Ribet's Converse to Herbrand: Part II - Cuspstruction
02 Apr 2017So around a month back I posted the first post in this 2 (or more who knows?) part series on Ribet’s converse to Herbrand’s theorem. This is the sequel, Cuspstruction, it is basically just my personal notes from my STAGE talk with the same name. We were following Ribet’s paper and this is all about section 3. The goal is to construct a cusp form with some very specific properties, which we can then take the corresponding Galois representation and use that to obtain the converse to Herbrand. In this post though we’ll be focussing on constructing the cusp form, hence, Cuspstruction.
Cuspstruction
We will make use the following building blocks, some specific modular forms of weights 2 and type ϵ G2,ϵ=L(−1,ϵ)/2+∞∑n=1∑d|ndϵ(d)qns2,ϵ=∞∑n=1∑d|ndϵ(n/d)qn the latter is not a cusp form (not cuspidal at the other cusp of Γ1(p)) we call such forms semi-cusp forms, denote the space of such by S∞ (not standard notation) we will also use G1,ϵ=L(0,ϵ)+∞∑n=1∑d|nϵ(d)qn the Eisenstein series are all hecke eigenfunctions for Tn n coprime to p.
Fix a prime ideal p|p of Q(μp−1), can think of μp⊆Q∗p and take ω:(Z/pZ)∗∼→μp−1 the unique character with ω(d)≡d(modp) for all d∈Z.
We start with a key lemma, will use this repeatedly.
Lemma 3.1
Let k be even 2≤k≤p−3, then G2,ωk−2,G1,ωk−1 have p-integral q-expansions in Q(μp−1) which are congruent mod p to −Bk2k+∞∑n=1∑d|ndk−1qn (this is the q-expansion of Gk).
By our choice of ω we get the desired result for the non-constant terms of the q-expansion.
So it remains to prove that L(−1,ωk−2)≡−Bkk(modp) L(0,ωk−1)≡−Bkk(modp) we make use of the following expressions (see probably Washington) L(0,ϵ)=−1pp∑n=1ϵ(n)(n−p2) L(−1,ϵ)=−12pp∑n=1ϵ(n)(n2−pn+p26) ω(n)≡np(modp2) so pL(0,ωk−1)=−p∑n=1ωk−1(n)(n−p/2) ≡−p∑n=1np(k−1)+1(modp2) and pL(−1,ωk−2)=−12p∑n=1ωk−2(n)(n2−pn+p26) ≡−12p∑n=1np(k−2)+2(modp2) but for all t>0 even we have the congruence p−1∑n=1nt≡pBt(modp2) and so pL(0,ωk−1)≡−pBp(k−1)+1(modp2) cancelling L(0,ωk−1)≡−Bp(k−1)+1(modp) ≡−Bp(k−1)+1(modp) ≡−Bkk(modp) as p(k−1)+1≡k(modp−1) using Kummer's congruence. Similarly L(−1,ωk−2)≡−122kBk(modp).
Corollary 3.2
Let k be as before and 2≤n,m≤p−3 both even with n+m≡k(modp−1). Then the product G1,ωn−1G1,ωm−1∈M2(Γ1(p),ωk−2) with q-expansion coefficients still p-integers in Q(μp−1).
Note: the constant term is a p unit unless Bn or Bm is divisible by p. We need to remove this condition.
Theorem 3.3
Let k be as before, then there exists g∈M2(Γ1(p),ωk−2) whose q-expansion coefficients are p-integers and whose constant coefficient is 1.
Proof
It suffices to find a form with constant coefficient a p-unit. If p∤Bk then we can use G2,ωk−2 by lemma 3.1.
If p|Bk try the possible products G1,ωm−1G1,ωn−1 with 2≤m,n≤p−3 even as above with m+n≡k(modp−1). We want to claim that at least one of these must work (i.e. we have a pair m,n with p∤Bm,p∤Bn). If this isn't the case if we let t=#{2≤n even≤p−3:p|Bn}, we must have t≥(p−1)/4, assume otherwise, we will derive a contradiction from this.
Greenberg showed that hphQ(μp)+=h∗p=2?pp−2∏k=2evenL(0,ωk−1) (this is obtained by taking a quotient of the analytic class number formulas for Q(μp),Q(μp)+) but by lemma 3.1 we know that pt will divide the product of L-values. And so pt|h∗p, we will get a contradiction if we show h∗p≤p(p−1)/4.
Work of Carlitz-Olson '55, Maillet's determinant shows that h∗p=±Dp(p−3)/2 where D is the determinant of a (p−1)/2×(p−1)/2 matrix with entries in [1,p−1]. So recalling Hadamard's inequality |det(v1⋯vn)|≤n∏i=1||vi||, or the simpler corollary |Aij|≤B⟹|det(A)|≤nn/2Bn and applying with B=p,n=(p−1)/2 gives |D|≤(p−12)(p−1)/4p(p−1)/2<2−(p−1)/2p(3p−3)/4 so h∗p<p(p+3)/42−(p−1)/4. And we are done as h∗p=1 for p≤19 and as p≤2(p−1)/4 for p>19.
Now we fix 2≤k≤p−3 even with p|Bk and let ϵ=ωk−2, k must really be at least 4 (or even 10) so ω is a non-trivial even character, we will work in weight 2, type ϵ from now on.
Proposition 3.4
There exists f=∞∑n=1anqn∈S∞2(Γ1(p),ϵ) with p-integer ap∈Q(μp−1) with f≡Gk≡G2,ϵ(modp).
Let f=G2,ϵ−cg with c=L(−1,ϵ)/2. This is a semi-cusp form by construction. We get f≡G2,ϵ(modp) because c≡−Bk/2k≡0(modp) by lemma 3.1 (again!) as we assume p|Bk now. Additionally by the same lemma Gk≡G2,ϵ(modp).
Let's take stock, we have a semi-cuspidal form which mod p looks like Gk and is hence an eigenform mod p but we want an actual eigenform, bro do you even lift?
Proposition 3.5
There exists 0≠f′∈S2(Γ1(p),ϵ) which is an eigenform for all Tn with (n,p)=1. With the eigenvalue λ(ℓ) for Tℓ (ℓ≠p) satisfying λ(ℓ)≡1+ϵ(ℓ)ℓ≡1+ℓk−1(modP) for some prime P|p in Q(μp−1,λ(n):(n,p)=1).
We start with f from the proposition above it's a mod p eigenform and so we can use Deligne-Serre lifting lemma (6.11 in Formes modulaires de poids 1) to obtain a semi-cusp form f′, that is an eigenvalue for the Hecke operators stated.
To promote the semi-cusp form to a full blown cusp form we observe that the space S∞2(Γ1(p),ϵ) is generated by the cusp forms and s2,ϵ which is also an eigenform we only have to check that f′ isn't s2,ϵ (or it's scalar multiple). So we check the eigenvalues mod p. ϵ(ℓ)+ℓ≡1+ℓϵ(ℓ)(modp) implies ϵ(ℓ)=1, but ϵ is non-trivial!
The final challenge is to ensure that f′ is also an eigenform for Tpi.
Proposition 3.6
f′ is an eigenform for all Hecke operators, so we can normalise as f′=∞∑n=1λ(n)qn.
Use the theory of newforms. There are no oldforms for Γ1(p) as M2(SL2(Z))=0. A newform that is an eigenform for all hecke operators coprime to the level p is also an eigenform for the remaining Hecke operators.
So in conclusion:
Theorem 3.7
Assume p|Bk then there exists f=∞∑n=1anqn∈S2(Γ1(p),ϵ) which is an eigenform for all Tn and p|p an ideal of Q(an) such that aℓ≡1+ℓk−1≡1+ϵ(ℓ)ℓ(modp) for all ℓ≠p.
Remark
Word on the internet is that Mazur, Mazur-Wiles' proof of the Main conjecture of Iwasawa theory is modelled on this.
That's all for now, in the remainder of Ribet's paper he constructs a Galois representation from this and use it to prove the theorem.