Minkowski sum of polytopes

Remarks

The set $P + Q$ is again a convex polytope. If $P$ has vertex set $V$ and $Q$ has vertex set $W$, then $P + Q$ is the convex hull of the finite set $\set{v + w : v \in V, \ w \in W}$. For a linear functional $u$, the face of $P + Q$ on which $u$ is largest is the sum of the faces of $P$ and of $Q$ on which $u$ is largest. So each vertex of $P + Q$ is $v + w$ for a unique vertex $v$ of $P$ and a unique vertex $w$ of $Q$.

Example: let $T$ be the triangle with vertices $(0,0)$, $(1,0)$, $(0,1)$ in $\R^2$, and let $-T$ be its image under $x \mapsto -x$. The sums of vertices of $T$ and $-T$ are the differences of vertices of $T$, namely $(0,0)$, $(\pm 1, 0)$, $(0, \pm 1)$, $(1,-1)$ and $(-1,1)$. Their convex hull is the hexagon with vertices $(1,0)$, $(1,-1)$, $(0,-1)$, $(-1,0)$, $(-1,1)$, $(0,1)$, so $T + (-T)$ is this hexagon.

Non-example: the closed unit disk $D$ is convex but not a polytope, and $[0,1]^2 + D$ is a square with rounded corners, which is not a polytope. So the closure of polytopes under the sum needs both summands to be polytopes.

The operation is commutative and associative, and $\set{0}$ is its neutral element. It is the general Minkowski addition of sets, restricted to polytopes.