Is the poset of affine subspaces of a vector space highly connected? The question is in the title.  Fix a field $k$.  Let $P_n$ be the poset of proper nonempty affine subspaces of $k^n$ under inclusion.  The geometric realization $|P_n|$ is $n$-dimensional.  Is it $(n-1)$-connected?  Or at least highly connected?
If I removed the word “affine” and thus required the subspaces to pass through the origin, this would be the usual Tits building, which is $(n-1)$-dimensional and by the Solomon-Tits theorem is $(n-2)$-connected.
 A: If $k$ is finite, then I'm pretty sure that $P_n$ is the (proper part of) the lattice of flats of a matroid. Such lattices are known to be shellable, which implies that the order complex of $P_n$ is "homotopy Cohen-Macaulay". This last condition implies that all the homotopy groups of $|P_n|$ vanish below its dimension.
A: Let's prove $P_n$ has homology only in top dimension $n$.
Let $A^n$ be the affine space of dimension $n$. For a point $P$ in $A^n$, let $U_P$ be the sub-poset of $P_n$ of those subspaces of $A^n$ that contain $P$. It is contractible since it is has a smallest element.
Moreover, if $S$ is a finite set of points in $A^n$, the intersection of the $U_P$'s for $P$ in $S$ is nothing but the poset of subspaces containing the affine span of $S$. Thus it is either contractible (when $S$ doesn't span the whole $A^n$) or empty (when $S$ contains a subset of $n+1$ points in general position).
The spectral sequence of the covering is thus quite simple in the $n$ first columns, and shows that the reduced homology $\tilde H_k(P_n)$ is  $0$ for $k\leq n-1$. Since the poset has dimension $n$, the claim is proved.
