Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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 $w^\perp \cap \Lambda = 0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.