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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.

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

We can get injectivity explicitly too:

$Ai = Bi \iff AB^{-1} \in \mathrm{SO}(2) \iff A \cdot \mathrm{SO}(2) = B \cdot \mathrm{SO}(2)$, showing that $A=B$ modulo $\mathrm{SO}(2)$ as required.

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$$\gamma . i = \beta . i \\\implies \beta^{-1} \gamma.i =\beta^{-1}.\beta.i= i\\ \beta^{-1}\gamma. i = \frac{ai+b}{ci+d}= i \implies ai+b = di-c \implies (c,d) = (-b,a)\\ \implies \beta^{-1} \gamma \in SO_2(\Bbb{R})\\ \implies \beta^{-1} \gamma SO_2(\Bbb{R}) =SO_2(\Bbb{R}) \\ \implies \gamma SO_2(\Bbb{R})=\beta SO_2(\Bbb{R}) $$

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