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

From Diffgeom

(Created page with '==Statement== Suppose <math>M</math> is a differential manifold and <math>g</math> is a fact about::Riemannian metric or fact about::pseudo-Riemannian metric and <ma…') |
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Line 7: | Line 7: | ||

Then: | Then: | ||

− | <math>\! R(X,Y,Z,W) = R(X,Y,W,Z)</math>. | + | <math>\! R(X,Y,Z,W) = -R(X,Y,W,Z)</math>. |

− | == | + | ==Proof== |

+ | |||

+ | We consider the expression <math>R(X,Y,Z,W) + R(X,Y,W,Z)</math>: | ||

+ | |||

+ | <math>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)</math> | ||

+ | |||

+ | By the bilinearity of <math>g</math>, this simplifies to: | ||

− | + | <math>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)</math> | |

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

− | == | + | <math>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)</math>. |

+ | |||

+ | We now show <math>\dagger</math>. Since <math>g</math> is a [[metric connection]], the left side simplifies to: | ||

+ | |||

+ | <math>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)</math>. | ||

+ | |||

+ | Simplifying each of the two terms on the right side of <math>\tag{\dagger\dagger}</math>, we get: | ||

+ | |||

+ | <math>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)</math>. | ||

+ | |||

+ | And: | ||

+ | |||

+ | <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>. |

## Revision as of 01:50, 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 **Failed to parse (unknown function "\tag"): \tag{\dagger\dagger}**
, we get:

.

And:

.

Substituting (1) and (2) in yields .