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.
 A: 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.
