The dual of a regular polyhedral cone is regular 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.
 A: 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 $\{e^*_1,\ldots,e^*_n\}$.
