Let \(\sigma \subseteq N_\RR\) be a full-dimensional convex polyhedral cone, \(\dim N_\RR = n\). The assignment \[ \tau \longmapsto \tau^\star \da \dualof{\sigma} \intersect \tau^\perp \] is an inclusion-reversing bijection from the faces of \(\sigma\) to the faces of \(\dualof{\sigma}\), with inverse of the same shape, and \[ \dim \tau + \dim \tau^\star = n . \]
(The two ends, and what they pin down)
The bijection sends \(\sigma\) itself to \(\ts{0}\) and \(\ts{0}\) to \(\dualof{\sigma}\), and in between it exchanges the two extremes of the face lattice:
| face of \(\sigma\) | image in \(\dualof{\sigma}\) |
|---|---|
| facet \(\tau\), \(\dim = n-1\) | ray spanned by the facet normal \(u_\tau\) |
| ray \(\rho\), \(\dim = 1\) | facet of \(\dualof{\sigma}\), \(\dim = n-1\) |
So the rays of \(\dualof{\sigma}\) are the facet normals of \(\sigma\) and vice versa, which is exactly what makes the plane computation work: in \(N_\RR = \RR^2\) facets and rays coincide, so dualising a two-dimensional cone is rotating its two generators a quarter turn.
Strong convexity is visible here too. \(\sigma\) contains no line exactly when \(\ts{0}\) is a face of \(\sigma\), which under the bijection says \(\dualof{\sigma}\) is full-dimensional, that is \(\dim \dualof{\sigma} = n\). That is the condition needed for \(S_\sigma\) to generate \(M\) and for \(U_\sigma\) to have the right dimension.
Dependencies
PR-FULFACETFacet normals and the half-space presentation of a coneusesPR-FULSEPSeparation for convex polyhedral cones, and the double dualuses