Can the Ricci tensor twist relative to the Levi-Cevita connection? I know that in a Riemannian manifold, torsion is an artifact of a connection, and we can always choose the Levi-Cevita connection to eliminate torsion.  So in that sense, the geodesics of a manifold cannot "twist around each other" in any fundamental way.
But in 3 or more dimensions, the Ricci tensor can have different values in different directions.  Let's say, in the coordinate system where $R_{\mu\nu}$ is diagonal, that $R_{xx}$ is positive while $R_{yy}$ is negative.
At some point, take a vector that points along the direction of the positive curvature.  As you move in the z-direction from that point, the vector will be transported according to the Levi-Cevita connection.
But, is it possible that the actual direction of the positive curvature along that same path does not stay in line with the transported vector?  That would mean that the Ricci tensor is effectively twisting in the truest possible sense, that is, relative to the Levi-Cevita connection.
Is there an example of such a manifold?  Or at least an explanation as to why it should be possible?
 A: It seems to me that your question boils down to asking if the eigenvector fields of the Ricci tensor are parralel, i.e. if $V$ is a (local) vector field such that $g^{ij}R_{jk}V^k=\lambda V^i$ for some fixed $\lambda\in\mathbb{R}$ (such fields generically exist locally), then $V$ is parallel w.r.t. the Levy-Civita connection.
The answer is in general no, since many Riemannian manifolds admit no parallel vector fields, even locally. Put another way, a "generic" Riemannian manifold has local holonomy $SO(N)$, so if $R_{ij}$ has two distinct eigenvectors, it will generically be possible to transform one into the other via parallel transport.
Unfortunately a simple concrete example is hard to come by, since the Ricci tensor has no interesting behavior until dimension 3, and even then only in particularly asymmetric cases. the family of submanifolds of Euclidean space
$$
M=\left\{(x,y,z,w)\in\mathbb{E}^4:w=\frac{a}{2}x^2+\frac{b}{2}y^2+\frac{c}{2}z^2\right\}
$$
with $a,b,c\in\mathbb{R}$ such that $ab,ac,bc$ are all distinct, should illustrate the relevent behavior at $0$. I haven't done all the computations, but if I'm not mistaken, the Ricci tensor will have three distinct eigenvalues at $0$ and and parallel transportation can take any eigenvector to any other.
