How is $PSL(2, \mathbb{R})$ explicitly identified with the unit tangent bundle $T^1(\mathbb{H})$?

Question

Let's say I have a given matrix $\begin{pmatrix} 1 & s \\ 0 & 1 \end{pmatrix}$, what point and tangent direction is it identified with?

Answer 1

Use the upper half plane model $\mathbb{H} = \{z = x + iy \bigm| y > 0 \}$. At each point $z = x+iy \in \mathbb{H}$, its unit tangent bundle is $$T^1_z(\mathbb{H}) = \{V \in \mathbb{C} \approx \mathbb{R}^2 \, \bigm| \, |V| = y\} $$

Consider a matrix $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in SL(2,\mathbb{R})$, so $ad-bc=1$. Let $f_M : \mathbb{H} \to \mathbb{H}$ be the fractional linear transformation $$f_M(z) = \frac{az+b}{cz+d} $$ The point $z \in \mathbb{H}$ and tangent direction $V \in T^1_z(\mathbb{H})$ that correspond to $M$ are given by $$z = f_M(i) = \frac{ai+b}{ci+d} $$ and $$V = \frac{df_M}{dz} \biggm|_{\, i} = \frac{1}{(ci+d)^2} $$