What is the distance, in the spherical metric on \(\widehat{\mathbb C}\), from \(1\) to \(1+i\)?
Solution
1. The spherical distance is\[ d_{\mathrm{sph}}(1,1+i) =2\arccos\sqrt{\frac56}. \]
For points of \(\widehat{\mathbb C}\cong\mathbb P^1(\mathbb C)\) represented by nonzero vectors \(v,w\in\mathbb C^2\), the spherical distance \(\theta\) satisfies \[ \cos\frac\theta2 =\frac{|\langle v,w\rangle|}{\|v\|\,\|w\|}. \] Represent \[ 1=[1:1], \qquad 1+i=[1+i:1], \] so take \[ v=(1,1), \qquad w=(1+i,1). \] Then \[ |\langle v,w\rangle| =|1(1-i)+1| =|2-i| =\sqrt5, \] while \[ \|v\|=\sqrt2, \qquad \|w\|=\sqrt3. \] Therefore \[ \cos\frac\theta2 =\frac{\sqrt5}{\sqrt6} =\sqrt{\frac56}, \] and hence \[ \theta=2\arccos\sqrt{\frac56}. \] Numerically, this is approximately \(0.841069\) radians.