[Math] What makes Langlands for n=2 easier than Langlands for n>2

langlands-conjecturesnt.number-theory

I must confess a priori that I haven't read the proof of Taniyama-Shimura, and that my familiarity with Langlands is at best tangential.

As I understand it Langlands for $n=1$ is class field theory. Not an easy theory, but one that was known for a long time.

Langlands for $n=2$ is the Taniyama-Shimura conjecture, proven recently by Andrew Wiles and others (some of whom participate in this forum).

Clearly Taniyama-Shimura required new ideas. What special property of the $n=2$ case made the proof of Taniyama-Shimura possible, that doesn't exist for Langlands with $n\geq 3$?

Best Answer

"Langlands for $n = 2$", to the extent that such a notion is defined, is more than just Shimura--Taniyama, and for even Galois representations/Maass forms, it is still very much open. (See here for more on this.) For odd Galois representations of dimension $2$, though, it is completely (or almost completely, depending on exactly what you mean by "Langlands") resolved at this point, with the proof of Serre's conjecture (by Khare, Wintenberger, and Kisin) playing a pivotal role.

Much is known for $n > 2$ (see the web-pages of e.g. Michael Harris, Richard Taylor, and Toby Gee). A key point is that it is hard to say anything outside the essentially self-dual case (and this is a condition which is automatic for $n = 2$). A second is that Serre's conjecture is not known in general.

If one restricts to the regular (corresponding to weight $k \geq 2$ when $n = 2$), essentially self-dual case (automatic when $n = 2$), then basically everything for $n = 2$ carries over to $n > 2$, with the exception of Serre's conjecture. (See e.g. the recent preprint of Barnet-Lamb--Gee--Geraghty--Taylor.)

So really, what is special for $n = 2$ is that Serre's conjecture was able to be resolved. And the reason that this has (so far) been possible only for $n = 2$ is that the proof depends on certain special facts about $2$-dimensional Galois representations.

More specifially:

In the particular case of Shimura--Taniyama, the Langlands--Tunnell theorem allowed Wiles to resolve a particular case of Serre's conjecture (for $p = 3$). To then get all the necessary cases of Serre's conjecture, Wiles introduced the $3$-$5$ switch.

The general proof of Serre's conjecture uses a massive generalization of the $3$-$5$ switch (along with many other techniques), and although (unlike with Wiles's argument) it doesn't build specifically on Langlands--Tunnell, it does build on a result of Tate which is a special fact about $2$-dimensional representations of $G_{\mathbb Q}$ over a finite field of characteristic $2$.