Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The Jacobi sum of $n$ multiplicative character $\chi_1,\dots,\chi_n$ on a finite field $\mathbb F_q$ is defined as $$J(\chi_1,\dots,\chi_n) = \sum_{x_1,\dots,x_n \in \mathbb F_q, x_1+\dots+x_n=1} \chi_1(x_1) \chi_2(x_2) \dots \chi_n(x_n).$$ One also considers the variant: $$J_0 (\chi_1,\dots,\chi_n) = \sum_{x_1,\dots,x_n \in \mathbb F_q, x_1+\dots+x_n=0} \chi_1(x_1) \chi_2(x_2) \dots \chi_n(x_n)$$

It is well-known (see e.g. Ireland and Rosen) that one can compute the complex modulus of $J(\chi_1,\dots,\chi_n)$ and $J_0(\chi_1,\dots,\chi_n)$ (when the $\chi_i$ are "general" in some precise sense, $|J(\chi_1,\dots,\chi_n)|=q^{n-3/2}$, and the other cases are not hard to determine as well).

The definition of the Jacobi sum can be rewritten in a more compact way using the maximal diagonal torus $T$ of $Gl_n$: for $\chi=(\chi_1,\dots,\chi_n)$ a character of the maximal torus, $$J(\chi) = \sum_{x \in T(\mathbb F_q), \ tr\ x = 1} \chi(x)$$ $$J_0(\chi)= \sum_{x \in T(\mathbb F_q), \ tr\ x = 0} \chi(x)$$ Now it is clear that the definition above of $J(\chi)$ makes sense when $T$ is replaced by any subtorus of $Gl_n$ defined over $\mathbb F_q$, not necessarily maximal or split. My question is

Is it possible to determine, or at least estimate, $J(\chi)$ for general character of $T$ when $T$ is a general torus of $Gl_n$ defined over $\mathbb F_q$?

In particular, by how much (if anything) is it possible to improve on the trivial bound $|J(\chi)| \leq \sum_{x \in T(\mathbb F_q), \ tr\ x = 1} 1$?

If the general question is too hard, here is one case I am especially interested: $T$ is the torus of $Gl_4$ of diagonal matrices $(x,y,y^{-1},x^{-1})$. So in this case, $T$ is still a split torus, but is not maximal (actually it it is the maximal torus of the symplectic group $Sp_4$ seen as a torus of $Gl_4$ through the natural inclusion). So in this case, in down-to-earth terms, $J_0(\chi)=\sum_{x,y \in \mathbb F_q, x + y + x^{-1} + y^{-1} =0} \chi_1(x) \chi_2(y)$.

share|improve this question
add comment

3 Answers 3

The case where the sum is zero should be "more degenerate". In Felipe Voloch's expression, the sum $S(\chi,c)$ has at most two terms for a given $c$ (solving a quadratic equation), so it gives a "trivial" bound of size $q$. (It might well be smaller in fact.)

In the case where the trace is non-zero, I would expect that a fairly direct application of Deligne's theory will give $q^{3/2}$ (up to a multiplicative factor, that is concretely a sum of Betti numbers, and for which very general bounds can be found in a paper of Katz, "Bounds for sums of Betti numbers in arbitrary characteristics"). Basically, this will hold unless the function summed is constant on some irreducible component of the parameter variety, and this is well explained in SGA 4 1/2 "Sommes trigonométriques".

It's then a natural question whether the bound is in fact of size $q$ (square-root cancellation) -- there are general estimates in a number of cases in works of Katz and Adolphson-Sperber or Denef-Loeser, and this general setting might fit in one of them.

share|improve this answer
    
Dear Denis, thank you for your answer. Before I assimilate it, I have the same remark as for Felipe: do you mean $q^{1/2}$ instead of $q^{3/2}$? –  Joël Apr 10 '13 at 17:14
1  
Right, I saw the two variables but forgot the equation... It is a one-variable sum, and so Weil's techniques should give you $q^{1/2}$ "except for trivial reasons". And the multiplicative constant is now basically just the dimension of some $H^1_c$ space and can be estimated using the Euler-Poincaré formula. If you look at higher-dimensional tori, the ideas I mentioned become relevant again, since the restriction on the trace just drops the dimension by one. –  Denis Chaperon de Lauzières Apr 10 '13 at 19:27
add comment

For general tori this problem gets really weird. For instance, given any polynomial equation $f(x_1,..,x_n)$ defined over $\mathbb F_p$, I can encode the point-counting problem $|\{x_1,\dots,x_n \in \mathbb F_q | x_1,\dots x_n \neq 0, f(x_1,\dots,x_n)=0 \}|$ as one of these. This point-counting problem is usually at least as hard as counting solutions to $f$ which is a tough problem that we can usually only approximate the answer to.

Indeed, just consider a diagonal matrix where each monomial in $x_1,\dots,x_n$ occurs with multiplicity congruent modulo $p$ to its coefficient in $f$. As $x_1$ and $x_n$ vary, this defines a torus. Choose a trivial character of this torus, and you recover the point-counting problem.

Or choose a nontrivial, generic one - it shouldn't make the problem any easier.

Of course, in this specific case, Felipe Voloch has the solution.

share|improve this answer
add comment

$$\sum_{x+y+x^{-1}+y^{-1}=0} \chi_1(x)\chi_2(y) = \sum_{c} S(\chi_1,c)S(\chi_2,-c)$$ where $S(\chi,c) = \sum_{x+x^{-1}=c} \chi(x)$. [Stuff deleted]

Edit: As Denis points out in his answer, this is a bit silly, as the sum $S(\chi,c)$ has only two terms.

Edit 2: Again, as Denis points out, $x+y+x^{-1}+y^{-1}=0$ is a curve. However, it is a reducible curve with components $y=-x, y = -1/x$. So $\sum_{x+y+x^{-1}+y^{-1}=0} \chi_1(x)\chi_2(y) = \chi_2(-1)\sum_x \chi_1\chi_2(x) + \chi_1\chi_2^{-1}(x)$. If $\chi_1 = \chi_2$ is a character of order two, the sum is $2\chi_2(-1)q$. If $\chi_1 = \chi_2$ or $\chi_2^{-1}$ is not a character of order two, then the sum is $\chi_2(-1)q$ and otherwise the sum is $0$.

share|improve this answer
    
Dear Felipe, I am not following you. The set of $(x,y)$ such that $x+y+x^{-1}+y^{-1}=0$ has cardinality at most $2q$, because for a fixed $x$, $y$ satisfies a quadratic equation. Hence $|\sum_{x+y+x^{-1}+y^{-1}=0} \chi_1(x) \chi_2(y) | \leq |\sum_{x+y+x^{-1}+y^{-1}=0} \leq 2q$. Perhaps you mean $O(q^{1/2})$? –  Joël Apr 10 '13 at 17:12
    
My answer is messed up but I am leaving it up for now, to think about some more. –  Felipe Voloch Apr 10 '13 at 17:15
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.