# Do kissing numbers with distance $d$ always grow polynomially or exponentially in dimension?

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 $$d\in(\sqrt 2,\sqrt 3)$$?

• 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/… – j.c. Apr 1 '19 at 18:08
• @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. – Alex Meiburg Apr 1 '19 at 18:10
• By Rankin's bound (see mathoverflow.net/questions/208484/…), A_d(n) is asymptotically linear for all d < √2 – Yoav Kallus Apr 2 '19 at 2:41
• @YoavKallus Is there a typo in your comment or did I screw something up in my comment? – j.c. Apr 3 '19 at 1:09
• @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. – Yoav Kallus Apr 3 '19 at 13:05