Can every symmetric, unimodular and positive definite $G\in\mathbb{Z}^{n\times n}$ be written as $G=U^TU$?

Question

Let $G\in\mathbb{Z}^{n\times n}$ be symmetric, unimodular and positive definite. Does there exist a unimodular matrix $U\in\mathbb{Z}^{n\times n}$ such that $G=U^TU$? I now that the result is true if $n=2$, but I have a feeling that it fails if $n$ gets larger.

Answer 1

No. This is essentially asking whether every unimodular lattice is isometric to $\Bbb Z^n$. This is false from $n=8$ onwards, and the $n=8$ counterexample is the $E_8$ root lattice. I think an appropriate $G$ is $$\pmatrix{2&0&0&0&0&0&0&1\\0&2&1&1&1&1&1&-2 \\0&1&2&1&1&1&1&-2 \\0&1&1&2&1&1&1&-2 \\0&1&1&1&2&1&1&-2 \\0&1&1&1&1&2&1&-2 \\0&1&1&1&1&1&2&-2 \\1&-2&-2&-2&-2&-2&-2&4 }.$$

I should add that this matrix is, or at least is intended to be $VV^T$ where $$V=\pmatrix{1/2&1/2&1/2&1/2&1/2&1/2&1/2&1/2\\ 0&1&0&0&0&0&0&-1\\ 0&0&1&0&0&0&0&-1\\ 0&0&0&1&0&0&0&-1\\ 0&0&0&0&1&0&0&-1\\ 0&0&0&0&0&1&0&-1\\ 0&0&0&0&0&0&1&-1\\ 0&0&0&0&0&0&0&2}$$ so that being unimodular and positive definite is clear. Also it cannot have the form $UU^T$ for $U$ a unimodular integer matrix, since such a matrix would have at least one odd diagonal entry.