Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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:

The sum of the exterior angles at each vertex is a full turn ($2\pi$ radians).

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:

The sum of the exterior angles of a polytope is $4\pi$ radians.

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.

My question is: is there a general n-dimensional version of these properties?

share|improve this question
    
I think that for "non-weird" polygons, you want convex polygons. Non-convex polygons, even if they have no self-crossings, can have exterior angles summing to more than $2\pi$. –  Gerry Myerson Aug 24 '10 at 6:52
    
Gerry: For a simple polygon (no self-crossings), if you define the exterior angle measure to be π minus the interior angle measure at the same vertex, with the understanding that the interior angle measure is greater than π at a concave vertex, then the sum of the exterior angle measures will always be 2π. –  Jack Lee Aug 24 '10 at 16:37
    
@Jack, your TeX isn't working on my screen, but I get the drift, and of course you are right, if one is willing to accept the notion of the measure of an (exterior) angle being negative. –  Gerry Myerson Aug 25 '10 at 0:39

3 Answers 3

up vote 1 down vote accepted

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 this paper (theorem 4.8).

share|improve this answer
    
I do not understand what is the higher-dimensional generalization of this. –  Benoît Kloeckner Aug 24 '10 at 11:26
    
The idea I had is to present the higher-dimensional generalization of Gauss-Bonnet for the polyhedral case. Bloch has a paper ("The angle defect for arbitrary polyhedra") where all of this is discussed especially the higher dimensional analogue of discrete curvature, I will edit my answer to include more information later. I understand that this answer is not very informative as it is now... –  Gjergji Zaimi Aug 24 '10 at 12:23

The standard terminology is ``non-crossing polygon'', I think.

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.

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.

share|improve this answer

Here's something a bit less numerical, but more satisfying, I think.

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.

Theorem: if $P$ is a compact convex polytope, then the cones {$v'$} exactly cover space (at least, their interiors don't intersect).

The set of cones $v^*$ is called the "dual fan" to $P$. You can read about it in Fulton's book on toric varieties.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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