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Two vectors from Leech lattice - as defined on wikipedia - have scalar product $\pm 32,\pm 16, \pm 8$ or $0$. Do there exist 24 vectors having scalar product 8 pairwise ? When we consider unit vectors in Leech lattice then scalar product is $\frac{1}{4}$ pairwise.

Motivation for this question is the fact that for 2b involution there are 196560 involutions in Monster group 2a conjugacy class which generate extraspecial group $2^{1+24}$. Modulo its center elements in this group correspond to Leech vectors. I believe vectors having scalar product 8 correspond to not commuting elements.

What I was able to achieve is that for $\Lambda_{16}$ sublattice there are at most 9 such vectors. The arguments are following. This sublattice can be seen as union of 18 disjoint $E_8$ lattices (see Wilson definition with octonions which refer to earlier Dixon definition). Given two vectors having scalar product 8 ($\frac {1}{2}$ in Wilson version where Leech vector is of legth $\sqrt 2$) must lie in two different E8s.

For Baby monster 2b element there are 4600 elements in 2a commuting with it. In Leech lattice language for fixed vector there are 4600 vectors having scalar product 16 with it.

Regards,

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    $\begingroup$ The scalar product can also be 0. (The lattice is usually normalized to be self-dual, so the inner products are ±4, ±2, ±1, and 0.) $\endgroup$ – Noam D. Elkies Mar 16 '17 at 20:34
  • $\begingroup$ It seems that permutation $p=(1..23)$ preserve both Leech lattices - classical one (C) and one defined below in the answer (E). For (C) lattice vector $(1^{23},-3)$ is fixed. For (E) lattice vector $(1^{23},-5)$ is fixed. I tried to map both lattices using 24 vectors having $\frac{1}{2}$ scalar product pairwise (unit vectors in lattice) but it does not work. There are 2300 vectors having $\frac{1}{2}$ scalar product with $v_0$ fixed. In this set there is one 23-size orbit under action of $Normalizer(p)=23:11$. I take $v_0$ and these 23 vectors but not sure what should be the order in this set. $\endgroup$ – Marek Mitros Mar 29 '17 at 12:15
  • $\begingroup$ Note question math.stackexchange.com/questions/2211637/… I asked on MathStackExchange. $\endgroup$ – Marek Mitros Mar 31 '17 at 12:25
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Yes, such a configuration exists, even with all inner products positive (as the title of the question requires, even though the text allows either sign).

We shall use the standard normalization of the Leech lattice $\Lambda$ that makes it unimodular, so the $196560$ minimal vectors $(v,v) = 4$, not $32$. Thus we claim that we can choose $24$ vectors $v_i$ from those $196560$ vectors such that $(v_i,v_j) = 1$ if $i \neq j$.

The Gram matrix of those vectors has determinant $3^{23} (3+24\cdot 3) = 3^{26}$. Instead of finding them in some standard model of $\Lambda$, we find a lattice $L$ containing the ${\bf Z}$-span of the $v_i$ with index $3^{13}$, and verify that $L$ is an even unimodular lattice with no vectors of norm $2$; it is known that this suffices to prove $L \cong \Lambda$. I do not know whether the resulting construction of $\Lambda$ is new or equivalent to one of the many previously known ones (though the ingredients are certainly familiar).

We have not yet named the $24$ indices $i$ of our $v_i$. Let us call them $1,2,3,\ldots,23,\infty$, with all but $\infty$ taken mod $23$. We generate $\Lambda$ by the $v_i$ together with $u_\infty := \frac19(v_\infty + \sum_{i \bmod 23} v_i)$ and the $23$ vectors $u_j := \frac13 \sum_{a=0}^{11} v_{a^2+j}$.

It is straightforward to check that all these vectors generate an even lattice $L$. The fact that $L$ is unimodular requires the dimension of ternary quadratic-residue codes, and then checking that $L$ has no vectors of norm $2$ would take rather more work $-$ so instead we write and run a few lines of gp code:

G1 = matrix(24,24,i,j,if(i==j,4,1))
B = matrix(24,24,i,j,if(j==24,1,if(i<24,3*(kronecker(i-j,23)>=0))))
B = concat(9*matid(24), B);
A = matrixqz(B,-1)
A /= 9;
G = A~ * G1 * A
matdet(G)
qfminim(G,,0)

The first line constructs the Gram matrix of the $v_i$; the second line constructs a $24\times 24$ matrix $B$ with columns $9u_i$ and $9u_\infty$; the third line adds the $24$ columns $9v_i$ by extending $B$ by the matrix $9I_{24}$; the next line finds a $24 \times 24$ matrix $A$ whose columns generate the ${\bf Z}$-span of those 48 vectors. We then divide by $9$ and compute the Gram matrix $G$ of this span with respect to the $A$ generators. Finally we verify thet $\det G = 1$ and ask for the number and norm of minimal nonzero vectors (the concluding ",,0" suppresses the output of those minimal vectors). The result is

[196560, 4, [;]]

confirming $L \cong \Lambda$, QED.

P.S. Is it clear that there is no such configuration with more than $24$ vectors? There certainly cannot be 26, because their Gram matrix would be invertible even modulo $2$; but conceivably a 25th vector might be added if we allow both $+1$ and $-1$ inner products.


Added later: Marek Mitros notes in a comment that the inner products $(v_i,v_j) = 3 \delta_{ij} + 1$ are realized by the $24$ vectors of shape $12^{-1/2} (-5, 1^{23})$ in ${\bf R}^{24}$. In those coordinates my description of $\Lambda$ amounts to the following simpler construction, which should already be known $-$ indeed it seems to be the same as doubling the lattice obtained by "construction B$_3$" from the ternary QR (quadratic-residue) code of length $24$, see pages 148--149 (Chapter 5, $\S\S$5.5--5.7) in Conway and Sloane's SPLAG = Sphere Packings, Lattices and Groups.

Let $e_i$ be the $24$ pairwise orthogonal unit vectors (with $i=1,2,\ldots,23,\infty$ as before), and let $f_i$ be the rows of a $24 \times 24$ Hadamard matrix constructed using quadratic residues mod $23$: $f_{ij} = -1$ unless $i,j \neq \infty$ and $i-j$ is a square mod $23$ (including zero), in which case $f_{ij} = +1$. Then $12^{1/2}\Lambda$ is generated by the $24+24$ vectors $6(e_i + e_\infty)$ and $f_i - 6e_\infty$ (NB $i = \infty$ is allowed), with $v_i = -(f_\infty - 6e_\infty) - 6(e_i+e_\infty)$.

Here's the gp code:

E = 6 * matrix(24,24,i,j,(i==j)+(i==24));
F = matrix(24,24,i,j, if((i<24) && (j<24) && kronecker(i-j,23) >= 0, 1, -1));
for(j=1,24,F[24,j]-=6);
A = matrixqz(concat(E,F),-1)
G = A~ * A / 12
matdet(G) \\ = 1
qfminim(G,,0)

which again outputs

[196560, 4, [;]]

SPLAG reports that all that this construction requires of the QR code is that it be self-dual with parameters $[24,12,9]$, and thus that the ternary Pless (double circulant) code of the same length works as well; this should give an inequivalent configuration of $24$ Leech vectors of norm $4$ with all inner products equal $1$.

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  • $\begingroup$ Thank you very much. I am impressed that you found the answer so quickly. I do not follow all the details yet. I will try to digest. As I understand you have constructed new lattice from 24 vectors having pairwise scalar product 1 and you have proved that it is Leech lattice. I wonder why it is so difficult to find such 24 vectors in given Leech lattice. I tried in GAP but I found 17 vectors as someone said in comment. $\endgroup$ – Marek Mitros Mar 17 '17 at 7:45
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    $\begingroup$ The example 24 unit vectors having scalar product $\frac {1}{4}$ is following. Take $-5,1^{23}$ and all permutations and divide by length $\sqrt 48$. $\endgroup$ – Marek Mitros Mar 17 '17 at 7:58
  • $\begingroup$ Finally I was able to convert your code to GAP and using $LLLReducedBasis$ function I obtained 24 vectors of length 48 with proper scalar products, so indeed they should generate Leech lattice ! So I don't understand why, but it works :) Next I need to map this lattice to my previous Leech lattice I am working with, but that should be easy. $\endgroup$ – Marek Mitros Mar 17 '17 at 13:44
  • $\begingroup$ The comes with a symmetry by a 23-cycle (and indeed its 11*23 element normalizer); once you choose such an element of Aut(Leech) it should be a manageable computation to find an isomorphism that commutes with it because there are two minimal vectors whose orbits generate everything else over Q. $\endgroup$ – Noam D. Elkies Mar 17 '17 at 13:51
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    $\begingroup$ When I've done this and I filter all vectors having scalar product 1/4 and take random vector from obtained set then I receive following number of vectors: 47104,15400,5892(orbits 5832+60),2455,1072,482,218,103,44,20,7. Currently I am interested what dimension in Leech mod 2 this set of 24 vectors span. $\endgroup$ – Marek Mitros Mar 19 '17 at 13:07
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Here are 24 vectors in classic Leech lattice having pairwise scalar product 8. I obtained it from $(-5, 1^{23})$ using this.

$\begin{bmatrix} -1& 1& 1& 1& 1&-1& 1&-1& 1& 1&-1&-1& 1& 1&-1&-1& 1&-1& 1&-1&-1&-1&-1&-3 \\ \dots\\ 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1&-3 \end{bmatrix}$

The dots $\dots$ should be replaced by 22 vectors obtained from the first one by permutation $(1,2,...,23)$.

As a consequence we can obtain following useful

Fact

For each element $p$ of order $23$ in $M_{23}$ there exist dodecad $d$ such that dodecads $d, pd, p^2d,...,p^{22}d$ intersect in six points pairwise.

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