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 $\nabla$ 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:

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

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 Failed to parse (unknown function "\tag"): {\displaystyle \tag{\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 $(\dagger )$.