Second Bianchi identity

Statement

Let $\mathrm{\nabla }$ be a linear connection and $R$ be the Riemann curvature tensor of $\mathrm{\nabla }$. Then $R$ can itself be differentiated via $\mathrm{\nabla }$, since $R$ is a $(1,3)$-tensor and we can define the connection on all $(p,q)$-tensors. With this meaning, the following cyclic summation is zero:

$({\mathrm{\nabla }}_{X}R)(Y,Z)+({\mathrm{\nabla }}_{Y}R)(Z,X)+({\mathrm{\nabla }}_{Z}R)(X,Y)=0$

Proof

To prove this we look more closely at what ${\mathrm{\nabla }}_{X}R$ means.

${\mathrm{\nabla }}_{X}R$ must satisfy the following compatibility condition:

$(({\mathrm{\nabla }}_{X}R)(Y,Z))(W)+R({\mathrm{\nabla }}_{X}Y,Z)W+R(Y,{\mathrm{\nabla }}_{X}Z)W+R(Y,Z){\mathrm{\nabla }}_{X}W={\mathrm{\nabla }}_X(R(Y,Z)W)$

We now concentrate on all the other terms.