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 $1$).

  • $\begingroup$ This action has to be the extension by scalars of the action on $H_{n-1}(\Delta(P), \mathbb{Z})$, right? Aren't such representations unlikely to be irreducible? $\endgroup$ Aug 27, 2022 at 22:57
  • $\begingroup$ @QiaochuYuan, I general no but sometimes this happens like for the Steinberg representation for a finite group of Lie type. Maybe it doesn’t happen here but the situation seems close enough to the general linear group that one could hope. $\endgroup$ Aug 27, 2022 at 23:01
  • $\begingroup$ There are results sort of like what you're looking for in the literature on matching complexes. See, i.a., Dong--Wachs, Prop 2.1, Sigg, Prop 9, Friedman--Hanlon, Prop 1: If $G$ acts simplicially on a simplicial complex $\Delta$, then $\tilde{H}_p(\Delta;k) \cong_G \ker \Lambda_p$, where $\Lambda_p$ is the combinatorial Laplacian $\partial \delta + \delta \partial$. This might not be any easier to compute, though... $\endgroup$
    – Alex Lazar
    Aug 28, 2022 at 13:59
  • $\begingroup$ @AlexLazar I'm really looking for results in this specific group acting on this specific poset rather than general stuff on groups acting on posets $\endgroup$ Aug 28, 2022 at 14:22

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.

  • $\begingroup$ Thanks. My understanding is Solomon is looking at that representation but I didn't know how to compute the dimension to see if Solomon was picking out a subrepresentation or the whole thing. I didn't know enough about this kind of geometric lattice stuff. All I could find was counting nonbroken circuits. The recursion is a good approach. Thanks. I'm glad this representation is irreducible $\endgroup$ Nov 1, 2022 at 12:56
  • $\begingroup$ By the way would it not make more sense to combine your two answers into one? $\endgroup$ Nov 1, 2022 at 12:57
  • $\begingroup$ Yes, I combined the two answers into one. $\endgroup$ Nov 1, 2022 at 13:01
  • $\begingroup$ It seems like the paper On the characters of the affine group over a field sciencedirect.com/science/article/pii/0021869392901337 by Siegel, who was a student of Solomon, says the affine group has a unique irreducible character of that dimension and it is a constituent in the module of n-1-chains. Moreover, if we restrict that character to the general linear group you get a character with the Steinberg representation as a constituent with multiplicity one. To be ctd $\endgroup$ Nov 1, 2022 at 16:39
  • $\begingroup$ This irreducible rep of the affine group appears once as a constituent of the n-chains and carries the Steinberg rep of GL_n as a constituent. Moreover, if I understand the proof, then n-1-chains on this geometric lattice as an affine group rep is induced from the n-1-chains of the Tits building as a GL_n-rep and the unique copy of the Steinberg rep in this restriction is in the kernel of the boundary map on the Tits building. It follows that the Affine group module generated by this copy of the Steinberg representation is the unique irreducible of our desired dimension and is in the homology $\endgroup$ Nov 1, 2022 at 16:43

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