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Fix an odd prime $p$. Then for a positive integer $a$, I can look at the quadratic Legendre symbol Gauss sum

$$ G_p(a) = \sum_{n \,\bmod\, p} \left( \frac{n}{p} \right) e^{2 \pi i a n / p}$$

where I use $\left( \frac{\cdot}{\cdot} \right)$ to be the Legendre Symbol. We know how to calculate it explicitly, and that's all well and good.

Recently, I've had reason to consider a "prime power" Gauss Sum

$$G_{p^k}(a) = \sum_{n\,\bmod\,p^k}\left(\frac{n}{p^k}\right)e^{2\pi i a n/p^k}$$

and I've noticed that as long as $k > 1$, $G_{p^k}(a) = 0$.

A sketch of why this is true:

  1. If $k$ is even, this is very easy to see as the Legendre symbol goes away, so we are summing the $p^k$th roots of unity.
  2. If $k$ is odd, the way I see it is a bit less obvious: reduce the Legendre symbol to be $\left( \frac{n}{p} \right)$ so that it's $p$-periodic, write $n = pn' + n''$ fpr $0 \leq n' < p^{k-1}$ and $0 \leq n'' \leq p$. Reorder the sum to sum over the $p^{k-1}$ roots of unity first, which for $k > 1$ will still give $0$.

My question is this: there are many ways of viewing Gauss sums. They can be viewed as discrete Fourier transforms, generators of quadratic extensions in $\mathbb{Q(e^{2\pi i / p})}$, as eigenfunctions, natural character sums, etc. I suspect that there is some 'obvious' or 'clear' reason why these prime power Gauss sums should be zero.

Is there an obvious or clear reason why $G_{p^k}(a) = 0$?

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I don't see any more "obvious" reason than what you gave. –  i707107 Apr 17 '13 at 5:50

2 Answers 2

What follows is essentially a way of rewording the argument you gave above.

Let $\chi$ be a Dirichlet character modulo $N$, and consider the Gauss sum $$\tau(\chi)=\sum_{k=1}^{N}\chi(k)e^{2\pi ik/N}.$$ Suppose that $q|N$ is the conductor of $\chi$ so that $\chi$ is induced by a primitive character modulo $q$. Call this primitive character $\chi^{\star}.$ Then Theorem $9.10$ of Montgomery and Vaughn states that $$\tau\left(\chi\right)=\mu\left(\frac{N}{q}\right)\chi^{\star}\left(\frac{N}{q}\right)\tau\left(\chi^{\star}\right).$$ When $k\geq2$ is even, the primitive character inducing $\left(\frac{n}{p^{k}}\right)$ is $\chi_{0},$ the principle character, and so $\frac{N}{q}=p^{k}.$ This is certainly not squarefree, so the $\mu\left(\frac{N}{q}\right)$ term is zero. Similarly, when $k\geq3$ is odd, the primitive character inducing $\left(\frac{n}{p^{k}}\right)$ is $\left(\frac{n}{p}\right),$ and so $\frac{N}{q}=p^{k-1}.$ Again this not squarefree, so the $\mu\left(\frac{N}{q}\right)$ term is zero implying that for $k\geq2$ $$\sum_{n=1}^{p^{k}}\left(\frac{n}{p^{k}}\right)e^{2\pi in/p^{k}}=0.$$

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The value of the sum is determined by $a$ and to a lesser extent $k$. Set $e(q):= e^{2\pi \imath q}$ for any rational $q$. We consider the sum \begin{align*} G_{p^k}(a) &:= \sum_{x=0}^{p^k-1} \left(\frac x{p^k}\right) e\left(\frac {ax}{p^k}\right) = 0 + \sum_{x=1}^{p^k-1} \left(\frac xp\right)^k e\left(\frac{ax}{p^k}\right)\\ &= \sum_{\substack{x=1\\(x,p)=1}}^{p^k-1} \left(\frac xp\right)^k e\left(\frac{ax}{p^k}\right) + \sum_{\substack{x=1\\ p \mid x}}^{p^k-1} \left(\frac xp\right)^k e\left(\frac{ax}{p^k}\right)\\ &= \sum_{\substack{x=1\\(x,p)=1}}^{p^k-1} \left(\frac xp\right)^k e\left(\frac{ax}{p^k}\right) \end{align*}

where $\displaystyle \left(\frac x{p^k}\right) = \left(\frac xp\right)^k$ is a well known property of the Jacobi symbol.

We re-index by mapping $x \mapsto u+pv$ where $u$ runs from $1$ to $p-1$ and $v$ runs from $0$ to $p^{k-1} -1$. Hence, \begin{align*} \sum_{\substack{x=1\\(x,p)=1}}^{p^k-1}& \left(\frac xp\right)^k e\left(\frac{ax}{p^k}\right) = \sum_{u=1}^{p-1} \sum_{v=0}^{p^{k-1}-1} \left(\frac up\right)^k e\left(\frac{a(u+pv)}{p^k}\right)\\ &= \sum_{u=1}^{p-1} \left(\frac up\right)^k e\left(\frac{au}{p^k}\right) \sum_{v=0}^{p^{k-1}-1} e\left(\frac{av}{p^{k-1}}\right)\\ &= \sum_{u=1}^{p-1} \left(\frac up\right)^k \begin{cases} 0 &\mbox{ if } a \not\equiv 0 \mod {p^{k-1}}\\ p^{k-1} &\mbox{ if } a \equiv 0 \mod {p^{k-1}} \end{cases}\\ &= \begin{cases} 0 &\mbox{ if } a \not\equiv 0 \mod {p^{k-1}} \mbox{ or } k \mbox{ odd}\\ (p-1)p^{k-1} &\mbox{ if } a \equiv 0 \mod {p^{k-1}} \mbox{ and } k \mbox{ even} \end{cases} \end{align*}

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