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We are considering $\mathbb{R}^n$ with the inner product (the usual one), and $L$ an unimodular lattice (so $covolume(L)=1$ and $<u,v> \in \mathbb{Z}$ for all u,v in L).

I have to show that, for $n \geq 4$, we can find $v \in L$ such that : $<v,v>=1$. Then, there is some intermediary step, and finally we want to show that if $n \geq 4$, it exists $g \in O_n(\mathbb{R})$ such that : $L = g(\mathbb{Z}^n)$.

Actually, I didn't succeed to prove the first question about $<v,v>$. But for the final result, there is also something I don't understand. Just after, we are considering : $\mathbb{Z} \frac{1}{2}(1,1,1,1,1,1,1,1) + \{ \; (x_1, ..., x_8) \in \mathbb{Z}^8 \; | \; x_1 + ... + x_8 = 0 \;(\mod 2) \; \}$, and we show that it's a unimodular lattice, but it's not of the form $g(\mathbb{Z}^n)$ for some $g \in O_n(\mathbb{R})$.

I think there is a contradiction right here... Does the result right for $n \geq 4$ or only $4 \leq n \leq 7$ ?

Thank you !

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1 Answer 1

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There are lattices in $\mathbb{R}^n$ which are unimodular but do not have a lattice vector with norm 1. For example, consider the lattice you have mentioned. This lattice has a name - the $E_8$ lattice and the shortest vector in this lattice has norm $\sqrt{2}$. From this fact, there does not exist a $g \in O_n(\mathbb{R})$ such that $E_8 = g(\mathbb{Z}^n)$.

Also, all unimodular lattices for $n \leq 7$ are isomorphic to $\mathbb{Z}^n$. The proof can be found here.

This problem of checking whether an unimodular lattice is isomorphic to $\mathbb{Z}^n$ is an interesting problem in the area of computational complexity (the problem is shown to be in $\mathsf{NP} \cap \mathsf{coNP}$) and cryptography (paper). The problem is called $\mathbb{Z}^n$-isomorphism problem.

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