User l. j. p. kilford - MathOverflow

Answer by L. J. P. Kilford for Finitely many spaces generated by eta-products

2012-03-12T23:28:15Z

There's a brief answer to this on page 3 of the paper; there are only finitely many eta-products with q-valuation 1 (one can write them all down), these all have a given level, so other levels will have a form with q-valuation 1 which can't be written as an eta-product. Note that I defined eta-products to be products of eta-functions with non-negative exponents.

Fields of definition for p-adic overconvergent modular eigenforms

2010-08-19T16:11:36Z

If we consider the action of the $U_p$ operator on overconvergent $p$-adic modular forms, then we can get some information about the field over which the eigenforms are defined by looking at the slopes. For instance, my paper in Math Research Letters (MR2106238) proves that the slopes of $U_2$ acting on 2-adic overconvergent modular forms of level 4 with primitive Dirichlet character are distinct, so the field of definition has to be $\mathbf{Q}_2$. However, there are cases when the slopes fail to be distinct; for instance, in Emerton's thesis it is proved that the lowest slopes of T_2 acting on level 1 forms of weight congruent to 14 modulo 16 are 6 and 6.

For classical modular forms of level 1, we have Maeda's Conjecture which says that the field of definition is essentially as large as it can be; the Hecke polynomial is irreducible with Galois group $S_n$ where $n$ is the dimension. However, there is no reason that this should be true for overconvergent modular forms, and in fact it isn't. Discussions with Robert Coleman led me to the concrete example of 2-adic overconvergent modular forms of tame level 1 and weight 142, where there are two eigenforms of slope 6 which are both defined over the ground field $\mathbf{Q}_2$.

The question is, what should one expect here? Can one tell any more about the field of definition from the slopes than the absolute minimum?

Finding zeroes of classical modular forms

2010-05-06T16:58:44Z

There are several papers which compute zeroes of modular forms for genus 0 congruence subgroups, such as "Zeros of some level 2 Eisenstein series" by Garthwaite et al published in Proc AMS and work of Shigezumi and others in levels 3,5 and 7. However, there don't seem to be generalizations of this to higher genus subgroups.

I know a few examples of modular forms for higher genus subgroups where one can compute all the zeroes; for instance, the unique normalized cusp form of weight 1 and level $\Gamma_0(31)$ with character the Legendre character modulo 31 has simple zeroes at the two cusps and the two elliptic points because the valence formula forces them to be there. Similar ideas work for levels 17, 19, 21 and 39.

My question is this: is there a more general way to find the zeroes of modular forms in an explicit way for congruence subgroups?

Answer by L. J. P. Kilford for What out-of-print books would you like to see re-printed?

2010-07-06T08:51:33Z

Mathematics Made Difficult, by Carl Linderholm. A great underground classic.

Comment by L. J. P. Kilford

2010-07-30T15:49:54Z

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.