# Derivative of the christoffel symbol

Consider $$\partial _dC_{ab}^c$$, where $$C_{ab}^c$$ is a field of Christoffel symbols. Is it not true that the tensor field $$\partial _dC_{ab}^c$$, anti-symmetrized over $$d$$ and $$a$$, is $$0$$ if and only if there is no curvature? $$\partial_a$$ is the co-ordinate derivative operator.

Not quite. You missed one further antisymmetric term. The Riemann curvature tensor is zero if and only if \begin{align} {R^\rho}_{\sigma\mu\nu}&=\Gamma^\rho_{\nu\,\mu\,,\sigma}-\Gamma^\rho_{\mu\,\sigma\,,\nu}+\Gamma^\rho_{\mu\,\lambda}\Gamma^\lambda_{\nu\,\sigma}-\Gamma^\rho_{\nu\,\lambda}\Gamma^\lambda_{\mu\,\sigma}\\[3mm] &=2\,\Gamma^\rho_{\nu\,[\mu,\sigma]}+2\,\Gamma^\rho_{[\mu\,|\lambda|}\,\Gamma^{|\lambda|}_{\nu]\,\sigma} \end{align} vanishes. For example, it is known that the Rindler metric for Minkovski space time $$g_{\mu\nu}=\left(\begin{matrix}-\alpha^2x^2 & 0\\0&1\end{matrix}\right)$$ has zero curvature. Its non zero Christoffel symbols are $$\Gamma^t_{tx}=\Gamma^t_{xt}=\frac{1}{x}\,,\quad \Gamma^x_{tt}=\alpha^2x\,.$$ Therefore, the only non zero derivatives are $$\Gamma^t_{tx,x}=\Gamma^t_{xt,x}=-\frac{1}{x^2}\,,\quad \Gamma^x_{tt,x}=\alpha^2\,.$$ Therefore, \begin{align} {R^t}_{xxt}&=\underbrace{\Gamma^t_{tx,x}}_{-1/x^2}-\underbrace{\Gamma^t_{xx,t}}_{0}+\underbrace{\Gamma^t_{xt}\Gamma^t_{tx}}_{1/x^2}+\underbrace{\Gamma^t_{xx}\Gamma^x_{tx}}_{0}- \underbrace{\Gamma^t_{tt}\Gamma^t_{xx}}_{0}-\Gamma^t_{tx}\underbrace{\Gamma^x_{xx}}_{0}\,. \end{align} We see that the antisymmetric partial derivatives term is not zero, $$\Gamma^t_{tx,x}-\Gamma^t_{xx,t}=-1/x^2$$, but the entire expression is zero as it should.