Ideal quadrilateral in $\mathbb{H^2}$ can be mapped to triangle with vertices $-1,0,\infty, x$ where $x \in \mathbb{R}$

Question

Why can we always map vertices of an ideal quadrilateral in $\mathbb{H^2}$ to $-1,0,\infty, x$ where $x \in \mathbb{R}$?

I'm not realising why this can always be done? I.e why $x$ is always real.

Answer 1

Ideal quadrilaterals have vertices on the boundary of hyperbolic space, so in the upper half-plane model the vertices are real (or at $\infty$). Three of those you can always normalize by a hyperbolic isometry (i.e., a linear fractional transformation), the fourth one is then uniquely determined.

Answer 2

The only possibility for $x$ not to be real would be $x=\infty$. But this is excluded because already the third point is sent to infinity.

To put the question in a more general context: $x$ can actually be determined by using the cross ratio $$(z_1,z_2;z_3,z_4) = \frac{(z_1-z_3)(z_2-z_4)}{(z_2-z_3)(z_1-z_4)}.$$ Namely $$(-1,0,\infty,x)=\frac{-x}{-1-x}=1-\frac{1}{1+x}$$ and because hyperbolic isometries preserve the cross ratio you can compute $x$ from the formula $$(z_1,z_2;z_3,z_4) = 1-\frac{1}{1+x}$$ for the vertices $z_1,z_2,z_3,z_4$ of your original quadrilateral. (And you will see from the formula that it's a real number :-).)