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 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(XY(Z)YX(Z)[X,Y](Z),W)g(XY(W)YX(W)[X,Y](W),Z)

By the bilinearity of g, this simplifies to:

g(XY(Z),W)g(YX(Z),W)g([X,Y](Z),W)+g(XY(W),Z)g(YX(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(XY(Z),W)+g(XY(W),Z)g(YX(Z),W)g(YX(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(XY(Z),W)+g(Y(Z),X(W))+g(Z,XY(W))+g(X(Z),Y(W))(1).

And:

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

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