- Here are two properties enjoyed by automorphisms of the upper half-plane.
Suppose \(\Phi\) is an automorphism of H that fixes three distinct points on the real axis. Then Φ is the identity.
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
\[ x _ {1} < x _ {2} < x _ {3} \quad \text { and } \quad y _ {1} < y _ {2} < y _ {3}. \]
Prove that there exists (a unique) automorphism Φ of H so that \(\Phi ( x _ { j } ) = y _ { j }\) ， \(j = 1 , 2 , 3\) . The same conclusion holds if \(y _ { 3 } < y _ { 1 } < y _ { 2 }\) or \(y _ { 2 } < y _ { 3 } < y _ { 1 }\)
Solution
Every automorphism of the upper half-plane has the form \[ \Phi(z)=\frac{az+b}{cz+d}, \qquad a,b,c,d\in\mathbb R, \qquad ad-bc>0. \]
For part (a), a fixed point satisfies \[ cz^2+(d-a)z-b=0. \] If three distinct real numbers are fixed, this quadratic polynomial has three distinct roots and hence is identically zero. Thus \(c=b=0\) and \(d=a\), so \(\Phi(z)=z\).
For part (b), given three distinct real numbers \((x_1,x_2,x_3)\), define \[ M_x(z)=\frac{(x_3-x_1)(z-x_2)}{(x_3-x_2)(z-x_1)}. \] Then \[ M_x(x_1)=\infty,\qquad M_x(x_2)=0,\qquad M_x(x_3)=1. \] Its determinant has the sign of \[ (x_3-x_1)(x_3-x_2)(x_2-x_1). \] For \(x_1<x_2<x_3\) this sign is positive, so \(M_x\) maps \(\mathbb H\) to itself.
Define \(M_y\) analogously. Under each of the three stated cyclic orderings of \((y_1,y_2,y_3)\), \[ (y_3-y_1)(y_3-y_2)(y_2-y_1)>0, \] so \(M_y\) also belongs to \(\operatorname{Aut}(\mathbb H)\). Hence \[ \Phi=M_y^{-1}\circ M_x \] is an automorphism of \(\mathbb H\) satisfying \(\Phi(x_j)=y_j\) for \(j=1,2,3\).
If \(\Psi\) is another such automorphism, then \(\Psi^{-1}\circ\Phi\) fixes \(x_1,x_2,x_3\), so part (a) implies \(\Psi^{-1}\circ\Phi=\mathrm{id}\). Therefore \(\Phi\) is unique.