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In this paper by Viazovska, she said that:

  • "The E8-lattice sphere packing 𝒫E8 is the union of open Euclidean balls with centers at the lattice points and radius $1/\sqrt{2}$." So I think the distance between centers of the two balls is $$D=2 \cdot 1/\sqrt{2}=\sqrt{2}.$$

She then said:

  • The Leech lattice Ξ›24 was constructed by J. Leech in 1967. This lattice is an even unimodular lattice of rank 24. There exist 24 isomorphism classes of such lattices. Among these 24, the Leech lattice is the unique one having the shortest non-zero vector of length 2. In the other 23 classes, the shortest vector has minimal possible for even lattices length $$D=√2.$$

As the minimal distance between two points in Ξ›24 is $$D=2,$$ it is a good candidate for a dense sphere packing

My question is this:

  1. Leech lattice $Ξ›24$ has larger distance between centers of the two balls (namely $D=2$), in contrast with other 23 classes of 24-dimensional lattice which as $D=√2$, also in contrast with the E8 lattice which as $D=√2$. Am I correct to summarize this way?

  2. What are the sizes of unit balls? Of radius $R=1/\sqrt{2}$ for the case of E8 and of 23 other 24-dimensional lattices? Of radius also $R=1/\sqrt{2}$ or $R=1$ for the Leech lattice? How come the unit ball sizes are different with different radius $R$ for these cases??

Is the relation: shortest vector $$D=2R?$$ as twice of the radius of unit ball always hold?

  1. If so, how can larger $D$ give rise to denser sphere packing? It is very counter intuitive! Can we explain in some intuitive ways then on the larger $D$ denser sphere packing?

Thanks for replying and answering!

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  1. Leech lattice $Ξ›24$ has larger distance between centers of the two balls (namely $D=2$), in contrast with other 23 classes of 24-dimensional lattice which as $D=√2$, also in contrast with the E8 lattice which as $D=√2$. Am I correct to summarize this way?

Yes - that looks right to me.

  1. What are the sizes of unit balls? Of radius $R=1/\sqrt{2}$ for the case of E8 and of 23 other 24-dimensional lattices?

Yes.

Of radius also $R=1/\sqrt{2}$ or $R=1$ for the Leech lattice?

$R=1$.

How come the unit ball sizes are different with different radius $R$ for these cases??

One chooses the largest ball that will fit, which is different for different lattices.

Alternatively, one can use balls of a fixed radius, but then rescale the lattice so the balls just barely touch without overlapping. If we choose balls of radius $1$, we must scale the other lattices up by a factor of $\sqrt{2}$ to make the balls fit.

Is the relation: shortest vector $$D=2R?$$ as twice of the radius of unit ball always hold?

Yes, this is the condition that means the balls just barely touch without overlapping.

  1. If so, how can larger $D$ give rise to denser sphere packing? It is very counter intuitive! Can we explain in some intuitive ways then on the larger $D$ denser sphere packing?

The density refers to the proportion of space taken up by balls (we can also think of this as the literal physical density, mass of spheres divided by volume of space, if the spheres have some fixed mass per unit volume). The larger $D$ is, the larger $R$ is, and thus the more space each ball takes up.

Alternately, if we fix the radius of the ball, we can define the density to be the average number of balls in a given volume of space. In this case, we need to scale the lattice by a factor of $2/D$. Thus, the larger $D$ is, the more we can scale the lattice down (or the less we need to scale that lattice up), and scaling it down will make it more dense.

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  • $\begingroup$ Many thanks -- I appreciate your answer. Any refs are welcome! $\endgroup$
    – zeta
    May 29, 2023 at 17:32
  • $\begingroup$ Thanks -- "Alternatively, one can use balls of a fixed radius, but then rescale the lattice so the balls just barely touch without overlapping. If we choose balls of radius 1, we must scale the other lattices up by a factor of √2 to make the balls fit." in the case, why not just choose $R=1$ and $D=2$ universally? This seems to work for the Leech lattice, but we multiply $√2$ for other 23 cases and E8 cases? so $√2 \times$ ($R=1/√2$ and $D=√2$)? $\endgroup$
    – zeta
    May 29, 2023 at 20:36
  • $\begingroup$ are you able to confirm it -- i did vote it up. Otherwise I could not make sure to accept the answer 100%? thanks $\endgroup$
    – zeta
    Nov 25, 2023 at 21:31
  • $\begingroup$ @zeta The only reason for not choosing $R=1$ and $D=2$ universally is that the lattices in question can be expressed via quadratic form with nice properties (integer-valued, unimodular) with these values of $R$ and $D$. When studying the sphere-packing problem analytically, one does usually fix $R=1$. $\endgroup$
    – Will Sawin
    Nov 29, 2023 at 21:37

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