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](http://www.maths.qmul.ac.uk/~raw/pubs_files/octoLeech1rev.pdf) with octonions which refer to earlier [Dixon](https://arxiv.org/abs/hep-th/9506080) 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,



24 vectors in Leech lattice having scalar product $\frac{1}{4}$ pairwise
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](http://en.wikipedia.org/wiki/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 $1$.


