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共変微分とリーマン曲率テンソルの関係 📂幾何学

共変微分とリーマン曲率テンソルの関係

定理1

f:A⊂R2→Mf : A \subset \mathbb{R}^{2} \to Mをパラメータ付き曲面としよう。(s,t)(s, t)をR2\mathbb{R}^{2}の標準座標としよう。V=V(s,t)V = V(s,t)をffに従うベクトル場としよう。各点(s,t)(s, t)で次が成り立つ。

D∂tD∂sV−D∂sD∂tV=R(∂f∂s,∂f∂t)V \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

証明

微分可能多様体の多様体MM上の点ppで座標系(U,x)(U, \mathbf{x})を一つ選ぼう。接空間TpMT_{p}Mの基底を{Xi=∂∂xi}\left\{ X_{i} = \dfrac{\partial }{\partial x_{i}} \right\}としよう。そしてV=∑iviXi,vi=vi(s,t)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、

D∂sV=D∂s(∑iviXi)=∑iviD∂sXi+∑i∂vi∂sXi \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D∂t\dfrac{D }{\partial t}を適用すると次のようになる。

D∂t(D∂sV)=D∂t(∑iviD∂sXi+∑i∂vi∂sXi)=∑iviD∂tD∂sXi+∑i∂vi∂tD∂sXi+∑i∂vi∂sD∂tXi+∑i∂2vi∂t∂sXi \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D∂s(D∂tV)\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、

D∂tD∂sV−D∂sD∂tV=∑i(viD∂tD∂sXi−viD∂sD∂tXi)=∑ivi(D∂tD∂sXi−D∂sD∂tXi) \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)=x(x1(s,t),…,xn(s,t))p = f(s,t) = \mathbf{x}(x_{1}(s,t), \dots, x_{n}(s,t))としよう。∂f∂s\dfrac{\partial f}{\partial s}を計算すると、

∂f∂s:=df(∂∂s)=[∂x1∂s∂x1∂t⋮⋮∂xn∂s∂xn∂t][10]=[∂x1∂s⋮∂xn∂s]=∂xj∂sXj \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∂f∂t=∂xk∂tXk\dfrac{\partial f}{\partial t} = \dfrac{\partial x_{k}}{\partial t}X_{k}だ。今、D∂tD∂sXi\dfrac{D }{\partial t}\dfrac{D }{\partial s}X_{i}を計算すると、

D∂sXi=∇∂f/∂sXi=∇(∂xj/∂s)XjXi=∂xj∂s∇XjXi \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

D∂tD∂sXi=D∂t(∂xj∂s∇XjXi)=∂2xj∂t∂s∇XjXi+∂xj∂s∇∂f/∂t(∇XjXi)=∂2xj∂t∂s∇XjXi+∂xj∂s∇(∂xk/∂t)Xk(∇XjXi)=∂2xj∂t∂s∇XjXi+∂xj∂s∂xk∂t∇Xk∇XjXi \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。

D∂tD∂sXi−D∂sD∂tXi=∂xj∂s∂xk∂t(∇Xk∇XjXi−∇Xj∇XkXi) \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、[Xi,Xj]=0[X_{i}, X_{j}]=0なので、リーマン曲率は

R(Xj,Xk)Xi=∇Xk∇XjXi−∇Xj∇XkXi+∇[Xj,Xk]Xi=∇Xk∇XjXi−∇Xj∇XkXi 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、

D∂tD∂sXi−D∂sD∂tXi=∂xj∂s∂xk∂tR(Xj,Xk)Xi \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(1)\eqref{1}に代入すると、RRが線形であるため、

D∂tD∂sV−D∂sD∂tV=∑ivi(D∂tD∂sXi−D∂sD∂tXi)=∑ivi∂xj∂s∂xk∂tR(Xj,Xk)Xi=∑iR(∂xj∂sXj,∂xk∂tXk)viXi=∑iR(∂f∂s,∂f∂t)V \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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  1. Manfredo P. Do Carmo, Riemannian Geometry (Eng Edition, 1992), p98-99 ↩︎