6
$\begingroup$

The book Mathematical Evolutions contains the following excerpt:

A last, famous, example is the following. It is known that in the space of one hundred and ninety six thousand eight hundred and eighty three dimensions there is a wonderful lattice whose symmetry group has order

808 017 424 794 512 875 886 459 904 961 710 757 005 754 368 000 000 000

= $2^{46} \cdot 3^{20} \cdot 5^9 \cdot 11^2 \cdot 13^3 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 \cdot 41 \cdot 47 \cdot 59 \cdot 71$

Of course, no-one has really seen this lattice: we know that it exists, but an explicit construction is lacking. Nevertheless, one can construct the symmetry group of the lattice, the so-called monster group $M$ of R. Griess and B. Fischer.

The existence of such a lattice is not immediate: for example, the simple group $A_5$ has a faithful three-dimensional representation, but there is no lattice in $\mathbb{R}^3$ with an (orientation-preserving) point symmetry group isomorphic to $A_5$.

However, a remark in a paper by Conway reveals the existence of a (probably unimodular) integral lattice in 196884 dimensions which is closed under the Griess algebra multiplication:

Norton has shown that the lattice $L$ spanned by vectors of the form $1, t, t \ast t'$, where $t$ and $t'$ are transposition vectors, is closed under the algebra multiplication and integral with respect to the doubled inner product $2(u, v)$. The dual quotient $L^{\star} / L$ is cyclic of order some power of 4, and we believe that in fact $L$ is unimodular. The vectors $t, u, v, w$ of Table 3 all lie in $L$.

Now, if we take the intersection of the 196884-dimensional lattice $L$ with the orthogonal complement of the fixed axis $1$, we obtain a 196883-dimensional integral lattice $L'$ with Gram determinant $|L'| = 3|L|$. This is an explicit construction of a lattice with the properties alluded to in the Mathematical Evolutions excerpt.

The orientation-preserving point symmetry group of this lattice $L'$ is isomorphic to the Monster group $M$, and the full point symmetry group is isomorphic to the direct product $M \times C_2$ (since the ambient dimension is odd, so the negated identity matrix $-I$ is an orientation-reversing symmetry which commutes with everything).

I'd like to understand this 'wonderful lattice' $L'$ in greater detail. So, now for a pair of concrete questions:

  • Question 1: what are the shortest nonzero vectors in the lattice $L'$?

The shortest vectors I can find are those of the form $x := t - t'$ (where $t$ and $t'$ are the axes of two type-2A involutions whose product is also a type-2A involution), which each have doubled inner product $2(x, x) = 448$. The number of such vectors is 2639459181687194563957260000000, one for each arc (oriented edge) in the Monster graph. If these are indeed the shortest vectors, then the sphere packing corresponding to this lattice would be moderately dense (but not record-breakingly so, and well below Minkowski's lower bound on lattice density).

  • Question 2: does the set of shortest nonzero vectors generate the lattice?

The fact that the Monster graph is connected implies that any vector of the form $t - t'$ (where $t, t'$ are any pair of axes of type-2A involutions) is generated by the aforementioned norm-448 vectors. But I'm not immediately convinced that these generate the lattice $L'$, because Norton's lattice $L$ was described as being generated by vectors of the form $1, t, t \ast t'$, rather than just vectors of the form $1, t$.

$\endgroup$
  • 1
    $\begingroup$ I don't know if this helps, but I have uploaded a python package (with optimized C subroutines) that can do some calulations in the moster and also generate random 2A transposition axes $t$. See: github.com/Martin-Seysen/mmgroup $\endgroup$ – Martin Seysen Jun 17 '20 at 21:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.