# $\text{SL}(2, \mathbb{R})$ mapping on the boundary and interior of the disk

I am trying to show that $$\operatorname{SL}(2, \mathbb{R})$$ can be used to map an arbitrary point in the interior of the disk and an arbitrary point on its boundary to any pair of points, one in the interior and one on the boundary. This is in the context of string theory, so I am thinking about the disk as the conformal mapping of the upper half plane, so that the boundary is the real line and the interior is the set $$\{z \in \mathbb{C} | \Im(z) > 0\}$$.

I have tried to set up a system of equations where $$y_2 = \frac{ay_1+b}{cy_1+d}$$ $$z_2 = \frac{az_1+b}{cz_1+d}$$ $$ab - cd = 1$$ and $$a,b,c,d \in \mathbb{R}$$.

However, I cannot seem to find way to make this work. I think that this brute force approach cannot work and there must be something that I am missing. Any hints would be greatly appreciated!

Here is a hint:

Rather than send a general pair (interior point and boundary point) to another general pair, it is easier to try to send a general pair to a specific one. If we can do this, then we can use this specific point like a transit center to send general points to general points. Take this specific point to be, say, $$i$$ (interior point) and $$0$$ (boundary point).

So given $$z\in \{ Im(z)>0\}$$ and $$x\in\mathbb{R}$$, we want $$\begin{bmatrix}a & b \\ c & d\end{bmatrix}\in SL_2(\mathbb{R})$$ with $$\frac{az+ b}{cz+ d} = i$$ and $$\frac{ax+b}{cx+d} = 0$$. Which should easier to solve.

Geometrically, you can think of applying a translation and rotation. Take some translation $$T$$ sending $$z$$ to $$i$$. If $$\gamma$$ is the geodesic through $$z$$ hitting the boundary at $$x$$, then $$T(\gamma)$$ will be a geodesic through $$i$$ hitting the boundary in some point. Then one applies a rotation around $$i$$ so that $$T(\gamma)$$ becomes the vertical geodesic through $$i$$ (hitting the boundary at $$0$$).

For example, we can take $$T = \begin{bmatrix}1 & -\xi\\ 0 & \eta\end{bmatrix}$$ to send $$z= \xi + i\eta$$ to $$i$$. Then $$x$$ gets sent to $$x_1 =(x-\xi)/\eta$$ and a rotation around $$i$$ sends $$x_1$$ to $$\frac{\cos\theta x_1 - \sin\theta}{\sin\theta x_1 + \cos\theta}$$ which we want to be $$0$$. So take $$\tan\theta =x_1$$. The composition of these is just by matrix multiplication:

$$\begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}T$$, where $$\tan\theta = (x-\xi)/\eta$$

Finally, to get it in $$SL_2(\mathbb{R})$$, we can divide by the root of the determinant (multiply by $$\frac{1}{\sqrt{\eta}}$$).