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 $M$?



When does a cone contain its dual cone?
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 $f|\_{\partial A}\leq0$.


