Let $V$ be a finite-dimensional vector space with an inner-product $(,)$ and let $C\subset V$ be a cone in $V$. Let $C^\vee$ denote the dual of $C$ with respect to $(,)$, i.e., the set of vector $v\in V$ having non-negative intersection with all of $C$. Of course there is a natural identification of $V$ with its dual $V^\vee$, given by $v\mapsto (w\mapsto (v,w))$. In this way it makes sense to compare the two cones $C$ and $C^\vee$.

Is there some condition on $C$ that would allow one to conclude that $C^\vee\subseteq C$?.

For example, if $C$ is generated by finitely many vectors $v_1,\ldots,v_s$ and I know the matrix $M=[(v_i,v_j)]$ of all intersection numbers, can I see the above containment in terms of some kind of property of $M$?


The cone $C$ contains its dual cone if and only if for every $x\in\partial C$ there exists $y\in C$ so that $(x,y)\leq0$. (This is quite obvious in two dimensions.)

Let $S\subset V$ denote the unit sphere. The cone $C$ is uniquely determined by $A=C\cap S$ and similarly $C^V$ by $B=C^V\cap S$, so let us restrict our attention to $A$ and $B$.

If $B\neq\emptyset$, then $B\cap A\neq\emptyset$. Let $f:S\to\mathbb [-1,1]$, $f(x)=\min(x,A)$, where $(x,A)=\{(x,y);y\in A\}$. Now $f$ is continuous and reaches the value $-1$ in $S\setminus A$. The function $f$ reaches its maximum in $A$ and $B=f^{-1}([0,1])$ is connected (the dual cone is a cone). Thus $B\subset A$ is equivalent with $f|_{\partial A}\leq0$.


If your cone $C$ is generated by the vectors $(v_i)$, then its transverse sections are polyhedral. Lets, without losing much generality, assume that this polyhedron is closed and convex, that the $(v_i)$ are its vertices and that each face is a simplex. Each of the faces of the polyhedron, and hence of the cone, corresponds to an $(n-1)$-tuple of the vertex vectors, those generating the simplex. Let the vectors $(w_j)$ denote a set of oriented normal vectors to the faces (the normal vectors should be facing inside) of the cone $C$. These are the vectors that generate the dual cone $C^\vee$. Now, to check that $C^\vee \subseteq C$, you need to check that each of the $(w_j)$ belongs to $C$. With the $(-)^\vee$ being self-dual, this means that the inner products $(w_j,v_i)$ all need to be non-negative. If you unwind the definitions, you'll find that these inner products are determinants of matrices whose columns consist of the vector $v_i$ and the $(n-1)$-tuple of the vectors $(v_i)$ corresponding to the $w_j$, with some fixed column ordering and normalization. These determinants can also be interpreted as the signed volumes of the simplices spanned by these $n$-tuples of vectors.

In principle, once all the normalizations have been fixed, you could compute the absolute values of these volumes using only the inner products $(v_i,v_j)$, but not their signs. The reason is that any orientation reversing isometry of $V$ keeps these inner products the same, but negates all the signed volumes.


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