Homotopy type of some lattices with top and bottom removed The only reaction to this question on math.SE was 21 views in 4 days, so I decided to repost it here. I am not changing anything.
There was an interesting question on MO which OP removed by some reason. Here is a (more or less) equivalent form.
Take a finite cartesian product of finite linear orders, and remove top and bottom. What is the homotopy type of the obtained poset?
When all linear orders have two elements, this is a sphere. In that question, it was conjectured that if at least one of them has more than two elements, then it is contractible. I think in fact from $3\times2\times2\times2$ one gets a (thickened) 2-sphere but I am not sure.
In general, is it known which homotopy types may occur when one removes top and bottom from a finite lattice?
Later:
In the course of answers, the following additional question became relevant (I think).
Call an element of a bounded lattice dense if it has nonbottom meet with every nonbottom element. Top is obviously such; call a dense element nontrivial if it is not top.
It is known that a finite distributive lattice (more generally a not necessarily finite pseudocomplemented lattice) does not have nontrivial dense elements if and only if it is a Boolean algebra.
Is a generalization of this known for arbitrary lattices? That is, which general lattices do not have any nontrivial dense elements?
Still later:
I've posted a question about the last one, Lattices without nontrivial dense elements; it was answered by Tri. So seems like correct type of lattices to consider in this context are the geometric lattices.
 A: Let $C(P)$ be the poset obtained by removing top and bottom from $P$ (where $P$ is a poset having a top and a bottom, not equal). Then $C(P\times Q)$ is homotopy equivalent to $\Sigma(C(P)\ast C(Q))$, the suspension of the join. Thus if one of the linear orders in your product has at least three elements then $C$ of the product will indeed be contractible, since the join of $X$ with a contractible space is always contractible. Note that the join of $X$ with the empty space is $X$.
A: Comments to the question contain several contributions to alternative approaches by Uri Bader. Since the latter seems to be reluctant to collect them into an answer, I decided to do the part I understand.
Approach 1.
Suppose given $P$ with top and bottom; then removing top and bottom from $P\times\{1,...,n\}$ is the union of $B:=(P\setminus\{\text{bottom}\})\times\{1\}$, $M:=P\times\{2,...,n-1\}$ and $T:=(P\setminus\{\text{top}\})\times\{n\}$. The hypothesis is that $\{2,...,n-1\}$ is not empty, so all three are obviously contractible, $B$ attached to $M$ along $(P\setminus\{\text{bottom}\})\times\{1,2\}$ and $M$ to $T$ along $(P\setminus\{\text{top}\})\times\{n-1,n\}$. If I understand Uri Bader's comment correctly, one just contracts $B$ and $T$ to $M$ separately.
Sort of an illustration, with $P=\{\text{top},\text{bottom}\}$ and $n=4$:

Another one, with $P=2\times2\times2$ and $n=3$:

Approach 2.
This I turn upside down since I am more used to it. For any element $a$ of a (bounded (finite)) lattice $L$ the embedding $[\text{bottom},a]\hookrightarrow L$ has a right adjoint $a\land\_$. Call an element $d$ of a lattice $L$ $\textit{dense}$ if $\forall x\in L\ (d\land x=\text{bottom})\Rightarrow(x=\text{bottom})$ holds. For any nontrivial ($\ne\text{top}$) such element the restriction of the adjoint $d\land\_$ to $L\setminus\{\text{top},\text{bottom}\}$ lands on $(\text{bottom},d]$, so the latter (which has top $d$, hence is contractible) becomes a deformation retract of $L\setminus\{\text{top},\text{bottom}\}$.
Now invoking the comment by Richard Stanley - in case $L$ is distributive, it has no nontrivial dense elements if and only if it is a Boolean algebra.
A natural question here is whether a characterization is known of those general (non-distributive) lattices which do not possess nontrivial dense (neither codense) elements.
Note that for non-distributive lattices generality of this approach is somehow orthogonal to that in Tom Goodwillie's answer: the latter works for products of not necessarily lattices having tops and bottoms, while here one approaches lattices which might not decompose into products.
In this connection there was a very interesting comment by Dan Petersen which has been elucidated by Uri Bader, but unfortunately I do not know this area well enough to say anything definite. The way I understand the idea is to consider homotopy types arising from lattices constructed in the same way in different characteristics as sort of "$q$-analogues", that is, families of homotopy types depending on a "modular" parameter encoded in $q$. I don't really know what that means that I wrote.
Finally, - it might be true that face lattices of polytopes with top and bottom removed actually have the same homotopy types as the polytopes themselves. Does anybody know anything about it? Just a couple of considerations: if this is the case then obviously the answer to my second question is that any homotopy type of a finite CW-complex may occur. Indeed if I am not mistaken already simplices of the second barycentric subdivision of any CW-complex form a (topless) lattice. Note also that for a poset one possible version of the barycentric subdivision is formed by linearly ordered subposets, and maximal such are not altered if one removes top and bottom...
A: Let me make a long comment here regarding the "q-analogue" of this question.
Let me start with a product of $n$ linear orders of length 2, $\{0,1\}^n$ (in fact, it be better to think of this as length 1, as we start at 0).
There is an obvious identification of this poset with the power set of $[n]$, $P([n])$. The simplicial complex associated with the latter is the $n-1$-simplex, and the chopped one is homeomorphic to the sphere $S^{n-1}$.
Now for the "$q$-analogue": take a finite field of order $q$, $\mathbb{F}_q$ and consider $V=\mathbb{F}_q^n$. Consider the lattice of linear subspaces of $V$. The chopped lattice is known also as "the spherical Tits building of $\text{PGL}_n(\mathbb{F}_q)$". Its homotopy type is considered in the Solomon-Tits Theorem which tells that it is a bouquet of $S^{n-1}$ spheres.
Note that for $n=1$ the $q$-case degenerates to the first: you get the same lattice.
Keep $n=1$, take $q=p$ a prime for simplicity, pick a "length" $k$ and consider the ring $R=\mathbb{Z}/p^k$. Consider the lattice of ideals in $R$. It is a chain of length $k+1$. That is, it coincides with the $n=1$"combinatorial" question considered in the original post, which we may think of as the "$q=1$ case".
Repeat this with arbitrary $n$: Let $M=\oplus_{i=1}^n \mathbb{Z}/p^{k_i}$ and consider the lattices of subgroups (or $\mathbb{Z}_p$-submodules). One may think of this lattice as the "$p$-analuge" of the one considered in the original post.
It turns out (and this is shown in comments and in other answers) that this lattice will be contractible provided for some $i$, $k_i>1$.
