Correlation of a quadratic character with the Mobius function

Question

Let $p$ be a prime number and consider the sum $S(x)=\sum_{n\le x}\left(\frac{n}{p}\right)\mu(n)$. For how small an $x$ in terms of $p$ is it known that $S(x)=o(x)$? I am especially interested in unconditional results.

Answer 1

In general, we can take $x>\exp\{c_\epsilon p^\epsilon\}$ by the Prime Number Theorem for arithmetic progressions. More generally, one can use $L$-functions methods to relate $x$ to zero-free regions. In doing so it is hard to avoiding `losing logarithms'. The following elementary argument though, essentially due to Granville, does the job in an easier way.

Let $f$ be a completely multiplicative such that $|f(n)|\le1$ for all $n$ (so one can think that $f$ is a Dirichlet character), and assume that we know that $$ \left|\sum_{n\le x} \Lambda(n)f(n) \right| \le Cx\cdot \frac{\log Q}{\log x} \tag{*} $$ for all $x>Q$ (the size of $Q$ will depend on the available zero-free regions). Then we claim that $(*)$ holds for $\mu(n) f(n)$ too (with a different constant). Note that if suffices to show the result for $g(n)=\prod_{p^e\|n}(-f(p))^e$ (then one can use a convolution argument to pass to $f$). In order to show that $(*)$ holds with $g$ in place of $f\Lambda$, possibly with another constant $C'$ in place of $C$, we use induction, with the induction hypothesis being that $$ \left|\sum_{n\le x} g(n) \right| \le C' x\cdot \frac{\log Q}{\log x} \tag{**} $$ for all $x\le 2^m$. If $2^m\le Q$, this holds trivially (choosing $C'$ appropriately). Next, assume that $2^m>Q$ (and that $Q$ is large). Suppose also that that $(**)$ holds for $x\le 2^m$, and consider $x\in(2^m,2^{m+1}]$.Then $$ \sum_{n\le x} g(n) \log n = \sum_{n\le x} g(n) \sum_{d|n} \Lambda(d) = \sum_{dm\le x} \Lambda(d) g(d)g(m) . $$ We apply Dirichlet's hyperbola method: \begin{align*} \sum_{n\le x} g(n) \log n &= \sum_{m\le x^{1-\epsilon}} g(m) \sum_{d\le x/m} g(d)\Lambda(d) + \sum_{1<d\le x^{\epsilon}} \Lambda(d) g(d) \sum_{x^{1-\epsilon}<m\le x/d} g(m) \\ &\ll \sum_{m\le x^{1-\epsilon}} \frac{Cx}{m}\cdot \frac{\log Q}{\epsilon \log x} + \sum_{d\le x^{\epsilon}} \Lambda(d) \frac{C'x}{d} \cdot \frac{\log Q}{\log x} \\ &\ll \left( \frac{C}{\epsilon} + \epsilon C'\right) x\log Q \end{align*} Then applying partial summation and choosing $\epsilon$ and $C'$ appropriately completes the inductive step and thus the proof of $(**)$.

In the special case that $f(n)=(n/p)$, we know that $$ \sum_{n\le x}\Lambda(n) \left(\frac{n}{p}\right) = -\sum_{\substack{\rho=\beta+i\gamma\\L(\rho,(\cdot/p))=0,\,|\gamma|\le p}} \frac{x^{\rho}}{\rho} + O\left(xe^{-c\sqrt{\log x}}\right), $$ (see e.g. eq (13), p. 120 in Davenport's book "Multiplicative Number Theory"). There is a $c>0$ such that the first sum has at most one summand with $\beta\ge1-c/\log p$, for which one then necessarily has that $\gamma=0$ (i.e. $\rho=\beta$ is a Siegel zero). The sum over the zeroes with $\beta\le 1-c/\log p$ can be shown to be $\ll x^{1-c'/\log q}$ for some absolute constant $c'>0$, using zero-density estimates (see e.g. equation (18.9) in p. 428 of the book "Analytic Number Theory" by Iwaniec and Kowalski). We conclude that $$ \sum_{n\le x}\Lambda(n) \left(\frac{n}{p}\right) = -\frac{x^{\beta}}{\beta} + O\left( x^{1-c'/\log p} + xe^{-c\sqrt{\log x}}\right) . $$ So $(*)$ holds with $Q=1/(1-\beta)$ if $\beta$ exists and with $Q=p$ otherwise.

Answer 2

Wirsing's Theorem tells us that if $f$ is multiplicative and each $f(p)=-1,0 $ or $ 1$ then $\sum_{n\leq x} f(n) = o(x)$ as $\sum_{p\leq x} (1-f(p))/p \to \infty$ (and one cannot do much better). Moreover one can get an explicit upper bound: $ \sum_{n\leq x} f(n) \ll x \exp( -.32\sum_{p\leq x} (1-f(p))/p)$. In your case $f(n)=\mu(n)(n/p)$ so that $\sum_{p\leq x} (1-f(p))/p = 1/p+ 2\sum_{q\leq x, (q/p)=1} 1/q $. Therefore to get the bound $o(x)$ you need that a significant number of the $q\leq x$ satisfy $(q/p)=1$. So your question becomes: For what $x$ can we guarantee this? Or, in other words, is it possible that $(q/p)=-1$ for "most" of the primes $\leq x$ (as may well be the case of one has a Siegel zero)?

If we use quadratic reciprocity, then $(q/p)=-1$ is equivalent to demanding $(p/q)=$ something fixed, and we can find such $p$ for which this holds for all but one prime $q\ll \log p$, by Dirichlet's Theorem. But then, by smooth number estimates, one knows that for almost all such $p$ one has $\sum_{n\leq x} \mu(n)(n/p) \gg \rho(A)x$ for $x=(\log p)^A$ (for each $A$).

So we have "proved" that for any fixed $A>0$, the estimate $\sum_{n\leq x} \mu(n)(n/p) = o(x)$ does not hold uniformly for $x=(\log p)^A$.

The same ideas give, assuming GRH, that $\sum_{n\leq x} \mu(n)(n/p) = o(x)$ does hold uniformly provided $\log x/\log\log p \to \infty$ as $p\to \infty$.

These ideas can be found in my paper "Large Character Sums" with Soundararajan, though there we looked at character sums $\sum_{n\leq x} \chi(n)$; it should not take much to modify those ideas for this situation.