Polar polytope

Remarks

If $v_1, \dots, v_m$ are the vertices of $P$, then $P^\circ = \set{y : v_i \cdot y \le 1, \ i = 1, \dots, m}$. $P^\circ$ is again a convex polytope with $0$ in its interior, and $(P^\circ)^\circ = P$. The map $F \mapsto \set{y \in P^\circ : x \cdot y = 1 \text{ for all } x \in F}$ is an inclusion-reversing bijection from the faces of $P$ to the faces of $P^\circ$, sending $k$-faces to $(d - 1 - k)$-faces.

Example: let $P = [-1, 1]^2$. Its vertices are $(\pm 1, \pm 1)$, so $y \in P^\circ$ means $\pm y_1 \pm y_2 \le 1$ for all signs, that is $\abs{y_1} + \abs{y_2} \le 1$. So $P^\circ$ is the square with vertices $(\pm 1, 0)$ and $(0, \pm 1)$.

Non-example: for $P = [0,1]^2$ the origin is a vertex, not an interior point, so the definition does not apply. The same formula gives the unbounded set $\set{y : y_1 \le 1, \ y_2 \le 1, \ y_1 + y_2 \le 1}$, which is not a polytope.