# What makes Langlands for n=2 easier than Langlands for n>2?

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 $n\geq 3$?

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One reason is that the automorphic forms relevant to Galois representations for $n = 2$ are much easier to understand than the corresponding automorphic forms (conjecturally) associated to Galois representations when $n = 3$. See, for example, this question: mathoverflow.net/questions/43377/automorphic-forms-on-gl3 –  Michael Sep 4 '11 at 2:15
Dear Nicole, You should probably register, and ask the moderators (at the appropriate place on meta) to consolidate your various accounts. This will give you a unified profile for your questions and answers on MO. Regards, Matthew –  Emerton Sep 4 '11 at 4:22
One could also quibble a little with the premise of the question. It's taken almost 100 years (more or less) to go from the case $n = 1$ to the case $n = 2$, and the latter <i> still </i> has some unresolved problems which seem very hard. It's not impossible that the even Artin conjecture for $n = 2$ requires tools powerful enough to do everything. –  Michael Sep 4 '11 at 18:02
$SL(2) \simeq SO(2,1) \simeq Sp(2)$ (isogenies) –  David Hansen Sep 4 '11 at 22:50

"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 $2$.

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what is the $3-5$ switch? Is there a layman's exposition? Can we do $p-q$ switch where $p$ and $q$ are primes? Is there a easy reading on this? –  J.A Aug 10 '13 at 23:25
@Arul: Dear Arul, You can find a discussion in the introduction to Wiles's 1994 Annals paper (the one on FLT), and a discussion of similar such prime switches in the Pacific Journal paper of Richard Taylor from around the same time. Basically, it is a method in which one has a mod $p$ Galois representation for some particular prime $p$, and one finds a compatible family of $\ell$-adic representations such that the given representation is the mod $p$ reduction of the $\ell = p$ member of the family, and one then argues at a different prime $q$ to deduce properties (e.g. modularity) of ... –  Emerton Aug 11 '13 at 0:22
... the entire compatible family, and so in particular of the original mod $p$ representation. Regards, –  Emerton Aug 11 '13 at 0:22

Most (if not all) results in the Langlands program use, at some point, the cohomology of Shimura varieties. The linear group gives rise to Shimura varieties when $n=2$ (modular curves), but not for $n>2$. However, unitary groups furnish Shimura varieties in any dimension and this permits to obtain the automorphy of self-dual Galois representations.

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One could (and sometimes would) argue the opposite of your claim/question. Namely, that "Langlands for $n = 2$" is more difficult than "Langlands for $n > 2$". Or that more specifically (since I really can't agree with my previous sentence in full generality), "Langlands for $n = 2$" is more difficult than "Langlands for $n > 2$ an odd prime". Here are two key reasons :

1) Suppose $F$ is a $p$-adic field, $n$ is prime, and $p$ doesn't divide $n$. Then the part of the local Langlands correspondence of $GL(n,F)$ dealing with supercuspidal representations (proven by Bushnell/Henniart) is given by $$Ind_{W_E}^{W_F}(\chi) \mapsto \pi(\chi \Delta_{\chi})$$ where $W_F$ is the Weil group of $F$, $E/F$ is a degree $n$ separable extension, $\chi : W_E \rightarrow \mathbb{C}^*$ is a certain type of character ("admissible" to be precise, see Bushnell/Henniart papers/book), $\Delta_{\chi} : W_E \rightarrow \mathbb{C}^*$ is a "twisting character" associated to $\chi$ (again see Bushnell/Henniart), and $\pi(\chi \Delta_{\chi})$ is the supercuspidal representation of $GL(n,F)$ attached to $\chi \Delta_{\chi}$ via the "Howe construction". If $n$ is an odd prime, then $\Delta_{\chi}$ is either trivial or the unramified quadratic character of $W_E^{ab} \cong E^*$ (note that any character of $W_E$ factors through the abelianization $W_E^{ab}$). If $n = 2$, then $\Delta_{\chi}$ is much more complicated (see page 217 of Bushnell/Henniart's recent book on $GL(2)$).

2) Let $\pi$ be a supercuspidal representation of $GL(n,F)$ (same assumptions as in the beginning of 1) above). If $n$ is an odd prime, then the distribution character $\theta_{\pi}$ of $\pi$ is much simpler to write down on elliptic tori than if $n = 2$. In particular, if $n = 2$, a non-trivial Gauss sum arises in $\theta_{\pi}$, whereas no such term arises if $n$ is an odd prime. So the theory for $GL(2,F)$ contains some nontrivial arithmetic information that just doesn't arise for $GL(n,F)$, $n$ an odd prime.

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