(The facet normal)
Let \(\sigma \subseteq N_\RR\) span \(N_\RR\), and let \(\tau \leq \sigma\) be a facet. Then there is \(u_\tau \in \dualof{\sigma}\), unique up to a positive scalar and unique on the nose once taken primitive in \(M\), with \[ \tau = \sigma \intersect u_\tau^\perp . \] It spans the ray \(\dualof{\sigma} \intersect \tau^\perp\), and the hyperplane \(H_\tau \da u_\tau^\perp\) is the hyperplane spanned by \(\tau\).
(Two presentations)
If \(\sigma\) spans \(N_\RR\) and \(\sigma \neq N_\RR\), then \[ \sigma = \Intersect_{\tau \text{ a facet of } \sigma} H_\tau^+ , \qquad H_\tau^+ \da \ts{ v \in N_\RR \st \inp{u_\tau}{v} \geq 0 } . \] Equivalently, if \(u_1, \ldots, u_t\) generate \(\dualof{\sigma}\), then \[ \sigma = \ts{ v \in N_\RR \st \inp{u_1}{v} \geq 0, \ \ldots, \ \inp{u_t}{v} \geq 0 } . \]
(Generators against inequalities)
This is the working content of double duality. A cone arrives in one of two forms, and every computation needs the other.
Given by generators, \(\sigma = \Cone(v_1, \ldots, v_r)\): the inequalities are \(\inp{u}{v_i} \geq 0\), and solving them produces \(\dualof{\sigma}\).
Given by inequalities, \(\sigma = \Intersect H_{u_j}^+\): the \(u_j\) are generators of \(\dualof{\sigma}\).
So passing from generators to inequalities on one side is passing from inequalities to generators on the other, and dualising twice returns the start. The facet normals are the minimal such list: \(\dualof{\sigma}\) needs exactly the primitive \(u_\tau\) as its ray generators, one per facet of \(\sigma\).
Dependencies
PR-FULSEPSeparation for convex polyhedral cones, and the double dualuses
Backlinks
PR-FULLOCA face gives a principal open subset, which is how the affine charts glueusesPR-FULFACEDUALFaces of a cone and faces of its dual correspond contravariantlyuses