The functional equation for exponential sums

I am continuing my class on exponential sums (and the current notes are still available on the web page). Before spring break, and a dive into Stepanov’s method to prove the Riemann Hypothesis for one-variable sums, the last important thing to do was to set up the L-functions associated to those sums, so that the fact that it is a polynomial (in q-s, if seen as a complex-analytic function) can be used to bootstrap Stepanov’s upper bound of the right-order of magnitude to the “right” upper bound. (Roughly, one can show using this method results like

|S|=|\sum_{x\in\mathbf{F}_{p^n}}{\chi(N_{\mathbf{F}_{p^n}/\mathbf{F}_p}(x^3+ax+b))|\leq Cp^{n/2},

for some large constant C, assuming the cubic polynomial X3+aX+n is squarefree, and knowing a priori that there exists a complex number α such that

S=\alpha+q\alpha^{-1}

is enough to show that C can be replaced by 2, which is then optimal).

There are a number of proofs of the functional equation that I know, but I ended up using an ad-hoc argument which has its charm. This allowed me to avoid mentioning either adèles (and the Poisson summation formula thereupon) or the Riemann-Roch theorem, in keeping with my focus this semester on elementary methods. Once more, I seriously doubt that the technique is new, but I noticed that Schmidt’s book (which I use as a guiding text for this part of my lectures) does not prove the functional equation for exponential sums in any great generality. He just computes the leading term of the polynomial in some cases. In fact, part of the reason I wanted a different argument is that, by analogy with classical Dirichlet characters, one knows (or expects) this leading term to be more or less a Gauss sum, and although Schmidt’s argument immediately reminded me of the computation of the modulus (square) of a Gauss sum, he does not interpret it explicitly in this way.

More or less, the argument I used is as follows: one has a non-trivial character

\eta\,:\, \mathbf{F}_q[X]\rightarrow \mathbf{C}

of a polynomial ring over a finite field, analogous to a Dirichlet character, defined modulo some principal ideal

I=g\mathbf{F}_q[X],\quad\quad\quad \deg(g)=d\geq 1,

(meaning that η(f) only depends on the class of f modulo g), which is primitive. Then the L-function is defined by the formal power series

L(\eta,T)=\sum_{f\text{ monic}}{\eta(f)T^{\deg(f)}},

and a very simple orthogonality argument (based on the non-triviality of the character) shows that in fact

L(\eta,T)=1+c_1(\eta)T+\cdots + c_{d-1}(\eta)T^{d-1}

is a polynomial of degree at most d-1. Then one “computes” the coefficients by

c_j(\eta)=\quad\sum_{f\text{ monic, } \deg(f)=j}{\ \ \eta(f)},

and the question is to simplify this coefficient — in fact, the functional equation means that it is related to

c_{d-1-j}(\bar{\eta}),

the coefficient of complementary degree for the dual character. The idea is then to express the conditions

f\text{ monic},\quad  \deg(f)=j\leq d-1,

which are conditions on the coefficients of the polynomial f, using additive characters of the finite field to create a sum over all f in the finite ring

R=\mathbf{F}_q[X]/(g),

twisted by the various additive characters. The latter involve

d-1-j

variables (corresponding to the number of conditions to detect among polynomials of degree at most d-1), and those can be parameterized by polynomials of degree d-1-j over the finite field. Such a mixture of additive and multiplicative characters is of course rather conducive to the blooming of Gauss sums, and indeed, it turns out to lead, without much difficulty, to the result… (This is Proposition 4.15, page 45, of my notes).

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Kowalski

I am a professor of mathematics at ETH Zürich since 2008.

2 thoughts on “The functional equation for exponential sums”

  1. Dear Emmanuel,
    the bootstraping argument in your notes (lemma 4.18)
    completely makes my day worth it !
    Thanks for it.

    Pierre

  2. Thanks! I should of course say that this nice trick is very old. It is in Schmidt’s book (Lemma 6A p. 57), for instance, and although he doesn’t add any comments, I presume it was known before.

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