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In geometry, a kissing number is defined as the number of non-overlapping unit spheres that can be arranged such that they each touch another given unit sphere.

Let $\tau_n$ be the kissing number in $n$ dimensions. Kabatiansky and Levenshtein proved the following asymptotic upper bound (ppi1518, mr514023, 1978): $$\tau_n \le 2^{0.401n(1+o(1))} = (1.32\dots)^{n(1+o(1))}$$

Question: What is the smallest $\alpha$ such that $\tau_n \le \alpha^n$, for all $n$?
($\alpha := \min_{n \ge 1} \tau_n^{1/n}$)

By using volume, we can prove that $\tau_n \le \frac{Vol(B(3))-Vol(B(1))}{Vol(B(1))}=3^n-1$, so $\alpha \le 3$.

Now $\tau_2 = 6$, so $\alpha \ge \sqrt 6 \simeq 2.45$. Moreover, for $n \le 24$, $\tau_n^{1/n} \le \sqrt 6$. Is it true that $\alpha = \sqrt 6$?

This post is motivated by arXiv:1710.00285, Section 5.

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It’s almost certainly true, and provable, that $\alpha=\sqrt{6}$, although I haven’t worked out the details rigorously. Kabatiansky and Levenshtein give an exact upper bound (not just an asymptotic expression), which is equation (52) in their paper. Numerical calculations indicate that it improves on $\sqrt{6}$ in dimensions 8 and higher, while other kissing bounds cover dimensions 3 through 7 (see Chapter 1 of Conway and Sloane). So all one has to do is prove this inequality for the Kabatianski-Levenshtein bound in dimensions 8 and up. It suffices to prove an asymptotic bound with explicit constants to handle high dimensions, and then check any remaining dimensions not covered by that bound. I haven’t worked it out, but standard techniques should suffice to do this. (I.e., there’s nothing ineffective about the bounds, so one can obtain explicit constants.) This approach looks ugly and tedious enough that I haven’t worked up the energy to actually do it, but it should work in principle. The main technical input needed is explicit bounds for the largest roots of Gegenbauer polynomials.

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  • $\begingroup$ Do you think that $\sqrt{5}$ is provable as well in dimension $ \ge 4$ ? $\endgroup$ – Sebastien Palcoux Oct 5 '17 at 11:09
  • $\begingroup$ Can we expect $(\tau_n^{1/n})_{n \ge 2}$ to be decreasing ? $\endgroup$ – Sebastien Palcoux Oct 5 '17 at 11:19
  • $\begingroup$ That sequence probably isn’t always decreasing (for example, from 23 to 24 it seems not to be). As for $\sqrt{5}$, I haven’t looked at the numbers carefully, but it sounds plausible. Kabatiansky-Levenshtein will certainly prove this for high enough dimensions, and I believe other upper bounds will cover all the remaining cases before that bound kicks in (but I haven’t looked at this carefully, so I’m not as confident as for $\sqrt{6}$). $\endgroup$ – Henry Cohn Oct 5 '17 at 11:43

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