Curvature is antisymmetric in last two variables: Difference between revisions

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


==Related facts==
* [[Curvature is tensorial]]
* [[Curvature is antisymmetric in first two variables]]
* [[Curvature is symmetric in the pairs of first and last two variables]]
==Proof==
==Proof==


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And:
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>.
<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>.

Latest revision as of 01:52, 24 July 2009

Statement

Suppose M is a differential manifold and g is a Riemannian metric or pseudo-Riemannian metric and ∇ is the Levi-Civita connection for g. Consider the Riemann curvature tensor R of ∇. 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(∇X∘∇Y(Z)−∇Y∘∇X(Z)−∇[X,Y](Z),W)−g(∇X∘∇Y(W)−∇Y∘∇X(W)−∇[X,Y](W),Z)

By the bilinearity of g, this simplifies to:

g(∇X∘∇Y(Z),W)−g(∇Y∘∇X(Z),W)−g(∇[X,Y](Z),W)+g(∇X∘∇Y(W),Z)−g(∇Y∘∇X(W),Z)−g(∇[X,Y](W),Z)

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

g(∇[X,Y](Z),W)+g(∇[X,Y](W),Z)=g(∇X∘∇Y(Z),W)+g(∇X∘∇Y(W),Z)−g(∇Y∘∇X(Z),W)−g(∇Y∘∇X(W),Z)(†).

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

g(∇[X,Y](Z),W)+g(∇[X,Y](W),Z)=[X,Y]g(Z,W)=XYg(Z,W)−YXg(Z,W)(††).

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

XYg(Z,W)=Xg(∇Y(Z),W)+Xg(Z,∇Y(W))=g(∇X∘∇Y(Z),W)+g(∇Y(Z),∇X(W))+g(Z,∇X∘∇Y(W))+g(∇X(Z),∇Y(W))(1).

And:

YXg(Z,W)=Yg(∇X(Z),W)+Yg(Z,∇X(W))=g(∇Y∘∇X(Z),W)+g(∇X(Z),∇Y(W))+g(∇Y(Z),∇X(W))+g(Z,∇Y∘∇X(W))(2).

Substituting (1) and (2) in (††) yields (†).