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Ribet's Converse to Herbrand: Part II - Cuspstruction

So 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=1d|ndϵ(d)qns2,ϵ=n=1d|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=1d|nϵ(d)qn the Eisenstein series are all hecke eigenfunctions for Tn n coprime to p.

Fix a prime ideal p|p of Q(μp1), can think of μpQp and take ω:(Z/pZ)μp1 the unique character with ω(d)d(modp) for all dZ.

We start with a key lemma, will use this repeatedly.

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,ωk2)Bkk(modp) L(0,ωk1)Bkk(modp) we make use of the following expressions (see probably Washington) L(0,ϵ)=1ppn=1ϵ(n)(np2) L(1,ϵ)=12ppn=1ϵ(n)(n2pn+p26) ω(n)np(modp2) so pL(0,ωk1)=pn=1ωk1(n)(np/2) pn=1np(k1)+1(modp2) and pL(1,ωk2)=12pn=1ωk2(n)(n2pn+p26) 12pn=1np(k2)+2(modp2) but for all t>0 even we have the congruence p1n=1ntpBt(modp2) and so pL(0,ωk1)pBp(k1)+1(modp2) cancelling L(0,ωk1)Bp(k1)+1(modp) Bp(k1)+1(modp) Bkk(modp) as p(k1)+1k(modp1) using Kummer's congruence. Similarly L(1,ωk2)122kBk(modp).

Note: the constant term is a p unit unless Bn or Bm is divisible by p. We need to remove this condition.

Proof

It suffices to find a form with constant coefficient a p-unit. If pBk then we can use G2,ωk2 by lemma 3.1.

If p|Bk try the possible products G1,ωm1G1,ωn1 with 2m,np3 even as above with m+nk(modp1). We want to claim that at least one of these must work (i.e. we have a pair m,n with pBm,pBn). If this isn't the case if we let t=#{2n evenp3:p|Bn}, we must have t(p1)/4, assume otherwise, we will derive a contradiction from this.

Greenberg showed that hphQ(μp)+=hp=2?pp2k=2evenL(0,ωk1) (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|hp, we will get a contradiction if we show hpp(p1)/4.

Work of Carlitz-Olson '55, Maillet's determinant shows that hp=±Dp(p3)/2 where D is the determinant of a (p1)/2×(p1)/2 matrix with entries in [1,p1]. So recalling Hadamard's inequality |det(v1vn)|ni=1||vi||, or the simpler corollary |Aij|B|det(A)|nn/2Bn and applying with B=p,n=(p1)/2 gives |D|(p12)(p1)/4p(p1)/2<2(p1)/2p(3p3)/4 so hp<p(p+3)/42(p1)/4. And we are done as hp=1 for p19 and as p2(p1)/4 for p>19.

Now we fix 2kp3 even with p|Bk and let ϵ=ωk2, 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.

Let f=G2,ϵcg with c=L(1,ϵ)/2. This is a semi-cusp form by construction. We get fG2,ϵ(modp) because cBk/2k0(modp) by lemma 3.1 (again!) as we assume p|Bk now. Additionally by the same lemma GkG2,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?

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 S2(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()+1+ϵ()(modp) implies ϵ()=1, but ϵ is non-trivial!

The final challenge is to ensure that f is also an eigenform for Tpi.

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:

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.