2
$\begingroup$

Given two Coxeter groups $W(\Gamma)$ and $W(\gamma)$ of equal rank and their Artin groups $A(\Gamma)$ and $A(\gamma)$, is $W(\Gamma)\supset W(\gamma)$ (as reflection groups) equivalent to $A(\Gamma)\supset A(\gamma)$ (as abstract groups)?

$\endgroup$
3
  • $\begingroup$ Do you mean containment as abstract groups? $\endgroup$ May 27, 2022 at 0:08
  • 1
    $\begingroup$ For the Artin groups, yes. For the Coxeter groups, no. $\endgroup$ May 27, 2022 at 1:30
  • $\begingroup$ I guess that you do not want just an embedding as abstract groups at the level of Artin groups, but you at least want compatibility with the quotient maps to the Coxeter group ? For instance, the free group on two generators (which is an Artin group of rank two) embeds into the three strand braid group, while the universal Coxeter group of rank two is infinite, hence cannot embed into $S_3$. $\endgroup$ Jun 3, 2022 at 15:13

1 Answer 1

5
$\begingroup$

No, there are counterexamples already in the right-angled case.

Let $C_n$ denote the cycle of length $n \geq 5$ and let $C_n^\ast$ denote the graph obtained by gluing two distinct copies of $C_n$ along a path of length two. Observe that, in $C_n^\ast$, there exists induced tree $T_n$ that has $n$ vertices but that is not a line. On the one hand:

Proposition. A Coxeter group is isomorphic to a subgroup of the right-angled Coxeter group $C(C_n)$ if and only if it is isomorphic to a right-angled Coxeter group $C(\Gamma)$ where $\Gamma$ is either a disjoint union of segment or a single cycle of length divisible by $n-4$.

See Proposition 4.21 in the article Morphisms between right-angled Coxeter groups and the embedding problem in dimension two. As a particular case, the right-angled Coxeter group $C(T_n)$ does not embed (abstractly) in $C(C_n)$. (Observe that, by construction, the two Coxeter groups have equal ranks.)

On the other hand, the right-angled Artin group $A(C_n^\ast)$ (and a fortiori $A(T_n)$) is isomorphic to a subgroup of $A(C_n)$. See for instance Corollary 5 in Embedability between right-angled Artin groups.

It is possible to construct many other examples. For instance by using the fact that, if $\Gamma$ has diameter $\geq 3$, then $A(F)$ embeds in $A(\Gamma)$ for every finite forest $F$. On the other hand, given two finite trees $R$ and $S$, $C(R)$ is isomorphic to a subgroup of $C(S)$ if and only if there exists a graph morphism $\varphi : R → S$ that sends a vertex of degree $2$ to a vertex of degree $\geq 2$ and a vertex of degree $\geq 3$ to a vertex of degree $\geq 3$.

$\endgroup$
2
  • $\begingroup$ Does the equivalence at least hold when restricting to finite Coxeter groups? $\endgroup$ May 30, 2022 at 0:49
  • $\begingroup$ I don't know. Maybe something to check is the dihedral case: there are non-trivial embeddings between dihedral Coxeter groups, and they have the same rank, but what about dihedral Artin groups? $\endgroup$
    – AGenevois
    May 30, 2022 at 5:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.