An equivalent D4 lattice?

Question

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

Answer 1

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]

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

Answer 2

@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 $\delta$.

Answer 3

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).

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

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

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

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

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

Answer 4

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.