Curvature is antisymmetric in last two variables: Difference between revisions
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<math>YXg(Z,W) = Yg(\nabla_X(Z),W) + Yg(Z,nabla_X(W)) = g(\nabla_Y \circ \nabla_X(Z),W) + g(\nabla_X(Z),\nabla_Y(W)) + g(\nabla_Y(Z),\nabla_X(W)) + g(Z,\nabla_Y \circ \nabla_X(W)) \qquad (2)</math>. | <math>YXg(Z,W) = Yg(\nabla_X(Z),W) + Yg(Z,\nabla_X(W)) = g(\nabla_Y \circ \nabla_X(Z),W) + g(\nabla_X(Z),\nabla_Y(W)) + g(\nabla_Y(Z),\nabla_X(W)) + g(Z,\nabla_Y \circ \nabla_X(W)) \qquad (2)</math>. | ||
Substituting (1) and (2) in <math>(\dagger\dagger)</math> yields <math>(\dagger)</math>. | Substituting (1) and (2) in <math>(\dagger\dagger)</math> yields <math>(\dagger)</math>. | ||
Revision as of 01:51, 24 July 2009
Statement
Suppose is a differential manifold and is a Riemannian metric or pseudo-Riemannian metric and is the Levi-Civita connection for . Consider the Riemann curvature tensor of . In other words, is the Riemann curvature tensor of the Levi-Civita connection for . We can treat as a -tensor:
.
Then:
.
Proof
We consider the expression :
By the bilinearity of , this simplifies to:
To prove that this is zero, it thus suffices to show that:
.
We now show . Since is a metric connection, the left side simplifies to:
.
Simplifying each of the two terms on the right side of , we get:
.
And:
.
Substituting (1) and (2) in yields .