# Transitive Action of $SO(2)$ on $TH^+$

We define two dimensional hyperbolic space via $\mathbb{H}^2=(H^+,(dx^2+dy^2)/y^2)$ with $$H^+=\{z\in\mathbb{C}|Im(z)\ge 0\}$$ and the action of $SL(2,\mathbb{R})$ via $$(\begin{pmatrix} a & b \\ c & d \end{pmatrix},z)\mapsto \frac{az+b}{cz+d}$$ It is now an exercise to show that $\operatorname{Iso}(\mathbb{H}^2)\subseteq SL(2,\mathbb{R})$, meaning that every Isometry of Hyperbolic two-space is contained in $SL(2,\mathbb{R})$.

The proof goes by showing that $SL(2,\mathbb{R})$ acts transititive on $\mathbb{H}^2$ and that the stabilizer of $i$ is $SO(2)$. This is used to calculate the derivative of the action of $SO(2)\le SL(2,\mathbb{R})$ via the above action on $H^+$. This gives $$D_if_A(v)=e^{-2\varphi i}v\text{ for } A=\begin{pmatrix} \cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi\end{pmatrix}$$ using that $D_zf_A=\frac{1}{(cz+d)^2}$ for some $A\in SL(2,\mathbb{R})$.

Here in the solution of this exercise it is concluded, that $SO(2)$ acts transitvely on $T_iH^+$ and thus using the transitive action of $SL(2,\mathbb{R})$ on $H^+$ we have a transitive action on the whole tangent bundle $TH^+$. I really don't get how one is able to conclude this.

The rest of the proof than once uses transitivity of $SL(2,\mathbb{R})$ on $H^+$ to construct $f_T\circ f(i)=i$ for some isometry $f$ and than uses transititity of $SO(2)$ on $TH^+$ to construct $D_if_s(Df_T\circ f(1))=1$. This is than used together with geodesical completeness to conclude that $f_S\circ f_T\circ f=Id_2$ and thus that $f$ is fractional linear.

Edit:

Corrected typos:

• Missing $-$ sign
• Changed absolute value to $()$

To prove that $SO(2)$ acts transitively on $T_i H^+$, using the formula $$D_if_A(v)=e^{2\varphi i}v\text{ for } A=\begin{pmatrix} \cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi\end{pmatrix}$$ all you have to do is prove that for each unit vector $w$ the equation $D_i f_A(v)=w$ can be solved for $v$. Since $D_i f_A$ is an orthnormal matrix, this should be straightforward: simply left multiply by the inverse of that matrix.
By the way, your formula $$D_z f_A=\frac{1}{|cz+d|^2}$$ is odd. You can think of $D_z f_A$ either as an orthogonal transformation of $\mathbb{R}^2$ or as a complex number, but either way it is not a real number in general.
• Thank you for your answer. The point I don't get is about the solution of the equation you state. As I understand $e^{i\varphi}v$ is just a rotation leaving the length of the vector unchanged, so how should I ever reach any vector which doesn't have the same length as the one I started with. The other question is about your comment that my formula is odd. If i regard $Az$ for some $A\in SL(2,\mathbb{R})$ I get $\frac{az+b}{cz+d}$ which if i don't misunderstand things using the quotient rule $\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}$ gives the formula I obtained. – PascExchange Aug 24 '16 at 14:06
• Okay, I still don't get how one gets transitivity if setting $z=i$ and $A\in SO(2)$. It gives $D_if_A(v)=e^{-2\varphi i}v$ which is just "running circles" and won't cover $T_iH^+$, or what am I missing? – PascExchange Aug 24 '16 at 15:45
• Hmm... I'm beginning to see where your question may really lie. The correct statement is that $SL(2,\mathbb{R})$ acts transitively on the unit tangent bundle, meaning the circle sub-bundle of the tangent bundle consisting of all tangent vectors of unit length. I had presumed that your question was (correctly) formulated in terms of only unit tangent vectors, but perhaps it was instead (incorrectly) formulated in terms of all tangent vectors? – Lee Mosher Aug 24 '16 at 19:15