The second Bianchi Identity is
$$
\nabla_{[a}R_{bc]de}=0
$$
As far as I know, the proof (say, Walfram Mathword) start by stating the representation of Riemann tensor in local inertial coordinates
$$
R_{abcd}=\frac{1}{2}(\partial_a\partial_cg_{bd}-\partial_a\partial_dg_{bc}-\partial_b\partial_cg_{ad}+\partial_b\partial_dg_{ac}).
$$
Then we calculate
$$
\partial_aR_{bcde}
$$
accordingly. Then we say that it is true in a local inertial coordinate, and after changing partial derivative into covariant derivative, it is true in general.
My concern is, I think we cannot express the Riemann tensor and the covariant derivative into local frame one by one, but should simultaneously. Say
$$
\nabla_{a}R_{bcde}=\frac{1}{2}(\partial_a+\Gamma_1)(\partial_a\partial_cg_{bd}-\partial_a\partial_dg_{bc}-\partial_b\partial_cg_{ad}+\partial_b\partial_dg_{ac}+\Gamma_2)
$$
where $\Gamma_1$ and $\Gamma_2$ are some terms involving the Christoffel symbol. When we only concern $R_{bcde}$ in a local frame, $\Gamma_2$ vanishes. But now we get a new term
$$
\partial_a\Gamma_2
$$
which I cannot see to vanish because it involves derivative of the Christoffel symbol. So I think in a local frame $\nabla_aR_{bcde}$ is not $\partial_aR_{bcde}$.
Is there anything wrong?
Best Answer
The Riemann tensor terms involving the Christoffel symbols are a product of two Christoffel symbols. So if you take the derivative of the product, you end up with products of the Christoffel symbol with its derivative. But since the Christoffel symbol is zero, so is the product.