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Let $\Lambda$ be the Leech lattice. There is a nice set of coset representatives for $\Lambda/2 \Lambda$ given by short vectors [Conway and Sloane, Ch. 10, Theorem 28 or Ch. 23, Theorem 3]. The proof uses the identity: $\lvert \Lambda(0)\rvert + \lvert \Lambda(4)\rvert/2 + \lvert \Lambda(6)\rvert/2 + \lvert \Lambda(8)\rvert/48 = \lvert \Lambda/2 \Lambda \rvert$. Here $\Lambda(m)$ is the set of norm $m$ vectors of $\Lambda$.

Here is the corresponding easier statement for the $E_8$ root Lattice (This is assuming my calculations are correct, because I have not seen this in print):

A full set of coset representatives for $E_8/2E_8$ consists of the zero vector, norm $2$ vectors chosen up to sign, and one norm $4$ vector chosen from each ``coordinate frame" of size $16$. Here a coordinate frame means a set of eight orthogonal norm $4$ lattice vectors and their negatives. The proof uses the identiy: $\lvert E_8(0)\rvert + \lvert E_8(2)\rvert/2 + \lvert E_8(4)\rvert/16 = \lvert E_8/2E_8 \rvert$.

A similar statement is known for the Complex Leech lattice [R. Wilson, J. Alg. 84 (1983) 151-- 188].

Here are my questions:

  1. Is there any ``reason" behind these facts?

  2. Are there other examples known where similar statements can be made?

  3. Finally, is the statement for $E_8$ written out anywhere?

Thanks.

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  • $\begingroup$ Two reasons to expect something like this might be true: First, there are many short vectors (because the lattices meet packing bounds) and two short vectors are unlikely to be congruent modulo 2 (because half their difference would have to be quite short as well) so it is not surprising that every congruence class mod 2 has a short vector. Second, the lattice is highly symmetric, so it is not surprising that this can be done in a regular way. $\endgroup$
    – Will Sawin
    Commented Jun 27, 2017 at 20:00

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