Curvature is symmetric in the pairs of first and last two variables

Statement

Suppose $M$ is a differential manifold and $g$ is a Riemannian metric or pseudo-Riemannian metric on $M$. Suppose $\nabla$ is the Levi-Civita connection on $M$ and $R$ is the Riemann curvature tensor for $\nabla$. (In other words, $R$ is the Riemann curvature tensor of the Levi-Civita connection).

We can view $R$ as a $(0,4)$-tensor as follows:

$\!R(X,Y,Z,W)=g(R(X,Y)Z,W)$.

Then, we have:

$R(X,Y,Z,W)=R(Z,W,Y,X)$.

Note that since curvature is antisymmetric in first two variables and curvature is antisymmetric in last two variables, this essentially shows that $R$ is a symmetric bilinear form on $\bigwedge ^{2}(TM)$.

Related facts

• Curvature is tensorial
• Curvature is antisymmetric in first two variables
• Curvature is antisymmetric in last two variables
• First Bianchi identity
• Second Bianchi identity

Facts used

1. Curvature is antisymmetric in first two variables
2. Curvature is antisymmetric in last two variables
3. First Bianchi identity: This states that if $\nabla$ is a torsion-free linear connection, and $R$ is its Riemann curvature tensor, then:

$\!R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0$.

For the corresponding $(0,4)$-tensor, we have:

$\!R(X,Y,Z,W)+R(Y,Z,X,W)+R(Z,X,Y,W)=0$.

Proof

Applying fact (3), we get:

$\!R(X,Y,Z,W)+R(Y,Z,X,W)+R(Z,X,Y,W)=0\qquad (1)$

Similar statements, permuting the variables, are:

$\!R(Y,Z,W,X)+R(Z,W,Y,X)+R(W,Y,Z,X)=0\qquad (2)$

$\!R(Z,W,X,Y)+R(W,X,Z,Y)+R(X,Z,W,Y)=0\qquad (3)$

$\!R(W,X,Y,Z)+R(X,Y,W,Z)+R(Y,W,X,Z)=0\qquad (4)$

Consider (1) + (2) - (3) - (4) and uses facts (1) and (2). We get:

$\!2R(X,Y,Z,W)-2R(Z,W,Y,X)=0$

This completes the proof.