Descartes-like formula in 4D

Question

In 3D, Descartes' formula states that the total angular defect of a convex polyhedron must always be equal to $4\pi$. In particular, this implies that if we have a bound on the "bluntness" of a polyhedron's vertices, we can also obtain a bound on its total number of vertices.

It seems intuitive that something similar should be the case for a 4-polytope: if its edges (or perhaps vertices) are too "sharp", it should close up "faster". As an example, the dihedral angles around a tesseract are all equal to $270^\circ$, admittedly not that much, and correspondingly, it doesn't have a lot of constituents. On the other hand, the ginormous omnitruncated 120-cell has edges with total dihedral angles ranging from $322^\circ$ up to $356^\circ$. And of course, the honeycombs, with complete vertices and edges, are able to have an infinite amount of components.

These are all completely empirical observations, but I'd like to make claims on these grounds. So, is there any formula, perhaps like Descartes' one, that can make these observations precise? Is there any relation between the angles in a 4-polytope and the amount of elements it can have?

Answer 1

This paper gives a few results on angular defects in higher dimensions

Peter Hilton and Jean Pedersen, Descartes,Euler,Poincaré,PólyaandPolyhedra Séminaire de Philosophie et Mathématiques, 1982 http://www.numdam.org/article/SPHM_1982___8_A1_0.pdf

The total angle defect in dimension $n>3$ is no longer a topological invariant. In particular, polytopes (polyhedra which are topologically equivalent to $n$ dimensional spheres) don't all have to have the same total angle defect.

Corollary $1$ on page $12$ of the paper states that the total angle defect $\Delta P_1$ of a polytope obtained by subdividing the $n$ dimensional sphere $S^{n-1}$ as the boundary of an $n$-simplex is $\Delta P_1=-\frac{\pi}6(n-4)(n+1)(n+3)$.

When $n=4$, $\Delta P_1=0$ no matter how many vertices the polytope has.

And no, I'm not kidding, convex polytopes can have a negative total angle defect.

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Aside: Descartes theorem on total angle defect is a discrete version of the Gauss-Bonnet theorem. There is a (seriously non-trivial) generalisation of the Gauss-Bonnet theorem to manifolds with higher even dimension called the Chern-Gauss-Bonnet theorem. It might be possible (?) to come up with a discrete version of the Chern-Gauss-Bonnet theorem for polytopes of odd dimension (remember that an $n$ dimensional polytope has an $n-1$ dimensional surface)???

Answer 2

The angular defect at a vertex of a polyhedron is the area of the dual of the vertex figure, which is a spheric polygon. The dual of the polyhedron can be constructed as a spheric tiling; see Gaussian sphere, or Gauss map. The sum of areas of the tiles must equal the area of the sphere; hence Descartes' formula.

In the same way, the dual of any Euclidean $n$-polytope can be constructed as a tiling of $\mathbb S^{n-1}$. The sum of spheric measures of duals of vertex figures must equal the total measure of $\mathbb S^{n-1}$.

The problem is that the measure of a spheric polytope is harder to calculate in higher dimensions. (It may be easier when the dimension $n-1$ is even, but I haven't looked into that.)