I am wondering if there are any references that would help me with the following problem:

Let $p>2$ be a prime number, $n \in \mathbb{Z}^+$ odd such that $(n,p)=1$, $\chi$ an imaginary quadratic character of conductor $n$, and $\omega$ the Teichmuller character.

I'm trying to calculate the sum \begin{equation} \sum_{a=1}^{(np-1)/2} \chi\omega^{-1}(a)(-1)^a. \end{equation}

This sum almost reminds me of a quadratic Gauss sum (I know this is a big stretch) which I've seen can be computed using Fourier analysis. From my basic understanding (see Rodgers - Fourier series, Gauss sums, and quadratic reciprocity), the computation relies on the fact that \begin{equation} \sum_{a=0}^{p-1}\left(\frac{a}{p}\right)e^{2\pi i a/p} = \sum_{a=0}^{p-1} e^{2\pi i a^2/p} \end{equation} (here $(a/p)$ is the Legendre symbol) and defining $f(x) = \sum_{a=0}^{p-1} e^{2\pi i (a+x)^2/p}$, which is periodic on $\mathbb{R}$, we compute $\hat{f}(x)$ and find that \begin{equation} \sum_{a=0}^{p-1}\left(\frac{a}{p}\right)e^{2\pi i a/p}= \sum_{a = -\infty}^{\infty}\hat{f}(a) = \text{a computable integral}. \end{equation}

In my naive attempt to mimic this process, I think I fail (or just can't see how) to write $\sum_{a=1}^{(np-1)/2} \chi\omega^{-1}(a)(-1)^a$ as a sum involving only exponentials and then coming up with a periodic function $f(x)$ that is my character sum when evaluated at 0. Because it is quite a stretch to compare my sum to a Gauss sum, I'm now thinking it is a bad idea to try to mimic this computation.


  • Is there a more general method out there for computing character sums like the one I have using Fourier analysis?

  • Are there any other methods/theory that might help me out in calculating this sum?

I would also be happy if I could somehow factor this sum in $\mathbb{Q}(e^{2\pi i/(p-1)})$. I'm reminded of Stickelberger's theorem, where part of the proof involves finding the prime factorization of a certain Gauss sum (here I go again with the comparison). I know that the sum I'm looking at will not behave as nicely as the Gauss sum in the proof of Stickelberger's, but I am wondering,

  • Is there is a general theory for finding the $\mathcal{P}$-adic valuation of a character sum, where $\mathcal{P}$ lies above $p$ in whatever field the character sum lives in.
  • $\begingroup$ Using the Chinese remainder theorem, doesn't your sum factor into a product of two independent Gauss sums ? $\endgroup$ Jan 6, 2021 at 17:33
  • $\begingroup$ @HenriCohen Doesn't summing up to $(np-1)/2$ mess up the Chinese remainder theorem since it is an isomorphism from $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}/np\mathbb{Z}$? $\endgroup$ Jan 6, 2021 at 22:50
  • $\begingroup$ Since both $\chi$ and $\omega$ are odd characters, the sum to $np-1$ is twice that up to $(np-1)/2$, or do I misunderstand ? $\endgroup$ Jan 7, 2021 at 10:16
  • $\begingroup$ @HenriCohen I think the alternating part messes that up since $(-1)^{np-a} = (-1)(-1)^a$, so summing up to $np-1$ gives you zero. $\endgroup$ Jan 7, 2021 at 17:36
  • $\begingroup$ sorry, indeed since you assume $n$ odd. $\endgroup$ Jan 7, 2021 at 17:46


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