I am tasked to compute the curvature of the metric $$g=\frac{4}{(1-(u^2+v^2))^2}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ on the unit disk. I'm told this is a model of hyperbolic space, so I'm supposed to get $-1$. I'm using the formulas for the curvature tensor and covariant derivative $$R(\frac{\partial}{\partial{x_1}},\frac{\partial}{\partial{x_2}})\frac{\partial}{\partial{x_2}}=\nabla_{\frac{\partial}{\partial{x_1}}}\nabla_{\frac{\partial}{\partial{x_2}}}\frac{\partial}{\partial{x_2}}-\nabla_{\frac{\partial}{\partial{x_2}}}\nabla_{\frac{\partial}{\partial{x_1}}}\frac{\partial}{\partial{x_2}}$$
$$ \nabla_{\frac{\partial}{\partial{x_i}}}\frac{\partial}{\partial{x_j}}=\Gamma_{ij}^1\frac{\partial}{\partial{x_1}}+\Gamma_{ij}^2\frac{\partial}{\partial{x_2}}$$and to get $R$ I know the properties of the covariant derivatives. I also have formulas for the Christoffel symbols in terms of the metric $g$. Writing out $R$, it ends up being a sum of Christoffel symbols and their first derivatives times the vectors $\frac{\partial}{\partial{x_1}}$ and $\frac{\partial}{\partial{x_2}}$ Is that the right idea? Then I'm supposed to find the curvature
$$k(g)=\frac{R(\frac{\partial}{\partial{x_1}},\frac{\partial}{\partial{x_2}})\frac{\partial}{\partial{x_2}}\bullet\frac{\partial}{\partial{x_1}}}{det(g)}$$ and I think the dot on top uses the metric $g$. I think the curvature is $-1$ everywhere, but I can't seem to get that from what I have. I tried computing it all in cartesian coordinates and got $-1$ times some second degree polynomial in x and y. So I think I maybe computed something wrong, but plugging in 0 it's just -1. I also tried it in polar coordinates, but I got a messy answer that probably coincidentally was $-1$ when I plugged in $r=0$, but it seemed kind of hard to totally simplify. I'm not sure if I missed something when changing the metric or taking derivatives in polar coordinates. Using $u=rcos\theta,v=rsin\theta$ is this the right? $$g=\frac{4}{(1-r^2)^2}\begin{pmatrix} 1 & 0 \\ 0 & r^2 \end{pmatrix}$$ I've just been trying this for a while and can't seem to find $-1$ everywhere. Next I'm going to try again in polar coordinates because more of the Christoffel symbols were zero because every derivative wrt theta is zero when I did it that way. Should I just try again or can anyone offer some insight? Is there something with the chain rule that I might be missing with polar coordinates? Am I at least on the right track? I'm pretty new to curvature. Here's what I have tried. link
You should have gotten the Christoffel symbols
$$ \Gamma^1_{ij} = \frac{2}{1-u^2-v^2}\left( \begin{array}{cc} u & v \\ v & -u \end{array} \right) $$ $$ \Gamma^2_{ij} = \frac{2}{1-u^2-v^2}\left( \begin{array}{cc} -v & u \\ u & v \end{array} \right) $$
The numerator in your curvature formula is
$$ \left< R \left( \frac{\partial}{\partial u}, \frac{\partial}{\partial v} \right) \frac{\partial}{\partial v}, \, \frac{\partial}{\partial u} \right> = R^1_{212} $$
Use the formula for the curvature coefficients (see here for example):
$$ R^1_{212} = \frac{\partial \Gamma^1_{22}}{\partial u} - \frac{\partial \Gamma^1_{21}}{\partial v} + \sum_{k=1}^2 \Gamma^1_{1k}\Gamma^k_{22} - \Gamma^1_{2k} \Gamma^k_{21} $$
Using the Christoffel symbols above, this should come out to be $\frac{-4}{(1-u^2-v^2)^2}$. Now you still have to divide by the determinant of the metric, which cancels except for the $-1$.