Is the poset $\mathrm{Idl}_{\neq \emptyset, P}(P)$ of nonempty, proper ideals in a finite connected poset $P$ (empty or) weakly contractible?

Question

$\DeclareMathOperator\Idl{Idl}$Let $P$ be a finite, connected poset with at least two elements, and let $\Idl_{\neq \emptyset, P}(P)$ be the set of downward closed sets $S \subset P$ such that $S \neq \emptyset$ and $S \neq P$, ordered under inclusion.

Question: Is $\Idl_{\neq \emptyset, P}(P)$ weakly contractible?

Notes:

• If $P$ has less than two elements, then $\Idl_{\neq \emptyset, P}(P)$ is empty. If $P$ is a discrete poset with $\geq 2$ elements, then $|\Idl_{\neq \emptyset, P}(P)| \simeq S^{|P|-2}$. Hence the restriction to connected posets with $\geq 2$ elements.

• If $P$ has four elements $a,b < c,d$, so that $|P| \simeq S^1$, then $\Idl_{\neq \emptyset, P}(P)$ has 5 elements arranged in an "$X$" shape, and is weakly contractible.

• In general, if we add $\emptyset$ and $P$ back in, the poset $\Idl(P)$ of all ideals is always contractible (for at least 4 reasons -- it has an initial and a terminal object, and binary meets and binary joins). Similarly, $\Idl_{\neq \emptyset}(P)$ and $\Idl_{\neq P}(P)$ are contractible if $P \neq \emptyset$.

• I checked a few random examples with Sage, which at least turned out to have vanishing reduced homology. (Although I'm not sure how much I trust my code!)

Answer 1

Let me be a little more historical. As mentioned, it is well known that (open intervals of) distributive lattices are shellable. This is mentioned in Corollary 3.2 of Björner's classic paper "Shellable and Cohen-Macaulay partially ordered sets" (https://doi.org/10.1090/S0002-9947-1980-0570784-2) but is an older result I believe.

The homotopy type of a shellable complex is a wedge of spheres (of the same dimension). How many spheres do we get? By a basic homology computation, we get $|\mu(\hat{0},\hat{1})|$ of them, where $\mu$ is the Möbius function and $\hat{0}$/$\hat{1}$ are the minimal/maximal elements of the lattice. In a distributive lattice $J(P)$ of order ideals, the Möbius function is given by $$\mu(I,I') = \begin{cases} (-1)^{\# I' \setminus I} &\textrm{if $I' \setminus I$ is an antichain}, \\ 0 &\textrm{otherwise}. \end{cases}$$ See e.g. Example 3.9.6 of Stanley's EC1, 2nd ed. So your restriction to connected $P$ with $\geq 2$ elements gives $\mu(\hat{0},\hat{1})=0$, hence the open interval $(\hat{0},\hat{1})$ in $J(P)$ is a (homotopy) ball.

Answer 2

Here is another proof. Let $M$ be the set of minimal elements of $P$, and let $2_{\ne \emptyset}^M$ be the poset of non-empty subsets of $M$. There is an obvious inclusion of posets $2_{\ne \emptyset}^M \hookrightarrow Idl_{\ne \emptyset}(P)$. Assuming that $P$ is not a disjoint union of points, this restricts to an inclusion $$i\colon 2_{\ne \emptyset}^M \hookrightarrow Idl_{\ne \emptyset, P}(P).$$

There also is a map of posets $$r\colon Idl_{\ne \emptyset, P}(P) \to 2_{\ne \emptyset}^M,$$ which sends an ideal $I$ to $I\cap M$. The composition $ri$ is the identity, while $ir$ is related to the identity by the inequality $ir(I)\le I$ for every ideal $I$. It follows that $i$ and $r$ induce homotopy equivalences of geometric realizations $$|Idl_{\ne \emptyset, P}(P)| \simeq |2_{\ne \emptyset}^M|.$$ But $2_{\ne \emptyset}^M$ has a maximum element $M$, so its geometric realization is contractible.