Quadratic Gauss sum


Quadratic Gauss sum

:"For the general type of Gauss sums see" Gaussian period, Gauss sum

In mathematics, quadratic Gauss sums are certain sums over exponential functions with quadratic argument. They are named after Carl Friedrich Gauss, who studied them extensively.

Definition

Let "a","b","c" be natural numbers. The generalized Gauss' sum "G"("a","b","c") is defined by

:G(a,b,c)=sum_{n=0}^{c-1} e((a n^2+bn)/c),

where "e"("x") is the exponential function exp(2πi"x"). The classical Gauss sum is the sum G(a,c)=G(a,0,c).

Properties of Gauss sums

*The Gauss sum "G"("a","b","c") depends only on the residue class of "a","b" modulo "c".

*Gauss sums are multiplicative, i.e. given natural numbers "a", "b", "c" and "d" with gcd("c","d") =1 one has

:"G"("a","b","cd")="G"("ac","b","d")"G"("ad","b","c").

This is a direct consequence of the Chinese remainder theorem.

*One has "G"("a","b","c")="0" if gcd("a","c")>1 except if gcd("a","c") divides "b" in which case one has:G(a,b,c)= gcd(a,c) cdot Gleft(frac{a}{gcd(a,c)},frac{b}{gcd(a,c)},frac{c}{gcd(a,c)} ight)

Thus in the evaluation of quadratic Gauss sums one may always assume gcd("a","c")="1".

*Let "a","b" and "c" be integers with ac eq 0 and "ac+b" even. One has the following analogue of quadratic reciprocity for (even more general) Gauß sums:sum_{n=0}^{|c|-1} e^{pi i (a n^2+bn)/c} = |c/a|^{1/2} e^{pi i (|ac|-b^2)/(4ac)} sum_{n=0}^{|a|-1} e^{-pi i (c n^2+b n)/a}.

*Define varepsilon_m = egin{cases} 1 & mequiv 1mod 4 \ i & mequiv 3mod 4 end{cases} for every odd integer "m".

All Gauss sums with "b=0" and gcd("a","c")="1" are explicitely given by

:G(a,0,c) = egin{cases} 0 & cequiv 2mod 4 \ varepsilon_c sqrt{c} left(frac{a}{c} ight) & c ext{odd} \(1+i) varepsilon_a^{-1} sqrt{c} left(frac{c}{a} ight) & a ext{odd}, 4mid cend{cases}.

Here left(frac{a}{c} ight) is the Jacobi symbol. This is the famous formula of Carl Friedrich Gauß.

* For "b">"0" the Gauss sums can easily be computed by completing the square in most cases. This fails however in some cases (for example "c" even and "b" odd) which can be computed relatively easy by other means. For example if "c" is odd and gcd("a","c")="1" one has

:G(a,b,c) = varepsilon_c sqrt{c} cdot left(frac{a}{c} ight) e^{-2pi i psi(a) b^2/c}

where psi(a) is some number with 4psi(a)a equiv 1 ext{mod} c . As another example, if "4" divides "c" and "b" is odd and as always gcd("a","c")="1" then G("a","b","c")="0". This can, for example, be proven as follows: Because of the multiplicative property of Gauss sums we only have to show that G(a,b,2^n)=0 if "n">"1" and "a,b" are odd with gcd("a","c")=1. If "b" is odd then a n^2+bn is even for all 0leq n < c-1 . By Hensel's lemma, for every "q", the equation an^2+bn+q=0 has at most two solutions in mathbb{Z}/2^n mathbb{Z} . Because of a counting argument an^2+bn runs through all even residue classes modulo "c" exactly two times. The geometric sum formula then shows that G(a,b,2^n)=0 .

*If "c" is odd and squarefree and gcd("a","c")="1" then

:G(a,0,c) = sum_{n=0}^{c-1} left(frac{n}{c} ight) e^{2pi i a n/c}.

If "c" is not squarefree then the right side vanishes while the left side does not. Often the right sum is also called a quadratic Gauss sum.

*Another useful formula is

:"G"("n","pk")="pG"("n","p""k"-2)

if "k"&ge;2 and "p" is an odd prime number or if "k"&ge;4 and "p"=2.

ee also

*Kummer sum

References

*cite book | author = Ireland and Rosen | title = A Classical Introduction to Modern Number Theory | publisher = Springer-Verlag | year = 1990 | id=ISBN 0-387-97329-X
*cite book | author = Bruce C. Berndt, Ronald J. Evans and Kenneth S. Williams | title = Gauss and Jacobi Sums | publisher = Wiley and Sons, Inc. | year = 1998 | id=ISBN 0-471-12807-4
*cite book | author = Henryk Iwaniek, Emmanuel Kowalski | title = Analytic number theory | publisher = American Mathematical Society | year = 2004 | id=ISBN 0-8218-3633-1


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