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The norm-2 count does not distinguish the two rank-16 lattices

Example

The two lattices $E_{8} \oplus E_{8}$ and $D_{16}^{+}$ are positive even unimodular rank-16 lattices with the same number, $480$ , of vectors of norm squared $2$ , but they are not isometric. Thus the norm-2 count is an isometry invariant that does not distinguish these two classes.

Facts & Assumptions

Given: The coordinate definitions of $E_{8} \oplus E_{8}$ and $D_{16}^{+}$ with the standard positive-definite inner products.

[F1]

Both coordinate models are full positive even unimodular rank-16 lattices (The E8 and D16+ models are positive even unimodular lattices).

[F2]

In coordinates, $E_{8} = D_{8} \cup (D_{8} + h_{8})$ , where $D_{8} = {z \in Z^{8} : \sum_{i} z_{i} \in 2 Z}$ and $h_{8} = (1 / 2, \dots, 1 / 2)$ (E8 ⊕ E8 and D16+ rank-16 lattice models).

[F3]

The rank-16 classification states that the two models are distinct isometry classes (Classification of positive even unimodular lattices of rank 16).

Verification

Given: The two coordinate lattices and facts [F1]-[F3].

1.1F1F2algebra

An integer vector in $D_{8}$ has norm squared $2$ exactly when two coordinates are $\pm 1$ and the other six are zero; every such vector has even coordinate sum, so there are $4 (\binom{8}{2}) = 112$ . A vector in $D_{8} + h_{8}$ has eight half-integer coordinates, each of absolute value at least $1 / 2$ , so its norm squared is at least $2$ ; equality requires every coordinate to be $\pm 1 / 2$ . Such a sign pattern belongs to $D_{8} + h_{8}$ exactly when it has an even number of minus signs, giving $2^{7} = 128$ vectors. Hence $E_{8}$ has $112 + 128 = 240$ norm-2 vectors. In $E_{8} \oplus E_{8}$ , a vector $(x, y)$ has norm squared $∥ x ∥^{2} + ∥ y ∥^{2}$ ; by [F1], the squared norms $∥ x ∥^{2} = ∥ (x, 0) ∥^{2}$ and $∥ y ∥^{2} = ∥ (0, y) ∥^{2}$ are even integers, and positive definiteness makes each strictly positive when its component is nonzero. Thus each nonzero component contributes at least $2$ , so their sum is $2$ exactly when one component is zero and the other is an $E_{8}$ norm-2 vector. Thus there are $240 + 240 = 480$ such vectors.

1.2F1algebra

An integer vector in $D_{16}$ has norm squared $2$ exactly when two coordinates are $\pm 1$ and the rest are zero; all four sign choices for every coordinate pair have even coordinate sum, so there are $4 (\binom{16}{2}) = 480$ . Every vector in $D_{16} + h_{16}$ has sixteen half-integer coordinates, each of absolute value at least $1 / 2$ , hence norm squared at least $16 / 4 = 4$ ; that coset contributes no norm-2 vectors. Therefore $D_{16}^{+}$ also has exactly $480$ norm-2 vectors.

2.1F3step 1.1step 1.2∎

By [F3], the two lattices are not isometric despite the equal norm-2 counts established in steps 1.1 and 1.2. The counts are an invariant because an isometry preserves the inner product, but they are not a complete invariant.

The norm-2 count does not distinguish the two rank-16 lattices

Example

Facts & Assumptions

Verification

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