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?


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.

  • $\begingroup$ It is quite easy to see that the chopped product of linear orders, not all with $\leq 2$ elements, will be contractible. Pick a vertex for which (say) the first instance is intermidiate and work out the contraction to it coordinatewise, starting with the first. $\endgroup$ – Uri Bader Jul 26 '16 at 11:44
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    $\begingroup$ Here is a general observation: assume you have in your lattice an element $x\neq 0$ s.t for all $y\neq 1$, $x\vee y\neq 1$. Then the chopped lattice is contractible. Indeed, $y\mapsto x\vee y$ gives a deformation retract onto the sublattice $[x,1)$, which is clearly contractible as it is coned over $x$. In a product of linear orders, every element $x$ which at every coordinates is not maximal will do. $\endgroup$ – Uri Bader Jul 26 '16 at 13:39
  • $\begingroup$ @UriBader Seems like a competing answer emerges ;) $\endgroup$ – მამუკა ჯიბლაძე Jul 26 '16 at 13:48
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    $\begingroup$ @UriBader it sounds like the question asked here is the "$q\to 1$" specialisation of the result in your comment! $\endgroup$ – Dan Petersen Jul 26 '16 at 16:55
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    $\begingroup$ @მამუკა ჯიბლაძე The earliest reference I know is the 1977 Ph.D. thesis of Scott Provan, but I don't know how to access it online. A more general result appears as Corollary 2.2 of arxiv.org/pdf/math/0505576.pdf. $\endgroup$ – Richard Stanley Aug 12 '16 at 16:17

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 $X$.

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    $\begingroup$ A reference to this result is Theorem 5.1 (d) in the paper by J. Walker, Canonical homeomorphisms of posets, European J. Combin. 9 (1988), no. 2. $\endgroup$ – Gregory Arone Jul 26 '16 at 12:12
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    $\begingroup$ What's your favorite definition of $\Sigma X$? For example, you can defined it as the double mapping cylinder of the pushout diagram $*\leftarrow X \to *$. Now substitute $X=\emptyset$. $\endgroup$ – Gregory Arone Jul 26 '16 at 13:22
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    $\begingroup$ By the way, for the result to hold you have to assume that $P$ and $Q$ each have more than one element. You can include the one-element poset in the result by by decreeing that if $P$ has one element then $C(P)=S^{-2}$ is the $-2$-dimensional sphere. $\endgroup$ – Gregory Arone Jul 26 '16 at 13:26
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    $\begingroup$ Oh I see you collapse them to fresh points, yes? $\endgroup$ – მამუკა ჯიბლაძე Jul 26 '16 at 13:36
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    $\begingroup$ To show that $C(P\times Q)\simeq\Sigma(C(P)\ast C(Q))$ first notice that the Grothendieck construction of the diagram of posets $X\leftarrow X\times Y\to Y$ is the poset $Z(X,Y):=(X\cup\{1\})\times(Y\cup\{1\})\setminus\{(1,1)\}$ where $1$ denotes a new top element to adjoin. So $Z(X,Y)$ has the homotopy type of $X \ast Y$. Then observe that $C(P\times Q)$ is the union of $A:=Z(C(P),C(Q)\cup\{0\})$ and $B:=Z(C(P)\cup\{0\}), C(Q))$ which intersect along $Z(C(P),C(Q))$ ($0$ is a new bottom element to adjoin). Since $A$ and $B$ are contractible, $C(P\times Q)\simeq\Sigma Z(C(P),C(Q))$. $\endgroup$ – Omar Antolín-Camarena Jul 29 '16 at 3:14

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 $i$, $k_i>1$.


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 $n=4$:

enter image description here

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

enter image description here

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...


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