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Every right-angled Artin group and Coxeter group has a graph $\Gamma$ associated to it. The vertices of the graph stayfor the generators of the group and the edges correspond to the relations of the generators (two  vertices are connected by the edge if they do not commute).

My first question is if anything is known about the Artin groups whose graphs differ just by one edge - only one additional relation.

The second question is concerning ramified coverings. Monodromy group of the Hurwitz covering (covering of the 2-dim sphere that has only simple ramification points except one, where ramification profile is arbitrary)  turns out to be a simple case of Artin group - generators are just permutations. Is there a continuation of this fact to a general Artin group?

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"My first question is if anything is known about the Artin groups whose graphs differ just by one edge - only one additional relation." What sort of thing do you have in mind? –  HJRW Sep 27 '12 at 13:42
    
Also, what's 'the Hurwitz covering'? (A google search seems to find various definitions of a Hurwitz covering, but nothing obvious about the Hurwitz covering.) –  HJRW Sep 27 '12 at 13:45
    
By the way, the definition in your first paragraph is the definition of a right-angled Artin group - more generally, the edges may correspond to relations of the form $xyx=yxy$ etc. –  HJRW Sep 27 '12 at 13:46
    
Alexey, as HW noted, you should ask more specific questions. Also, at first you should try to answer your own question in the case of finite dihedral groups of different order. If you find a relation between them that you like, then you might come up with a precise question in general. –  Misha Sep 27 '12 at 17:07
    
Thank you very much for comments. I mean indeed right-angled Artin group. By the Hurweitz cover I take those counted by simple Hurwitz numbers. In fact this is the most interesting for me - I am missing geometric object behind the Artin group that is consistent with the Hurwitz cover and edge dropping of the graph makes sense. –  Alexey Basalaev Sep 27 '12 at 17:50

1 Answer 1

Let $A_{\Gamma'}$ be the group asoociated to any graph $\Gamma'$. Concerning your first question I can show that there is a short exact sequence $$1\rightarrow F \rightarrow A_{\Gamma \setminus e}\rightarrow A_\Gamma\rightarrow 1.$$

Let me work in the case of graph products (which includes both the right angled Artin and the right andgled Coxeter cases).

We obtain a presentation for $A_\Gamma$ by adding a relation to a presentation for $A_{\Gamma\setminus e}$.

So there is a canonical group homomorphism from $A_{\Gamma\setminus e}$ to $A_\Gamma$. So maybe you are wondering what the Kernel looks like.

Instead of answering this question let me instead consider the homomorphism $A_{\Gamma\setminus e}\rightarrow A_{\Gamma\setminus v}$ where $v$ is one of the endpoints of $e$ (Of course we also have to remove all edge to $v$). By the argument of Section 4 in Holt,Rees its Kernel is a free product of copies of the vertex group. We have a factorization $A_{\Gamma\setminus e}\rightarrow A_\Gamma \rightarrow A_{\Gamma\setminus v}$. So especially the Kernel of the map we were originally interested in is contained in that Kernel.

In the Artin case we get immediately that this kernel is free (as it is a subgroup of a free group) and in the Coxeter case it is also a free group. To show this it is enough to show that it is torsionfree (using the action of a free product on a tree). But this is also clear since any torsion element in a free product is conjugate to an element in one of the factor groups. Such elements cannot lie in the Kernel of $A_{\Gamma\setminus e}$ to $A_\Gamma$ (but to show this I would have to make the isomorphism from the kernel to such a free product explicit. It is in the paper mentioned above but writing it down would take quite some notation).

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Henrik: Some of your arrows go in wrong direction, for instance, there is a homomorphism $A_{\Gamma \setminus e} \to A_{\Gamma}$ but not the other way around. You should correct this. –  Misha Sep 27 '12 at 17:05
    
@Misha: thank you. –  HenrikRüping Sep 27 '12 at 19:51

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