It is well known that a convex polyhedron $D$ can be written as the Minkowski sum of a polytope $P$ and a positive cone $C$. It is equally well-known that the $C$ in the decomposition is unique ; in fact, $C$ is necessarily the so-called characteristic cone of $D$ :
$$ C=\bigg\lbrace c \ \bigg| \ \forall x\in D, x+c\in D \bigg\rbrace $$
All the presentations I've seen note that $P$ is obviously nonunique (since we may add an arbitrary finite number of elements of $C$ to $P$ and take the convex hull, and still get a polytope making the decomposition true) but none wonder if one can make it unique by adding an extra constraint.
A natural (to me) definition is as follows : let
$$ M=\bigg\lbrace x\in D \bigg| \forall x'\in D, \ x+C \subseteq x'+C \Rightarrow x'=x \bigg\rbrace $$
and let $\Pi$ be the convex hull of $M$.
Question. Is $\Pi$ a polytope, and does the decomposition $D=\Pi+C$ hold ?
If the answer to both those questions is yes, then it makes sense to consider $D=\Pi+C$ as the "canonical" decomposition.
My thoughts :
When $D$ is an "obelisk" (a pyramid with square base sitting on top of an infinite parelliped), then $M$ is the frontier of the pyramid minus the interior of the base, so that $M$ is not convex or closed in this case.
If we start with an arbitrary decomposition $D=P+C$ where $P$ is the convex hull of a finite number of points $p_1,p_2,\ldots,p_m$, it would seem that $\Pi$ is the convex hull of the $p_i$'s that are contained in $M$, but I have been unable to prove this so far.
Here are the (standard) definitions I use : a polyhedron is an intersection of a finite number of half-spaces (i.e. sets of the form $\lbrace x | \phi(x)\geq c\rbrace$ where $\phi$ is a linear form and $c$ is a constant), a polytope is the convex hull of a finite set of points, and a positive cone is the "positive hull" (i.e. the sets of all linear combinations with nonnegative coefficients) of a finite set of vectors.