Reelles Vektorbündel/Trivial/Linearer Zusammenhang/Krümmungsoperator/Beschreibung/Fakt/Beweis

Beweis

Es ist

${}{\begin{aligned}R(\partial _{i},\partial _{j})&=\nabla _{\partial _{i}}\circ \nabla _{\partial _{j}}-\nabla _{\partial _{j}}\circ \nabla _{\partial _{i}}-\nabla _{[\partial _{i},\partial _{j}]}\\&=\nabla _{\partial _{i}}\circ \nabla _{\partial _{j}}-\nabla _{\partial _{j}}\circ \nabla _{\partial _{i}},\end{aligned}}$

da ja die partiellen Ableitungen kommutieren und daher ihre Lie-Klammer gleich ${}0$ ist. Für einen Standardschnitt ${}s_{k}$ ist daher unter Verwendung von Fakt

${}{\begin{aligned}R(\partial _{i},\partial _{j})(s_{k})&=\nabla _{\partial _{i}}(\nabla _{\partial _{j}}(s_{k}))-\nabla _{\partial _{j}}(\nabla _{\partial _{i}}(s_{k}))\\&=\nabla _{\partial _{i}}{\left(\sum _{\ell =1}^{r}\Gamma _{jk}^{\ell }s_{\ell }\right)}-\nabla _{\partial _{j}}{\left(\sum _{\ell =1}^{r}\Gamma _{ik}^{\ell }s_{\ell }\right)}\\&=\sum _{\ell =1}^{r}{\left(\nabla _{\partial _{i}}{\left(\Gamma _{jk}^{\ell }s_{\ell }\right)}-\nabla _{\partial _{j}}{\left(\Gamma _{ik}^{\ell }s_{\ell }\right)}\right)}\\&=\sum _{\ell =1}^{r}{\left({\partial _{i}}\Gamma _{jk}^{\ell }s_{\ell }+\Gamma _{jk}^{\ell }\sum _{\rho =1}^{r}\Gamma _{i\ell }^{\rho }s_{\rho }-{\partial _{j}}\Gamma _{ik}^{\ell }s_{\ell }-\Gamma _{ik}^{\ell }\sum _{\rho =1}^{r}\Gamma _{j\ell }^{\rho }s_{\rho }\right)}\\&=\sum _{\ell =1}^{r}{\left({\partial _{i}}\Gamma _{jk}^{\ell }-{\partial _{j}}\Gamma _{ik}^{\ell }\right)}s_{\ell }+\sum _{\ell =1}^{r}\sum _{\rho =1}^{r}{\left(\Gamma _{jk}^{\ell }\Gamma _{i\ell }^{\rho }s_{\rho }-\Gamma _{ik}^{\ell }\Gamma _{j\ell }^{\rho }s_{\rho }\right)}\\&=\sum _{\ell =1}^{r}{\left({\partial _{i}}\Gamma _{jk}^{\ell }-{\partial _{j}}\Gamma _{ik}^{\ell }\right)}s_{\ell }+\sum _{\ell =1}^{r}{\left(\sum _{a=1}^{r}\Gamma _{jk}^{a}\Gamma _{ia}^{\ell }-\sum _{a=1}^{r}\Gamma _{ik}^{a}\Gamma _{ja}^{\ell }\right)}s_{\ell }.\end{aligned}}$

Zur bewiesenen Aussage