Polyhedral cones

Question

Suppose there are two disjoint (non-overlapping edges) polyhedral cones with common vertex at the origin and having three faces each in $\mathbb{R}^{3}$. Is the space between them also a polyhedral cone ?

Answer 1

There's still a bit of vagueness in the definition of 'between' here, but if I understand correctly, then the answer should be yes. Briefly, let's say that the first cone has faces $A, B, C$ and the second has faces $D, E, F$, with $A$ and $D$ opposed. Consider the lines $\ell_{AB}$, etc, defined by the intersections of the relevant faces. Now, take the convex hull of $\ell_{AB}$, $\ell_{AC}$, $\ell_{DE}$ and $\ell_{DF}$. This will be a polyhedral cone, and if none of these lines is within it (that is, if the none of the lines is in the convex hull of the other three) then it will be your 'space between'. It's possible that one of the lines will be within that convex hull; that arguably strains the definition of opposing faces, but in such a case, you can look at the planes $P$ defined by $\ell_{AB}$ and $\ell_{DE}$ and $Q$ defined by $\ell_{AC}$ and $\ell_{DF}$, or the planes $P'$ defined by $\ell_{AB}$ and $\ell_{DF}$ and $Q'$ defined by $\ell_{AC}$ and $\ell_{DE}$. Then either $APDQ$ or $AP'DQ'$ will form a (non-convex) polyhedral cone.

This construction can break down in particularly egregious cases (e.g., if your cones intersect), but for anything where opposing faces can (IMHO) be reasonably defined, I think it makes the case that there's a 'between' region that is a polyhedral cone.

Answer 2

Let the edges be designated $ (a,b,c), (p,q,r)$ Any three edges make a polyhedral cone.

From given definition from the possible combinations $C(6,3)= 6!/(3!)^2=20, $ 15 combinations are given..

$$ abp,abq,abr,\, bcp,bcq,bcr,\, cap,caq,car. $$ $$ pqa,pqb,pqc, pra, prb,prc $$