1
$\begingroup$

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 $()$
$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – PascExchange Aug 24 '16 at 14:06
  • $\begingroup$ There is just no absolute sign in the formula. Otherwise ok (when normalized so det=1) $\endgroup$ – H. H. Rugh Aug 24 '16 at 14:34
  • $\begingroup$ 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? $\endgroup$ – PascExchange Aug 24 '16 at 15:45
  • $\begingroup$ 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? $\endgroup$ – Lee Mosher Aug 24 '16 at 19:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.