Auxiliary results

Equation (3.1.6) of Section 3.1.2 will be our main tool for the evaluation of the scalar product of theta functions. In this section, we make this formula easier to evaluate by giving some simple lemmas that will be used latter on. Since we have attached a theta function to each essential cusp of a groupΓ⊂Γ₂, the next step will be to determine allΓ^∗₀ π^h

-inequivalent essential cusps. We achieve this in Lemma 3.5. In Lemma 3.6, we work out the summation condition appearing in the formula (3.1.6), then in Lemma 3.7 we deal with some Gauß sums similar to those appearing in the inner sum of (3.1.6) and finally, in Lemma 3.8 we evaluate a Dirichlet series.

For06`6h, let us defineP<sub>`</sub>as

P_`=







1 if`= 0,

R/π^h−`R×

if16l6h−1,

1 if`=h.

(3.2.1) Since we assume π 6= λ, we shall assume that the elements of P_` are chosen congruent to 1 modulo3. Besides, for a givenm ∈ Rspecified in the context, we shall use the notation xfor a representative of the inverse modulom of an element x coprime with m. Finally, we define Tm = (_m^{1 0}₁)andT^m= (¹_{0 1}^m).

Lemma 3.5. A set of essential cusps (not necessarily inequivalent) ofΓ^∗₀ π^h

is given by 1

α π<sup>`</sup>ν : α∈ {±1,3},06`6h, ν∈ P_`

∪ 1

π<sup>`</sup>ν −1 : 06`6h, ν∈ P_`

. Proof. Let P = ^a_c be a cusp of Γ^∗₀ π^h

, with a, c ∈ Z[ω], gcd(a, c) = 1. Since _3π¹h⁰1

∈ Γ^∗₀ π^h

, the cuspP is equivalent to the cusp _3π¹h⁰1

(P) = _3πh^aa+c, and this allow us to assume that ord_π(c)6h. According to this, we definej=ord_π(c),06j 6h. First of all, we determine the three types of cusps ofΓ^∗₀ π^h

, and then rule out some of them which are not essential.

Case 1: π|c and λ6 |a. Multiplying by some unit ofZ[ω]if necessary, we assume that a ≡ 1 (modλ²). Letb, d, l such thatad−bcλ²π^l = 1, for a parameterl > 0to be chosen latter on.

Then, we remark thatd≡1 (mod 3)and that γ :=

a bλ²π^l

c d

∈SL₂(R).

Define

3.2 Relations between theta functions

- a representativeαof the class ofcmoduloλ²such thatαis coprime toπ,

- a representativeνof the class ofαcπ^−jamoduloπ^h−jsuch thatν ≡1 (mod 3).

Then

γ T_−απjν =

a−αbλ²π^l+jν bλ²π^l c−αdπ^jν d

≡ 1 0

0 1

(mod 3),

and c − αdπ^jν ≡ c −cad (modπ^h); but c − cad = c(1 −ad) = −bc²λ²π^l, and thus ordπ(c−cad) = 2j+l. We now choosel > h−2j and conclude that γ T_−απjν ∈ Γ^∗₀ π^h

. Therefore, we have thatP = ^a_c = γ(∞) = γ T_−απ^j_ν(T_απ^j_ν(∞))is equivalent toT_απ^j_ν(∞) = (α π^jν)⁻¹.

Case 2: π|cetλ|a. Multiplying by some unit if necessary, we assume thatc≡1 (mod λ²). Let b, dbe such thatad−bc= 1andd≡1 (mod 3). Then we remark thata−b≡1 (mod 3)and that

γ :=

a b c d

∈SL₂(R).

Defineνto be a representative of the class ofc π^−jdmoduloπ^h−j, then γ T_−π^j_ν =

a−bπ^jν b c−dπ^jν d

≡ 1 b

0 1

(mod 3),

andc−dπ^jν ≡ 0 (modπ^h). Letu be the unit ofR representative of the class ofbmodulo3;

thenγ T_−πjνT^−u ∈Γ^∗₀ π^h

and thusP =γ(∞) =γ T_−πjνT^−u(T^uT_πjν(∞))is equivalent to (π^jν)⁻¹+u. Finally we remark thatb ≡ a−1 (mod 3), thus ifais divisible byλ², we have u=−1, and otherwise (sinceais at least divisible byλ) we haveu=−ωor=−ω².

Case 3:π6 |candλ6 |a. As before, assumea≡1 (modλ²). Letαbe a representative ofcmodulo 3such thatα is coprime withπ. Considerb, dsuch thatad−bcλ² = 1and such thatd ≡ αc (modπ^h). Thend≡1 (mod 3),γ := ^{a bλ}_{c d}²

∈SL₂(R)and γ T−α=

a−bαλ² bλ² c−dα d

∈Γ₁, andc−αd≡0 (mod π^h). Thusγ T−α ∈Γ^∗₀ π^h

andP =γ(∞) =γ T−α(Tα(∞))is equiva-lent to _α¹.

Case 4:π6 |candλ|a. Assumec≡ −1 (mod λ²). Letαbe a representative of−amodulo3, and considerb, dsuch thatadπ^hλ²−bc= 1. Thenb≡1 (mod 3),γ := ^a_{c dπ}h^bλ²

∈SL2(R)and γS =

b −a dπ^hλ² −c

≡ 1 α

0 1

(mod 3).

Finally,P =γST^−α(T^αS(∞))is equivalent toα.

Define the setC = {±1,±ω,±ω²,±λ,3} of representatives modulo3. From the previous discussion, we conclude that a cuspP belongs to one of the following types:

type 1, ifP ∼ _απ¹<sub>`ν</sub>, with06`6h−1,ν ∈ P_{`</sub>, andα∈ C, (case 1 and case 3), type 2, ifP ∼u+<sub>π</sub><sup>1</sup>`ν, with06`6h−1,ν∈ P<sub>`}andu∈ {−1,−ω,−ω²}(case 2), 92

type 3, ifP ∼α, withα∈ {±λ,3}(case 4).

We now have to specify which of these cusps are essential. Recall that σ⁻¹(∞) is an essen-tial cusp ifκ(γσ) = 1, for all γσ ∈ Γσ; we saw in Lemma 3.1 that such γσ can be written as By periodicity we have

c²λσ (see Theorem 2.1 of Chapter 2), we obtain

c²λσ which, by reciprocity law, amounts to

λ 1+3π^jνr(1+π^jνu)

3 = 1, and by the complementary reci-procity law, we need1 +u=x1+ωx2, withx1 ≡0 (mod 3); among the set−1,−ω,−ω², only u=−1satisfies this condition.

Ifσ⁻¹(∞)is of type 3, i.e.σ⁻¹ = ^α_{1 0}⁻¹

To finish the proof of Lemma 3.5, we merely have to notice that the only essential cusp of the third type, namelyP = 3, is equivalent to the cuspP = 0, which correspond to the cusp of second type(π<sup>`</sup>ν)<sup>−1</sup>−1for`= 0.

3.2 Relations between theta functions

After having determined the set of essential cusps of the groupΓ^∗₀ π^h

= Γ1∩Γ0 π^h , and before coming back to our initial problem, i.e. to work out the scalar product of two theta functions, let us define some new notations. To any parameterξ = (`, ν) with0 6 ` 6 handν ∈ P_` (as defined in (3.2.1)), correspond four essential cusps as defined in Lemma 3.6, and associated to them four theta functions, namely

θαξ :=θσ, where σ=

1 0 απ^`ν 1

, forα∈ {±1,3} (3.2.3) and

θ_ξ⁰ :=θσ, where σ=

1 −1 π^`ν 1

1 0

απ^`ν 1

. (3.2.4)

Therefore, a generating system of the spaceΘ π^h

is given by V

π^h

:=

θ_ξ, θ−ξ, θ_3ξ, θ⁰_ξ : ξ= (`, ν)with06`6handν ∈ P_` .

The method for extracting a basis from this set is to come to simple relations between theta func-tions, using the formula (3.1.6) of Section 3.1.2. This will be achieved in various steps; at the present, we give some useful results that we shall apply in the next section. We have seen that the latticeΛ_σassociated to the cuspσ⁻¹(∞)only depends on ord_π(c(σ)). In particular, it is the same for all four theta functions attached to some couple(`, ν) in (3.2.3) and (3.2.4). More precisely, we showed in Lemma 3.1 of Section 3.1 thatΛ<sub>`</sub>= 3πMax(0,h−2`)R.

Lemma 3.6. Letα1, α2 ∈ {±1,3},`1, `2 ∈Zsuch that0 6`1, `2 6h, andν1 ∈ P_{`1</sub>, ν2 ∈ P<sub>`2}. Assume that`1 6`2. Then

(i) hθ_α

1π<sup>`1</sup>ν1, θ<sub>α2π</sub>`2ν2i = Res_s=4/3 X⁰

c≡α₂−α₁(3)

N(c)^−s X(∗)

d(Λ`2c) d≡1 (3)

c d

3

! ,

(ii) hθ_π`1ν1, θ<sup>0</sup><sub>π</sub>`1ν1i = Res_s=4/3 X⁰

c≡1 (3)

N(c)^−s X(∗)

d(Λ`1c) d≡−1 (3)

c d

3

! ,

where(∗)means

c6≡α₂d,

if`1=`2 = 0, and otherwise, i.e. (`1`2)6= (0,0),(∗)means the conditions

ordπ(c)>`<sub>1</sub> and dα<sub>2</sub>ν<sub>2</sub>π<sup>`2−`1</sup> −dα<sub>1</sub>ν<sub>1</sub> ≡cπ<sup>−`1</sup> (modπ^{min(`1,h−`1)}).

Moreover, the first condition is an equality, i.e.ordπ(c) =`<sub>1</sub>, in the case where`₁< `₂. Proof. Letσ⁻¹(∞)andτ⁻¹(∞)be two essential cusps. We start by proving that

κ σ^{−1 a b}_{c d} τ

= c

d

3. 94

The computation of the Kubota symbol is easy, since one can assume (see Lemma 3.5) thatσ, τ ∈

This proves the result about the Kubota symbol. We now have to interpret the summation condition in (3.1.6).

Letc6= 0, and(d, c) = 1. Writeβ_i=α_iν_i, fori= 1,2and look at the formula (3.1.6). In (i) the condition on(c, d)is

1 0

3.2 Relations between theta functions

Similarly, in (ii)the condition is 1 0

β1π^`1 1

a b c d

1 0

−β₁π^`1 1

1 1 0 1

∈Γ^∗₀

π^h

, i.e.

a−bβ1π^{`1</sup> b+a−bβ1π<sup>`1}

c+aβ₁π^{`1</sup>−dβ<sub>1</sub>π<sup>`1}−bβ₁²π^{2`1</sup> d+bβ<sub>1</sub>π<sup>`1} +c+aβ₁π^{`1</sup> −dβ<sub>1</sub>π<sup>`1} −bβ₁²π^2`1

∈Γ^∗₀ π^h

, i.e.











a b c d

≡

0 −1 1 −1

(mod 3) c+aβ1π^{`1</sup> −dβ1π<sup>`1} −bβ₁²π^2`1 ≡0 (modπ^h)

.

Thus(c, d)has to satisfyc ≡ α₂−α₁ (mod 3)andd ≡ 1 (mod 3), in the case (i), orc ≡ 1 (mod 3)and d ≡ −1 (mod 3), in the case (ii). Since gcd(π,3) = 1, the chinese remainder theorem asserts that we only have to study the congruence condition moduloπ^h. Leta0, b0such thata₀d−b₀c= 1. Then(a, b)is of the form(a₀+kc, b₀+kd), and we ask if there existsksuch that

c+a0β1π^{`1</sup>−dβ2π<sup>`2}−b0β1β2π^`1+`2 +kβ1π^{`1</sup>(c−dβ2π<sup>`2})≡0 (modπ^h). (3.2.5) This is the equation in case(ii), but forβ₁=β₂and`<sub>1</sub>=`₂, this corresponds to case(i).

If`<sub>1</sub>=`₂= 0, this equation has a solution if and only ifc−dβ₂6≡0 (modπ), i.e.c6≡α₂d (modπ), becauseP_{`1</sub> =P<sub>`2} ={1}).

If(`<sub>1</sub>, `₂)6= (0,0), we see from (3.2.5) thatc≡0 (mod π^`1), and (3.2.5) becomes

cπ^{−`1</sup> +a<sub>0</sub>β<sub>1</sub>−dβ<sub>2</sub>π<sup>`2−`1</sup>−b<sub>0</sub>β<sub>1</sub>β<sub>2</sub>π<sup>`2}+kβ₁π^{`1</sup>(cπ<sup>−`1} −dβ₂π^`2−`1)≡0 (modπ^h−`1).

(3.2.6) We remark that if `1 < `2, thencπ^{−`1</sup>(1 +kβ1π<sup>`1}) +a0β1 ≡ 0 (mod π), what shows that ord_π(c) =`₁, since otherwiseπ|a₀, what contradictsa₀d−b₀c= 1.

Now (3.2.6) impliescπ^{−`1</sup> +a0β1−dβ2π<sup>`2−`1</sup> ≡0 (mod π<sup>min(`1,h−`1)</sup>), and we claim that this is sufficient to prove the existence ofk. Actually, if`₁< `<sub>2</sub>, thenπdividesdβ<sub>2</sub>π<sup>`2−`1</sup> but not cπ<sup>−`1}, thuscπ^{−`1</sup> −dβ<sub>2</sub>π<sup>`2−`1</sup> is invertible, and if`₁ =`<sub>2</sub>>0, thenπ|c, thusa<sub>0</sub>d≡1 (mod π) anda0β1 6≡0 (mod π), and in view ofcπ<sup>−`1} +a0β1−dβ2π^`2−`1 ≡0 (mod π^{min(`1,h−`1)})we also getcπ^{−`1</sup> −dβ<sub>2</sub>π<sup>`2−`1} 6≡0 (mod π).

We are now interested in the Gauß sum relative to the integer3π^εc with some extra congruence conditions.

Lemma 3.7. Letc ∈ R and definec⁰ = cN(λ)^−ordλ(c)π^−ordπ(c). Letk, ε ∈ Zwithk > 1and ε>0. Letα∈ {0,±1}, and letv∈Rbe some integer.

Ifc≡0 (mod 3)andα6= 0, then 96

(i) X

For the first factor, we use the reciprocity law, and for the second, the hypothesisi>2allows us to use the consequence of the complementary reciprocity law. We obtain that

X

3.2 Relations between theta functions

We conclude the proof of(iii)and(iv)as we concluded the proof of(i)and(ii).

Letf ∈ R be an integer and letχ be a primitive character modulof. The Dirichlet series associated toχis

L(χ, s) = Y

Proof. Since we take the sum over thec ≡1 (mod 3), we can write them asc = _{m π}π , and the lemma is proved.

Im Dokument Twisted Kloosterman sums and cubic exponential sums (Seite 91-99)