First Bianchi identity

Statement

Let $\mathrm{\nabla }$ be a torsion-free linear connection. The Riemann curvature tensor $R$ of $\mathrm{\nabla }$ satisfies the following first Bianchi identity or algebraic Bianchi identity:

$R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0$

for any three vector fields $X,Y,Z$.

Notice that since this proof is applicable for any torsion-free linear connection, it in particular holds for the Levi-Civita connection arising from a Riemannian metric or pseudo-Riemannian metric.

Proof

Using repeated simplication and the Jacobi identity

Let us plug the definition of the Riemann curvature tensor:

${\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}Z+{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{Z}X-{\mathrm{\nabla }}_Z{\mathrm{\nabla }}_{Y}X+{\mathrm{\nabla }}_Z{\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Z}Y-{\mathrm{\nabla }}_{[X,Y]}Z-{\mathrm{\nabla }}_{[Y,Z]}X-{\mathrm{\nabla }}_{[Z,X]}Y$

This can be regrouped as:

${\mathrm{\nabla }}_X({\mathrm{\nabla }}_{Y}Z-{\mathrm{\nabla }}_{Z}Y)+{\mathrm{\nabla }}_Y({\mathrm{\nabla }}_{Z}X-{\mathrm{\nabla }}_{X}Z)+{\mathrm{\nabla }}_Z({\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X)-{\mathrm{\nabla }}_{[X,Y]}Z-{\mathrm{\nabla }}_{[Y,Z]}X-{\mathrm{\nabla }}_{[Z,X]}Y$

Now, since $\mathrm{\nabla }$ is torsion-free, we have ${\mathrm{\nabla }}_{Y}Z-{\mathrm{\nabla }}_{Z}Y=[Y,Z]$ and similar simplifications yield:

${\mathrm{\nabla }}_X[Y,Z]+{\mathrm{\nabla }}_Y[Z,X]+{\mathrm{\nabla }}_Z[X,Y]-{\mathrm{\nabla }}_{[X,Y]}Z-{\mathrm{\nabla }}_{[Y,Z]}X-{\mathrm{\nabla }}_{[Z,X]}Y$

again using the fact that $\mathrm{\nabla }$ is torsion-free, this simplifies to:

$[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]$

this becomes zero by the Jacobi identity.

Using the differential Bianchi identity

Fill this in later