Justify the term "face lattice": When is an abstract polytope an order-theoretic lattice?

Question

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.

Answer 1

• 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].

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

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

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

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

5. Connelly, Robert, Asia Ivić Weiss, and Walter Whiteley, eds. Rigidity and Symmetry. Vol. 70. Springer, 2014.

Answer 2

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 $\{P_i: i\in S_1\cap S_2\}$.

Answer 3

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 ($A=S$) is clearly true when the number of vertices is $0$ or $1$.

For induction, suppose the result is true for smaller sets of vertices. Considering $A\geq S$, let's assume $A>S$ for contradiction. Then by the dyadic/diamond property there is some $T\neq S$ but with the same rank, so $T\not\leq S$ and $T\not\geq S$.

If the vertex set of $T$ is all of the vertices ($T\geq V_i$ for all $i$), then $T\geq\sup\{V_i\}=S$, a contradiction.

If the vertex set of $T$ is smaller ($T\not\geq V_i$ for some $i$), then by induction $T$ is the supremum of this set. But since $S\geq V_i$ for all $i$, $S$ is an upper bound for this smaller set, so $S\geq T$, a contradiction. This concludes the proof.