Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I could have made this question very brief but instead I've maximally gone the other way and explained a huge amount of background. I don't know whether I put off readers or attract them this way. The question is waay down there.

Let $f$ be a cuspidal modular eigenform of level $\Gamma_0(N)\subseteq SL_2(\mathbf{Z})$ (for example $f$ could be the weight 2 modular form attached to an elliptic curve) and let $p$ be a prime. In the theory of modular forms, one Hecke operator at $p$ is singled out, namely $T_p$, sometimes called $U_p$ if $p$ divides $N$, and defined by the double coset attached to the matrix $\left(\begin{array}{cc}p& 0\\ 0&1\end{array}\right)$. Now $f$ is an eigenform for $T_p$, and $f$ has an eigenvalue for this operator---a Galois-theoretic interpretation of this eigenvalue is that it is (modulo fixing embeddings of $\overline{\mathbf{Q}}$ in $\overline{\mathbf{Q}}_\ell$ and $\mathbf{C}$) the trace of the geometric Frobenius on the inertial invariants of the $\ell$-adic representation attached to $f$, for $\ell\not=p$ a prime.

Now here is a very naive question that I don't know the answer to, and I really should, and I'm sure it's very well-known to people who do this sort of stuff. Say $N=p^rM$ with $M$ prime to $p$. One can approach the theory of Hecke operators entirely locally. Let $K:=U_0(p^r)$ denote the subgroup of $GL_2(\mathbf{Z}_p)$ consisting of matrices whose bottom left hand entry is $0$ mod $p^r$. Now there is an "abstract Hecke algebra" of locally left- and right-$K$-invariant complex-valued functions on $G:=GL_2(\mathbf{Q}_p)$ with compact support. As a complex vector space this algebra has a basis consisting of the characteristic functions $KgK$ as $KgK$ runs through the double cosets of $K$ in $G$. But this space also has an algebra structure, given by convolution.

If $r=0$ then $K$ is maximal compact, and the structure of this Hecke algebra is well-known and easy. Via the Satake isomorphism, the abstract Hecke algebra is isomorphic to $\mathbf{C}[T,S,S^{-1}]$, with $S$ and $T$ independent commuting polynomial variables. The interpretation is that $T$ is the usual Hecke operator $T_p$ attached to the matrix $\left(\begin{array}{cc}p& 0\\ 0&1\end{array}\right)$ and $S$ is the matrix attached to $\left(\begin{array}{cc}p& 0\\ 0&p\end{array}\right)$. One doesn't always see this latter Hecke operator explicitly in elementary developments of the theory because it acts in a very dull way---it acts by scalars on forms of a given weight and level $\Gamma_0(N)$, typically (depending on normalisations) as the scalar $p^{k-2}$ on forms of weight $k$. In particular the "abstract Hecke algebra" doesn't give us any more information than that which classical texts explain, as it's generated by $T_p$, $S_p$ and $S_p^{-1}$.

The next case is $r=1$ and this case I also understand. The abstract Hecke algebra now is non-commutative, "because of oldforms": I don't think the operators attached to $\left(\begin{array}{cc}p& 0\\ 0&1\end{array}\right)$ and $\left(\begin{array}{cc}0& p\\ 1&0\end{array}\right)$ (that is, the operators corresponding to these double coset spaces) commute, but if $f$ has level $Mp$ and is old at $p$ then we should be working at level $M$, and if it's new at $p$ then we get two invariants---the $T_p$ (or $U_p$) eigenvalue, which is classical, and the $w$-eigenvalue, which is the local sign for the functional equation. Again both of these numbers are classical and a lot is known about them. I am pretty sure that the abstract Hecke algebra in this case is generated by these operators $T_p$, $w$, and the uninteresting $S_p$ and $S_p^{-1}$, the latter two still acting by scalars on forms of a given weight. Am I right in thinking that these operators generate the local Hecke algebra? I think so.

The next case is $r=2$ and this I am not 100 percent sure I understand. The classical theory gives us $T_p$, $S_p^{\pm1}$ and $w$. Note that on a newform of level $\Gamma_0(Mp^2)$, $T_p$ is zero in this situation, $S_p$ acts by a scalar, and $w$ is some subtle sign which people have clever ways of working out.


Finally then, the question! Let $K$ be the subgroup of matrices in $GL_2(\mathbf{Z}_p)$ consisting of matrices for which the bottom left hand corner is $0$ mod $p^2$. Let $H$ denote the abstract double coset Hecke algebra of compactly supported $K$-bivariant functions on $GL_2(\mathbf{Q}_p)$.

Is this abstract Hecke algebra generated (as a non-commutative algebra) by the characteristic functions of $KgK$ for $g$ in the set {$\left(\begin{array}{cc}p& 0\\ 0&1\end{array}\right)$, $\left(\begin{array}{cc}0& p^2\\ 1&0\end{array}\right)$, $\left(\begin{array}{cc}p& 0\\ 0&p\end{array}\right)$, $\left(\begin{array}{cc}p^{-1}& 0\\ 0&p^{-1}\end{array}\right)$}?

In the language I've been using in the waffle above: modular forms of level $p^2$ have an action of the Hecke operators $T_p$, $w$, and the invertible $S_p$. Are there any more, lesser known, Hecke operators that we're missing out on?

share|improve this question
    
Kevin- I took the liberty of adding a set-off box for your question (I think it deserved to pop more). I hope it wasn't too forward. –  Ben Webster Apr 10 '10 at 21:11
    
I would have done it but I'd forgotten how to and was exhausted from typing all the background---I just wanted to get back to the dishes at that point ;-). I also couldn't be bothered to figure out how to turn (a,b;c,d) into the 2x2 matrix... –  Kevin Buzzard Apr 10 '10 at 21:35
1  
+1 for long background! –  Theo Johnson-Freyd Apr 11 '10 at 0:13
    
I stuck in the 2x2 matrices, and don't think that I changed any of them. Also, good question, looking forward to answers. –  Charles Siegel Apr 11 '10 at 0:54
1  
Just a remark: a much easier related question is to look at forms of level not $\Gamma_0(p^2)$, but the subgroup of $\Gamma_0(p^2)$ consisting of matrices whose diagonal entries are 1 mod p. Then your level is conjugate to $\Gamma(p)$, and for any level of the form $\Gamma(p^r)$ it's enough to take the Hecke operators corresponding to $T_p$, $S_p^{+-1}$, Atkin-Lehner $w$, and the two standard generators of ${\rm SL}_2(\mathbb{Z})$. –  David Loeffler Apr 11 '10 at 8:54
show 4 more comments

2 Answers

up vote 7 down vote accepted

Just to expand on a comment I made above: I'm not exactly sure what operators generate the Hecke algebra of $\Gamma_0(p^2)$, but the Hecke algebras of the principal congruence subgroups $\Gamma(p^r)$ are easier to handle.

Let's write $K_n$ for the principal congruence subgroup of level $p^n$ in $G = {\rm GL}_2(\mathbb{Z}_p)$.


CLAIM: For any $n > 0$, the Hecke algebra $H(G // K_n)$ is generated by the double cosets $K_n x K_n$ for $x$ in the set

$S = \left\{ \begin{pmatrix} 1 & 0 \\\ 0 & p \end{pmatrix}, \begin{pmatrix} p & 0 \\\ 0 & p \end{pmatrix}, \begin{pmatrix} p^{-1} & 0 \\\ 0 & p^{-1} \end{pmatrix}, \begin{pmatrix} 1 & 1 \\\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & -1 \\\ 1 & 0 \end{pmatrix}, \begin{pmatrix} a & 0 \\\ 0 & 1 \end{pmatrix} \right\}$

where a is your favourite generator of $\mathbb{Z}_p^\times$.

(These correspond, classically, to $T_p$, $S_p$, $S_p^{-1}$, a "twisting operator at p", something close to the Atkin-Lehner $w$, and the diamond operator $\langle a \rangle$.)


Proof: It suffices to show that the subalgebra generated by these operators contains the double coset $[K_n g K_n]$ for any given $g \in G$. Let $X$ be the monoid of elements of the form $\begin{pmatrix} p^r & 0 \\\ 0 & p^s\end{pmatrix}$ with $r \le s \in \mathbb Z$. The Cartan decomposition tells us that any $g \in G$ can be written as $g = k x k'$ for some $k, k' \in K_0$.

We now write $K_n\ g\ K_n = K_n\ k\ x\ k'\ K_n$

$ = K_n\ k\ K_n\ x\ K_n\ k'\ K_n$ (using the normality of $K_n$ in $K_0$)

$ = [K_n\ k\ K_n]\ [K_n\ x\ K_n]\ [K_n\ k'\ K_n]$

The first and last terms are obviously in the subalgebra $H(K_0 // K_n)$ of $H(G // K_n)$, which is isomorphic to the group algebra of the finite group $K_0 / K_n = {\rm GL}_2(\mathbb Z / p^n)$. This is clearly generated by the images of the last three elements of $S$, since these are topological generators of ${\rm GL}_2(\mathbb{Z}_p)$.

Meanwhile, the middle term is in the subalgebra of $H(G // K_n)$ generated by $X$, and it's easy to see that for $x, y \in X$ we have $K_n\ x\ K_n\ y\ K_n = K_n\ xy\ K_n$. Hence this subalgebra is just the monoid algebra of $X$, which is generated by the first three elements of $S$.

Now, as for your original question, the subgroup $U_0(p^2) \subseteq {\rm GL}_2(\mathbb{Z}_p)$ of matrices that are upper triangular modulo $p^2$ contains a conjugate of $K_1$, so its Hecke algebra is isomorphic to a subalgebra of the Hecke algebra of $K_1$. So although I can't give generators for your algebra, I can exhibit it as a subalgebra of something we know generators for.

share|improve this answer
    
Although this doesn't strictly speaking answer the question, it does of course answer the question behind the question :-) so I think I should accept it really. Thanks! –  Kevin Buzzard Apr 12 '10 at 16:42
add comment

I am not sure to completely understand the question, so I will not elaborate too much. But maybe this is helpful.

The local data at $p$ (i.e. the isomorphism class of the local restriction of the automorphic representation) of a modular form $f$ of level $Np^r$ (and possibly non trivial nebentypus) is determined by

(1) the Euler factor at $p$ (whose coefficients are the eigenvalues of $T_p$ and $S_p$) and

(2) the (pseudo)-eigenvalues of the Atkin Lehner operator at $p$

for all the twists of $f$ by local characters of level dividing $p^r$ (in fact if you have this data for these characters you have it for all characters). This has been shown, I believe, by Atkin and Winnie Li in the classical language and by Winnie Li in the adelic language. (Their goal was apparently to establish converse theorems. Perhaps it was in germs in Jacquet-Langlands.) If $r=0$ or $r=1$, it is not hard to show that the data for the twists can be deduced from the data (1) and (2) for f itself, so there is no need to twist.

To convince you that this might be the right point of view, consider an elliptic curve $E$ with good reduction at $p$. Twist it by the Legendre symbol mod $p$. You get another elliptic curve which has additive reduction at $p$, attached to a modular form $f$. Ideally you would like to recover from $f$ the Euler factor at $p$ of $E$. But the application of $T_p$, $w$ on $f$ can't give you that. You need the data above for the twists.

Conclusion: the missing operators are the twisting operators (combined with $w$ and the Hecke operators). If you are dissatisfied by the fact that they do not preserve individual eigenforms, consider the fact that when the nebentypus is not trivial, $w$ does not preserve individual eigenforms either.

share|improve this answer
    
This is a very interesting answer! I didn't know the result you assert here. I knew it was true without the ramification bound (if I can twist by an arbitrary finite order character). I wonder if I can see these twists in the abstract Hecke algebra? That sounds like a much more accessible question. I am also not entirely sure whether that question, when resolved, answers my original question!... –  Kevin Buzzard Apr 12 '10 at 5:34
    
...Somehow I am looking for an exotic Hecke operator, you are telling me that a huge amount of information can be obtained from non-exotic ones, and I'm now wondering whether this can be turned into a proof that no exotic one of the type I'm looking for can exist. Thanks for the ideas! –  Kevin Buzzard Apr 12 '10 at 5:35
    
The Atkin-Li statement can be proved along the lines of the answer of David Loeffler, which shows (1) how twists appear in your "Hecke algebra" (2) that it might not be necessary to twist by characters of level as high as $p^r$, ($p^{r/2}$ might suffice -- I am not sure anymore) and (3) that the setting of forms with non trivial nebentypus is more natural. –  user4186 Apr 12 '10 at 17:04
    
I think the correct statement is to use twists by characters modulo $p^{\min(\lfloor r/2 \rfloor, r - s)}$, where s is the power of p dividing the conductor of the character of f. This is the most you can twist by without the level going up. –  David Loeffler Apr 15 '10 at 9:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.