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I'm interested in knowing to what extent is level lowering known to hold in weight 1. Specifically, let's say I have an eigenform $f$ in $S_1(N,\chi)$ and a prime $p$ which doesn't divide $N$. Let's assume that the mod $p$ Galois representation attached to $f$ is unramified at some prime $q$ dividing $N$. Is there then an eigenform $g$ in $S_1(N/q,\chi)$ with the same mod $p$ Galois representation as $f$?

I think I can show this if $p>5$, but the argument relies on Artin's conjecture and feels like overkill. Let me sketch the argument, and then ask some more specific questions at the end.

Here's the argument: if $f$ is CM, check directly that no such $q$ can exist. So the projective image of $\overline{\rho}_f$ must be $A_4$, $S_4$, or $A_5$. In particular, if $p>5$ then one can lift $\overline{\rho}_f$ to a $p$-adic representation with the same image. In particular, the ramification properties are unchanged by the lift. Take this $p$-adic representation and view it as a complex representation (i.e. an Artin representation). By Artin's conjecture (which I guess is now a theorem in this 2-dimensional case by Serre's conjecture), there is a weight 1 eigenform $g$ giving rise to this representation. But since the representation is unramified at $q$, we can take $g$ to have level $N/q$, and we're done.

So questions:

0) Does this argument look okay?

1) Is there a more direct argument that doesn't rely on Artin's conjecture to achieve this. (I'm ultimately interested in the Hilbert modular case, and so I don't want to be using relying on Artin's conjecture).

2) Is there a way to handle $p=2,3,5$ even assuming Artin? Is this even true with $p=2$??

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  • $\begingroup$ I am not trained in Galois representations. Will the level lowering have the same factorization as automorphic representation? If yes, it might be easier to analyze which invariant vectors the representation at $q$ has? What is the conductor of $\chi$ btw? $\endgroup$
    – Marc Palm
    Oct 18, 2012 at 17:07
  • $\begingroup$ In math.ubc.ca/~vatsal/research/pseudo.pdf Cho and Vatsal study the behavior of specialization of Hida families in weight one from a deformation perspective (in a paper with Ghate, Vatsal said it was being done with Greenberg, but the paper did not appear). Relying on that, can't you deduce your result from the analogous in weight bigger than $1$ - I guess you tried, what goes wrong? $\endgroup$ Oct 19, 2012 at 2:25
  • $\begingroup$ Dear Robert, if you assume that $a_p(f)^2 \ne \chi(p)$ mod $p$ then I think you can deduce level lowering in weight one from level lowering in weight $p$ using Corollary 13.11 of Gross's companion forms paper. At least, this will give you a mod $p$ weight one form of level $N/q$ - I don't know how easy it is to show that this form will lift to characteristic zero when the form of level $N$ you started with lifts, which seems to be the setting of your question. The weight one case of Gross's result was generalised to the Hilbert setting recently by Gee and Kassaei (arXiv:1206.6631). $\endgroup$
    – jnewton
    Oct 19, 2012 at 9:53

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Everything you wish for is true for modular forms over $\mathbb Q$ even at $p=2$, as it follows from refined forms of Serre's conjecture; here I am assuming of course that $\bar{\rho}$ is absolutely irreducible (I think that you meant to include this explicitly in your set-up, of course otherwise level-lowering can fail). In particular your argument is correct. I don't think there is a significantly more direct general argument incorporating $p=2$ as it is my understanding that it is only with Khare-Wintenberger's proof that the last cases of "Weak Serre implies Refined Serre" were proved.

Over totally real field, this is the whole business of Serre's weight; a topic which has seen exponential development these last 10 years. I am vey far from the most superficial understanding of the current literature but I guess Buzzard-Diamond-Jarvis is a good place to start. Other key players are Toby Gee, Matt Emerton and Florian Herzig. I don't think much is actually known about level-lowering in weight 1, though I could very well be wrong.

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