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What is the best known result concerning the existence of companion forms for classical modular forms? Gross' tameness criterion paper is always mentioned with a "unchecked compatibility" caveat? Are the results in this paper bona-fide theorems? The Coleman-Voloch paper on the topic (which does not rely on these unchecked compatibilities) seems not to allow k=2 (but Gross' paper is fine with k=2). Does Khare-Winterberger's proof of Serre's conjecture trump all of this? If so, is Serre's conjecture known unconditionally now or are there still unresolved cases?

flag	asked May 20 2011 at 19:41 akula 103●4

	PS I am pretty sure that Brian Conrad once told me that the unchecked compatibilities are now checked. – Kevin Buzzard May 21 2011 at 8:08
	@Kevin: Yes, in Cais's thesis. – Felipe Voloch May 21 2011 at 11:18
	@Felipe: you're right. Thanks. – Kevin Buzzard May 21 2011 at 19:39

Kevin Buzzard · Answer 1 · 2011-05-20 20:53:04Z UTC

It might all depend on precisely what you mean by Serre's conjecture. Various versions are in print. Serre's original conjecture stayed away from $k=1$ and K-W resolved this version of the conjecture completely I believe, including all companion forms issues. Did you take a look at the K-W papers? They surely give a precise statement of what they prove. Edixhoven made the most optimistic conjecture, allowing weight 1, and I might be wrong but at the back of my mind I suspect that in some cases where the mod $p$ representation is trivial on a decomposition group at $p$, K-W only produce a weight $p$ form whereas what is conjectured is that there's a mod $p$ weight 1 form (in the sense of Katz).

A. Pacetti · Answer 2 · 2011-05-20 23:57:26Z UTC

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There is also a paper of Toby Gee which, as in K-W, uses tools of modularity lifting theorems. He proves the existence of a weight p form as mentioned by Kevin which works as well for Hilbert modular forms. You can check the arxiv paper http://arxiv.org/abs/math/0507507

answered May 20 2011 at 23:57

A. Pacetti
261●5

	Yeah---the sticky issue is going from weight $p$ to weight 1---both in the classical and the Hilbert case. If $\rho$ is unramified at $p$ and the eigenvalues are distinct then one can build two weight $p$ forms and then, I think, a weight 1 form, using tricks. But if the eigenvalues are the same then I think the existence of the weight 1 form is still open. K-W/Gee are always working in weight $k\geq2$. – Kevin Buzzard May 21 2011 at 8:10
	@Kevin: Wiese's paper quoted in my answer gets weight one forms, but I can't tell just by glancing at it whether some additional assumptions is still needed. – Felipe Voloch May 21 2011 at 11:16
	@Felipe: Wiese starts with weight 1 forms---I think he's going the other way. – Kevin Buzzard May 21 2011 at 19:40

Felipe Voloch · Answer 3 · 2011-05-20 21:30:52Z UTC

-1

In addition to what Kevin already said, there is some work of Bryden Cais and of Gabor Wiese that have removed some and perhaps all of the additional hypotheses. I can't guarantee for sure that it's all done, you should ask them. See http://arxiv.org/abs/1102.2302 and http://www.math.wisc.edu/~cais/Papers/PhDThesis/PhD.html

edited May 20 2011 at 21:30

answered May 20 2011 at 21:23

Felipe Voloch
16.6k●2●36●61

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Felipe: from what I can see Wiese's paper is irrelevant for the question at hand. The question we're talking about here is: "if rho-bar is unramified, does it come from a weight 1 form?". Wiese's paper is dealing with the question "if rho-bar comes from a weight 1 form, is it unramified?". I am still going to stick with my assertion that the question we're talking about (in the special case where the eigenvalues of Frob_p are the same) is still open. – Kevin Buzzard May 21 2011 at 19:37

Companion forms

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