2
$\begingroup$

Let $G$ be a finitely generated group and $\mathcal{A}$ a collection of subgroups closed under conjugation and taking subgroups.

I'm interested in the problem of classifying all deformation spaces of JSJ decompositions of $G$ over subgroups in $\mathcal{A}$. Can it ever happen that the deformation space is unique? Perhaps if $G$ is such that any splitting of $G$ over $\mathcal{A}$ does not contain any quadratically hanging vertices (in the sense of Guirardel-Levitt)?

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ The question isn't clear. For a given $G$ and $\mathcal{A}$, there is one deformation space of JSJs, namely the one defined by Guirardel and Levitt: the space of maximal, universally elliptic $G$-trees. Are you varying $\mathcal{A}$ and wondering how the deformation space changes? Or are you wondering about situations in which the deformation space is a single point?... (tbc) $\endgroup$
    – HJRW
    Commented Sep 28, 2021 at 16:25
  • 2
    $\begingroup$ I think Generalised Baumslag--Solitar groups, as studied by Forester, provide examples of groups with interesting (i.e. not just a point) deformation spaces of JSJ decompositions despite having no quadratically hanging vertices; indeed, all the vertex groups are cyclic. See §I.3.5 of Guirardel--Levitt's monograph, which refers to Forester's paper arxiv.org/abs/math/0110176 . $\endgroup$
    – HJRW
    Commented Sep 28, 2021 at 16:35
  • $\begingroup$ Apologies for making myself unclear. In my question, $\mathcal{A}$ is the set of subgroups containing edge stabilisers of the splitting. So for instance, if $\Sigma$ is a large enough closed surface, $G = \pi_1(\Sigma)$ and $\mathcal{A}$ is the set of cyclic subgroups of $G$, then any two non-isotopic pair of pants decompositions of $\Sigma$ induce two cyclic splittings which live in two distinct deformation spaces (which are singletons). $\endgroup$
    – 24601
    Commented Sep 29, 2021 at 7:58
  • $\begingroup$ Even if $\mathcal{A}$ is only allowed to contain cyclic subgroups, I don’t think this ever happens unless every cyclic splitting of $G$ is trivial. If $T$ is a non-trivial $G$-tree, then taking $\mathcal{A}$ to contain $\langle g\rangle$ where $g$ acts hyperbolically on $T$ will force $T$ not to appear in the deformation space relative to $\mathcal{A}$. $\endgroup$
    – HJRW
    Commented Sep 29, 2021 at 15:47
  • $\begingroup$ Are you asking the groups in $\mathcal A$ to be slender or are you considering more general edge groups? Not much is known when the edge groups are not slender. $\endgroup$
    – NWMT
    Commented Sep 30, 2021 at 13:10

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .