Welcome to MathOverflow

Question

0

1

A group $G$ has Serre's property $FA$ if any isometric action of $G$ on a simplicial tree has a global fixed point. Let $n\geq 3$. It is well-known that $SL_n(\mathbb{Z} )$ has property $FA$. Now my question is that are there nontrivial group actions of $SL_n(\mathbb{Z} )$ on a simplicial tree by isometries? Here "nontrivial" means the fixed point set is NOT the whole tree.

edited Nov 1 at 17:30

asked Oct 29 at 21:52

yeshengkui
883●3●11

Yves Cornulier · Accepted Answer · 2012-10-29 22:05:41Z UTC

5

Any non-trivial group $G$ has a non-trivial action on a "star" tree $T$ whose vertex set is $G\cup\{\infty\}$ (where $\infty\notin G$) and edges are $\{\infty,g\}$ for $g\in G$. Thus any group admits a faithful action on a tree.

Any residually finite countable group has a faithful action on a locally finite tree. If $(H_n)$ is a decreasing sequence of finite index subgroups with trivial intersection with $H_0=G$, this tree is the disjoint union of cosets $G/H_n$, with an edge between $gH_n$ and $gH_{n+1}$ for all $g$ and all $n$; the root is just the point $G/H_0$ and is fixed by the action. This applies to $\text{SL}_d(\mathbf{Z})$; in this precise case the $H_n$ can be chosen so that the index $H_n/H_{n+1}$ is bounded and thus the tree has bounded valency.

answered Oct 29 at 22:05

Yves Cornulier
3,045●1●6●11

	Dear Yves, that's very nice. Thanks a lot. – yeshengkui Oct 29 at 22:25
	There is a generalization of both examples, where one has an action on a tree associated to a nested sequence of subgroups. Any action on a tree has such a filtration, coming from orbits of stabilizers of arcs going from a fixed point to any other point of the tree (essentially Bass-Serre theory). – Ian Agol Oct 29 at 22:44
	@Agol I see a generalization to any nested sequence of subgroups $(H_n)_{n\in K}$ with $K$ a convex subset of the chain $\mathbf{Z}$ such that $\bigcup_{n\in K} H_n=G$. But I can't guess which kind of generalization you have in mind even in the case of the action of $\mathbf{Z}$ on a linear tree. – Yves Cornulier Oct 29 at 22:54
	right, I meant in the case of a group with property FA, of course. – Ian Agol Oct 30 at 1:41
	I guess also the quotient should be an infinite ray, otherwise one would have to consider a partially ordered set of subgroups, corresponding to stabilizers of arcs coming from a basepoint fixed by the group. – Ian Agol Oct 30 at 2:16

show 1 more comment

Non-trivial action of $SL_n(\mathbb{Z} )$ on a simplicial tree

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1 Answer

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.