Let $M$ be a smooth, pseudo-Riemannian manifold with $\dim(M) \ge 2.$ Let $\nabla$ be *any affine connection* on $M$. No reason for it to be the Levi-Civita connection. All we assume is that it has zero torsion.

Given two smooth vector fields $X,Y \in \mathfrak{X}(M),$ The curvature tensor, with respect to $\nabla$, is given by

$R(X,Y) := \nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]},$

where $R(X,Y) : \mathfrak{X}(M) \to \mathfrak{X}(M).$ The Ricci curvature tensor is given by the trace:

$ \mbox{Ric}(Y,Z) := \mbox{trace} \left[ X \mapsto R(X,Y)Z \right].$

I've read that the Ricci curvature tensor measures the second order deviation between the volume of a $\nabla$-geodesic ball and a standard Euclidean geodesic ball. This explanation causes me problems. A geodesic ball, centre $x \in M$ and radius $r$ is given by following each $\nabla$-geodesic, that passes through $x$, a distance $r$ with respect to the pseudo-Riemannian metric on $M$. Besides $\nabla$, this depends only on $x \in M$ and $r \ge 0$.

The volume element is expressed in terms of the *symmetric* bilinear form $h$ that is the pseudo-Riemannian metric. We have:

$\mbox{Vol}_h(X_1,\ldots,X_n) := \sqrt{|\det\left( h_{i,j} \right)|}, \ \mbox{ where } \ h_{i,j} := h(X_i,X_j).$

Again, recall that $\nabla$ *need not be the Levi-Civita connection* on the pseudo-Riemannian manifold $M$. In other words $\nabla h$ need not be identically zero.

I know how to manipulate the tensor and connection notation. But my geometrical insight is lacking. I don't see how the symmetry, or non-symmetry, of $\mbox{Ric}$ should have any relation to the volume of a ball, which is determined with respect to a *symmetric* bilinear form.

I would appreciate some information and some references as to how to improve my geometrical intuition.