What makes Langlands for n=2 easier than Langlands for n>2? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T08:10:41Z http://mathoverflow.net/feeds/question/74472 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74472/what-makes-langlands-for-n2-easier-than-langlands-for-n2 What makes Langlands for n=2 easier than Langlands for n>2? Nicole 2011-09-03T23:32:20Z 2011-09-06T21:37:41Z <p>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.</p> <p>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.</p> <p>Langlands for $n=2$ is the Taniyama-Shimura conjecture, proven recently by Andrew Wiles and others (some of whom participate in this forum).</p> <p>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 $n\geq 3$?</p> http://mathoverflow.net/questions/74472/what-makes-langlands-for-n2-easier-than-langlands-for-n2/74489#74489 Answer by Emerton for What makes Langlands for n=2 easier than Langlands for n>2? Emerton 2011-09-04T04:12:42Z 2011-09-04T04:32:44Z <p>"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 <a href="http://math.stackexchange.com/questions/55449/in-what-sense-is-taniyama-shimura-the-n-2-case-of-langlands/55597#55597" rel="nofollow">here</a> 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. </p> <p>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.</p> <p>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.)</p> <p>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.</p> <p>More specifially:</p> <p>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 <em>all</em> the necessary cases of Serre's conjecture, Wiles introduced the $3$-$5$ switch. </p> <p>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 $2$. </p> http://mathoverflow.net/questions/74472/what-makes-langlands-for-n2-easier-than-langlands-for-n2/74506#74506 Answer by ACL for What makes Langlands for n=2 easier than Langlands for n>2? ACL 2011-09-04T09:41:32Z 2011-09-04T09:41:32Z <p>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.</p> http://mathoverflow.net/questions/74472/what-makes-langlands-for-n2-easier-than-langlands-for-n2/74688#74688 Answer by Moshe Adrian for What makes Langlands for n=2 easier than Langlands for n>2? Moshe Adrian 2011-09-06T21:37:41Z 2011-09-06T21:37:41Z <p>One could (and sometimes would) argue the opposite of your claim/question. Namely, that "Langlands for $n = 2$" is <em>more</em> 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 : </p> <p>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 $GL(2)$).</p> <p>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.</p>