A finite Boolean lattice is a lattice isomorphic to the subset lattice of a finite set. Every Boolean lattice is Eulerian, namely, a graded lattice $L$ such that $\mu(a,b) = (-1)^{|b|-|a|}$ for all $a,b \in L$ such that $a \le b$, with $a \mapsto |a|$ the rank function and $\mu$ the Mobius function on $L$.

All the lattices is this post are assumed finite.

By Birkhoff's representation theorem, the sublattices of the Boolean lattices are exactly the distributive lattices, namely, the lattices $L$ such that $\forall a,b,c \in L$ then $a \wedge(b \vee c) = (a \wedge b) \vee (a \wedge c) $.

Question: What are the sublattices of the Eulerian lattices?

By "what" I mean a direct definition (as above for the distributive lattice). Then, we would get (by construction) an Eulerian generalization of Birkhoff's representation theorem.

Every interval or reverse of an Eulerian lattice is Eulerian. By the crosscut theorem, every Eulerian lattice is atomistic and coatomistic, so it is its own bottom interval $[\hat{0}, b]$ and top interval $[t,\hat{1}]$, where $b$ is the join of the atoms and $t$ the meet of the coatoms. A distributive lattice is top and bottom Boolean, namely, its top and bottom intervals are Boolean. Now some sublattices of some Eulerian lattices are neither top nor bottom Eulerian.

The smallest non-Boolean Eulerian lattice is the following:
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It is the face lattice of the square polytope (see here). It admits the pentagon lattice as sublattice, which is its own top and bottom interval, but is not Eulerian. Note that the face lattice of the hexagon polytope admits the diamond lattice as sublattice. So maybe every finite lattice is the sublattice of some Eulerian lattice.

Bonus question: Is there a finite lattice which is not isomorphic to a sublattice of an Eulerian lattice?

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    $\begingroup$ I suspect that every finite lattice is isomorphic to a sublattice of an Eulerian lattice. It is known (link.springer.com/article/10.1007/BF02482893) that every finite lattice is a sublattice of the lattice of partitions of some finite set. Thus it suffices to show that every finite set partition lattice is isomorphic to a sublattice of an Eulerian lattice. Perhaps this makes the problem easier. $\endgroup$ Mar 5 at 13:33

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