0
$\begingroup$

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?

$\endgroup$
6
  • $\begingroup$ What then is your definition of the Ricci tensor? $\endgroup$ Commented Jun 25, 2023 at 5:29
  • $\begingroup$ 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. $\endgroup$
    – Plantation
    Commented Jun 25, 2023 at 5:44
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Jun 25, 2023 at 5:59
  • $\begingroup$ 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. $\endgroup$
    – Plantation
    Commented Jun 25, 2023 at 6:19
  • 1
    $\begingroup$ Summing on $k$ in $\sum R^k_{k11}$ is the trace. $\endgroup$ Commented Jun 25, 2023 at 6:26

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .