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Let $\bar{\rho} : Gal(\bar{\mathbb{Q}}/ \mathbb{Q} ) \rightarrow GL_2(\bar{\mathbb{F}}_p)$ be an odd, irreducible Galois representation mod $p$ which is unramified outside $S$, where $S$ is a finite set of primes which contains $p$. Fix an integer $k \geq 2$ and a local Galois representation $\rho _p ' : Gal(\bar{\mathbb{Q}} _p/ \mathbb{Q} _p ) \rightarrow GL_2(\bar{\mathbb{Q}} _p)$

Question: Is there a way to compute precisely a number (which is finite) of modular lifts $\rho : Gal(\bar{\mathbb{Q}}/ \mathbb{Q} ) \rightarrow GL_2(\bar{\mathbb{Q}}_p)$ of $\bar{\rho}$, such that

1) $\rho _{|Gal(\bar{\mathbb{Q}} _p/ \mathbb{Q} _p )} \simeq \rho _p '$

2) $\rho$ comes from a modular form of weight $k$.

3) $\rho$ is unramified outside $S$.

I'd like to understand it even in the simplest (?) case, when $S =$ {$p$ }, $k=2$ so that $\bar{\rho}$ actually comes from a reduction of some level 1 form (after Serre's conjecture)

Is it true that in this case, a modular lift $\rho$ will be unique (if it exists)?

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2 Answers

up vote 4 down vote accepted

As Kevin said, I'm not sure if you can get a formula in any concrete sense, but there is a way to tell if the modular form that gives rise to $\bar\rho$ is the unique form (of a specific level). Say that $\bar\rho$ takes values in $GL_2(k)$ and that $f \in S_k(N,\mathcal{O})$ gives rise to $\bar\rho$ (so we have that $k$ is the residue field of $\mathcal{O}$). Then you can show (see for instance Section 4.1 of Darmon, Diamond, Taylor- Fermat's Last Theorem) that the number of forms $g$ which are congruent to $f$ mod $p$ (i.e. $f$ and $g$ have the same residual representation), is equal to the rank (as an $\mathcal{O}$-algebra) of a certain completed Hecke ring, $\mathbb{T}$. In particular, if this rank is 1, then $f$ is the unique form which gives rise to $\bar\rho$.

Thanks to the general $R=T$ theorems, one can study the rank of $\mathbb{T}$ by studying the deformation theory of the representation $\bar\rho$. From this point of view, the rank is something you can almost get your hands on (and by "almost", I mean, "isn't completely hopeless"). That's because it is a standard fact that $R$ is a quotient of a power series ring over $\mathcal{O}$ in $d$ variables, where $d$ is the dimension of specific subgroup $H\subset H^1(G_S, ad^0\bar\rho)$. This subgroup $H$ is a so-called "Selmer group" since it is the kernel of a global-to-local map on cohomology, defined by specifying local conditions at all primes in $S$.

There are lots of choices of local conditions, and each choice will lead to a different Selmer group. You can read "Deforming Galois representations and the conjectures of Serre and Fontaine-Mazur" by Ravi Ramakrishna for local conditions that deal with "modular" deformations (the same conditions appear in "On icosahedral Artin representations, II" by Richard Taylor). Now, since you have a local-at-$p$ representation in mind, you might have to alter these local conditions a bit (but not too much).

The point though, is that, if you can show that $H=0$, then we have that $R$ is a quotient of of a power series ring over $\mathcal{O}$ in 0 variables. Since completed Hecke rings are known to be finite flat complete intersections, we conclude that $R = \mathbb{T} = \mathcal{O}$ when $H=0$. As mentioned above, the $\mathcal{O}$-rank of $\mathbb{T}$ corresponds to the number of modular forms giving rise to $\bar\rho$. Thus, we get to the punchline:

$f$ is the unique modular form (of weight $k$ and level $N$) giving rise to $\bar\rho$ if and only if $H=0$.

The natural question to ask now is, "When is $H=0$?" This, of course, is the hard part. If you are willing to allow finitely many additional primes into your ramification set, then using the methods of Ramakrishna (Op. cit.) you can force $H=0$. If you want a modular form of optimal Serre conductor, then I don't know if anybody knows how to do this.

Studying whether or not $H =0$ as you vary the set $S$ is the subject of a paper of mine with Ramakrishna (whose contents are basically my thesis), "New Parts of Hecke Rings". You can look at that for ideas on how to determine (in specific settings) if $H=0$.

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Thank you very much! That's more or less what I need in my case. –  Przemyslaw Chojecki Feb 9 '12 at 10:55
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I can give you a "formula" in the sense that I can give you an algorithm to compute the number in any given case. If $\ell\not=p$ is prime then an old result of Carayol and Livn\'e says that the conductor of a lift $\rho$ of $\overline{\rho}$ at $\ell$ can only jump by a factor of $\ell$ or $\ell^2$. The conductor $N_p$ at $p$ can be read off from $\rho_p'$. So here's the algorithm: let $N$ be $N_p.cond(\overline{\rho}).\prod_{\ell\in S}\ell^2$, and now run through the finite set of newforms of level dividing $N$ and weight $k$, and for each such form check to see whether the associated Galois representation has the right properties. Now just count the number of times that it worked.

This might not be what you wanted for a "formula" but I would be very surprised if you could get anything more concrete than this. Even in the simple case of dihedral lifts you're asking for something like the dimension of certain global cohomology groups generalising class groups: even deciding if the answer is 0 or positive will depend on some delicate global cohomology groups. A question very closely related to yours is this: I'll give you a $p$-adic number $a_p$ with $|a_p|<1$, and a finite set of positive integers $X$; what is a "formula" for the number of weight 2 normalised newforms with level in $X$ and coefficient of $q^p$ equal to $a_p$? (the point being that $k$ and $a_p$ determine the local crystalline lift). I say: just count 'em, because you'll surely not get any better than this.

As for the "simple" case -- why would you expect a lift to be unique? But I can't answer the question. On the other hand perhaps the following might give you problems: perhaps one can find some large $p$ and a weight 2 modular form of level $p^3$ or whatever, with coefficients in some big number field, but with the property that the associated local $p$-adic Galois representation is defined over a smaller $p$-adic field. Then perhaps you can just look at two Galois conjugates of the form that are globally non-isomorphic but which both happen to satisfy your conditions. I can't immediately rule this phenomenon this out...

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Thank you for an answer. Indeed, I'd like to have more "theoretic" rather than "algorithmic" formula, because in general, I'm more interested in the question of whether there exists a unique modular lift. As for the uniqueness in the "simple" case, I can't well describe while I suspect it to be like it. It comes from a completely different problem and it'd simplify many things if it'd be like that. –  Przemyslaw Chojecki Feb 8 '12 at 0:04
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