Given a polyhedral cone, its intersection with any two-dimensional plane is either a polygon or a region enclosed by a polygonal curve. Is it a characterization of polyhedral cones? Does there exists a convex cone, which is not polyhedral, but such that all its two-dimensional sections are regions enclosed by polygonal curves?

  • $\begingroup$ In what space is your cone? $\endgroup$ – Alexandre Eremenko Feb 13 '15 at 21:06
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    $\begingroup$ Presumably your definition of a polyhedral cone is the convex hull of a finite set of halflines (as opposed to an infinite set). Then I think that all the 2-dim intersections with parallel planes (say, orthogonal to the halflines median) determine the cone. $\endgroup$ – Joseph O'Rourke Feb 14 '15 at 0:35
  • $\begingroup$ Convex cones are subsets of euclidean space that are closed under positive linear combinations, polyhedral cones are intersections of a finite number of halfspaces. Intuition says that given a set included in the n-dimensional euclidean space, if its intersection with any two-dimensional plane is enclosed by a polygonal curve, then the set is a polyhedral cone. It is true if n=3, but, strange things could happens at higher dimensions $\endgroup$ – ANDRES Feb 14 '15 at 1:20

Victor Klee, "Some characterizations of convex polyhedra." Acta Mathematica 1959, Volume 102, Issue 1-2, pp 79-107.

I haven't read the whole paper, but Klee proved that if all $j$-dimensional sections of an $n$-dimensional convex region are polyhedral for $2\le j \le n-1$, then the region is polyhedral. His definition for polyhedral includes polyhedral cones and other unbounded polyhedra. For $j=2$ and a closed convex cone, Klee cites Mirkil for a slightly earlier proof.


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