Abstract polytope

Remarks

Conditions 1 and 2 make $\mathcal{P}$ a graded poset with ranks $-1, 0, \dots, n$. Faces of rank $0$ are vertices, faces of rank $1$ are edges, and faces of rank $n - 1$ are facets.

Example: the faces of a triangle with vertices $a, b, c$, together with $\varnothing$ as $F_{-1}$ and the triangle $T$ as $F_2$, ordered by inclusion, form an abstract polytope of rank $2$. Every flag has the form $\varnothing < a < ab < T$, with $4 = 2 + 2$ faces. Between $\varnothing$ and an edge there are exactly two vertices, and between a vertex and $T$ there are exactly two edges, so the diamond condition holds. More generally, the faces of any convex polytope of dimension $n$, with $\varnothing$ added, form an abstract polytope of rank $n$.

Non-example: take two triangles that share one vertex $c$, with all their vertices and edges, a least face, and one greatest face of rank $2$. The vertex $c$ lies on four edges, so between $c$ and the greatest face there are four faces of rank $1$, and the diamond condition fails.

The hemicube is obtained from the cube by identifying antipodal faces. It is an abstract polytope of rank $3$ with $4$ vertices, $6$ edges and $3$ square faces, and it is not the face lattice of any convex polytope.