Interpretation of Ricci rotation coefficients in tetrad formalism Given an orthonormal frame (tetrad in 4 dimensions, vielbein more generally) $\{(e_{\mu})^{a}\}$ with $g(e_{\mu}, e_{\nu}) = \eta_{\mu\nu}$, the Ricci rotation coefficients are defined as $$\omega_{\lambda\mu\nu} = (e_{\lambda})^{a}(e_{\mu})^{b}\nabla_{a}(e_{\nu})_{b}$$ and you can extract information about the curvature of the spacetime in question if you have all the relevant Ricci rotation coefficients. My question is: what is, exactly, the geometrical/physical interpretation of the Ricci rotation coefficients? In a vague sense, I can see that they account for "non-inertial" effects that a local Lorentz observer would detect if them chose this respective vielbein as their frame, which in turn would give information about the local curvature, but that is very hand-wavy; I wanted something more illuminating. So is there any intuitive sense I can give to these coefficients?
 A: I think what you said in the comments is right. I don't know if what follows give you more intuition. In your notation, $\big\{e_{\nu} \, : \, \nu = 1...n \big\}$ is a frame of $n$ linearly independent vector fields $$e_{\nu}(x) = \big(\,e_{\nu}(x)\,\big)^b \,\frac{\partial}{\partial x^b}$$ that determine the frame. The part 
$$(e_{\lambda})^a \nabla_a \, (e_{\nu})_b$$ should more accurately be written as
$$\Big((e_{\lambda})^a \nabla_a \, e_{\nu} \Big)_b$$
and so 
$$\Big((e_{\lambda})^a \nabla_a \, e_{\nu} \Big)^b \frac{\partial}{\partial x^b}= \nabla_{e_{\lambda}} \, e_{\nu}$$
is the covariant derivative of the frame vector field $e_{\nu}$ in the direction of the frame vector $e_{\lambda}$. The derivative  $\nabla_{e_{\lambda}} \, e_{\nu}$ is a vector that describes how the frame vector $e_{\nu}$ deviates from its parallel transport when moved slightly in the direction of the frame vector $e_{\lambda}$. In other words, it measures the "rotation" of vector $e_{\nu}$ when moved along vector $e_{\lambda}$. Then you take the "projections" of these deviation vectors $\nabla_{e_{\lambda}} \, e_{\nu}$ onto the axes of the frame
$$\Big(e_{\mu}\circ \big(\,\nabla_{e_{\lambda}} \, e_{\nu}\,\big)\Big) = 
(e_{\mu})^b \, g_{bs} \, \Big((e_{\lambda})^a \nabla_a \, e_{\nu} \Big)^s = (e_{\mu})^b \Big((e_{\lambda})^a \nabla_a \, e_{\nu} \Big)_b = \omega_{\lambda \mu \nu}$$ where $g_{bs}$ is the Riemannian (or Lorentz) metric. 
So the Ricci rotation coefficients are (related to) the coordinates of the derivative vectors that measure the deviation (rotation) of all the frame vectors when moved in various directions. Observe that the Ricci coefficients carry information about how the frame twists and rotates (so how non-inertial it is) and not just information about the curvature of space-time. Even in a flat space-time the coordinates may be non-linear, and hence non-inertial, so the Ricci rotation coefficients may still be non-zero. The point is that in flat space-time there is a system of coordinates in which they can all be made zero. In curved space-time that's not the case. 
Remark: Christoffel symbols encode how the differentiation of an object on a curved space (a space with non-trivial geometry) in the direction of a tangent vector differs from the usual differentiation of the object in the direction of the vector in flat space (trivial geometry).      
