Is there a polytope with face poset given by the lattice of (unordered) partitions?

Question

The permutohedron has as it's face poset the lattice of ordered set partitions (I've also seen this called the lattice of set compositions). I'm wondering if there is a polytope with face poset given by the lattice of set partitions https://en.wikipedia.org/wiki/Partition_of_a_set.

Answer 1

The lattice of partitions of an $n$-element set fails the diamond property (if $n \ge 3$), so according to this Wikipedia definition it is not (even) an abstract polytope.

The diamond property says that if the ranks of two faces $a>b$ differ by 2, then there are exactly 2 faces strictly between $a$ and $b$.

In the partition lattice of $3$ elements ($1, 2, 3$) the diamond property fails between the rank-$2$ face $a=\{\{1,2,3\}\}$ and the rank-$0$ face $b=\{\{1\},\{2\},\{3\}\}$, since strictly between them there are three faces, the two-block partitions. You can easily generalize this to bigger partition lattices.

As an example of the diamond property in geometric terms, in a three-dimensional polytope: (i) any edge is incident with exactly two faces; (ii) a vertex of a face is incident with exactly two edges of that face.

With $n=2$ the partition lattice has just two elements, $\{\{1\},\{2\}\}$ and $\{\{1,2\}\}$, and in this case you do get a polytope (a single point).