We know that Mobius transformations, $z\to\frac{az+b}{cz+d}$, permutes circles and lines in the Euclidean plane, $(\mathbb{R}^2, dx^2 + dy^2 )$.
It may even be possible to write an explicit formula for the general Mobius action on any given circle:
$$ |z-z_0|=r \mapsto \left|\frac{az+b}{cz+d} -z_0 \right|=r$$
Such a space can be generated by translations $z \mapsto z + z_1$, rotations $z \mapsto \omega z$, dilations $z \mapsto rz$ and inversions $\displaystyle z = \frac{1}{\overline{z}}$. TheThe action on all circles is clear except for the last case:
$$ |z-z_0|=r \mapsto \left| z - \frac{\tfrac{1}{2}z_0}{|z_0|^2 - r^2} \right|= \frac{\tfrac{1}{2}r}{|z_0|^2 - r^2} $$
Is there a more succinct way to write this transformation as a Lie group action?
Here, the Mobius group $G = PSL(2, \mathbb{C}) \simeq SO(1,3)^+$ should act on on the space of circles (and lines) in the plane $\mathbb{R}^2$.
This question also leads me to wonder what the space of circles in the Euclidean plane should be. NaivelyNaively, the circles are a copy of $\mathbb{C} \times \mathbb{R}^+$ which looks like it could possibly be Hyperbolic space $\mathbb{H}^3$, in which case there would even be a natural metric.
However, the lines all have infinite radius. InIn which case, we still have 3three parameters, now a copy of $\mathbb{C} \times S^1$ identifying each line with the closest point to the origin and its direction. TheThe $S^1$ behaving something like the circle at infinity (YoutubeYouTube).
So now my space of circles is $\big(\mathbb{C}\times \mathbb{R}^+ \big) \cup \big(\mathbb{C}\times S^1 \big)$. HowHow does that happen?
EDIT In fact this space should be $\big(\mathbb{C}\times \mathbb{R}^+ \big) \cup \big(S^1 \times \mathbb{R}^+ \big)$.