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Quoting Do Carmo's 'Differential Geometry of Curves and Surfaces':

"We have only to think of all possible shapes of a surface homeomorphic to a sphere to find it very surprising that in each case the curvature function distributes itself in such a way that the total curvature, i.e. $\int\int Kd\sigma$, is the same for all cases."

If I am not wrong, the classical version of Gauss-Bonnet Theorem is applied to smooth surfaces embedded in $\mathbb{R}^{3}$ with the induced metric.

Taking into account that the Gaussian curvature of a surface coincides with its sectional curvature, and the Gauss-Bonnet-Chern theorem does not require the metric to be induced by the metric of some $\mathbb{R}^{n}$ (I am not really sure about the veracity of these assertions) , I would like to know if we could restate Do Carmo's sentence to obtain a new striking one as follows:

"We have only to think of all possible metrics of a two dimensional differentiable manifold to find it very surprising that in each case the sectional curvature function distributes itself in such a way that the total sectional curvature, i.e. the surface integral of that function, is the same for all cases."

Notice that, if I am not wrong, we need the generalized version of Gauss-Bonnet theorem for this to be true, because in the original version only the induced metric is considered.

Am I misunderstanding something?

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  • $\begingroup$ You are right that the metric on the surface do not need to come from the induced metric. On the other hand, all surfaces can be isometrically embedded into $\mathbb R^N$ (this fact is independent of Gauss Bonnet) , so indeed all metrics are induced metric. . $\endgroup$
    – user99914
    Dec 12 '15 at 12:25
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    $\begingroup$ @JohnMa some $\mathbb{R}^N$, not necessarily $\mathbb{R}^3$, one should add. And if you pull out the heavy machinery you should also give it's name (the embedding theorem of John Nash). $\endgroup$
    – Thomas
    Dec 12 '15 at 12:28
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Note that in your rephrasing, you replaced "a surface homeomorphic to a sphere" by "a two dimensional differentiable manifold" which makes your statement wrong. Indeed, the Gauss-Bonnet theorem holds for any two dimensional Riemannian manifold and not necessarily for embedded manifolds of $\mathbb{R}^3$ with the induced metric, but integral of the curvature does depend on the topology of the surface involved, so you need to keep it fixed.

That is, you can write instead "We have only to think of all possible Riemannian metrics on a smooth, orientable, compact surface of genus $g$ to find it..." (where the case of $g = 0$ corresponds to the sphere).

For a nice proof of Gauss-Bonnet that doesn't assume the surface is embedded, I refer to Chapter 9 of Lee's book "Introduction to Riemannian Manifolds".

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