ビアンキの恒等式
恒等式の導出
\[[\nabla_\lambda, [\nabla_\mu, \nabla_\nu]] = \nabla_\lambda(\nabla_\mu \nabla_\nu - \nabla_\nu \nabla_\mu) - (\nabla_\mu \nabla_\nu - \nabla_\nu \nabla_\mu) \nabla_\lambda\] \[[\nabla_\mu, [\nabla_\nu, \nabla_\lambda]] = \nabla_\mu(\nabla_\nu \nabla_\lambda - \nabla_\lambda \nabla_\nu) - (\nabla_\nu \nabla_\lambda - \nabla_\lambda \nabla_\nu) \nabla_\mu\] \[[\nabla_\nu, [\nabla_\lambda, \nabla_\mu]] = \nabla_\nu(\nabla_\lambda \nabla_\mu - \nabla_\mu \nabla_\lambda) - (\nabla_\lambda \nabla_\mu - \nabla_\mu \nabla_\lambda) \nabla_\nu\]より
\[([\nabla_\lambda, [\nabla_\mu, \nabla_\nu]]+[\nabla_\mu, [\nabla_\nu, \nabla_\lambda]] + [\nabla_\nu, [\nabla_\lambda, \nabla_\mu]]) A^\alpha = 0\]これと
\[\begin{aligned} [\nabla_\lambda, [\nabla_\mu, \nabla_\nu]] A^\alpha &= \nabla_\lambda ([\nabla_\mu, \nabla_\nu] A^\alpha) -[\nabla_\mu, \nabla_\nu] (\nabla_\lambda A^\alpha) = \nabla_\lambda (R^\gamma_{\lambda \mu \nu} A^\gamma) - (-R^\gamma_{\lambda \mu \nu} \nabla_\gamma A^\alpha + R^\alpha_{\gamma \mu \nu} \nabla_\lambda A^\gamma) \\ &= (\nabla_\lambda R^\alpha_{\gamma \mu \nu}) A^\gamma + R^\gamma_{\lambda \mu \nu} \nabla_\gamma A^\alpha \end{aligned}\]より
\[(\nabla_\lambda R^\alpha_{\gamma \mu \nu} + \nabla_\mu R^\alpha_{\gamma \nu \lambda} + \nabla_\nu R^\alpha_{\gamma \lambda \mu}) A^\gamma + (R^\gamma_{\lambda \mu \nu} + R^\gamma_{\mu \nu \lambda} + R^\gamma_{\nu \lambda \mu}) \nabla_\gamma A^\alpha = 0\]\(A^\gamma, \nabla_\gamma A^\alpha\)は任意のベクトルなのでこれが恒等的に成り立つためには、この2つの係数は0でなければなりません。
\[\therefore \ \nabla_\lambda R^\alpha_{\gamma \mu \nu} + \nabla_\mu R^\alpha_{\gamma \nu \lambda} + \nabla_\nu R^\alpha_{\gamma \lambda \mu} = 0\] \[\therefore \ R^\gamma_{\lambda \mu \nu} + R^\gamma_{\mu \nu \lambda} + R^\gamma_{\nu \lambda \mu} = 0\]この2つをビアンキの恒等式と呼びます。