리만 곡률 텐서

접속 (connection)\(\nabla\)이 정의되어 있다고 하자
세 개의 벡터장 X,Y,Z 가 주어지면, 새로운 벡터장 R(X,Y)Z 를 얻는다
\(R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z\)
covariant tensor

\({R^\rho}_{\sigma\mu\nu} = dx^\rho(R(\partial_{\mu},\partial_{\nu})\partial_{\sigma})\)
크리스토펠 기호 를 이용한 성분의 계산
\({R^\rho}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma}\)
\({R^l}_{kij} = \partial_i\Gamma^l_{jk} - \partial_j\Gamma^l_{ik} + \Gamma^l_{is}\Gamma^s_{jk} - \Gamma^l_{js}\Gamma^s_{ik}\)
\(R_{\rho\sigma\mu\nu} = g_{\rho \zeta} {R^\zeta}_{\sigma\mu\nu} .\)

제1기본형식이 \(E=e(u,v),F=0,G=g(u,v)\) 로 주어진 경우, 리만 곡률 텐서는 다음과 같다 (이외의 \( R_{jkl}^i\)는 0이다)
\( R_{212}^1 = \frac{e(u,v) \left(e^{(0,1)}(u,v) g^{(0,1)}(u,v)+g^{(1,0)}(u,v)^2\right)+g(u,v) \left(e^{(1,0)}(u,v) g^{(1,0)}(u,v)-2 e(u,v) \left(e^{(0,2)}(u,v)+g^{(2,0)}(u,v)\right)+e^{(0,1)}(u,v)^2\right)}{4 e(u,v)^2 g(u,v)}\)
\(R_{112}^2 = -\frac{e(u,v) \left(e^{(0,1)}(u,v) g^{(0,1)}(u,v)+g^{(1,0)}(u,v)^2\right)+g(u,v) \left(e^{(1,0)}(u,v) g^{(1,0)}(u,v)-2 e(u,v) \left(e^{(0,2)}(u,v)+g^{(2,0)}(u,v)\right)+e^{(0,1)}(u,v)^2\right)}{4 e(u,v) g(u,v)^2}\)
\(R_{221}^1 = -\frac{e(u,v) \left(e^{(0,1)}(u,v) g^{(0,1)}(u,v)+g^{(1,0)}(u,v)^2\right)+g(u,v) \left(e^{(1,0)}(u,v) g^{(1,0)}(u,v)-2 e(u,v) \left(e^{(0,2)}(u,v)+g^{(2,0)}(u,v)\right)+e^{(0,1)}(u,v)^2\right)}{4 e(u,v)^2 g(u,v)}\)
\(R_{121}^2 = \frac{e(u,v) \left(e^{(0,1)}(u,v) g^{(0,1)}(u,v)+g^{(1,0)}(u,v)^2\right)+g(u,v) \left(e^{(1,0)}(u,v) g^{(1,0)}(u,v)-2 e(u,v) \left(e^{(0,2)}(u,v)+g^{(2,0)}(u,v)\right)+e^{(0,1)}(u,v)^2\right)}{4 e(u,v) g(u,v)^2}\)

목차