Bounding sums of the Möbius function over arithmetic progressions

preprint2014arXivOpen access

Abstract

Let M(x)=∑1≤n≤xμ(n)M(x)=\sum_{1\le n\le x}μ(n) where μμ is the Möbius function. It is well-known that the Riemann Hypothesis is equivalent to the assertion that M(x)=O(x1/2+ε)M(x)=O(x^{1/2+ε}) for all ε>0ε>0. There has been much interest and progress in further bounding M(x)M(x) under the assumption of the Riemann Hypothesis. In 2009, Soundararajan established the current best bound of M(x)≪xexp⁡((log⁡x)1/2(log⁡log⁡x)c) M(x)\ll\sqrt{x}\exp\left((\log x)^{1/2}(\log\log x)^c\right) (setting cc to 1414, though this can be reduced). Halupczok and Suger recently applied Soundararajan's method to bound more general sums of the Möbius function over arithmetic progressions, of the form M(x;q,a)=∑n≤xn≡a(modq)μ(n). M(x;q,a)=\sum_{\substack{n\le x \\ n\equiv a\pmod{q}}}μ(n). They were able to show that assuming the Generalized Riemann Hypothesis, M(x;q,a)M(x;q,a) satisfies M(x;q,a)≪εxexp⁡((log⁡x)3/5(log⁡log⁡x)16/5+ε) M(x;q,a)\ll_ε\sqrt{x}\exp\left((\log x)^{3/5}(\log\log x)^{16/5+ε}\right) for all q≤exp⁡(log⁡22⌊(log⁡x)3/5(log⁡log⁡x)11/5⌋)q\le\exp\left(\frac{\log 2}2\lfloor(\log x)^{3/5}(\log\log x)^{11/5}\rfloor\right), with aa such that (a,q)=1(a,q)=1, and ε>0ε>0. In this paper, we improve Halupczok and Suger's work to obtain the same bound for M(x;q,a)M(x;q,a) as Soundararajan's bound for M(x)M(x) (with a 1/21/2 in the exponent of log⁡x\log x), with no size or divisibility restriction on the modulus qq and residue aa.

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