Let $\chi$ be a Dirichlet character modulo $q$ induced by a primitive character $\chi^*$ modulo $d$ for some divisor $d$ of $q$. Let $n$ be a positive integer, and consider the generalised Gauss sum \[c_{\chi}(n) = \sum_{a \pmod{q}} \chi(a) e\left(\frac{an}{q}\right).\] Note that I do not assume that $(n,q) = 1$. When $n = 1$, this is just the usual Gauss sum $\tau(\chi)$.

In Multiplicative Number Theory by Montgomery and Vaughan, it is shown (Theorem 9.12) that \[c_{\chi}(n) = \overline{\chi}^* \left(\frac{n}{(q,n)}\right) \chi^* \left(\frac{q}{d(q,n)}\right) \mu\left(\frac{q}{d(q,n)}\right) \frac{\varphi(q)}{\varphi\left(\frac{q}{(q,n)}\right)} \tau(\chi^*)\] if $d \mid \frac{q}{(q,n)}$, while $c_{\chi}(n) = 0$ otherwise.

Is it true that \[c_{\chi}(n) = \tau(\chi^*) \sum_{c \mid \left(\frac{q}{d},n\right)} c \overline{\chi}^* \left(\frac{n}{c}\right) \chi^* \left(\frac{q}{cd}\right) \mu\left(\frac{q}{cd}\right) ?\] If not, is there some other way to write $c_{\chi}(n)$ as a sum over divisors?

In the simplest case when $d = 1$, so that $\chi^*$ is the trivial character, this states that \[c_{\chi}(n) = \sum_{c \mid (q,n)} c \mu\left(\frac{q}{c}\right) = \mu\left(\frac{q}{(q,n)}\right) \frac{\varphi(q)}{\varphi\left(\frac{q}{(q,n)}\right)},\] which is Theorem 4.1 of Montgomery and Vaughan.


Wouldn't you know, I found a reference: a proof of this is given in Modular Forms by Miyake, Lemma 3.1.3(2).


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.