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

Hello everyone,

In ${\mathbb R}^n$, we say that a linear subspace is \emph{rational} if it admits a basis in ${\mathbb Q}^n$ (or equivalently in ${\mathbb Z}^n$). It means that $E\cap {\mathbb Z}^n$ is a submodule of {\mathbb Z}^n of rank equal to the dimension of the given subspace. By convention, $\{0\}$ is rational.

Now, if $E$ is any linear subspace of ${\mathbb R}^n$, we denote $b(E)$ (resp. $c(E)$ ) the dimension of the biggest rational subspace of ${\mathbb R}^n$ contained in $E$ (resp. the smallest rational subspace of ${\mathbb R}^n$ containing $E$).

My questions: 1) How can one compute $b(E)$ and $c(E)$? I am looking mainly for an algorithm (if it is possible). My entries for the algorithm are either a basis of $E$, or a system of equations (or any equivalent characterization).

2) It is clear that $0\leq b(E)\leq dim(E)\leq c(E)\leq n$. From an article, every case is possible. How one can construct such examples of different cases? I have some ideas but I am stuck in verifying because of the first question.

Thanks in advance.

(Maybe the question is simple for specialists. In this case, I would be grateful if someone could provide me a reference).

share|improve this question
    
If $b(E)=\dim E$, then $c(E)=\dim E$ also. So, not all cases of the inequality in (2) can occur. –  Felipe Voloch Oct 3 '12 at 9:55
    
Exact. Thank you for the remark. The article states that $E$ is rational iff $b(E)=c(E)$. Otherwise, $2+b(E)\leq c(E)$ and all cases can occur. –  Taladris Oct 3 '12 at 13:53
    
What is this article you allude to darkly? –  Igor Rivin Oct 4 '12 at 0:43
    
The article is "Foliations modelling nonrational simplicial toric varieties" by Battaglia and Zaffran (yet unpublished). ArXiv link: arxiv.org/abs/1108.1637 and the passage I am quoting is (similar to) the beginning of section 1. –  Taladris Oct 4 '12 at 1:41

1 Answer 1

up vote 1 down vote accepted

Given a matrix $M$ whose kernel is $E$, let $\{\alpha_1, \ldots, \alpha_k\}$ be a basis for the linear span over $\mathbb Q$ of the entries of $M$. Then $M = \sum_{j=1}^k \alpha_j M_k$ where $M_j$ are matrices with rational entries. Let $N$ be the matrix obtained by stacking the $M_j$ vertically.
Any member of $E \cap {\mathbb Q}^n$ is in the kernel of $N$, while the kernel of $N$ has a basis consisting of vectors in ${\mathbb Q}^n$. Thus $b(E)$ is the dimension of the kernel of $N$.

EDIT: In the other direction, consider the annihilator $E^\perp = \{y \in {\mathbb R}^n: y \cdot x = 0 \ \text{for all}\ x \in E\}$. Note that the annihilator of a rational subspace is rational (a basis in ${\mathbb Q}^n$ can be obtained using Gaussian elimination). It follows that the smallest rational subspace containing $E$ is the annihilator of the largest rational subspace contained in $E^\perp$. Thus $c(E) = n - b(E^\perp)$.

share|improve this answer
    
Thank you for the answer. I already knew the trick with the orthogonal. O should have mentionned it. –  Taladris Oct 5 '12 at 8:57

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.