Flattening a corner in a convex $d$-polytope (into $d-1$ dimensions, without overlap)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T02:40:13Z http://mathoverflow.net/feeds/question/91944 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91944/flattening-a-corner-in-a-convex-d-polytope-into-d-1-dimensions-without-over Flattening a corner in a convex $d$-polytope (into $d-1$ dimensions, without overlap)? Emily Peters 2012-03-22T20:18:15Z 2012-03-23T16:07:49Z <p>I'm interested in the following question, which seems to be assumed all over the place (at least for 3 dimensions) in convex geometry, and which I cannot find a proof of.</p> <blockquote> <p>Suppose we have a corner of a convex polytope in $\mathbb{R}^d$. How do we show that we can 'flatten' the surrounding facets into $d-1$ dimensions without overlap? </p> </blockquote> <p>What do I mean by that? Well, for the three-dimensional case, it means that if you sum the angles around a given vertex of a convex polyhedron, you get a sum of less than $2 \pi$. Here's another way to say that: </p> <p>(0) Given a collection of 2-dim cones $C_i$ with angles $a_i$, if the sum of the $a_i$ is greater than $2 \pi$, then $C_i$ cannot be facets of a 3-dim cone. </p> <p>(I'm defining a cone to be the convex hull of a collection of rays; so, cones are assumed to be convex.) </p> <p>And in general dimensions: </p> <p>(1) Given a collection of $(d-1)$-dim cones $C_i$ with total $(d-1)$-angle measures $a_i$, if the sum of the $a_i$ is greater than the total angle surrounding a point in $\mathbb{R}^{(d-1)}$, then the $C_i$ cannot be facets of a d-dim cone.</p> <p>This fact can be restated in a lot of other ways. Maybe one of these is easier to prove?</p> <p>(2) If a collection of $(d-1)$-dim cones $C_i$ with total $(d-1)$-angle sum greater than the total angle surrounding a point in $\mathbb{R}^{(d-1)}$ is configured in $\mathbb{R}^d$ with all cone points set at the origin and every $(d-2)$-face identified with a $(d-2)$-face of some other cone (i.e., the cones are glued to make a simplicial complex), and if this configuration lies on one side of a hyperplane, then the configuration is not convex.</p> <p>(3) The facets of any d-cone can be isometrically mapped (unfolded!) into a $(d-1)$-hyperplane, retaining the coincidence of the cone point and without overlap.</p> <p>(4) The convex spherical polygon of largest perimeter is a great circle, i.e. any spherical polygon with perimeter larger than $2 \pi$ is not convex.</p> <hr> <p>It seems to me like there should be a straightforward, convex-geometry proof of this fact, but I can't find it. If you know another way to prove it (say, using ideas from curvature?) I'd be very interested in that too!</p> http://mathoverflow.net/questions/91944/flattening-a-corner-in-a-convex-d-polytope-into-d-1-dimensions-without-over/91946#91946 Answer by Igor Rivin for Flattening a corner in a convex $d$-polytope (into $d-1$ dimensions, without overlap)? Igor Rivin 2012-03-22T20:33:48Z 2012-03-23T16:07:49Z <p>One way of doing this (in $d=3,$ say) requires the "Archimedes axiom": if one convex body $K$ contains another body $L,$ then the perimeter of $K$ is greater than that of $L$ (the nicest proof uses Crofton's formula, which says that the perimeter is proportional to the measure of the set of lines which intersect the set). Then, you set $K$ to the hemisphere, and $L$ to your spherical convex polygon, and you are done. (Archimedes actually introduced this as an axiom in the Euclidean case when computing the perimeter of the circle, since he needed to know that the inscribed polygons provided a lower bound).</p> <p><strong>EDIT</strong> To answer the OP's question: Crofton's (or kinematic) formulas work in all dimension in all constant curvature spaces. The canonical reference is L. Santalo's integral geometry and geometric probability. A nice survey of the generalizations <a href="http://www.math.kit.edu/stoch/~hug/media/20.pdf" rel="nofollow">this paper by Hug and Schneider</a>, but it does not cover the non-Euclidean case.</p>