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I have two definitions of the Riemann tensor which are $$\Large R^{\rho}_{\,\,\,\sigma\mu\nu}=\partial_{\mu}\Gamma^{\rho}_{\nu\sigma}-\partial_{\nu}\Gamma^{\rho}_{\mu\sigma}+\Gamma^{\rho}_{\mu\lambda}\Gamma^{\lambda}_{\nu\sigma}-\Gamma^{\rho}_{\nu\lambda}\Gamma^{\lambda}_{\mu\sigma}=R_{\nu\mu\sigma}^{\,\,\,\,\,\,\,\,\,\rho}$$

But can someone demonstrate how these two ways of writing the Riemann tensor are equivalent?

MRT
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1 Answers1

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The fact follows from the two identites of the Riemann tensor: $R_{ijkl}=R_{klij}=-R_{lkij}=R_{lkji}$. Now, $R_{jkl}^i =g^{im} R_{mjkl}$ and ${R_{jkl }}^i=g^{im}R_{jklm }$.

With that you should be able to get the result.

You can use the Riemann if you want. However, for 3D spaces, the Ricci is enough. Anyway, given the metric, remember that the expression for the Christoffel symbols is $$\Gamma_{ijk}=\frac{1}{2}\left( \frac{\partial g_{ij}}{\partial x^k} + \frac{\partial g_{ik}}{\partial x^j} - \frac{\partial g_{jk}}{\partial x^i} \right) .$$ Moreover, your metric is diagonal ($g=diag(u^2+v^2,u^2+v^2,u^2v^2$). Hence some of the Christoffel symbols are $0$.

By the way, your space seems a 2D surface. In this case, according to Wiki, the Riemann is given in coordinates just by $$R_{ijkl}= K(g_{ik}g_{kl}-g_{il}g_{jk}),$$ where $K$ is the Gaussian curvature

J.G.
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Dog_69
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  • Oh thank you very much for your answer. It makes perfect sense now. I have that $$\Large R^{\rho}{,,,\sigma\mu\nu} = g^{\rho\lambda}R{\lambda\sigma\mu\nu}=g^{\rho\lambda}R_{\mu\nu\lambda\sigma}=-g^{\rho\lambda}R_{\nu\mu\lambda\sigma}=g^{\rho\lambda}R_{\nu\mu\sigma\lambda} = R_{\nu\mu\sigma}^{,,,,,,,,,,,,\rho}$$ – MRT Oct 24 '18 at 11:57
  • So now if I have the line element $$\Large ds^2 = (u^2+v^2)(du^2+dv^2) + u^2v^2d\phi^2$$ how would I find the Riemann tensor for this $3$ dimensional problem? I know that the fourth index repeats but I just wanna know how it looks so I can do the computations. Would it be something like $$\Large R^{u}{,,,uv\phi},\quad R^{v}{,,,uv\phi},\quad R^{\phi}_{,,,uv\phi}$$ – MRT Oct 24 '18 at 12:11
  • Wait apparently I need to use the Ricci tensor for a $3$-D tensor problem but there is no explanation how to choose the indices to find the the non-zero components... – MRT Oct 24 '18 at 13:51
  • @MRT: I have added mor information to my answer traying to clarify your comments. – Dog_69 Oct 24 '18 at 15:25
  • oh it is in $3$-D because it's in $\mathbb{E}^3$ space. I read your answer but it doesn't quite get to what I was asking. I do have all of the Christoffel components but I just don't know how to find the Ricci tensor. – MRT Oct 24 '18 at 15:45
  • Oh, sorry. The Ricci tensor is the thrace of the Riemann: $$R_{ij}= R_{ikj}^k = g^{kl}R_{likj} .$$ – Dog_69 Oct 24 '18 at 16:45