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Are there infinitely many (linearly independent) cuspidal eigenforms for $\Gamma(1)$ with integer coefficients?

Someone told me that the Hecke algebra is conjectured to act irreducibly on the space of modular forms of level 1, so there would be no eigenforms if $\mathrm{dim} S_k > 1$, i.e. for $k \geq 12$.

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You might want to provide a few more definitions than that. (I cannot readily find a definition of "eigenform," for example, online.) –  Qiaochu Yuan Jan 13 '10 at 18:34
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@Qiachu - The Hecke operators are acting on the space S_k, and by the Spectral Theorem there exists a basis of simultaneous eigenvectors for the Hecke operators. These eigenvectors are the 'eigenvorms' norondion is referring to. –  Ben Linowitz Jan 13 '10 at 18:38
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I just want to remark that for those who know the theory of modular forms (who are the ones who are likely to answer this question) the question is completely clear. –  Emerton Jan 13 '10 at 18:57
    
Sure, but I enjoy reading questions and answers I have no expertise in (I hope I'm not the only one) and it's hard to do so without at least a little background. –  Qiaochu Yuan Jan 13 '10 at 19:08
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@Qiaochu: with all due respect (which is a tremendous amount!), I don't think questioners should be prompted to provide sufficient background for their question so as to make it comprehensible to a general mathematical audience (or to an undergraduate, no matter how brilliant and knowledgeable). That's what tags are for, and that's what experts are for. MO was all over the question as written, and to someone who knows about modular forms it was unimpeachably clear. –  Pete L. Clark Jan 13 '10 at 22:35
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5 Answers

up vote 16 down vote accepted

I will replace modular form by cuspform in your question, just to avoid the trivial answer of yes, because of the Eisenstein series. Given this, numerical evidence suggests that the eigenforms of weight $k$ are all conjugate (in the sense that their $q$-expansions are all algebraically conjugate under the action of $Gal(\overline{\mathbb Q}/{\mathbb Q})$). If this were the case, then the answer to your question would be no. Since this question is open, I'm pretty sure the answer to your question is not known.

There is a lot of evidence for this conjugacy statement, but the only suggestion I know for why it's true is essentially that it's the simplest possibility, given that no reason is known for it to be false (in contrast to the case of say fixing the weight to be 2, but letting the level grow; in that case elliptic curves give rise to integral eigenforms, and there are elliptic curve of arbitrarily high conductor).

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Let me remark that in the case of Sp_4 there is this terrifying example due to Skoruppa: some of the cusp forms are Saito-Kurokawa lifts and these naturally form a Hecke-stable subspace, but if you let k get big enough then eventually you see some isobaric forms appearing, and one might conjecture that these forms give you an irreducible Hecke module at level 1. The first time the space is 2-dimensional, it's a sum of two 1-dimensional Hecke-stable subspaces!! And then after that every space he computed was irreducible as a Hecke module. –  Kevin Buzzard Jan 13 '10 at 21:33
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The statement that the Hecke algebra acts irreducibly on $S_k(\Gamma(1))$ is known as Maeda's conjecture, and it is still open. So an affirmative answer to your question about eigenforms with integer coefficients would provide a negative answer to Maeda's conjecture. The negation of your question--that there are only finitely many integral eigenforms of level 1--is a weaker form of Maeda's conjecture, but one which nevertheless seems very hard to me.

There's lots of computational evidence in support of Maeda's conjecture, as Google will reveal. For instance, I don't think there is a weight $k$ known for which the operator $T_2$ has reducible characteristic polynomial (let alone has a linear factor).

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Even more: there is a conjecture of Maeda that every single T_p acts irreducibly on the space of cuspforms -- and more still, that the characteristic polynomial of T_p on the space S_k(1) has as Galois group the full symmetric group on dim S_k(1) letters! Whether there's good evidence for this conjecture I can't really say; as Emerton says, it's the simplest thing that could happen and no one can think of a good reason it shouldn't be that way.

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oops, crossed answers with Jared! –  JSE Jan 13 '10 at 19:05
    
I checked this statement for T_2 and k<=2048. I talked to Hida about it though and my impression was that he was skeptical. He said (and I never understood this comment so feel free to fill me in) that S_k(1;Q) being irreducible as a Hecke module was related to (equivalent to?) a certain L-value not vanishing, and L-values tend to vanish occasionally when you look hard enough. –  Kevin Buzzard Jan 13 '10 at 21:30
    
I think when the justification for a conjecture is along the lines of "For each k, it seems quite unlikely that anything else would happen," and the evidence is "I checked lots of k and nothing else happened," I think a better conjecture is "It happens except for finitely many k" or even "It happens with probability 1 as k -> infty." I have no insight as to Hida's remark and hope someone else will chime in! –  JSE Jan 14 '10 at 1:33
    
Re my comment above: please see Idoneal's comment for an excellent unpacking of Hida's remark. –  JSE Jan 14 '10 at 16:44
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This is a (far too long) comment on Buzzard's comment about Hida's remark.

I think I can guess what Hida was saying. He was probably talking about non-vanishing of L-functions of Hecke eigenforms of level one and weight $k \equiv 0$ (mod 4). This is a long-standing (folklore? ) conjecture in its own right, well-known among analytic number-theorists.

Here is how such a thing can be proven using Maeda's conjecture. There is a result of Shimura that says that the Galois group acts nicely on the central values (in fact any critical value) L-function of eigenforms. In particular, if one of them is zero then all the Galois twists are also zero and hence their sum is also zero. Now, even though it may be difficult to show that an L-function doesn't vanish at the centre, it is often easy to show that the sum of the central values of L-functions in a family is non-zero (see, for example, the work of Rohrlich and Rodriguez-Villegas on non-vanishing of L-functions of Hecke characters).

In the case in question, Maeda's conjecture will imply that if one central L-value is zero then the sum of all the central L-values over the whole basis must be zero and I think a contradiction will ensue after one uses the approximate functional equation to write the central value in terms of the Fourier coefficients and then using the Petersson formula ( I need to check this up).

Note 1: There is an article by Conrey and Farmer titled "Hecke operators and nonvanishing of L-functions" (Ahlgreen et al. (eds.), Topics in Number Theory, 1999) where they prove the above mentioned result along a different line.

Note 2: I think the following is easier. One can think of $f\rightarrow L(f,k/2)$ as a linear functional on the space of cusp forms $S_k(\Gamma(1))$ and indeed it is possible to explicitly write a function $G$ such that

$L(f,k/2)=\langle f,G \rangle$

for all Hecke eigenform $f$ in $S_k(\Gamma(1))$. Now Maeda + Shimura's result will imply that $G$ is orthogonal to the whole space and therefore zero. So it is just a matter of checking that $G$ is not identically zero, which shouldn't be too hard.

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Yes, I'm sure this is what Hida meant! Indeed, this was always the context in which people like Sarnak would talk about the conjecture. –  JSE Jan 14 '10 at 16:44
    
Ooh I only just spotted this answer. Hey thanks a lot Idoneal! I love the way that when a question is raised at MO the resulting comments might enlighten far more than just the questioner. –  Kevin Buzzard Jan 20 '10 at 22:12
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"He said (and I never understood this comment so feel free to fill me in) that S_k(1;Q) being irreducible as a Hecke module was related to (equivalent to?) a certain L-value not vanishing, and L-values tend to vanish occasionally when you look hard enough."

I dispute the impression of Hida with vanishing L-values. To precise this, a density statement is needed. The standard L-function technology whizzes from random matrices should expect that it doesn't vanish ever. In the same vein, Conrey conjectures that quadratic twists of weight 6+ forms never vanish aside from sign, though he kindly phrases it as "finitely many" as pointed out above.

http://www.aimath.org/~aimath/WWN/rmtapplications/rmtapplications.pdf

For weight 6 we have rank 2 vanishing for a few forms, as Dummigan lists: 95k6, 122k6, 260k6.

http://neil-dummigan.staff.shef.ac.uk/dsw_13.dvi

I expect no vanishing for weight 8+. To my knowledge, no rank 3 vanishing exists for weight 4+. My recollection (Stein 2000) is that, outside with Gamma1(N), there is one at level 122 (sic, as above) weight 2 form with quadratic sign that vanishes to order 1 with no self-dual functional equation sign (eps = -0.76822128 + 0.6401844i).

I am editing this now to explain L-function methods. The right random matrix idea is that L-values have cumulative distribution with $\sqrt t$ for small $t$. It is probably unnecessary though.

For rather look at the BSD analogue. There is $L(centre)/\Omega$ and the other side is up to few rational factors an integer. It is also a square. So it is "like" a random integral square up to size $\Omega$ as the Tamagawa and torsion and much smaller. The "probability" of an (even signed) L-function vanishing centrally can be thought as $\sqrt\Omega$ as a chance that a random integral square up to size $\Omega$ is 0 is just 1 in $\sqrt\Omega$.

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