21

12

Galois deformations are an important tool in Wiles' arsenal for proving FLT. Are there any more elementary aspects (I'm thinking of 1-dimensional Galois representations attached to number fields) that would help the novice in better understanding what's going on?

Here's what I have in mind. Let $\rho: G_{\mathbb Q} \longrightarrow {\mathbb C}^\times$ be a 1-dimensional representation of the absolute Galois group of the rationals factoring over some finite extension. Given a Dirichlet character $\chi: GL_1({\mathbb Z}/N{\mathbb Z}) \longrightarrow {\mathbb C}^\times$, we can find representations $\rho_\chi: Gal(K/{\mathbb Q}) \longrightarrow {\mathbb C}^\times$ for any cyclotomic extension $K = {\mathbb Q}(\zeta_N)$. Call $\rho$ modular if there is a $\chi$ such that $\rho = \rho_\chi$. The statement that every $\rho$ coming from an abelian extension is modular is the theorem of Kronecker-Weber, and in this form it can be proved using Galois deformations along the lines of Wiles' proof (see Tunnell's proof in Kowalski's notes). BTW if anyone knows a source for this result that is more readable than Kowalski's notes (which I discovered just a couple of days ago and haven't studied in detail yet) I'm all ears.

Question: Are there other similarly "elementary" questions, for example in embedding problems or inverse Galois theory, that can be described in terms of Galois deformations?

flag
3 
normally when we talk about Galois deformation, we talk about p-adic representations, right? – natura Mar 2 2010 at 1:19
2 
In the 1-d case the universal deformation rings that show up are the ones usually called "Lambda" in Iwasawa theory (and this is trivial to check---you're just basically forming profinite group rings in this case). If I remember correctly Mazur makes some remarks about this in his original paper on the subject. By the way I think it's better to let rho be a p-adic Galois representation and say it's modular if it's a product of an integral power of the cyclo char by a finite order char. Then you get modular iff de Rham iff comes from an algebraic automorphic form, which is the analogy you want. – Kevin Buzzard Mar 2 2010 at 12:37
2 
Someone who writes out all the details of the proof of Kronecker-Weber using this approach will do a great service to the community (and to himself, in the bargain). – Chandan Singh Dalawat Mar 3 2010 at 8:06
2 
"...Kronecker-Weber...". I'm not sure it's as easy as that. Wiles/Taylor use the Poitou-Tate exact sequence as an input into proving their R=T theorems and this exact sequence encodes most of global CFT already. So you might find the argument is circular. In fact if I remember correctly I think Larry Washington wrote an article in Cornell-Silverman-Stevens explaining all this. Nowadays one needs weaker numerical criteria to get R=T results but I'm still not at all convinced that you can get around CFT going in as an input. – Kevin Buzzard Mar 3 2010 at 9:59
8 
Dear Kevin and Brian, I've looked over the Tunnell notes before, and I don't think they are circular. Of course, they don't give the quickest proof of Kronecker--Weber by any means; but nor were they intended to. Rather, they were supposed to give a blueprint for the arguments of Wiles and Taylor in a simpler setting, where the deformation-theoretic arguments could be made by hand (hence, no cohomological theorems needed, and no circularity). – Emerton Mar 3 2010 at 14:15
show 2 more comments

2 Answers

5

Late answer in continuation to #4: if I remember right, at first Tunnell thought he would need Poitou-Tate to complete the Kronecker-Weber proof in this manner (which would make things rather unsatisfactory, as people have said), but in the end -- for Kronecker-Weber only -- he saw that he only needed the local Kronecker-Weber theorem to complete the argument, which he then proved in the class (or maybe he just sketched the proof?).

link|flag
0

Nigel Boston has a set of notes on the proof of FLT: http://www.math.wisc.edu/~boston/ (the notes is the link "Spring 2003 Math 869 Fermat's Last Theorem notes" under "Courses"). I haven't finished reading it, but it looks good.

link|flag

Your Answer

Get an OpenID
or

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