nt.number theory – Victor Miller basis for higher $N$ // why is this bilinear form perfect? – MathOverflow

I am trying to understand the proof of Thm 9.23 in http://wstein.org/books/modform/modform/newforms.html#congruences-between-newforms . Let Sk(Γ)Sk(Γ) be the cusp forms for a subgroup Γ1(N)⊆Γ⊆Γ0(N)Γ1(N)⊆Γ⊆Γ0(N). Let T:=Z[T1,T2,T3,...] be the hecke algebra viewed as a subset of End(Sk(Γ)) (not as abstract double coset stuff!). Let Sk(Γ,Z):=Sk(Γ)∩Z[[q]] then the author claims that ⟨⋅,⋅⟩:T×Sk(Γ,Z)↦Z, ⟨T,f⟩=a1(T(f)) is a perfect bilinear form and because of finite-rank-arguments i already know that it suffices to show that the Z-module homomorphism T↦⟨T,⋅⟩∈HomZ(Sk(Γ,Z),Z) is surjective but i am unable to understand why this is the case. Only for N=1 i have found a workaround (see below). As was pointed out to me here 'Hecke algebra' finitely generated? , the set Sk(Γ,Z) is a finitely generated free Z-module. Let us take a basis Sk(Γ,Z)=Zf1⊕...⊕Zfn. The HomZ(Sk(Γ,Z))=Zϕ1⊕...⊕Zϕn where ϕi(fj)=δij. Let us assume that we select Γ0(N) with trivial character. Then, the ring T is nothing else than the Z-module (because Tpe=TpTpe−1−pk−1T1 so that every polynomial in the Tm is actually a sum in the Tm). We want to show that there is a T such that ⟨T,fj⟩=ϕ1(fj)=δ1j so that we search for a Z linear combination of the Tm mapping to ϕ1, say T=v1T1+...+vlTl. Then, ⟨T,fj⟩=ϕ1(fj)=δ1j is equivalent to Let us denote the matrix on the left hand side by A, then the question is: is A a surjective map seen as a map A:Zl↦Zn (where we can select l as big as we want)? Rephrased i ask for the following: For a subgroup Γ that satisfies some properties (i.e. Γ a congruence subgroup or so) is there a C-basis f1,...,fn of Sk(Γ) with fourier coefficients in Z such that the matrix A becomes surjective as a map from Zl to Zn (for example: when selecting n columns, does the matrix consisting of these columns satisfy det(Matrix)=±1? I.e. is there something like a Victor Miller basis for modular forms of higher level)? For Γ=SL2(Z), this is easy, because one has a Miller basis, i.e. a C-basis f1,...,fn of Sk(SL2(Z)) with integral fourier coefficients such that ai(fj)=δij so that A=(Id ...). Note: if f1,...,fn are a C-basis for Sk(SL2(Z)) with integral fourier coefficients then this does not necessarily imply that f1,...,fn form a Z-basis for Sk(Γ,Z) but this suffices to see that the Hecke algebra is finitely generated by the first r Hecke operators since one can do the same argument with Zf1⊕...⊕Zfn in place of Sk(Γ,Z) since the behavior of an operator in End(Sk(Γ)) is determined by its actions on f1,...,fn. Note that for N>,1 this seems to be true in 'many' cases: For example, typing the following lines in magma Also note the following: Using the Sturm bound and selecting l≥max(n,SturmBound) one can show that the matrix A has a full Q-rank, i.e. there are v1,...,vl∈Q satisfying the above relation but can they be somehow found to be in Z? You pose your questions for a general Γ, but I'm not sure that quite makes sense, in general the Hecke algebra won't be commutative and will have a very different structure, and Sk(Γ,Z) won't necessarily span Sk(Γ,C). So let's assume Γ is Γ0(N) or Γ1(N) for some N. Let S=Sk(Γ,Z) and T⊆EndZ(S) the Hecke algebra. Then (as I think you know) it's not too hard to show that T→Hom(S,Z) and S→Hom(T,Z) are injections. This alone is enough to imply that the pairing is nondegenerate, hence becomes perfect after extending scalars to Q. So it suffices to show either that T→Hom(S,Z) is surjective, or that S→Hom(T,Z) is surjective (because the cokernels of these maps are finite groups of the same order, equal to the det of the matrix of the pairing with respect to Z-bases of either side). You seem to be trying to do the first, but the second is (I think) easier. Let ϕ∈Hom(T,Z) and consider the formal power series f=∑n≥1ϕ(Tn)qn. It's clear that f∈Z[[q]], but also that f∈Sk(Γ,Q), so f∈S. Clearly we have (Tn,f)=an(f)=ϕ(Tn) and thus we're done. Now you can construct a 'Miller-like' basis as follows. [EDIT: This doesn't actually work, see comments.] The Tn's generate T, so there is some finite subset which is a Z-basis of T, let these be Tn1,…,Tnd. Then there is a dual basis f1,…,fd of S such that (Tni,fj)=δij, and there are your Miller-like basis. (You can't necessarily take the ni to be {1,…,rk(S)} though, as your example above shows.) Source.


Яндекс.Метрика Рейтинг@Mail.ru Free Web Counter
page counter
Last Modified: April 18, 2016 @ 1:05 am