4
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ PS I am pretty sure that Brian Conrad once told me that the unchecked compatibilities are now checked. $\endgroup$ May 21, 2011 at 8:08
  • $\begingroup$ @Kevin: Yes, in Cais's thesis. $\endgroup$ May 21, 2011 at 11:18

3 Answers 3

2
$\begingroup$

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).

$\endgroup$
0
1
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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 $k\geq2$. $\endgroup$ May 21, 2011 at 8:10
  • $\begingroup$ @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. $\endgroup$ May 21, 2011 at 11:16
  • $\begingroup$ @Felipe: Wiese starts with weight 1 forms---I think he's going the other way. $\endgroup$ May 21, 2011 at 19:40
0
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$ May 21, 2011 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.