Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I think that the problems below are open, but maybe I missed something. Definitions follow the problems. All groups are assumed to be finitely generated.

Question 1. Is every Artin group (resp. Shephard group, resp. generalized von Dyck group) residually finite (or even linear)?

Note that linearity is well-known for RAAGs (Right Angled Artin groups); linearity was proven for the Artin Braid groups by Steven Bigelow.

Question 2. Is there a "local" characterization of hyperbolic groups among Shephard group and generalized von Dyck group anologous to the one given by Moussong in the case of Coxeter groups?

Question 3. Is there at least a characterization of finite groups among generalized von Dyck groups, analogous to the one for Shephard groups (given by Shephard and Todd in "Finite unitary reflection groups", Canad. J. Math. 6 (1954), 274–304)?

Definitions. Let $\Gamma$ be a simple finite graph (i.e., a 1-dimensional finite simplicial complex) where every edge $e$ and vertex $x$ is given a label $m(e)\ge 2$, resp. $n(x)\ge 2$. All three classes of groups will be given presentations where vertices $x$ of $\Gamma$ are generators. Below are the relators:

a. For the Artin group $A_\Gamma$ relators are of the form $$ \underbrace{x y x\dots }_{m(e)~\hbox{terms}}=\underbrace{y x y\dots }_{m(e)~ \hbox{terms}} $$ whenever vertices $x, y$ are connected by an edge $e$.

b. For the Shephard group $S_\Gamma$ in addition to Artin relators one has relators of the form $$ x^{n(x)}=1 $$ for all vertices $x$ of $\Gamma$.

c. For the generalized von Dyck group $D_\Gamma$, presentation has the form $$ (xy)^{m(e)}=1 $$ whenever $x, y$ are connected by an edge $e$, together with the Shepard relators $$ x^{n(x)}=1. $$ If $\Gamma$ consists of a single edge, then $D_\Gamma$ is called a von Dyck group, in which case everybody (in group theory) knows answers to all of the above questions, they are all positive.

Of course, if all vertex-labels are equal to $2$, then the Shephard group (or generalized von Dyck group) is just a Coxeter group.

Moussong's Theorem. Gabor Moussong proved in his thesis that a Coxeter group $G$ is Gromov-hyperbolic if and only if none of the parabolic subgroups of $G$ is virtually a direct product of two infinite groups.

Edit: As noted by Derek, not only linearity, but even solvability of the word problem is open for Artin groups, although the latter is known for some large classes of Artin groups, see for instance the paper by Eddy Godelle and Luis Paris "$K(\pi,1)$ and word problems for infinite type Artin–Tits groups, and applications to virtual braid groups", Math. Z. 272 (2012), no. 3-4, p. 1339–1364.

share|improve this question
1  
What is the question? Is it the status of Q1-Q3? –  Victor Protsak Feb 12 '13 at 0:20
    
Victor: Yes. I do not even know if these questions were stated somewhere (except, maybe linearity of general Artin groups, but I am not sure about this). –  Misha Feb 12 '13 at 0:25
    
I do not think it is known if all Artin groups are linear; at least, this is stated as an open problem in Charney's survey paper on RAAGs. In fact, I think the only Artin groups known to be linear are precisely those with finite Eilenberg-Maclane space. –  Steve D Feb 12 '13 at 9:09
2  
I don't believe it is known even whether all Artin groups have solvable word problem. It might be more realistic to ask these questions for specific classes of Artin groups, such as large type or extra-large type. –  Derek Holt Feb 12 '13 at 9:16
    
Derek: You are clearly right, once can infer this much from the paper "$K(\pi,1)$ and word problems for infinite type Artin–Tits groups, and applications to virtual braid groups" by Godelle and Paris published in 2012. On the other hand, the problem might be easier for Shephard and generalized von Dyck groups, since they are "closer" to Coxeter groups. –  Misha Feb 13 '13 at 0:54
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.