Seeking a sharp bound for a Dirichlet character sum

Question

Cross-posting notice: This question is also posted on Math Stack Exchange: "Bounding a sum over primitive character"

I'm trying to obtain the upper bound of the sum $$ S= \sum_{\chi_s^{\star}} {\overline{\chi_s^{\star}}(\alpha)} \sum_{k=1}^{s} k\, \chi_s^{\star}(k) \prod_{p \mid t} \left( 1 - \chi_s^{\star}(p) \right), $$ where:

$s$ and $t$ are two coprime square-free integers,
$\alpha$ is an integer coprime to $s$, and
$\chi_s^{\star}$ denotes a primitive Dirichlet character modulo $s$.

Numerically, this sum appears to be much smaller than expected.
Question: Is it possible to obtain a sharp bound for this sum?

Below is my attempt:

Attempt 1:

By the Cauchy–Schwarz inequality, $$ \lvert S \rvert \le \left(\sum_{\chi_s^{\star}} \Bigl|\sum_{k=1}^{s} k\,\chi_s^{\star}(k)\Bigr|^2\right)^{\frac{1}{2}} \left(\sum_{\chi_s^{\star}} \Bigl|\sum_{d \mid t} \mu(d)\,\chi_s^{\star}(d)\Bigr|^2\right)^{\frac{1}{2}}, $$ which leads to, after some treating of the bilinear sum, the bound $$ \lvert S \rvert \le \frac{1}{\sqrt{3}}\, d(t)^{\frac{1}{2}}\, s^{\frac{3}{2}}\, \varphi(s) + o\Bigl(d(t)^{\frac{1}{2}}\, s^{\frac{3}{2}}\, \varphi(s)\Bigr). $$

However, this bound is not sharp enough.
Attempt 2:

By the orthogonality of Dirichlet characters and Möbius inversion, $$ \begin{aligned} S &= \sum_{d \mid t} \mu(d) \sum_{k=1}^s k \sum_{\chi_s^{\star}} {\overline{\chi_s^{\star}}(\alpha)}\, \chi_s^{\star}(k d) \\ &= \sum_{d \mid t} \mu(d) \sum_{k=1}^s k \sum_{\substack{q \mid s \\ \alpha \equiv k d \, (\bmod\, q)}} \mu(q)\, \varphi(q), \end{aligned} $$ which leads to $$ S = \sum_{d \mid t} \mu(d) \sum_{k=1}^{s} k \prod_{p \mid (s,\, \alpha - k d)} \left(2 - p\right) $$ or $$ S = \sum_{q \mid s} \mu(q) \varphi(q) \sum_{d \mid t} \mu(d) \sum_{\substack{k=1 \\ (k,s)=1 \\ k \equiv d^{-1}\alpha \, (\bmod\, q)}}^{s} k. $$

However, I have not been able to further treat either of these forms.

Will Sawin · Accepted Answer · 2025-02-14 17:00:03Z

I think in attempt 2 the $\mu(q)$ should be $\mu(s/q)$, so after "which leads to" the $2-p$ should be $p-2$. For example if $\alpha=1$ the $d=1, k=1$ term should certainly be $\prod_{p\mid s} (p-2)$. In the corrected version:

$$ S = \sum_{d \mid t} \mu(d) \sum_{k=1}^{s} k \prod_{p \mid (s,\, \alpha - k d)} \left(p-2\right) $$

the sum over $k$ has no cancellation. So the only possible cancellation in the sum is the sum over $d$, which might sometimes have very little cancellation (e.g. if $t$ is prime.)

We can bound the inner sum $ \sum_{k=1}^{s} k \prod_{p \mid (s,\, \alpha - k d)} \left(p-2\right)$ by $s \sum_{k=1}^{s} \prod_{p \mid (s,\, \alpha - k d)} \left(p-2\right)$, and this should be sharp to within a factor of $2$. In this sum, since $d$ is coprime to $s$, $\alpha-kd$ hits each residue class modulo $s$ exactly once so this is the same as $s \prod_{p \mid s} ( p-2+ p-1) = \prod_{p \mid s} ( p (2p-3)) \leq \sigma_0(s) s^2$.

Then bounding trivially the sum over $d$ gives a bound of $\sigma_0(d) \sigma_0(s) s^2$ which is a little sharper than your Cauchy-Schwarz bound (which grows like $s^{5/2}$).

Is that sharp enough for your purposes? If not, is it really true that the sum empirically always has more cancellation than that?

Thanks! That was my oversight; however, I think it still leads to μ(s)∏(2−p), not ∏(p−2), which might allow for additional cancellation. Actually, this sum arises when calculating the remainder term R=∑g(s)S in a sieve method, where g is multiplicative and P(z)=s⋅t. The bound you proposed is indeed sharp, but it still doesn't let me control that remainder term directly; I'd need to account for the extra cancellation to bound it properly. — baxter001225
– baxter001225, Commented Feb 15 at 8:46

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