Consider the unit sphere Sn−1 embedded in Rn with is center at the origin. Consider a polyhedral region of Sn−1 whose boundary is a collection of flat faces. A flat face is the intersection of Sn−1 with an n−1 dimensional subspace of Rn thru the origin. The region is of dimension n−1 and the region boundaries are of dimension n−2.
Exterior trihedral angles are depicted nicely towards the bottom of the above theorem reference. In our terminology the exterior angle at a vertex of a polyhedron is the content of the set of tangents there. If the vertex is nearly flat there will be few tangents and correspondingly a small exterior angle. The sum of the exterior angles of a convex polyhedron in Rn is clearly one sphere’s worth, or 4π.
Here is a generalization of exterior angles.
This has some connection with Minkowski Functionals which in turn lead to a generalized Gauss Bonnet theorem. I think my supplementary angles are discreet versions of the functionals.
I am now stumped. (2004 Nov)
2005 February. I have realized that it is easy to compute unit normals to faces and that these contain all of the information to compute general “spherical excess”. (It is possible that for irregular simplexes edge lengths may be required too.)
The conjecture is that the content of the region is just 4π − Σ (exterior angle at a vertex). This is a straight forward extrapolation of the classic spherical trigonometry rule for area of spherical triangles.
I can write a computer program to test this! Too I can test it by hand with these figures. Take for instance the initial four cornered room; what is its volume? It may described in terms of rectangular coordinates in an R4 in which S3 may be embedded; it is those places on S3 each of whose four coordinates are positive. The volume of S3 is c4 = 2π2 and the room is just c4/16 = π2/8. Calculating from our formula, each of the four corners are locally congruent to the corner of a cube whose exterior angle is π/2; together exterior angles of the four corners of our room sum to 2π. The formula yields 4π − 4(π/2) = 2π. This is not nearly the correct answer: π2/8.
Insight into the classic S2 spherical formula, c = 2π − Σ (exterior angle), may be had by a sliver of a triangle with sides of length π, π, ε. The respective exterior angles are π/2, π/2, π − ε, and we get c = ε. This does not depend on ε being small but taking ε = π/180 leads to a sliver that is sometimes defined as one spherical degree, 720 of which compose the entire sphere.
Stepping up one dimension we consider the tall skinny room with edge lengths π/2, π/2, π/2, meeting at vertex t (top), and remaining edges ε, ε, √2ε, that bound the floor opposite t. Now we do require ε to be small. The exterior angle at v is 2π − ε2/2. The three other exterior angles are 3π/4, 3π/4, π/2. These last three values depend only 3D visualization! Our conjecture yields c of room = ε2/2. The floor (small face) is in an “equator” for S3 which is itself S2 with content 4π. t is the pole which is π/2 distant from this equator. This equator bisects S3.