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First of all, let me apologize in advance for the terseness of this question.

It seems that by now there are well-developed techniques (the "Taylor-Wiles-Kisin" method) for proving modularity lifting theorems over totally real fields. For example, it is now known that many elliptic curves over totally real fields are modular. I am curious: what exactly is the stumbling block to proving such results over non-totally real fields? It seems to be common knowledge among experts that the usual techniques for comparing a universal deformation ring and a Hecke algebra in this situation break down quite badly, but I cannot find a reference for this in print. It would be great if someone could illustrate the problem here.

I understand that there are some obvious difficulties. For example, for $GL_2$ over an imaginary quadratic field, the locally symmetric spaces on which the relevant modular forms live is $SL_2(\mathbb{C})/SU(2)$, and arithmetic quotients of this are certainly not Shimura varieties. I am asking about more fundamental obstructions to applying the Taylor-Wiles-Kisin method, i.e strange behavior in the Hecke algebra and the universal deformation ring...

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1 Answer

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Note: This is a fairly precise and detailed question about an important but technical aspect of algebraic number theory. My answer is written at a level that I think is appropriate for the question; it assumes some familiarity with the topic at hand.


The most basic difficulty is that there is non a map $R \rightarrow {\mathbb T}$ in general (i.e. one typically doesn't know how to create Galois representations attached to automorphic forms).

The second difficulty is that in the TWK method, one must argue with auxiliary primes (the primes typically labelled $Q$), and show that as you add these primes, ${\mathbb T}$ grows in a reasonable way (basically, is free over $\mathcal O[\Delta_Q],$ where $\Delta_Q$ is something like the $p$-Sylow subgroup of $({\mathbb Z}/Q{\mathbb Z})^{\times}.)$

One shows this (or some variant of it) by considering the analogous queston about cohomology of the arithmetic quotients. Suppose for a moment we are in the Shimura variety context, or perhaps the compact at infinity context. Then it will be the middle dimensional cohomology that is of interest, and if we localize at a non-Eisenstein maximal ideal we might hope to kill all other cohomology. Then we can replace a comuptation of middle dimensional cohomology by an Euler characteristic computation, and its easy to see that the Euler char. will multiply by $|\Delta_Q|$ when we add the auxiliary primes $Q$.

But in more general contexts, there won't be a single middle dimension in which the maximal ideal of interest is supported (even if it is non-Eisenstein), and computing Euler characteristics will just give $0$, which is not much use. It's not clear that it's even true that adding the auxiliary primes forces the approriate growth of cohomology, and possible torsion in the cohomology just adds to the complication.

There is much current work, by various groups of researchers, with various different approaches, aimed at breaking this barrier.


I should add that one can now handle certain questions about non-totally real field, say question related to conjugate self-dual Galois reps. over CM fields, because these are still related to a Shimura variety context. This plays a role in the recent progress on Sato--Tate for higher weight forms by Barnet-Lamb--Geraghty--Harris--Taylor and Barnet-Lamb--Gee--Geraghty, and is also the basis for a recent striking theorem of Calegari showing that if $\rho:G_{\mathbb Q} \to GL_2({\mathbb Q}_p)$ is ordinary at $p$ and de Rham with distinct Hodge--Tate weights (and probably $\overline{\rho}$ should satisfy some technical conditions), then $\rho$ is necessarily odd!


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Thanks for the great answer! The conjugate self dual over a CM field cases you mention do all require that the gal. rep. in question has distinct Hodge-Tate weights, right? – David Hansen Jan 20 2010 at 19:27
Yes; this seems to be a difficult barrier to break. (Certainly people have tried to generalize Buzzard--Taylor, but nothing has yet worked out, at least that people have made public.) – Emerton Jan 20 2010 at 20:16
@Emerton: even if one could prove R=T over an im quad field, isn't another problem that the numerical criterion probably won't pan out? Because the contribution at infinity doesn't add up (I think). As for Buzzard-Taylor, my former student Sasaki has something for totally real fields when p splits completely, and I think I can do something in some ramified cases---I'll write it up when I finish Buzzard-Gee (so expect it to appear in about 2014 :-/ ) – Kevin Buzzard Jan 20 2010 at 22:04
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@KB: You are stuck in a time warp ~2006. If one works with (appropriately defined) spaces built out of integral homology (including torsion), say Hida families or Emerton's completed cohomology, the numerical criterion works out perfectly, as it should, because R = T. There is something that "looks" like a numerical criterion, and which is obviously false, because it would imply that T was much bigger than R, which it isn't. Another way of saying this is that the real numerical criterion is A = B - l_0, rather than A = B; the second equality having the deficiency that it is false if l_0=/=0. – Lavender Honey Jan 20 2010 at 23:09
@KB: Oberwolfach 2011, can we say yes? – Lavender Honey Jan 20 2010 at 23:09
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