Timeline for Odd powers of the theta function as eigenforms
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 11, 2015 at 18:02 | history | edited | Peter Humphries | CC BY-SA 3.0 |
added tag
|
| Apr 25, 2011 at 19:34 | comment | added | GH from MO | Underflow, just a very quick answer to your second question: up to constant multiples there are only finitely many eigenforms in your space, and these can be found in the same way as you would diagonalize a matrix which can be diagonalized. Again by finite dimensionality, a form in your space equals any of these eigenforms iff the first few Hecke eigenvalues match for the two forms. So you have an algorithm for sure. Implementing it properly is another matter. | |
| Apr 25, 2011 at 19:20 | answer | added | Dror Speiser | timeline score: 9 | |
| Jul 30, 2010 at 15:49 | comment | added | L. J. P. Kilford | This isn't an answer really either, but there is a paper of Ghate (journals.impan.pl/cgi-bin/doi?aa102-1-3) which gives conditions on when the product of two eigenforms both of integral weight can again be an eigenform, and essentially the answer is as Buzzard says; almost always only when forced by dimension considerations. | |
| Jul 30, 2010 at 14:58 | comment | added | Emerton | I don't know how much it will help you for this question, but an interesting reference about half-integral weight modular forms and theta series, written from a classical point of view ($q$-expansions and so on) is the paper of Stark and Serre in one of the Antwerp volumes. | |
| Jul 30, 2010 at 12:25 | comment | added | Kevin Buzzard | I don't know a theorem about powers of theta. Here might be one way to go about it: the power of theta will necessarily be an Eisenstein series, and specific formulae for these will surely be known. Check that for a sufficiently big power, the coefficient of $q^2$ isn't what it is supposed to be. This might be ugly and might even be so ugly as to be unworkable, but it might work. As for being an eigenform for all Hecke operators---you'll certainly only have to check for $p$ up to a certain point because the space is finite-dimensional, right? But I don't know what that point is :-/ | |
| Jul 30, 2010 at 11:04 | comment | added | Underflow | Yes, I agree with the reasoning behind the guess, but I wondered if there is a theorem. It seems to happen for weight 3/2, 5/2, and 7/2 but then no more. For the last part of the question I should have specified eigenform for ALL of the Hecke operators. Is it true that I only have to check that it works for T(p^2) for p up to a certain point? | |
| Jul 30, 2010 at 10:15 | comment | added | Kevin Buzzard | If you want a guess then here we go: the moment the space of modular forms of level 4 (or 1 or whatever your convention is) and weight n+1/2 has dimension bigger than one, why would you expect a random product of eigenforms to be an eigenform? So I would guess that it would only happen for very small powers. As for an algorithm---given a modular form, hit it with a Hecke operator and compare q-expansions. You'll quickly guess the eigenvalue from the first few terms, if it's an eigenform. The Sturm bound tells you when you have to stop checking q-expansion coeffts. Did I misunderstand? | |
| Jul 30, 2010 at 9:46 | history | asked | Underflow | CC BY-SA 2.5 |