Definitions: $SL_2 (\mathbb{R})$ is the group of $2 \times 2$ matrices with real valued entries with determinant $1$ and $SO(2)$ is the group of orthogonal $2 \times 2$ matrices with determinant $1$.
Question: Let $SL_2 (\mathbb{R})$ act on $\hat{\mathbb{C}}$ by Möbius transformations. Find the orbit and stabiliser of $i$ and $\infty$. By considering the orbit of $i$ under the action of the stabiliser of $\infty$, show that every $g \in SL_2(\mathbb{R})$ can be written as $g=hk$ with $h$ upper triangular and $k \in SO(2)$. In how many ways can this be done?
I think that
$\mathrm{Orb}_{SL_{2}\mathbb{R}}(i)=\{z \in\mathbb{C}: \mathrm{Im}(z)>0\}$
$\mathrm{Stab}_{SL_{2}\mathbb{R}}(i)=SO(2)$
$\mathrm{Orb}_{SL_{2}\mathbb{R}}(\infty)=\mathbb{R}$
$\mathrm{Stab}_{SL_{2}\mathbb{R}}(\infty)=\Big\{\begin{pmatrix} a & b \\ 0 & d \end{pmatrix}: a,b,d \in \mathbb{R}, ad=1\Big\}$
$\mathrm{Orb}_{\mathrm{Stab_{SL_2\mathbb{R}(\infty)}}}(i)= \mathrm{Orb}_{SL_{2}\mathbb{R}}(i)$.
But I can't figure out how these relate to the next part. Some guidance would be much appreciated.
Best Answer
HINT(S):
Let $g\in{\rm SL}_2({\Bbb R})$ and let $\tau=g\cdot i$. Then there exists an upper triangular matrix $h\in{\rm SL}_2({\Bbb R})$ such that $h\cdot i=\tau$.
If $g$ and $h$ are as above, then $h^{-1}g\in{\rm SO}(2)$.