The dual of a regular polyhedral cone is regular

Question

A rational polyhedral cone in $\mathbb{R}^n$ is a set of the form $$\sigma=\{\lambda_1x_1+\dots+\lambda_kx_k\in \mathbb{R}^n\mid \lambda_i\in \mathbb{R}_{\geq 0} \; \;\forall \, 1\leq i\leq k\}$$ for some $x_1,\dots,x_k\in\mathbb{Z}^n$. If the set $\{x_1,\dots,x_k\}$ can be extended to a basis of $\mathbb{Z}^n$ we say that $\sigma$ is a regular (or sometimes smooth) cone.

The dual cone of $\sigma$ is the set defined as $$\sigma^\vee=\{y\in\mathbb{R}^n \mid \langle x,y\rangle\geq 0, \; \forall \,x\in \sigma\}$$ and it is also a rational polyhedral cone.

How can I show that $\sigma$ regular $\implies$ $\sigma^\vee$ regular? If someone has a reference would be good also.

Answer 1

The claimed implication is not true, unless you suppose $\sigma$ to be of maximal dimension. Basically, a regular cone is necessarily strongly convex, but $\sigma^\vee$ is strongly convex only if $\sigma$ is of maximal dimension.

If you want to convince yourself with a simple counterexample, let \begin{equation} \sigma=\big\{(\lambda,0)\in\mathbb{R}^2:\lambda\geq0\big\} \end{equation} be the cone generated by $(1,0)$ in $\mathbb{R}^2$ (it is the horizontal positive semi-axis). According to the definition, it is clearly regular, as the canonical basis of $\mathbb{R}^2$ contains $(1,0)$. On the contrary, its dual cone is \begin{equation} \sigma^\vee=\big\{(y_1,y_2)\in\mathbb{R}^2:y_1\geq0\big\}, \end{equation} which is the cone with generators $(1,0), (0,1)$ and $(0,-1)$.

Suppose instead that $\sigma$ is regular of maximal dimension, which means it is generated by a basis $\{e_1,\ldots,e_n\}$ of $\mathbb{Z}^n$. It is an exercise to check that $\sigma^\vee$ is generated by its dual basis $\{e^*_1,\ldots,e^*_n\}$.