PART TWO : FOLKLORE DRAFT
PAVLE MICHKO
Abstract. In this note1, I report all the basic algebra that can be used in the context of Weil sums.
Let L be a finite field of characteristic p and order q. One defines the Fourier coefficient of a polynomial mapping f ∈L[X] at a point a∈L by means of the canonical additive character of Lby :
fb(a) =X
x∈L
µ(f(x)−ax)
Remark 1. After several hesitation, I finally adopted the the minus sign in the definition of the Fourier coefficient.
Strictely speaking,fb(a) is the Fourier coefficient of the complex mapµ◦f at the additive characterµa:x7→µ(ax).
Contents
1. Weil sum 2
2. Characterization 2
3. Kernel trick 3
4. Symmetry 3
5. Waring formula 4
6. Kloosterman sum 4
7. Gauss sums 5
8. Divisibility 6
9. J-set connection 7
10. Rayleigh quotient 7
11. More orthogonality relation 8
12. Dedekind’s determinant 8
13. Moment of Weil sums 10
14. Asymptotic 11
15. Hyperplane section 11
16. Representation 12
17. sum over subfields 13
Date: start january 2013, last revision September 16, 2014.
1
18. Partial sums 13
19. Open problem 14
20. Multiplicities 14
21. Turyn’s result 15
22. vanishing conjecture 15
References 16
1. Weil sum
We are mainly interested by the values of these sums in the case of f is a monomial. The Fourier coefficients falls in Weil sums with binomial argument. Given a positive integer d, the Fourier coefficient of the power mappingxdis denoted
WL,(a, d) =X
x∈L
µ(xd−ax)
Remark 2. In the paper, I use d for any exponent, and s for invertible exponent. In this case, t denotes the inverse ofs moduloq−1:
st= 1 (modq−1).
In that case, I also use the notation f:x7→xs and g:x7→xt.
The Fourier coefficient at 0 is said in phase, the other are out phase.
Remark 3. Like for any permutation π, the phase Fourier coefficient of any power permutation is null,
π(0) =b X
x∈L
µ(π(x)) =X
x∈L
µ(x) = 0.
An exponentsis saidr-valued if the number of distinct out-phase Fourier coefficients isr.
2. Characterization
Several conjecture are proposed in the litterature concerning Weil sums with binomial argument. A possible argument resides in the following ele- mentary fact
fbb(a) =X
x∈L
µ(bxs+ax)
=X
x∈L
µ(xs+ab1−tx)
=f(abb 1−t
It appears that the spectrum ofbf is a permutation of those of f. Problem 1. Characterize the map f such that the spectrum of f is equal to the spectrum of bf for allb∈L×.
3. Kernel trick Assuming a r-valued spectrum,
r
Y
i=1
(f(a)b −Ai)×fb(a) = 0
whence (f−A1δ0)∗ · · · ∗(f−Arδ0) is in the kernel of the convolution byf i.e. the space generated by the µz withz∈Z ={a|fb(z) = 0}.
(1)
r
X
i=0
σi(A1, . . . , Ar)f[r−i]=X
z∈Z
xzµz
Using Fourier transformation (2)
r
X
i=0
σi(A1, . . . , Ar) ˆf(a)r−i =qX
z∈Z
xzδz(a) one sees that all the coefficient xa are equal toσr(A1, . . . , Ar)/q.
(3)
r
X
i=0
σi(A1, . . . , Ar)f[r−i]= σr(A1, . . . , Ar) q
X
z∈Z
µz
It is possible to show that whenr = 2 thenxa is an integer.
Remark 4. I do not know if thexa = 1qQr
i=1Ai is an integer.
4. Symmetry Now, we symply check our formulas.
bb
F(a) = X
x∈K
X
y∈K
F(y)¯µ(xy)¯µ(ax) =q X
a+y=0
F(y) =qF(−a)
In odd characteristic disymmetry appears ! Let us denote byZthe indicating function of Z.
First we write (3) as :
r
X
i=0
σi(A1, . . . , Ar)f[r−i](a) = σr(A1, . . . , Ar) q Z(−a)ˆ Remark 5. No trouble minus signing here !
And, we re-apply the Fourier transform again :
r
X
i=0
σi(A1, . . . , Ar) ˆf(a)r−i= σr(A1, . . . , Ar) q
ˆˆ Z(a)
=σr(A1, . . . , Ar)Z(a)
Now, iffb(a) = 0 then the left-hand side is equal to the productσr(A1, . . . , Ar) and thus :
∀a∈K, Z(a) =Z(a).
5. Waring formula
An+Bn=
n/2
X
j=0
(−1)j n n−j
n−j j
(AB)j(A+B)n−2j In general, denoting by
Sk=xk1+xk2+· · ·+xkn then
Sk= X
i1+2i2+···+nin=k
(−1)i2+i4+···×(i1+i2+· · ·+in−1)!k
i1!i2!· · ·in! σi11σi22· · ·σinn This fact can be used to proof that if f ∈ K[X] has degree d then there existsd−1 algebraic numbers ωi such that for any extension L of degreer of K, one has
S(f, L) =X
x∈L
µL(f(x)) =−[ω1r+ωr2+. . .+ωrd−1]
In fact these numbers are Weil numbers and they have absolute value√ q.
6. Kloosterman sum kloosL(a) = X
x∈L×
µ(1 x −ax) and one has the Weil bound
||kloosL(a)|| ≤2√ q.
It is now known that Kloosterman sum does not vanish in characteristic p >3. At the opposite, according to my knowledge, there is no simple proof of the fact that Kloosterman sums vanish when p= 2 or p= 3.
Letf(x) =xs be a pwer mapping and let h(x) =x−d X
a∈L
fb(ab)bh(a)∗=X
a
X
x
X
y
µ(xd−y−d)µ(ax−aby)
=qX
y
µ(bdyd−y−d)
=qkloosL(bd)
7. Gauss sums We denote by τK(χ) the Gauss sum
τK(χ) = X
x∈K×
µ(x)χ(x)
by Fourier inversion, for x∈K×: µ(x) = 1
q−1 X
χ
τK(χ) ¯χ(x) By a direct calculation
f(a) = 1 +b 1 (q−1)2
X
x
X
χ
τK(χ) ¯χ(xd)X
χ0
¯
τK(χ0)χ0(ax)
= 1 + 1 (q−1)
X
χ0=χd
τK(χ)¯τK(χ0)χ0(a)
= 1 + 1 (q−1)
X
χ
τK(χ)¯τK(χd)χd(a)
In the case of an invertible exponents, we continue : fb(a) = 1 + 1
(q−1) X
χ
τK(χ)¯τK(χs)χs(a)
= q
q−1+ 1 q−1
X
χ6=1
τK(χt)¯τK(χ)χ(a)
= 1
q−1 X
χ
F(χ)χ(a)
where we use the notation F(χ) =
(q, χ= 1;
τ(χt)¯τ(χ), else.
It is well known that flat spectrum implies orthogonality : X
a∈L×
fb(ab)fb(a)∗ = X
x,y∈L×
µ(xd−yd+a(bx−y)
=q X
bx=y
µ(xd−yd)
=q2δ1(b)
8. Divisibility
By Stickelgerber’scongruence theorem, for any prime ℘ above p in the appropriate field, we know the existence of a generatorω of Kd× such that :
τK(¯ωj)≡ (1−ζ)S(j)
R(j) mod ℘(p−1)S(j)+1
wherej =Pm−1
k=0 jkpk,S(j) =P
kjk and R(j) =Q
kjk!.
Let us debote byν the minimal value : ν = 1
p−1 min
0<j S(−jt) +S(j) and define theJ-set ofs :
(4) J(s) ={0< j|S(−jt) +S(j) =ν(p−1)}
Indeed,
fb(a) = 1 q−1
X
χ
F(χ)χ(a)
≡ 1 q−1
X
χ6=1
τ(χt)τ(χ)χ(a)
≡ 1 q−1
X
j>0
τ(ωjt)τ(¯ωj)ωj(a)
≡(1−ζ)νX
j∈J
ωj(a)
R(−jt)R(j) (mod (1−ζ)1+ν) It follows
VK(s) = 1
p−1 min
0<j<q−1S(−jt) +S(j) Remark 6. Using the relation
wp(s) + wp(−s) = [K :Fp](p−1)
one gets that the weight of sis smaller or equal to(m−v)(p−1) + 1. That is also the degree of the map x7→TK/Fp(xs).
Remark 7. If we write
fb(a) =
∞
X
i=ν
fi(a)pi the degree of fν is less or equal toν.
Lemma 1. Let K be a subfield of L.
VL(s)≤[L:K]×VK(s)
Proof. It is a direct consequence of Hasse-Davenport relation.
9. J-set connection LetJ be the J-set of s.
wp(−jt) + wp(j) = wp(−jt) + wp(−(−jt)s) It follows that
J(s)3j7→ −jt∈J(t) This is a manifestation of the Rule
fb(a) =X
x∈K
µ(xs+ax) = X
x∈K
µ(x+axt) =bg(a−s) Since
X
j∈J(s)
ωj(a)
R(−jt)R(j) ≡ X
k∈J(t)
ωk(a−s)
R(−ks)R(k) ≡ X
k∈J(t)
ω−ks(a) R(−ks)R(k)
The independance of characters concludes.
−sJ(t) =J(s), −tJ(s) =J(t).
10. Rayleigh quotient
The eigenvalues of the Fourier operator f 7→ fbare ±√
q. The Rayleigh quotient of a vector x satisfy
−√
q≤R(F, x) = Ax.x x.x ≤√
q
In particular, the Rayleigh quotient of a mappingf is :
−q√
q ≤fb×f(a) =X
a∈K
fb(a)f(a)≤q√ q
We are interested by the
fb∗f(t) = X
a+b=t
f(a)fb (a)
=X
a
f(−a)f(a)µ(at)b
Problem 2. What is the Rayleigh quotient of a power mapping ?
Forb6= 0, by a direct calculation X
a∈L
fb(ab)¯µ(as) = 1 q−1
X
a6=0
X
χ
F(χ)χ(ab)¯µ(as)
= 1
q−1 X
χ
F(χ)χ(b)X
a6=0
χ(a)¯µ(as)
= 1
q−1 X
χ
F(χ)χ(b)τ(χt)χt(−1)
= −q
q−1 + 1 q−1
X
χ6=1
τ( ¯χ)τ(χt)2χ(−b)
The above relation can be interpreted like next. First of all the fb are orthogonal :
fb.fc=qδb(c).
The decomposition offbin the basis gb is more simple fb(a) =X
b
µ(bt)gb(a) as we can check very easely
X
b
µ(bt+ba−s) =X
b
µ(a.b/a+ (b/a)s) =f(a)b 11. More orthogonality relation Letf be a power permutation.
X
a∈L
fb(a+t)fb(a)∗ =qδ0(t).
All these orthogonality relation could be make hard the existence of not symmetic small spectrum.
12. Dedekind’s determinant
The convolutional endomorphism diagonalizes in the basis of addidive characters of L, the eigenvalues are precisely Fourier coefficients
µa∗F(z) = X
x+y=z
F(x)µ(ay)
=X
x
F(x)µ(a(z−x)
=Fb(a)µa(z) The convolution byF on the Dirac basis
F∗δa(b) = X
x+y=a
F(x)δa(y)
=F(b−a) We recover Dedekind’s determinant
(5) Y
a∈L
Fb(a) = det[F(b−a))]a,b∈L
Remark 8. Let κ be an arbitrary complex number. Using the map F +κ, we get
(6) (Fb(0) +qκ) Y
a∈L×
Fb(a) = det[µ(F(b−a)) +κ]a,b∈L For example, withκ=−1 A direct calculation gives
(fb(0)−q) Y
a∈L×
fb(a) =X
σ
sgn(σ)Y
a∈L
µ(f(σ(a)−a))−1
=X
σ∈Φ
sgn(σ)P(σ)
where
Φ ={σ| ∀x∈L,traceL f(σ(x)−x)
6= 0} and P(σ) =Y
a∈L
µ(f(σ(a)−a))−1 .
Remark 9. If f(0) = 0then a permutation of Φ as no fixed point.
Remark 10. In the case p = 2, with κ = −1 and f(x) ∈ L[X]. A direct calculation gives
(fb(0)−q) Y
a∈L×
fb(a) =X
σ
sgn(σ)Y
a∈L
µ(f(σ(a)−a))−1
= 2qX
σ∈Φ
sgn(σ)
Remark 11. One can recover a classical divisibility result using this deter- minantal approach. Again, with
Φ ={σ | ∀x∈L,traceL f(σ(x) +x) 6= 0}, we get
(fb(0)−q) Y
a∈L×
f(a) =b X
σ∈Φ
sgn(σ)P(σ)
where thep-adic valuation of theP(σ) is greater than q/(p−1).
Considering (6) as a polynomial in κ, by identification : qD(f) =−X
y
X
π
sgn(π)µ(X
x6=y
f(π(x)−x))
As in [13], one can isolate the phase using the space orthogonal to the trivial character. It is generated by ∆a=δa−δ0. a∈L×.
F∗∆a=F∗δa−F∗δ0
=X
b
F(b−a)δb−X
b
F(b−0)δb
=X
b6=0
F(b−a)−F(b)
∆b
In this space
(7) Y
a∈L×
Fb(a) = det[F(b−a)−F(a)]a,b∈L×
13. Moment of Weil sums We introduce the moment of Weil sums
Sr= X
a∈K
fb(a)r The first values are well known
S1=q, S2 =q2
Assuming a three valued spectrum, we get a recursion formula : Sk= (A+B)Sk−1−AB Sk−2
S3 = (A+B)q2−ABq S4 = (A+B)S3−ABq2
The sums Sk are connected by convolution to the character sums :
Sk=q X
x1+x2+···+xk−1+z=0
µ(xs1+xs2+· · ·+xsk−1+zs)
=Sk−1+qX
z6=0
X
x1+x2+···+xk−1+z=0
µ(xs1+xs2+· · ·+xsk−1+zs)
=Sk−1+q2Nk−1−qk−1
whereNk is the number of solutions of
{ x1 + x2 + · · · + xk + 1 = 0 xs1 + xs2 + · · · + xsk + 1 = 0 In particular, using Daniel’s notationV :=N2 :
S3 = (A+B)q2−ABq S3 =S2+q2N2−q2
=q2V
N2 =A+B−AB
q =V =⇒ [K:Fp] +(p)≤a+b where(p) = 1 ifp= 2 (immediate) or 3 (more tricky).
14. Asymptotic
LetAbe the maximal absolute value in the spectrum, and let NA0 be the number of occurences. Recall the relation (??)
Pk =Pk−1+q2Nk−1−qk−1(q2Nk−1−qk−1)∼AkNA0 one should compare to Deligne bound.
15. Hyperplane section
Letλ1,λ2,. . . ,λn benelements ofL×. By convolution, or a direct calcu- lation
X
a∈L n
Y
i=1
fcλi(a) = X
x1,x2,...
µ(X
i
xsi −aX
i
λixi)
=q X
λ1x1+···+λnxn=0
µ(xs1+· · ·+xsn)
Since s is invertible, the last character sum does not depend on µ, its valuesSis a non trivial contribution in the character counting of the number of solutionsN the hyperplane section of the Fermat hypersurface
0 = xs1 + xs2 + . . . + xsn 0 = λ1x1 + λ2x2 + . . . + λ1xn. whenceqN = (q−1)S+qn−1 and
(8) q2N = (q−1)X
a∈L n
Y
i=1
fcλi(a) +qn
Remark 12. Using (8) with n = q −1 and all the λi are distincts, the existtence of zero outphase Fourier coefficient is equivalent to say that N = qn−2.
16. Representation To anyf, we introduce the notation
fbb(a) = X
x∈K
µ bf(x)−ax and the matrix
R(f) := 1
q fbb(a)
b,a
It is easy to check that
R(f ◦g) =R(f)×R(g) In particular, we have a representation of groups, and
χ(f) = 1 q
X
a∈L
fba(a) = 1 q
X
a∈L
fb(a1−t) = fix(f) Note that
X
a∈L
|fba(a)|2 = X
a,x,y
µa(f(x)−f(y) +x−y)
=q ]{(x, y)|f(x)−f(y) =x−y}
=q(q+N(q−1)) In particular,
sup
a∈L×
|fba|2 ≥N q
Letℵ(t) the number of preimages oftby x7→xs−x.
ℵ(t) = 1 q
X
a∈L
fba(a)¯µ(at) qX
t∈L
ℵ(t)2 =X
a∈L
fba(a)2
Remark 13. The set of bijections having an integral spectrum is a group.
LetG the cyclic group generated byf.
Note the pseudo-cyclic structure ofM(f) =qR(f) :
q 0 0 . . . 0
0 fcγ0(γ0) fcγ0(γ1) . . . fcγ0(γq−1) 0 fcγ1(γ0) fcγ1(γ1) . . . fcγ1(γq−1)
... ... ... ... ... 0 f[γq−1(γ0) f[γq−1(γ1) . . . f[γq−1(γq−1) One has the relation
M(f)X
g∈G
M(g) =X
g∈G
M(g) =: Σ(f) Remark 14. What is the ring generated by the M(g) ?
17. sum over subfields LetF be a subfield ofL,
X
a⊥F
fbb(a) =|F|X
x∈F
µb(xd)
=qδF⊥(b) LetG be a subgroup ofL×,
X
x∈G
fb(ax) = |G|
q−1 X
χ⊥G
F(χ) ¯χ(a)
In particular, if G=K× then the Gauss sums are equal to √
q and we get X
x∈K
fb(ax) = q
√q+ 1 X
χ∈K×
χ(a)
=qδK×(a)
18. Partial sums
Since s ≡ 1(p−1), the distribution of x 7→ TK/Fp(xs) is well balanced over any subspace. Writing
nr(S) =]{x∈S|TK/Fp(xs) =r}, n0(S) + (p−1)n1(S) =pk
X
x∈S
µ(xs) =
p−1
X
i=0
ni(S)ζi =n0(S)−n1(S)
=pn1(S)
= 1
pm−k X
s⊥S
fb(s) 19. Open problem
From now and on, sis a three valued invertible exponent : it takes three values 0, A, and B over a finite field L of order q = pm, p prime. Recall thats is congruent to 1 modulo (p−1).
A=paα, B =pbβ, A−B =pcγ withα,β and γ coprime withp.
fbb(a) = X
x∈K
µ(bf(x)−ax)
Let NA the multiplicity of A, NB those of B in the spectrum of f. We solve the system
(9) { NAA + NBB = q
NAA2 + NBB2 = q2
(10) NA=q(B−q)/A(B−A) NB=q(q−A)/B(B−A) and
NA+NB= q(B2−A2) +q2(A−B)
AB(B−A) =qA+B−q AB
20. Multiplicities
Let us denotes by Mr(s) the number of v having r pre-images by f.
Actually, we don’t know if it is possible to find some exponentsssuch that M2 = 0 !
I run a program to find all the exponent ssatisfying the condition s≡1 (modp−1), ]{r|Mr(s)>0} ≤3, q ≤232.
all of them have the same shape M2(s) > 0. Of course, for such s the spectrum cannot be three-valued.
Actually, one can detailN(u, v) whenv6=us. Let us denote by n(v) the number ofx,y inK× such thatxs+ys=v and x+y =u.
We use
n(u, v) = 1 (q−1)2
X
x+y=u
X
X+Y=v
X
χ,ψ
χ(xs/X)ψ(ys/Y)
= 1
(q−1)2 X
χ,ψ
X
x+y=u
X
X+Y=v
χ(xs)ψ(ys) ¯χ(X) ¯ψ(Y)
= 1
(q−1)2 X
χ,ψ
J(χs, ψs)J( ¯χ,ψ)χψ(u¯ −s)χψ(v)
Problem 3. Letns(v)be the number of preimages ofv byx7→xs+ (1−x)s. We are interested by the exponents s such that ns(v) takes only two values over K\ {0,1}.
21. Turyn’s result
I recall here charcater sums divisibility results that can be useful in this context.
Proposition 1. Let S be a cyclic group of order N. Let f be an integral mapping over S. Letχ be acharacter of order b. If fb(χ)6= 0 is divisible by an positive integer m then
m≤ 2rad(b)Nkfk∞
b .
Moreover, if f is positive
m≤ 2rad(b)Nkfk∞
2b .
where rad(b) denotes the number of prime divisors ofb.
22. vanishing conjecture A few words, on the vanishing conjecture :
Conjecture 1 (Helleseth). Let L be a field of cardinal q > 2. If f is a power permutation of L of exponents≡1 mod (p−1)then it exists a6= 0 such that f(a) = 0.b
The congruence modulop−1 is essential. It ensures that the Weil sums are rational integers. Moreover, one knows bu [11] that the Kloostermann sums do not vanish when p >3. Aubry and Langevin got a very small step in this direction
Theorem 1. Let Lbe a field of cardinalq >2. Iff is a power permutation of L of exponent s≡1 mod (p−1)then it exists a6= 0 such that fb(a)≡0 (mod 3).
Observing this statement I recently propose the following optimist version :
Conjecture 2. Let L be a field of cardinal q > 2. Let s be coprime with q−1, t the inverse ofs moduloq−1. There exists a6= 0 such that
X
x∈L
µFp traceL(xs)t+ traceL(ax)
= 0.
This strange idea come naturally by introducing the new additive law x⊕y= (xs+ys)t
Hence, the exponential sums of argument traceL(xs)t−traceL(ax) is nothing but the scalar product of two additive characters of a “bifield”. Finaly, it false, the smaller counter exemple has parmeters : p= 5,s= 3,q= 53, and the distribution of the sums are:
-30 -20 -15 -10 -5 5 10 15 20
1 3 10 21 21 28 21 10 9
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