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
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.