Proving that inversions are isometries with respect to the hyperbolic metric.

Question

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.

Answer 1

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 $r,\theta$:

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