Let $\Lambda$ be a lattice with a quadratic form $q$ of signature (3,19).

Let $\Lambda_{\mathbb{R}}:=\Lambda\otimes \mathbb{R}$ and $W\subset \Lambda_{\mathbb{R}}$ a positive subspace of dimention 3.

I have found this proposition:

$W^\perp\cap\Lambda=0$ if and only if $\exists w \in W$ such as $w^\perp \cap \Lambda =0$ (of course orthogonality is referred to the quadratic form $q$)

clearly the implication $\Leftarrow$ is trivial as $W^\perp \subset w^\perp$.

I have difficulties formalizing the other implication, this is how i'm thinking, but i don't know if it's right:

In the worst case it will be $\Lambda\cap W\simeq \mathbb{Z}^3$ and so in $W$ i'm able to find a 2-dimensional plane $\widetilde{W}$ such as $\widetilde{W}\cap \Lambda=\{0\}$ (thinking in $\mathbb{R}^3$ with the canonical quadratic form i can take the plane $\langle e_1,e_2\rangle$ with $e_1=(1,0,\sqrt{2})$ and $e_2=(0,1,\sqrt{3})$). $\widetilde{W}$ is the orthogonal space of a vector in $W$, say $\widetilde{w}$. So now i should have $\widetilde{w}^\perp=\widetilde{W}\oplus W^\perp$ and $\widetilde{w}^\perp\cap \Lambda=\{0\}$.

Does it convince you?


For each $\lambda \neq 0$ in $\Lambda$, consider the subspace $H_\lambda$ of $W$ given by $\langle \lambda, x \rangle = 0$. Saying $W^\perp \cap \Lambda = 0$ means these are all proper subspaces of $W$. Now, since $\mathbb{R}$ is uncountable, it's not too hard to show that a countable union of proper subspaces can't be the whole vector space. So there's an element $w \in W$ which is outside the union, and therefore $w^\perp \cap \Lambda = 0$.


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