# 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.

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.
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.