Let \(a,b\in\Delta\subset\mathbb P^1\) be represented by vectors \(A,B\in\mathbb C^2\), and let \(\delta\) be the distance between \(a\) and \(b\) in the hyperbolic metric \[ \rho_\Delta=\frac{2|dz|}{1-|z|^2}. \] Prove that \[ \cosh(\delta/2)=\frac{|A\cdot B|}{\sqrt{|A^2|\,|B^2|}}, \] where \[ A\cdot B=A_0\overline{B_0}-A_1\overline{B_1}, \qquad A^2=A\cdot A, \qquad B^2=B\cdot B. \]
Solution
1. The quantity\[ \frac{|A\cdot B|}{\sqrt{|A^2|\,|B^2|}} \]depends only on the projective points represented by \(A\) and \(B\).
If \(A\) is replaced by \(\lambda A\) and \(B\) by \(\mu B\), with \(\lambda,\mu\neq0\), then \[ (\lambda A)\cdot(\mu B) =\lambda\overline\mu(A\cdot B), \] while \[ (\lambda A)^2=|\lambda|^2A^2, \qquad (\mu B)^2=|\mu|^2B^2. \] The common factor \(|\lambda||\mu|\) therefore cancels from the ratio.
2. We may consequently take\[ A=(a,1), \qquad B=(b,1). \]Then\[ \frac{|A\cdot B|}{\sqrt{|A^2|\,|B^2|}} =\frac{|1-a\overline b|} {\sqrt{(1-|a|^2)(1-|b|^2)}}. \]
For these representatives, \[ A\cdot B=a\overline b-1, \] and \[ A^2=|a|^2-1, \qquad B^2=|b|^2-1. \] Since \(a,b\in\Delta\), both squared norms are negative, so \[ |A^2|=1-|a|^2, \qquad |B^2|=1-|b|^2. \] Taking absolute values gives the stated expression.
3. Let\[ \phi_b(z)=\frac{z-b}{1-\overline b z}. \]Then \(\phi_b\in\operatorname{Aut}(\Delta)\), \(\phi_b(b)=0\), and if\[ r=|\phi_b(a)|, \]then\[ \delta=d_\Delta(a,b)=d_\Delta(r,0). \]
Automorphisms of the unit disk are isometries for the hyperbolic metric. Thus \[ d_\Delta(a,b) =d_\Delta(\phi_b(a),0). \] A rotation about the origin is also a disk automorphism, so it is an isometry and sends \(\phi_b(a)\) to the positive real number \(r=|\phi_b(a)|\). Hence the distance equals \(d_\Delta(r,0)\).
4. The radial hyperbolic distance from \(0\) to \(r\in[0,1)\) is\[ d_\Delta(0,r) =\log\frac{1+r}{1-r} =2\operatorname{artanh}r. \]Hence\[ \cosh(\delta/2)=\frac1{\sqrt{1-r^2}}. \]
Along the radial segment \([0,r]\), the metric is \[ \frac{2\,dt}{1-t^2}. \] Therefore \[ d_\Delta(0,r) =\int_0^r\frac{2\,dt}{1-t^2} =\log\frac{1+r}{1-r}. \] Thus \(r=\tanh(\delta/2)\), and the identity \(1-\tanh^2x=\operatorname{sech}^2x\) gives \[ \cosh(\delta/2)=\frac1{\sqrt{1-r^2}}. \]
5. For \(r=|\phi_b(a)|\),\[ 1-r^2 =\frac{(1-|a|^2)(1-|b|^2)}{|1-a\overline b|^2}. \]
By definition, \[ r^2=\frac{|a-b|^2}{|1-\overline b a|^2}. \] Hence \[ \begin{aligned} 1-r^2 &=\frac{|1-\overline b a|^2-|a-b|^2} {|1-\overline b a|^2}\\ &=\frac{(1-|a|^2)(1-|b|^2)}{|1-\overline b a|^2}. \end{aligned} \]
6. Therefore\[ \boxed{ \cosh(\delta/2) =\frac{|A\cdot B|}{\sqrt{|A^2|\,|B^2|}}.} \]
Combining 4 and 5 gives \[ \cosh(\delta/2) =\frac{|1-a\overline b|} {\sqrt{(1-|a|^2)(1-|b|^2)}}. \] By 2 this is precisely the projective expression in the statement.