21
$\begingroup$

Given the march 2016 breakthrough concerning sphere packings by Viazovska for the case of dimension 8, and by Cohn, Kumar, Miller, Radchenko and Viazovska for the case of dimension 24, it follows that the known cases $\Delta_1=1$, $\Delta_2=\frac{\pi}{\sqrt{12}}$, $\Delta_3=\frac{\pi}{\sqrt{18}}$, $\Delta_8=\frac{\pi^4}{384}$ and $\Delta_{24}=\frac{\pi^{12}}{12!}$ all have a lattice solution with neat number-theoretic properties, yet the densities obtained do not seem to follow any obvious pattern, except for the presence of $\pi$.

Question A: Is there a consensus among experts that one should not hope for a synthetic formula for densities, or are there still reasonable candidate formulas standing ?

and

Question B: are there workshops planned on the recent papers soon (say, this fall, maybe even this summer) ?

$\endgroup$
6
  • $\begingroup$ Yes, apologies, now edited I hope ! $\endgroup$
    – Archie
    Commented Apr 4, 2016 at 18:58
  • 12
    $\begingroup$ Any integer lattice packing in dimension $n$ has a density of the form $\pi^{\lfloor \frac{n}{2} \rfloor} $ times the square root of a rational number, because of the formula for the volume of the sphere. Because the optimal packings in dimensions $1,2,3,8,24$ are integer lattices, this explains the occurrence of $\pi$. $\endgroup$
    – Will Sawin
    Commented Apr 4, 2016 at 19:05
  • $\begingroup$ I think it is generally believed that the optimal packing in most dimensions is not a lattice packing... in fact, dimension 24 may be the last time where the optimal packing is a lattice packing. I believe in dimension 10 it is known that the best packing is not a lattice packing $\endgroup$ Commented Apr 5, 2016 at 11:10
  • $\begingroup$ @StanleyYaoXiao Then there would have to be a reason that says why 24 is the final number (perhaps grounded in fixed point theory, no idea)... $\endgroup$
    – Suvrit
    Commented Apr 5, 2016 at 13:56
  • $\begingroup$ @Suvrit, I think there is some reason grounded in group theory and having to do with the Monster (perhaps someone knowledgeable can elaborate). $\endgroup$
    – usul
    Commented Apr 5, 2016 at 18:19

1 Answer 1

18
$\begingroup$

Here's an attempt at answering your Question A:

Currently, one of the most powerful methods for proving upper bounds on sphere packing densities is the linear programming bound of Cohn and Elkies (which is what Viazovska used).

Exact answers

According to the numerical computations of Cohn and Elkies, we do not know any dimensions other than 1, 2, 8, and 24 where their linear program has a chance of proving the exact bound (see their Figure 1). There are exceptionally good lattices in those four dimensions that match the bounds given by the Cohn-Elkies linear program. It could be the case (likely?) that these are the only dimensions where the bounds match, although we don't know how to rule out other possibilities. There could be some deep reason (perhaps related to finite simple groups) why these four dimensions are special.

Hales' proof of Kepler's conjecture (sphere packing in dimension 3) uses a completely different method.

Asymptotic bounds

An important open question, for which there is still an exponential amount of room for improvement, is the problem of sphere packing bounds in very high dimensions.

The current best asymptotic upper bound to the packing density $\Delta_n$ in $\mathbb{R}^n$ is by Kabatiansky and Levenshtein (1977) $$ \Delta_n \le 2^{-(0.5990\dots + o(1))n} $$ (see, e.g., my paper with Henry Cohn on the matter; in particular, this asymptotic bound lies within the scope of the Cohn-Elkies linear program). The current best lower bound is due to Venkatesh. For all $n$, $$ \Delta_n \ge c n 2^{-n} $$ for some explicit $c$ (that has been improved over time), and for a specific infinite set of $n$, $$ \Delta_n \ge c n (\log\log n) 2^{-n}. $$

Closing the gap seems to be a difficult problem.

Further reading

$\endgroup$
2
  • $\begingroup$ Thanks for this answer. I was wondering, is there an intuitive answer for why the densest packing density decreases with dimension? $\endgroup$
    – user88381
    Commented Apr 12, 2016 at 14:43
  • $\begingroup$ I don't have a good answer I'm afraid. Our intuition for how packings work in high dimensions is rather poor. As far as I know, there is no proof that $\Delta_{n+1} \le \Delta_n$ for all sufficiently large $n$. Even if you know a good packing in $n$ dimensions, you might not get a good packing by tiling it in one higher dimension (although this strategy works well in low dimensions). See Conway and Sloane for a lot more. $\endgroup$
    – Yufei Zhao
    Commented Apr 12, 2016 at 20:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .