AlgTop16: Rational curvature of polytopes and the Euler number

We show that the total curvature of a polyhedron is equal to its Euler number. This only works with the rational formulation of curvature, using an analog of the turn angle suitable for the 2 dimensional sphere. We treat Harriot's theorem on the area of a spherical polygon.

This is the first video in the 16th lecture of this beginner's course in Algebraic Topology, given by N J Wildberger at UNSW.
Video Contents:
00:00 Introduction
00:40 Harriott's theorem
02:14 Ex. A has degree 3
04:20 Ex. A has degree 4
07:00 Tangles of opposite Cone at vertex A has internal tangles
10:47 Theorem. IF A polygon P has vertex A with tangles of facesat A.
11:47 Tetrahedron
14:40 Ex.2 Cube
15:50 Ex.3 Octahedron
16:29 Ex.4 Cosahedron
17:40 Ex.5 Dodecahedron
20:00 Problem 18. Computing directly the curvature at a vertex
32:09 Problem 19. Compute directly the curvatures

************************

Here are the Insights into Mathematics Playlists:

Here are the Wild Egg Maths Playlists (some available only to Members!)

************************

[hellenakinyi]

Video Content
00:00 Introduction
00:40 Harriott's theorem
02:14 Ex. A has degree 3
04:20 Ex. A has degree 4
07:00 Tangles of opposite Cone at vertex A has internal tangles
10:47 Theorem. IF A polygon P has vertex A with tangles of facesat A.
11:47 Tetrahedron
14:40 Ex.2 Cube
15:50 Ex.3 Octahedron
16:29 Ex.4 Cosahedron
17:40 Ex.5 Dodecahedron
20:00 Problem 18. Computing directly the curvature at a vertex
32:09 Problem 19. Compute directly the curvatures

[Reliquancy]

That's really nice the way that the curvature works out to the Euler number!