Gauss-Bonnet for complex valued edge lengths[repost]
Gerard Westendorp
dimensions.
If we think of an "ordinary" triangulated surfaces, (ie embedded in 3D
Euclidean space) we would take the edge lengths of all triangles,
calculate all angles at the vertices, and find that the sum of deficits
equals 2 pi times the Euler characteristic. (Gauss Bonnet theorem.)
In (2+1)D, the lengths are not necessarily positive real numbers.
A----c----B
\ /
\ a
b /
\ /
C
To calculate angles for the triangles, we can use the cosine law:
cos(A) = (b2+c2-a2)/(2bc)
I tried extending this to triangles with arbitrary complex-valued
lengths, extending the formula to complex numbers. It turns out that the
rule that;:
A+B+C = pi
continues to work. The imaginary part of the angles appears to sum to
zero, and the real part continues to sum to pi.
(I do not have a proof, but I tried it out on random numbers)
Apparently,
acos((b2+c2-a2)/(2bc))
+ acos((a2+a2-b2)/(2ca))
+ acos((c2+b2-c2)/(2ab))
=pi
For all (a,b,c) triples in complex numbers
Also, it appears that Gauss Bonnet continues to work.
(I checked this by hand for a tetrahedron with arbitrary complex-valued
lengths)
So the sum of all complex-valued deficits continues to be 2pi times the
Euler characteristic.
This seems very mysterious. I was wondering, maybe there are other ways
to see this, that make it more obvious.
Maybe it also works for quaternions, or even more general case for the
edge lengths?
Gerard
John Baez
In article <hehmp9$22l$1...@crackerjack.ma.ic.ac.uk>,
Gerard Westendorp <wes...@xs4all.nl> wrote:
>To calculate angles for the triangles, we can use the cosine law:
>
> cos(A) = (b^2+c^2-a^2)/(2bc)
>
>I tried extending this to triangles with arbitrary complex-valued
>lengths, extending the formula to complex numbers. It turns out that the
>rule that:
>
> A+B+C = pi
>
>continues to work.
>This seems very mysterious.
In general, rules involving complex-analytic functions tend to
still work for complex values if they hold for all real values.
But your thoughts remind me of a lot of work relating the 6j symbols
to tetrahedra in Euclidean and also Minkowskian space. It's a long
story, so I urge you to check out the first volume of Biedenharn and
Eliot's tome... or if that's too hard to find, maybe this:
http://arxiv.org/abs/gr-qc/9310016
I'm afraid the connection won't be obvious, but I think there is one!
Gerard Westendorp
Learned 2 things in the mean time.
First, Gauss Bonnet (total angular deficit = 2pi*Euler characteristic)
is an easy consequence of the fact that for any triangle, the sum of
internal angles is pi. This can obtained by combining:
angluar deficit = 2pi* vertices - sum (internal angles)
= 2pi* vertices - pi * faces
edges=faces*3/2
Euler characteristic= vertice+faces-edges.
So we only need to see why the sum=pi relationship holds.
Second, you can assign a complex number to each vertex with angle A
cos(A) + i sin(A)
= (b^2+c^2-a^2 + i Area)/bc
with Area computed using Heron's formala.
Then, the following is indeed an identity for any 3 complex numbers {a,b,c}:
-1 =
(b^2+c^2-a^2 + i Area)*(a^2+b^2-c^2 + i Area)
*(c^2+b^2-a^2 + i Area)/(abc)^2
the argument of the above equation is the sum=pi relationship.
This formula avoids the use of the arccos function. It still has an
ambiguous sign convention because of the square root in Heron's formula.
But if you take either the positive or the negative root, the identity
holds. (In other formulations, choosing the sign of the square root
leads to multivalued answers:{pi, pi-2A, pi-2B, pi-2C}
The link you sent talks about using spinors assigned to the edges of a
triangle. That seems interesting. If you start with real valued lengths
and angles, you can assign complex number to each edge (simply by
drawing the triangle on the complex plane. So I am trying to figure out
how this fits in with the idea of a complex-valued lengths and
complex-valued angles: maybe they relate in a similar way to
spinors/quaternions.
Gerard