How to understand the Möbius transform as a group action?

Question

The group $SL(2,R)$ acts on the upper half-plane by the formula

$$ \left(\begin{array}{cc} a & b \\ c & d \end{array} \right) z = \frac{az + b}{cz + d} .$$

It is indeed straightforward to check that it is an action. But is there any smart way to see it without calculation?

It is kind of miracle for me.

Answer 1

The Cayley transform $$ z\mapsto\frac{z-i}{z+i} $$ sends the complex upper half plane ${\cal H}$ conformally onto the unitary disc $D$. Under this transform, the Möbius transformations corresponding to elements in the maximal compact group $K={\rm SO}_2$ become rotations of $D$. In order to understand the remaining tranformations recall the decomposition ${\rm SL}_2=NAK$ where $N$ are the unitary upper triangular matrices and $A$ are the diagonal matrices.

When the "translation" is completed it becomes quite clear that we have an action, although it may be argued that the direct verification is actually simpler.

Answer 2

You can view $\widehat{\Bbb{C}}=\Bbb{C} \cup \{\infty\}$ as $\Bbb{CP}^1$, from which you get a natural action of $SL_2(\Bbb{C})$ on $\Bbb{\widehat{C}}$. You can check that this action coincides with Mobius transformations, which explains the "miracle". Now if you restrict attention to the upper half plane $H$, it is easy to see that at least $SL_2(\Bbb{R})$ preserves the boundary (that is, the real line). Its image turns out to be exactly the Mobius transformations preserving $H$.