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Let P be a finite set of points on the unit sphere $S^2$ such that

  • for every $p\in P$, there exists a closed curve $\gamma_p \subset S^2$ which has a self intersection at $p$ and passes through $-p$. Moreover, every plane passing through $p$ and the origin intersect $\gamma_p$ in at most 2 points, excluding $-p$ (see the graph below by Neil Strickland).

  • At the self intersection point $p$, $\gamma_p$ is orthogonal to itself.

  • $\gamma_{p_1} \cap \gamma_{p_2} \subset P$ for every $p_1,p_2 \in P$

  • If $A_i \subset S^2$ be such that $\partial A_i \subset \gamma_{p_i}$, $p_i\in P$, i=1,2, then neither $A_1 \subset A_2$ nor $A_2 \subset A_1$.

  • $P \subset \gamma_p$ for all $p\in P$.

Can we prove that $P$ has either one or infinite elements? It seems that more than one element leads to infinite number of elements in $P$. It is easy to see that if $P$ has more than one point, then it has at least 6 points.

Please see the picture provided by Neil Strickland for a visualization of $\gamma_p$.

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    $\begingroup$ Please write couple of words on motivation. $\endgroup$ Sep 5, 2017 at 20:51
  • $\begingroup$ Is it OK to assume that $\gamma_p$ is smooth? $\endgroup$ Sep 5, 2017 at 23:01
  • $\begingroup$ Yes, $\gamma_p$ is indeed smooth. $\endgroup$ Sep 5, 2017 at 23:24
  • $\begingroup$ I find it hard to imagine even two of these curves satisfying all assumptions. Could you draw a picture with two or three curves, please? $\endgroup$ Sep 6, 2017 at 16:26
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    $\begingroup$ No, just excluding -p. It is indeed at most three points including all points. $\endgroup$ Jan 21, 2023 at 14:25

1 Answer 1

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This is just a comment, really. One can write a nice formula for a family of curves of the indicated type, as follows. Let $p$, $q$ and $r$ be an orthonormal basis of $\mathbb{R}^3$, and suppose that $a>0$. Put $$ C = \{x\in S^2: (q.x)\,(r.x)=a((q-r).x)(1 - p.x)\} $$ enter image description here

This can be parametrised as $$ \gamma(t) = \frac{(a^2(c-s)^2-s^2c^2)p+2asc(c-s)(cq+sr)}{a^2(c-s)^2+s^2c^2}, $$ where $s=\sin(t/2)$ and $c=\cos(t/2)$. This passes through $p$ with derivative $q/a$ at $t=0$ and through $p$ again with derivative $r/a$ at $t=\pi$. It also passes through $-p$ with derivative $-2a(q+r)$ at $t=\pi/2$. Maple code is as follows:

g := unapply(subs({s = sin(t/2),c=cos(t/2)},((a^2*( c-s)^2-s^2*c^2) *~ 
 [1,0,0] +~ (2*a*( c-s)*( s*c)) *~ [0, c, s])/~(a^2*( c-s)^2+s^2*c^2)),t);

Obviously this does not answer the question, but you might gain some intuition by plotting curves of this type.

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  • $\begingroup$ Thank you Neil. The problem is still not solved and I have offered another bounty. $\endgroup$ Jan 21, 2023 at 16:34

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