Let $\Sigma$ be a compact surface with boundary $\partial \Sigma$. Fix a Riemannian metric on $\Sigma$ and let $D \in \Omega^1(SO(\Sigma),\mathfrak{o}(2) )$ be a principal connection on the associated oriented orthonormal frame bundle $p:SO(T\Sigma) \to \Sigma$. There exists a unique $\Omega \in \Omega^2(\Sigma)$ such that $p^* \Omega = \text{Pf}(\text{Curv}_D)$, the Pfaffian of the curvature form (up to some constant involving $\pi$ which I ignore). Let $\pi_0 : S(T\Sigma) \to \Sigma $ be the unit sphere bundle with $N: \partial \Sigma \to S(T \Sigma)$ the unit normal and let $\Phi \in \Omega^1(S(T\Sigma))$ satisfy $d \Phi = \pi_0^* \Omega$. Then the Chern-Gauss-Bonnet theorem for manifolds with boundary says:

$$ \int_{\partial \Sigma}N^*\Phi = \int_{\Sigma} \Omega - \chi(\Sigma). $$

$T\Sigma$ is a trivial bundle so $SO(T\Sigma)$ is a trivial principal $SO(2)$-bundle so pick the canonical flat connection $A=\theta^L =g^{-1}dg$ on $\Sigma \times SO(2)$ (i.e. the left-invariant Maurer-Cartan form). The curvature of $A$ is $0$ so the Pfaffian of the curvature of $A$ is $0$ hence $\Omega=0$ and $\Phi=0$. Putting this into the Chern-Gauss-Bonnet theorem we conclude $\chi(\Sigma)=0$ which is certainly not true for all $\Sigma$.

  • 2
    $\begingroup$ With a flat metric on $\Sigma$, the geodesic curvature integral on $\partial\Sigma$ will most definitely not be $0$. $\endgroup$ Commented Oct 6, 2022 at 18:15
  • $\begingroup$ Why is $T\Sigma$ a trivial bundle? $\endgroup$ Commented Oct 6, 2022 at 23:01
  • $\begingroup$ @Andreas That's a good question to ask! It hinges on $2$-dimensional manifold with (non-empty) boundary. $\endgroup$ Commented Oct 6, 2022 at 23:04
  • $\begingroup$ @Math: please do not delete your question after others have spent time and effort helping you. Others can benefit from their answers, as well. $\endgroup$
    – robjohn
    Commented Oct 7, 2022 at 2:23

1 Answer 1


As I commented, when you put a flat metric on $\Sigma$, the geodesic curvature integral on $\partial\Sigma$ will definitely not vanish. (Think about classical Gauss-Bonnet for planar regions.)

Your claim that $T\Sigma$ is trivial may not be obvious to everyone. But you can put a nowhere-zero vector field on a manifold with nonempty boundary (for example, double the manifold, take a generic vector field, and then move its zeroes to "the other half"). On a (Riemannian) surface, once you have a nowhere-zero vector field, rotating it $\pi/2$ gives another, and so you've trivialized the tangent bundle.


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