What is the minimal dimension of a complex realising a group representation? This question is inspired by this one, which was about representations that can be realised homologically by an action on a graph (i.e., a 1-dimensional complex).
Many interesting integral representations of groups arise from a group acting on a simplicial complex that is homotopy equivalent to a wedge of spheres, by applying homology. A classical example is the action of groups of Lie type on spherical buildings. On homology this gives an integral form of the Steinberg representation.
One may ask if there exists a complex of lower dimension than the Tits building that realises the (integral) Steinberg representation in this way. I expect that the answer is No, but how to prove it? 
More generally, given an integral $G$-representation that can be realised as the homology of a spherical complex with an action of $G$, is there an effective lower bound on the dimension of such a complex? One obvious lower bound is given by the minimal length of a resolution by permutation representations. Is this something that has been studied?
 A: This does not answer Greg's question, but it is related.  You can realize any $\mathbb{Z}G$-module you that like as $H_1$ of a based 2-complex, or as $H_2$ of a 3-complex if you insist that the complex should be simply-connected.  Furthermore, you can require $G$ to act freely on the complex except for fixing the base point.  
Given a $\mathbb{Z}G$-module $M$, take a presentation for $M$, i.e. an exact sequence $F_1\rightarrow F_0\rightarrow M\rightarrow 0$ in which each $F_i$ is a free $\mathbb{Z}G$-module.  Now you can realize $F_0$ as the 2nd homology of a wedge of 2-spheres with $G$ acting freely except on the basepoint.  
You can attach a disjoint union of 3-balls permuted freely by $G$ (with $H_0$ isomorphic to $F_1$) in such a way that the cellular chain complex is just $F_1\rightarrow F_0\rightarrow 0\rightarrow \mathbb{Z}$, with the given map from degree 3 to degree 2.  $H_3$ of this space is of course the kernel of the map $F_1\rightarrow F_0$, while $H_2$ is isomorphic to $M$.  
With a wedge of 1-spheres and attached 2-cells everything works the same way (the Hurewicz theorem tells you that $\pi_1$ surjects onto $H_1$), except that the 2-dimensional complex will probably have fundamental group a lot bigger than the abelian group $M$.  
