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Let $A_d(n)$ be the largest number of points that can be packed on the $n$-unit sphere, such that every point is at least $d$ apart. Compare with, for instance, https://arxiv.org/abs/1507.03631

When d=1, these are the standard kissing numbers, and they are known to grow exponentially in n. For any smaller d we can see that it must be exponential as well, because there is a simple exponential upper bound based on solid angles subtended by $A$-many circles on the surface.

For d=√2, these are vectors that are "at least" orthogonal (nonpositive inner product), and $A_d(n)$ is just 2n, in an octahedral packing. This implies that for any d>√2, the growth in n is at most linear.

On a smaller note, for d≥√3, A_d(n) is a constant in n.

Question: What happens for values of $d \in (1,\sqrt 2)$? Is it true that $A_d(n)$ is always either asymptotically linear, or at least exponential? If so, is the transition at d=√2 exactly?

And a side question, is $A_d(n)$ always either asymptotically linear or constant on $d\in(\sqrt 2,\sqrt 3)$?

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  • $\begingroup$ I think the answer to the main question is that $A_d(n)$ is exponential in $n$ for all $d<\sqrt{2}$, see e.g. mathoverflow.net/questions/158575/… $\endgroup$ – j.c. Apr 1 '19 at 18:08
  • $\begingroup$ @j.c. Mm, yes, that definitely answers the first question! Sorry for such a near duplicate, then, I hadn't been able to find an answer but this is definitely is. $\endgroup$ – Alex Meiburg Apr 1 '19 at 18:10
  • $\begingroup$ By Rankin's bound (see mathoverflow.net/questions/208484/…), A_d(n) is asymptotically linear for all d < √2 $\endgroup$ – Yoav Kallus Apr 2 '19 at 2:41
  • $\begingroup$ @YoavKallus Is there a typo in your comment or did I screw something up in my comment? $\endgroup$ – j.c. Apr 3 '19 at 1:09
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    $\begingroup$ @j.c. My screw up. It should read "By Rankin's bound, A_d(n) is asymptotically constant for all d > √2". Thanks for catching this. $\endgroup$ – Yoav Kallus Apr 3 '19 at 13:05

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