A positive even unimodular rank-16 lattice has 480 norm-2 vectors

Choose orthonormal coordinates and write $\mathrm{\Lambda }=A{\mathbb{Z}}^{16}$ using [F1]. Write $B=A^{-1}=(b_{ij})$. For any $x$, $\mathrm{\mid }(Bx{)}_i\mathrm{\mid }\le ({\sum }_j\mathrm{\mid }{b}_{ij}\mathrm{\mid })\mathrm{\mid }x\mathrm{\mid }$, so $\mathrm{\mid }Bx\mathrm{\mid }\le 4{\max }_i{\sum }_j\mathrm{\mid }{b}_{ij}\mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }$ by summing the sixteen squared component bounds. Put $C:=4{\max }_i{\sum }_j\mathrm{\mid }{b}_{ij}\mathrm{\mid }$. For $m\ge 0$ let $r_m:=\mathrm{\#}\{x\in \mathrm{\Lambda }:⟨x,x⟩=2m\}$. If $x=Ak$ is counted, $\mathrm{\mid }k\mathrm{\mid }=\mathrm{\mid }{A}^{-1}x\mathrm{\mid }\le C\sqrt{2m}$, hence $r_m\le (2C\sqrt{2m}+1{)}^{16}\le C^{\mathrm{\prime }}(1+m{)}^8$ for a constant $C^{\mathrm{\prime }}$. This proves every coefficient is finite and ${\sum }_{m\ge 0}{r}_m{q}^m$ converges for $\mathrm{\mid }q\mathrm{\mid }<1$ by [F9]. Define ${\mathrm{\Theta }}_{\mathrm{\Lambda }}(\tau ):={\sum }_{m\ge 0}{r}_m{e}^{2\pi im\tau }$; by [F6] it is holomorphic on $\mathfrak{H}$ and holomorphic at the cusp.

Let $c=\mathrm{covol}(\mathrm{\Lambda })$. By [F1] and unimodularity, $c=c^{-1}$; since $c>0$, $c=1$. Apply [F2] with $n=16$ and $x=0$ to obtain, for every $y>0$, ${\mathrm{\Theta }}_{\mathrm{\Lambda }}(iy)={\sum }_{\lambda \in \mathrm{\Lambda }}{e}^{-\pi y\mathrm{\mid }\lambda {\mathrm{\mid }}^2}=y^{-8}{\sum }_{\mu \in {\mathrm{\Lambda }}^{\ast }}{e}^{-\pi \mathrm{\mid }\mu {\mathrm{\mid }}^2\mathrm{/}y}=y^{-8}{\mathrm{\Theta }}_{\mathrm{\Lambda }}(i\mathrm{/}y),$ where $\mathrm{\Lambda }={\mathrm{\Lambda }}^{\ast }$. Thus ${\mathrm{\Theta }}_{\mathrm{\Lambda }}(-1\mathrm{/}(iy))=(iy{)}^8{\mathrm{\Theta }}_{\mathrm{\Lambda }}(iy)$ because $i^8=1$.

Each lattice vector has squared norm $2m$ for an integer $m\ge 0$ because $\mathrm{\Lambda }$ is even. Hence $e^{\pi i⟨x,x⟩}=1$ and ${\mathrm{\Theta }}_{\mathrm{\Lambda }}(\tau +1)={\mathrm{\Theta }}_{\mathrm{\Lambda }}(\tau )$. Positive definiteness makes $0$ the only vector with squared norm zero, so the constant coefficient is $r_0=1$.

The functions $\tau \mapsto {\mathrm{\Theta }}_{\mathrm{\Lambda }}(-1\mathrm{/}\tau )$ and $\tau \mapsto {\tau }^8{\mathrm{\Theta }}_{\mathrm{\Lambda }}(\tau )$ are holomorphic on $\mathfrak{H}$ by [F6], since $-1\mathrm{/}\tau \in \mathfrak{H}$ there. Their difference vanishes at all $iy$ with $y>0$ by step 1.2, and these zeros accumulate at $i\in \mathfrak{H}$. By [F8], $\mathfrak{H}$ is a complex domain; [F7] gives ${\mathrm{\Theta }}_{\mathrm{\Lambda }}(-1\mathrm{/}\tau )={\tau }^8{\mathrm{\Theta }}_{\mathrm{\Lambda }}(\tau )$ on all of $\mathfrak{H}$.

The matrices $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ generate $S{L}_2(\mathbb{Z})$: left multiplication by powers of $T$ and by $S$ performs the Euclidean algorithm on the coprime first-column entries $(a,c)$ of a determinant-one integer matrix until they are $(\pm 1,0)$; the remaining matrix is $\pm T^j$, and $-I=S^2$. The weight factors compose by $(c_1({\gamma }_2\tau )+d_1)(c_2\tau +d_2)=c_{12}\tau +d_{12}$. Therefore steps 2.1 and 2.2 imply the weight-$8$ transformation law for every $S{L}_2(\mathbb{Z})$ matrix. By [F5] and steps 1.1-2.1, ${\mathrm{\Theta }}_{\mathrm{\Lambda }}\in M_8$.

By [F3], $M_8$ is spanned by $E_4^2$. Its constant coefficient and that of ${\mathrm{\Theta }}_{\mathrm{\Lambda }}$ are both $1$ by step 2.1, so ${\mathrm{\Theta }}_{\mathrm{\Lambda }}=E_4^2$. By [F4], the coefficient of $q$ in $(1+240q+O(q^2){)}^2$ is $480$. The coefficient of $q$ in ${\mathrm{\Theta }}_{\mathrm{\Lambda }}$ is $r_1$, exactly the number of vectors of squared norm $2$, so $r_1=480$.