Are there $p$-adic modular forms for non-congruence subgroups? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T18:52:04Z http://mathoverflow.net/feeds/question/104535 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104535/are-there-p-adic-modular-forms-for-non-congruence-subgroups Are there $p$-adic modular forms for non-congruence subgroups? Ramsey 2012-08-12T03:44:19Z 2013-05-02T23:22:00Z <p>My answer to the question:</p> <p><a href="http://mathoverflow.net/questions/104362/moduli-interpretations-for-modular-curves" rel="nofollow">http://mathoverflow.net/questions/104362/moduli-interpretations-for-modular-curves</a></p> <p>led me to wonder about the question in the present title.</p> <p>It seems that modular forms for non-congruence subgroups may not "arise" from algebraic geometry in the same way modular forms for congruence subgroups do (as is evidenced by the lack of a good moduli problem and this whole unbounded denominators thing). Nonetheless, they are clearly complex-analytic objects. I wonder if there's a $p$-adic analytic analog to be had.</p> <p>Here are two more precise questions:</p> <ol> <li><p>(less ambitious) Are there interesting congruences to be had between such modular forms? Interesting $p$-adic limits? Interesting $p$-adic families?</p></li> <li><p>(more ambitious) Is there some (perhaps inherently analytic) description of non-congruence forms in terms of moduli of elliptic curves that can be mimicked in the $p$-adic analytic setting?</p></li> </ol> <p>My rather vague feeling is that the Atkin and Swinnerton-Dyer congruences suggest that there's something to be said here, but I haven't been able to dig up much on these questions in particular. Does anyone know of such work?</p> http://mathoverflow.net/questions/104535/are-there-p-adic-modular-forms-for-non-congruence-subgroups/128025#128025 Answer by Will Chen for Are there $p$-adic modular forms for non-congruence subgroups? Will Chen 2013-04-18T23:07:26Z 2013-04-18T23:07:26Z <p>Perhaps I'm not understanding what you're asking, but there's a paper of Scholl's (Modular Forms and de Rham Cohomology) where he proves a version of the ASD congruences using some relatively sophisticated machinery.</p> <p>Specifically, if $\Gamma$ is any finite index subgroup of $\text{SL}_2(\mathbb{Z})$, and suppose $k > 2$ is even, then if $d = \text{dim }S_k(\Gamma)$, then there exists integers $A_0,\ldots,A_{2d}$ such that $A_0 = 1$, $A_{2d} = p^{(k-1)d}$, for which we have the congruence:</p> <p>$$a_{np^d}(f) + A_1a_{np^{d-1}}(f) + \cdots + A_da_n(f) + A_{d+1}a_{n/p}(f) + \cdots + A_{2d}a_{n/p^d}(f)\equiv 0\mod p^{(k-1)(1+\text{ord}_p n)}$$</p> <p>If $d = 1$, then this is exactly the ASD congruence. The $A_i$'s are actually the coefficients of the characteristic polynomial of Frobenius acting on a certain $2d$-dimensional $\ell$-adic sheaf.</p>