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



"Hecke algebra" finitely generated?
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 $S\_k(N)$.


