Even unimodular lattices with root system $32 A_1$

Question

I'm studying Venkov's proof of the classification of even unimodular rank 24 lattices, and it prompted the following question.

For an even unimodular lattice $L$, let $R(L)= \{ x \in L : (x,x) =2\}$ be its root system.

Question. Can one "roughly" classify, up to isomorphism, all rank $32$ even unimodular lattices with root system $32 A_1$? Or are there too many of these? (Note: there are at least $10^8$ even unimodular lattices of rank 32.)

Remark. There are no even unimodular lattices of rank 8 and 16 with root system $8 A_1$ and $16A_1$, respectively. There is a unique rank $24$ even unimodular lattice with root system $24 A_1$ by the work of Niemeier (and Venkov).

Any comments are appreciated!

Answer 1

Even unimodular lattices of rank $n$ with root system $n\cdot A_1$ correspond bijectively with Type II codes of length $n$ and minimal weight at least $8$. ("Type II" = self-dual and doubly even.) The results you quote for $n=8,16,24$ correspond to the fact that there's no such code for $n=8$ and $n=16$, and a unique one for $n=24$ (the extended binary Golay code). For $n=32$ there are five such codes according to

John H. Conway and Vera S. Pless, On the enumeration of self-dual codes, J. Combin. Theory Ser. A 28 (1980), 26$-$53.

Therefore there are $5$ even unimodulars of rank $32$ with roots $32A_1$. For $n=40$ the count is at least 17493 (Oliver King, 2001).

The correspondence is via what's called "Construction A" in SPLAG =

John H. Conway and Neil J. A. Sloane, Sphere Packings, Lattices and Groups. New York: Springer-Verlag, 1998.

This construction associates to any length-$n$ binary linear code $C$ the lattice $$ L_C := 2^{-1/2} \{ v \in {\bf Z}^n : v \bmod 2 \in C \}. $$ This lattice is even unimodular iff $C$ is Type II; then the root system contains $n \cdot A_1$, and is exactly $n \cdot A_1$ iff $C$ has no vectors of weight $4$.

To recover $C$ from $L$: Given an integral lattice $L$ of rank $n$ with root system containing $n \cdot A_1$, we have $L \subseteq L^* \subseteq (n \cdot A_1)^* = n \cdot \frac12 A_1$. Lattices sandwiched between $n \cdot \frac12 A_1$ and $n \cdot A_1$ correspond bijectively with subgroups of the quotient group $(n \cdot \frac12 A_1) / (n \cdot A_1) \cong ({\bf Z}/2{\bf Z})^n$. A subgroup of $({\bf Z}/2{\bf Z})^n$ is a linear code $C$, and we soon confirm that $L \cong L_C$.

Answer 2

Another construction (basically, different terminology, but the same outcome) is the maximal (of rank 16) isotropic subgroups in the discriminant group (in the sense of Nikulin) of $32A_1$, which is $32\langle\frac12\rangle$, not containing a sum of four generators. This is a finite problem, but I wouldn't do that manually.