Examples of q-expansions in a Hida family - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:03:27Z http://mathoverflow.net/feeds/question/106498 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106498/examples-of-q-expansions-in-a-hida-family Examples of q-expansions in a Hida family Victor Rotger 2012-09-06T11:10:31Z 2012-09-09T01:53:06Z <p>Let $p$ be a prime number and $N$ a positive integer not divisible by $p$. </p> <p>For some easy choices of $p$ and $N$, can anybody provide me with explicit examples of collections $${f_k,\quad 2\leq k \leq k_0 }$$ of q-expansions such that, for each $k$, $f_k$ is the $q$-series of a classical eigenform of weight $k$ and level $Np$, all being members of an ordinary $p$-adic Hida family of tame level $N$? Here $k_0$ is some reasonable upper bound for the weight, and I leave to the reader the meaning of the term "reasonable". Even if $k_0=4$ I will already be happy, and it is also fine if $k$ only runs among some of the integers between $2$ and $k_0$. But of course, the cardinal of ${ f_k}$ should be greater than $1$!</p> <p>I am interested in Hida families which are neither Eisenstein not CM, as explicit examples of those are already well-known.</p> <p>The simplest example seems to arise when one takes $N=1$ and $p=11$. In this case the associated Hida Hecke algebra is $\mathbf{T}=\Lambda:=\mathbb{Z}_p[[T]]$. I guess that people like Kevin Buzzard, Robert Pollack, William Stein and many others have computed the $q$-expansions $f_k$ in this case, up to some reasonable $k_0$, and I would be happy to have access to these computations.</p> <p>Yet another interesting example: reading the paper "How can we construct abelian Galois extensions of basic number fields?" by Barry Mazur, he reports on an example computed by Citro and Stein, where $N=1$ and $p=691$. They compute explicitly ${f_2,f_{12}}$ where $f_{12}$ is the $p$-stabilization of the discriminant modular form $\Delta$ and $f_2$ turns out to be the only weight 2 newform of level $p$ and character $\omega^{-10}$, where $\omega$ is the Teichmuller character. I would like to know the $q$-expansions $f_k$ in this case for some other integers in between $2$ and $12$. (For integers $k$ larger than $12$ one can possibly compute $f_k$ by multiplying $f_{12}$ by a suitable Eisenstein series of weight $k-12$ and applying Hida's ordinary projector.) </p> <p>And of course, any other collection of examples is welcome!</p> http://mathoverflow.net/questions/106498/examples-of-q-expansions-in-a-hida-family/106509#106509 Answer by David Loeffler for Examples of q-expansions in a Hida family David Loeffler 2012-09-06T13:54:06Z 2012-09-06T13:54:06Z <p>Here is a Sage session which computes, approximately, the (unique) 13-adic cuspidal Hida family of tame level 1 over the identity component of weight space, evaluated at weights 0, 12 and 24. (This uses an explicit parametrization of $X_0(13)$, so it only works because $X_0(13)$ has genus 0.)</p> <blockquote> <p>sage: OverconvergentModularForms(13, 0, 0, Qp(13), prec=200).eigenfunctions(14)[1].q_expansion() * (1 + O(13^5))</p> <p>(1 + O(13^5))*q + (2 + 11*13 + 13^2 + 12*13^3 + 3*13^4 + O(13^5))*q^2 + (5 + 5*13 + 10*13^2 + 12*13^3 + 11*13^4 + O(13^5))*q^3 + (10 + 11*13 + 4*13^2 + 8*13^3 + 3*13^4 + O(13^5))*q^4 + (7 + 12*13 + 11*13^2 + 8*13^3 + 6*13^4 + O(13^5))*q^5 + (10 + 7*13^2 + 10*13^3 + 7*13^4 + O(13^5))*q^6 + (8*13 + 6*13^2 + 5*13^3 + 2*13^4 + 11*13^5 + O(13^6))*q^7 + (6 + 4*13 + 5*13^2 + 13^3 + 7*13^4 + O(13^5))*q^8 + (3 + 4*13 + 3*13^2 + 13^4 + O(13^5))*q^9 + (1 + 11*13 + 12*13^2 + 11*13^3 + 8*13^4 + O(13^5))*q^10 + (12*13 + 6*13^2 + 4*13^3 + 12*13^4 + 5*13^5 + O(13^6))*q^11 + (11 + 4*13 + 13^2 + 5*13^3 + 9*13^4 + O(13^5))*q^12 + (8 + 10*13 + 11*13^2 + 9*13^3 + 11*13^4 + O(13^5))*q^13 + O(q^14)</p> <p>sage: OverconvergentModularForms(13, 12, 0, Qp(13), prec=200).eigenfunctions(14)[1].q_expansion() * (1 + O(13^5))</p> <p>(1 + O(13^5))*q + (2 + 11*13 + 12*13^2 + 12*13^3 + 12*13^4 + O(13^5))*q^2 + (5 + 6*13 + 13^2 + O(13^5))*q^3 + (10 + 3*13 + 4*13^2 + 12*13^3 + 12*13^4 + O(13^5))*q^4 + (7 + 7*13 + 2*13^2 + 2*13^3 + O(13^5))*q^5 + (10 + 2*13 + 3*13^2 + 10*13^3 + 12*13^4 + O(13^5))*q^6 + (12*13 + 4*13^2 + 5*13^3 + 12*13^4 + 12*13^5 + O(13^6))*q^7 + (6 + 11*13 + 5*13^2 + 12*13^3 + 2*13^4 + O(13^5))*q^8 + (3 + 7*13 + 3*13^2 + 9*13^4 + O(13^5))*q^9 + (1 + 13 + 3*13^2 + 12*13^3 + 8*13^4 + O(13^5))*q^10 + (5*13 + 4*13^2 + 9*13^3 + 5*13^4 + 13^5 + O(13^6))*q^11 + (11 + 2*13^2 + O(13^5))*q^12 + (8 + 5*13 + 10*13^3 + 5*13^4 + O(13^5))*q^13 + O(q^14)</p> <p>sage: OverconvergentModularForms(13, 24, 0, Qp(13), prec=200).eigenfunctions(14)[1].q_expansion() * (1 + O(13^5))</p> <p>(1 + O(13^5))*q + (2 + 11*13 + 6*13^2 + 2*13^3 + 13^4 + O(13^5))*q^2 + (5 + 7*13 + 9*13^2 + 6*13^3 + 13^4 + O(13^5))*q^3 + (10 + 8*13 + 2*13^2 + 12*13^3 + 9*13^4 + O(13^5))*q^4 + (7 + 2*13 + 10*13^2 + 4*13^3 + 5*13^4 + O(13^5))*q^5 + (10 + 4*13 + 4*13^3 + 11*13^4 + O(13^5))*q^6 + (3*13 + 11*13^2 + 8*13^3 + 12*13^4 + 8*13^5 + O(13^6))*q^7 + (6 + 5*13 + 9*13^2 + 6*13^3 + 12*13^4 + O(13^5))*q^8 + (3 + 10*13 + 2*13^2 + 10*13^3 + 8*13^4 + O(13^5))*q^9 + (1 + 4*13 + 12*13^2 + 7*13^3 + 6*13^4 + O(13^5))*q^10 + (11*13 + 13^2 + 4*13^3 + 2*13^4 + 5*13^5 + O(13^6))*q^11 + (11 + 9*13 + 8*13^2 + 10*13^3 + O(13^5))*q^12 + (8 + 7*13^2 + 11*13^3 + 7*13^4 + O(13^5))*q^13 + O(q^14)</p> </blockquote> <p>This uses some code I wrote during the Harvard eigenvarieties semester back in 2006; it's included in all recent Sage versions. It'll compute eigenforms of small non-zero slope as well if you like (but you get less and less precision as the slope gets large).</p> http://mathoverflow.net/questions/106498/examples-of-q-expansions-in-a-hida-family/106549#106549 Answer by Robert Pollack for Examples of q-expansions in a Hida family Robert Pollack 2012-09-06T21:42:14Z 2012-09-07T03:34:10Z <p>In the case when the Hida algebra is simply $\Lambda$, one can use <em>families</em> of overconvergent modular symbols to compute the $q$-expansion of the Hida family where the coefficients are functions of the weight. </p> <p>The idea is the following: form a "random" family of overconvergent modular symbols -- that is modular symbols with values in a distribution module tensor power series in the weight. Then iterate $U_p$ on this family until you are in the ordinary subspace (modulo the accuracy at which you are working). Since the Hida algebra is just $\Lambda$, the result will be an eigensymbol. Its Hecke-eigenvalues will then be functions of the weight and these are the coefficients of the formal $q$-expansion I mentioned above.</p> <p>With this $q$-expansion in hand, just plug in your favorite weight and you'll get the form in the Hida family of that weight (of course, computed modulo your fixed accuracy). </p> <p>This approach actually came out of an Arizona Winter School 2011 student project. Currently three graduate students at University of Madison (Lalit Jain, Marton Hablicsek, and Daniel Ross), together with Rob Harron and myself, are writing up a paper on this. </p> <p>Our code is still very much in beta form, but hopefully this example works. Take $p=3$ and $N=5$. Then the first few Hecke-eigenvalues (mod $3^3$) of the associated Hida family are:</p> <p>$$a_2(w) = 12w^2 + 7w - 1,$$ $$a_3(w) = 21w^2 + 11w - 1,$$ and $$a_5(w) = 16w^2 + 17w + 1.$$ Here $w$ is not exactly the weight variable. Instead, to specialize these eigenvalues to weight $k$, one simply sets $w = 4^{k-2} - 1$. (Here is 4 is a topological generator of $1+3{\mathbb Z}_3$.)</p> <p>Here's the example with $N=1$ and $p=11$ -- below I'm working modulo $11^8$:</p> <p>$$a_2(w) = 37w^7 + 880w^6 + 12388w^5 + 151975w^4 + 385840w^3 + 10344442w^2 + 40094463w + 857435522,$$ $$a_3(w) = 71w^7 + 681w^6 + 8325w^5 + 94314w^4 + 220797w^3 + 2758794w^2 + 16210985w + 428717761,$$ and $$a_5(w) = 12w^7 + 472w^6 + 2245w^5 + 44445w^4 + 30443w^3 + 14355835w^2 + 180294533w + 1.$$ As before, setting $w = (1+11)^{k-2}-1$ specializes to weight $k$. For instance, taking $k=12$ one should recover the first few Fourier coefficients of $\Delta$ modulo $11^8$.</p>