STATEMENT: Suppose $(x_1,x_2,x_3)$ and $(y_1,y_2,y_3)$ are two pairs of three distinct points on the real axis with$$x_1<x_2<x_3 \;\;\;\;\text{and} \;\;\;\;\;y_1<y_2<y_3$$
Prove that there exists (a unique) automorphism $\Phi$ of $\mathbb{H}$ so that $\Phi(x_j)=y_j$, $j=1,2,3$.
QUESTION: I am uncertain about how to proceed with this problem. I know that the group of automorphisms of the upper half plane is transitive, but I don't see how this will help me. Any hints would be appreciated.
Best Answer
We can identify the group of automorphisms of the upper half plane with $PSL(2,\mathbb{R})$, the group of $2\times 2$ matrices with determinant 1, up to $\pm I$. That is, they are transformations of the form $$z \rightarrow \frac{az + b}{cz + d},$$ where $ad - bc = 1$ and $(a,b,c,d) \sim (\alpha a, \alpha b, \alpha c, \alpha d)$ for $\alpha \neq 0$, $a, b, c, d \in \mathbb{R}$. Observe that this is a three-parameter family.
If there were two such automorphisms $\Phi_1$ and $\Phi_2$, then $\Phi_1^{-1}\circ\Phi_2(x_j) = x_j$ for $j = 1, 2, 3$, i.e. there are three fixed points. The only automorphism which has three fixed points is the identity, so $\Phi_1^{-1}\circ\Phi_2 = I$. (Why? If $(az+b)/(cz+d) = z$ is a fixed point, then unless $c = a = 1$ and $b = d = 0$, we obtain at best a quadratic with at most two roots, by the fundamental theorem of algebra.)