クリストッフェルの記号と平行移動の関係

平行移動における変化分 $d(g^{\mu\nu}A_\mu A_\nu)$ は， $$\begin{aligned} d(g^{\mu\nu}A_\mu A_\nu) &= g^{\mu\nu}A_\mu dA_\nu + g^{\mu\nu}dA_\mu A_\nu + dg^{\mu\nu} A_\mu A_\nu\\ &= A^\nu dA_\nu + A^\mu dA_\mu + A_\mu A_\nu \cdot \dfrac{\partial{g^{\mu\nu}}}{\partial{x^\sigma}} dx^\sigma\\ &= A^\nu A^\mu \Gamma_{\mu\nu\sigma}dx^\sigma + A^\mu A^\nu \Gamma_{\nu\mu\sigma}dx^\sigma + A_\mu A_\nu \dfrac{\partial{g^{\mu\nu}}}{\partial{x^\sigma}} dx^\sigma\\ &= A^\nu A^\mu dx^\sigma (\Gamma_{\mu\nu\sigma} + \Gamma_{\nu\mu\sigma}) + A_\nu A_\mu \dfrac{\partial{g^{\mu\nu}}}{\partial{x^\sigma}} dx^\sigma\\ &= (A^\nu A^\mu \dfrac{\partial{g_{\mu\nu}}}{\partial{x^\sigma}} + A_\mu A_\nu \dfrac{\partial{g^{\mu\nu}}}{\partial{x^\sigma}}) dx^\sigma \end{aligned}$$ さて， $$\dfrac{\partial{(g^{\alpha\mu} g_{\mu\nu})}}{\partial{x^\sigma}} = \dfrac{\partial{g^{\alpha\mu}}}{\partial{x^\sigma}} g_{\mu\nu} + g^{\alpha\mu} \dfrac{\partial{g_{\mu\nu}}}{\partial{x^\sigma}} = \dfrac{\partial{\delta^\alpha_\nu}}{\partial{x^\sigma}} = 0$$ において，$g^{\beta\nu}$ をかければ $$\begin{aligned} g^{\beta\nu}g_{\mu\nu}\dfrac{\partial{g^{\alpha\mu}}}{\partial{x^\sigma}} + g^{\beta\nu}g^{\alpha\mu} \dfrac{\partial{g_{\mu\nu}}}{\partial{x^\sigma}} &= 0\\ \therefore \dfrac{\partial{g^{\alpha\beta}}}{\partial{x^\sigma}} + g^{\beta\nu}g^{\alpha\mu} \dfrac{\partial{g_{\mu\nu}}}{\partial{x^\sigma}} &= 0 \end{aligned}$$ この両辺に $A_\alpha A_\beta$ をかけて， $$\begin{aligned} A_\alpha A_\beta \dfrac{\partial{g^{\alpha\beta}}}{\partial{x^\sigma}} + A^\mu A^\nu \dfrac{\partial{g_{\mu\nu}}}{\partial{x^\sigma}} &= 0\\ A_\mu A_\nu \dfrac{\partial{g^{\mu\nu}}}{\partial{x^\sigma}} + A^\mu A^\nu \dfrac{\partial{g_{\mu\nu}}}{\partial{x^\sigma}} &= 0 \end{aligned}$$ が成立するから， $$d(g^{\mu\nu}A_\mu A_\nu) = 0 \tag{1}$$