Representation of $\mathrm{AGL}(V)$ on the homology of the poset of affine subspaces of $V$

Question

Let $V$ be an $n$-dimensional vector space over a finite field $F$ (of order $q$). Denote by $\mathrm{AGL}(V)$ the group of invertible affine transformations of $V$; so $\mathrm{AGL}(V)$ consists of all mappings $x\mapsto Tx+v$ where $T\in GL(V)$ and $v\in V$.

Clearly, $\mathrm{AGL}(V)$ acts on the poset of all affine subspaces of $V$ where an affine subspace is, of course, a coset of a linear subspace. Let $P$ be the subposet of proper affine subspaces and let $\Delta(P)$ be the order complex of $P$. Then $\mathrm{AGL}(V)$ acts simplicially on $\Delta(P)$ and hence acts on the reduced homology of $\Delta(P)$.

Now $P$ is, it seems to me, the proper part of a geometric lattice. If we declare a subset of $V$ to be independent if its elements are affinely independent, then I believe this gives a matroid whose corresponding lattice of flats is the set of affine subspaces (and I guess the empty set). So $P$ should be the proper part of this and hence its reduced homology should be concentrated in the top dimension, which I guess is $n-1$ if I count correctly. Does anybody know any literature describing the decomposition of $\widetilde{H}_{n-1}(\Delta(P),\mathbb C)$ into irreducible $\mathbb C\mathrm{AGL}(V)$-modules? Is it irreducible?

Of course, the easy case is when dimension $n$ of $V$ is $1$. Then $AGL(V)$ consists of all invertible $ax+b$-maps and the proper affine subspaces are points. So the reduced homology in dimension $0$ is the augmentation submodule of the permutation module of $F\rtimes F^\times$ acting on $F$. This action is doubly transitive, and so the augmentation submodule is irreducible of degree $q-1$ (and is the unique irreducible representation of degree greater than $1$).

Answer 1

I suspect that

Solomon, Louis The affine group. I. Bruhat decomposition.

proves what you are looking for.

Let $A_n(q)$ denote the poset of proper affine subspaces of $\mathbf{F}_q^n$. The only non-vanishing reduced homology group of $A_n(q)$ is in degree $n-1$ and it is free abelian of rank $|E_n(q)|$ where $E_n(q) = -\widetilde{\chi}(A_n(q))$ (minus the reduced Euler characteristic). This is because $A_n(q)$ is the proper part of a geometric lattice with $\widehat 0 = \emptyset$ and $\widehat 1 = \mathbf{F}_q^n$. See papers by Folkman and Björner on homology of geometric lattices.

The poset $A_0(q)$ is empty and $A_1(q)$ is discrete on $q$ points so $E_0(q)=1$ and $E_1(q)=1-q$. The recursion \begin{equation*} 1 = \sum_{0 \leq k \leq n} E_k(q) \binom{n}{k}_q q^{n-k} \end{equation*} can be obtained from Corollary 3.8 in Homotopy equivalences between $p$-subgroup categories applied to $A_n(q)$. This formula features a Gaussian binomial coefficient. The solution to the recursion is \begin{equation*} E_n(q) = \prod_{1 \leq j \leq n} (1-q^j), \qquad n >0 \end{equation*} This shows that your representation and the one discussed by Solomon are representations of the affine group of the same degree. I suspect the two representations are identical. See also Section 8.2 in Brown: The coset poset and Probabilistic Zeta function of a Finite Group, Journal of Algebra 225 (2000) for the case where $q$ is a prime.