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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.

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.

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.

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.

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?

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.