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I'd like to attain the identity in the title. Before we start, let me introduce some definitions. Thank you. The book I currently use is the one written by John M. Lee on Riemannian manifolds.

... The most important such tensor is the Ricci curvature or Ricci tensor, denoted by Rc (or often Ric in the literature), which is the covariant $2$-tensor field defined as the trace of curvature endomorphism on its first and last indices. Thus for vector fields $X,Y$, $$Rc(X,Y)=\mathrm{tr}(\ Z\mapsto R(Z,X)Y\ ).$$ The components of Rc are usually denoted by $R_{ij}$, so that $$R_{ij}=R_{kij}\ ^k=g^{km}R_{kijm}.$$

I'm not worried about the second equality: just substitute $g_{ml}R_{kij}\ ^l$ for the components $R_{kijm}$ of the Riemann curvature tensor given by $$Rm(X,Y,Z,W)=\left<R(X,Y)Z,W\right>_g.$$ We have used $R$ to denote the $(1,3)$-curvature tensor given by $$R(X,Y)Z=\nabla_X\nabla_Y Z-\nabla_Y\nabla_X Z-\nabla_{[X,Y]}Z,$$ and its components can be determined by $R(\partial_i,\partial_j)\partial_k=R_{ijk}\ ^l\partial_l$. Now let us go back to the first equality $R_{ij}=R_{kij}\ ^k$. How could I attain this result? Thank you.

Boar
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    Do you know what is $tr$ and how to calculate that? – Arctic Char Sep 14 '21 at 15:37
  • The operator $\mathrm{tr}$ stands for the action of taking the trace of a $(1,1)$-tenor, but that's all I know about it. – Boar Sep 14 '21 at 15:52
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    My suggestion is to view the first equality as the definition of the Ricci tensor. You then need to verify that it is well-defined, independent of the tangent basis used. That follows by the fact that the trace of a $(1,1)$-tensor is independent of the basis. – Deane Sep 14 '21 at 15:54
  • Definition here. – Arctic Char Sep 14 '21 at 15:56

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Some ideas came into my mind, but I'm not sure if they work. I'd appreciate it if someone could help me proofread. Thank you.

In terms of local coordinates $\{x^i\}_{i=1}^n$, we write $$Rc=R_{ij}dx^i\otimes dx^j.$$ Then $\forall p\in M$, \begin{align} R_{ij}(p)&=Rc_p(\partial_i|_p,\partial_j|_p)\\ &=\textrm{tr}(\ [Z\mapsto R(Z,\partial_i)\partial_j]_p\ ). \end{align} To evaluate the trace, we identify the $(1,1)$-tensor in the parentheses with a linear operator $F$ on $T_p M$ defined by $$F:u^k\partial_k|_p\mapsto u^k R_{kij}\ ^\ell(p)\partial_\ell|_p.$$ Now the task can be done by finding the trace of this operator. And we agree that \begin{align} \text{tr}(F)&=\text{tr}\left( \begin{matrix} R_{1ij}\ ^1(p)&\cdots&R_{nij}\ ^1(p)\\ \vdots&\ddots&\vdots\\ R_{1ij}\ ^n(p)&\cdots&R_{nij}\ ^n(p) \end{matrix} \right)\\ &=\sum_{k=1}^n R_{kij}\ ^k(p)\\ &=R_{kij}\ ^k(p).\tag{summation convention} \end{align} We then assert that $$R_{ij}=R_{kij}\ ^k.$$

Boar
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