# Vanishing of the Riemann tensor

The Riemann tensor in a coordinate basis is $$R^{i}_{\,jkl} = \partial_k \Gamma^i_{jl} - \partial_l \Gamma^i_{jk} + \Gamma^m_{jl}\Gamma^i_{mk} - \Gamma^m_{jk}\Gamma^i_{ml}$$

Consider $\mathbb{R}^2$ in polar coordinates $(r,\theta)$ with a connection with non vanishing components $\Gamma^r_{\theta \theta} = -r$ and $\Gamma^{\theta}_{\theta r} = r^{-1}$. SHow the Riemann tensor vanishes.

Attempt: It is clear that it vanishes since the Riemann tensor is a measure of the deviation of the curvature of the manifold from Euclidean space. Since we are dealing with (flat) Euclidean space in two dimensions, the tensor vanishes. I am just not completely sure of what all these indices mean on the Riemann tensor. I see it is a $(1,3)$ tensor so I think I can write $R^i_{\,jkl} \equiv R(f^a, e_1, e_2, e_3)$ where the arguments are basis for covectors and vectors defined in tangent spaces and co-tangent spaces. Do the values $\left\{ijkl\right\}$ take on values $r$ and $\theta$? I am just not sure what the indices mean and to show it vanishes I would try to evaluate all possible elements $R^i_{\,jkl}$ and show they all vanish?

Many thanks!

Your instincts here are correct. The general notation is for a coordinate basis $(\partial_1,\partial_2,\ldots,\partial_n)$. In your case, this is $(\partial_r,\partial_\theta)$, so the indices $i,j,k,l$ take on all possible combinations of $r$ and $\theta$.
So the components of the tensor are $$R^r_{rrr},\ R^r_{rr\theta},\ R^r_{r\theta r},\ R^r_{\theta rr},\ R^r_{r\theta\theta},\cdots$$ etcetera.
Another way of interpreting the question is, "Show that for all possible combinations $(i,j,k,l)\in\{r,\theta\}^4$ the curvature tensor component $R^i_{jkl} = 0$." I'll leave the actual computations to you.
• Many thanks! The only symmetry I have seen is that $R^{i}_{\,jkl} = -R^i_{\,jlk},$ which looks like will reduce the number of component computations to $2^4/2!$ – CAF Mar 8 '15 at 11:57