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In what follows I always use Einstein summation convention.

The Riemann curvature is defined as $$ R(X,Y)Z = \nabla_{X}\nabla_{Y}Z - \nabla_{Y}\nabla_{X}Z - \nabla_{[X,Y]}Z $$ Now, I want to calculate the $k$-th component. The first and second terms of the above equation are quite clear to me. For $X=X^i\partial_i$, $Y=Y^j\partial_j$ and $Z=Z^k\partial_k$ ($\partial_k \equiv \partial/\partial x^k$ in a local coordinate system) For example

$$\nabla_{X}\nabla_{Y}Z = \nabla_X ( Y^i\nabla_{\partial_j}Z^k) = X^i\nabla_{\partial_i}( Y^i\nabla_{\partial_j}Z^k)= X^i\partial_iY^j\nabla_{\partial_j}Z^k + X^iY^j\nabla_{\partial_i}\nabla_{\partial_j}Z^k $$ and similarly for the middle term of the definition of the curvature. My problem is in the last term, where the Lie bracket arises. In that term we have $$ \nabla_{[X,Y]}Z^k = \nabla_{XY}Z^k - \nabla_{YX}Z^k $$ if I am not mistaken. Then, let us take for example the left term:

$$ \nabla_{XY}Z = X^i\nabla_{\partial_i Y^j \partial_i}Z^k = X^i\partial_iY^j \nabla_j Z^k + X^iY^j \nabla_{\partial_i \partial_j}Z^k $$ but here I am almost sure something is wrong! I do not understand what the $\nabla_{\partial_i \partial_j}$ operator is and how to continue. What is my mistake here?

In the end I just want to prove that $$R(X,Y)Z^k = X^iY^jZ^m( \partial_i \Gamma_{jm}^k - \partial_j \Gamma_{im}^k + \Gamma_{im}^k \Gamma_{jm}^l - \Gamma_{jl}^k \Gamma_{im}^l ) $$

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  • $\begingroup$ Do you know that $R$ is a tensor? $\endgroup$
    – Holonomia
    Jun 21, 2015 at 11:00
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    $\begingroup$ Note that the expression $\nabla_{XY}Z$ has no meaning, because $XY$ is not a vector field. $\endgroup$
    – Jack Lee
    Jun 21, 2015 at 21:49
  • $\begingroup$ @JackLee, so then how does one properly write out the reduction of $\nabla_[X,Y]?$ $\endgroup$
    – Lemon
    Sep 6, 2018 at 9:58

1 Answer 1

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If you just want to prove the equation in the last line of your post then it is enough to compute $$R(\partial_i,\partial_j) \partial_k \, .$$ Notice that the last term (which bother you) does not appear since the Lie bracket $[\partial_i,\partial_j] = 0$.

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  • $\begingroup$ It seems that you have assumed that $R$ is a tensor. I am not sure if the OP knows about that. $\endgroup$
    – user99914
    Jun 21, 2015 at 10:58
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    $\begingroup$ I already asked the OP about it, thank's! $\endgroup$
    – Holonomia
    Jun 21, 2015 at 11:02

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