Re: [math-fun] [math>-fun] Negative polyhedra
Since the Gauss-Bonnet theorem states that for a compact surface M, Integral K dA = 2π X(M) (K = Gaussian curvature, X(M) = Euler characteristic of M), we could in the case of constant negative curvature K = -1 (say), divide both sides by K. Then using the face that X(M) = V - E + F, we get area(M) = 2π (-V + E - F). So for a hyperbolic surface, that is, for a surface whose Euler characteristic is negative (the connected sum of ≥ 2 tori or of ≥ 3 projective planes), we could interpret the V, E, and F as negative (V' = -V, E' = -E, F' = -F) to get area(M) = 2π (V' - E' + F') if we like. —Dan ----- Is it in any way meaningful to think of this thing as a polyhedron with -8 pentagonal faces, four of which meet at each of the -10 vertices, joined by -20 edges? -----
On 10/06/2020 23:06, Dan Asimov wrote:
Since the Gauss-Bonnet theorem states that for a compact surface M,
Integral K dA = 2π X(M)
(K = Gaussian curvature, X(M) = Euler characteristic of M),
we could in the case of constant negative curvature K = -1 (say), divide both sides by K. Then using the face that X(M) = V - E + F, we get
area(M) = 2π (-V + E - F).
So for a hyperbolic surface, that is, for a surface whose Euler characteristic is negative (the connected sum of ≥ 2 tori or of ≥ 3 projective planes), we could interpret the V, E, and F as negative (V' = -V, E' = -E, F' = -F) to get
area(M) = 2π (V' - E' + F')
if we like.
Aha, yes. That's the kind of thing I meant by
Something something Gauss-Bonnet something
in what I originally wrote. Except that you actually know what you're talking about and remembered the details of what Gauss-Bonnet says, whereas I just had a vague idea that it said the right sort of thing to make something like this work :-). Thanks! -- g
participants (2)
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Dan Asimov -
Gareth McCaughan