4
$\begingroup$

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 :)

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $r_0 = 1$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.