Classification of positive even unimodular lattices of rank 16

By [F1], choose an orthonormal basis of $V$ and identify $V$ isometrically with ${\mathbb{R}}^{16}$. In these coordinates $\mathrm{\Lambda }=A{\mathbb{Z}}^{16}$ by [F2]. The isometry preserves pairings, norms, integrality, evenness and the dual, so it suffices to classify this coordinate lattice.

Define $R:=\{x\in \mathrm{\Lambda }:⟨x,x⟩=2\}$. It is finite: if $x=Ak\in R$, then $k=A^{-1}x$ has bounded coordinates, leaving only finitely many integer vectors $k$. It is a reduced crystallographic root system in ${\mathrm{span}}_{\mathbb{R}}R$: it has no zero vector; all roots have the same norm, so only $\pm x$ lie on each root line; for $x,y\in R$, $⟨y,x⟩\in \mathbb{Z}$ by integrality and $s_x(y)=y-⟨y,x⟩x\in \mathrm{\Lambda }$ has the same norm $2$, so reflection preserves $R$. By construction its roots all have squared norm $2$.

By [F3], $\mathrm{\mid }R\mathrm{\mid }=480$. The hypotheses of [F4] now hold by step 1.2, so $R$ has rank $16$ and its root lattice $Q(R)$ is isometric to either $D_{16}$ or $E_8\oplus E_8$.

If $Q(R)$ is isometric to $E_8\oplus E_8$, it is a full sublattice of $\mathrm{\Lambda }$ because every root lies in $\mathrm{\Lambda }$. Both lattices have covolume $1$ by unimodularity [F5]; their inclusion therefore has index $1$ by [F6], so $Q(R)=\mathrm{\Lambda }$ up to that isometry. Hence $\mathrm{\Lambda }$ is isometric to $E_8\oplus E_8$.

If $Q(R)$ is isometric to $D_{16}$, transport coordinates by that lattice isometry and regard $D_{16}\subseteq \mathrm{\Lambda }$. The coordinate-sum parity map ${\mathbb{Z}}^{16}\to \mathbb{Z}\mathrm{/}2\mathbb{Z}$ is onto with kernel $D_{16}$, so $[{\mathbb{Z}}^{16}:D_{16}]=2$ and [F6] gives $\mathrm{covol}(D_{16})=2$; unimodularity gives $\mathrm{covol}(\mathrm{\Lambda })=1$. Hence $[\mathrm{\Lambda }:D_{16}]=2$ by the covolume ratio in [F6]. Integrality of $\mathrm{\Lambda }$ gives $\mathrm{\Lambda }\subseteq D_{16}^{\ast }$ because every $x\in \mathrm{\Lambda }$ pairs integrally with every element of $D_{16}\subseteq \mathrm{\Lambda }$. Thus [F7] applies and $\mathrm{\Lambda }$ is isometric to $D_{16}^{+}$.

Both candidate models satisfy the hypotheses by [F5]. They are not isometric: the norm-2 vectors of $D_{16}^{+}$ are exactly the roots $\pm e_i\pm e_j$ of the irreducible system $D_{16}$, since every vector in the half-integer coset has norm at least $16\mathrm{/}4=4$. By [F5], every nonzero vector of $E_8$ has positive even integer squared norm, hence squared norm at least $2$. Therefore a vector $(x,y)\in E_8\oplus E_8$ has squared norm $2$ if and only if exactly one component is zero and the other is an $E_8$ norm-2 root. These give the two orthogonal $E_8$ root components; each is nonempty because $e_1-e_2\in D_8\subset E_8$ has squared norm $2$, so $(e_1-e_2,0)$ and $(0,e_1-e_2)$ lie in the respective components. An inner-product-preserving bijection preserves the norm-2 root system and its orthogonal decomposition, so it cannot identify these two types. Steps 3.1 and 3.2 prove exhaustiveness, and this nonisometry proves uniqueness of the listed class.