According to Wikipedia the global Gauss-Bonnet theorem concludes,

$\int_{M}KdA + \int_{\partial M}k_{g}ds=2\pi\chi(M)$.

The lecture notes I am using however does not have the integral with the geodesic curvature but they seem to have the same hypotheses. Here is the proofs from the notes;

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Either Wikipedia is wrong or the notes must be wrong at some point. In particular if a geodesic triangulation always is possible for any compact surface then there would never be a geodesic curvature integral imo.

The notes can be found at http://www.matematik.lu.se/matematiklu/personal/sigma/Gauss.pdf


Both are correct but your lecture notes cover only a subset of the cases the wikipedia formula covers. Namely you lecture notes assume that $M$ is a surface without boundary, whereas wikipedia allows surfaces with smooth boundary.

  • $\begingroup$ If it is compact then I think it has a boundary? $\endgroup$ – Lobsided Aug 13 at 9:26
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    $\begingroup$ The compactness is unrelated. Your lecture notes define a surface without boundary, ie each point has a neighborhood that looks like the open disk. Wikipedia uses the more general notion of manifolds with boundary. Your lecture notes actually contain a version of Gauss-Bonnet that includes a boundary, namely theorem 8.5. $\endgroup$ – quarague Aug 13 at 10:42
  • $\begingroup$ To me it looks like the boundary belongs to $M$ but vanish since it is parametrised by a geodesic, somthing I find hard to belive work unless the boundary is a geodesic to begin with. If it is compact and in $\mathbb{R}^n$ it is closed which means that the boundary belongs to the set. Compactness is needed to have finite cover. $\endgroup$ – Lobsided Aug 13 at 14:28
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    $\begingroup$ The surface needs to be closed for Gauss-Bonnet, so if there is a boundary it is part of the surface. Your lecture notes talk about closed compact surfaces without boundary, like the 2-sphere. The upper half sphere, including the equator would be an example of a manifold with boundary. $\endgroup$ – quarague Aug 14 at 6:53

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