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?

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