# Unimodular lattice

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

I have to show that, for $$n \geq 4$$, we can find $$v \in L$$ such that : $$=1$$. 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 $$$$. But for the final result, there is also something I don't understand. Just after, we are considering : $$\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 !

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.