# JSJ-decompositions of hyperbolic groups and elementary vertices

My question is the following:

In Bowditch's JSJ-decomposition of hyperbolic groups, can elementary (virtually-cyclic) vertices have degree 1? If not, why not?

I had thought for a long time that this was not possible, as it seems to be pointless (with respect to Outer automorphism groups). For example, if you have a hyperbolic group $A\ast_CB$ where $A$ is virtually cyclic then sinking the vertex corresponding to $A$ into $B$ does not change the information we gain about the outer automorphism group.

Have I been thinking about this incorrectly? Do rigid vertices actually have all their possible virtually cyclic splittings protruding from them like this (as these splittings are not allowed so we have to push them out)? So rigid vertices are surrounded by elementary vertices of degree one which are possible attaching points for other hyperbolic groups? Or is there some subtlety of rigid hyperbolic groups which I have not understood?

EDIT: As the commenter has pointed out, the source of my confusion is the difference between Sela's and Bowditch's definitions. Sela only allows edge groups to be maximal, so I presumed that when you add in the elementary vertices this maximality is somehow maintained. But, seemingly, this is not the case (edge groups have to be maximal in the adjacent non-elementary vertex groups, but not in the elementary ones). Now, I have a single silly question before I think I understand this all better - can I simply attach roots? More precisely, say I have a hyperbolic group and every relator which contains the word $w$ contains it as part of $w^{in}$ for some fixed $n$, so $G=\langle X, w; R_1(X, w^n), \ldots, R_k(X, w^n)\rangle$. Then the word $w$ can be taken out as a root, and we obtain the group $\langle X, u, w; R_1(X, u), \ldots, R_k(X, u), u=w^n\rangle=G_1\ast_{u=w^n}\langle w\rangle$. Does Bowditch's definition encode this?

• The phrase 'it seems to be pointless (with respect to Outer automorphism groups)' perhaps gives a hint about your confusion. Rips--Sela's JSJ is designed for studying the outer automorphism group. Bowditch's, on the other hand, is not, and the two are often distinct. – HJRW Feb 15 '14 at 20:38
• The question in your last paragraph is a bit vague (what does 'encode' mean?) but if you mean 'does the splitting obtained when you attach a root appear in Bowditch's JSJ?' the answer is 'yes'. If $G=G_0*_{n\mathbb{Z}}\mathbb{Z}$ then the endpoints of $\mathbb{Z}$ separate $\partial G$ into $n$ components. In particular, $\mathbb{Z}$ can't be in a rigid vertex and, if $n>2$ then it can't be in a surface vertex either. Non-orientable surfaces give example where square roots are contained in surface vertices. – HJRW Feb 16 '14 at 20:01
• Okay, thanks. So Sela's rigid vertices are not really rigid, while Bowditch's are, yes? – user1729 Feb 17 '14 at 11:36
• Well, let's just say that Bowditch's are more rigid than Sela's. Another key distinction is that Sela's is not canonical up to isomorphism, but only up to equivalence. If you really want to understand the difference, I suggest you look at Guirardel--Levitt's papers about JSJ decompositions and trees of cylinders. – HJRW Feb 17 '14 at 11:42

This is quite possible. Take an infinite cyclic group $G_1$, its index 2 subgroup $C_1$ and, say, a closed surface hyperbolic group $G_2$. Now, take a maximal cyclic subgroup $C_2<H$ and consider the amalgam $$G_1\star_{C_1=C_2} G_2.$$
• It will be JSJ if the loop corresponding to $C_2$ fills the hyperbolic surface (i.e., if you remove the corresponding closed geodesic, you only have disks left). If you do not like this condition, replace closed surface with a closed hyperbolic n-manifold, $n\ge 3$. – Misha Feb 15 '14 at 15:55