Rich writes:
<<
There's a plane geometry theorem that the sum of the vertex angles of
a polygon is pi*(#vertices-2). Is there a 3D analog?
>>
In short, the external spherical angles at all vertices of a polyhedron
K must add up to 4*pi.
(As in 2D this can be thought of as a limiting case of the
embedded Gauss-Bonnet theorem for smooth surfaces in 3D (resp.
smooth curves in 2D).
Approximate a polyhedron K by an (internal) smooth surface M.
M will have a Gauss map G: M -> S^2 -- G(x) = the point of
S^2 having its outward unit normal = the same vector as the
outward unit normal of M at x.
Let JDG: M -> R be the Jacobian determinant, i.e. the factor
by which G magnifies area at each point of. Then Gauss-Bonnet
tells us that the (integral over M of JDG dA) = 2pi*X(M),
where X(M) = its Euler characteristic = 2 for K a topological
sphere.
Back to K: As M approaches K smoothly (at least in the C^3 topology)
the Gauss map is no longer well-defined at edges or vertices of M.
But its image can be replaced by the set of outward unit normals to
all planes P in space that (assuming K convex) keep K inside of
one of the closed half-spaces defined by P. The area of the spherical
image of all the unit normals must still be 2pi*X(K) = 4*pi.
This is the sum over all vertices of the external spherical angle.
It's a nice little exercise to calculate the formula for the spherical
angle at a vertex where 3 edges meet, whose unit edge vectors
(i.e., vectors leaving the vertex in the directions of the edges it's in)
are say u,v,w.
--Dan
