Christoffel symbols for Poincare metric on unit disk

Question

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.

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.