Convex cones and self-duality

2012-12-10T08:41:54Z

Consider the Euclidian space $E_n={\mathbb R}^n$, with standard scalar product $$x\cdot y=x_1y_1+\cdots+x_ny_n.$$ A closed convex cone $\Gamma\subset E_n$ defines an order by $y\ge x$ iff $y-x\in\Gamma$. An order is compatible with the Euclidian structure if

$x,y\in\Gamma$ implies $x\cdot y\ge0$,
conversely, if $x\in\Gamma$ implies $x\cdot y\ge0$, then $y\in\Gamma$.

Cones satisfying these properties are usually called self-dual. Examples of self cones are $({\mathbb R}^+)^n$, a circular cone with an appropriate aperture angle (which depends on $n$), and the cone of semi-positive definite symmetric $d\times d$ matrices if $n=\frac{d(d+1)}2$. Self-dual cones are also present in the theory of Jordan algebras.

I have two questions.

If $n=2$, the angle of a cone and of its dual are related by the formula $\alpha+\beta=\pi$. In particular, a self-dual cone has angle $\frac\pi2$. In dimension $n=3$, there is no such formula. If the cone is circular, its solid angle $\Omega$ and $\Omega'$, that of the dual cone are related by $$\left(1-\frac{\Omega}{2\pi}\right)^2+\left(1-\frac{\Omega'}{2\pi}\right)^2=1$$ But for the positive orthant, the left-hand side above equals $\frac98$. Is it true that for every convex cone, the solid angles of the cone and of its dual are constrained by $$\left(1-\frac{\Omega}{2\pi}\right)^2+\left(1-\frac{\Omega'}{2\pi}\right)^2\ge1?$$ In particular, what are the possible values for the solid angle of a self-dual convex cone ? Is there a similar inequality (with equality for circular cones) in higher dimension ?

The side question is whether the set of self-dual convex cones form a compact metric space, where we may take the Hausdorff metric on the intersections with the unit sphere. I should bet so.

Answer by Anton Petrunin for Convex cones and self-duality

2012-12-10T14:58:21Z

For convex figure $\Sigma$ in $\mathbb S^2$, the isoperimetrical inequality should look like $$\left(\frac{\mathop{\rm perim}\Sigma}{2\cdot\pi}\right)^2+\left(1-\frac{\mathop{\rm area}\Sigma}{2\cdot\pi}\right)^2\ge 1.$$

If $\Sigma$ and $\Sigma'$ are the intersections of $\mathbb S^2$ with your cones then by Crofton formula we get $$\frac{\mathop{\rm perim}\Sigma}{2\cdot\pi}+\frac{\mathop{\rm area}\Sigma'}{2\cdot\pi}=1$$ Hence te result.

P.S. The extreme values should be for round cone and positive octant, but I do not see a proof in higher dimensions.

Convex cones and self-duality - MathOverflow

Convex cones and self-duality

Answer by Anton Petrunin for Convex cones and self-duality