Bounding sums of the Möbius function over arithmetic progressions

Lynnelle Ye

Published 2014-06-27Version 1

Let $M(x)=\sum_{1\le n\le x}\mu(n)$ where $\mu$ is the M\"obius function. It is well-known that the Riemann Hypothesis is equivalent to the assertion that $M(x)=O(x^{1/2+\epsilon})$ for all $\epsilon>0$. There has been much interest and progress in further bounding $M(x)$ under the assumption of the Riemann Hypothesis. In 2009, Soundararajan established the current best bound of \[ M(x)\ll\sqrt{x}\exp\left((\log x)^{1/2}(\log\log x)^c\right) \] (setting $c$ to $14$, though this can be reduced). Halupczok and Suger recently applied Soundararajan's method to bound more general sums of the M\"obius function over arithmetic progressions, of the form \[ M(x;q,a)=\sum_{\substack{n\le x \\ n\equiv a\pmod{q}}}\mu(n). \] They were able to show that assuming the Generalized Riemann Hypothesis, $M(x;q,a)$ satisfies \[ M(x;q,a)\ll_{\epsilon}\sqrt{x}\exp\left((\log x)^{3/5}(\log\log x)^{16/5+\epsilon}\right) \] for all $q\le\exp\left(\frac{\log 2}2\lfloor(\log x)^{3/5}(\log\log x)^{11/5}\rfloor\right)$, with $a$ such that $(a,q)=1$, and $\epsilon>0$. In this paper, we improve Halupczok and Suger's work to obtain the same bound for $M(x;q,a)$ as Soundararajan's bound for $M(x)$ (with a $1/2$ in the exponent of $\log x$), with no size or divisibility restriction on the modulus $q$ and residue $a$.