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

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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

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.