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.

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.

  • 5
    $\begingroup$ The kind of explanation sought in the 2nd paragraph of this question does not seem to exist. In the simply connected case it is comparatively elementary that Pic vanishes, but there is no direct link between Tamagawa number and Sha -- it is just very deep that both "happen" to equal 1 in this case. There is no "simple" explanation for (or interpretation of) this equality. A good analogue is Tate's Euler characteristic formula in Galois cohomology for $p$-adic fields: in one model case ($\mu_n$) we can compute, and everything else links back to it by brilliant use of homological methods. $\endgroup$
    – BCnrd
    Commented Oct 30, 2010 at 2:42

2 Answers 2


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).

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 $G$.

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".

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 absolutely simple 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 simply connected semisimple group is a UFD. Not obvious!)

  • $\begingroup$ Thanks for the sketch and references, but I wasn't really looking for a proof. I was hoping for a slightly nicer statement of the theorem, even if it doesn't make it easier to prove. In the semisimple case, Weil's conjecture is nice, but there's still the torus case. Perhaps the applications of class field theory can be phrased as the Tamagawa number of the ideles? $\endgroup$ Commented Oct 29, 2010 at 23:57
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    $\begingroup$ Dear Ben: The Tamagawa number of the ideles is 1 almost by definition. As I mentioned in my answer, Oesterle explains the entire proof of the Pic/Sha formula in the case of tori in his great paper. It's a somewhat nontrivial argument with class field theory, and I don't know any intuition for why it works (apart from reading the proof). In Ono's original work, this is not how he expressed the formula; the Pic/Sha formula is much nicer than Ono's version, since it unifies the torus and semisimple cases. It's not beautiful enough for you? :) $\endgroup$
    – BCnrd
    Commented Oct 30, 2010 at 2:17
  • $\begingroup$ Oesterlé paper: springerlink.com/content/r3r642748w7m4682 And Voskresenskii: springerlink.com/content/738031m792181j54 $\endgroup$
    – Junkie
    Commented Apr 9, 2011 at 6:28

I found the Sha part of the answer in Voskresenskii's book.

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.

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 $Q_G$?

I still have some hope for a geometric explanation for the Picard group.

  • $\begingroup$ Dear Ben: I don't understand where your proposed exact sequence is coming from. One can speak of a connected reductive group being "split" by a field (in the sense of admitting a split maximal torus over that field), but apart from the case of tori there's often a huge amount of nontrivial cohomology over such a field. So (for several reasons) your diagram doesn't make sense; e.g., explain exactness at ${\rm{H}}^1(K,G)$. As for the torus case, unfortunately I don't follow the notation being used. Have you looked at Oesterle's treatment of the torus case in his Tamagawa number paper? $\endgroup$
    – BCnrd
    Commented Nov 1, 2010 at 7:16
  • $\begingroup$ Oops...I forgot that the quotient isn't a group. But don't we still get a 5-term exact sequence? I'll think about that. Don't split reductive groups have vanishing Sha? In the torus case, I think $Q_T$ is called the ideles of the torus, but tradition dictates that I make up my own notation. The point is that the ideles are a Galois module and fixed points are not $T(A)/T(K)$. $\endgroup$ Commented Nov 1, 2010 at 21:29
  • $\begingroup$ Dear Ben: Yes, the quotient not being a group is one thing I had in mind. You are right that Sha is trivial in the split connected reductive case (over any global field); I had not realized before that it is true in general (but it's clear now that you have pointed it out). Since Sha is not part of a low-degree $\delta$-functor, I don't think that this feature helps much for computing it in general (but maybe I am overlooking something). $\endgroup$
    – BCnrd
    Commented Nov 2, 2010 at 3:34
  • $\begingroup$ I probably failed to make clear that the point about splitting reductive groups was (a) an aside and (b) out of logical order. If you buy the exact sequence, so that the sheafification of $G(A)/G(K)$ has the effect of multiplying by Sha, then to compute the sheafification you only need to pass to the splitting field, which is nice for concreteness. $\endgroup$ Commented Nov 2, 2010 at 16:59

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