Why are Tamagawa numbers equal to Pic/Sha? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T22:35:30Z http://mathoverflow.net/feeds/question/44184 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44184/why-are-tamagawa-numbers-equal-to-pic-sha Why are Tamagawa numbers equal to Pic/Sha? Ben Wieland 2010-10-29T21:48:58Z 2011-04-09T01:38:36Z <p>For a connected algebraic group $G$ over a global field $K$ with adeles $A$, the Tamagawa number of $G$ is the volume of $G(A)/G(K)$. It is conjectured (and often known) to be rational, namely the quotient of the order of the Picard group divided by the order of the Tate-Shafarevich set. </p> <p>Is there some simple explanation for the appearance of these integers? For example, is there some natural object that turns out to have volume 1 that divides into equal pieces, indexed by TS, and then copies indexed by Pic reassemble to form $G(A)/G(K)$? You can throw in bundles where the fibers have volume 1, if necessary. </p> http://mathoverflow.net/questions/44184/why-are-tamagawa-numbers-equal-to-pic-sha/44188#44188 Answer by BCnrd for Why are Tamagawa numbers equal to Pic/Sha? BCnrd 2010-10-29T22:30:04Z 2010-10-30T17:31:46Z <p>I assume $G$ is affine. The quick answer is that in the simply connected case it says $1 = 1/1$ by various hard ingredients, and then it is a kind of (not easy) game with Galois cohomology and structure theory of semisimple groups to check both sides have the same behavior as we build up a general $G$ from the simply connected case (with the help of class field theory to deal with tori). </p> <p>Let's address number fields $K$ in more detail (the case of global function fields has a variety of serious complications). Since $K$ is perfect, the geometric unipotent radical descends to a smooth connected unipotent normal $K$-subgroup $U$ in $G$, with $G/U$ reductive. Now $U$ is $K$-split (composition series over $K$ with successive quotients $\mathbf{G}_a$), so its underlying variety is an affine space over $K$. Because we're in characteristic 0, the quotient map $G \rightarrow G/U$ admits a homomorphic section, which is to say that $G = U \rtimes (G/U)$ as $K$-groups (for a suitable semi-direct product structure). Thus, ${\rm{Pic}}(G) = {\rm{Pic}}(G/U)$. Likewise, the Tate-Shafarevich sets for $G$ and $G/U$ match because the semidirect product structure ensures that the pullback on ${\rm{H}}^1$'s is bijective over $K$ and its completions. The Tamagawa numbers also match, by behavior of Tamagawa numbers in exact sequences (see Oesterle's masterpiece paper) and the fact that Tamagawa number of $\mathbf{G}_a$ is rigged to be 1 by definition. OK, so we can focus on the case with content, which is reductive $G$. </p> <p>For tori, one uses work of Ono and its refinements (building on Tate-Nakayama duality for tori, etc.) This is all in Oesterle's paper too. In general there's an etale (central) isogeny $Z \times G' \rightarrow G$ where $G'$ is semisimple and simply connected. By arguments with Galois cohomology and class field theory, one has to show that the validity of the desired formula can be pulled down to $G$ from $Z \times G'$ (the key case being isogenies between connected semisimple groups); this sort of thing is addressed a bit in Voskresenskii's survey paper "Adele groups and Siegel-Tamagawa formulas". </p> <p>So then finally we're brought to the case when $G$ is semisimple and simply connected. Thus, $G = \prod G_i$ for $K$-simple factors, and then $G_i = {\rm{Res}}_{K_i/K}(H_i)$ for finite (separable) extensions $K_i/K$ and <em>absolutely simple</em> and simply connected $H_i$. Tamagawa numbers are invariant under Weil restrictions (once again, see Oesterle's paper) and commute with products, so the assertion $\tau_G = 1$ reduces to the absolutely simple case, which was a conjecture of Weil solved by Langlands, Lai, and Kottwitz. By Shapiro's Lemma, the triviality of Tate-Shafarevich also reduces to the absolutely simple case, where it is the famous "Hasse principle" due to many people over many years. Finally, the triviality of Pic is handled by relating line bundles on connected semisimple groups to central extensions by $\mathbf{G}_m$ (this requires some input from the structure theory of semisimple groups, with help of Galois descent to pass to the case of split groups, for which the structure of the open cell allows us to copy some arguments used to study Pic of abelian varieties). We exploit simple connectedness by the following elementary observation: if $E$ is a central extension of simply connected $G$ by $\mathbf{G}_m$ then it is reductive and hence $D(E) \rightarrow G$ is a central isogeny, thus an isomorphism because $G$ is simply connected. So voila, the central extension splits and thus Pic($G$) = 1 in the simply connected case. (That's actually quite remarkable: the coordinate ring of a <em>simply connected</em> semisimple group is a UFD. Not obvious!)</p> http://mathoverflow.net/questions/44184/why-are-tamagawa-numbers-equal-to-pic-sha/44360#44360 Answer by Ben Wieland for Why are Tamagawa numbers equal to Pic/Sha? Ben Wieland 2010-10-31T18:11:26Z 2010-10-31T18:11:26Z <p>I found the Sha part of the answer in Voskresenskii's book.</p> <p>I'm looking at a quotient and I want cohomology to appear. Cohomology is the obstruction to invariants being exact, so I should lift this to a quotient of Galois modules. $G(\bar K)$ has a Galois action, as does $G(A_K\otimes \bar K)$. Their quotient $Q$ has fixed points $Q(K)=(G(A_L)/G(L))^{Gal(L/K)}$, where $L$ is a splitting field. The long exact sequence in cohomology $$0\to G(K)\to G(A_K)\to Q(K)\to H^1(K;G)\to H^1(K;G(A_{-}))=\prod H^1(K_\nu;G)$$ shows that $Q(K)$ differs from $G(A_K)/G(K)$ by exactly the Tate-Shafarevich group. </p> <p>In the torus case, one studies the failure of multiplicativity of the Tamagawa number in extensions $T'\to T\to T''$. The sheafified quotient $Q_{T'}(K)$ naturally appears as the (maximum possible) kernel of $T(A)/T(K)\to T''(A)/T''(K)$. Thus it is better to look at $Q_{T'}(K)\to Q_T(K)\to Q_{T''}(K)$. This is still not right exact, so $H^1(K;Q_T)$ enters, which Nakayama duality says is Pontrjagin dual to $Pic(T)$. Perhaps this is saying that we should form a $Pic(G)^*$ gerbe over $Q_G$?</p> <p>I still have some hope for a geometric explanation for the Picard group.</p>