1
$\begingroup$

Let $G$ be a subgroup in $SL_2(\mathbb{Z})$ and $S_k(G)$ be the space of cusp (automorphic?) forms invariant by any element of $G$ of weight $k$.

Question 1: Generally for two arithmetic subgroups $G < G'$ in $SL_2(\mathbb{Z})$, is there a surjection $\operatorname{End}(S_k(G)) \twoheadrightarrow \operatorname{End}(S_k(G'))$?

For example, what is the relation between $\operatorname{End}(S_k(\Gamma_0(N)))$ and $\operatorname{End}(S_k(\Gamma_1(N)))$?

Question 2: In Wiles's Theorem $R = T$, which is the right one between $T = \operatorname{End}(S_2(\Gamma_0(N)))$ or $T = \operatorname{End}(S_k(\Gamma_1(N)))$?

Pierre MATSUMI

$\endgroup$
4
  • 1
    $\begingroup$ By "End", you mean endomorphisms as complex vector spaces, or preserving some additional structure? If you just mean $\mathbf{C}$-endomorphisms, then the answer to Q1 is trivially "yes" and the answer to Q2 is "obviously neither". $\endgroup$ Commented Dec 10, 2013 at 12:15
  • $\begingroup$ Please use TeX. $\endgroup$
    – GH from MO
    Commented Dec 10, 2013 at 12:32
  • $\begingroup$ Then, what is T in Wiles' Theorem? $\endgroup$
    – Pierre
    Commented Dec 10, 2013 at 20:11
  • $\begingroup$ @Pierre The best way to find out is to read Wiles' paper, or any of the numerous excellent accounts of it that exists, or discuss the topic with someone around you that knows about this. $\endgroup$
    – Olivier
    Commented Dec 11, 2013 at 10:32

2 Answers 2

4
$\begingroup$

For any space $S$ of modular forms one has a Hecke algebra $T$ associated to $S$, defined as the $\mathbf{Z}$-subalgebra of $\operatorname{End}_{\mathbf{C}} S$ generated by the Hecke operators $T_n$. This is much smaller than the matrix algebra $\operatorname{End}_{\mathbf{C}} S$ itself. In particular it's commutative (which is a prerequisite for an $R = T$ theorem, since universal deformation rings are commutative by definition).

If $\Gamma \le \Gamma'$ then $S_k(\Gamma') \subset S_k(\Gamma)$ (for any $k$) but that doesn't give any map between the corresponding Hecke algebras $T'$ and $T$ in general (because the Hecke operators $T_n$ of level $\Gamma$ and $\Gamma'$ aren't necessarily compatible). However, if $\Gamma$ and $\Gamma'$ are both congruence, of levels $N$ and $N'$ say, and $N$ and $N'$ have the same set of prime factors, then one gets a surjection $T \to T'$. For instance, this applies to $\Gamma_1(N)$ and $\Gamma_0(N)$ for any $N$.

In Wiles' original $R = T$ theorem, his $T$ was not quite a Hecke algebra in the above sense, but the completion of the Hecke algebra associated to $S_2(\Gamma_0(N))$ at a "non-Eisenstein maximal ideal".

$\endgroup$
0
$\begingroup$

Regarding your first question, I might know where your confusion is coming from. Some people also would refer to a Hecke algebra by the space of complex valued function $f$ on $SL_2(\mathbb{Q})$ or $GL_2^+(\mathbb{Q})$ biinvariant under $G$ respective $G'$ (the algebra multiplication defined via convolution), especially if you are coming from an adelic point of view and $G, G'$ are congruence. If $G$ is finite index in $G'$, there is a inclusion (by restriction) as well as a projection (by averaging over double cosets).

But the usual Hecke operators due to Hecke are really only a subalgebra of these algebras. But it is e.g. a commutative algebra if $G = SL_2(\mathbb{Z})$ or $G=GL_2^+(\mathbb{Z})$, because the $( GL_2(\mathbb{Q}_p), GL_2(\mathbb{Z}_p))$ is a Gelfand pair for each $p$.

$\endgroup$
1
  • $\begingroup$ Thanks for all teachings. I really appreciate your explanations. Pierre MATSUMI $\endgroup$
    – Pierre
    Commented Dec 12, 2013 at 18:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .