2.1. Right-angled Artin groups

Let $\Gamma$ be a finite graph with vertex set $V(\Gamma )$ . The right-angled Artin group determined by the graph $\Gamma$ is the finitely presented group $A_\Gamma$ with the presentation:

\begin{equation*} A_\Gamma = \langle v \in V(\Gamma ) \, | \, vw=wv \text { if $v$ and $w$ span an edge in $\Gamma $.} \rangle \end{equation*}

We will always assume subgraphs $\Delta \subset \Gamma$ are full, so that two vertices in $\Delta$ are connected by an edge if and only if they are connected by an edge in $\Gamma$ . We will also assume that our graphs are simple, so that there are no loops and no double edges in $\Gamma$ (this is safe to do as relations given by loops or double edges do not change the group obtained from the above presentation). We make use of the following facts:

• The centre of a right-angled Artin group $A_\Gamma$ is generated by the vertices $v$ that are adjacent to every other vertex.

• A RAAG $A_\Gamma$ is one-ended if and only if $\Gamma$ is connected and contains at least two vertices ([Reference Groves and Hull18] proves something much stronger than this - probably the simplest way to show this directly is by using [Reference Groves and Hull18, Lemma 1.1]).

For introductions to RAAGs and their automorphisms, we recommend the survey papers of Charney [Reference Charney13] and Vogtmann [Reference Vogtmann28].

2.2. Flag complexes, the Cohen–Macaulay condition, and duality for RAAGs

For a simple graph $\Gamma$ , we use $\hat{\Gamma }$ to denote the flag complex determined by $\Gamma$ . One can define $\hat{\Gamma }$ as being obtained from $\Gamma$ by filling in any ‘visible’ simplices, or as the the largest simplicial complex on the vertex set $V(\Gamma )$ with the same edge set as $\Gamma$ . The star of a simplex $\sigma$ is the subcomplex spanned by simplices containing $\sigma$ , and the link of $\sigma$ is the subcomplex consisting of simplices $\tau \in \mathrm{st}(\sigma )$ with $\sigma \cap \tau = \emptyset$ .

A group $G$ is a duality group of dimension $n$ if there exists a $G$ –module $D$ and an element $e \in H_n(G;\;D)$ such that the cap product with $e$ induces an isomorphism

\begin{equation*}H^{n-k}(G;\; M) \cong H_k(G;\; D \otimes M) \end{equation*}

for all $k$ and all $G$ –modules $M$ . If we can take $D=\mathbb{Z}$ in the above then $G$ is a Poincaré duality group (Bieri and Eckmann allow a nontrivial action on $\mathbb{Z}$ in this definition). We do not work with the definition in this note, instead relying on the following theorem of Brady and Meier.

A group $G$ is a virtual duality group if some finite-index subgroup $H$ of $G$ is a duality group. In this case, $H$ is torsion-free, and every finite-index torsion-free subgroup of $G$ is also a duality group. This is well-known and follows directly from results in [Reference Bieri and Eckmann5] but as it is important in what follows we record it below.

Proof. As $G$ is a virtual duality group, there exists a finite-index subgroup $H_0$ of $G$ that is a duality group. Let $H'=H\cap H_0$ . As $H'$ is finite-index in $H_0$ , it is also a duality group by [Reference Bieri and Eckmann5, Theorem 3.2]. By [Reference Bieri and Eckmann5, Theorem 3.3], any torsion-free, finite-index overgroup of a duality group is also a duality group. As $H'$ is also finite-index in $H$ , it follows that $H$ is a duality group.

2.3. Finiteness conditions for ${\rm{Out}}(A_\Gamma )$ and the join Lemma

In this section, we show that for our example graph in Figure 1 the group ${\rm{Out}}(A_\Gamma )$ has a finite index subgroup isomorphic to $A_{\Gamma _1} \oplus A_{\Gamma _2}$ . For experts, this is the subgroup of ${\rm{Out}}(A_\Gamma )$ generated by partial conjugations, and it is finite index as $\Gamma$ is chosen in a way so that ${\rm{Out}}(A_\Gamma )$ contains no transvections. We break this down into two steps, starting with conditions that describe when ${\rm{Out}}(A_\Gamma )$ is finite:

Going back in the other direction, finiteness of the outer automorphism group ${\rm{Out}}(A_\Gamma )$ imposes the following restrictions on the graph $\Gamma$ and its associated RAAG.

Proof. Let $\Gamma$ be a graph such that ${\rm{Out}}(A_\Gamma )$ is finite. The first condition of Proposition 2.4 implies that $\Gamma$ has at most two connected components, and if there are exactly two connected components then the star of each vertex $u$ is equal to its own component. Suppose there are exactly two components $C_1$ and $C_2$ such that for each $u \in C_i$ we have $\mathrm{st}(u)=C_i$ . This contradicts the second bullet point from Proposition 2.4: either we can find two distinct vertices $u$ and $v$ in the same component that satisfy $\mathrm{lk}(u) \subset \mathrm{st}(v)$ , or there is an isolated vertex $u$ whose link is empty and therefore contained in the star of every other vertex. Hence, $\Gamma$ is connected. Furthermore, if there are at least two vertices in $\Gamma$ then $Z(A_{\Gamma })$ must be trivial, otherwise the star of some vertex is the whole graph and we would have a contradiction to the second bullet point from Proposition 2.4.

The second proposition we use is a bit more general and follows from Guirardel and Levitt’s work on automorphism groups of free products [Reference Guirardel and Levitt19].

Proposition 2.6. Let $A$ and $B$ be one-ended groups with centres denoted $Z(A)$ and $Z(B)$ , respectively. If $A$ and $B$ have finite outer automorphism groups, then ${\rm{Out}}(A \ast B)$ has a finite-index subgroup isomorphic to

\begin{equation*} A/Z(A) \oplus B/Z(B). \end{equation*}

Sketch proof. As both $A$ and $B$ are one-ended, $G=A\ast B$ is the Grushko decomposition of $G$ . In this case, the Outer space of the free product (see [Reference Guirardel and Levitt19]) reduces to a single point: the Bass–Serre tree given by the splitting $A\ast B$ is invariant under the whole of ${\rm{Out}}(G)$ . By looking at the stabiliser of this tree ([Reference Guirardel and Levitt19, Section 5] or alternatively [Reference Bass and Jiang3, Reference Levitt24]) one obtains a subgroup ${\rm{Out}}^0(G)$ of ${\rm{Out}}(G)$ of index at most two that splits as the following short exact sequence:

\begin{equation*} 1 \to A/Z(A) \oplus B/Z(B) \to {\rm {Out}}^0(G) \to {\rm {Out}}(A) \oplus {\rm {Out}}(B) \to 1. \end{equation*}

When both ${\rm{Out}}(A)$ and ${\rm{Out}}(B)$ are finite, the kernel of this exact sequence is finite index in ${\rm{Out}}(G)$ .

Combining the above allows us to prove the Join Lemma from the introduction:

Proof of the Join Lemma. Let $\Gamma =\Gamma _1 \sqcup \Gamma _2$ be the disjoint union of two graphs with the property that their associated RAAGs $A_{\Gamma _i}$ are noncyclic and have finite outer automorphism groups. By Lemma 2.5, both $A_{\Gamma _1}$ and $A_{\Gamma _2}$ are one-ended and have trivial centres. As $A_\Gamma \cong A_{\Gamma _1} \ast A_{\Gamma _2}$ , Proposition 2.6 tells us that ${\rm{Out}}(A_\Gamma )$ has a finite index subgroup isomorphic to $A_{\Gamma _1} \oplus A_{\Gamma _2}$ .

Applying the Join Lemma to our specific example, we have:

Sketch proof. In order to apply the Join Lemma we check the conditions of Proposition 2.4 vertex-by-vertex: that is, for each $u \in \Gamma _i$ , the star of $u$ does not separate $\Gamma _i$ and if $\mathrm{lk}(u) \subset \mathrm{st}(v)$ then $u=v$ . For $\Gamma _1$ this is straightforward. For $\Gamma _2$ this is a little harder, but we feel that a line-by-line proof is not beneficial to the paper or the reader. To convince oneself that this holds, we recommend that one looks at the following cases for a vertex $u \in \Gamma _2$ : $u$ is a vertex on one of the two strings attached to the hexagon, $u$ is the endpoint of a string, $u$ is one of the two vertices in the middle of the hexagon, and lastly $u$ is one of the two points on the boundary of the hexagon that are not endpoints of a string. These cases cover all vertices in $\Gamma _2$ , and in each case the star of $u$ does not separate the graph and the link of $u$ is not contained in the star of any other vertex.

2.4. The proof of Theorem A and the Aut case

Proof of Theorem A. Let $\Gamma$ be the graph from Figure 1. By Lemma 2.7, the group ${\rm{Out}}(A_\Gamma )$ has a finite-index subgroup isomorphic to $A_{\Gamma _1} \oplus A_{\Gamma _2}$ . This is the right-angled Artin group on the graph $\Delta = \Gamma _1 \star \Gamma _2$ formed by taking the join of the two graphs $\Gamma _1$ and $\Gamma _2$ . A one-dimensional maximal simplex in $\widehat{\Gamma _2}$ (i.e. an edge on one of the strings) is contained in simplices of dimension at most three in the join $\widehat{\Delta } \cong \widehat{\Gamma _1} \star \widehat{\Gamma _2}$ , whereas $\widehat{\Delta }$ is 4-dimensional. As the maximal simplices of $\widehat{\Delta }$ are not of uniform dimension, the flag complex $\widehat{\Delta }$ is not Cohen–Macaulay. Therefore, $A_\Delta$ is not a duality group. As every finite-index, torsion-free subgroup of a virtual duality group is a duality group (Lemma 2.3), the group ${\rm{Out}}(A_\Gamma )$ is not a virtual duality group.

For completeness, we note that we can obtain a similar result in the Aut case:

Proof. As ${\rm{Out}}(A_{\Gamma _2})$ is finite and the centre of $A_{\Gamma _2}$ is trivial, the group of inner automorphisms is finite-index in ${\rm{Aut}}(A_{\Gamma _2})$ and is isomorphic to $A_{\Gamma _2}$ . As $\widehat{\Gamma _2}$ is not Cohen–Macaulay (the maximal simplices of $\widehat{\Gamma _2}$ do not all have the same dimension), the group $A_{\Gamma _2}$ is not a duality group, so that ${\rm{Aut}}(A_{\Gamma _2})$ is not a virtual duality group.

3.1. Obstructions to duality: Fouxe–Rabinovitch groups

Recall that if

\begin{equation*} \mathcal {G}=G_1 \ast G_2 \ast \cdots \ast G_k \ast F_n \end{equation*}

is a (not necessarily maximal) free factor decomposition of a group, the associated Fouxe–Rabinovitch group is the subgroup of ${\rm{Out}}(G)$ consisting of outer automorphisms $\Phi$ that have representatives $\phi _1, \ldots, \phi _k \in \Phi$ such that each representative $\phi _i$ acts as the identity when restricted to $G_i$ . This is written as ${\rm{Out}}(G;\; \mathcal{G}^t)$ .

The decomposition series constructed in our work with Day [Reference Day and Wade17] break up ${\rm{Out}}(A_\Gamma )$ into consecutive quotients that are either free-abelian, ${\rm{GL}}(n,\mathbb{Z})$ , or certain Fouxe–Rabinovitch groups. If all the consecutive quotients are virtual duality groups, then so is ${\rm{Out}}(A_\Gamma )$ (when working with more general subnormal series one has to be a little bit careful about the passage to finite index subgroups, however here we can use congruence subgroups). This leads us to ask the following question, which also appeared briefly in [Reference Day, Sale and Wade15].

For the graph in Figure 1, the RAAG subgroup $A_\Delta$ of ${\rm{Out}}(A_\Gamma )$ appears as the Fouxe–Rabinovitch group ${\rm{Out}}(A_\Gamma ;\;\{A_{\Gamma _1},A_{\Gamma _2}\}^t)$ . Here, the failure of duality comes from failure of the factor groups to be duality groups.

However, we conjecture that there is another possible obstruction. In comparison with known examples in the literature, we expect natural classifying spaces for duality groups to look uniformly of the same dimension as the group (i.e. all maximal simplices/cells are of dimension $d={\rm{cd}}(G)$ ). For Fouxe–Rabinovitch groups, classifying spaces of minimal dimension can be obtained as a blow-up of the spine of relative Outer space by replacing each simplex $\sigma$ with a copy of $\sigma \times \textrm{E}{\rm{Stab}}(\sigma )$ [Reference Day, Sale and Wade15] (at least after passing to an appropriate torsion-free f.i. subgroup). However, as we will see below, one can have simplices in relative Outer space of the same dimension whose stabilisers have different (geometric/cohomological) dimensions. As the spine is uniform, the resulting blow-up will not be uniform - some maximal simplices will be of dimension strictly less than ${\rm{cd}}(G)$ .

The group of pure symmetric outer automorphisms (also called basis-conjugating automorphisms) ${\rm{PSO}}(A_\Gamma )$ is the subgroup of ${\rm{Out}}(A_\Gamma )$ given by all outer automorphisms $\Phi$ whose representatives $\phi \in \Phi$ send every generator $v\in V(\Gamma )$ to a conjugate of itself. For $A_\Gamma =\mathbb{Z}^2 \ast \mathbb{Z}^3 \ast \mathbb{Z}^4$ , the group ${\rm{PSO}}(A_\Gamma )$ forms a Fouxe–Rabinovitch group. Either using the work in [Reference Day and Wade16] or by working directly from a presentation, one can show that

\begin{equation*}{\rm {PSO}}(\mathbb {Z}^2 \ast \mathbb {Z}^3 \ast \mathbb {Z}^4) \cong \mathbb {Z}^2 \ast \mathbb {Z}^3 \ast \mathbb {Z}^4. \end{equation*}

This gives another example of a Fouxe–Rabinovitch group that is not a virtual duality group. In this case, ${\rm{Out}}(A_\Gamma )$ fits in a short exact sequence

\begin{equation*} 1 \to \mathbb {Z}^2 \ast \mathbb {Z}^3 \ast \mathbb {Z}^4 \to {\rm {Out}}(A_\Gamma ) \to {\rm {GL}}_2(\mathbb {Z}) \oplus {\rm {GL}}_3(\mathbb {Z}) \oplus {\rm {GL}}_4(\mathbb {Z})\to 1, \end{equation*}

so unlike our first example the group ${\rm{PSO}}(A_\Gamma )$ is not finite-index in ${\rm{Out}}(A_\Gamma )$ . It would be interesting to know how this decomposition is reflected in the geometry of the outer space for the RAAG.

The following simplification of Question 3.2 is interesting in its own right:

This is true when $\mathcal{F} = \emptyset$ (by Bestvina and Feighn [Reference Bestvina and Feighn4]), and in the case where $\mathcal{F}=\mathbb{Z} \ast \mathbb{Z} \ast \cdots \ast \mathbb{Z}$ is the free factor decomposition that determines the pure symmetric automorphism group (by Brady, McCammond, Meier, and Miller [Reference Brady, McCammond, Meier and Miller7]). In both of the above cases, simplex stabilisers behave well in the spine of the (relative) outer space, and the problem illustrated in Figure 2 does not occur. However, this is not true for an arbitrary free factor system of $F_N$ .

Figure 2.

Three points in the spine of the relative outer space for $A_\Gamma =\mathbb{Z}^2 \ast \mathbb{Z}^3 \ast \mathbb{Z}^4$ , whose ${\rm{Out}}(A_\Gamma )$ -stabilizers are isomorphic to $\mathbb{Z}^2$ , $\mathbb{Z}^3$ , and $\mathbb{Z}^4$ , respectively.

3.2. Commensurability problems

Given the construction in Theorem A, it seems worthwhile to repeat the following question, a version of which appeared as Question 1.1 in [Reference Day and Wade16].

One can also ask similar questions up to quasi-isometry. The above problem is discussed at some length in the introduction of [Reference Day and Wade16], so we will limit ourselves to mentioning more recent developments. Notably, the work of Aramayona and Martinez–Perez [Reference Aramayona and Martínez-Pérez2] on when ${\rm{Out}}(A_\Gamma )$ can have property (T) has been recently extended by Sale [Reference Sale27]. Through this work, as well as Guirardel and Sale’s work on vastness properties and ${\rm{Out}}(A_\Gamma )$ [Reference Guirardel and Sale20], we now have much better control over the behaviour of outer automorphism groups of RAAGs that, roughly speaking, do not look like ${\rm{Out}}(F_n)$ or ${\rm{GL}}(n,\mathbb{Z})$ . These results give reasons to be more optimistic about the tractability of the first part of Question 3.4. The second part of this question seems much harder, given the fact that quasi-isometry and commensurability classification problems for RAAGs themselves are incredibly difficult (see [Reference Huang23, Reference Margolis25]). However, the Join Lemma does provide a way to construct families of examples $A_\Delta$ that are finite index in ${\rm{Out}}(A_\Gamma )$ for some $\Gamma$ (and now the work of Wiedmer greatly extends this [Reference Wiedmer29]). It is also worth noting that [Reference Day and Wade16, Question 1.2] gave a more general recognition problem about RAAGs, which was later answered in the negative by Bridson [Reference Bridson10].