The following question emerged from thinking about the Erdős discrepancy problem. I don't know whether an answer would be directly helpful, but it might, and in any case I find the question quite a nice one.

Suppose we have a character on the additive group of the rationals. That is, we have a function $\chi$ from $\mathbb{Q}$ to $\mathbb{T}$ such that $\chi(r+s)=\chi(r)\chi(s)$ for any two rational numbers $r$ and $s$. The following argument shows that for every $\epsilon$ there exists a positive integer $n$ such that $|\chi(n^{-1})-1|<\epsilon$. You choose a positive integer $r$ that's bigger than $2\pi/\epsilon$, then set $N=r!$, and let $\alpha$ be such that $\chi(1/N)=e(\alpha)$ (which is standard notation for $\exp(2\pi i\alpha)$). Then you use Dirichlet's pigeonhole argument to find $n\leq r$ such that the distance from $n\alpha$ to the nearest integer is at most $1/r$ and therefore less than $\epsilon/2\pi$. From that it follows that $|\chi(n/N)-1|<\epsilon$, and we are done, since $N/n$ is an integer.

This argument gives a pretty disappointing bound: $n$ has an $x^x$ type dependence on $1/\epsilon$. Can more sophisticated techniques (Kloosterman sums perhaps?) give much better bounds?

  • 2
    $\begingroup$ Minor improvement: take $N$ to be the LCM of $1$, $2$, ..., $r$. Then $N \approx e^r$ instead of $r!$, and you get $e^x$ instead of $x^x$. $\endgroup$ – David E Speyer Jul 14 '10 at 12:11
  • 2
    $\begingroup$ Nice question. The character group of $\mathbf{Q}$ is $\mathbf{A}_ {\mathbf{Q}}/\mathbf{Q}$: there's an adele $a = (a_{\infty}, a_2, a_3, \dots)$ unique mod $\mathbf{Q}$ such that $\chi(x) = e(-xa_{\infty}) \cdot \prod_p e_p(xa_p)$ where $e_p(t) = e(\langle t\rangle_ p)$, with $\langle t\rangle_ p$ the image of $t \in \mathbf{Q}_ p$ under $\mathbf{Q}_ p \rightarrow \mathbf{Q}_ p/\mathbf{Z}_p \hookrightarrow \mathbf{Q}/\mathbf{Z}$ ("$p$-adic Laurent tail"). Can arrange all $a_p \in \mathbf{Z}_p$ and $a_{\infty} \in [0,1)$ to make $a$ unique. Not deep, but maybe allows a more hands-on approach? $\endgroup$ – BCnrd Jul 14 '10 at 12:25

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.