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I recently came across some lattices which really interested me and, during my computations and trials, I've found one which seems equivalent to the D4 Checkerboard lattice.
I was looking to a similar lattice but with an higher packing radius, i.e.: $$\rho = 1$$
, while D4 is known to have $\rho = \frac{\sqrt{2}}{2}$, like other Dn lattices.
I started taking all couples $$(\pm1^{4})$$
which currently are far from the origin exactly 2, i.e. can be all put in tangency with a unitary sphere centred in the origin. This actually gives us a kissing number of 16.
Knowing the highest kissing number in 4D is 24, I've tried to reach that threshold, adding some further vectors having form:$$(\pm2, 0^{3})$$
They are in tangency with the same unitary centred sphere in the origin, and they are kissing the other 16 spheres as well. Actually this lead to a total of 24 as kissing number, which is optimal in 4D.
I then tried to compute the packing density for this lattice. In 4D the generic formula reads:$$\Delta = \frac{\pi^{2}}{2}\rho^{4}\frac{1}{det(L_4)^{\frac12}}$$
, where $L_4$ is the lattice I'm taking into account. Looking to SPLAG, we can compute the generating matrix, which squared will lead to the lattice determinant. I've taken:$$M = \begin{matrix} 2 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ -1 & 1 & 1 & -1 \\ -1 & -1 & 1 & 1\end{matrix}$$
, which has $det(M) = 8$, leading to $det(L_4) = 64$. Putting all stuff together and taking $\rho=1$, we finally got: $$\Delta=\frac{\pi^{2}}{16}$$
Actually, this is exactly the same packing density we got from the densest D4. How this is possible? If I'm right, we raised the packing radius having an overall improvement due to it by $\sqrt{2}^{4}=4$ but at the same time we lowered the center density from $\frac12$ to $\frac18$, i.e. reducing by $\frac14$. So, this two 2 contributions cancel out, leading to the same packing density as D4.

The first question is: are my computations leading to meaningful results, from your POV?

After a while, I found a similar lattice in Thompson's From Error Correcting Codes through sphere packings to simple groups, precisely at page 73, quoting it:

One orientation of the densest lattice packing in $E^{4}$ has as sphere centers the set of all quadruples of either all even or all odd integers

The matrix is just another sequence of independent vectors similar to mine. Actually this explains also why we got a less dense center density, since $L_4$ lattice can be obtained extracting a subset of D4, with the rule pointed out by Thompson above.

The other two questions are:
does $L_4$ has a name? Actually it's not D4, even it has the same density and kissing number. Does $L_4$ can be obtained by a transformation of D4? I've seen in SPLAG, pag. 10, a lattice can be defined equivalent or similar if the two generator matrix can be put in this relation: $$M^{`} = c U M B$$
, where c is a nonzero constant, U is a matrix with integer entries and $det(U) = \pm1$ and B is an orthogonal matrix. More than this, if c=1 they can be called congrunet lattices.

Actually, given the density and the kissing numbers I'm conjecturing they are equivalent, or even congruent, but I'm not able to find those matrices and constant to prove this. Can you give me any hint with this, please?

Thank for your support

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  • $\begingroup$ math.rwth-aachen.de/~Gabriele.Nebe/LATTICES/d4to457.html $\endgroup$ – Will Jagy Aug 6 '17 at 19:03
  • $\begingroup$ Thanks @Will, I already encountered this site during my navigation through lattices. I'm studying quadratic forms both using the amazing Conway's "The sensual quadratic form" and directly on SPLAG. However, my knowledge on them is still at the beginning so, possibly, may you tell me exactly how this page can answer to my question? Thanks in advance $\endgroup$ – riccardoventrella Aug 6 '17 at 21:47
  • $\begingroup$ Well, if you can render the two candidate lattices in Nipp's format, ten integers in order, I can tell you whether they are the same. Hmmm. Reading again, you have a lattice (form) and you don't know where it fits in Nipp's list. Carefully put it in Nipp's format and I can do that. It does not need to be "reduced." I do, however, need the Gram matrix and its determinant. $\endgroup$ – Will Jagy Aug 6 '17 at 22:04
  • $\begingroup$ I got your point @Will. I will try to put it in Nipp's format ASAP, thanks. $\endgroup$ – riccardoventrella Aug 7 '17 at 9:01
  • $\begingroup$ @Will, I post here the Gram matrices for both the reference D4 and "my" $L_4$, together with a computation of center densities of both. $\endgroup$ – riccardoventrella Aug 7 '17 at 11:26
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I am trying this: if your $M$ rows are a lattice basis, the Gram matrix is $MM^T.$ The unusual aspect here is that everything is divisible by another factor of 2. So, $$ H = M M^T / 2 = \left( \begin{array}{rrrr} 2 & 1 & -1 & -1 \\ 1 & 2 & 0 & 0 \\ -1 & 0 & 2 & 0 \\ -1 & 0 & 0 & 2 \end{array} \right) $$

The Gram matrix for $D_4$ in INDEX is

$$ G = \left( \begin{array}{rrrr} 2 & 0 & 1 & 0 \\ 0 & 2 & -1 & 0 \\ 1 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{array} \right) $$

These two have the same characteristic polynomial, $x^4 - 8x^3 + 21x^2 - 20x + 4.$ They are congruent, by symmetric integer matrices.

===============================

? g
%17 = 
[2 0 1 0]

[0 2 -1 0]

[1 -1 2 -1]

[0 0 -1 2]

? h
%18 = 
[2 1 -1 -1]

[1 2 0 0]

[-1 0 2 0]

[-1 0 0 2]

? p
%19 = 
[-1 0 1 0]

[0 -1 0 0]

[1 0 0 0]

[0 0 0 1]

? matdet(p)
%20 = 1
? p * g * p 
%21 = 
[2 1 -1 -1]

[1 2 0 0]

[-1 0 2 0]

[-1 0 0 2]

? p * g * p  - h
%22 = 
[0 0 0 0]

[0 0 0 0]

[0 0 0 0]

[0 0 0 0]

================================================

? 
? q = matadjoint(p)
%23 = 
[0 0 1 0]

[0 -1 0 0]

[1 0 1 0]

[0 0 0 1]

? q * p
%24 = 
[1 0 0 0]

[0 1 0 0]

[0 0 1 0]

[0 0 0 1]

? q * h * q
%25 = 
[2 0 1 0]

[0 2 -1 0]

[1 -1 2 -1]

[0 0 -1 2]

? q * h * q - g
%26 = 
[0 0 0 0]

[0 0 0 0]

[0 0 0 0]

[0 0 0 0]

=================================================

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  • $\begingroup$ Sorry @Will, you lost me here. How did you compute p? EDIT: OK now I see your point. So, this is basically the U matrix in the SPLAG formual above, I guess. So they are congruent, as I was conjecturing. Does my result to have them isomorphic thanks to their $\delta$ is coherent with this? $\endgroup$ – riccardoventrella Aug 7 '17 at 19:08
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@Will, I'm answering here since the comment would be quite messy. Here the Gram matrices you asked:

D4 (got from SPLAG at pag 9): $$ M = \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \end{array} \right)$$
$$ A = MM^T = \left( \begin{array}{rrrr} 4 & 2 & 2 & 2 \\ 2 & 2 & 1 & 1 \\ 2 & 1 & 2 & 1 \\ 2 & 1 & 1 & 2 \end{array} \right),Det(A) = 4$$

"My" $L_4$:$$ M =\left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ -1 & 1 & 1 & -1 \\ -1 & -1 & 1 & 1 \end{array} \right)$$
$$ A = MM^T = \left( \begin{array}{rrrr} 4 & 2 & -2 & -2 \\ 2 & 4 & 0 & 0 \\ -2 & 0 & 4 & 0 \\ -2 & 0 & 0 & 4 \end{array} \right), Det(A) = 64$$

I'd like to compute better the center densities, since I found a mistake in my original post. In SPLAG, you can see:$$ \delta = \rho^n(det(\Lambda))^\frac12$$

So, for the D4, we have:$$\delta = 2^{-\frac42}4^{-\frac12} = 1/8$$

For $L_4$ we have:$$\delta = 64^{-\frac12} = 1/8$$

So the two lattices have the same center density and, if I'm not wrong, they are isomorphic for this reason. Do you agree with that?

That said, I'm still interested in porting the two lattices in the format you were talking about,since I'd like to get more involved with quadratic forms, and you may eventually check my results for this.

EDIT: @Will, I think your matrix is the one I was searching for stating the equivalence. Looking at SPLAG pag. 10, we can use the formula (23), i.e: $$A^` = c^2*U*A*U^T$$. So, U is exactly your P matrix. However, as you noticed, the original Gram matrix for $L_4$ is actually divided by 2, so $c^2$ in the formula must match 2, i.e. $c=2^\frac12$. This factor should be exactly the scale factor applied between the two bases and reflected by the packing radius as well. However, as stated in SPLAG, your matrix has perfectly det(U) = 1, but c is different than 1, so we cannot state the two lattices are congruent, but we can just just they are equivalent or similar. Actually this creates some doubts to me about my logic of having them isomorphic thank to their $\delta$.

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  • $\begingroup$ The $D_4$ lattice has all sums of coordinate entries even. You have a sublattice, in which all entries are the same $\pmod 2.$ There are, indeed, 24 points in your sublattice at distance $2$ from the origin, so that is the kissing number you wanted. In $D_4$ you also get kissing number 24, pairwise center distances now $\sqrt 2.$ So, your sublattice is also scaled by $\sqrt 2$ from $D_4.$ $\endgroup$ – Will Jagy Aug 7 '17 at 22:44
  • $\begingroup$ Thanks @Will, as told above $\sqrt{2}$ is exactly the c constant depicted in the Conway's formula. So the resulting lattice is generated by a basis which is a scaled permutation (as you found) of D4 one. However, your D4 Gram matrix is smarter than mine (i.e. the one I've taken from SPLAG) since let us to find the congruent matrix quite quickly, while mine would have been a bit more tricky. Finally, since I don't know exactly the exact meaning of isomorphism between lattices, I'm not sure we can claim their are isomorphic thank to their $\delta$. $\endgroup$ – riccardoventrella Aug 8 '17 at 7:16
  • $\begingroup$ @Will, I was trying to figure out how you claimed "in which all entries are the same (mod2)". I've tried to apply the modding to all entries, but I was not able to achieve your result. More than this I was trying to apply your reasoning about the modding to the centres per unit volume: actually D4, since it's a checkerboard lattice, has the centre density = $\frac12$. My $L_4$ lattice has centre density = $\frac18$ (the inverse of its determinant). So, the density is not halved but lowered to $\frac14$. I cannot make consistent this ratio with the mod 2 you are mentioning. Where am I wrong? $\endgroup$ – riccardoventrella Aug 9 '17 at 10:18
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On pages 4 and 5 of SPLAG we find that a generator matrix need not be square or have rational entries. The rows of $M$ are the basis vectors (page 4) $$ M = \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ -1 & 1 & 1 & -1 \\ -1 & -1 & 1 & 1 \end{array} \right) $$
Your basis vectors are always all even or all odd integers. Constant $\pmod 2.$ Can we write any integer vector, constant $\pmod 2,$ in the form $\xi M?$ Page 4, formula(2).

$$ \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ -1 & 1 & 1 & -1 \\ -1 & -1 & 1 & 1 \end{array} \right) = \left( \begin{array}{rrrr} 1 & 1 & 1 & 1 \end{array} \right) $$

$$ \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ -1 & 1 & 1 & -1 \\ -1 & -1 & 1 & 1 \end{array} \right) = \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \end{array} \right) $$

$$ \left( \begin{array}{rrrr} -1 & 1 & 0 & -1 \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ -1 & 1 & 1 & -1 \\ -1 & -1 & 1 & 1 \end{array} \right) = \left( \begin{array}{rrrr} 0 & 2 & 0 & 0 \end{array} \right) $$

$$ \left( \begin{array}{rrrr} 1 & 0 & 1 & 1 \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ -1 & 1 & 1 & -1 \\ -1 & -1 & 1 & 1 \end{array} \right) = \left( \begin{array}{rrrr} 0 & 0 & 2 & 0 \end{array} \right) $$

$$ \left( \begin{array}{rrrr} -1 & 1 & -1 & 0 \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ -1 & 1 & 1 & -1 \\ -1 & -1 & 1 & 1 \end{array} \right) = \left( \begin{array}{rrrr} 0 & 0 & 0 & 2 \end{array} \right) $$

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  • $\begingroup$ Thank @Will, actually you are using xM = y, where x are the components of the vector y evaluated in the M basis. So, if I'm correct, in the first equation you expressed the (1, 1, 1, 1) vector in my basis, i.e. mod 2. However, I'm still struggling in understanding how this can be a mod 2, since I have -1 as well (I'm a Math developer and I'm used to the % 2 operation, giving 0 for even and 1 for odd). I think I'm still missing something, probably in how you are taking into accounts even and odd components. $\endgroup$ – riccardoventrella Aug 10 '17 at 16:56
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Thanks @Will, I see your point, i.e. the way in which you are using formula (2) of page 4 and makes sense to me. However, I'm struggling a bit with "Constant (mod 2)". As far as I know (I'm a math developer), the mod 2 operation ( % 2 in C language) actually can lead to 0 with even input or 1 with odd input. I'm trying to understand this with your formula. When you say: "Can we write any integer vector, constant (mod2)", I guess you are talking about the right hand side, so for example:$$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ -1 & 1 & 1 & -1 \\ -1 & -1 & 1 & 1 \end{array} \right) = \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \end{array} \right)$$ the integer vector is (2, 0, 0, 0). If I have understood well, the left row vectors are taken from the D4 basis, and you are trying to get from them their mod 2 counterpart passing through the M, matrix isn't it? So, in this case the row vector (2, 0, 0, 0) can be written mod 2 as (1, 0, 0, 0) thanks to the M matrix. However, I see some -1 in between so it's strange to me interpret them as mod 2 remainders.
For example, I've tried to compute (3, 1, 3, 1) on the right, getting on the left (2, 1, 1, 1). Actually a 2 has appeared, making even more difficult to understand the mod 2. Does this apply to the fundamental region only?

EDIT: "If I have understood well, the left row vectors are taken from the D4 basis". This is wrong indeed. Basis from D4 must sum to an even number, so (1, 0, 0, 0) cannot belong to them. (1, 0, 0, 0) is the (2, 0, 0, 0) vector evaluated in my basis, but I cannot say more than this.

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