共変微分とリーマン曲率テンソルの関係
定理1
$f : A \subset \mathbb{R}^{2} \to M$をパラメータ付き曲面としよう。$(s, t)$を$\mathbb{R}^{2}$の標準座標としよう。$V = V(s,t)$を$f$に従うベクトル場としよう。各点$(s, t)$で次が成り立つ。
$$ \dfrac{D }{\partial t} \dfrac{D }{\partial s}V - \dfrac{D }{\partial s} \dfrac{D }{\partial t}V = R(\dfrac{\partial f}{\partial s}, \dfrac{\partial f}{\partial t})V $$
説明
- アインシュタインの記法が使用されている。
証明
微分可能多様体の多様体$M$上の点$p$で座標系$(U, \mathbf{x})$を一つ選ぼう。接空間$T_{p}M$の基底を$\left\{ X_{i} = \dfrac{\partial }{\partial x_{i}} \right\}$としよう。そして$V = \sum_{i}v^{i}X_{i}, v^{i} = v^{i}(s, t)$としよう。すると、共変微分の性質により、
$$ \dfrac{D }{\partial s}V = \dfrac{D }{\partial s}(\sum_{i} v^{i}X_{i}) = \sum_{i}v^{i}\dfrac{D }{\partial s}X_{i} + \sum_{i} \dfrac{\partial v^{i}}{\partial s}X_{i} $$
ここに再び$\dfrac{D }{\partial t}$を適用すると次のようになる。
$$ \begin{align*} \dfrac{D }{\partial t}\left( \dfrac{D }{\partial s}V \right) &= \dfrac{D }{\partial t}\left( \sum_{i}v^{i}\dfrac{D }{\partial s}X_{i} + \sum_{i} \dfrac{\partial v^{i}}{\partial s}X_{i} \right) \\ &= \sum_{i}v^{i}\dfrac{D }{\partial t}\dfrac{D }{\partial s}X_{i} + \sum_{i}\dfrac{\partial v^{i}}{\partial t}\dfrac{D }{\partial s}X_{i} + \sum_{i} \dfrac{\partial v^{i}}{\partial s}\dfrac{D }{\partial t}X_{i} + \sum_{i} \dfrac{\partial^{2} v^{i}}{\partial t\partial s}X_{i} \end{align*} $$
同じ方法で$\dfrac{D }{\partial s}\left( \dfrac{D }{\partial t}V \right)$を計算して互いに引くと、上の式から最後の三項は互いに打ち消しあうことが分かる。したがって、
$$ \begin{equation} \begin{aligned} \dfrac{D }{\partial t} \dfrac{D }{\partial s}V - \dfrac{D }{\partial s} \dfrac{D }{\partial t}V &= \sum_{i}\left( v^{i}\dfrac{D }{\partial t}\dfrac{D }{\partial s}X_{i} - v^{i}\dfrac{D }{\partial s}\dfrac{D }{\partial t}X_{i} \right) \\ &= \sum_{i}v^{i}\left( \dfrac{D }{\partial t}\dfrac{D }{\partial s}X_{i} - \dfrac{D }{\partial s}\dfrac{D }{\partial t}X_{i} \right) \end{aligned} \label{1} \end{equation} $$
今、$p = f(s,t) = \mathbf{x}(x_{1}(s,t), \dots, x_{n}(s,t))$としよう。$\dfrac{\partial f}{\partial s}$を計算すると、
$$ \begin{align*} \dfrac{\partial f}{\partial s} &:= df(\dfrac{\partial }{\partial s}) \\ &= \begin{bmatrix} \dfrac{\partial x_{1}}{\partial s} & \dfrac{\partial x_{1}}{\partial t} \\ \vdots & \vdots \\ \dfrac{\partial x_{n}}{\partial s} & \dfrac{\partial x_{n}}{\partial t}\end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} \dfrac{\partial x_{1}}{\partial s} \\ \vdots \\ \dfrac{\partial x_{n}}{\partial s} \end{bmatrix} = \dfrac{\partial x_{j}}{\partial s}X_{j} \end{align*} $$
同様に$\dfrac{\partial f}{\partial t} = \dfrac{\partial x_{k}}{\partial t}X_{k}$だ。今、$\dfrac{D }{\partial t}\dfrac{D }{\partial s}X_{i}$を計算すると、
$$ \dfrac{D }{\partial s}X_{i} = \nabla_{\partial f/\partial s}X_{i} = \nabla_{(\partial x_{j}/\partial s)X_{j}}X_{i} = \dfrac{\partial x_{j}}{\partial s}\nabla_{X_{j}}X_{i} $$
そして
$$ \begin{align*} \dfrac{D }{\partial t}\dfrac{D }{\partial s}X_{i} &= \dfrac{D }{\partial t}\left( \dfrac{\partial x_{j}}{\partial s}\nabla_{X_{j}}X_{i} \right) \\ &= \dfrac{\partial^{2} x_{j}}{\partial t\partial s}\nabla_{X_{j}}X_{i} + \dfrac{\partial x_{j}}{\partial s}\nabla_{\partial f / \partial t}(\nabla_{X_{j}}X_{i}) \\ &= \dfrac{\partial^{2} x_{j}}{\partial t\partial s}\nabla_{X_{j}}X_{i} + \dfrac{\partial x_{j}}{\partial s}\nabla_{(\partial x_{k} / \partial t})X_{k}(\nabla_{X_{j}}X_{i}) \\ &= \dfrac{\partial^{2} x_{j}}{\partial t\partial s}\nabla_{X_{j}}X_{i} + \dfrac{\partial x_{j}}{\partial s}\dfrac{\partial x_{k}}{\partial t}\nabla_{X_{k}}\nabla_{X_{j}}X_{i} \end{align*} $$
したがって次を得る。
$$ \dfrac{D }{\partial t}\dfrac{D }{\partial s}X_{i} - \dfrac{D }{\partial s}\dfrac{D }{\partial t}X_{i} = \dfrac{\partial x_{j}}{\partial s}\dfrac{\partial x_{k}}{\partial t}\left( \nabla_{X_{k}}\nabla_{X_{j}}X_{i} - \nabla_{X_{j}}\nabla_{X_{k}}X_{i}\right) $$
しかし、$[X_{i}, X_{j}]=0$なので、リーマン曲率は
$$ R(X_{j}, X_{k})X_{i} = \nabla_{X_{k}}\nabla_{X_{j}}X_{i} - \nabla_{X_{j}}\nabla_{X_{k}}X_{i} + \nabla_{[X_{j}, X_{k}]}X_{i} = \nabla_{X_{k}}\nabla_{X_{j}}X_{i} - \nabla_{X_{j}}\nabla_{X_{k}}X_{i} $$
そして、上の式は、
$$ \dfrac{D }{\partial t}\dfrac{D }{\partial s}X_{i} - \dfrac{D }{\partial s}\dfrac{D }{\partial t}X_{i} = \dfrac{\partial x_{j}}{\partial s}\dfrac{\partial x_{k}}{\partial t}R(X_{j}, X_{k})X_{i} $$
これを$\eqref{1}$に代入すると、$R$が線形であるため、
$$ \begin{align*} \dfrac{D }{\partial t} \dfrac{D }{\partial s}V - \dfrac{D }{\partial s} \dfrac{D }{\partial t}V &= \sum_{i}v^{i}\left( \dfrac{D }{\partial t}\dfrac{D }{\partial s}X_{i} - \dfrac{D }{\partial s}\dfrac{D }{\partial t}X_{i} \right) \\ &= \sum_{i}v^{i}\dfrac{\partial x_{j}}{\partial s}\dfrac{\partial x_{k}}{\partial t}R(X_{j}, X_{k})X_{i} \\ &= \sum_{i}R(\dfrac{\partial x_{j}}{\partial s}X_{j}, \dfrac{\partial x_{k}}{\partial t}X_{k})v^{i}X_{i} \\ &= \sum_{i}R(\dfrac{\partial f}{\partial s}, \dfrac{\partial f}{\partial t})V \end{align*} $$
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Manfredo P. Do Carmo, Riemannian Geometry (Eng Edition, 1992), p98-99 ↩︎