Deriving Riemann curvature tensor in local coordinates [closed]

Question

The Riemann curvature tensor is

$$ R(X,Y,Z) \enspace = \enspace \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z $$

I want to calculate its representation in local coordinates, i.e.

$$ R^{\ell}{}_{kij} \enspace = \enspace \partial_i \Gamma^{\ell}_{kj} - \partial_j \Gamma^{\ell}_{ki} + \Gamma^m_{kj} \, \Gamma^{\ell}_{mi} - \Gamma^{m}_{ki} \, \Gamma^{\ell}_{mj}$$

So i thought I just plug in my vectors into the equation above. I have

$$ \begin{align} \nabla_X \nabla_Y Z^{\ell} \enspace &= \enspace X^i \, \nabla_i Y^j \, \nabla_j Z^{\ell} \\ &= \enspace X^{i} \, (\partial_i Y^j)(\nabla_j Z^{\ell}) + X^i Y^j \, \partial_i (\nabla_jZ^{\ell}) + X^i Y^j \, \Gamma^{\ell}_{im} \nabla_j Z^m \end{align} $$

As well as

$$ \begin{align} \nabla_Y \nabla_X Z^{\ell} \enspace &= \enspace Y^i \nabla_i X^j \nabla_j Z^{\ell} \\ &= \enspace Y^{i} \, (\partial_i X^j)(\nabla_j Z^{\ell}) + Y^i X^j \, \partial_i (\nabla_j Z^{\ell}) + Y^i X^j \, \Gamma^{\ell}_{im} \nabla_j Z^m \end{align} $$

and

$$ \begin{align} \nabla_{[X,Y]} \enspace &= \enspace \nabla_X \nabla_Y Z^{\ell} - X^i Y^j \, \Gamma^{\ell}_{im} \nabla_j Z^m - \Big( \nabla_Y \nabla_X Z^{\ell} - Y^i X^j \, \Gamma^{\ell}_{im} \nabla_j Z^m \Big) \end{align} $$

So I am left with

$$R(X,Y,Z)^{\ell} \enspace = \enspace X^i Y^j Z^k \Big( \Gamma^{\ell}_{im} \, \Gamma^{m}_{jk} - \Gamma^{\ell}_{jm} \, \Gamma^m_{ik} \Big) + X^i Y^j \Big( \Gamma^{\ell}_{im} \, \partial_j Z^m - \Gamma^{\ell}_{jm} \, \partial_i Z^m \Big)$$

What is the next step to do here? How do I get rid of the $\partial_j Z^m$ and $\partial_i Z^m$ and get some $\partial \Gamma$ instead, such that I am finished?

Answer 1

The correct expansions are

$$ \nabla_X \nabla_Y Z^{\ell} \enspace = X^i \, \nabla_i \Big(Y^j \, \nabla_j Z^{\ell}\Big) = X^i \, \nabla_i \Big(Y^j ( \partial_j Z^{\ell}+\Gamma_{jm}^\ell Z^m)\Big)=\\=X^i\partial_i\Big(Y^j( \partial_j Z^{\ell}+\Gamma_{jm}^\ell Z^m)\Big)+X^iY^j\,\Gamma_{in}^\ell( \partial_j Z^n+\Gamma_{jm}^n Z^m)=\\=X^i\partial_iY^j( \partial_j Z^{\ell}+\Gamma_{jm}^\ell Z^m)+X^iY^j\Big(\partial_i\partial_jZ^\ell +Z^m\partial_i\Gamma_{jm}^\ell+\Gamma_{jm}^\ell\partial_i Z^m+\Gamma_{in}^\ell \partial_j Z^n+\Gamma_{in}^\ell\Gamma_{jm}^n Z^m\Big), $$

similarly, you get

$$ \nabla_Y \nabla_X Z^{\ell} \enspace = Y^j \, \nabla_j \Big(X^i \, \nabla_i Z^{\ell}\Big) = Y^j \, \nabla_j \Big(X^i ( \partial_i Z^{\ell}+\Gamma_{im}^\ell Z^m)\Big)=\\=Y^j\partial_j\Big(X^i( \partial_i Z^{\ell}+\Gamma_{im}^\ell Z^m)\Big)+Y^jX^i\,\Gamma_{jn}^\ell( \partial_i Z^n+\Gamma_{im}^n Z^m)=\\=Y^j\partial_jX^i( \partial_i Z^{\ell}+\Gamma_{im}^\ell Z^m)+Y^jX^i\Big(\partial_j\partial_iZ^\ell +Z^m\partial_j\Gamma_{im}^\ell+\Gamma_{im}^\ell\partial_j Z^m+\Gamma_{jn}^\ell \partial_i Z^n+\Gamma_{jn}^\ell\Gamma_{im}^n Z^m\Big). $$

If you now subtract them, you are left with

$$\nabla_X \nabla_Y Z^{\ell} \enspace-\nabla_Y \nabla_X Z^{\ell} \enspace=\\=(X^i\partial_iY^j-Y^i\partial_iX^j)( \partial_j Z^{\ell}+\Gamma_{jm}^\ell Z^m)+X^iY^jZ^m\Big(\partial_i\Gamma_{jm}^\ell-\partial_j\Gamma_{im}^\ell+\Gamma_{in}^\ell\Gamma_{jm}^n-\Gamma_{jn}^\ell\Gamma_{im}^n\Big)=\\=[X,Y]^j\nabla_jZ^\ell+X^iY^jZ^m\Big(\partial_i\Gamma_{jm}^\ell-\partial_j\Gamma_{im}^\ell+\Gamma_{in}^\ell\Gamma_{jm}^n-\Gamma_{jn}^\ell\Gamma_{im}^n\Big).$$

Now, the term $\nabla_{[X,Y]}Z$ is just

$$\nabla_{[X,Y]}Z^\ell=[X,Y]^j\nabla_jZ^\ell,$$

so if you subtract it too, you obtain the right expression.