Convex cones and self-duality - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T20:59:52Z http://mathoverflow.net/feeds/question/115958 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115958/convex-cones-and-self-duality Convex cones and self-duality Denis Serre 2012-12-10T08:41:54Z 2012-12-10T15:14:06Z <p>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 <em>compatible</em> with the Euclidian structure if </p> <ul> <li>$x,y\in\Gamma$ implies $x\cdot y\ge0$,</li> <li>conversely, if $x\in\Gamma$ implies $x\cdot y\ge0$, then $y\in\Gamma$.</li> </ul> <p>Cones satisfying these properties are usually called <em>self-dual</em>. 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.</p> <p>I have two questions.</p> <blockquote> <p>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 ?</p> </blockquote> <p>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.</p> http://mathoverflow.net/questions/115958/convex-cones-and-self-duality/115985#115985 Answer by Anton Petrunin for Convex cones and self-duality Anton Petrunin 2012-12-10T14:58:21Z 2012-12-10T15:14:06Z <p>For convex figure $\Sigma$ in $\mathbb S^2$, the isoperimetrical inequality should look like $$\left(\frac{\mathop{\rm perim}\Sigma}{2\cdot\pi}\right)^2+\left(1-\frac{\mathop{\rm area}\Sigma}{2\cdot\pi}\right)^2\ge 1.$$</p> <p>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> <p><strong>P.S.</strong> The extreme values should be for round cone and positive octant, but I do not see a proof in higher dimensions. </p>