The Gauss–Bonnet theorem says 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, is $K$ the curvature sectional, or is it the product of principal curvatures of the surface?

Can someone suggest a reference? Thank you!


2 Answers 2


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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    $\begingroup$ 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. $\endgroup$ Nov 14, 2012 at 10:13

I think a proof of Gauss-Bonnet for non-orientable setting is very simple. Here we follow John Lee's idea in his book (page 169).

Let $M$ be a compact two dimensional non-orientable manifold. Then a double cover of $M$ must a compact, orientable two dimensional manifold. The double cover $\tilde{M}$ should 'lift up' the piece corresponding to the $\mathbb{RP}^{2}$ or $K$ in the classification theorem. Now for $\tilde{M}$ we can endow the push-back metric defined by $$ \tilde{g}=\pi^{*}g $$ such that the projection map is a local isometry. Then it is obvious that $\pi^{*}\tilde{K}=K$ since sectional curvature is only defined locally. Now we have $$ \chi(\tilde{M})=\frac{1}{2\pi}\int_{\tilde{M}} \tilde{K}d\tilde{\sigma} $$ But we know that that $\chi(\tilde{M})=2\chi(M)$, and similarly $\tilde{K}=K$'s integral would appear twice when we do the integral on $M$ using a compact-supported density. So dividing by 2 on both sides we get the desired formula for $M$: $$ \chi(M)=\frac{1}{2\pi}\int_{M} Kd\sigma $$ and this finished the proof.

  • $\begingroup$ The issue here is that your last formula technically doesn't make sense, assuming by $d\sigma$ you mean the volume form, since the volume form doesn't exist globally on a non-orientable manifold. One can skirt this issue by defining the concept of densities on a manifold. $\endgroup$ Nov 16, 2017 at 20:55
  • $\begingroup$ The formula $d\overline{\sigma}$ means densities. $\endgroup$ Nov 16, 2017 at 20:57
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    $\begingroup$ Okay. There wasn't any overbar in the formula in the answer. $\endgroup$ Nov 16, 2017 at 21:03

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