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

I have been reading Levitt's paper Automorphisms of Hyperbolic groups and Graphs of Groups. I am having some trouble trying to fit all the bits together, and would appreciate some help with this last step.

In the paper, Levitt considers minimal graphs of groups, and gives results regarding (a specific subgroup of) the outer automorphism group of the fundamental group of an arbitrary graph of groups. A graph of groups is minimal if edge groups are proper subgroups of adjacent vertex groups.

Levitt applies this work to the (canonical) JSJ-decomposition of a (one-ended) hyperbolic group. The "canonical" means that we can forget that we are looking at a specific subgroup of the outer automorphism group: his analysis can be applied to the whole outer automorphism group. Which is good, and is the point.

Now, I (believe that I) understand all of the fiddly bits of this paper. However, I have come across a stumbling block when trying to pin it all together. You see, the canonical JSJ-decomposition Levitt is using is due to Bowditch (from Bowditch's paper Cut points and canonical splittings of hyperbolic groups). This decomposition is entirely canonical, unlike the one Sela obtains for the torsion-free case (where it is canonical up to certain "moves"). Bowditch's decomposition is canonical because of the addition of "elementary" vertices, that is, vertices whose stabiliser is virtually cyclic. My problem is as follows: I believe that these vertices imply that the graph of groups is not necessarily minimal.

For example, pin together two hyperbolic triangle groups $H$ and $K$ across an infinite cyclic subgroup, so $G=H_{C_1=C_2}K$, and ensure that these subgroups are malnormal infinite cyclic. Then the JSJ-decomposition of this (one-ended hyperbolic group) looks like $H-C-K$, where the edge stabilisers are $C_1$ and $C_2$, and where $C_1=C=C_2$. A contradiction. No?

I am sure I am just missing something silly, something obvious, something I should have seen a long time ago. However, I have no idea what that could possibly be!

share|improve this question
1  
I don't have the paper in front of me, but are you sure that's what Levitt means by 'minimal'? Minimal usually means that the corresponding Bass--Serre tree doesn't contain a proper invariant subtree. The property you describe is usually called 'reduced'. –  HJRW Sep 1 '13 at 22:02
    
@HJRW Ah, okay, thanks. You are right. He defines a graph of groups as minimal if every $G_e$ is a proper subgroup of $G_v$ where $\tau(G_e)=G_e$. In my mind, every vertex has two edges between them (positive and negative), but he only has one. Thanks for pointing that out. –  user1729 Sep 2 '13 at 8:56

1 Answer 1

up vote 4 down vote accepted

This has been dealt with in comments, but since MO works better if answers are given, I'll elaborate a little bit here.

A graph of groups satisfying your definition of 'minimal' is usually called 'reduced'. You're absolutely correct that Bowditch's JSJ is often not reduced (though note that, in your example, the graph of groups that you write down is only the JSJ if neither $H$ nor $K$ splits over a 0- or 2-ended subgroup relative to $C$).

The relevant definition in Levitt's paper is on page 53:

We assume that $\Gamma$ is minimal: $G_e$ is a proper subgroup of $G_v$ for every terminal vertex $v$ (equivalently, $\pi_1(\Gamma')$ is a proper subgroup of $\pi_1(\Gamma)$ for every proper connected subgraph $\Gamma'$).

The non-parenthetical part of the sentence is indeed very easy to misunderstand in the way you have; I suppose by 'terminal vertex' he must (unusually) mean 'leaf'.

The parenthetical part, though, makes it absolutely clear: his definition of `minimal' coincides with the usual one. That is, a splitting is minimal if the the Bass--Serre tree has no proper invariant subtrees. Bowditch's JSJ is always minimal, as can be seen, for instance, from the fact that orbits in the boundary are dense.

share|improve this answer
    
Thanks for your answers, and for pointing out the omission in the example (they were going to be triangle groups so couldn't split, but then thought torsion-free would be easier. I'll change it). What do you mean by "leaf"? By a terminal vertex, I believe he means that an edge and an origin $\iota(e)$ and a terminal $\tau(e)$, which I thought was standard? –  user1729 Sep 3 '13 at 8:25
    
Also, do you know of a good reference for the equivalent definitions of "minimal"? I have a "feel" for the proof, but am struggling to nail it. –  user1729 Sep 3 '13 at 8:26
    
@user1729 - by 'leaf' I mean, as usual, a vertex of valence 1. If you substitute 'leaf' for 'terminal vertex' in his definition, then you get the usual notion of 'minimal'. This also chimes with the fact that he never says that $v$ is the terminal vertex *of $e$*. On the other hand, if he meant '$G_e$ is a proper subgroup of $G_{\tau(e)}$' then this would indeed be a definition of a reduced graph of groups. For instance, in your minimal-but-not-reduced example, if $e$ is the edge pointing from $H$ to $C$, we have $C=G_e=G_{\tau(e)}=C$. –  HJRW Sep 3 '13 at 9:38
    
By the way, some support for this theory is given by the wikipedia page on trees, where it is written that 'A terminal vertex ... is a vertex of degree 1'. See: en.wikipedia.org/wiki/Tree_%28graph_theory%29 . –  HJRW Sep 3 '13 at 9:42
    
I'm not sure I know of a reference where the three equivalent definitions of 'minimal' under discussion are treated. It's a fairly easy exercise. I suppose you could ask it as a question here if you continue to have difficulty with it! –  HJRW Sep 3 '13 at 9:47

Your Answer

 
discard

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

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