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?
Just an additional comment:
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.