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For clarity, let me recall: By a graph of finite groups, I mean a finite graph $\Gamma$ with edge set $E$ and vertex set $V$, together with, for each vertex $v$, a finite group $G_v$, and for each edge $e$ a finite group $G_e$. Additionally, we have monomorphisms $G_e \to G_{i(e)}$ and $G_e \to G_{t(e)}$, where $i(e)$, $t(e)$ denote the initial and terminal vertices of $e$, respectively. (See Serre's Trees or the Wikipedia article on Bass–Serre theory) The finiteness assumption on the groups $G_v$, $G_e$ is part of my question, not the usual definition.

Write $F(\Gamma,G_v,E)$ for the free product of the $G_v$ and the free group on the set $E$. If $T$ is a maximal tree in $\Gamma$, define $G = \pi_1(\Gamma,T)$, the fundamental group of the graph of groups $\Gamma$ with respect to $T$ to be the quotient of $F(\Gamma,G_v,E)$ adding the relations

  • $\bar e = e^{-1}$, where $\bar e$ is the edge $e \in E$ with the opposite orientation.
  • $ei_e(x)e^{-1} = t_e(x)$, where $e \in E$ and $x \in G_e$.
  • $e = 1$ for $e$ an edge in $T$.

The group $G$ is virtually free, hence word-hyperbolic, and thus has finitely many conjugacy classes of finite subgroups. As such, $\operatorname{Out}(G)$ permutes these conjugacy classes. Some questions:

  1. Do the $G_v$ form a set of representatives for the conjugacy classes of maximal (with respect to inclusion) finite subgroups of $G$?

  2. Does each $\Phi\in\operatorname{Out}(G)$ have a representative $\varphi$ that preserves the "splitting type"? In the sense that there is a graph of groups decomposition with vertex groups $\varphi(G_v)$, edge groups $\varphi(G_e)$ and a graph isomorphism between the underlying graphs that respects $G_v \mapsto \varphi(G_v)$?

The answer to both questions is yes in the case that $\Gamma$ has all edge groups trivial, by the Grushko decomposition theorem. I suspect if $\Gamma$ is a tree it should also be true, but I worry about the case of nontrivial HNN extensions.

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    $\begingroup$ Regarding Question 2: People have studied the subgroup of $\operatorname{Out}(G)$ (for certain $G$) which satisfies this condition. See, for example, Section 3 of this recent paper which called it the "Pettet group". The paper also contains relevant citations to related papers. There is also a paper of Pettet, but its written in terms of $\operatorname{Aut}(G)$ rather than $\operatorname{Out}(G)$. $\endgroup$ – user1729 Jun 6 '19 at 13:30
  • $\begingroup$ Isn't related to a MO-question of mine: mathoverflow.net/q/269936/11856 ? $\endgroup$ – draks ... Sep 6 '19 at 10:28
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    $\begingroup$ @draks it is not. the fundamental group of a graph of groups is a concept designed to understand the structure of groups acting on trees without global fixed point. Not all Coxeter groups (not even all triangle groups) admit such actions. $\endgroup$ – Rylee Lyman Sep 6 '19 at 11:25
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The answer to 1 is: almost yes. The "almost" is needed because of the possibility that there is an edge $e$ and a finite order element $g \in G$ such that $ge = \bar e$, in which case $g$ does not fix any vertex. The solution to that is to subdivide each edge $e$ at its midpoint, if there exists such a $g$. Having done that, one reduces to the case under the action of $G$ on $T$, no such edge $e$ exists, and now the answer is an unqualified yes. The proof is that every finite subgroup must fix a point, and therefore it must fix a vertex.

The answer to 2 is no, and it's still no for example where all edge groups are trivial. I think that you mis-applied Grushko's theorem, as I'll explain below.

For a simple counterexample, take $\Gamma$ to be the rose graph with $n$ petals, and take each $G_v$ and $G_e$ to be the trivial group. In this case $G$ is a rank $n$ free group, and the petals represent a particular free basis of $G$. But $\text{Out}(G)$ is a large and complicated group in that situation, and there are many elements which are not represented by any graph isomorphism as you suggest.

Similar counterexamples with nontrivial vertex groups are easily obtained. For example, just take the same rose graph $\Gamma$ but take $G_v$ to be your favorite finite group. In this case $\text{Out}(G)$ is still a large and complicated group that does not preserve the splitting represented by $\Gamma$.

As for Grushko's Theorem applied in the case that the edge groups are trivial, that theorem only tells you that an automorphism permutes the vertex groups, it makes no guarantee about preserving the splitting itself represented by $\Gamma$.

Added after comments: Although it remains unclear to me exactly what is the intent of your question 2, let me add a few remarks.

One thing I can think of to say is that after properly formulating an equivalence relation on the set of actions of $G$ on simplicial trees, there is a natural action of the group $\text{Out}(G)$ on the set of equivalence classes. One could equivalently describe this action with a proper formulation of an equivalence relation on the set of graph-of-groups descriptions of $G$. This action can be roughly described as in your alternative answer.

This was first done for the case of a finite rank free group $G=F_n$ and free actions on trees, in the Culler-Vogtmann paper "Moduli of graphs". Much more was done in that paper, including applications to the group $\text{Out}(F_n)$, by packaging these equivalence classes into a kind of "deformation space" for a free group (as we now think of it more generally; see below), known as the outer space of the free group.

The completely general theory was done in the Guirardel-Levitt paper "Deformation spaces of trees".

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  • $\begingroup$ I think we may be talking at cross purposes—probably I slightly misspoke when I said "preserve the splitting"—I don't want to require that the stable letters be fixed. If $x_1,\dots,x_n$ are the petals of the rose $R_n$, (thought of as group elements), then for $\varphi \in \operatorname{Aut}(F_n)$, I would like the graph of groups with underlying graph $R_n$ and edges $\varphi(x_1),\dots,\varphi(x_n)$ to qualify as an example of a situation satisfying 2). $\endgroup$ – Rylee Lyman Jun 6 '19 at 3:28
  • $\begingroup$ Thank you for the reminder that I need a "without inversion" condition! $\endgroup$ – Rylee Lyman Jun 6 '19 at 3:28
  • $\begingroup$ I've added a few more comments regarding your question 2. $\endgroup$ – Lee Mosher Jun 6 '19 at 13:18
  • $\begingroup$ Thanks! That's helpful. $\endgroup$ – Rylee Lyman Jun 6 '19 at 13:54
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The answer to 2) is yes, for perhaps silly reasons. Let $\varphi \in \operatorname{Aut}(G)$ and $(\Gamma,G_v,G_e)$ as in the question. What I had in mind was a graph of groups structure on $\Gamma$ with edges labeled by $\varphi(e)$ in $G$ (abusing notation, I will continue writing $e$, rather than $\varphi(e)$, for edges at this stage), with vertex groups $\varphi(G_v)$, edge groups $G_{e}$, monomorphisms $\varphi\circ i_e\colon G_e \to \varphi(G_v)$, $\varphi\circ t_e\colon G_e \to \varphi(G_{t(e)})$.

One checks that the data $F(\Gamma,\varphi(G_v),\varphi(E))$ and the same tree $T \subset \Gamma$ yields the isomorphism $\varphi$ on fundamental groups of graphs of groups, simply because $\varphi$ is an automorphism.

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