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The face lattice of a convex polytope $P\subset\Bbb R^d$ is the partially ordered set whose elements are the faces of $P$ ordered by inclusion. We can turn it into a graph by considering its Hasse diagram. Let's focus on the $d$-dimensional cube $Q_d:=[0,1]^d$:

Quetion: Is the (Hasse diagram of the) face lattice of the $d$-cube $Q_d$ the edge graph of some polytope?

For $d=0,1$ the answer is easily seen to be yes. For $d=2$ the answer is yes as well, and the polytope is the tetragonal trapezohedron:

I don't know the answer for $d\ge 3$. Since the Hasse diagram for $Q_d$ has vertices of degree $d+1$, the dimension of a polytope with this edge graph is at most $d+1$.

I should note that I am not aware of a single polytope whose Hasse diagram is not a polytope graph, though I have not looked at too many examples. One other easy example is the Hasse diagram of the $d$-simplex, which is is the edge graph of the $(d+1)$-cube.

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2 Answers 2

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Yes. The Hasse diagram of any polytope is the edge graph of the dual of its antiprism.

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  • $\begingroup$ What is the antiprism of a general polytope? And do you have a source for this claim? $\endgroup$
    – M. Winter
    Commented May 13 at 20:12
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    $\begingroup$ Here is my guess for what "antiprism" of $P$ could mean: take $P$ and its polar dual $P^\circ$, then the antiprism is the convex hull of $P\times\{1\}$ and $P^\circ\times \{-1\}$ in $\Bbb R^{d+1}$. Is this what you meant? I think it is at least plausible that its dual has the right edge graph, though I would really prefer to see a source or rough argument for this claim. $\endgroup$
    – M. Winter
    Commented May 17 at 18:32
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    $\begingroup$ researchgate.net/publication/… $\endgroup$
    – Daniel Sebald
    Commented Jun 9 at 3:43
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Let me expand on Daniel's answer. The paper Daniel cites defines the antiprism (p.4) of an abstract polytope $\mathcal P$ to be

$$\operatorname{Ant}(\mathcal P):=\{(F,G)\mid F,G\in\mathcal P \text{ with } F\le_{\mathcal P} G\}\cup\{P\}$$

where $P$ is a symbol for the new top-most element. The order relation is defined as

\begin{align} (F,G)\le(H,K) &\;:\Longleftrightarrow\; F\le_{\mathcal P} H\le_{\mathcal P} K\le_{\mathcal P} G,\\ (F,G)\le P&\;:\Longleftrightarrow\; F,G\in\mathcal P. \end{align}

In the paper the authors show that this given indeed an abstract polytope. Since we are interested in the edge graph of the dual, we determine the faces of $\operatorname{Ant}(\mathcal P)$ of codimension 0,1 and 2.

  • clearly $P$ is the unique codimension 0 element. It corresponds to $\varnothing$ in the dual of the antiprism.
  • the faces of $\operatorname{Ant}(\mathcal P)$ that are dominated by only $P$ are of the form $(F,F),F\in\mathcal P$. So the vertices of the dual of the antiprism correspond precisely to the faces of $\mathcal P$.
  • the faces of $\operatorname{Ant}(\mathcal P)$ that are dominated by only $P$ and $(F,F),F\in\mathcal P$ are of the form $(F,G)$, where $F\lessdot G$ is a cover relation. These correspond to the edges in the Hasse diagram. Hence, the edges of the dual of the antiprism are precisely the edges of the Hasse diagram.

As an abstract polytope, the dual of the antiprisms does exactly what I asked for. I might have been unclear in my question, but I was actually hoping for a convex polytope with this property. The paper states however (p.1) "For higher dimensions [i.e. higher than two], the concept of a convex antiprism is not always defined", so antiprisms might not provide an answer here in general. At least for the cube (in every dimension) the following simple strategy suffices to realize the antiprisms as a convex polytope (see the comment by David): if $C$ is the unit $d$-cube and $O$ is the corresponding polar dual $d$-crosspolytope, then

$$\operatorname{Ant}(\mathcal P) \simeq \operatorname{conv}\big(C\times\{-1\}\cup O\times\{1\}\big).$$

Update - 19/10/2024

In the comments David provides a reference for an article that was inspired by the exact question of obtaining face lattices as edge graphs, and also constructs a polytope whose antiprism cannot be realized as a convex polytope.

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  • $\begingroup$ The "simple strategy" in your last paragraph will work. We can see this using ideas from arxiv.org/abs/1307.0071 . Let $P$ be a polytope in $\mathbb{R}^n$. We call $P$ "perfectly centered" if $0 \in P$ and for every facet $F$ of $P$, the orthogonal projection of $0$ onto the hyperplane bounding $F$ lies in $F$. Theorem 3.2 and Lemma 3.3 of the linked paper shows that, if $P$ is perfectly centered, then $\text{Hull}(P \times \{ 1 \} \ \cup\ P^{\ast} \times \{ -1 \})$ realizes the antiprism of $P$. $\endgroup$
    – David E Speyer
    Commented Aug 25 at 13:40
  • $\begingroup$ Clearly, the cube $[-1, 1]^n$ is perfectly centered. $\endgroup$
    – David E Speyer
    Commented Aug 25 at 13:40

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