Automorphism of Upper half plane

Question

Let $M=\left\{\left.\displaystyle z\mapsto\frac{az+b}{cz+d}\ \right|\ \ ad-bc\not =0\right\}$,$$p:GL(2,\mathbb C)\to M, \begin{bmatrix}a & b \\ c & d \end{bmatrix}\mapsto\frac{az+b}{cz+d}.$$ Let $\mathbb H=\left\{z\in\mathbb C|Im(z)>0\right\}$ be the upper half plane. Proof that $p(SL(2,\mathbb R))=Aut(\mathbb H)$.

I tried to show $p(SL(2,\mathbb R))\subset Aut(\mathbb H)$ but failed, because $Im\left(\frac{az+b}{cz+d}\right)$ is not necassary bigger then zero. Is there another way to show $p(SL(2,\mathbb R))=Aut(\mathbb H)$?

Answer 1

In fact $$\operatorname{Im}\frac{az+b}{cz+d}=\frac{(ad-bc)\cdot\operatorname{Im}z}{\mid cz+d\mid^2}.$$
So $p(SL(2,\mathbb R))\subseteq \operatorname{Aut}(\mathbb H).$

Hope this helps.

Answer 2

In fact, $Im (g(z))=\frac{Im (z)}{|cz+d|^2}>0$ for all $z\in \mathbb{H}$ and all $g\in SL_2(\mathbb{R})$. So we have a well-defined action of $SL_2(\mathbb{R})$ on the upper half plane. It is perhaps useful to have a reference for the basics here, see for example these notes.