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Let $\ell$ be a prime number, denote by $K_\ell$ the maximal algebraic extension of $\Bbb{Q}$ ramified only at $\ell$. Let $f = \sum a_n q^n$ be a Hecke eigenform of level $1$ with integer coefficients. By a theorem of Serre and Deligne, there is a continuous homomorphism $$ \rho_\ell: \mathrm{Gal}(K_\ell / \Bbb{Q}) \to GL_2(\Bbb{Z}_\ell) $$ such that $\rho_\ell(\mathrm{Frob}_p)$ has characteristic polynomial $$ x^2 - a_p x + p^{k-1} $$ for each $p \neq \ell$. This induces the mod $\ell$ representation $\tilde{\rho}_\ell: \mathrm{Gal}(K_\ell / \Bbb{Q}) \to GL_2(\Bbb{F}_\ell)$.

Question: How to find the conductor of $\tilde{\rho}_\ell$?

We know that the conductor is a power of $\ell$ since $K_\ell$ is ramified only at $\ell$. But I don't know if we have control on the exponent.

We've looked at section 5.5 of the notes by Bryden Cais on Serre's Conjectures (http://math.stanford.edu/~conrad/vigregroup/vigre05/Serre05.pdf), which seems to be related to my question, but I'm not familiar with the topic enough to understand the relation.

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Did you mean to ask about the conductor of the mod $\ell$ representation, instead of $\ell$--adic? –  Jesse Silliman Jul 18 '13 at 13:14
    
Yes, I mean the mod $\ell$ representation. Thanks. –  Ping Ngai Chung Jul 18 '13 at 13:28
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The $\ell$-part of the conductor of the mod $\ell$-representation comes out of the recipe for the weight in Serre's conjecture. A special case is that the representation is unramified at $\ell$ if and only if the weight of the representation is $1$. This should be in Cais's notes, or in Edixhoven's paper Inv. Math. 109 (1992) 563-594. –  Felipe Voloch Jul 18 '13 at 14:54
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1 Answer 1

It is difficult to even define the conductor at $\ell$. The problem is that rho is infinitely wildly ramified at $\ell$ (it must be, since its determinant is a power of the $\ell$-adic cyclotomic character); so the naive definition of the conductor would be infinity.

The morally right way to define the conductor at p of a p-adic Galois representation is to use Fontaine's p-adic Hodge theory. In your case the representation is crystalline at $\ell$, so the conductor is 1, matching the level of the modular form.

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Oh I'm sorry. I mean the mod $\ell$ representation (edited above) instead of the $\ell$-adic one. Sorry for the typo. –  Ping Ngai Chung Jul 18 '13 at 13:30
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