The metric is $$g=\frac{4}{(1-(u^2+v^2))^2}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ I have tried this like ten times and I just need someone to help me out. I don't think I totally understand this metric. I know I take derivatives of the entries to get the Christoffel symbols, yet everytime I put it together in the tensor $$k(g)=\frac{R(\frac{\partial}{\partial{x_1}},\frac{\partial}{\partial{x_2}})\frac{\partial}{\partial{x_2}}\bullet\frac{\partial}{\partial{x_1}}}{det(g)}$$ I can't get $-1$. Someone elsewhere said I should get the Christoffel symbols $$\Gamma^1_{ij}=\frac{2}{1-u^2-v^2}\begin{pmatrix} u & v \\ v & -u \end{pmatrix}$$ $$\Gamma^2_{ij}=\frac{2}{1-u^2-v^2}\begin{pmatrix} -v & u \\ u & v \end{pmatrix}$$ But I don't understand how it's a matrix, I thought it was like a scalar function, so I have no clue how he got those. I'm uploading a pic of my formulas for the Christoffels and my calculations. This seems like a direct calculation, but I can't get it right. Hopefully someone can tell me what I'm doing wrong. Riemann Tensor my work excuse my handwriting.

  • $\begingroup$ I'm having trouble following your computations. Did you mean to link to two copies of the same picture? Where are your values for the Christoffel symbols? Wikipedia has the easiest formula for the Christoffel symbols in terms of the metric. (en.wikipedia.org/wiki/…) $\endgroup$
    – nkm
    May 7, 2018 at 21:11
  • $\begingroup$ @NathanielMayer yeah my bad lol I fixed it $\endgroup$
    – N Dizzle
    May 7, 2018 at 21:22

1 Answer 1


The usual formula for the Christoffel symbols is $$ \Gamma^{k}_{ij} = \frac{1}{2}g^{km}(g_{ik,j}+g_{jk,i}-g_{ij,k}) $$ The inverse metric is just $$ g^{-1} = \frac{(1-(u^2+v^2))^2}{4} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, $$ so we only need to calculate $$ g_{ik,j}+g_{jk,i}-g_{ij,k} $$ and multiply it by this. We have $$ g_{11,1} = g_{22,1} = +2u \frac{8}{(1-(u^2+v^2))^3} $$ and $$ g_{11,2} = g_{22,2} = +2v \frac{8}{(1-(u^2+v^2))^3}, $$ while $g_{12,i}=g_{21,i}=0$ for $i=1,2$. So, $$ \Gamma^1_{11} = \frac{(1-(u^2+v^2))^2}{8} ( g_{11,1}+g_{11,1} - g_{11,1} ) = \frac{2u}{1-(u^2+v^2)} \\ \Gamma^1_{12} = \Gamma^1_{21} = \frac{(1-(u^2+v^2))^2}{8} ( g_{11,2}+g_{21,1} - g_{12,1} ) = \frac{2v}{1-(u^2+v^2)} \\ \Gamma^1_{22} = \frac{(1-(u^2+v^2))^2}{8} ( g_{21,2}+g_{21,2} - g_{22,1} ) = \frac{-2u}{1-(u^2+v^2)}. $$ This is summarised in the matrix form you quote, which has no particular meaning beyond being convenient notation: the Christoffel symbols are neither a transformation matrix nor a tensor, so it has no meaning beyond the current coordinates.

Similarly, $$ \Gamma^2_{11} = \frac{(1-(u^2+v^2))^2}{8} ( g_{12,1}+g_{21,1} - g_{11,2} ) = \frac{-2v}{(1-(u^2+v^2))} \\ \Gamma^2_{12} = \Gamma^2_{21} = \frac{(1-(u^2+v^2))^2}{8} ( g_{12,2}+g_{22,1} - g_{12,1} ) = \frac{2u}{1-(u^2+v^2)} \\ \Gamma^2_{22} = \frac{(1-(u^2+v^2))^2}{8} ( g_{22,2}+g_{22,2} - g_{22,2} ) = \frac{2v}{1-(u^2+v^2)}. $$

The nonzero components of the Riemann tensor are all related to $$ R_{1212} = g_{11}R^1{}_{212} = g_{11} \left( \Gamma^1_{22,1} - \Gamma^1_{21,2} + \Gamma^1_{11} \Gamma^1_{22} + \Gamma^1_{12} \Gamma^2_{22} - \Gamma^1_{21} \Gamma^1_{21} - \Gamma^1_{22} \Gamma^2_{21} \right) \\ = \frac{4}{(1-(u^2+v^2))^4} \left([-2(1-(u^2+v^2))-4u^2] - [2(1-(u^2+v^2))+4v^2] -[4u^2] + [4v^2] - [4v^2] + [4u^2] \right) \\ = \frac{-16}{(1-(u^2+v^2))^4} = -\det{g}, $$ since almost everything cancels.

  • $\begingroup$ Thanks a lot for answering. That's very clear and I can follow it all, up until the second to last line at the end, right before the ending parenthesis in the last square bracket. I cannot see where the $+4u^2$ comes from. I keep looking at the $\Gamma^2_{12}$ and for me it always is equal to zero when I compute the Christoffel, so that term isn't there for me. Can you explain that a little bit? Because that term is important to get the right cancellation, but every time I have tried this problem that guy is zero. $\endgroup$
    – N Dizzle
    May 8, 2018 at 5:06
  • $\begingroup$ I've added calculations of the other Christoffel symbols. They should be simply related to the $\Gamma^1_{ij}$, since the metric is symmetric in $u$ and $v$. $\endgroup$
    – Chappers
    May 8, 2018 at 11:53
  • $\begingroup$ I'll have to figure out how the ones are related to the twos. The matrix/vector formula I had written down was a little off in the second component which is why I kept getting zero for that one. Weird how all the others were fine though. Thanks! $\endgroup$
    – N Dizzle
    May 8, 2018 at 14:26

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