Curvature is antisymmetric in last two variables

Statement

Suppose $M$ is a differential manifold and $g$ is a Riemannian metric or pseudo-Riemannian metric and $\mathrm{\nabla }$ is the Levi-Civita connection for $g$. Consider the Riemann curvature tensor $R$ of $\mathrm{\nabla }$. 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:

$R(X,Y,Z,W)=-R(X,Y,W,Z)$.

Proof

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

$g({\mathrm{\nabla }}_X\circ {\mathrm{\nabla }}_Y(Z)-{\mathrm{\nabla }}_Y\circ {\mathrm{\nabla }}_X(Z)-{\mathrm{\nabla }}_{[X,Y]}(Z),W)-g({\mathrm{\nabla }}_X\circ {\mathrm{\nabla }}_Y(W)-{\mathrm{\nabla }}_Y\circ {\mathrm{\nabla }}_X(W)-{\mathrm{\nabla }}_{[X,Y]}(W),Z)$

By the bilinearity of $g$, this simplifies to:

$g({\mathrm{\nabla }}_X\circ {\mathrm{\nabla }}_Y(Z),W)-g({\mathrm{\nabla }}_Y\circ {\mathrm{\nabla }}_X(Z),W)-g({\mathrm{\nabla }}_{[X,Y]}(Z),W)+g({\mathrm{\nabla }}_X\circ {\mathrm{\nabla }}_Y(W),Z)-g({\mathrm{\nabla }}_Y\circ {\mathrm{\nabla }}_X(W),Z)-g({\mathrm{\nabla }}_{[X,Y]}(W),Z)$

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

$g({\mathrm{\nabla }}_{[X,Y]}(Z),W)+g({\mathrm{\nabla }}_{[X,Y]}(W),Z)=g({\mathrm{\nabla }}_X\circ {\mathrm{\nabla }}_Y(Z),W)+g({\mathrm{\nabla }}_X\circ {\mathrm{\nabla }}_Y(W),Z)-g({\mathrm{\nabla }}_Y\circ {\mathrm{\nabla }}_X(Z),W)-g({\mathrm{\nabla }}_Y\circ {\mathrm{\nabla }}_X(W),Z)(†)$.

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

$g({\mathrm{\nabla }}_{[X,Y]}(Z),W)+g({\mathrm{\nabla }}_{[X,Y]}(W),Z)=[X,Y]g(Z,W)=XYg(Z,W)-YXg(Z,W)(††)$.

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

$XYg(Z,W)=Xg({\mathrm{\nabla }}_Y(Z),W)+Xg(Z,{\mathrm{\nabla }}_Y(W))=g({\mathrm{\nabla }}_X\circ {\mathrm{\nabla }}_Y(Z),W)+g({\mathrm{\nabla }}_Y(Z),{\mathrm{\nabla }}_X(W))+g(Z,{\mathrm{\nabla }}_X\circ {\mathrm{\nabla }}_Y(W))+g({\mathrm{\nabla }}_X(Z),{\mathrm{\nabla }}_Y(W))(1)$.

And:

$YXg(Z,W)=Yg({\mathrm{\nabla }}_X(Z),W)+Yg(Z,nabl{a}_X(W))=g({\mathrm{\nabla }}_Y\circ {\mathrm{\nabla }}_X(Z),W)+g({\mathrm{\nabla }}_X(Z),{\mathrm{\nabla }}_Y(W))+g({\mathrm{\nabla }}_Y(Z),{\mathrm{\nabla }}_X(W))+g(Z,{\mathrm{\nabla }}_Y\circ {\mathrm{\nabla }}_X(W))(2)$.

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