14
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ Perhaps you should explicitly say that $\Omega$ is the solid angle of the cone and $\Omega'$ that of the dual cone. $\endgroup$ Dec 10, 2012 at 10:36

1 Answer 1

10
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Nice, but could you give a reference for the isoperimetric inequality in the $2$-sphere? $\endgroup$ Dec 10, 2012 at 15:34
  • $\begingroup$ @Danis, Wikipedia says Paul Lévy (1919) $\endgroup$ Dec 10, 2012 at 16:14
  • 1
    $\begingroup$ You can also find a proof by symmetrization in Burago-Zalgaller. $\endgroup$ Dec 10, 2012 at 19:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.