Let $H ⊂ C$ be the upper half plane with the hyperbolic metric. Prove than any matrix $M ∈ SL(2, \mathbb{R})$ representing a Möbius transformation is an isometry of $H$.
How do I prove this statement? I know $SL(2,\mathbb{R})$ is the group of 2 x 2 matrices with entries in $\mathbb{R}$, and that the Möbius transformation is defined as: $$A(z) = \frac{az+b}{cz+d}$$
We identify the tangent space $T_zH$ with $T_p\mathbb{R}^2 $.
We will use the metric $g_z(u, v) = \frac{\langle u, v\rangle}{Im(z)^2}$ for $u, v \in T_zH$. What we need to show is, $g_z(u, v) = g_{A(z)}(A'(z)u, A'(z)v)$.
Notice that $A'(z) = \frac{1}{(cz+d)^2}$ and $\text{Im}(A(z)) = \frac{\text{Im}(z)}{|cz+d|^2}$ for $M \in SL(2, \mathbb{R})$.
Then we have $g_{A(z)}(A'(z)u, A'(z)v) = \frac{\langle \frac{1}{(cz+d)^2} u, \frac{1}{(cz+d)^2} v \rangle}{(\frac{\text{Im}(z)}{|cz+d|^2})^2} = \frac{\frac{1}{(cz+d)^2}\frac{1}{\overline{(cz+d)}^2}\langle u, v \rangle}{\frac{\text{Im}(z)^2}{|cz+d|^4}} = \frac{\frac{1}{|cz+d|^4} \langle u, v \rangle}{\frac{\text{Im}(z)^2}{|cz+d|^4}} = \frac{\langle u, v \rangle}{\text{Im}(z)^2} = g_z(u,v)$.