# Why is the map $\mathrm{SL}_2(\mathbb{R})/\mathrm{SO}(2) \rightarrow \mathbb{H} : A \mapsto Ai$ injective?

First some notation:

$$\mathrm{SL}_2(\mathbb{R})=\left\lbrace\begin{pmatrix} a&b \\ c &d \end{pmatrix}\;\Biggm|\;a,b,c,d \in \mathbb{R} ,ad-bc=1 \right\rbrace$$

$$\mathrm{SO}(2)=\lbrace K\in \mathrm{SL}_2(\mathbb{R}) : A^TA=AA^T=I \rbrace$$.

$$\mathbb{H}$$ denotes the upper half plane.

The map $$\mathrm{SL}_2(\mathbb{R})/\mathrm{SO}(2) \rightarrow \mathbb{H} : A \mapsto Ai$$

is bijective.

My problem is to understand why the map is injective.

My idea is to take $$N\in \mathrm{SO}(2)$$ and consider $$Mi=Ni$$.

Since $$\mathrm{SO}(2)$$ is the stabilizer of $$i$$ it follows that $$M=N$$.

Thanks for the help.

• How do you mean $Ai$? (I guess, $i$ is the imaginary unit in $\Bbb H\subseteq \Bbb C$.) Jul 31 '19 at 19:20
• sorry , the operation should be the moebius transformation . Jul 31 '19 at 19:32
• Yes , i is the imaginary unit Jul 31 '19 at 19:34

If $$X$$ is any set and $$G$$ is a group acting transitively on it then we can identify $$X$$ with $$G/G_x$$ where $$G_x$$ is the stabilizer of some fixed point $$x \in X$$.
The upper triangular matrices act transitively on $$\Bbb H$$, so $$\mathrm{SL}_2(\Bbb R)$$ does too, and $$\mathrm{SO}(2)$$ is the stabilizer of $$i$$.
$$Ai = Bi \iff AB^{-1} \in \mathrm{SO}(2) \iff A \cdot \mathrm{SO}(2) = B \cdot \mathrm{SO}(2)$$, showing that $$A=B$$ modulo $$\mathrm{SO}(2)$$ as required.
$$\gamma . i = \beta . i \\\implies \beta^{-1} \gamma.i =\beta^{-1}.\beta.i= i\\ \beta^{-1}\gamma. i = \frac{ai+b}{ci+d}= i \implies ai+b = di-c \implies (c,d) = (-b,a)\\ \implies \beta^{-1} \gamma \in SO_2(\Bbb{R})\\ \implies \beta^{-1} \gamma SO_2(\Bbb{R}) =SO_2(\Bbb{R}) \\ \implies \gamma SO_2(\Bbb{R})=\beta SO_2(\Bbb{R})$$