Gauss-Bonnet Theorem in dimension four

Question

I've read that the generalized Gauss-Bonnet theorem states that

$$\int\limits_{M}Pf(\Omega)=(2\pi)^n\chi(M)$$

where, $M$ is a 2n-dimensional compact orientable Riemannian manifold without boundary $\Omega$ is the curvature form and $Pf(\Omega)$ is the Pfaffian of $\Omega$, $Pf(\Omega)$ is a 2n-form.

How can I prove that in dimension four is valid:

$$\chi(M)=\frac{1}{32\pi^2}\int\limits_M(|Rm|^2-4|Ric|^2+R^2)\,d\mu$$

where, $Rm$ is the Riemannian curvature tensor, $Ric$ is the Ricci curvature tensor ans $R$ is then scalar curvature.

Tahnks in advice.

Answer 1

Here some steps to achieve the solution:

1. Remember the curvature form $\Omega$ is a 2-form with values in $\mathfrak{o}(2n)$. If $(e_i)$ is a ortonormal basis and $(\theta^i)$ its dual, we have $$\Omega_i^j=\frac{1}{2}\theta^k\wedge\theta^lR_{kli}^j.$$

2. $$Pf(\Omega)=\frac{1}{2^nn!}\sum_{\sigma\in S_{2n}}{\rm sgn}\,(\sigma)\prod_{i=1}^n\Omega_{\sigma(2i-1)}^{\sigma(2i)}.$$

3. Check that, for $n=2$, $$Pf(\Omega)=\frac{1}{8}(|Rm|^2-4|Ric|^2+R^2).$$

4. Thus, $$\chi(M)=\frac{1}{32\pi^2}\int\limits_M(|Rm|^2-4|Ric|^2+R^2)\,d\mu $$