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Let $\chi$ be the quadratic residue character in $\mathbb{F}_q$, where $q$ is a prime such that $q \equiv 1 \text{ (mod 4)}$, defined by \begin{align*} \chi(x) = \left\{ \begin{array}{ll} 0, & \text{if $x=0$}, \\ 1, & \text{if $x \neq 0$ is a quadratic residue in $\mathbb{F}_q$}, \\ -1, & \text{otherwise}. \\ \end{array} \right. \end{align*}

It is known, through Weil's bound, that for any degree-$d$ polynomial $P(x)$ over $\mathbb{F}_q$ that is not in the form $P(x) = Q^2(x)$ for another polynomial $Q$, \begin{align*} \sum_{x \in \mathbb{F}_q} \chi\left( P(x)\right) \leq (d-1) \sqrt{q}. \end{align*}

I have a character sum in the form \begin{align*} \sum_{\left(x_1,\dots,x_k\right) \in \mathbb{F}_q^k} \chi\left( (x_1 - x_2)(x_2 - x_3)\dots(x_k - x_1)\right). \end{align*}

Is there a multivariate version of Weil's bound that can work here?

I thought that perhaps the isomorphism between $\mathbb{F}_q^k$ and the extension field $\mathbb{F}_{q^k}$ can be used to map this character sum to the setup of Weil's bound, but I don't know much about finite fields, so I am not sure if when we treat $\left(x_1,\dots,x_k\right)$ as an element of $\mathbb{F}_{q^k}$, what's inside $\chi(\cdot)$ can still be considered a polynomial.

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    What is the actual formula/pattern giving your product $(x_1-x_2)(x_2-x_3)\cdots (x_k-x_1)$? Is it a product like $\prod \limits_{i – 2017-01-13
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    I have a $q\times q$ matrix $A$ whose $(i,j)$th element is $\chi(i-j)$, and this sum appears by expanding $\text{trace}\left( A^k\right)$. – 2017-01-13
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    So is $q$ a prime number? – 2017-01-13
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    Yes, sorry just edited the question. – 2017-01-13
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    So you have some sort of circulant matrix $A$, and you want to investigate the quadratic residue of the elements in the main diagonal of $A^k$, right? Do you have any condition on $k$? like $p\nmid k$, or something? – 2017-01-14

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