Recall that the (geodesic) distance on the unit sphere $S^n$ is given by $$ d(p, q) = \arccos \langle p, q \rangle. $$

Let $f_r = f : \mathbb{R}^2 \to S^3$ be defined by

$$f(\theta, \phi) = \left(r \cos \theta, r \sin \theta, \sqrt{1-r^2} \cos \phi, \sqrt{1-r^2} \sin \phi \right),$$

where $r \in (0,1)$ is a parameter, and consider $M = M_r$ to be the image of $f$, which is a 2-torus. My question is: what is (are) the point(s) in the sphere that is (are) the furthest away from $M$? Due to the symmetry of the problem, I suspect it is either $e_1 = (1,0,0,0)$ or $e_3 = (0,0,1,0)$, depending on whether $r \leq \sqrt{1-r^2}$ or $r \geq \sqrt{1-r^2}$. However, I am unable to prove it. Any help would be appreciated :)

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    $\begingroup$ I think symmetry should suggest to you that it'll always be one of the circles $(\cos \theta, \sin \theta, 0, 0)$ or $(0,0,\cos \phi, \sin \phi)$, no? $\endgroup$ – Anthony Carapetis Feb 20 '16 at 3:47

First note that the tori foliate all of $S^3$ (with degeneracy to circles at $r=0,1$), so we can use $r,\theta,\phi$ as coordinates. In these coordinates the inner product appearing in the distance function is (using an angle difference identity): $$\langle (r,\theta,\phi),(r',\theta',\phi') \rangle = r r' \cos(\theta - \theta') + \sqrt{1-r^2}\sqrt{1-r'^2}\cos(\phi - \phi').$$

Let's call the $r$ from your question $R$, so that we're trying to find the point $p = (r_0, \theta_0, \phi_0)$ maximizing distance from the torus given by $r=R$. The inner product formula makes it clear that the closest point on the torus to $p$ will be $(R,\theta_0, \phi_0)$, so we just need to choose $r_0 \in [0,1]$ minimizing $r_0 R + \sqrt{1-r_0^2}\sqrt{1-R^2}$.

If you differentiate this expression you'll see that the only critical point on $[0,1]$ is the maximum at $r_0=R$; so as suspected the solution is either the circle at $r_0 = 0$ or the circle at $r_0 = 1$.

  • $\begingroup$ So the distance of the furthest point to the torus $r = R$ is $\arccos( \min \{ R, \sqrt{1-R^2}\})$, correct? $\endgroup$ – Eduardo Longa Feb 21 '16 at 0:43
  • $\begingroup$ @EduardoLonga: correct. $\endgroup$ – Anthony Carapetis Feb 21 '16 at 0:57
  • $\begingroup$ Thank you very much. Nice answer, by the way. $\endgroup$ – Eduardo Longa Feb 21 '16 at 0:58

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