Sum of angle defects of polyhedrons

In this short note we will discuss a generalization of the well-known fact that the sum of the exterior angles of a polygon is always {2\pi}, which can also be regarded as a very special case of the Gauss-Bonnet theorem. We would like to have a similar result for a polyhedron in {\mathbb{R}^3} (in this note, a polyhedron actually refers to the polyhedral surface rather than the whole solid). After discussing with Hon Leung and Raymond etc, we have the following

Theorem 1 (Hon Leung-Raymond’s theorem, 2011 (Discrete Gauss Bonnet theorem))
For a polyhedron {P\subset\mathbb{R}^3}, the sum of its angle defects (see definitions below) is {2\pi} times its Euler characteristic:

\displaystyle \sum_{v:\;\text{vertices}} K(v)=2\pi\chi(P).

First of all we have to generalize the notion of the exterior angle of a triangle. It can be regarded as the defect, or failure for two adjacent edges (of a triangle) to form a straight angle (i.e. {\pi}, or {180^\circ}). Motivated from this, we have the following:

Definition 2 For a vertex {v} of a polyhedron, suppose there are exactly {k} faces sharing {v} as a common vertex, then the angle defect {K(v)} is defined as

\displaystyle K(v)= 2\pi - \sum_{i=1}^k \theta_i

where {\theta_i} are the (interior) angles of that {k} faces at the vertex {v}.

For example, if all the faces which shares {v} as a vertex all lies on a plane, then the angle defect at {v} is {0}. Intuitively, the angle defect measures how the faces meeting at {v} deviates from the plane. If {K(v)>0}, then a small circle with “radius” {r} (in the obvious sense) around {v} will have length smaller than {2\pi r}, thus it looks “round” near that vertex. On the other hand, if {K(v)<0}, then the small circle of radius {r} will be longer then the Euclidean circle of the same radius (i.e. {2\pi r}), so it looks something like a saddle near here, we then say that the curvature at that point is negative.

Definition 3 For a polyhedron {P}, if the number of its vertices is {V}, that of its edges is {E} and that of its faces is {F}, then its Euler characteristic {\chi(P)} is defined as

\displaystyle \chi(P)=V-E+F.

Note that {\chi} is a topological invariance, it depends on its shape (topological/homotopy type) only, in particular if we regard {P} as a surface and joining some vertices of some faces if necessary (without multiple or intersecting edges of course), so that each face now becomes a triangle (say), its Euler charcteristic is still unchanged, although each of the {V,E} and {F} changes.

Triangulating faces


With this remark in mind, we are now ready to prove the main result.

Proof: By the previous remark, we can assume that each face is a triangle (i.e. consists of 3 edges) for simplicity (note that each K(v) is unchanged).

Now,

\displaystyle  \begin{array}{rcl} \displaystyle \sum_{v:\;\text{vertices}} K(v)	 &= &\displaystyle\sum_{v:\;\text{vertices}}(2\pi - \text{sum of interior angles at }v)\\ &= &2\pi V - \displaystyle\sum_{v:\;\text{vertices}}\text{sum of interior angles at }v\\ &= &2\pi V - \displaystyle\sum_{f:\;\text{faces}}\text{sum of interior angles of }f\\ &= &2\pi V - \displaystyle\sum_{f:\;\text{faces}}\pi\\ &= &2\pi V - \pi F. \end{array}

By double counting (as in handshaking lemma), or more precisely, since each edge is shared by exactly two faces, and each face has exactly three edges, counting the number w.r.t. the edge orientation and w.r.t. the faces gives

\displaystyle  3F=2E.

Plugging this into the above equation, we have

\displaystyle \sum_{v:\;\text{vertices}} K(v)	=2\pi (V-E+F)=2\pi \chi.

\Box

Remark: I believe there is still a higher dimensional generalization of this result, but I haven’t thought about it too deeply yet.

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One Response to Sum of angle defects of polyhedrons

  1. Sasa Kariz says:

    It would be interesting to read similar concepts for triangles/polygons embedded into the non-euclidean geometrical spaces, spheres..

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