Right-angled Artin groups that split as direct products

Question

For a finite graph $X$, let $A_X$ denote the associated right-angled Artin group. Thus $A_X$ is generated by the vertices of $X$ subject to the relations $[v,w]=1$ whenever vertices $v$ and $w$ are connected by an edge.

I have seen references to the following theorem in several places, but I can neither figure out a proof myself not find a reference that proves it:

Let $X$ be a finite graph. Assume that $A_X \cong G \times H$, where $G$ and $H$ are nontrivial groups. Then $X$ decomposes as a nontrivial join, i.e., we can partition the vertices of $X$ into two nonempty sets $V_1$ and $V_2$ such that each vertex in $V_1$ is connected by an edge to each vertex in $V_2$ and also each vertex in $V_2$ is connected by an edge to each vertex in $V_1$.

Can anyone provide me a proof or a reference for this?

Answer 1

The place I've seen this is in Koberda's RAAG notes here, see Corollary 2.15. This relies on the description of centralizers in Proposition 2.14, which is also proved in Behrstock and Charney's paper here (if you want an officially-published reference). Given Proposition 2.14, Corollary 2.15 is immediate because if $A_X\cong G\times H$ for non-trivial $G$ and $H$ then every non-trivial element $g$ has non-cyclic centralizer, so picking some $g$ that "really uses" every vertex of $X$, Proposition 2.14 implies $X$ is a non-trivial join.

Answer 2

Matt Zaremsky commented above that there should be a CAT(0) cubical proof that's "easier" modulo some hard result; I think that's exactly right, and one such proof uses the rank-rigidity theorem for CAT(0) cube complexes.

The Salvetti complex $S_X$ is formed from the presentation complex of the given presentation of $A_X$ by adding an $n$--cube in the natural way whenever its $2$--skeleton appears (more precisely, see Section 2.6 in [A].)

By construction, $\pi_1S_X \cong A_X$, and the universal cover $\widetilde S_X$ is a CAT(0) cube complex on which $A_X$ acts freely, cocompactly, and without an invariant proper convex subcomplex.

Suppose that $A_X$ splits nontrivially as a direct product $G\times H$. Since, for example, finite groups of isometries of CAT(0) spaces have fixed points, $A_X$ is torsion-free and hence $G$ and $H$ are infinite. Hence no asymptotic cone of $A_X$, and hence no asymptotic cone of $\widetilde S_X$, has a cut-point (this follows from a general property of asymptotic cones, namely that any asymptotic cone of $G\times H$ is bilipschitz equivalent to the product of an asymptotic cone of $G$ with one of $H$).

By Corollary I in [B], $\widetilde S_X$ is reducible, i.e., there are CAT(0) cube complexes $C,D$ such that $\widetilde S_X$ is isomorphic to the complex $C\times D$, and neither $C$ nor $D$ is a single vertex. (This step is the application of rank rigidity.)

Label the (oriented) edges of $S_X$ by the vertices of $X$ (generators of $A_X$) in the usual way, and pull back the labels to $\widetilde S_X$. From the presentation, two edges lying on opposite sides of a square have the same label, and from one of the characterisations of hyperplanes in a CAT(0) cube complex, this gives a well-defined labelling of the hyperplanes of $\widetilde S_X$: the label of a hyperplane $h$ is the label of some (hence any) edge $e$ such that $h\cap e$ is the midpoint of $e$.

If $h_1, h_2$ are distinct hyperplanes that intersect, then $h_1\cap h_2$ contains the barycentre of a square whose edges are labelled by the label $v_1$ of $h_1$ and the label $v_2$ of $h_2$. Projecting this square down to $S_X$ tells you that $[v_1,v_2]=1$. Moreover, intersecting hyperplanes have different labels, since $S_X$ contains no square whose edges all have the same labels.

Fix vertices $c\in C,d\in D$ and consider the subcomplexes $C\times\{d\}$ and $\{c\}\times D$. By Lemma 2.5 in [B], the hyperplanes of $\widetilde S_X$ are partitioned into two subsets: $H_C$ --- those that intersect $C\times\{d\}$ --- and $H_D$ --- those that intersect $\{c\}\times D$. By the same lemma, every hyperplane in $H_C$ intersects every hyperplane in $H_D$.

In summary, we have a surjective map $\ell:H_C\sqcup H_D\to V(X)$ that sends each hyperplane to its label. Since intersecting hyperplanes have different labels, and hyperplanes in $H_C$ intersect hyperplanes in $H_D$, we get a partition $V(X) = \ell(H_C) \sqcup \ell(H_D)$ with all elements of the first subset commuting with all elements of the second. It remains to observe that $H_C$ and $H_D$ are nonempty, since $C$ and $D$ are CAT(0) cube complexes each having at least one edge.

(So from that point of view, the statement about product RAAGs is a manifestation of the fact that product decompositions of cube complexes correspond to partitions of the hyperplanes into two disjoint crossing sets, and the role of "RAAGness" is to translate that into an algebraic statement.)

[A]: Charney, Ruth, An introduction to right-angled Artin groups., Geom. Dedicata 125, 141-158 (2007).

[B]: Caprace, Pierre-Emmanuel; Sageev, Michah, Rank rigidity for CAT(0) cube complexes., Geom. Funct. Anal. 21, No. 4, 851-891 (2011).

Answer 3

Assuming some geometry, here is an argument avoiding rank-one rigidity.

Assuming that our graph $\Gamma$ is not a join and contains at least two vertices, we want to construct an element in the right-angled Artin group $A(\Gamma)$ whose centraliser is infinite cyclic. Saying that $\Gamma$ is not a join amounts to saying that its opposite graph $\Gamma^{\mathrm{opp}}$, namely the graph whose vertex-set is $V(\Gamma)$ and whose edges connect two vertices whenever they are not adjacent in $\Gamma$, is connected. Thus, we can fix a path $\gamma$ in $\Gamma^{\mathrm{opp}}$ passing through all the vertices at least once and whose endpoints are adjacent. Let $g \in A(\Gamma)$ denote the element given by the product of the generators successively encountered along $\gamma$, say $g=u_1 \dots u_k$.

The Cayley graph $M(\Gamma)$ of $A(\Gamma)$ given by the canonical generators is a median graph$^\ast$. We will use some median geometry to prove that the centraliser of $g$ is infinite cyclic.

Consider the bi-infinite path $\alpha = \bigcup_{i \in \mathbb{Z}} g^i (1,u_1,u_1u_2, \ldots, u_1 \cdots u_k)$. Observe that no two consecutive edges span a $4$-cycle (which amounts to saying that no two consecutive letters in $g^\infty$ commute), which implies that $\alpha$ is convex. In particular, this is a geodesic on which $g$ acts as a translation. In other words, it's an axis for $g$. The key point is that $\alpha$ is the unique axis of $g$. Indeed, two axes of an isometry either have the same convex hull or are separated by a hyperplane. Because $\alpha$ is convex, if there exists another axis it must be separated from $\alpha$ by some hyperplane, say $J$. And, since any two axes crosse exactly the same hyperplanes, this hyperplane must be transverse to all the hyperplanes crossing $\alpha$. However, all the edges of a hyperplane are labelled by the same generator, so the hyperplanes of $M(\Gamma)$ are naturally labelled by vertices of $\Gamma$, and any two transverse hyperplanes must be labelled by adjacent vertices. Thus, since every vertex of $\Gamma$ labels edges of $\alpha$, the vertex of $\Gamma$ labelling $J$ must be adjacent to all the vertices of $\Gamma$, which is of course impossible since a vertex of $\Gamma$ cannot be adjacent to itself. We conclude that $\alpha$ is indeed the unique axis of $g$.

Now, the centraliser $C(g)$ of $g$ has to preserve the union of all the axes of $g$, which is reduced to $\alpha$ here. Therefore, $C(g)$ acts freely on a line, which implies that $C(g)$ has to be infinite cyclic, as desired. In particular, we conclude that $A(\Gamma)$ cannot decompose as a product of two non-trivial groups.

$^\ast$There is a natural equivalence between median graphs and CAT(0) cube complexes, since the one-skeleton of every CAT(0) cube complex is median and that the cube completion of every median graph is CAT(0). However, I milit to replace the terminology "CAT(0) cube complex" with "median graph". The reason is that nobody uses the CAT(0) geometry anymore in this context, but only the graph metric of the one-skeleton (aka the combinatorial or $\ell^1$ metric). In fact, I am not aware of a single statement that can be proved using CAT(0) geometry but that cannot be proved naturally using only median geometry.