# Riemann tensor given the space/metric

## Homework Statement

Given two spaces described by
$ds^2 = (1+u^2)du^2 + (1+4v^2)dv^2 + 2(2v-u)dudv$
$ds^2 = (1+u^2)du^2 + (1+2v^2)dv^2 + 2(2v-u)dudv$

Calculate the Riemann tensor

## Homework Equations

Given the metric and expanding it $~~~g_{τμ} = η_{τμ} + B_{τμ,λσ}x^λx^σ + ...$
We have the Riemann tensor, $R_{τρμν} = B_{τν,ρμ} + B_{ρμ,τν} - B_{ρν,τμ} - B_{τμ,ρν}$

## The Attempt at a Solution

I'm having problems calculating the B's for the above spaces.
There is also one space given by $ds^2 = (1+(ax+cy)^2)dx^2 + (1+(by+cx)^2)dy^2 + 2(ax+cy)(by+cx)dxdy$ where I know how to compute the B's. Since in 2D, the Riemann tensor has only one component, that is $R_{1212}$. For $g_{11}$ we have $B_{1122} = c^2$, for $g_{22}$ we have $B_{2211} = c^2$, for $g_{12} = g_{21}$ we have $B_{1212} = ½(ab + c^2)$

For the above spaces, it seems every B is 0. I can't seem to understand why.

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Orodruin
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Your metric tensor is not of the given form, the off diagonal components are linear and not quadratic in the coordinates.

Your metric tensor is not of the given form, the off diagonal components are linear and not quadratic in the coordinates.
Yes, that is why I'm wondering. The book stated that the first space is flat and the second space has a nonzero curvature.

Orodruin
Staff Emeritus
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Yes, that is why I'm wondering. The book stated that the first space is flat and the second space has a nonzero curvature.
This just means you have to find you curvature tensor through other means, not that it is impossible to compute it.

This just means you have to find you curvature tensor through other means, not that it is impossible to compute it.
I tried finding the geodesic equation and read off the Christoffel symbol then derive the Riemann tensor through the Christoffel symbol. It worked but it's insanely long, that is why I'm thinking of another way to do it.

In a 2 dimensional (semi)-Riemannian manifold there is only $\mathbf{one}$ linearly independent component of the Riemann tensor (I think in Wald's General Relativity there's a proof of this). That should make it easier ($\simeq$ shorter)!

In a 2 dimensional (semi)-Riemannian manifold there is only $\mathbf{one}$ linearly independent component of the Riemann tensor (I think in Wald's General Relativity there's a proof of this). That should make it easier ($\simeq$ shorter)!
Yes I already stated that in my first post. My question is the method that I've done in post#1 is not working but I've done it in another way (post#5). Maybe there is another way.

Yes I already stated that in my first post. My question is the method that I've done in post#1 is not working but I've done it in another way (post#5). Maybe there is another way.
Oh, sorry, it seems i missed that.
You can also compute the components of $R^{a}{}_{bcd}$ in a orthonormal frame, via the connection one-forms and Cartan's structure equations.