Subgroups of RAAGs vs. subgroups of RACGs

Question

Is a (finitely generated) torsion-free subgroup of a right-angled Coxeter group isomorphic to a subgroup of a right-angled Artin group?

It is well-known from the theory of special cube complexes that a subgroup $H$ of a right-angled Coxeter group $G$ contains a finite-index subgroup isomorphic to a subgroup of a right-angled Artin group (e.g. $H \cap [G,G]$), but is it true for the whole group if we assume in addition that $H$ is torsion-free?

The difficulty when applying the theory of special cube complexes comes from the fact that the induced action of $H$ on the usual CAT(0) cube complex $X$ of $G$ may invert hyperplanes (i.e. some isometries may stabilise a hyperplane and switch its halfspaces). This can be avoided by replacing $X$ with its barycentric subdivision, but then hyperplane-inversions yield self-osculations (i.e. there exist an element $h \in H$ and an oriented hyperplane $\vec{J}$ such that $\vec{J}$ and $g \vec{J}$ contain two intersecting oriented edges pointing to their common vertex and that do not span a square).

NB: In all the question, right-angled Artin/Coxeter groups are finitely generated. Recall that, given a (finite) simplicial graph $\Gamma$, the associated right-angled Coxeter group is defined by the presentation $$\langle u \in V(\Gamma) \mid u^2=1 \ (u \in V(\Gamma)), \ [u,v]=1 \ (\{u,v\} \in E(\Gamma)) \rangle$$ and the associated right-angled Artin group by the presentation $$\langle u \in V(\Gamma) \mid [u,v]=1 \ (\{u,v\} \in E(\Gamma)) \rangle$$ where $V(\Gamma)$ and $E(\Gamma)$ denote respectively the vertex- and edge-sets of $\Gamma$.

Answer 1

The fundamental group of the (non-orientable) closed surface of Euler characteristic -1 provides a counterexample.

On the one hand, it’s a subgroup of index 4 in the reflection group on the right-angled pentagon, so it embeds in a RACG, and of course it’s torsion-free.

On the other hand, Crisp—Wiest proved that it never embeds in a RAAG.

Answer 2

I finally found a elementary example: $$BS(1,-1):= \langle x,y \mid yxy^{-1}=x^{-1} \rangle.$$ It embeds into $\mathbb{Z} \oplus \mathbb{D}_\infty$, which itself embeds in the right-angled Coxeter group $\mathbb{D}_\infty \oplus \mathbb{D}_\infty$. Indeed, if $\mathbb{Z} = \langle t \mid \ \rangle$ and $\mathbb{D}_\infty = \langle a,b \mid a^2=b^2=1 \rangle$, then $\langle ab,ta \rangle$ is isomorphic to $BS(1,-1)$.

But $BS(1,-1)$ does not embed into a right-angled Artin group. To see this, one can use a theorem due to Baudisch that claims that a $2$-generated subgroup in a right-angled Artin group is abelian or free. One can also use the fact that right-angled Artin groups are bi-orderable, which is not the case for $BS(1,-1)$: if $x>1$, then $x^{-1}=yxy^{-1}>yy^{-1}=1$ hence $x<1$.

Answer 3

Just for the record, I would like to add a few details and references regarding Henry's answer.

First, the right-angled Coxeter group $C$ defined by a cycle of length five can be realised as the reflection group generated by the reflections along the sides of a right-angled pentagon in the hyperbolic plane. As described in Scott's article Subgroups of surface groups are almost geometric, if $x_1,\ldots, x_5$ denote the five reflections generating our Coxeter group (see the figure below), then $x_1x_2x_5$, $x_1x_4$, $x_3x_5$, and $x_1x_3x_1x_5$ generate a subgroup with a fundamental domain that is a union of four pentagons.

By looking at how the sides of this fundamental domain are identified, we deduce that the subgroup $G$ under consideration is the fundamental group of a non-orientable surface of Euler characteristic $-1$.

Now, the fact that $G$ is not isomorphic to a subgroup of a right-angled Artin group is proved, as mentioned by Henry, by Crisp and Wiest in their article Embeddings of graph braid and surface groups in right-angled Artin groups and braid groups. It is also a consequence of a strong Tits alternative satisfied by right-angled Artin groups: every subgroup is abelian or it surjects onto a non-abelian free group. An algebraic proof can be found in Antolin and Minasyan's article Tits alternative for graph products; and a topological argument can be found in Bregman's article Automorphisms and homology of non-positively curved cube complexes.