# In search of a quick proof that groups acting freely on $\mathbb R$-trees are linear

Many years ago I had the idea to use non-standard analysis to prove that a group acting freely on an $$\mathbb R$$-tree must be linear. The heuristic went like this:

• A non-standard model $$G^*$$ of the group $$G$$ is acting on a non-standard model of the $$\mathbb R$$-tree.

• We approximate the non-standard tree by a $$\mathbb Z^*$$ -tree (where $$\mathbb Z^*$$ denotes a non-standard model of the integers) in such a way that there is still an action of $$G$$ (although we accept that the action of $$G^*$$ will be corrupted).

• Now if the original group $$G$$ is finitely generated then it sits inside a 2-generator group and that in turn is a subgroup of finite index in $$\mathrm{SL}_2(\mathbb Z^*)$$.

• Since $$\mathbb Z^*$$ is an integral domain of characteristic zero we deduce that $$G$$ embeds in $$\mathrm{GL}_2(\mathbb C)$$.

There was always going to be some work to do because already a non-standard model of a $$\mathbb Z$$-tree falls into disconnected components when viewed externally. I naively hoped that some useful formulae involving length functions could be brought into play.

My question is:

Can someone with real expertise in model theory tell me whether this argument could be made into something mathematically sound or whether it is doomed to failure perhaps for some very simple reason?

• IMHO, it is extremely unlikely to work. To begin with, it's unclear how would you use freeness of the action. Another problem is that the group of automorphisms of a tree (even simplicial, of constant valence) is much larger than the group of "algebraic" automorphisms of that tree. How could the model theory help you to "deform" a general free action to an algebraic one? Say, suppose you have a 2-generator group $G$ acting freely and simplicially on the tree associated with $PGL(2,Q_p)$. Of course, we know that $G$ is free but how would this proposal prove linearity of $G$? Dec 21, 2022 at 16:11
• @MoisheKohan I was using the term 'linear' loosely. I would regard the projective general linear group as close enough to linear. The idea of model theory is really inspired by the Wilkie -- van den Dries proof of Gromov's polynomial growth theorem using non-standard analysis. So $\mathbb Z^*$ is a non standard model of the integers and could be used to approximate ordinary real numbers. Dec 21, 2022 at 20:35
• The group $PGL(n,K)$ is always $K$-linear, that's not an issue. Dec 21, 2022 at 20:47
• Yes, sorry, what I meant to say was that if a group $G$ is a subgroup of a non-standard free group and is finitely generated (in the standard sense) then $G$ has a faithful 2 dimensional linear representation over $\mathbb C$. Dec 21, 2022 at 20:56
• I understand, the issue is that the group of automorphisms of a simplicial tree is way bigger that a nonstandard free group. Dec 21, 2022 at 21:02