Curvature is antisymmetric in last two variables

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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).

Related facts

Proof

We consider the expression R(X,Y,Z,W) + R(X,Y,W,Z):

g(\nabla_X \circ \nabla_Y(Z) - \nabla_Y \circ \nabla_X(Z) - \nabla_{[X,Y]}(Z),W) - g(\nabla_X \circ \nabla_Y(W) - \nabla_Y \circ \nabla_X(W) - \nabla_{[X,Y]}(W),Z)

By the bilinearity of g, this simplifies to:

g(\nabla_X \circ \nabla_Y(Z),W) - g(\nabla_Y\circ \nabla_X(Z),W) - g(\nabla_{[X,Y]}(Z),W) + g(\nabla_X \circ \nabla_Y(W),Z) - g(\nabla_Y\circ \nabla_X(W),Z) - g(\nabla_{[X,Y]}(W),Z)

To prove that this is zero, it thus suffices to show that:

g(\nabla_{[X,Y]}(Z),W) + g(\nabla_{[X,Y]}(W),Z) = g(\nabla_X \circ \nabla_Y(Z),W) + g(\nabla_X \circ \nabla_Y(W),Z) - g(\nabla_Y \circ \nabla_X(Z),W) - g(\nabla_Y \circ \nabla_X(W), Z) \qquad (\dagger).

We now show \dagger. Since g is a metric connection, the left side simplifies to:

g(\nabla_{[X,Y]}(Z),W) + g(\nabla_{[X,Y]}(W),Z) = [X,Y]g(Z,W) = XYg(Z,W) - YXg(Z,W) \qquad (\dagger\dagger).

Simplifying each of the two terms on the right side of (\dagger\dagger), we get:

XYg(Z,W) = Xg(\nabla_Y(Z),W) + Xg(Z,\nabla_Y(W)) = g(\nabla_X \circ \nabla_Y(Z),W) + g(\nabla_Y(Z),\nabla_X(W)) + g(Z,\nabla_X \circ \nabla_Y(W)) + g(\nabla_X(Z),\nabla_Y(W)) \qquad (1).

And:

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).

Substituting (1) and (2) in (\dagger\dagger) yields (\dagger).