Ricci tensor from Riemann tensor I have been studying differential geometry and general relativity, and I have a question about the Ricci tensor.
So as I understand, to set things up: one defines the Riemann curvature operator as $R: (U,V,W) \rightarrow R(U,V)W$ mapping three vector fields to a fourth one:
$$ R(U,V)W = \nabla_{U}\nabla_{V}W - \nabla_{V}\nabla_{U}W - \nabla_{[U,V]}W $$
with $[U,V]$ being the commutator. This has a nice geometrical interpretation as a measure of the noncommutativity of successive parallel transport along directions $U$ and $V$, versus $V$ and $U$. Then, in some coordinate system one computes:
$$R(\partial_{\alpha},\partial_{\beta})\partial_{\gamma} = \nabla_{\partial_{\alpha}}\nabla_{\partial_{\beta}}\partial_{\gamma} - \nabla_{\partial_{\beta}}\nabla_{\partial_{\alpha}}\partial_{\gamma} $$
where the commutator term disappears; this has a nice expression in terms of the Christoffel symbols, and then $ R^{\delta}_{\gamma \alpha \beta} = dx^{\delta} \left(R(\partial_{\alpha},\partial_{\beta})\partial_{\gamma}\right) $ is the Riemann tensor in component notation. Everything is wonderful and beautiful thus far.
And so now, enter the Ricci tensor. I know it is defined as the contraction of the Riemann tensor:
$$ R_{\mu \nu} = R^{\delta}_{\mu \delta \nu} $$
My questions are the following:
1) What is the geometrical motivation behind this particular contraction of the Riemann tensor? (we are contracting in the index corresponding to the first "test direction" used in the Riemann curvature operator, is there a purpose for this choice?) 
2) What is the geometrical interpretation of the Ricci tensor? Is there a coordinate-free way of defining it as a curvature operator, like the Riemann tensor?
3) Why contract the second covariant index and not another one? Is there a reason for this, or is it just a convention?
Thanks in advance!
 A: A nice answer for second question is in this post. for first ans third, A Lemma (page 124) of Riemannian geometry: An Introduction to curvature written by John M. Lee states that
Lemma 7.6: The Ricci curvature is a symmetric $2$-tensor field. It can be
expressed in any of the following ways:
$$R_{ij}=R_{kij}{\ }^k=R_{ik}^{\;\;k}{\ }_j=-R_{ki}^{\;\;\ k}{}_j=-R_{ikj}^{\;\;\;\;k}.$$
I hope this would be helpful.
A: 1) Sure. You contract with the test direction and one of the directions of parallel transport. You could use the other and introduce a minus sign. If you contract with respect to the $W$ direction then you lose meaning. That is to say: I don't think that contraction is interesting.
2) Ricci tensor represents the change in Volume of a geodesic ball. Say, $\textrm{Ric}(X,W)$ is the change in volume along $W$ as $X$ changes. So You can write it in coordinate free notation but with the use of the trace map notation. (Think of the contraction as adding over directions, to make a volume. This is a crude way to think of it but it is how I always picture it in my head.)
3) Um. I think I answered this with 1). It is convention.
Does that help?
Edit: You can see 2) easily if you expand the lowest terms of the volume element given small deviations from a flat metric.
A: The discussion is old but short and I would like to add a precision about 3). I suppose there you can choose either. They give opposite Ricci tensors. The convention that you mention has the advantage that when you contract further Ricci to get scalar curvature, this gives positive curvature to the sphere.
More: Someone wrote on Wikipedia that, although there are two opposite convention around for the Riemannian tensor, people seem to agree on the Ricci. I noted that in Arthur L. Besse's book Einstein manifold the Riemann is the opposite of yours and they contract the other index of it so that they indeed get the same Ricci and the same scalar.
(Of course you could decide to take the other Ricci and define scalar to be the opposite of the contracted Ricci. Or you could adopt a convention where the sphere has negative curvature. I do not recommend taking that path.)
