User l. j. p. kilford - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T10:56:00Z http://mathoverflow.net/feeds/user/4555 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90983/finitely-many-spaces-generated-by-eta-products/91036#91036 Answer by L. J. P. Kilford for Finitely many spaces generated by eta-products L. J. P. Kilford 2012-03-12T23:28:15Z 2012-03-12T23:28:15Z <p>There's a brief answer to this on page 3 of the paper; there are only finitely many eta-<em>products</em> 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.</p> http://mathoverflow.net/questions/36091/fields-of-definition-for-p-adic-overconvergent-modular-eigenforms Fields of definition for p-adic overconvergent modular eigenforms L. J. P. Kilford 2010-08-19T16:11:36Z 2010-08-19T20:46:03Z <p>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 (<a href="http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;co4=AND&amp;co5=AND&amp;co6=AND&amp;co7=AND&amp;dr=all&amp;pg4=AUCN&amp;pg5=RVCN&amp;pg6=ALLF&amp;pg7=ALLF&amp;pg8=ET&amp;review_format=html&amp;s4=kilford&amp;s5=&amp;s6=&amp;s7=&amp;s8=All&amp;vfpref=html&amp;yearRangeFirst=&amp;yearRangeSecond=&amp;yrop=eq&amp;r=9&amp;mx-pid=2106238" rel="nofollow">MR2106238</a>) 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.</p> <p>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 $\mathbf{Q}_2$.</p> <p>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?</p> http://mathoverflow.net/questions/23747/finding-zeroes-of-classical-modular-forms Finding zeroes of classical modular forms L. J. P. Kilford 2010-05-06T16:58:44Z 2010-08-16T04:47:01Z <p>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.</p> <p>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.</p> <p>My question is this: is there a more general way to find the zeroes of modular forms in an explicit way for congruence subgroups?</p> http://mathoverflow.net/questions/18271/what-out-of-print-books-would-you-like-to-see-re-printed/30743#30743 Answer by L. J. P. Kilford for What out-of-print books would you like to see re-printed? L. J. P. Kilford 2010-07-06T08:51:33Z 2010-07-06T08:51:33Z <p>Mathematics Made Difficult, by Carl Linderholm. A great underground classic.</p> http://mathoverflow.net/questions/33905/odd-powers-of-the-theta-function-as-eigenforms Comment by L. J. P. Kilford L. J. P. Kilford 2010-07-30T15:49:54Z 2010-07-30T15:49:54Z This isn't an answer really either, but there is a paper of Ghate (<a href="http://journals.impan.pl/cgi-bin/doi?aa102-1-3" rel="nofollow">journals.impan.pl/cgi-bin/doi?aa102-1-3</a>) 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.