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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

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    $\begingroup$ It will be easier if you write down what you have calculated from here it is hard to see what you did wrong. It's a direct calculation, after all. $\endgroup$
    – user99914
    May 6, 2018 at 1:13
  • $\begingroup$ @JohnMa I uploaded my calculations if you care to take a look $\endgroup$
    – N Dizzle
    May 6, 2018 at 21:18
  • $\begingroup$ I don't think you need to go beyond calculating with u and v, i.e. you can leave u and v as your "cartesian coordinates," no need to convert to polar or x,y. I'm doing this problem too, but my only issue is the way that dot product is defined. I'll let you know when I get further. $\endgroup$
    – Michael M
    May 6, 2018 at 21:45

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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 $-1$.

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  • $\begingroup$ You are the man Nick thank you. I checked and I have the right formulas for the curvature coefficients, I guess I screwed up the Christoffel symbols, it seems like I don't totally understand them. I don't really get your formula, I didn't know it could be expressed as a matrix, and then I wouldn't know how to take it's derivative. I think I'm confused about g in the first place. Can you explain how you got those, and what I did that was wrong? $\endgroup$
    – N Dizzle
    May 7, 2018 at 4:13
  • $\begingroup$ isn't the determinant 16/(1-(u^2+v^2))^4? $\endgroup$
    – N Dizzle
    May 7, 2018 at 5:51
  • $\begingroup$ I don't understand your representation of the Christoffel symbols $\endgroup$
    – N Dizzle
    May 7, 2018 at 18:49
  • $\begingroup$ When I say $\Gamma^1_{ij}$, I mean for instance that $\Gamma^1_{21}$ is the element of that matrix in the 2nd row and 1st column. I was just collecting all the Christoffel symbols with the same upper index into one matrix. As for how I got the Christoffel symbols, see the standard formulas in the link in my post. $\endgroup$
    – Nick
    May 9, 2018 at 2:50

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