I am trying to prove that the action of $SL_2(\mathbb{R})$ on $\mathbb{H}$ via $$ \left( \begin{array}{ c c } a & b \\ c & d \end{array} \right)z\rightarrow \frac{az+b}{cz+d} $$ is transitive and the Stabilizer of $i$ is $SO_2(\mathbb{R})$. For stabilizer of $i$ I took an arbitrary matrix and through some algebra I got that $a^2+b^2=1$ and $c^2+d^2=1$. It is clear that $b$ or $c$ has to be negative but I cannot find a reason to eliminate the case when $a,d<0$. How can we eliminate the case as I want to show that $a=cos\theta$, $b=-sin\theta$, $c=sin\theta$, $d=cos\theta$.

Next I am stuck at proving the part that the action is transitive. I am trying the similar strategy as above but the calculations just get messier


For the stabilizer calculation, you need as well to use that, by virtue of the matrix being in $SL(2, \mathbb{R})$, you have $$ad - bc = 1.$$ It may also be helpful to note that because, e.g., $a^2 + b^2 = 1$, there is some angle $\theta$ such that $$a = \cos \theta, b = -\sin \theta.$$

To show that the action of a group $G$ on a space $X$ is transitive, it's enough to pick an element $x_0 \in X$ and show that for all $x \in X$ there is some $g \in G$ such that $x = g \cdot x_0$; then, one can map any element $x$ to any other element $y$ by mapping the $x$ to $x_0$ and then $x_0$ to $y$. Often, this turns out to be cleaner than showing directly that one can map an arbitrary element to another arbitrary element, especially if one can choose an element $x_0$ on which the group acts particularly nicely.

  • $\begingroup$ It worked out quite nicely for transitive action when I pick $x_0=i$. However, out of curiosity is there any other element such that $G$ acts nicely on it and we can still prove transitive action $\endgroup$ – Rutherford Mark Oct 23 '14 at 5:31
  • $\begingroup$ You should in principle be able to do this for any element, but $i$ is the most natural choice, because its real and imaginary components are especially simple, which leads to comparatively simple equations when constructing a group element that maps it to another point. $\endgroup$ – Travis Willse Oct 23 '14 at 6:13
  • $\begingroup$ Another motivation for taking $x_0 = i$ is that it corresponds to taking $x_0 = 0$ in the Poincare disk model of hyperbolic space; we can pass between the upper half-plane and that model via the Cayley transform, $z \mapsto \frac{z - i}{z + i}$. (You might find it interesting to redo the problem in that setting.) $\endgroup$ – Travis Willse Oct 23 '14 at 6:18
  • $\begingroup$ Now that you've worked out your problem, you might like to post your solution as an answer here. $\endgroup$ – Travis Willse Oct 23 '14 at 6:19
  • $\begingroup$ How can we show that for all $z \in \mathbb H$, there exists some $g \in SL_2(\mathbb R)$ such that $gz=i$? $\endgroup$ – Al Jebr Nov 20 '17 at 18:51

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