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.



Is the poset of affine subspaces of a vector space highly connected?
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.


