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Let $V \subset \mathbb P^n$ be a nice (smooth, projective?) variety over a finite field $\mathbb F_q$. Let $\chi_0,\chi_1,\dots,\chi_{n}: \mathbb F_q^\times \to \mathbb Q(\mu_{q-1})$ be multiplicative characters that take the value $0$ on $0$ if they are nontrivial.

What can be said about the sum: $$J_V(\chi_0,\dots,\chi_n) = \sum_{x_0,\dots,x_n \in V(\mathbb F_q)}\chi_0(x_0)\dots\chi_n(x_n)?$$

I am really summing over the cone of the projective variety.

A variant of the classical Jacobi sum is obtained from the $\mathbb P^1$ defined by $x+y+z = 0$ in $\mathbb P^2$.

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    $\begingroup$ There's something strange with your wording. $J_V(\chi_1, \chi_2)$ is stated as a function of $\chi_1, \chi_2$, but the right hand side depends on $\chi_0, \cdots, \chi_n$. $\endgroup$
    – Stanley Yao Xiao
    Nov 4, 2019 at 14:45
  • $\begingroup$ Oh, sorry. Let me fix that. $\endgroup$
    – Asvin
    Nov 4, 2019 at 15:05

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