I would appreciate a reply to my question: let' s consider the Euclidean d-dimension real vector space. A polyhedral cone K is said to be simplicial if K is generated by linearly independent vectors (and non simplicial otherwise).
If d=2, if I'm nor wrong, every cone is simplicial. This implies that, if d=3, for an arbitrary cone K, its faces are always simplicial cones.
Question: In dimension d>= 4, the last statement is always true? That is, does there exist non simplicial cones with faces that are non simplicial?
Thank you very much for your reply.
a polyhedral cone is the non-negative hull of a given set of points. E.g. for two given points $x$ and $y$ the nonnegative hull is $\{ \lambda x+\mu y : \lambda, \mu\ge0\}$. You well might visualize this set by means of the convex hull of two points is $\{\lambda x + (1-\lambda) y : 0\le\lambda\le 1\}$, as any ray from the (also given) origin, intersecting that convex full, would be contained within that cone. So, in other words, that convex hull is an intersection of that cone, and conversely that cone is like an infinite pyramid of that hull.
Thence your question will be answered negatively, by providing a $(d-1)$-dimensional polytope, which itself is not simplicial and the facets of which are neither simplicial. Here you might think of a 3-dimensional polyhedron, serving as that convex hull, and using that one within an affine hyperplane of a 4-dimensional space: each of the six facet cones (of that cube-supported cone) will be supported by a square. None of these is simplicial for sure.
--- rk