# Deriving Riemann curvature tensor in local coordinates [closed]

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?

• you incorrectly expanded $\partial_{i}\left(\nabla_{j}Z^{k}\right)$ to omit the terms with derivatives of the christoffel symbols. I'm unclear if there are more errors. Mar 26, 2021 at 7:31
• But I did not expand them at all. They cancel each other out due to $\nabla_{[X,Y]} Z$. Mar 26, 2021 at 7:44
• Why would you leave out terms that you think might cancel if you're not completely sure your calculcations are right? Mar 26, 2021 at 8:46
• I am not sure whether I understand what you mean. I did not expand them in the sense that I did not go further than $\partial_i ( \nabla_j Z^{\ell} )$ because the same term occurs within $\nabla_{[X,Y]}$ - but since $R(...) = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z$ they indeed cancel. Mar 26, 2021 at 8:56
• @Octavius: the same term does not occur. $X^{i}Y^{j}\partial_{i}\nabla_{j}Z^{k}\neq X^{j}Y^{i}\partial_{i}\nabla_{j}Z^{k}$ Mar 26, 2021 at 14:28

## 1 Answer

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.

• Thank you for your answer. Are the Lie-Brackets $[X,Y]^j (f)$ to be understood as $X^i (\partial_i Y^j \big) f - Y^i \big( \partial_i X^j \big) f$ or as $X^i \partial_i \big( Y^j f \big) - Y^i \partial_i \big( X^j f \big)$? Because in the latter case your terms do not add up, since $[X,Y]^j \nabla_j Z^{\ell}$ then contains a term $\propto \partial_i \Gamma^{\ell}_{jm}$ Mar 26, 2021 at 9:26
• $[X,Y]^j f=X^i(\partial_i Y^j)f-Y^i(\partial_i X^j)f$, but you could have $[X,Y]f=[X,Y]^j\partial_j f=\big(X^i(\partial_i Y^j)-Y^i(\partial_i X^j)\big)\partial_j f$. The first one is just multiplying a component by a function, the second one is applying the $[X,Y]$ vector to a function.
– AFG
Mar 26, 2021 at 9:40
• This confuses me so much right now. I thought applying $[X,Y]$ to a function $f$ is $$[X,Y](f) = X^i \partial_i ( Y^j \partial_j f ) - Y^i \partial_i ( X^j \partial_j f )$$ i.e. $$[X,Y](f) = X\big(Y(f)\big) - Y \big( X (f) \big)$$ Mar 26, 2021 at 9:55
• It is equivalent, you can see it if you expand the derivatives.
– AFG
Mar 26, 2021 at 10:00
• I see. Thank you so so much for your answer, you helped me a lot! Mar 26, 2021 at 10:05