# Bounding the level for eigenforms satisfying a deformation condition

Let $k$ be a finite field of char $p \geq 3$. Given an absolutely irreducible, continuous, odd representation $\overline{\rho}: G_\mathbb{Q} \longrightarrow GL_2(k)$ and a deformation condition $D$ for $\overline{\rho}$, let $S(D)$ be the collection of all newforms with associated $p$-adic representation in $D$. If $f \in S(D)$ then is its level bounded? I remember reading somewhere that one might work out the level using local Langlands but do not recall the reference or the argument.

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For an arbitrary $D$ the answer is "no". For some specific $D$'s the answer is "yes"; it mainly depends on what local condition at $p$ you impose. What sort of $D$ did you have in mind? –  David Loeffler Sep 3 '13 at 15:39
Yes, I would require that $\overline\rho$ is ordinary at $p$ and $D$ consists of ordinary lifts with fixed determinant. –  unramified Sep 3 '13 at 16:53
Do you also fix the Hodge--Tate weights at $p$? Are you imposing any conditions at primes away from $p$? –  David Loeffler Sep 3 '13 at 18:05
The Hodge-Tate weights are $0$ and $k-1$ for some integer $k \geq 2$. Away from $p$ the lifts have fixed determinant (given by $k$.) –  unramified Sep 3 '13 at 18:15

Let's bound the level of such an $f$ in two stages. Firstly, let's look at a prime $\ell \ne p$. Here there is a theorem of Livne and (independently) Carayol which says that if $\rho$ is a lifting of $\bar\rho$, the exponent of $\ell$ dividing the Artin conductor of $\rho$ is bounded (it's at most 2 more than the $\ell$-conductor of $\bar\rho$). That leaves just the power of $p$ dividing the level to be controlled; and it's easy to see that if you require $\rho|_{D_p}$ to be upper-triangular, with fixed determinant and Hodge--Tate weights, then the conductors of the characters occuring along the diagonal are bounded above and this gives you a bound on the level of $f$ at $p$.
Under these stipulations, couldn't $\rho|D_p$ be of the form $0 \to \psi \to \rho \to \psi^{-1} \det \rho \to 0$ for any $\psi$ lifting the unramified line in $\overline{\rho}$? Such things don't have bounded conductor. –  David Hansen Sep 3 '13 at 20:39
Then $\rho$ won't be ordinary. Depending on your conventions, one of $\psi$ and $\psi^{-1} \det(\rho)$ needs to be an unramified character times a power of cyclotomic. –  David Loeffler Sep 4 '13 at 6:39