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.
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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.
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$\begingroup$ Dear Yves, that's very nice. Thanks a lot. $\endgroup$ Oct 29, 2012 at 22:25
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$\begingroup$ 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). $\endgroup$– Ian AgolOct 29, 2012 at 22:44
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$\begingroup$ @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. $\endgroup$– YCorOct 29, 2012 at 22:54
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$\begingroup$ right, I meant in the case of a group with property FA, of course. $\endgroup$– Ian AgolOct 30, 2012 at 1:41
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$\begingroup$ 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. $\endgroup$– Ian AgolOct 30, 2012 at 2:16