The Gauss–Bonnet theorem say that: If $\Sigma \subset M =\mathbb{R}^3$ is a compact 2-dimensional Riemannian manifold without boundary, then $$\int_{\Sigma} K = 2\pi\chi_{\Sigma}$$

where $K$ is the Gaussian curvature of $\Sigma$ and $\chi_{\Sigma}$ is the Euler characteristic of $\Sigma$.

In Proposition 7 of http://arxiv.org/abs/0909.1665, $M^3$ is compact (thus $M\ne \mathbb{R}^3$), but they use Gauss-Bonnet. Why?

Moreover, in this case, $K$ is the curvature sectional or the product of principal curvatures of the surface?

Can someone suggest a reference? Thank you!

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Good question.

The answer is that Gauss-Bonnet does not actually require the hypothesis that $\Sigma$ is embedded (or even immersed) in $\mathbb{R}^3$. Rather, it is an intrinsic statement about abstract Riemannian 2-manifolds. See Robert Greene's notes here, or the Wikipedia page on Gauss-Bonnet, or perhaps John Lee's Riemannian Manifolds book.

Many texts on elementary differential geometry include the hypothesis that $\Sigma \subset \mathbb{R}^3$ because that is the context they're working in. That is, some books don't define abstract manifolds.

In this case, I would interpret $K$ as the sectional curvature. To be honest, I've never seen a definition of "principal curvature" outside the setting of $\mathbb{R}^N$, though my ignorance shouldn't be taken as definitive proof that the concept doesn't exist in more general settings.

Aside: More interesting to me, by the way, is the fact that $\Sigma$ is non-orientable (it is homeomorphic to $\mathbb{RP}^2$). I don't think I've seen a proof of Gauss-Bonnet for the non-orientable setting, but it's apparently not a difficult modification.

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Hi Jesse. Indeed, integration on manifolds can be defined for compactly supported densities without the need of an orientation (but then, without the use of volume forms), cf. Nicolaescu - "Lectures on the Geometry of Manifolds" sec.3.4.1. Then for compact surfaces, since Gauß curvature is locally defined and a scalar field, there can be an analogue of Gauß-Bonnet in the nonorientable case. Nevertheless, it is not then a remarkable corollary to Atiyah-Singer or Grothendieck-Riemann-Roch, and other index results, since they depend on the surface to have a complex structure, thus an orientation. –  Javier Álvarez Nov 14 '12 at 10:13