modular eigenforms with integral coefficients [Maeda's Conjecture] - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:18:02Z http://mathoverflow.net/feeds/question/11677 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11677/modular-eigenforms-with-integral-coefficients-maedas-conjecture modular eigenforms with integral coefficients [Maeda's Conjecture] norondion 2010-01-13T18:19:18Z 2012-09-22T04:41:40Z <p>Are there infinitely many (linearly independent) cuspidal eigenforms for $\Gamma(1)$ with integer coefficients?</p> <p>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 $k \geq 12$.</p> http://mathoverflow.net/questions/11677/modular-eigenforms-with-integral-coefficients-maedas-conjecture/11681#11681 Answer by Emerton for modular eigenforms with integral coefficients [Maeda's Conjecture] Emerton 2010-01-13T18:55:10Z 2012-09-22T04:41:40Z <p>I will replace <I> modular form</I> by <I>cuspform</I> in your question, just to avoid the trivial answer of <I>yes</I>, 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 <I>no</I>. Since this question is open, I'm pretty sure the answer to your question is not known.</p> <p>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). </p> http://mathoverflow.net/questions/11677/modular-eigenforms-with-integral-coefficients-maedas-conjecture/11685#11685 Answer by Jared Weinstein for modular eigenforms with integral coefficients [Maeda's Conjecture] Jared Weinstein 2010-01-13T19:04:30Z 2010-01-13T19:04:30Z <p>The statement that the Hecke algebra acts irreducibly on <code>$S_k(\Gamma(1))$</code> 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.</p> <p>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 <code>$T_2$</code> has reducible characteristic polynomial (let alone has a linear factor). </p> http://mathoverflow.net/questions/11677/modular-eigenforms-with-integral-coefficients-maedas-conjecture/11686#11686 Answer by JSE for modular eigenforms with integral coefficients [Maeda's Conjecture] JSE 2010-01-13T19:05:06Z 2010-01-13T19:05:06Z <p>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. </p> http://mathoverflow.net/questions/11677/modular-eigenforms-with-integral-coefficients-maedas-conjecture/11720#11720 Answer by Idoneal for modular eigenforms with integral coefficients [Maeda's Conjecture] Idoneal 2010-01-14T05:22:07Z 2010-01-14T08:15:49Z <p>This is a (far too long) comment on Buzzard's comment about Hida's remark.</p> <p>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. </p> <p>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). </p> <p>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). </p> <p>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. </p> <p>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 </p> <p>$L(f,k/2)=\langle f,G \rangle$ </p> <p>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. </p> http://mathoverflow.net/questions/11677/modular-eigenforms-with-integral-coefficients-maedas-conjecture/22055#22055 Answer by Junkie for modular eigenforms with integral coefficients [Maeda's Conjecture] Junkie 2010-04-21T13:18:37Z 2010-04-22T04:46:55Z <p>"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."</p> <p>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.</p> <p><a href="http://www.aimath.org/~aimath/WWN/rmtapplications/rmtapplications.pdf" rel="nofollow">http://www.aimath.org/~aimath/WWN/rmtapplications/rmtapplications.pdf</a></p> <p>For weight 6 we have rank 2 vanishing for a few forms, as Dummigan lists: 95k6, 122k6, 260k6.</p> <p><a href="http://neil-dummigan.staff.shef.ac.uk/dsw_13.dvi" rel="nofollow">http://neil-dummigan.staff.shef.ac.uk/dsw_13.dvi</a></p> <p>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).</p> <p>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.</p> <p>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 $\sqrt\Omega$.</p>