Evaluate integral (Chern article) My question is evaluate some integral of the article "A Simple Intrinsic Proof of the Gauss-Bonnet Formula for Closed Riemannian Manifolds" write by Chern. 
Let's go:
If $(M^n,g)$ is a closed even dimension Riemann manifold with $\nabla$ Levi-Civita connection, we can write locally $\nabla_X V = \theta^i(X)e_i$, where $V = v^ie_i$, $\theta^i = dv^i(X) + v^j\omega_j^i$ and $\omega_i^j$ connection forms. In the same way for Riemann curvature, we have $\Omega_i^j$ curvature forms, satisfying $d\omega_i^j = \omega_i^k \wedge \omega_k^j + \Omega_i^j$.
So, pulling-back $\theta_i$ and $\Omega_i^j$ by $\rho: SM \rightarrow M$, where $SM$ is the unit-sphere bundle, we define two kind of intrinsic forms in $SM$, namely 
$$\Phi_k = \sum_{\sigma \in S_n} sgn(\sigma)v_{\sigma(1)}\theta_{\sigma(2)} \wedge \cdots \wedge \theta_{\sigma(n-2k)}\wedge\Omega_{\sigma(n-2k+1)}^{\sigma(n-2k+1)}\wedge \cdots \wedge \Omega_{\sigma(n-1)}^{\sigma(n)}$$ 
and
$$ \Psi_k = \sum_{\sigma \in S_n} sgn(\sigma) \Omega_{\sigma(1)}^{\sigma(2)}\wedge\theta_{\sigma(3)}\wedge \cdots \wedge \theta_{\sigma(n-2k)}\wedge\Omega_{\sigma(n-2k+1)}^{\sigma(n-2k+1)}\wedge \cdots \wedge \Omega_{\sigma(n-1)}^{\sigma(n)}$$ 
$k = 0, \cdots \frac{n}{2} -1 $. It's not too hard to show the following recurrent relation: $$ d\Phi_k = \Psi_{k-1} + \frac{n - 2k - 1}{2(k+1)}\Psi_k$$
Where $\Psi_{-1} \equiv 0$. 
Define the form, in $M$, $\Omega = \displaystyle (-1)^{\frac{n}{2}-1}\frac{1}{(2\pi)^{\frac{n}{2}}}Pf(\Omega_i^j)$ (called Euler form), definition of Pfaffian polynomial here , obviously $\rho^{*}\Omega = \displaystyle (-1)^{\frac{n}{2}-1}\frac{1}{2^n\pi^{\frac{n}{2}}\left(\frac{n}{2}\right)!}\Psi_{\frac{n}{2}-1}$, write $\Psi_{\frac{n}{2}-1}$ in terms of $d\Phi_k's$ we obtain $d\Pi = \Omega$ in $SM$, with $\Pi =  \displaystyle \frac{1}{\pi^{\frac{n}{2}}}\sum_{t=0}^{\frac{n}{2}-1}(-1)^t \frac{1}{1 \cdot 3 \cdots (n - 2t - 1)t!2^{\frac{n}{2}+t}}\Phi_t$.
With some tricks and Stokes' theorem we show $$\displaystyle \int_M \Omega = \sum_{i=1}^{s}ind_{x_s}(\nabla_{g}f)\int_{SM_{x_s}}\Pi|_{SM_{x_s}} $$
for $x_1, \cdots, x_s$ singularities of $\nabla_{g} f$, $f$ a Morse's function. I'd like $\int_{SM_{x_s}} \Pi|_{SM_{x_s}} = 1$ to use the Hopf index theorem. 
In the paper, Chern claims $$ \int_{SM_{x_s}} \Pi|_{SM_{x_s}} = \frac{1}{1\cdot3 \cdots (n-1)(2\pi)^{\frac{n}{2}}} \int_{SM_{x_s}}\Phi_0|_{SM_{x_s}}$$
and of course it's equal 1, essentially, I don't know why $ \displaystyle \int_{SM_{x_s}} \Phi_k = 0$, for $k \geq 1$.
Thanks!
 A: It's not so scary, after all :) You're integrating the forms $\Phi_k$ over the unit sphere bundle at a fixed point $x_0$ of $M$ (your notation is different from his, since for him $M$ is the unit sphere bundle of the manifold $R$). For $k\ge 1$, the form $\Phi_k$ will involve at least one curvature form $\Omega_i^j$. The curvature forms are horizontal for the fibration $SM\to M$, and you're integrating over a fiber. So those integrals all vanish.
A: In a normal coordinate we have $\theta_i = dv_i$ and $\Omega_i^j = d\omega_i^j$, so at the center point of normal coordinate $\Phi_t = \displaystyle \sum_{\sigma \in S_n} \alpha_{\sigma} \wedge d\omega_{\sigma(n-1)}^{\sigma(n)}$, where $\alpha_{\sigma} = sgn(\sigma)v_{\sigma(1)} dv_{\sigma(2)} \wedge \cdots \wedge dv_{\sigma(n-2t)} \wedge d\omega_{\sigma(n-2t+1)}^{\sigma(n-2t+2)} \wedge \cdots \wedge d\omega_{\sigma(n-3)}^{\sigma(n-2)}$, we have $d(\alpha_{\sigma} \wedge \omega_{\sigma(n-1)}^{\sigma(n)}) = d\alpha_{\sigma}\wedge \omega_{\sigma(n-1)}^{\sigma(n)} - \alpha_{\sigma}\wedge d\omega_{\sigma(n-1)}^{\sigma(n)}$, because $deg(\alpha_\sigma) = (-1)^{n-3} = - 1$, since $\omega_i^j$ vanishes at the center of normal coordinate $\alpha_{\sigma} \wedge d\omega_{\sigma(n-1)}^{\sigma(n)} = - d(\alpha_{\sigma} \wedge \omega_{\sigma(n-1)}^{\sigma(n)})$ by stokes theorem $\displaystyle \int_{SM_x} \sum_{\sigma \in S_n} \alpha_{\sigma} \wedge d\omega_{\sigma(n-1)}^{\sigma(n)} = - \int_{\partial SM_x} \sum_{\alpha \in S_n} \alpha_\sigma \wedge \omega_{\sigma(n-1)}^{\sigma(n)} = 0$, therefore $\displaystyle \int_{SM_x} \Pi = \frac{1}{1 \cdot 3 \cdots (n-1)(2\pi)^{\frac{n}{2}}}\int_{SM_x} \Phi_0$, it's easy to see $\Phi_0$ in normal coordinate is $\displaystyle \sum_{\sigma \in S_n} sgn(\sigma)v_{\sigma(1)} dv_{\sigma(2)}\wedge \cdots \wedge dv_{\sigma(n)} = (n-1)!\sum_{i=1}^n (-1)^{i-1}v_i dv_1 \wedge \cdots \wedge \widehat{dv_i} \wedge \cdots dv_n$, then $\displaystyle \int_{SM_x} \Pi = \frac{(n-1)!2\pi^{\frac{n}{2}}}{1 \cdot 3 \cdots (n-1) 2^{\frac{n}{2}}\pi^{\frac{n}{2}}(\frac{n}{2} - 1)!} = 1.$
