The E8 lattice is even unimodular and is not diagonal

Let $A_7$ be the $7\times 7$ leading principal block on vertices $1,\dots ,7$. Its leading principal determinants $d_k$ satisfy $d_0=1$, $d_1=2$, and $d_k=2d_{k-1}-d_{k-2}$ for $2\le k\le 7$, by expansion along the last row of the tridiagonal chain matrix. Induction gives $d_k=k+1$, so these seven determinants are positive.

The vector $x=A_7^{-1}{e}_3$ has coordinates $x_i=\min (i,3)(8-\max (i,3))\mathrm{/}8$ for $1\le i\le 7$: substitution into $2x_i-x_{i-1}-x_{i+1}={\delta }_{i3}$ with $x_0=x_8=0$ verifies $A_{7}x=e_3$. Hence $e_3^{\mathsf{T}}{A}_7^{-1}{e}_3=15\mathrm{/}8$. The Schur-complement determinant of $C=\left(\begin{array}{cc}{A}_7& -e_3\\ -e_3^{\mathsf{T}}& 2\end{array}\right)$ is $\det (A_7)(2-15\mathrm{/}8)=8\mathrm{/}8=1$.

By Sylvester's criterion, the positive leading principal determinants $d_1,\dots ,d_7$ from step 1.1 and $\det C=1$ from step 2.1 make $C$ positive definite. For any $z\in {\mathbb{Z}}^8$, $z^{\mathsf{T}}Cz=2{\sum }_i{z}_i^2-2{\sum }_{\{i,j\}\text{ an edge}}{z}_i{z}_j$ is even. Every nonzero vector therefore has positive even integral norm, hence norm at least $2$; in particular there is no norm-one vector.

The standard lattice ${\mathbb{Z}}^8$ contains norm-one basis vectors, so no isometry from $L_{E_8}$ onto ${\mathbb{Z}}^8$ exists by step 3.1. If there were an isometric full-rank embedding, its Gram determinant would be $1$ by step 2.1, and [F2] would force its image to be all of ${\mathbb{Z}}^8$, which would be an isometry, again contradicting step 3.1.