First Bianchi identity

Statement

Let $\nabla$ be a torsion-free linear connection. The Riemann curvature tensor $R$ of $\nabla$ 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.

Related facts

• Second Bianchi identity (also called the differential Bianchi identity).
• Curvature is tensorial
• Torsion is tensorial

Proof

Using repeated simplication and the Jacobi identity

Let us plug the definition of the Riemann curvature tensor:

$\nabla _{X}\nabla _{Y}Z-\nabla _{Y}\nabla _{X}Z+\nabla _{Y}\nabla _{Z}X-\nabla _{Z}\nabla _{Y}X+\nabla _{Z}\nabla _{X}Y-\nabla _{X}\nabla _{Z}Y-\nabla _{[X,Y]}Z-\nabla _{[Y,Z]}X-\nabla _{[Z,X]}Y$

This can be regrouped as:

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

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

$\nabla _{X}[Y,Z]+\nabla _{Y}[Z,X]+\nabla _{Z}[X,Y]-\nabla _{[X,Y]}Z-\nabla _{[Y,Z]}X-\nabla _{[Z,X]}Y$

again using the fact that $\nabla$ 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

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