# Difference between revisions of "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 $\dagger\dagger$ , 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 $(\dagger)$.