by the conjugation action $N_G(H)\curvearrowright H$ (see the remark from the last post). If $C^G_H=\{1\}$ (for instance, if $N_G(H)=H$), then each $\phi\in\mathcal{O}$ has a uniquely defined restriction $\phi|_H\in\mathrm{Out}(H)$. As a consequence, there is a well-defined restriction homomorphism $\mathcal{O}\rightarrow\mathrm{Out}(H)$ in this case.

    In general, we can still always find a finite-index subgroup $\mathcal{O}'\leq\mathcal{O}$ with a well-defined restriction homomorphism $\mathcal{O}'\rightarrow\mathrm{Out}(H)$. In order to see this, let $\tilde{\mathcal{O}}\leq\mathrm{Aut}(G)$ be the group of automorphisms of $G$ that leave $H$ invariant and have outer class in $\mathcal{O}$. Each element of $\mathcal{O}$ is represented by at least one automorphism in $\tilde{\mathcal{O}}$. Let $r\colon\tilde{\mathcal{O}}\rightarrow\mathrm{Out}(H)$ denote the composition of the restriction homomorphism $\tilde{\mathcal{O}}\rightarrow\mathrm{Aut}(H)$ with the quotient projection $\mathrm{Aut}(H)\twoheadrightarrow\mathrm{Out}(H)$. Since $H$ is special, $\mathrm{Out}(H)$ is residually finite, and so there exists a finite-index subgroup $\mathrm{Out}'(H)\leq\mathrm{Out}(H)$ that intersects $C^G_H$ trivially. Let $\tilde{\mathcal{O}}'\leq\tilde{\mathcal{O}}$ denote the finite-index subgroup $r^{-1}(\mathrm{Out}'(H))$, and let $\mathcal{O}'$ be its projection to a finite-index subgroup of $\mathcal{O}$. By construction, any inner automorphism of $G$ that happens to lie in $\tilde{\mathcal{O}}'$ must restrict to an inner automorphism of $H$, and thus the restriction of $r$ to $\tilde{\mathcal{O}}'$ descends to a well-defined homomorphism $\overline r\colon\mathcal{O}'\rightarrow\mathrm{Out}(H)$.

We will also need the following:

Lemma 4:   

1. The collection $\mathcal{Z}_s(G)$ is $\mathrm{Aut}(G)$--invariant and closed under intersections.
2. The elements of $\mathcal{Z}_s(G)$ are precisely the elements of $\mathcal{Z}(G)$ that are contained in some element of $\mathcal{S}(G)$ (except for the trivial subgroup, which always lies in $\mathcal{Z}_s(G)$).  
3. If $Z\in\mathcal{Z}_s(G)$ and $Z\neq\{1\}$, then $N_G(Z)$ is contained in an element of $\mathcal{S}(G)$.

Proof:    Part~(1) follows from part~(2), so we only need to prove the latter. If $Z$ is non-abelian, then its parabolic part $P$ is nontrivial and we have $Z\leq N_G(P)\in\mathcal{VDP}(G)$; in this case, $Z$ is contained in an element of $\mathcal{S}(G)$. If $Z$ is abelian, we have $Z\leq Z_G(Z)$ and we can consider a maximal element $Z'\in\mathcal{Z}(G)$ containing $Z$. If $Z'$ is non-abelian, then $Z$ and $Z'$ are contained in an element of $\mathcal{S}(G)$ as above. If instead $Z'$ is abelian, then $Z'$ is an isolated abelian subgroup.

    This shows that each element $Z\in\mathcal{Z}(G)$ is contained in an element of $\mathcal{S}(G)$, except when it is an isolated abelian subgroup of rank $1$. The latter are precisely the centralisers of elements $g\in G$ with $Z_G(g)\cong\mathbb{Z}$, that is, the elements of $\mathcal{Z}_c(G)$. This completes the proof of parts~(1) and~(2).

    Part~(3) is now immediate from the fact that the normaliser $N_G(Z)$ has a finite-index subgroup of the form $Z\times Z'$, with $Z'$ trivial or infinite, by part 2 of Lemma 2 from JSJ IV. $\blacksquare$

The following result gives the complexity-reduction homomorphism mentioned above.

Proposition 5:  Let $G$ be special, $1$--ended, and not virtually a direct product. There exist a finite-index subgroup $\mathrm{Out}^1(G)\leq\mathrm{Out}(G)$ and finitely many convex-cocompact subgroups $H_1,\dots,H_k\leq G$ such that all the following hold:

• for each $i$, we have $N_G(H_i)=H_i$ and the $G$--conjugacy class of $H_i$ is $\mathrm{Out}^1(G)$--invariant;
• the kernel of the resulting restriction homomorphism $\rho\colon\mathrm{Out}^1(G)\rightarrow\mathrm{Out}(H_1)\times\dots\times\mathrm{Out}(H_k)$ is isomorphic to a special group; 
• each $H_i$ is either isomorphic to a free or surface group, or it has $\mathrm{ar}(H_i)<\mathrm{ar}(G)$.

Proof sketch:  For simplicity, denote by $\mathcal{S}(G)^*$ the union of $\mathcal{S}(G)$ with the collection of cyclic subgroups of $G$ whose conjugacy class has finite $\mathrm{Out}(G)$--orbit.  

As a warm-up, suppose that $G$ is $(\mathcal{Z}(G),\mathcal{S}^*(G))$--rigid. Let $\mathrm{Out}^0(G)\leq\mathrm{Out}(G)$ be the finite-index subgroup that preserves each $G$--conjugacy class of subgroups in $\mathcal{S}(G)$. For each $S\in\mathcal{S}(G)$, we have a well-defined restriction homomorphism $\mathrm{Out}^0(G)\rightarrow\mathrm{Out}(S)$ because $N_G(S)=S$ (recall remark 3). Choose representatives $S_1,\dots,S_k$ of the finitely many $G$--conjugacy classes in $\mathcal{S}(G)$ and consider the diagonal homomorphism $\rho\colon \mathrm{Out}^0(G)\rightarrow\prod_i\mathrm{Out}(S_i)$. We have $\mathrm{ar}(S_i)<\mathrm{ar}(G)$ for all $i$. Now, the fact that $G$ is rigid implies that $\ker\rho$ is finite: indeed, any infinite sequence in $\ker\rho$ would give a degeneration $G\curvearrowright\mathcal{X}_{\omega}$ in which all elements of $\mathcal{S}^*(G)$ are elliptic, and we would then be able to extract a $(\mathcal{Z}(G),\mathcal{S}^*(G))$--splitting of $G$, violating rigidity. Finally, since $\mathrm{Out}(G)$ is residually finite, a finite-index subgroup $\mathrm{Out}^1(G)\leq \mathrm{Out}^0(G)$ has trivial intersection with $\ker\rho$, and so the kernel of the restriction of $\rho$ to $\mathrm{Out}^1(G)$ is indeed special (it is trivial). This proves the proposition in the warm-up case.

  Now, we consider the general case. Theorem 1 of JSJ V provides an $\mathrm{Out}(G)$--invariant $(\mathcal{ZZ}(G),\mathcal{S}^*(G))$--tree $G\curvearrowright T$ (without loss of generality: a splitting) such that each vertex-stabiliser $V$ is either QH relative to $\mathcal{S}^*(G)$, or convex-cocompact and $(\mathcal{Z}(V),\mathcal{S}^*(G)|_V)$--rigid in itself. Let $\mathrm{Out}^1(G)\leq\mathrm{Out}(G)$ be the finite-index subgroup that acts trivially on the quotient graph $T/G$. Let $V_1,\dots,V_s$ be representatives of the $G$--conjugacy classes of (maximal) vertex-stabilisers of $T$. By the fact that $T$ is relative to $\mathcal{S}(G)$, each normaliser $N_G(V_i)$ is elliptic in $T$, and hence $N_G(V_i)=V_i$ for all $i$ (by maximality of $V_i$).

    Again, this shows that there is no ambiguity in restricting outer automorphisms, and we obtain a natural homomorphism $\rho\colon\mathrm{Out}^1(G)\rightarrow \prod_i\mathrm{Out}(V_i)$. 

 Up to passing to a further finite-index subgroup $\mathrm{Out}^2(G)\leq\mathrm{Out}^1(G)$, we can assume that we have well-defined restriction homomorphisms $\mathrm{Out}^2(G)\rightarrow\mathrm{Out}(E)$ for each edge group $E$ of $T$ (recall remark 3), and that the image of this homomorphism has trivial intersection with the finite subgroup $C^G_E\leq\mathrm{Out}(E)$ given by the conjugation action $N_G(E)\curvearrowright E$.The kernel is special by work of Levitt. We won't say more about this.  

Now, if $V_i$ is a QH vertex group of $T$, then $V_i$ is isomorphic to a free or surface group. However, when $V_i$ is a non-QH vertex group of $T$, we might still have $\mathrm{ar}(V_i)=\mathrm{ar}(G)$. In order to conclude the proof, we thus consider a vertex $x\in T$ whose stabiliser $V$ is not QH. Let $\mathcal{O}$ be the image of the restriction homomorphism $\mathrm{Out}^2(G)\rightarrow\mathrm{Out}(V)$. We wish to construct a homomorphism with trivial kernel from a finite-index subgroup of $\mathcal{O}$ to a finite product of outer automorphism groups of subgroups of $V$ with strictly lower ambient rank (this will prove the proposition). 

 Let $\mathcal{E}_x$ be the collection of $G$--stabilisers of edges of $T$ incident to the vertex $x$, and let ${\mathrm Cyc}(\mathcal{O})$ be the collection of cyclic subgroups of $V$ whose $V$--conjugacy class has finite $\mathcal{O}$--orbit. Recall that $V$ is $(\mathcal{Z}(V),\mathcal{S}^*(G)|_V)$--rigid in itself. Since all elements of $\mathcal{S}^*(G)$ are elliptic in $T$, it follows that the maximal elements of $\mathcal{S}^*(G)|_V$ are either elements of $\mathcal{E}_x$, or elements of $\mathcal{S}^*(G)$ that happen to be contained in $V$. The latter lie in $\mathcal{S}(V)\cup{\mathrm Cyc}(\mathcal{O})$. In conclusion, this shows that the group $V$ is $(\mathcal{Z}(V),\mathcal{S}(V)\cup\mathcal{E}_x\cup{\mathrm Cyc}(\mathcal{O}))$--rigid in itself.

We can now argue roughly as in the warm-up case. Let $\mathcal{O}_0\leq\mathcal{O}$ be a finite-index subgroup that preserves the $V$--conjugacy class of each subgroup in $\mathcal{S}(V)\cup\mathcal{E}_x$, and let $H_1,\dots,H_k$ be representatives of the finitely many $V$--conjugacy classes of non-cyclic, maximal subgroups in this collection. Each $H_i$ is self-normalising (by maximality), and we obtain a homomorphism $\rho'\colon\mathcal{O}_0\rightarrow\prod_i\mathrm{Out}(H_i)$. Each $H_i$ is a subgroup of a singular subgroup of $G$ (using Lemma 4), and hence $\mathrm{ar}(H_i)<\mathrm{ar}(G)$ for all $i$. Now, the kernel of $\rho'$ is finite: indeed, any infinite sequence in $\ker\rho'$ would give a degeneration $V\curvearrowright\mathcal{X}_{\omega}$ in which all elements of $\mathcal{S}(V)\cup\mathcal{E}_x\cup{\mathrm Cyc}(\mathcal{O})$ are elliptic, and we would be able to extract a $(\mathcal{Z}(V),\mathcal{S}(V)\cup\mathcal{E}_x\cup{\mathrm Cyc}(\mathcal{O}))$--splitting of $V$, violating rigidity. Finally, we use residual finiteness of $\mathrm{Out}(V)$ to choose a finite-index subgroup $\mathcal{O}_1\leq\mathcal{O}_0$ on which $\rho'$ is injective.

    

 Summing up, we have constructed a homomorphism $\rho\colon\mathrm{Out}^2(G)\rightarrow\prod_i\mathrm{Out}(V_i)$ with special kernel and, for each $V_i$ such that $\mathrm{ar}(V_i)=\mathrm{ar}(G)$ and $V_i$ is neither a free or surface group, a further injective homomorphism $\rho_i'\colon\mathcal{O}_1^i\rightarrow\prod_j\mathrm{Out}(H_j)$ where $\mathrm{ar}(H_j)<\mathrm{ar}(G)$ and $\mathcal{O}_1^i$ has finite index in the image of $\mathrm{Out}^2(G)\rightarrow\mathrm{Out}(V_i)$. Composing $\rho$ with the product of the $\rho_i'$ (and some identity homomorphisms), this concludes the proof of the proposition. $\blacksquare$

The next lemma is needed to extend results on $\mathrm{Out}(G)$, with $G$ special, to results on $\mathrm{Out}(U)$, with $U$ \emph{virtually} special. The proof is a mostly formal algebraic argument and hinges on remark 2.

Lemma 6:  Let $U$ be a finitely generated group and let $L\leq U$ be a finite-index subgroup.

• If $\mathrm{Out}(L)$ is boundary amenable, then $\mathrm{Out}(U)$ is boundary amenable.
•  If $\mathrm{Out}(L)$ satisfies the Tits alternative, then $\mathrm{Out}(U)$ satisfies the Tits alternative. If $L$ is normal, has finitely generated centre, and $\mathrm{Out}(L)$ satisfies the Tits$^*$ alternative, then $\mathrm{Out}(U)$ satisfies the Tits$^*$ alternative. 
•  If $U$ and $\mathrm{Out}(L)$ are virtually torsion-free and have finite ${\mathrm vcd}$, and if $\mathrm{Out}(U)$ is virtually torsion-free, then $\mathrm{Out}(U)$ has finite ${\mathrm vcd}$. 
• If $L$ is normal in $U$, if the centre of $L$ is finitely generated, if $\mathrm{Out}(L)$ is virtually torsion-free, and if $\mathrm{Out}(U)$ is residually finite, then $\mathrm{Out}(U)$ is virtually torsion-free.

Proof sketch of theorem 1: Special groups have finite cohomological dimension (as they have a finite classifying space) and their abelian subgroups have bounded rank. Moreover, it is known that for any virtually special group $U$, the group $\mathrm{Out}(U)$ is residually finite . Thus, using Lemma 5, it suffices to prove each of the three parts of the corollary for a finite-index special subgroup. We now show this by induction on the ambient rank.

First, by a Grushko argument, one can take a maximal free splitting with free factors either a free group or one-ended special subgroups and there are combination theorems for the relevant properties in this case. Hence we may assume that $G$ is 1-ended.

 Now, suppose for a moment that $G$ virtually splits as a direct product. Let $L\leq G$ be a finite-index subgroup of the form $L_1\times\dots\times L_k\times\mathbb{Z}^m$, where $k+m\geq 2$ and each $L_i$ is directly indecomposable and has trivial centre. By Lemma 5, it again suffices to prove the corollary for $\mathrm{Out}(L)$. Note that $\mathrm{ar}(L_i)<\mathrm{ar}(G)$ for all $i$. One can show that automorphisms of $L$ permute the subgroups $\langle L_i,\mathbb{Z}^m\rangle$ of $L$. Let $\mathrm{Out}^*(L)\leq\mathrm{Out}(L)$ be the group of outer classes of automorphisms of $L$ leaving each $G_i$ invariant and fixing pointwise the centre $\mathbb{Z}^m$. Denoting by $L_{\mathrm ab}$ the abelianisation of $L$, the natural homomorphism $\eta\colon\mathrm{Out}(L)\rightarrow\mathrm{Out}(L_{\mathrm ab})$ has $\ker(\eta)$ virtually contained in $\mathrm{Out}^*(L)$. 

The group $\mathrm{Out}(L_{\mathrm ab})$ is isomorphic to ${\mathrm GL}_p(\mathbb{Z})$ for some $p\in\mathbb{N}$, which enjoys all the relevant properties by classical work. On the other hand, we have $\mathrm{Out}^*(L)\cong\prod_i\mathrm{Out}(G_i)$, which satisfies all parts of the corollary by the inductive hypothesis; by remark 2, it follows that $\ker(\eta)$ is boundary amenable, satisfies the Tits$^*$ alternative, and has finite vcd. Again by remark 2, this implies that $\mathrm{Out}(L)$ is boundary amenable and satisfies the Tits$^*$ alternative, as this is true of both the image and the kernel of the homomorphism $\eta$. The same argument shows that $\mathrm{Out}(L)$ has finite vcd, provided that we find a finite-index subgroup of $\mathrm{Out}(L)$ that has torsion-free intersection with $\mathrm{Out}^*(L)$.

Consider a product of the form $G=H\times A$, where $H$ is a finitely generated group with trivial centre, and $A\cong\Z^N$ for some $N\geq 1$. The automorphism group $\mathrm{Aut}(G)$ can be described as follows. Consider the set $\mathcal{M}(H,A)$ of formal matrices 

\[\begin{pmatrix} \varphi & 0 \\ \alpha & \psi \end{pmatrix}\] 

where $\varphi\in\mathrm{Aut}(H)$, $\psi\in\mathrm{Aut}(A)\cong{\rm GL}_N(\Z)$ and $\alpha\in \mathrm{H}^1(H,A)$; in other words, $\alpha$ is a homomorphism $H\rightarrow A$. We can make $\mathcal{M}(H,A)$ into a group by endowing it with a natural product:

\[ \begin{pmatrix} \varphi_1 & 0 \\ \alpha_1 & \psi_1 \end{pmatrix}\cdot \begin{pmatrix} \varphi_2 & 0 \\ \alpha_2 & \psi_2 \end{pmatrix}= \begin{pmatrix} \varphi_1\varphi_2 & 0 \\ \alpha_1\varphi_2+\psi_1\alpha_2 & \psi_1\psi_2 \end{pmatrix}. \]

There is an action $\mathcal{M}(H,A)\curvearrowright G$ given by 

\[ \begin{pmatrix} \varphi & 0 \\ \alpha & \psi \end{pmatrix}\cdot (h,a)=(\varphi(h),\alpha(h)+\psi(a)), \]

which corresponds to a homomorphism $\iota\colon\mathcal{M}(H,A)\rightarrow\mathrm{Aut}(G)$. It is not hard to see that this is actually an isomorphism.

    

    The existence of a finite-index subgroup of $\mathrm{Out}(L)$ that has torsion-free intersection with $\mathrm{Out}^*(L)$ is now a straightforward consequence of the description of automorphisms of a product in terms of formal triangular matrices, using that $\mathrm{Out}^*(L)$ is virtually torsion-free by the inductive hypothesis. Specifically, the triangular matrices have one entry in $\mathrm{Out}^*(L)$, one in $\mathrm{Out}(\mathbb{Z}^m)={\mathrm GL}_m(\mathbb{Z})$ and one in $\hom(\prod_iG_i,\mathbb{Z}^m)$; if we only pick the first entry in a finite-index subgroup of $\mathrm{Out}^*(L)$, we get the required finite-index subgroup of $\mathrm{Out}(L)$.

    Summing up, this proves the entire corollary when $G$ virtually splits as a direct product.

    We are now in the setting of a 1-ended special group which doesn't virtually split as a direct product so we can invoke proposition and the fact that all the properties are known for outer automorphism groups of free and surface groups, as well as all special groups (the kernel of the complexity reduction homomorphism). We are then done by remark 2 $\blacksquare$.