It turns out that the curvature tensor associated to the connection from a Riemannian pseudo-metric ${g}$ has to satisfy certain conditions.  (As usual, we denote by $\nabla$ the Levi-Civita connection associated to $g$, and we assume the ground manifold is smooth.)

First of all, we have skew-symmetry

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

This is immediate from the definition.

Next, we have another variant of skew-symmetry:

Proposition 1 $\displaystyle g( R(X,Y) Z, W) = -g( R(X,Y) W, Z)$

This is equivalent to

$\displaystyle g(R(X,Y)Z,Z) = 0$

for vector fields ${X,Y,Z}$ (and ${W}$). If ${X,Y}$ commute, which we assume without loss of generality, then

$\displaystyle g(R(X,Y)Z,Z) = g( \nabla_X \nabla_Y Z,Z) - g(\nabla_Y \nabla_X Z,Z).$

Now we can use the connection derivative identity twice to get something like ${g(\nabla_X \nabla_Y Z,Z)}$, which is as follows.

$\displaystyle Y(X g(Z,Z)) =2 Y g(\nabla_X Z, Z) = 2g(\nabla_Y \nabla_X Z,Z) + 2g(\nabla_XZ, \nabla_YZ).$

We thus get an expression of ${g(\nabla_Y \nabla_X Z,Z)}$ as the difference of two quantities symmetric in ${X,Y}$, hence so is ${g(\nabla_Y \nabla_X Z,Z)}$. This implies that ${ g(R(X,Y)Z,Z) = 0}$.

The next is the algebraic Bianchi identity:

Theorem 2 (First Bianchi Identity) The symmetrization fo ${R(X,Y)Z}$ is ${0}$, i.e.$\displaystyle R(X,Y)Z + R(Z,X) Y + R(Y,Z) X = 0.\ \ \ \ \ (1)$

I guess this is Stigler’s law of eponymy again—Ricci discovered (1). The three quantities are tensors, so to check that the sum is zero, we need only evaluate on vector fields ${X,Y,Z}$ that commute with each other. In that case,

$\displaystyle \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z + \nabla_Z \nabla_X Y - \nabla_X \nabla_Z Y + \nabla_Y \nabla_Z X - \nabla_Z \nabla_Y X .$

This is a mess, but we can simplify it by reordering the terms into three pieces:

$\displaystyle \nabla_X( \nabla_Y Z - \nabla_Z Y)$

$\displaystyle \nabla_Y( \nabla_Z X - \nabla_X Z )$

$\displaystyle \nabla_Z( \nabla_X Y - \nabla_Y Z)$

Each of these terms vanishes, because ${X,Y,Z}$ commute and the connection ${\nabla}$ is symmetric. The Bianchi identity can be generalized to arbitrary connections, but they become much more cumbersome and messy. What we have shown is that (1) is true for an arbitrary symmetric connection.