Consider a Dirichlet character, $\chi(n)$, and the partial sum :

$$S(\chi,x)=\bigg |\sum_{n=1}^{x} \chi(n)\bigg|$$

There are many works to bound this sum when $\chi$ is a primitive character, but what can we say if $\chi$ is not primitive ?

More specifically if we fix a primitive Dirichlet character $\chi$ and then define from it a suite of non primitive character $\chi_N$ defined by:

$$\forall n, \chi_1(n)=\chi(n)$$

$$\forall n, \chi_N(n)=\chi_{N-1}(n).\chi^{P_N} (n)$$

Where $\chi^{P_N} (n)$ is the principal Dirichlet character associated to the N-th prime number (not considering 2).

(so $\chi^{P_N} (n)$ is the principal character simply defined by : $\chi^{P_N} (n) =0 $ if n is a multiple of the N-th prime number $P_N$ and 1 if not)

This suite of character is build from original character by "removing at each step a prime".

Question : how will evolute the max of $S(\chi^{P_N},x)$ for these charcaters ?

I would like to show that in this suite there are an infinity of characters with their $Max(S(\chi^{P_N},x))$ lower than a fix constant ? Is it realistic ?

Any reference on bounding partial sum of imprimitive characters ?

  • 1
    $\begingroup$ What does "Nieme prime" mean? All I've got so far is that Nieme is the name of a river. $\endgroup$ Feb 26, 2016 at 19:57
  • 3
    $\begingroup$ @Qiaochu Yuan, what language did you pick for your reading exam? That is French for N-th: the N-th prime $\endgroup$
    – KConrad
    Feb 26, 2016 at 21:19
  • 3
    $\begingroup$ The following version of the Pólya-Vinogradov bound holds for any non principal character $\chi \pmod{q}$: $\left| \sum_{M < n \leq M+N} \chi(n) \right| \leq 6\sqrt{q} \log q$ (cf. Iwaniec & Kowalski's Analytic Number Theory, p. 324). $\endgroup$ Feb 26, 2016 at 22:49
  • 1
    $\begingroup$ @KConrad: it was French, but in my defense the exam was very easy and I was allowed a French dictionary. Also, "nieme" is apparently Polish for "mute" as well. $\endgroup$ Feb 27, 2016 at 3:34
  • $\begingroup$ Sorry, Nième is the N-th prime number... $\endgroup$
    – Bertrand
    Feb 27, 2016 at 9:35

1 Answer 1


The original Pólya-Vinogradov inequality alredy works for non-primitive Dirichlet characters,

$$S(\chi,x)\leq c\sqrt {q} \log q$$

for some absolute constant $c$.

As J.H.S. mentions in the comments, apparently you can take $c=6$, see Iwaniec-Kowalski, theorem 12.5, p. 324, although they don't give a reference (I think that the Hildebrand improvement they mention is a stronger one, for primitive characters only).

Also, conditional on GRH, it is a result of Montgomery-Vaughan that

$$S(\chi,x)\ll \sqrt {q} \log\log q$$

Note that other than the GHR improvement, Pólya-Vinogradov is sharp, since $S(\chi,x)\geq \pi^{-1}\sqrt {q}$ for $\chi$ primitive. Also, $S(\chi,x)\gg \sqrt {q} \log\log q$ infinitely often for quadratic characters (Paley).

  • $\begingroup$ Do you mean that $S(\chi,x)\geq \pi^{-1}\sqrt {q}$ is always true even for non primitive Dirichlet characters ? $\endgroup$
    – Bertrand
    Feb 27, 2016 at 9:40
  • $\begingroup$ @Bertrand Definitely not, only for primitive characters. I meant that the bound was sharp for general $\chi$. I'll edit to make it clear. $\endgroup$
    – Myshkin
    Feb 27, 2016 at 9:48
  • 1
    $\begingroup$ Thanks for your help, I found that for all characters we have $S(\chi,x) \ge \frac{\tau(\chi)}{\pi}$ (see Page 311) Multiplicative Number Theory I: Classical Theory Par Hugh L. Montgomery,Robert C. Vaughan $\endgroup$
    – Bertrand
    Feb 27, 2016 at 10:11
  • $\begingroup$ In the case of the suite of induced characters mentionned in the question, using relation between Gauss sums of induced characters we have : $S(\chi^{P_N},x)\geq \pi^{-1} |\tau(\chi_1)|$, so the lower bound is a constant and we still do not know if the partial sums of an infinity of induced characters can be lower than a fixed constant. $\endgroup$
    – Bertrand
    Feb 29, 2016 at 10:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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