Every even index-2 overlattice of D16 is isometric to D16+

If $x\in D_{16}^{\ast }$, testing against the roots $e_i-e_j$ and $e_i+e_j$ of $D_{16}$ gives $x_i-x_j\in \mathbb{Z}$ and $x_i+x_j\in \mathbb{Z}$ for all $i<j$. Hence $2x_i\in \mathbb{Z}$ and all coordinates have the same fractional part, so either $x\in {\mathbb{Z}}^{16}$ or $x\in {\mathbb{Z}}^{16}+h$. Conversely, integer vectors pair integrally with $D_{16}$, and $(z+h)\cdot d=z\cdot d+\frac{1}{2}{\sum }_i{d}_i\in \mathbb{Z}$ for $z\in {\mathbb{Z}}^{16}$ and $d\in D_{16}$ because ${\sum }_i{d}_i$ is even. Therefore $D_{16}^{\ast }={\mathbb{Z}}^{16}\cup ({\mathbb{Z}}^{16}+h)$.

The quotient $D_{16}^{\ast }\mathrm{/}{D}_{16}$ has four distinct classes represented by $0,e_1,h,h+e_1$: the integer vectors split into even- and odd-coordinate-sum classes, and the half-integer vectors split into the two corresponding classes after subtracting $h$. The order-2 subgroup $L\mathrm{/}{D}_{16}$ is generated by one of the three nonzero classes.

The class of $e_1$ is excluded because $\mathrm{\parallel }{e}_1{\mathrm{\parallel }}^2=1$, contradicting evenness of $L$. The classes of $h$ and $h+e_1$ are even: their representative norms are $16\mathrm{/}4=4$ and $9\mathrm{/}4+15\mathrm{/}4=6$, and for any representative $u$ of either class and $d\in D_{16}$, $\mathrm{\parallel }u+d{\mathrm{\parallel }}^2=\mathrm{\parallel }u{\mathrm{\parallel }}^2+2⟨u,d⟩+\mathrm{\parallel }d{\mathrm{\parallel }}^2$ is even since $u\in D_{16}^{\ast }$, $D_{16}$ is even, and [F4] applies. Thus $L$ is one of $D_{16}\cup (D_{16}+h)$ or $D_{16}\cup (D_{16}+h+e_1)$.

The orthogonal sign change $S(x_1,x_2,\dots ,x_{16})=(-x_1,x_2,\dots ,x_{16})$ preserves $D_{16}$ because it changes the coordinate sum by $-2x_1$, an even integer. It sends $h$ to $h-e_1$, which is in the coset $h+e_1+D_{16}$ since $2e_1\in D_{16}$. Hence $S$ maps $D_{16}\cup (D_{16}+h)$ onto $D_{16}\cup (D_{16}+h+e_1)$ isometrically. Both allowed extensions are therefore isometric to $D_{16}^{+}$.