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Upper bound for an exponential sum involving characters of a finite field

MathOverflow Asked by nahila on December 14, 2020

Let $q = p^n $ be a prime power, $alphainmathbb{F}_{q} $
a primitive element of the finite field $mathbb{F}_q$ and denote by $chi $ a non-trivial additive character of $mathbb{F}_{q} $. Set
$omega = explbrace { ifrac{2pi} {q – 1} } rbrace $,
I am looking for an upperbound on the following sum
begin{equation}
leftvert sum _{k=0} ^{q – 2} omega ^{k^2} chi (alpha ^k ) rightvert.
end{equation}

If we denote by $psi _c $ the multiplicative character of
$mathbb{F}^* _{q} = mathbb{F}_{q}setminuslbrace 0rbrace =lbrace alpha ^0 ,cdots,alpha^{q-2}rbrace$ corresponding to $cinmathbb{F}_q ^* $
then the sum can be written as
begin{equation}
leftvert sum _{cinmathbb{F}_q^*} psi _c (c) chi (c) rightvert.
end{equation}

The second sum looks much like a Gaussian sum over finite fields, however in this one the multiplicative character changes as well.

ps: The multiplicative character is given by
$psi _{alpha ^l} (alpha ^k ) = omega ^{lk} = explbrace ifrac{2pi}{q-1} lk rbrace $ .
The additive character corresponding to an element $ainmathbb{F}_{p^n} $ is given by
$chi _a (b) = explbrace ifrac{2pi}{p} tr(ab) rbrace $ for all $binmathbb{F}_{p^n} $,
where the trace $tr : mathbb{F}_{p^n} rightarrow mathbb{F}_p $ is defined by
begin{equation}
tr(a) = a+a^p + cdots + a^{p^{n-1}}.
end{equation}

One Answer

I am going to assume that by an additive character you mean

an irreducible representation $chi_alpha : mathbb{F}^n_q longrightarrow mathbb{C}$, i.e. a group homomorphism from the additive group $(mathbb{F}^n_q ,+)$ to the multiplicative group $(mathbb{C},*)$

which we can prove must all take the form begin{equation}chi_alpha : beta mapsto expleft( {frac{2pi i leftlangle alpha ,beta rightrangle }{p }} right)end{equation} where $ leftlangle alpha ,beta rightrangle = sum_i alpha_i beta_i $, see chapter 4 of Tao for a proof of some of these statements and see ch.2 of Serre or ch.2 of Fulton & Harris for a general (non-abelian) overview of the representation theory perspective on characters. The point is the following

If we let begin{equation} f(x) = begin{cases} q psi_x(x) & text{if } x neq 0 \ 0 & text{if } x = 0 \ end{cases} end{equation} then the sum you are considering is equal to the Fourier transform of $f$ i.e. begin{equation} hat{f}(alpha) = frac{1}{q} sum_{c in mathbb{F} _q } f(c) chi_alpha(c) = sum_{c in mathbb{F} _q^* } psi_c (c) chi_alpha(c) end{equation} see definition 4.6 in Tao.

We apply the Hausdorff-Young inequality theorem 4.8 in Tao to get that begin{equation} left(sum_{alpha in mathbb{F} _q }left| hat f(alpha)right|^{p'} right)^{frac{1}{p'}} leq left(sum_{alpha in mathbb{F} _q } |f(alpha)|^pright)^{frac{1}{p}} = qleft( sum_{c in mathbb{F} _q^* } |psi_c (c) |^pright)^{frac{1}{p}} end{equation} where the LHS is the $l^{q}$-norm, the RHS is the $l^p$-norm, and $p$ satisfies the following $p^{-1} +q^{-1} = 1 land 1 leq pleq 2$. Plugging in $p = 2$ we get that

begin{equation} sum_{alpha in mathbb{F} _q }left| hat f(alpha)right|^{2} leq qsum_{c in mathbb{F} _q^* } |psi_c (c) |^2 end{equation} which is equivalent to saying that begin{equation} mathbb{Var}[hat f] = frac{1}{q}sum_{alpha in mathbb{F} _q }left| hat f(alpha)right|^{2} leq sum_{c in mathbb{F} _q^* } |psi_c (c) |^2leq q-1. end{equation}

Finally, if you can prove that at least $n$ many $alpha$ give a value $ | hat f(a)| geq sqrt b$ then you get that begin{equation} nb +sup_{alpha in mathbb{F} _q }left| hat f(a)right|^2 leq sum_{alpha in S}left| hat f(alpha)right|^{2} + sup_{alpha in mathbb{F} _q }left| hat f(a)right|^2 leq sum_{alpha in mathbb{F} _q }left| hat f(alpha)right|^{2} leq q(q-1) end{equation}

which gives you that the maximum value is at most

begin{equation} sup_{alpha in mathbb{F} _q }left|sum_{c in mathbb{F} _q^* } psi_c (c) chi_alpha(c) right| = sup_{alpha in mathbb{F} _q }left| hat f(a)right| leq sqrt{q(q-1)-nb} end{equation}

Essentially we reduced the problem of finding an upper bound to that of finding a lower bound.

Answered by Pedro Juan Soto on December 14, 2020

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