Let $\Bbb{F}_q$ be a finite field. Choose a non-square $\delta \in \Bbb{F}_q^*$ and form the quadratic extension $\Bbb{F}_q\big( \sqrt{\delta} \, \big)$. For an element $z \in \Bbb{F}_q\big( \sqrt{\delta} \, \big)$ let $\text{N}(z) := zz^q$ be its norm. Select a non-trivial multiplicative character $\chi: \Bbb{F}_q\big( \sqrt{\delta} \, \big)^* \longrightarrow \Bbb{C}^*$ together with a "radius" $r \in \Bbb{F}_q^*$ and consider the following "contour" sum:

\begin{equation} \sum_{\text{N}(z) = {\delta \over 4} (4 -r^2)} \, \chi \Big( z + (r-1) \sqrt{\delta}\Big) \, \overline{ \chi \Big( z + (r+1) \sqrt{\delta} \Big) } \end{equation}

What is the value of this sum ? Is it zero ? If not how does it depend on $r$ ?

kindly,

Ines

p.s. let's observe the convention that $\chi(0) = 0$.

p.p.s I've tried to use $\chi$´s (additive) Fourier expansion namely

\begin{equation} \chi(z) \ = \ \sum_{ w \ne 0} \, \widehat{\chi}(1) \, \overline{\chi(w)}\, \psi( - \text{tr} \, wz \big)\end{equation}

but without much success, even in the case when $r= 1$.

Let $\Bbb{F}_q$ be a finite field. Choose a non-square $\delta \in \Bbb{F}_q^*$ and form the quadratic extension $\Bbb{F}_q\big( \sqrt{\delta} \, \big)$. For an element $z \in \Bbb{F}_q\big( \sqrt{\delta} \, \big)$ let $\text{N}(z) := zz^q$ be its norm. Select a non-trivial multiplicative character $\chi: \Bbb{F}_q\big( \sqrt{\delta} \, \big)^* \longrightarrow \Bbb{C}^*$ together with a "radius" $r \in \Bbb{F}_q^*$ and consider the following "contour" sum:

\begin{equation} \sum_{\text{N}(z) = {\delta \over 4} (4 -r^2)} \, \chi \Big( z + (r-1) \sqrt{\delta}\Big) \, \overline{ \chi \Big( z + (r+1) \sqrt{\delta} \Big) } \end{equation}

What is the value of this sum ? Is it zero ? If not how does it depend on $r$ ?

kindly,

Ines

p.s. let's observe the convention that $\chi(0) = 0$.

p.p.s I've tried to use $\chi$´s (additive) Fourier expansion namely

\begin{equation} \chi(z) \ = \ \sum_{ w \ne 0} \, \widehat{\chi}(1) \, \overline{\chi(w)}\, \psi( - \text{tr} \, wz \big)\end{equation}

but without much success, even in the case when $r= 1$.

Let $\Bbb{F}_q$ be a finite field. Choose a non-square $\delta \in \Bbb{F}_q^*$ and form the quadratic extension $\Bbb{F}_q\big( \sqrt{\delta} \, \big)$. For an element $z \in \Bbb{F}_q\big( \sqrt{\delta} \, \big)$ let $\text{N}(z) := zz^q$ be its norm. Select a non-trivial multiplicative character $\chi: \Bbb{F}_q\big( \sqrt{\delta} \, \big)^* \longrightarrow \Bbb{C}^*$ together with a "radius" $r \in \Bbb{F}_q^*$ and consider the following "contour" sum:

\begin{equation} \sum_{\text{N}(z) = {\delta \over 4} (4 -r^2)} \, \chi \Big( z + (r-1) \sqrt{\delta}\Big) \, \overline{ \chi \Big( z + (r+1) \sqrt{\delta} \Big) } \end{equation}

What is the value of this sum ? Is it zero ? If not how does it depend on $r$ ?

p.s. let's observe the convention that $\chi(0) = 0$.

kindly,

Ines

Let $\Bbb{F}_q$ be a finite field. Choose a non-square $\delta \in \Bbb{F}_q^*$ and form the quadratic extension $\Bbb{F}_q\big( \sqrt{\delta} \, \big)$. For an element $z \in \Bbb{F}_q\big( \sqrt{\delta} \, \big)$ let $\text{N}(z) := zz^q$ be its norm. Select a non-trivial multiplicative character $\chi: \Bbb{F}_q\big( \sqrt{\delta} \, \big)^* \longrightarrow \Bbb{C}^*$ together with a "radius" $r \in \Bbb{F}_q^*$ and consider the following "contour" sum:

\begin{equation} \sum_{\text{N}(z) = {\delta \over 4} (4 -r^2)} \, \chi \Big( z + (r-1) \sqrt{\delta}\Big) \, \overline{ \chi \Big( z + (r+1) \sqrt{\delta} \Big) } \end{equation}

What is the value of this sum ? Is it zero ? If not how does it depend on $r$ ?

kindly,

Ines

p.s. let's observe the convention that $\chi(0) = 0$.

p.p.s I've tried to use $\chi$´s (additive) Fourier expansion namely

\begin{equation} \chi(z) \ = \ \sum_{ w \ne 0} \, \widehat{\chi}(1) \, \overline{\chi(w)}\, \psi( - \text{tr} \, wz \big)\end{equation}

but without much success, even in the case when $r= 1$.