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I'd like to prove that the standard inversion $$(r,\theta)\mapsto\left(\frac{1}{r},\theta\right)$$ is an isometry with respect to the hyperbolic metric on the upper half-plane, and it would be nice to me to do the calculations using polar coordinates. First of all, I compute the length element in polar coordinates: since $x=r\cos\theta$ and $y=r\sin\theta$ we get $dx=dr\cos\theta-r\sin\theta d\theta$ and $dy=dr\sin\theta+r\cos\theta d\theta$, so that $$\frac{dx^2+dy^2}{y^2}=\frac{dr^2+r^2d\theta^2}{r^2\sin^2\theta}.$$ Now, call $(u,v)=\left(1/r,\theta\right)$ and compute $du=-dr/r^2$ and $dv=d\theta$. Therefore, the new length element expressed in polar coordinate is $$\frac{du^2+dv^2}{v^2}=\left(\frac{dr^2}{r^4}+d\theta^2\right)\frac{1}{d\theta^2}=\frac{dr^2+r^4d\theta^2}{r^4d\theta^2},$$ which is not equal to the previous one. I guess that I'm wrong when I define $(u,v)$ directly in terms of polar coordinates, but I don't know how to fix it. Could you help me with that? Thanks.

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up vote 2 down vote accepted

You wouldn't compute $(dr^2+d\theta^2)/\theta^2$ for anything, so why would you compute $(du^2+dv^2)/v^2$ as you've defined the variables? Instead, you want your hyperbolic formula with $u,v$ replacing $r,\theta$:

$$u=\frac{1}{r},\; v=\theta:\quad \frac{du^2/u^2+dv^2}{\sin^2v} \;\text{ while }\; du^2\frac{1}{u^2}=\left(-\frac{dr}{r^2}\right)^2\frac{1}{(1/r)^2}=\frac{dr^2}{r^2} \;\text{ and }\; dv=d\theta. $$

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thanks! actually I just found out that the error was in starting with Euclidean coordinates on one hand and with polar coordinates on the other. If I define $$(u,v)=\left(\frac{\cos\theta}{r},\frac{\sin\theta}{r}\right)$$ instead, everything works smoothly! Many thanks again! – fatoddsun Mar 15 '12 at 18:48
@fatoddsun: I was going to mention that that's what I would have done, but you said it would be nice to me to do the calculations using polar coordinates. The issue wasn't in your using polar, it was in plugging polar terms into a Cartesian formula. – anon Mar 15 '12 at 18:52
yes, I see. Thank you again for the help! – fatoddsun Mar 16 '12 at 11:19

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