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

I have recently made the following observation:

Let $v_i := (v_{i1}, v_{i2})$, $1 \leq i \leq k$, be non-zero positive elements of $\mathbb{Q}^2$ such that no two of them are proportional. Let $M$ be the $k \times k$ matrix whose entries are $m_{ij} := \max${$v_{ik}/v_{jk}: 1 \leq k \leq 2$}. Then $\det M \neq 0$.

The above statement is equivalent to the basic case of a result I recently discovered about pull back of divisors under a birational mapping of algebraic surfaces. I was going to include it as a part of another paper, then noticed the equivalent statement stated above and found it a bit amusing. My question is: is it worthwhile to try to publish it in a journal (as an example of an application of algebraic geometry to derive an arithmetic inequality), and if it is, then which journal(s)?

It is of course also very much possible that it is already known, or has a trivial proof (or counterexample!) - anything along those directions would also be appreciated.

Edit: Let me elaborate a bit about the geometric statement. In the 'other' paper, I define, for two algebraic varieties $X \subseteq Y$, something called "linking number at infinity" (with respect to $X$) of two divisors with support in $Y \setminus X$. I can show that when $Y$ is a surface, (under some additional conditions) the matrix of linking numbers at infinity of the divisors with support in $Y \subseteq X$ is non-singular. In a special (toric) case, the matrix of linking numbers takes the form of $M$ defined above. So the question is if the result about non-singularity of the matrix and its corresponding implication(s) are publishable anywhere.

share|improve this question
    
Your assumption of $\mathbb{Q}^2$ can be replaced with any totally ordered field I suppose? Also is it still true if pairs are replaced by $k$-tuples? –  John Jiang Aug 29 '10 at 20:55
3  
If it's an application of something in another paper you're writing (you didn't explicitly say this, but that was the impression I got from your question), then the right place for it is probably in the paper in question! Is there a reason that it doesn't belong there? –  Andy Putman Aug 29 '10 at 22:03
    
@Andy: I edited the question in response to your comment. Hope it explains! The problem with putting it in the other paper is that it is a bit isolated from other results of that (longish) paper, so this result would probably not be very visible. –  auniket Aug 30 '10 at 3:35
    
@John: I have no idea about this being true for any fields other than $\mathbb{Q}$. For every variety (i.e. defined over an arbitrary field), the `linking number at infinity' is a rational number - so the geometry will not help, at least not right away. –  auniket Aug 30 '10 at 3:42
    
@john: About $k$-tuples: I have tried to prove it for $k$-tuples, but failed. The proof does not generalize. And, I also don't know of any counterexample. –  auniket Aug 30 '10 at 4:06

1 Answer 1

up vote 9 down vote accepted

It seems to be true, but relatively simple (unless I made a mistake). Let's see:

First of all, scaling any pair $(v_{i,1},v_{i,2})$ by a constant $c$ does not change the determinant (one row of the matrix is multiplied by $c$, and one column is divided by $c$). We can therefore assume without losing generality that $(v_{i,1},v_{i,2})=(v_i,1)$.

Also, permutation of $v_i$'s does not change the determinant; therefore, we may assume that $v_i$'s are strictly increasing. In this case, the matrix has the form $$m_{ij}=\begin{cases} 1,& i\le j\cr v_i/v_j,& i>j\end{cases}.$$ Now subtract the top row from all others, and shift it to the bottom. You end up with a triangular matrix whose determinant is non-zero.

share|improve this answer
    
It works! Thanks. –  auniket Aug 30 '10 at 5:13

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.