Is the SL(2,R) action on the upper half-plane transitive?

[Chris L T521]

Here's this week's problem.

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Problem: Let $G=\text{SL}_2(\mathbb{R})$ and let $\mathcal{H}=\{z\in\mathbb{C}:\text{Im}(z)>0\}$. Let $g=\begin{pmatrix}a & b\\ c & d\end{pmatrix}\in G$ and define an action $\cdot :G\times\mathcal{H}\rightarrow\mathcal{H}$ where $g\cdot z=\dfrac{az+b}{cz+d}$. Compute $\text{Orb}_G(i)$ and $\text{stab}_G(i)$. Is this action transitive?

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[Chris L T521]

No one answered this question correctly. You can find my solution below.

Spoiler

Let us first compute $\text{stab}_G(i)=\{g\in G:g\cdot i=i\}$. Let $g=\begin{pmatrix}a & b\\ c & d\end{pmatrix}\in\text{SL}_2(\mathbb{R})$. Then $\begin{pmatrix}a & b\\ c & d\end{pmatrix}\cdot i=i\implies \dfrac{ai+b}{ci+d}=i\implies ai+b=di-c\implies a=d\text{ and }b=-c$.
It now follows that $g$ is of the form $g=\begin{pmatrix} a & -c \\ c & a\end{pmatrix}$. Futhermore, since $g\in \text{SL}_2(\mathbb{R})$, we must have $\det g=1\implies a^2+c^2=1$. Thus, our matrix $g$ must represent a rotation on the unit circle. Hence, $g\in SO(2)$, and thus $\text{stab}_G(i)\cong SO(2)$.

To compute $\text{Orb}_G(i)$, we first observe that any element of our orbit must lie in the upper half plane $\mathcal{H}$; i.e. $g\cdot i=\dfrac{ai+b}{ci+d}\in\mathcal{H}$. Now, note that for $y>0$ and for any $x\in\mathbb{R}$, we have that $g=\begin{pmatrix}y^{1/2} & xy^{-1/2}\\ 0 & y^{-1/2}\end{pmatrix}\in \text{SL}_2(\mathbb{R})$. It now follows that $g\cdot i = \dfrac{y^{1/2}i+xy^{-1/2}}{y^{-1/2}}=y^{1/2}(y^{1/2}i+xy^{-1/2})=x+iy\in\mathcal{H}$. From this, it now follows that $g\cdot i$ generates the entire upper half plane; thus, $\text{Orb}_G(i)=\mathcal{H}$. Furthermore, $\text{Orb}_G(\mathcal{H})=\mathcal{H}$, and thus it follows that the action is transitive.