Several months ago I began to doubt several repetitions of the notion that the angle deficit of a bone told the whole story of the curvature at a bone in a simplical complex. It was obviously true in 2D and 3D but I suspected that the reasons for this broke down beyond 3. There were several intuitively arguments that this was so but they tended also to prove that circulating the bone left all vectors fixed.

I am now convinced that Regge and followers were right and I give a proof that convinces me. Also the code that I wrote to test this numerically told me that I was wrong.

Consider some bone. It is a simplex of n−2 dimensions. Several zones (simplex of n dimensions) include the bone and these occur in some cyclical sequence ‘about’ the bone. Between each successive pair of these zones is an n−1 dimensional simplex ‘facet’ that separates the two neighbors and belongs to both of them. Each facet also includes the bone. Going about the bone consists of going from zone to facet back to bone until we are back in the zone where we started. We will have traversed each zone and facet.

Each zone has a dihedral angle at each of its (n−2)-simplexes. Each zone that contains the bone has a dihedral angle for that bone. 2π−(the sum of these angles) is the deficit that is a measure of curvature. It is not generally 0 except in flat space. Choose some vertex of the bone. The vertex belongs to all of the zones and facets under consideration. Use that vertex for barycentric coordinates for each zone and facet. If we carry a test vector about the bone we begin by expressing the vector in bcc. If that vector is parallel to some vector in the bone then the only non zero coefficients of this component description will be for those basis vectors that are in the bone. Those n−2 coefficients will remain constant during the trip about the bone. That leaves room for only a ‘rotation’ about the bone involving the two remaining coefficients.

Q.E.D.

This proof probably works for affine complexes equipped with only an affine connection.