# Calculating Ricci Curvature $Rc_p(v,v)$

This question originates from proof of Proposition 8.32 of the John Lee's Introduction to Riemannian Manifolds book. It seems easy calculation but I don't understand more rigorously.

Let $$(M,g)$$ be a Riemannian $$n$$-manifold and $$p\in M$$. Let $$v \in T_pM$$ be a unit vector. Let $$(b_1, \dots , b_n)$$ be any orthonormal basis for $$T_pM$$ with $$b_1 =v$$. Then, why $$Rc_p(v,v) = R_{11}(p) = R^{k}_{k11}(p) = \Sigma_{k=1}^{n}Rm_p(b_k, b_1, b_1, b_k)$$

? I think that I am unfamiler to the definition of Ricci ( Riemannian ) curvature tensor-and its relation to component- Can any one give more detailed explanation?

• What then is your definition of the Ricci tensor? Commented Jun 25, 2023 at 5:29
• Uhm.. for vector fields $X,Y$, $Rc(X,Y) = \operatorname{tr}(Z \mapsto R(Z,X)Y)$. And I don't know how to cennect this abstract definition and the concrete calculation involving component of $Rc$ above. Commented Jun 25, 2023 at 5:44
• Take $X=Y=v=b_1$ and take $Z=b_k$, $k=1,\dots,n$. You still need to convert from the $(3,1)$-tensor definition you just gave me of curvature to the $(4,0)$-tensor form, which Lee is using here. Commented Jun 25, 2023 at 5:59
• Thank you.. And I still don't get it completely :) I don't know how to deal with the trace in the definition of $Rc$. I think that I am missing some key point. Commented Jun 25, 2023 at 6:19
• Summing on $k$ in $\sum R^k_{k11}$ is the trace. Commented Jun 25, 2023 at 6:26

Say, $$\mathbf{T}$$ is an arbitrary $$(1,3)$$-tensor. Its components are by definion the functions on $$M$$ which allow to express the tensor as linear combination $$\mathbf{T}={T^i}_{j\,k\,\ell}\;\partial_i\otimes dx^j\otimes dx^k\otimes dx^\ell\$$ (use summation over all indices).

From the fact that basis vector fields $$\partial_i$$ and basis one-forms $$dx^j$$ are dual, $$\partial_i(dx^j)={\delta_i}^j\,,\quad dx^j(\partial_i)= {\delta^j}_i\,,$$ it is obvious how to get the components: let $$\mathbf T$$ eat $$dx^i,\partial_j,\partial_k,\partial_\ell$$ and you have $${T^i}_{j\,k\,\ell}\,.$$

For simplicity I explained this in a coordinate basis but it works with your basis $$b_1,\dots,b_n$$ instead of $$\partial_1,\dots,\partial_n$$ as well. Just take the corresponding basis that is dual to $$b_1,\dots,b_n\,.$$ Of course the components in this basis are different but they are extracted in exactly the same way.