# On which space does $GL_n(F_p[X])$ act nicely?

The group $GL_n(\mathbb{Z})$ acts properly and isometrically on the space of homothety classes of scalar products on $\mathbb{R}^n$. This is a Riemannian manifold of nonpositive sectional curvature.

Is there a similar space for the case of $GL_n(F_p[x])$. Maybe one can construct a building or something like this. I guess if the space I am looking for exists it should be rather well known. Otherwise I apologize for the vagueness of this question.

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The group $GL_n(\mathbb{F}_p[x])$ acts on the Bruhat-Tits building for $GL_n$. The vertex set is $GL_n(\mathbb{F}_p((x^{-1})))/GL_n(\mathbb{F}_p[[x^{-1}]])$, and the higher simplices form sets of the form $GL_n(\mathbb{F}_p((x^{-1})))/I$ for various parahoric groups $I$. The action of $GL_n(\mathbb{F}_p((x^{-1})))$ on the left restricts to an action of $GL_n(\mathbb{F}_p[x])$.
When $n=2$ there is a nice exposition of the action in chapter 2 of Serre's Trees (as long as you're okay with $PGL_2$ instead of $GL_2$). You get an infinite sequence of 1-simplices that vaguely resembles the $SL_2(\mathbb{Z})$-quotient of the complex upper half-plane. In higher rank, I don't think things can be quite so explicit.
You can also find actions of $GL_n(\mathbb{F}_p[x])$ on many other spaces, namely those that are defined over rings containing $\mathbb{F}_p[x]$ and that come equipped with a natural algebraic or analytic action of $GL_n$. For example, the analytification of the flag variety for $GL_n$ over the completion of an algebraic closure of $\mathbb{F}_p((x^{-1}))$ admits such an action.
Perhaps it should be added that $GL_n(F_p[x])$ is inside a compact subgroup of $GL_n(F_p(x)_v)$ for all places $v$ except the $1/x$ place "at infinity", at which it really is a discrete subgroup. All places have affine buildings attached to them, but the actions of $GL_n(F_p[x])$ on almost all of them are "boring", for this reason. These are analogues of the p-adic buildings for $GL_n(Q_p)$'s, but the "place at infinity" of $Q$ behaves differently, obviously. –  paul garrett Jun 27 '11 at 16:44