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Know thy neighbours

Two lattices L and L in the same vector space are called neighbours if their intersection LL is of index two in both L and L.

In 1957, Martin Kneser gave a method to find all unimodular lattices (of the same dimension and signature) starting from one such unimodular lattice, finding all its neighbours, and repeating this with the new lattices obtained.

In other words, Kneser’s neighbourhood graph, with vertices the unimodular lattices (of fixed dimension and signature) and edges between them whenever the lattices are neighbours, is connected.



Martin Kneser (1928-2004) – Photo Credit

Last time, we’ve constructed the Niemeier lattice (A124)+ from the binary Golay code C24
L=(A124)+=C24×F2(A124)={12v | vZ24, v=v mod 2C24}
With hindsight, we know that (A124)+ is the unique neighbour of the Leech lattice in the Kneser neighbourhood graph of the positive definite, even unimodular 24-dimensional lattices, aka the Niemeier lattices.

Let’s try to construct the Leech lattice Λ from L=(A124)+ by Kneser’s neighbour-finding trick.



Sublattices of L of index two are in one-to-one correspondence with non-zero elements in L/2L. Take lL2L and mL such that the inner product l.m is odd, then
Ll={xL | l.x is even}
is an index two sublattice because L=Ll(Ll+m). By definition l.x is even for all xLl and therefore l2Ll. We have this situation
LlL=LLl
and Ll/LlF2F2, with the non-zero elements represented by {l2,m,l2+m}. That is,
Ll=Ll(Ll+m)(Ll+l2)(Ll+(l2+m))
This gives us three lattices
{M1=Ll(Ll+m)=LM2=Ll(Ll+l2)M3=Ll(Ll+(l2+m))
and all three of them are unimodular because
LlMiMiLl
and Ll is of index 4 in Ll.

Now, let’s assume the norm of l, that is, l.l4Z. Then, either the norm of l2 is odd (but then the norm of l2+m must be even), or the norm of l2 is even, in which case the norm of l2+m is odd.

That is, either M2 or M3 is an even unimodular lattice, the other one being an odd unimodular lattice.

Let’s take for l and m the vectors λ=12(1,1,,1)L2L and μ=2(1,0,,0)L, then
λ.λ=12×24=12andμ.λ=1
Because λ2.λ2=124=3 is odd, we have that
Λ=Lλ(Lλ+(λ2+μ))
is an even unimodular lattice, which is the Leech lattice, and
Λodd=Lλ(Lλ+λ2)
is an odd unimodular lattice, called the odd Leech lattice.



John Leech (1926-1992) – Photo Credit

Let’s check that these are indeed the Leech lattices, meaning that they do not contain roots (vectors of norm two).

The only roots in L=(A124)+ are the 48 roots of A124 and they are of the form ±2[1,023], but none of them lies in Lλ as their inproduct with λ is one. So, all non-zero vectors in Lλ have norm 4.

As for the other part of Λ and Λodd
(Lλ+λ2)(Lλ+μ+λ2)=(Lλ(Lλ+μ))+λ2=L+λ2
From the description of L=(A124)+ it follows that every coordinate of a vector in L+λ2 is of the form
12(v+12)or12(v+32)
with v2Z, with the second case instances forming a codeword in C24. In either case, the square of each of the 24 coordinates is 18, so the norm of such a vector must be 3, showing that there are no roots in this region either.

If one takes for l a vector of the form 12v=12[1a,024a] where a=8,12 or 16 and vC24, takes m=μ as before, and repeats the construction, one gets the other Niemeier-neighbours of (A124)+, that is, the lattices (A212)+, (A38)+ and (D46)+.

For a=12 one needs a slightly different argument, see section 0.2 of Richard Borcherds’ Ph.D. thesis.

Published in geometry groups