I was looking at the symmetries of the Riemann tensor, and tried to prove a couple of properties, namely
If $\nabla$ is torsion-free, then: (i) $R^a_{\,[bcd]}=0$, and (ii) $R^a_{\,b[cd;e]}=0$.
To prove these properties, I work in normal coordinates at a point $p$ (i.e. $\Gamma^\mu_{(\nu\rho)}(p)=0$) and use the fact that the connection is torsion-free (i.e. $\Gamma^{\mu}_{[\nu\rho]}=0$ everywhere), which simplifies the computation of the Riemann tensor greatly. Since the final result is that the components of (a particular form of) the tensor are vanishing, the result extends to any coordinate basis; and since $p$ is arbitrary, the result is true everywhere.
The last step of the computation involves antisymmetrizing an expression of the tensor over various indices, and showing that it vanishes. This is quite obvious from explicitly working out the antisymmetrization, but I'd like to know if there is a shortcut, or an intuition, that one could apply to any such problem.
Namely, the equations are:
$$R^\mu_{\,\nu\rho\sigma}=\partial_\rho\Gamma^\mu_{\nu\sigma}-\partial_\sigma\Gamma^\mu_{\nu\rho}\tag{1a}\label{1a}$$ $$R^\mu_{\,\nu\rho\sigma;\tau}=\partial_\tau\partial_\rho\Gamma^\mu_{\nu\sigma}-\partial_\tau\partial_\sigma\Gamma^\mu_{\nu\rho}\tag{2a}\label{2a}$$
In $\ref{1a}$, we go on to antisymmetrizing on $\nu\rho\sigma$; and in $\ref{2a}$, on $\rho\sigma\tau$. Is there a way of seeing the results (i) and (ii) directly, perhaps by writing $\ref{1a}$ and $\ref{2a}$ as
$$R^\mu_{\,\nu\rho\sigma}=2\partial_{[\rho|}\Gamma^\mu_{\nu|\sigma]}\tag{1b}\label{1b}$$ $$R^\mu_{\,\nu\rho\sigma;\tau}=2\partial_\tau\partial_{[\rho|}\Gamma^\mu_{\nu|\sigma]}\tag{2b}\label{2b}$$ using the fact that $\Gamma$ is symmetric in its lower indices, and antisymmetrizing?