クリストッフェル記号
∂ 2 x i ∂ u l ∂ u h = − 1 2 ( ∂ 2 { i j k } 0 ∂ u l ∂ u h u j u k + ∂ { i j k } 0 ∂ u h δ l j u k + ∂ { i j k } 0 ∂ u h u j δ l k ) − ∂ { i h k } 0 ∂ u l u k − { i h k } 0 δ l k {\displaystyle {\frac {\partial ^{2}x^{i}}{\partial u^{l}\partial u^{h}}}=-{\frac {1}{2}}\left({\frac {\partial ^{2}\left\{{{i} \atop {j\;k}}\right\}_{0}}{\partial u^{l}\partial u^{h}}}u^{j}u^{k}+{\frac {\partial \left\{{{i} \atop {j\;k}}\right\}_{0}}{\partial u^{h}}}\delta _{l}^{j}u^{k}+{\frac {\partial \left\{{{i} \atop {j\;k}}\right\}_{0}}{\partial u^{h}}}u^{j}\delta _{l}^{k}\right)-{\frac {\partial \left\{{{i} \atop {h\;k}}\right\}_{0}}{\partial u^{l}}}u^{k}-\left\{{{i} \atop {h\;k}}\right\}_{0}\delta _{l}^{k}}
となる。したがって、x = x0 のとき u = 0 であることから、
∂ x i ∂ u h = δ h i − 1 2 ( ∂ { i j k } 0 ∂ u h u j u k + { i j k } 0 δ h j u k + { i j k } 0 u j δ h k ) = δ h i − 1 2 ∂ { i j k } 0 ∂ u h u j u k − { i h k } 0 u k {\displaystyle {\frac {\partial x^{i}}{\partial u^{h}}}=\delta _{h}^{i}-{\frac {1}{2}}\left({\frac {\partial \left\{{{i} \atop {j\;k}}\right\}_{0}}{\partial u^{h}}}u^{j}u^{k}+\left\{{{i} \atop {j\;k}}\right\}_{0}\delta _{h}^{j}u^{k}+\left\{{{i} \atop {j\;k}}\right\}_{0}u^{j}\delta _{h}^{k}\right)=\delta _{h}^{i}-{\frac {1}{2}}{\frac {\partial \left\{{{i} \atop {j\;k}}\right\}_{0}}{\partial u^{h}}}u^{j}u^{k}-\left\{{{i} \atop {h\;k}}\right\}_{0}u^{k}}
となり、さらに u で偏微分を行うと