# 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

• For anyone else stumbling upon this question: The statement which "is easy to verify", $Q'(X, Y)= \langle Q(X), Y \rangle$, is only true for Riemannian metrics and crucially uses the fact that $\partial_i$ is an orthonormal basis (ONB) with respect to said metric. Moreover, the definition of the scalar curvature $S$ as $\mathop{tr} Q(X)$ only agrees with the usual one $S = R^i_i$ since we're using an ONB and thus $Q(X) = \sum_i \mathop{Ric}_{ii} = \mathop{Ric}^i_i$. (So the entire construction is a bit pointless because the contractions/traces with respect to the metric all become trivial.) – balu Jan 29 at 10:52

## 2 Answers

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 $\nabla_{\partial_a}\partial_b=\Gamma^c_{ab}\partial_c$).

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.