2
$\begingroup$

Let $M$ be a large positive integer, $d$ an odd positive integer and $f: \mathbb{Z}_{>0} \times \mathbb{Z}_{>0} \to \mathbb{R}$. For a non-principal character $\chi_d = \chi$ with modulus $d$, I am interested in the following type of sums. $$S(\chi, f) = \sum_{\substack{m = M \\ (m, d) = 1}}^{2M}f(m, d) \chi(m). $$

For $f \equiv 1$, we know that $\lvert S(\chi, f)\rvert \ll \sqrt{d}\log(d)$ by the Pólya–Vinogradov inequality. Now I am interested in bounding $\lvert S(\chi, f)\rvert$ when $$f(m, d) = \frac{\phi(m)}{m} \prod_{\substack{p \textrm{ prime} \\ (p, md) = 1}} \left(1 - \frac{\rho_{m}(p)}{p^2} \right)$$

where $\rho_m(p) = 1 + \genfrac(){}{}m p$ for primes $p$ that do not divide $m$. $\chi_d(m) = \genfrac(){}{}m d$, the Jacobi symbol.

I expect a similar result as the Pólya–Vinogradov inequality. Computational experiments for some $M$, $d$ values did agree with my expectation. Have these types of sums been studied earlier? Any help would be highly appreciated.

$\endgroup$
10
  • $\begingroup$ TeX notes: $\mathbb{Z_{> 0}}$ \mathbb{Z_{> 0}} puts everything in bold (note that $\mathbb 0$); prefer $\mathbb{Z}_{> 0}$ \mathbb{Z}_{> 0}. Also prefer $a \ll b$ a \ll b to $a << b$ a << b. I have edited accordingly. $\endgroup$
    – LSpice
    Mar 31, 2022 at 1:04
  • $\begingroup$ just some comments. at least for primitive characters the polya-vinogradov inequality amounts (through gauss' sum) to summing a linear exponential sum - more specifically showing $\sum _{n<N}e(an/q)\ll q/a$, or rather this with $N<q$. so, looking at $\sum _{n<N}$f(n)e(an/q)$ might be a non-terrible start $\endgroup$
    – tomos
    Mar 31, 2022 at 1:26
  • 1
    $\begingroup$ If you want to upper bound $\sum_{m \le M} f(m) \chi(m)$ and know how to bound $\sum_{m \le M} \chi(m)$, you can write $f$ as a convolution of the constant function $1$ and a function $g$: $f=1∗g$, and rewrite your sum as $\sum_{e \le M} g(e) \sum_{n \le M/e} \chi(m)$, which is bounded by $\ll\sum_{e \le M} |g(e)| \min\{M/e, \sqrt{d}\log d\}$. $\endgroup$ Mar 31, 2022 at 10:37
  • $\begingroup$ @OfirGorodetsky Thank you for your idea. Could you please refer me to a resource where I could learn the convolution theory involved in your answer? $\endgroup$
    – Melanka
    Mar 31, 2022 at 22:54
  • $\begingroup$ @OfirGorodetsky One more question. In this case, it appears that your method is the same as partial summation. Am I right? $\endgroup$
    – Melanka
    Apr 1, 2022 at 2:07

1 Answer 1

0
$\begingroup$

Based on the ideas posted by @OfirGorodetsky and @tomos as comments to this post, I managed to compile this solution.

$$\begin{align*} S &= \sum_{m = M}^{2M} \frac{\phi(m)}{m} \prod_{p \nmid md}\left(1 - \frac{1}{p^2} - \frac{\chi_p(m)}{p^2} \right) \; \chi_d(m) \\ &= \sum_{m = M}^{2M} \frac{\phi(m)}{m} \left( \sum_{r \nmid md} \prod_{p \nmid mdr} \left(1 - \frac{1}{p^2} \right) \mu(r) \frac{\chi_r(m)}{r^2} \right) \chi_d(m) \\ &= \sum_{m = M}^{2M} \prod_{p | m} \frac{1}{1 + \frac{1}{p}} \sum_{r \nmid d} \prod_{p \nmid dr} \left(1 - \frac{1}{p^2} \right) \mu(r) \frac{\chi_{rd}(m)}{r^2} \\ &= \sum_{r \nmid d} \left( \mu(r) \frac{1}{r^2} \prod_{p \nmid dr} \left(1 - \frac{1}{p^2} \right) \right) \sum_{m = M}^{2M} \left( \prod_{p | m} \frac{p}{1 + p} \right)\chi_{dr}(m) \end{align*} $$

Now setting $f(m) = \prod_{p | m} \frac{p}{1 + p}$ and $g = f*1$, by the Mobius inversion formula we have that $g(m) = \sum_{ab = m} f(a) \mu(b)$. It is easy to see by computation that $|g(m)| < \frac{1}{m}$. Hence we have,

$$\begin{align*} \left| \sum_{m = M}^{2M} \left( \prod_{p | m} \frac{p}{1 + p} \right)\chi_{dr}(m)\right| &= \left| \sum_{m = M}^{2M} \sum_{ab = m} g(a)\chi_{dr}(a) \chi_{dr}(b) \right| \\ &= \left| \sum_{a = 1}^{2M} g(a)\chi_{dr}(a) \sum_{b = M/a}^{2M/a} \chi_{dr}(b) \right| \\ &\leq \sum_{a = 1}^{2M} |g(a)| \left| \sum_{b = M/a}^{2M/a} \chi_{dr}(b) \right| \\ &\ll \sum_{a = 1}^{2M} \frac{1}{a} \sqrt{dr} \log{(dr)} \ll \log{(M)} \sqrt{dr} \log{(dr)} \end{align*}$$

So now, $$\begin{align*} |S| \ll \sum_{r \nmid d} \frac{1}{r^2} \log{(M)} \sqrt{dr} \log{(dr)} \ll \log{(M)} \sqrt{d} \log{(d)} \end{align*}$$

The Euler factor involving the $\rho_m$ function was too complicated to get a useful function applying Mobius inversion. So somehow had to expand out and dealt as shown.

$\endgroup$

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.