How do Brauer groups relate to zeta functions? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T14:55:22Z http://mathoverflow.net/feeds/question/79575 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79575/how-do-brauer-groups-relate-to-zeta-functions How do Brauer groups relate to zeta functions? Makhalan Duff 2011-10-31T02:30:32Z 2011-10-31T06:16:23Z <p>There are two approaches to class field theory that I was taught. The first, is the theory of $L$-functions, Dirichlet characters and so forth (which I described succintly in the question <a href="http://mathoverflow.net/questions/58901/what-are-the-pillars-of-langlands" rel="nofollow">http://mathoverflow.net/questions/58901/what-are-the-pillars-of-langlands</a> that I asked a long time ago), and the other is through Brauer groups.</p> <p>To be precise, I was taught that the following sketch is class field theory:</p> <p>Let $K$ be number field. Let $v$ be any non-archimedean place. Then there is some explicit isomorphism $inv_v:Br(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}$. If $v$ is an archimedean place, then there is an explicit isomorphism $inv_v:Br(K_v)\rightarrow \mathbb{Z}/2\mathbb{Z}$ if $K_v$ is $\mathbb{R}$, and $0$ if $K_v$ is $\mathbb{C}$.</p> <p>Furthermore, the following sequence is exact:</p> <p>$1\rightarrow Br(K)\rightarrow \bigoplus_v Br(K_v)\rightarrow \mathbb{Q}/\mathbb{Z}\rightarrow 1$</p> <p>where the first morphism is obvious, and the last morphism is $\sum_v inv_v$.</p> <p>I take both of these approaches have a lot of content, but it is not clear how people think of them as <em>equivalent</em>. How does one build a dictionary between the one approach and the other?</p> http://mathoverflow.net/questions/79575/how-do-brauer-groups-relate-to-zeta-functions/79588#79588 Answer by Alex Bartel for How do Brauer groups relate to zeta functions? Alex Bartel 2011-10-31T05:29:57Z 2011-10-31T05:29:57Z <p>The computation of the Brauer group is only the "first half" of class field theory. The main reason one is interested in the Brauer group is that for a finite Galois extension $L/K$, let's say of local fields, the cup product by a generator of $H^2(G_{L/K},L^\times)$ gives an isomorphism between $G^{ab}(L/K)$ and $K^\times/N_{L/K}L^\times$. This is the reciprocity map, the central part of (local) class field theory. Passing to the limit over all finite Galois extensions gives an isomorphism between $W_K^{ab}$ and $K^\times$, where $W_K$ is the Weil group of $K$. Dualising this establishes a bijection between the one-dimensional representations of $W_K$ (they are one-dimensional, because they are the ones that factor through the abelianisation), and irreducible representations of $K^\times=GL_1(K)$. This bijection turns out to have all the familiar nice properties and gives the local Langlands correspondence for $GL_1$.</p> <p>In other words, the link between the Brauer group and the "Langlands type" formulation of class field theory is that the computation of the Brauer group is used to define the reciprocity map and to prove all its nice properties. The computation of the Brauer group by itself does not establish the main results of class field theory.</p> http://mathoverflow.net/questions/79575/how-do-brauer-groups-relate-to-zeta-functions/79592#79592 Answer by BR for How do Brauer groups relate to zeta functions? BR 2011-10-31T06:16:23Z 2011-10-31T06:16:23Z <p>Global class field theory typically refers to the existence of a particular isomorphism (along with various compatibilities) $G(K/k)^{ab}\simeq C_k/N_{K/k} C_K$, where $K$ is a finite Galois extension. This can be recast, by dualizing, to say something about a correspondence between characters on $G(\bar k/k)$ and characters on $C_k$. These statements will be equivalent, but different proofs may not be...</p> <p>I think you can prove it via Taylor-Wiles method (Kowalski mentions this in a survey paper referring to unpublished notes from a class Tunnell taught). But the proofs of global class field theory that <em>I</em> know are essentially cohomological (though they may avoid saying it). The statement about Brauer groups that you made is essentially what is necessary to prove that $(G(\bar k/k),C_{\bar k})$ is a class formation, from which a bit more work is necessary to prove class field theory (though you can deduce reciprocity laws from it). </p>