# Making a character small at a reciprocal

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?

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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 $x^x$. – David Speyer Jul 14 '10 at 12:11
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? – BCnrd Jul 14 '10 at 12:25