# Integral of Euler class of tangent bundle and the generalized Gauss-Bonnet Theorem

Definition of Euler Class that I've been given Let $\Sigma$ be a compact orientable manifold of dimension 2 with metric $g$ on the tangent bundle $T\Sigma$, and connection $\nabla$ and curvature $F_{\nabla}$. For sections $s_1$, $s_2$ of $T\Sigma$, we can define the map $$\omega(s_1,s_2) = g(F_{\nabla}s_1,s_2)$$ It can be seen that $\omega\in \Omega^{2}(\Sigma; \Lambda^{2}T^{*}\Sigma)$ and that $\nabla\omega = 0$.

For any two local frames $\{s_1,s_2\}$ and $\{s'_1,s'_2\}$ of $T\Sigma$ with the same orientation, we find that $s_1\wedge s_2 = s'_1\wedge s'_2$, and hence $\Lambda^{2}T\Sigma$ admits a nowhere vanishing section, say $\sigma\in \Omega^{2}(\Sigma; \Lambda^{2}T\Sigma)$.

If we take $e$ to be the natural pairing $\langle\omega,\sigma\rangle$, then the Euler class of $T\Sigma$, $\varepsilon(T\Sigma)$, may be defined as the cohomology class of $e$, i.e. $\varepsilon(T\Sigma) = [e]\in H^{2}(\Sigma)$. It can be shown that this is independent of the choice of metric and connection.

Definition of Riemann Curvature For vector fields $X,Y,Z,W$, we may define $$R(X,Y,Z,W) = g(F_{\nabla}(X,Y)Z,W)$$ This satisfies the expected symmetry and anti-symmetry properties, as in http://mathworld.wolfram.com/RiemannTensor.html.

In particular, for the manifold $\Sigma$, we can relate the Riemann curvature to the Gauss curvature $K$ and a non-vanishing section $\sigma \in \Omega^{2}(\Sigma; \Lambda^{2}T\Sigma)$ by the following formula (for $x\in \Sigma$) $$R(X,Y,Z,W)\mid_{x} = K(x)\sigma(X,Y)\sigma(Z,W)$$ What I've been asked to show $$\int_{\Sigma}\varepsilon(T\Sigma) = \int_{\Sigma}K\sigma$$ What I'm trying to understand

1. This formula looks very like the Gauss-Bonnet formula for surfaces to me, especially if I consider the case where $\sigma = dS$, some surface element. In this case, I would expect that the left-hand-side of the equation should equal $2\pi\chi(\Sigma)$. Unfortunately, I don't fully understand how to reconcile my definition of the Euler class with a formula for the Euler characteristic. Different (presumably compatible) definitions have been mentioned in other questions on this site (e.g. the answer given by Elden Elmanto in Euler class of tangent bundle of the sphere, and also here: How to interpret the Euler class?), but I can't figure out how they tie back to this integral formula.
2. Would it be wiser to try to solve the problem using the pairing $\langle \omega, \sigma\rangle$, rather than going through the Gauss-Bonnet theorem?

Any help would be greatly appreciated.

• This should be a simple calculation, which looks to me more like part of a proof of Gauss-Bonnet than something that should be invoking it. I think you will in fact have the pointwise equality $e=K\sigma.$ – Anthony Carapetis Jan 1 '18 at 11:49
• Your formula $R(X,Y,Z,W)=K(x) \sigma (X,Y) \sigma (Z,W)$ is not correct as the rhs depends on the choice of $\sigma$, whereas the lhs is well defined. I think you must choose $\sigma$ to be the riemannian volume form, then $e=K\sigma$. – Thomas Jan 2 '18 at 6:42
• AnthonyCarapetis and Thomas, thank you both for sharing your insights. I will try to use the Riemannian volume form to obtain the pointwise equality. – An Coileanach Jan 2 '18 at 9:25