If a lattice can be embedded into $\mathbb Q^n,\langle-1\rangle^n$, then can it be embedded into $\mathbb Z^n,\langle -1 \rangle^n$?

Question

Given a graph with negative integers on each vertex $\Gamma$ there is a corresponding intersection lattice denoted $Q_\Gamma$, a free $\mathbb Z$ module generated by the vertices, endowed with a bilinear form (still denoted $Q_\Gamma$), where $Q_\Gamma(v,v)$ equals the decoration of the vertex $v$, and $Q_\Gamma(v,w)$ is 1 or 0 depending on if they are adjacent in $\Gamma$ or not. Let us denote the number of vertices (and so the rank of $Q_\Gamma$) by $n$.

Suppose, that $Q_\Gamma\otimes\mathbb Q$ is isomorphic to $(\mathbb Q,\langle -1 \rangle^n)$, does this imply, that $Q_\Gamma$ can be embedded into $(\mathbb Z^n,\langle -1 \rangle^n)$ (note, that the ranks are the same)?

The statement is true for all of the examples I've looked at, the problem seems to be finding a suitable orthonormal basis in $\mathbb Q^n$, which generate the points of my lattice over $\mathbb Z$, but it's not clear how one would go about doing this.

(Even some pointers towards relevant literature would be appreciated, I've looked through Gerstein's basic quadratic forms, but can't seem to find statements of this form.)

Answer 1

As Ian Agol points out in his comment, $-E_8$ is an example of a lattice that embeds in $\mathbb{Q}^8$ but not in $\mathbb{Z}^n$ for any $n$.

The embedding in $\mathbb{Q}^8$ is given explicitly in Conway and Sloane's Sphere packings, lattices, and groups (Chapter 4, Section 8, page 120–121 in my edition).

That it does not embed in $\mathbb{Z}^n$ follows from uniqueness of decompositions of definite unimodular lattices, or from an elementary argument of lattice embeddings (which I have learnt from work of Lisca and Lisca and Stipsicz on 3.5-dimensional topology).