TITLE: Why are Tamagawa numbers equal to Pic/Sha?
QUESTION [23 upvotes]: 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.

REPLY [3 votes]: 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.