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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.

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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.

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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.

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    $\begingroup$ Arent these a wedge of spheres? So maybe you mean that the first non vanishing homotopy group happens at the dimension of the sphere (but then lots more after that). $\endgroup$ Commented Feb 3, 2019 at 21:26
  • $\begingroup$ Oops! You're right. Edited, thanks. $\endgroup$
    – Alex Lazar
    Commented Feb 3, 2019 at 21:52

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