31
$\begingroup$

While reviewing the proof of Gauss-Bonnet in John Lee's book, I noticed the following paragraph:

" ...In a certain sense, this might be considered a very satisfactory generalization of Gauss-Bonnet. The only problem with this result is that the relationship between the Pfaffian and sectional curvature is obscure in higher dimensions, so no one seems to have any idea how to interpret the theorem geometrically! For example, it is not even known whether the assumption that $M$ has strictly positive sectional curvatures implies that $\chi(M)>0$.... (page 170) "

May I ask if this "interpretation problem" has been resolved? I felt much the same way when I read Milnor's proof of Chern-Gauss-Bonnet using Chern classes, and Chern's statement in his own book using Lipschitz-Killing curvature. Neither has a geometric meaning that is "self-transparent" to me. When I had a class in the index theorem, our proof basically showed Gauss-Bonnet is a special case of Atiyah-Singer using an appropriate Dirac operator. And I do not recall that it involved much geometry, but instead a lot of algebraic manipulations.

Chern suggested the following way to look at it in his book: Consider the exterior $2n$-form $$ \Omega=(-1)^{n}\frac{1}{2^{2n}\pi^{n}n!}\delta^{i_1 \cdots i_{2n}}_{1\cdots 2n}\Omega_{i_1i_2}\cdots \Omega_{i_{2n-1}i_{2n}}, \Omega=K d\sigma $$ Then the "key" to prove Chern-Gauss-Bonnet is to represent $\Omega$ on the sphere bundle of $M$ so that one has $\Omega=d\prod$, where $\prod$ is a $2n-1$-form. However, I still do not know how this shed any light on the picturesque side of the equation so that I can visualize it. So I decided to ask. I suppose that this might be one of those topics well known to experts but not written down in introductory level textbooks.

Reference:

John M. Lee: Riemannian Manifolds, page 170

Chern: Lectures on Differential Geometry, page 171

Milnor & Stasheff: Characteristic Classes, appendix A?

For a definition of Pfaffian, see here from wikipedia.

$\endgroup$
38
$\begingroup$

The thing you are missing is one further geometric property of the $(2n{-}1)$-form $\Pi$ that Chern constructs on the unit sphere bundle $\mathsf{S}(M)$ of the oriented $2n$-manifold $M$: The fact that the pullback of $\Pi$ to any unit sphere $\mathsf{S}_x(M)\subset T_xM$ is simply the induced volume form of $\mathsf{S}_x(M)$.

Once one establishes this, Chern's proof of the Gauss-Bonnet theorem is straightforward: Choose a vector field $X$ on $M$ that has isolated zeroes $z_1,\ldots, z_k\in M$, and let $$U = \frac{X}{|X|}:M\setminus\{z_1,\ldots,z_k\}\to \mathsf{S}(M)$$ be the corresponding unit vector field, defined and smooth away from the $z_i$. Let $\epsilon>0$ be sufficiently small that the geodesic $\epsilon$-balls $B_\epsilon(z_i)$ around the $z_i$ are disjoint and smoothly embedded. On the manifold with boundary $M_\epsilon\subset M$ that consists of $M$ with these $\epsilon$-balls removed, consider the section $U:M_\epsilon\to \mathsf{S}(M)$. By construction/definition, $U^*\Omega = U^*(\mathrm{d}\Pi)$ is the Gauss-Bonnet integrand over $M_\epsilon$. By Stokes' Theorem, $$ \int_{M_{\epsilon}}U^*\Omega = \sum_{i=1}^k \int_{\partial B_\epsilon(p_i)} U^*\Pi. $$ Now let $\epsilon$ go to zero. The left-hand side converges to the Gauss-Bonnet integrand over all of $M$ while the $i$-th summand on the right-hand side converges to the index of $X$ at $z_i$. (This is because $U^*\Pi$ on $\partial B_\epsilon(z_i)$ differs by a term vanishing with $\epsilon$ from the pullback of the unit volume form of $\mathsf{S}_{z_i}(M)$ to $\partial B_\epsilon(z_i)\simeq \mathsf{S}_{z_i}(M)$ under the indicial mapping induced by $U$ at $z_i$, whose degree is, by definition, the index of $X$ at $z_i$.) Thus, passing to the limit and using the Poincaré-Hopf theorem (that the sum of the indices of the vector field $X$ is equal to $\chi(M)$), one obtains Chern's proof of the Gauss-Bonnet Theorem.

As to why $\Pi$ pulls back to each $\mathsf{S}_{z_i}(M)$ to be the unit volume form, you need to look at Chern's definition of $\Pi$, which uses the Pfaffian, particularly its algebraic properties. This comes out of the computation that Chern does, and it is essentially a geometric fact, but it amounts to an explicit formula for the transgression operator defined in Chern-Weil theory for the Euler class. Another way to look at it would be to look at the generalized Gauss-Bonnet formula, a discussion of which you can find at the MO question A question on Generalized Gauss-Bonnet Theorem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.