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共変微分の交換関係メモ

Posted at

結論

Riemannテンソルを

R^\mu{}_{\alpha\beta\gamma}:=\Gamma^\mu{}_{\alpha\gamma,\beta}-\Gamma^\mu{}_{\alpha\beta,\gamma}+\Gamma^\mu{}_{\lambda\beta}\Gamma^\lambda{}_{\alpha\gamma}-\Gamma^\mu{}_{\lambda\gamma}\Gamma^\lambda{}_{\alpha\beta}

と定義すると以下が成り立つ.

\begin{align}
&A^{\mu}{}_{;\alpha\beta}-A^{\mu}{}_{;\beta\alpha}= R^{\mu}{}_{\nu\beta\alpha}A^\nu\\
&A^{\mu\nu}{}_{;\alpha\beta}-A^{\mu\nu}{}_{;\beta\alpha}=R^{\mu}{}_{\rho\beta\alpha}A^{\rho\nu}+R^{\nu}{}_{\rho\beta\alpha}A^{\mu\rho}
\end{align}

導出

1階テンソルの共変微分は

A^\mu{}_{;\alpha} = A^\mu{}_{,\alpha}+A^{\nu}\Gamma^{\mu}{}_{\nu\alpha}

である.これを$\beta$で共変微分する.

\begin{align}
A^\mu{}_{;\alpha\beta} &= (A^\mu{}_{;\alpha})_{;\beta}\\
&= (A^\mu{}_{;\alpha})_{,\beta}+A^\nu{}_{;\alpha}\Gamma^{\mu}{}_{\nu\beta}\\
&= (A^\mu{}_{,\alpha}+A^{\nu}\Gamma^{\mu}{}_{\nu\alpha})_{,\beta}+(A^\nu{}_{,\alpha}+A^{\rho}\Gamma^{\nu}{}_{\rho\alpha})\Gamma^{\mu}{}_{\nu\beta}\\
&= A^\mu{}_{,\alpha\beta}+A^{\nu}{}_{,\beta}\Gamma^{\mu}{}_{\nu\alpha}+A^{\nu}\Gamma^{\mu}{}_{\nu\alpha,\beta}+(A^\nu{}_{,\alpha}+A^{\rho}\Gamma^{\nu}{}_{\rho\alpha})\Gamma^{\mu}{}_{\nu\beta} 
\end{align}

$\alpha$と$\beta$を入れ替えて

A^\mu{}_{;\beta\alpha} = A^\mu{}_{,\beta\alpha}+A^{\nu}{}_{,\alpha}\Gamma^{\mu}{}_{\nu\beta}+A^{\nu}\Gamma^{\mu}{}_{\nu\beta,\alpha}+(A^\nu{}_{,\beta}+A^{\rho}\Gamma^{\nu}{}_{\rho\beta})\Gamma^{\mu}{}_{\nu\alpha} 

を得る.偏微分は交換するとしてこれらを辺々引くと

\begin{align}
A^\mu{}_{;\alpha\beta}-A^\mu{}_{;\beta\alpha} &= A^{\nu}\Gamma^{\mu}{}_{\nu\alpha,\beta}+A^{\rho}\Gamma^{\nu}{}_{\rho\alpha}\Gamma^{\mu}{}_{\nu\beta} - A^{\nu}\Gamma^{\mu}{}_{\nu\beta,\alpha}-A^{\rho}\Gamma^{\nu}{}_{\rho\beta}\Gamma^{\mu}{}_{\nu\alpha} \\
&=(\Gamma^{\mu}{}_{\nu\alpha,\beta}-\Gamma^{\mu}{}_{\nu\beta,\alpha}+\Gamma^{\mu}{}_{\rho\beta}\Gamma^{\rho}{}_{\nu\alpha}-\Gamma^{\mu}{}_{\rho\alpha}\Gamma^{\rho}{}_{\nu\beta})A^\nu\\
&=R^{\mu}{}_{\nu\beta\alpha}A^\nu
\end{align}

を得る.2階テンソルも同様である.

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