Difference between revisions of "Curvature is antisymmetric in last two variables"

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Revision as of 01:28, 24 July 2009

Statement

Suppose M is a differential manifold and g is a Riemannian metric or pseudo-Riemannian metric and \nabla is the Levi-Civita connection for g. Consider the Riemann curvature tensor R of \nabla. In other words, R is the Riemann curvature tensor of the Levi-Civita connection for g. We can treat R as a (0,4)-tensor:

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

Then:

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

Facts used

  1. First Bianchi identity: This states that if R is a torsion-free linear connection, then:

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

Proof

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