An exterior angle theorem for n-dimensional polytopes? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T22:17:56Z http://mathoverflow.net/feeds/question/36528 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36528/an-exterior-angle-theorem-for-n-dimensional-polytopes An exterior angle theorem for n-dimensional polytopes? tylern 2010-08-24T06:44:30Z 2010-08-25T03:45:56Z <p>In the plane, the exterior angle of a vertex is $\pi -$ the standard ("interior") angle, which may be negative in some cases. The following is true for non-weird polygons:</p> <blockquote> <p>The sum of the exterior angles at each vertex is a full turn ($2\pi$ radians).</p> </blockquote> <p>I am informally calling polygons with self-crossing edges or holes as "weird" -- please do let me know what the standard terminology is. I have seen an extension to 3-dimensional polytopes of this form:</p> <blockquote> <p>The sum of the exterior angles of a polytope is $4\pi$ radians.</p> </blockquote> <p>In this case, the exterior angle of a vertex is $2\pi -$ (the sum of the face angles at that vertex). I have not seen a proof, but I think it is true for non-weird polytopes, and a modified version is probably true for polytopes with nonzero genus.</p> <p>My question is: is there a general n-dimensional version of these properties?</p> http://mathoverflow.net/questions/36528/an-exterior-angle-theorem-for-n-dimensional-polytopes/36529#36529 Answer by Benoît Kloeckner for An exterior angle theorem for n-dimensional polytopes? Benoît Kloeckner 2010-08-24T06:56:13Z 2010-08-24T06:56:13Z <p>The standard terminology is non-crossing polygon'', I think.</p> <p>Concerning the generalization in arbitrary dimension, it is possible at least in the convex case. Simply define the exterior angle at a point to be the measure (with respect to the usual invariant measure on the unit sphere) of the set of exterior normal vectors to the polytope at this point. The sum is then the total volume of the unit sphere.</p> <p>The key word is Alexandrov measure, and it is a generalization of Gaussian curvature that can be defined for arbitrary convex compact sets. I do not know many references, but you can try the handbook of convex geometry. It is very likely that this can be generalized to more general polyedra.</p> http://mathoverflow.net/questions/36528/an-exterior-angle-theorem-for-n-dimensional-polytopes/36534#36534 Answer by Gjergji Zaimi for An exterior angle theorem for n-dimensional polytopes? Gjergji Zaimi 2010-08-24T08:55:42Z 2010-08-24T08:55:42Z <p>What you call exterior angles are the curvatures at the vertices, and the result you are referring to is the combinatorial Gauss-Bonnet theorem. It says that in an angled 2-complex $K$ the sum of the curvatures at the vertices plus the sum of the curvatures at the polygonal faces is $2\pi \chi(K)$ (Euler characteristic). When your complex is piecewise Euclidean then the curvature at each face is zero implying your result in the case of polyhedra. For polyhedra, another way to see this is to perform barycentric subdivision and sum the angles of all faces to reduce your identity to $V+F-E=2$. A possible reference is <a href="http://arxiv.org/abs/0910.5688" rel="nofollow">this</a> paper (theorem 4.8).</p> http://mathoverflow.net/questions/36528/an-exterior-angle-theorem-for-n-dimensional-polytopes/36619#36619 Answer by Allen Knutson for An exterior angle theorem for n-dimensional polytopes? Allen Knutson 2010-08-25T03:45:56Z 2010-08-25T03:45:56Z <p>Here's something a bit less numerical, but more satisfying, I think.</p> <p>For each vertex $v$ of the polytope $P$, let $v'$ be the set of vectors $\vec w$ such that $\langle \vec w, v-p \rangle \geq 0$ for all $p\in P$. This is a polyhedral cone. If $P$ is very round at $v$, then $v'$ is very sharp.</p> <p>Theorem: if $P$ is a compact convex polytope, then the cones {$v'$} exactly cover space (at least, their interiors don't intersect).</p> <p>The set of cones $v^*$ is called the "dual fan" to $P$. You can read about it in Fulton's book on toric varieties.</p>