# condition on rational polyhedral cone to guarantee dual cone is homogeneous

Let $$\sigma\subseteq \Bbb R^d$$ be a full-dimensional rational polyhedral cone which is strongly convex (i.e. $$\sigma\cap-\sigma=0$$).

Definition. The cone $$\sigma$$ is homogeneous if there are lattice points $$a_1,\ldots,a_n\in \sigma \cap \Bbb Z^d$$ and a vector $$\delta \in \Bbb Z^d$$ such that

1. the monoid $$\sigma\cap \Bbb Z^d$$ is generated by $$a_1,\ldots,a_n$$, and
2. $$\delta\cdot a_1 = \cdots = \delta \cdot a_n = 1$$,

where $$\cdot$$ denotes the usual Euclidean inner product.

My Question. Is there a condition on the cone $$\sigma$$ which guarantees that its dual cone $$\sigma^\vee:=\{u\in \Bbb R^d \mid u\cdot a \geq0 \text{ for all } a\in \sigma\}$$ is homogeneous?

Note that just being homogeneous is not enough: Let $$\sigma = \Bbb R_{\geq0}\{(1,0), (1,1), (1,2), (1,3)\}$$, which is homogeneous with $$\delta=(1,0)$$. Then $$\sigma^\vee$$ has Hilbert basis $$\{(1,0),(0,1),(3,-1)\}$$. Since the points $$(1,0)$$, $$(0,1)$$, and $$(3,-1)$$ are non-collinear, $$\sigma^\vee$$ is not homogeneous.