I learned that every $(k-2)$-face is contained in exactly two facets in a $k$-dimension polyhedra from 'Theory of linear and integer programming'. So every face lattice of polyhedra satisfies the "diamond property" (any interval of length 2 has exactly four elements). But can I show that the converse is true? I.e., any lattice with the "diamond property" can be obtained from the face lattice of some polyhedron?
2026-03-29 09:11:31.1774775491
Is there a polyhedron whose face lattice is a given lattice with the "diamond property"?
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No. The most direct example that comes to my mind is to take two face-lattices of two $d$-polytopes of the same dimension $d\ge 3$, and "glue" them at the maximal and the minimal element. The resulting lattice clearly has the diamond property but is not flag connected (see e.g. here), thus not the face-lattice of a polytope.
Other examples are provided by so-called simplicial spheres that have face-lattices that look just like the ones of polytopes, but that do not necessarily correspond to polytopes.