Talk:Generalized Gauss-Bonnet theorem

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Should we really redirect this article to Gauss-Bonnet theorem? That article doesn't describe its generalization. Phys 14:46, 8 Dec 2003 (UTC)

It generalizes to dimension 2n, not just for surfaces. Jholland 13:37, 10 Sep 2004 (UTC)

Depending on your conventions for the Pfaffian (bldy constants!) I think the constant is v2n / 2. It's probably worth posting a proof so nothing is left to chance. I have a nice one lying around somewhere from my grad school days, if only I could find it ;-) (My "proof" assumes the Nash embedding theorem, if memory serves.) Jholland 04:49, 11 Sep 2004 (UTC)

I do not think proof is needed here, maybe I calculated it wrong, feel free to change. Tosha

No, there is stadard conventions for the Pfaffian, for (S2)n we have χ(S2)n = 2n and Pf=1 vol and vol(S2)n = (4π)n. I do not know wherther it is the same answer as before, but hopefully it is. Tosha

Also the name for theorem is not quite correct, the first proof wa not by Chern, it were some guys who did it using Nash embedding (without having it) I would prefer to call in generalized Gauss-Bonnet. Tosha

I agree. Maybe you could handle the redirects? I also think your constant is now right. (Mine was wrong before too.)

To see that the constant is right, in my point of view, the Pfaffian and Euler class are both characteristic classes. Take a cartesian product k of S2's, and pullback the tangent bundle over each projection to S2. By the Whitney product formula, we have to have

e(T(S^2\times S^2\dots\times S^2))=\pi_1^*e(TS^2)\pi_2^*e(TS^2)\dots \pi_k^*e(TS^2)

and likewise with the Pfaffian. However, the Pfaffian and Euler class are related over a single S2 by the usual Gauss-Bonnet theorem

Pf(\Omega_{S^2})=2\pi e(TS^2)

so that by taking a cup product of n such cohomology classes produces a constant 2nπn.


Jholland 19:06, 11 Sep 2004 (UTC)

[edit] The Pfaffian in coordinates

I would also move this subsection to Pfaffian as yet an other definition Tosha

If you can suitably adapt it to the notation there, then by all means. Jholland 20:54, 11 Sep 2004 (UTC)

I moved it to talk page of Pfaffian, since I realized that ref to Kronecker tensor was a bit misleading. Tosha

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