3
$\begingroup$

I like very much the elementary property that if one cuts a geodesic triangle onto a sphere (one can use 3 plans that contain $0$). The cut surface of the sphere is given by the sum of the angles of this triangle minus $\pi$. More generaly, we have the Gauss Bonnet Theorem saying that on any surface this is equal to the integral of the Curvature of the cut surface.

Because Gauss Bonnet is valid on higher dimensional manifold, I wonder if we can state a similar "elementary" result on higher dimension. For exemple if one cut $\mathbb{S}^4$ using 5 hyperplan, can we obtained the volume of the cut with a formula using only the angles between the hyperplans?

My other motivation is that this summer I tried to explain (and almost prove) the Gauss Bonnet Theorem to 15 years old student with the anglular defects https://en.wikipedia.org/wiki/Angular_defect#Descartes.27_theorem . I would like to know if it is also possible to defined (such that a 15 years old boy can understand) an "angular defect" on higher dimensional polyhedron and such that we have a similar theorem : "the sum of the (high dimensional) angular defects are equal to the volume of the hypersphere"

$\endgroup$
1
2
$\begingroup$

There are two formulas of the sort that you describe.

The Poincare formula expresses the volume of a spherical polyhedron as an alternating sum of the interior angles. This holds for polyhedra of any dimension. In the Euclidean case this degenerates to the Gram formula saying that this alternating sum of angles equals 0.

The Allendoerfer-Weil formula (mentioned by Martin in a comment) holds for polyhedra of even dimension and expresses the volume as a combination of the exterior angles at the even-dimensional faces multiplied with the volumes of the faces.

See Sections 2.3 and 2.4 in Chapter 7 of

Gamkrelidze, R. V. (ed.); Vinberg, E. B. (ed.), Geometry II: spaces of constant curvature. Transl. from the Russian by V. Minachin, Encyclopaedia of Mathematical Sciences. 29. Berlin: Springer-Verlag. 254 p. (1993). ZBL0786.00008.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.