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I'm reading the Thompson's book about lattices and sphere packing and got stuck by a sentence of a kind of $Z_8$ he introduced to reach 2 pages later the full $E_8$ lattice. You can find this lattice defined at pages 73-74 and it's basically. To resume it, it's a lattice packing with 16 closest point to the origin having shape: $$(\pm2, 0^7)$$ The packing radius is then $1/2$ of their distance from the origin, i.e. $\rho=1$. So far, so good.

My problem is when he tries to compute the center density of the lattice. Notice this center density can be interpreted as the "real" sphere center density since $\rho=1$, as claimed in SPLAG when describing the formula for $\delta$.

Instead of using any formula, Thompson uses a clever idea to estimate it, which sound like this:

The center density is fairly easy to calculate. If all coordinate entries were written in binary form, then the lattice, by definition, would contain only those coordinates whose ones digits were either all 0's or 1's. In this case the only two out of every $2^8$ points with integer coordinates are acceptable. Thus, the center density = $1/2^7$

I've got 2 problems with this result.

The first is I can choose for this lattice a generating matrix made by only 2 in all diagonal entries, i.e. twice the identity matrix. The determinant of this would then $2^8$. Using SPLAG formula for center density, and keeping $\rho=1$, I would get $\delta=1/2^8$, which is smaller by a factor of 2 compared to the one claimed by Thompson.

To confirm this latest sentence: as far as I can see, the lattice defined above can be seen as $Z_8$ lattice, which density is (always from SPLAG), $\delta=1/2^8$

However Thompson is using this $1/2^7$ to derive the full $E_8$ lattice, so I'm not claiming it's wrong a priori. But I'd like to understand where my reasoning is wrong and how to express coordinates in that binary format (I'm a programmer, so used to binary digits) to emulates Thompson's idea.

Thanks in advance

riccardoventrella
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  • Link to the text: https://books.google.co.uk/books?id=ggqxuG31B3cC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q=The%20center%20density%20is%20fairly%20easy%20to%20calculate.%20If%20all%20coordinate%20entries%20were%20written%20in%20binary%20form%2C%20then%20the%20lattice%2C%20by%20definition%2C%20would%20contain%20only%20those%20coordinates%20whose%20ones%20digits%20&f=false – Heterotic Jul 02 '20 at 09:08
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    I think the answer is that twice the identity matrix does NOT describe the matrices that Tompson introduces at the bottom of page 73. You need 7 rows of twice the identity matrix but you also need a full row of 1s and this changes things. – Heterotic Jul 02 '20 at 09:10
  • I had the same feeling and I tried and of course the det was correct, as you mentioned, However, I cannot understand why the matrix cannot be made by all 2 in the diagonal. All in all, it should be made by basis vectors and I thought putting 2 in each slot would create a legitimate basis. Can you help me better in understanding that? More than this, why this could not be considered as a integral lattice in 8 dimension, i.e. $Z_8$? Since in that case the density should be $1/2^8$, as told in SPLAG at page 106. Finally, I cannot understand the "binary digits" trick Thompson used to claim it. – riccardoventrella Jul 02 '20 at 11:17
  • I was thinking all basis should be minimal contact vectors, while $(1, 1, 1, 1, 1, 1, 1, 1)$ is not, since with distance $2\sqrt2$ while the others $(2, 0^7)$ have distance 2 – riccardoventrella Jul 02 '20 at 11:31
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    Putting in 2 in each slot creates a legitimate basis, but for a different lattice! The lattice as introduced on page 73 is not the collection of vectors with all coordinates even. That would indeed be generated by 2xId$_8$. – Heterotic Jul 02 '20 at 14:27
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    However, the lattice the author wants to study is the collection of vectors with integer coordinates such that the sum of all coordinates is always even. 2xId$_8$ would be a sub-lattice of this lattice and there are other vectors such as (1,1,..,1) that belong to the former but not the latter. That's why he takes that particular basis. As to the "binary digits" trick, I have no idea either, probably because I cannot remember the definition of center density. – Heterotic Jul 02 '20 at 14:28
  • Yes I see now, I had the same feeling it was a sub-lattice, but your confirmation is fundamental, now. The $2xId_8$ is exactly a scaled version of the $Z_8$ lattice, i.e. the integral lattice generated by all versors, but scaled by 2. All is clear now, I was considering the wrong sub-lattice. The center densiy is defined in SPLAG as $\delta=\frac{\rho^n}{Det(\Lambda)^\frac{1}{2}}$, where $\rho$ is the radius and the denominator is volume of the fundamental region. As SPLAG points out, when $\rho=1$ this can be interpreted as sphere centers per unit volume. I will dump some digits in binary – riccardoventrella Jul 02 '20 at 15:15
  • @Heterotic, may you rank your comment higher to an answer, so I can vote for it? Your explanation was fundamental, thanks. – riccardoventrella Jul 02 '20 at 15:41

2 Answers2

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Putting in 2 in each slot creates a legitimate basis, but for a different lattice! The lattice as introduced on page 73 is not the collection of vectors with all coordinates even. That would indeed be generated by 2xId$_8$. However, the lattice the author wants to study is the collection of vectors with integer coordinates such that the sum of all coordinates is always even. 2xId$_8$ would be a sub-lattice of this lattice and there are other vectors such as (1,1,..,1) that belong to the former but not the latter. That's why he takes that particular basis. As to the "binary digits" trick, I have no idea either, probably because I cannot remember the definition of center density.

Heterotic
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I think to have found the explanation of the binary trick. First of all, let's consider this lattice has all coordinates even or all odd, by definition. The idea was suggested to me in the way in which Leech enriched the extended Golay code $C_{24}$. Let's write a generic point stacking the binary digits in vertical (let's consider positive numbers up to 15 for the sake of simplicity and without loss of generality).

$$(5, 9, 1, 3, 1,....) = \begin{bmatrix}0 & 1 & 0 & 0 & 0 &...\\1 & 0 & 0 & 0 & 0 &...\\0 & 0 & 0 & 1 & 0 &...\\1 & 1 & 1 & 1 & 1 &...\end{bmatrix}$$

, where the top row is the 8s digit while the bottom row is the 1s digit.

It's easy to see, for the case of 8 coordinates like for this lattice, the bottom row will be made by these couple of octuplets ONLY: $(0, 0, 0, 0, 0, 0, 0, 0)$ and $(1, 1, 1, 1, 1, 1, 1, 1)$. This is because we have claimed above the coordinates for this lattice are all odd or all even. If we consider 8 generic integer coordinates, it's easy to see the bottom row could span from all 0's to all 1's, passing through all the $2^8=256$ possibilities. As claimed by the book author, then, the density will be the ratio of the couple combinations over the overall possible ones, i.e. $\delta=\frac{2}{2^8}=\frac{1}{2^7}$. It's easy to extend this proof encompassing integral numbers > 15 (so having any kind of row numbers) and negative (considering a kind of 2-complement representation $-6=2(mod 8)$, like Leech did). Finally, we can use the same trick to estimate quickly the center density of the $D_4$ checkerboard lattice (i.e. the densest packing in 4 dimensions), which has the same property of having all coordinates even or all odd. Here we will have 4 coordinates to be represented in binary form, which forms an overall $2^4$ possible combinations for their 1s digit. But only the $(0, 0, 0, 0)$ and $(1, 1, 1, 1)$ out of them will host the $D_4$ lattice points, which nails the center density to $\delta=\frac{2}{2^4}=\frac{1}{8}$. This matches exactly the center density enumerated in SPLAG 1.2 Table pag.15.

riccardoventrella
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