5
$\begingroup$

I am working through the proof of the Bombieri-Vinogradov theorem in Analytic Number Theory (Iwaniec, Kowalski). My problem is that on page 424, it is said that $\mu(m)$ satisfies $D_f(x;q,a)\ll (\sum \limits_{n \le x} |f(n)|^2)^{1/2} \, \, x^{1/2}(\log x)^{-A}$ (17.12) by the Siegel-Walfisz theorem (so that we can apply Theorem 17.4). How does the Siegel-Walfisz theorem apply to the Möbius function?

$\endgroup$

1 Answer 1

7
$\begingroup$

The Siegel-Walfisz principle for a function $f(n)$ states that for all $A>0$ fixed then whenever $a$ modulo $q$ is a residue class with $a$ and $q$ coprime then one has $$\sum_{\substack{n\leq x \\ n \equiv a\hspace{-0,3cm}\mod{\hspace{-0,1cm}q}}}f(n)=\frac{1}{\phi(q)}\sum_{\substack{n\leq x\\ \gcd(n,q)=1}}f(n)+O_{A,f}\left(\sqrt{q}\frac{x}{(\log x)^A}\right).$$ Bear in mind that this is non--trivial only in the case that $q\leq (\log x)^B$ for some $B>0$ fixed. It has been proved for a lot of multiplicative functions, one example being $f(n)=\mu(n)$. The proof for this case is similar to the proof for primes, one simple exposition is Corollary 5.29, Equation 5.80, of the book you are reading. It gives a good upper bound for the quantities $$\sum_{n \leq x}\chi(n) \mu(n)$$ for all primitive Dirichlet characters and in order to deduce Siegel-Walfisz for $\mu(n)$ you use orthogonality of characters. Orthogonality means $$\sum_{\substack{n\leq x \\ n \equiv a\hspace{-0,3cm}\mod{\hspace{-0,1cm}q}}}\hspace{-0,5cm}\mu(n)=\frac{1}{\phi(q)}\sum_{\chi\hspace{-0,3cm}\mod{\hspace{-0,1cm}q}}\left(\sum_{n \leq x}\chi(n) \mu(n)\right).$$

Finally to answer your question, the bound $$D_f(x;q,a)\ll \left(\sum \limits_{n \le x} |f(n)|^2\right)^{1/2} \, \, x^{1/2}(\log x)^{-A}$$ for $q\leq (\log x)^B$ and $f=\mu$ is equivalent to $$D_\mu(x;q,a)\ll x (\log x)^{-A}$$ owing to $$\sum_{n\leq x}|\mu(n)|^2=\frac{6}{\pi^2}x+O(x^{\frac{1}{2}})\gg x.$$ Let me know if something is not clear.

$\endgroup$
2
  • $\begingroup$ In your first sentence, may one add (n,q)=1 to the right side? So instead of summing over all n less than x, only those coprime less than x? $\endgroup$ May 21, 2015 at 13:04
  • $\begingroup$ I am afraid you are right, I edited it accordingly. $\endgroup$
    – Dr. Pi
    May 21, 2015 at 17:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.