Justify the term "face lattice": When is an abstract polytope an order-theoretic lattice? The abstract combinatorial structure of a polytope is sometimes called its "face lattice". For example, see this or this.
But this is not always a lattice.
In a digon, the two edges are different upper bounds for the set of vertices, so there is no unique least upper bound. In a hemicube, a pair of vertices ($a,b$ in the wiki image) has an edge ($1$) and a face ($III$) as two incomparable upper bounds.
What conditions on the polytope are necessary or sufficient for it to be a lattice? For example, I suppose convexity is sufficient; and the "atomistic" property looks relevant.
 A: The argument in Wikipedia handles the case of a convex body fairly straightforwardly: if we have a (minimal) description of a $d$-dimensional body $B$ as the intersection of a set $H_1, H_2, \ldots, H_n$ of half-$d$-spaces defined as $H_i=\{v: v\cdot \mathbf{n}_i\leq c_i\}$, then each face is obtained by replacing some subset of the inequalities by equalities; i.e., by intersecting $B$ with some number of suitably defined $(d-1)$-planes $P_i=\{v: v\cdot\mathbf{n}_i=c_i\}$. Then if $\mathcal{F}_1$ is the face obtained by intersecting $B$ with $\{P_i: i\in S_1\}$ and $\mathcal{F}_2$ is the face obtained by intersecting $B$ with $\{P_i: i\in S_2\}$, then their meet and join in the face lattice are the faces obtained by intersecting $B$ with $\{P_i: i\in S_1\cup S_2\}$ and $\{P_i: i\in S_1\cap S_2\}$.
A: *

*Here is a necessary (but not sufficient) condition, [1, $\S$2A, pg. 29 ]:

If an $n$-polytope $\mathcal{P}$ is a lattice, then its edge-graph is $n$-connected,
  meaning that, for every pair of vertices of $\mathcal{P}$, there exist $n$ pairwise disjoint edge-paths having these vertices as end points.

It is evidently not sufficient since both the digon and the hemicube both satisfy this condition.

*As well, it is stated (without proof) that atomisticity -- or, as it seems to be more commonly known, vertex-describability -- is necessary in [5, pg. 291]. This is the property of each face being defined by a unique set of vertices. 

*A proof of the sufficiency of convexity can be found in [2, Thm. 2.7].

*Another set of polytopes for which their face posets are lattices are the universal regular polytopes [1, $\S$ 3D] which are a superset of regular convex polytopes. While they are stated to be lattices in [1, Thm. 3D7], no proof is provided there; instead, the reader is referred to [3] and [4]. 



*

*McMullen, Peter, and Egon Schulte. Abstract regular polytopes. Vol. 92. Cambridge University Press, 2002.

*Ziegler, Günter M.. Lectures on Polytopes. United States, Springer New York, 1995.

*Tits, Jacques. "Groupes et géométries de Coxeter, preprint Inst." Hautes Études Sci (1961).

*Tits, Jacques. "Géométries polyédriques et groupes simples." Atti della II Riunione del Groupement des Mathématiciens d'Expression Latine. Edizioni Cremonese, 1963.

*Connelly, Robert, Asia Ivić Weiss, and Walter Whiteley, eds. Rigidity and Symmetry. Vol. 70. Springer, 2014.
A: I'll try to show that any (finite) polytope that is also a lattice is atomistic.
In any polytope, for any face $A$, there is a well-defined set of all vertices $V_i\leq A$. The definition of a lattice gives the existence of $S=\sup\{V_i\}$, and clearly $A$ is an upper bound of $\{V_i\}$, so $A\geq S\geq V_i$. Now $S$ has its own set of vertices $W_i\leq S$. This contains $\{V_i\}$ since $V_i\leq S$; but it's also contained in $\{V_i\}$ since $W_i\leq S\leq A$. Therefore, $A$ and $S$ have the same vertex set $\{V_i\}$. We need to show that $A=S$.
The section $A/F_{-1}$ is itself a polytope, and also a lattice: if $X\leq A$ and $Y\leq A$, then $\sup\{X,Y\}\leq A$ and $\inf\{X,Y\}\leq A$. So, without loss of generality, $A=F_n$ is the greatest face, and $\{V_i\}$ is the set of all vertices in the polytope.
The result is clearly true when the rank $n\leq1$, or the rank of $S$ is $\leq0$, or the number of vertices is $\leq1$.
For $n=2$ we have the digon, but that's not a lattice; all larger polygons are lattices and satisfy $F_2=\sup\{V_i\}$.
Suppose the rank of $S$ is at least $1$, and, for induction, that the result is true for smaller sets of vertices. Considering $A\geq S$, let's assume $A>S$ for contradiction. Then for any vertex we can construct a flag
$$F_{-1}<V_i<\cdots<S<\cdots<A$$
and apply the diamond property to get a different flag
$$F_{-1}<V_i<\cdots<T_i<\cdots<A$$
where $T_i\neq S$ but they have the same rank, so $T_i\not<S$ and $T_i\not>S$.
If the vertex set of $T_i$ is all of the vertices ($T_i>V_j$ for all $j$), then $T_i\geq\sup\{V_j\}=S$, a contradiction.
If the vertex set of $T_i$ is smaller ($T_i\not>V_j$ for some $j$), then by induction $T_i$ is the supremum of this set. But since $S>V_j$ for all $j$, $S$ is an upper bound for this smaller set, so $S\geq T_i$, a contradiction. This concludes the proof.

This is rather messy. I may clean it up later, but feel free to edit it now.
