Timeline for How to derive from Gauss's results on the volume of hyperbolic orthoscheme tetrahedron the formula of Bolyai?
Current License: CC BY-SA 3.0
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| Dec 15, 2017 at 16:51 | comment | added | Igor Rivin | @user2554 The link of a vertex of a polyhedron in any geometry is a spherical triangle. As for Gauss-Bonnet, in its original form (due to Gauss) it is stated for triangles. In any case, area as angle defect is a fundamental fact of hyperbolic geometry. | |
| Dec 15, 2017 at 8:04 | comment | added | user2554 | In addition, since i don't understand well the rimannnian theory of 3-manifolds, i have a basic misunderstanding: in three-dimensional hyperbolic spaces with constant space curvature k, the Gauss-Bonnet theorem can be applied to figures with volume, but does it mean it can be applied to figures with finite area but zero volume? i ask because by analogy with the Gauss-Bonnet theorem for 2-manifolds, it cannot by applied to cross sections (sections of one dimension lower). | |
| Dec 15, 2017 at 7:55 | comment | added | user2554 | I voted your answer because it really helped me understand things about Gauss's formula (now i know it's form is similar to the one given by Schlafli, and that helps me a lot). Thanks!, but i still need further explanation. In particular, as far as i know, Gauss meant hyperbolic geometry when he talked about "non-euclidean geometry", but when you mentioned spherical triangle it made suspect that perhaps we are not talking about the same thing. | |
| Dec 15, 2017 at 7:48 | vote | accept | user2554 | ||
| Dec 15, 2017 at 3:40 | history | answered | Igor Rivin | CC BY-SA 3.0 |