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I think I'm misunderstanding how the Riemann tensor acts over general contravariant vectors.

I mean, as it behaves as a tensor, due to multilinear property, it should satisfy:

$$\begin{align} R(X^i\partial_i,Y^j\partial_j)(Z^k\partial_k)&=X^iY^jZ^kR(\partial_i,\partial_j)\partial_k\\ \tag{1}\label{eq:riemann} &=X^iY^jZ^k\left(\partial_i\Gamma^\alpha_{\ kj}- \partial_j\Gamma^\alpha_{\ ki}+ \Gamma^\alpha_{\ \sigma i}\Gamma^\sigma_{\ k j} - \Gamma^\alpha_{\ \sigma j}\Gamma^\sigma_{\ k i} \right)\partial_\alpha \end{align}$$

Nevertheless, when I try to compute this expression from its intrinsic definition:

$$R(\mathbf{X},\mathbf{Y})\mathbf{Z}=\left[ \nabla_\mathbf{X},\nabla_\mathbf{Y} \right]\mathbf{Z}-\nabla_{\left[ \mathbf{X},\mathbf{Y}\right]}\mathbf{Z}$$

Using coordinate basis in this way:

$$\begin{alignat}{3} R(X^i\partial_i,Y^j\partial_j)(Z^k\partial_k)&=\left[ \nabla_{X^i\partial_i},\nabla_{Y^j\partial_j} \right]\left(Z^k\partial_k \right)&&-\nabla_{\left[ X^i\partial_i,Y^j\partial_j\right]}\left(Z^k\partial_k \right)\\ &=\nabla_{X^i\partial_i}\left[ \nabla_{Y^j\partial_j}\left(Z^k\partial_k \right) \right]&&-\nabla_{Y^i\partial_i}\left[ \nabla_{X^j\partial_j}\left(Z^k\partial_k \right) \right]\\ &&\vdots \end{alignat}$$

I end up with something like this:

$$\tag{2}\label{eq:riemann_comp}R(X^i\partial_i,Y^j\partial_j)(Z^k\partial_k)=2X^{[i}Y^{j]}Z^k\left(\partial_i\Gamma^\lambda_{\ kj}+\Gamma^\alpha_{\ kj}\Gamma^\lambda_{\ \alpha i} \right)\partial_\lambda$$

Obviously \eqref{eq:riemann} and \eqref{eq:riemann_comp} are very diferent. Why? What's wrong with it?

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  • $\begingroup$ You can loose the $X^i, Y^i, Z^i$ By tensoriality it suffices to check $(1)$ for basis vectors $\partial_i$. $\endgroup$ – Carlos Esparza Oct 7 '19 at 13:39
  • $\begingroup$ yes, i know $(1)$ follows from applying $R$ to the basis vectors, but my question is why are not equivalent the two results? or, in other words, what's wrong with $(2)$? $\endgroup$ – Dani Oct 7 '19 at 13:50
  • $\begingroup$ Nothing, they're the same. $\endgroup$ – Carlos Esparza Oct 7 '19 at 13:58
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Both your equations are the same: $$ 2X^{[i}Y^{j]}Z^k\left(\partial_i\Gamma^\lambda_{\ kj}+\Gamma^\alpha_{\ kj}\Gamma^\lambda_{\ \alpha i} \right)\partial_\lambda = X^i Y^j Z^k \left( \partial_i \Gamma^\lambda_{\ kj} + \Gamma^\alpha_{\ kj} \Gamma^\lambda_{\ \alpha i} - \partial_j \Gamma^\lambda_{\ ki} - \Gamma^\alpha_{\ ki} \Gamma^\lambda_{\ \alpha j}\right) \partial_\lambda $$

(now just rename $\alpha \to \sigma$ and $\lambda \to \alpha$ and you get $(1)$).

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  • $\begingroup$ I don't see why that equality holds: $$ 2X^{[i}Y^{j]}Z^k\left(\partial_i\Gamma^\lambda_{\ kj}+\Gamma^\alpha_{\ kj}\Gamma^\lambda_{\ \alpha i} \right)\partial_\lambda = (X^i Y^j- X^jY^i)Z^k \left(\partial_i\Gamma^\lambda_{\ kj}+\Gamma^\alpha_{\ kj}\Gamma^\lambda_{\ \alpha i} \right)\partial_\lambda $$ which is not equal to the right hand side of your expression $\endgroup$ – Dani Oct 7 '19 at 14:08
  • $\begingroup$ Use the distributive law on the difference and then swap $i$ and $j$ in the second term. $\endgroup$ – Carlos Esparza Oct 7 '19 at 14:12
  • $\begingroup$ Ok now I see it, thank you for your reply! $\endgroup$ – Dani Oct 7 '19 at 14:16

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