# “Hecke algebra” finitely generated?

Hi all.

I have some question on Hecke operators and its relations to the Hecke algebra.

(1)

I want to understand why the "Hecke algebra" is finitely generated in some cases. I found a nice result in W.Stein, Modular Forms, a Computational Approach, Thm 9.23 (see http://wstein.org/books/modform/modform/newforms.html). This theorem states:

Let $\Gamma_0(N) \subset \Gamma \subset \Gamma_1(N)$ and $S_k(\Gamma)$ the cusp forms of weight $k$. Consider the Hecke algebra $\mathbb{T} = \mathbb{Z}[T_1, T_2, ...]$ not as an abstract Hecke ring but rather its image in $End(S_k(\Gamma))$, then for $$r := \frac{k[SL_2(\mathbb{Z}) : \Gamma]}{12} - \frac{[SL_2(\mathbb{Z}) : \Gamma] - 1}{N}$$ we have $$\mathbb{T} = \mathbb{Z}[T_j | j \leq r]$$ i.e. the image of the Hecke algebra in the endomorphisms is finitely generated.

I am a bit confused by the proof:

The first point concernes the pairing $\langle f, T_m \rangle := a_1(T_m f)$. I can see that it is non-degenerate in the second entry but in the first entry, quite the contrary is the case: We have to show that for every finite sequence $\lambda_1, ..., \lambda_s$ in $\mathbb{Z}$

$$\lambda_1 * a_1(f) + ... + \lambda_s * a_r(f) = 0$$ for all cusp forms $f$ implies $\lambda_1 = ... = \lambda_s = 0$.

I claim that we can always find a non-trivial vector $\lambda_1, ..., \lambda_s$ such that the relation above holds. Let $f_1, ..., f_n$ be a basis of $S_k(\Gamma)$. Every non-trivial vector in the kernel of the matrix $$a_1(f_1)~~ ... ~~a_s(f_1)$$ $$... ~~~~~~~~~~~~ ...$$ $$a_1(f_n) ~~ ... ~~ a_s(f_n)$$

of size $n \times s$ will do (and this matrix possesses a non-trivial kernel when $s > n$). Am i completely wrong here?

The second point is concerning the claim that $(\mathbb{T}/M) \otimes \mathbb{F}_p = 0$ implies that $\mathbb{T}/M = 0$. I only know one proof that works like this, namely if a module is finitely generated and contained in all its maximal modules then it must be the zero module (by localisation and Nakayamas lemma), but $(\mathbb{T}/M) \otimes \mathbb{F}_p = 0$ just asserts that $(\mathbb{T}/M) / p*(\mathbb{T}/M) \cong (\mathbb{T}/M) \otimes \mathbb{F}_p = 0$, i.e. $\mathbb{T}/M$ is contained in $p*(\mathbb{T}/M)$. How can one conclude that $(\mathbb{T}/M) = 0$? We do not even know that the quotient is finitely generated. If there was some theorem like "$M$ an $R$ module such that $M/pM = 0$ for all primes $p$ then $M = 0$" then we could also conclude that since $\mathbb{T}/p\mathbb{T} \cong (\mathbb{T}/M) / (p\mathbb{T}/M) \cong (\mathbb{T}/M) / p (\mathbb{T}/M) = 0$, also $\mathbb{T} = 0$ which is false...

(2)

Say we take a $\Gamma$ which does not satisfy the requirements above (i.e. $\Gamma = \Gamma(N)$ for example) but assume that we can still show that $S_k(\Gamma)$ possesses a basis of forms that have integral Fourier series expansions at the cusp $\infty$, does the above result still hold? (I cant see a reason why we need to assume that $\Gamma_0(N) \subset \Gamma \subset \Gamma_1(N)$ apart from the integrality of the Fourier coefficients)

(3)

In Steins book, there is a nice exercise on the Diamond operator: One should show that $\langle d \rangle$ is contained in $\mathbb{Z}[T_1, T_2, ...]$. Can somebody tell me how to solve this? Is it true that this relation does also hold in the abstract Hecke ring and not only as endomorphisms on $S_k(\Gamma)$?

(4)

In the situation of $\Gamma_1(N)$ without character, one has a decomposition $$M_k(\Gamma_1(N)) = \bigoplus_{\chi} M_k(\Gamma_0(N), \chi)$$ ($M_k(\Gamma, \chi) =$completely holomorphic modular forms with character $\chi$) where $\chi$ runs over all multiplicative characters $\chi : \mathbb{Z}_N^\times \mapsto \mathbb{C}^\times$ and the diamond operator $\langle d \rangle$ restricted to $M_k(\Gamma_0(N), \chi)$ acts as multiplication with the character $\chi$. In the situation of $\Gamma(N)$ vs. $\Gamma_1(N)$ one has $$M_k(\Gamma(N)) = \bigoplus_{\psi} M_k(\Gamma_1(N), \psi)$$ where $\psi$ runs over all additive characters $\psi : (\mathbb{Z}_N, +) \mapsto \mathbb{C}^\times$. The role of the diamond operator is now taken by the action of the element in the abstract Hecke algebra of $\Gamma(N)$, $$p * \Gamma(N) pR_p \Gamma(N)$$ where $R_p \equiv (p^{-1}, 0, 0, p) \mod N$ (i mean that $T_{p^e} = T_p T_{p^{e-1}} - p * \Gamma(N) pR_p \Gamma(N)$, see for example R. Rankin, Modular forms, formula (9.1.46) on p.285). Is it true that $p * \Gamma(N) pR_p \Gamma(N)$ acts on $M_k(\Gamma_1(N), \psi)$ as multiplication with $\psi(p)$?

Best,

Fabian Werner, Germany

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There are too many questions here and too much notation for any one person to tackle. But let me just point out two things: First, $(T/M)/p(T/M)$ is not equal to $T/pT$, unless $M\subset pT$. In general, the former quotient will be isomorphic to $T/(pT+M)$. Second, you can conclude $A=0$ from $A\otimes\mathbb{F}_p=0$ whenever $A$ is not $p$-divisible; that is, whenever $pA\neq A$. – Keerthi Madapusi Pera Jan 7 '12 at 23:51
Of course, if $A=0$, then $pA=A$, trivially! – Keerthi Madapusi Pera Jan 7 '12 at 23:52
Dear Fabian, regarding your first point, your formulation of non-degeneracy of the pairing is wrong. You have to prove that if $T$ satisfies $a_1(Tf)=0$ for every $f$, then $T=0$. This is true because $S_k(\Gamma)$ admits a basis consisting of newforms. Moreover, as a general remark, if you want to prove that the Hecke algebra is finitely generated, it is sufficient to prove that it is finitely generated as a $\mathbf{Z}$-module, and for this the non-degeneracy of the above pairing will essentially do. – François Brunault Jan 8 '12 at 0:29
@Keerthi Madapusi Pera: ok, then the question is: why is $p\mathbb{T} \neq \mathbb{T}$? Remember that we are not talking about the free $\mathbb{Z}$-module here (then this is clear) but rather about its image in the endomorphisms... – Fabian Werner Jan 8 '12 at 11:59
@François Brunault: Erm... why precisely does $S_k(\Gamma)$ admit a basis consisting of newforms? What if there are oldforms, then this is not true... but anyhow, let us restrict everything to the subspace of newforms. Every T can be written as $\sum \lambda_j T_j$ so we have to show that $\sum \lambda_j a_j(f) = \sum \lambda_j a_1(T_j f) = \langle f , \sum \lambda_j T_j \rangle = 0$ implies $T=0$ How does it help to know that the forms are newforms? – Fabian Werner Jan 8 '12 at 12:03

I will address point (1), namely why $\mathbf{T}$ is finitely generated as a $\mathbf{Z}$-module.
We have a pairing $\mathbf{T} \times S_k(N,\mathbf{Z}) \to \mathbf{Z}$ given by $\langle T,f \rangle = a_1(Tf)$. It is left-nondegenerate because if for every $f$ we have $\langle T,f \rangle =0$ then we also have $a_n(Tf)=a_1(T_n Tf) = a_1(T T_n f)=0$ so that $T=0$.
So we have an injection $\mathbf{T} \hookrightarrow \operatorname{Hom}_{\mathbf{Z}} (S_k(N,\mathbf{Z}),\mathbf{Z})$ and it suffices to prove $S_k(N,\mathbf{Z})$ is finitely generated. By definition we have an injection $S_k(N,\mathbf{Z}) \hookrightarrow \mathbf{Z}[[q]]$. But $S_k(N)$ is a finite dimensional complex vector space, so that there exists an integer $B \geq 1$ such that the elements of $S_k(N)$ are determined by their first $B$ Fourier coefficients. So we get $S_k(N,\mathbf{Z}) \hookrightarrow \mathbf{Z}^B$ which proves the claim.
In fact, it is Shimura who proved in general that $S_k(N)$ admits a basis consisting of forms with integer Fourier coefficients, so that in fact the rank of $S_k(N,\mathbf{Z})$ as a $\mathbf{Z}$-module is equal to the dimension of $S_k(N)$.
This is true but for computational things one needs to know precisely which ones of them generate the Ring of Endomorphisms. I think the proof of Stein works like this: $T/M$ is $p$-divisible for every $p$ $\Rightarrow$ it is divisible $\Rightarrow$ it is an injective module so that for $\phi:T\mapsto Hom_\mathbb{Z}(S_k(\Gamma), \mathbb{Z})$ as above and $\pi:T\mapsto T/M$ one finds a hom $\tau:Hom_\mathbb{Z}(S_k(\Gamma), \mathbb{Z})$ which satisfies $\tau\circ\phi=\tau$ hence,$\tau$ is surjective and $T/M$ is finitely generated. Now Nakayama tells us that $T/M$ must be trivial. – Fabian Werner Jan 10 '12 at 9:13
@Fabian : Yes, this is how the proof works. Two remarks : (1) In order to prove $\mathbf{T}/M$ is $p$-divisible, you need Thm 9.18 about congruences of modular forms. (2) Any divisible finitely generated $\mathbf{Z}$-module is trivial (you can see this with Nakayama, but maybe an easier way is with the structure theorem for finitely generated $\mathbf{Z}$-modules). – François Brunault Jan 11 '12 at 10:04