one substitution in Chern's intrinsic proof of Gauss-Bonnet-Chern theorem

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In Chern's proof for Gauss-Bonnet-Chern theorem, he claims that $$ \varepsilon_{i}u_{i_1}u_j\Omega_{ji_2}\theta_{i_3}\cdots\theta_{i_{2p-2k}}\Omega_{i_{2p-2k+1}i_{2p-2k+2}}\cdots\Omega_{i_{2p-1}i_{2p}}=P_k+2(p-k-1)\Sigma_k $$ where $$ P_k=\varepsilon_{i}u_{i_1}^2\Omega_{i_1i_2}\theta_{i_3}\cdots\theta_{i_{2p-2k}}\Omega_{i_{2p-2k+1}i_{2p-2k+2}}\cdots\Omega_{i_{2p-1}i_{2p}} $$ and $$ \Sigma_k=\varepsilon_{i}u_{i_1}u_{i_3}\Omega_{i_3i_2}\theta_{i_3}\cdots\theta_{i_{2p-2k}}\Omega_{i_{2p-2k+1}i_{2p-2k+2}}\cdots\Omega_{i_{2p-1}i_{2p}} $$

By direct computations:

  • $j=i_1$: we get $P_k$.
  • $j=i_3,\dots,i_{2p-2k}$: we get $\Sigma_k$.

My question is: why $$ \sum_{j=i_{2p-2k+1}}^{2p}\varepsilon_{i}u_{i_1}u_j\Omega_{ji_2}\theta_{i_3}\cdots\theta_{i_{2p-2k}}\Omega_{i_{2p-2k+1}i_{2p-2k+2}}\cdots\Omega_{i_{2p-1}i_{2p}}=0? $$ It seems that $$ \varepsilon_{i}u_{i_1}u_{j}\Omega_{ji_2}\theta_{i_3}\cdots\theta_{i_{2p-2k}}\Omega_{i_{2p-2k+1}i_{2p-2k+2}}\cdots\Omega_{i_{2p-1}i_{2p}}\\=\varepsilon_{i}u_{i_1}u_{i_{2p-2k+1}}\Omega_{i_{2p-2k+1}i_2}\theta_{i_3}\cdots\theta_{i_{2p-2k}}\Omega_{i_{2p-2k+1}i_{2p-2k+2}}\cdots\Omega_{i_{2p-1}i_{2p}} $$ for all $j\in\{2p-2k+2,\dots,2p\}$, then why $$ \varepsilon_{i}u_{i_1}u_{i_{2p-2k+1}}\Omega_{i_{2p-2k+1}i_2}\theta_{i_3}\cdots\theta_{i_{2p-2k}}\Omega_{i_{2p-2k+1}i_{2p-2k+2}}\cdots\Omega_{i_{2p-1}i_{2p}}=0? $$

Any help would be appreciated.

The link for Chern's original proof is: https://www.maths.ed.ac.uk/~v1ranick/papers/chern7.pdf

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Here's all that is going on. It is the skew symmetry in the $\epsilon_i$ symbol. Because $\theta_{i_3}\wedge\dots\wedge\theta_{i_{2p-2k}}$ is a form of even degree, we have for $\mu$ odd in $2p-2k+1,\dots,2p-1,2p$, $$\Omega_{i_\mu i_2}\wedge (\dots) \wedge \Omega_{i_\mu i_{\mu+1}} = \Omega_{i_\mu i_{\mu+1}}\wedge (\dots) \wedge \Omega_{i_\mu i_2},$$ but switching $2$ and $\mu+1$ introduces a factor of $-1$ in the $\epsilon_i$. Thus, the sum (for $j=i_\mu$ fixed) is zero. And if $\mu$ is even, first write $\Omega_{i_\mu i_2}=-\Omega_{i_2i_\mu}$ and make the same argument with $2$ and $\mu-1$.