Given two polytopes $P$ and $Q$, show that $(P^* \times Q^*)^* = P \oplus Q $

Question

Using definitions, I got so far as to express $(P^* \times Q^*)^*$ in the following form:

$$(P^* \times Q^*)^* = \left\{\,\begin{pmatrix} z^* \\ w^* \end{pmatrix} \in \mathbb{R}^{d+e} \,\middle|\, \forall \begin{pmatrix} z \\ w \end{pmatrix} \in (P^* \times Q^*), \left\langle \begin{pmatrix} z \\ w \end{pmatrix}, \begin{pmatrix} z^* \\ w^* \end{pmatrix} \right\rangle \leq 1\right\}$$

I think that I should continue trying to express $(P^* \times Q^*)^*$ in terms of $\begin{pmatrix} z^* \\ w^* \end{pmatrix}$ and $\begin{pmatrix} x \\ y \end{pmatrix}$, but I don't know how. Or should I look at another approach? Any tips would be greatly appreciated!

Answer 1

One way to see this is using the duality between facets and vertices. A polytope $P\subset\Bbb R^d$ containing the origin $0$ is given by either

• vertices $v_1,\dots,v_n$ such that $ P = \operatorname{conv}(v_1,\dots,v_n) $, or
• facet normals $a_1,\dots,a_m$ such that $P = \left\{\, x \in \Bbb R^d \,\middle|\, \langle x,a_i\rangle \le 1 \text{ for $i=1,\dots,m$}\,\right\}$.

The dual polytope $P^*$ then also contains the origin in its relative interior and has vertices $a_1,\dots,a_m$ and facet normals $v_1,\dots, v_n$.

Given two such polytopes $P$ and $Q$ what can we say about the descriptions of $P\times Q$ and $P\oplus Q$ in terms of the vertices and facets of $P$ and $Q$?

• The vertices of $P\oplus Q$ are $(v,0)$ for vertices $v$ of $P$ and $(0,w)$ for vertices $w$ of $Q$,
• The facet normals of $P\times Q$ are $(a,0)$ for facet normals $a$ of $P$ and $(0,b)$ for facet normals $b$ of $Q$.

Now what are the vertices of $(P^*\times Q^*)^*$? They are the facet normals of $P^*\times Q^*$, so of the form $(v,0)$ and $(0,w)$, where $v$ and $w$ are facet normals of $P^*$ and $Q^*$, respectively. However this just means $v$ and $w$ are vertices of $P$ and $Q$, respectively. Therefore, the vertices of $(P^*\times Q^*)^*$ are exactly those of $P\oplus Q$.