Ricci Tensor, Curvature and Scalar Curvature computation from definition I am studying a little of Riemannian geometry and I am having some problem in making the connection between two expressions of Ricci tensor, curvature and scalar curvature.
Well, in the book that I am studying I find that:
Let $L_{X,Y} : T_xM \to T_xM$, we define the Ricci tensor like:
$$Q'(X,Y) = \operatorname{trace} L_{X,Y}.$$
Where $L_{X,Y} = R(\cdot,X)Y$, where $R(Z,X)Y$  is the Riemann tensor.
Thats way, we define the Ricci curvature as $\operatorname{Ricci}(X) = Q'(X,X)$.
So, if we define $$Q(X) = \sum_i R(X,\partial_i)\partial_i$$ where $\{\partial_i\}$ is one orthonormal basis, it is easy to verify that $$Q'(X,Y) = \langle Q(X), Y \rangle,$$ for any $X,Y \in T_xM$.
In particular, if we use $X = \partial_j$ and compute $$Q'(X,X) = \operatorname{Ricci}(X) = \sum_i K(\partial_j, \partial_i),$$ that is, the sum of all secctional curvatures,
it is easy to see that if we denote $S$ the scalar curvature as $S = \operatorname{trace} Q(X)$, we find that $$S = \sum_i \operatorname{Ricci}(\partial_i).$$
My question is, how do I obtain the expression using Christoffel symbols from these definitions? What is the connection between these expressions and the expression in coordinates?
Thanks a lot for the help.
Cordially,
Leonardo
 A: Using the definition of the Riemann curvature tensor
$$R_{XY}Z=\nabla_{[X,Y]}Z-[\nabla_X,\nabla_Y]Z,$$
we can compute that
$$R^a_{\;bcd}=\partial_d\Gamma^a_{cb}-\partial_c\Gamma^a_{db}+\Gamma^a_{de}\Gamma^e_{bc}-\Gamma^a_{ce}\Gamma^e_{bd},$$
(note that my convention is $R_{\partial_c\partial_d}\partial_b=R^a_{\;bcd}\partial_a$).  Now using our definitions of the Ricci curvature
$$R_{ab}=R^c_{\;acb}$$
and of the scalar curvature
$$R=R^a_{\;a},$$
you would just substitute and calculate.  
I'm not sure I understand your second question, since anything expressed in terms of Christoffel symbols is also an expression in coordinates (recall that $\nabla_{\partial_a}\partial_b=\Gamma^c_{ab}\partial_c$).
A: These are all expressed in terms of the Riemann curvature tensor, the latter is expressed in terms of the connection, and the latter in terms of the $\Gamma$ symbols, so in principle it's only a calculation. In practice, one obtains rather complicated expressions even in the 2-dimensional case. In my class notes here I derive it from commutation of partials for imbedded surfaces. It's not a simple formula but it does prove theorema egregium.
