Simple Tamagawa number calculations As is well known, Euler proved the Basel identity $\displaystyle\sum\limits_{i=0}^{\infty} \frac{1}{n^2} = \frac{{\pi}^2}{6}$. By far the most illuminating explanation of this fact that I've seen is as follows:

By another theorem of Euler we can
  rewrite the identity as $\displaystyle
 \frac{{\pi}^2}{6}\cdot \prod_{p}
 \left(1 - \frac{1}{p^2}\right) = 1$,
  where $p$ ranges over all primes. The
  first term is the normalized volume of
$\displaystyle SL(2,\mathbb{R})/SL(2,
 \mathbb{Z})$ 
and the term corresponding to $p$ in
  the product is the normalized volume
  of 
$SL(2, \mathbb{Z}_p)$. 
With these replacements, the right
  hand side can be written as the
  normalized volume of 
$\displaystyle
 SL(2,\mathbb{\mathbb{A}_{\mathbb{Q}}})/SL(2,
 \mathbb{Q})$. 
where $\mathbb{A}_{\mathbb{Q}}$
  denotes the adeles of $\mathbb{Q}$ But
  this last volume is equal to 1: this
  is a special case of the Weil
  Conjecture on Tamagawa Numbers.

I've been fascinated by this result for years, but have never been able to understand a proof of it (from the adelic perspective) even in cases as simple the one above. In this way, my question contrasts with that of Ben Weiland who asked about the theorem in more general settings. 
I tried reading André Weil "Adeles and algebraic groups" with a view toward learning a proof but found the book unintelliglbe. I gathered that the idea of the proof is to show that the nonzero volume of some object is equal to the Tamagawa number multiplied by the original volume but beyond that understood nothing. My impression is that Marie-France Vigneras' book titled Arithmetique des algebres de quaternions has this material, but I don't read French.

What are some lucid sources that you
  would recommend for learning proofs of
  the some of the first few cases
  (including the case above) of the Weil
  Tamagawa Number conjecture from an
  adelic perspective?


1 I learned this material from Yuri Manin's "Reflections on Arithmetical Physics" and Maclachlan and Reid's "The Arithmetic of Hyperbolic 3-Manifolds"
 A: In order to talk about the volume you need to fix the measure first. Weil observed that there is a canonical choice for the Haar measure on $G(\mathbb{A})$. For a connected $n$-dimensional semisimple algebraic group $G$ over a global field $K$, there is a nonzero $n$-dimensional rational differential $K$-form which is invariant under left-translation. The differential form induces a Haar measure on $G(K_v)$ for each completion $K_v$ of $K$, hence it determines a Haar measure on the restricted direct product $G(\mathbb{A})$ as well. The differential form is unique up to scaling by $K^\times$, hence the obtained Haar measure on $G(\mathbb{A})$ is unique. When $G$ is simply connected and semisimple, Weil conjectured that $\mathrm{vol}(G(\mathbb{A})/G(K))=1$ for this particular measure.
In the case of $G=\mathrm{SL}_2$ and $K=\mathbb{Q}$, the invariant differential form is given by $\frac{dx\wedge dy\wedge dz}{x}$, where $\begin{pmatrix} x & y \\ z & t \end{pmatrix}$ are the usual coordinates on the group. For this form one obtains the numbers mentioned by Chandan Singh Dalawat (and also by Colmez's Un autre monde est possible), hence a proof of Weil's conjecture for $G=\mathrm{SL}_2$ and $K=\mathbb{Q}$. For details I recommend Platonov-Rapinchuk: Algebraic groups and number theory (Academic Press, 1994). See especially Example 3 on pp. 166-167, the Example on pp. 222-223, and the Example on p. 262.
Added. The OP is looking for a direct (global) proof of $\mathrm{vol}(\mathrm{SL}_2(\mathbb{A})/\mathrm{SL}_2(\mathbb{R}))=1$. For the analogous (local) statement that $\mathrm{vol}(\mathrm{SL}_2(\mathbb{R})/\mathrm{SL}_2(\mathbb{Z}))=\zeta(2)$ there are nice accounts via Eisenstein series, see for example here. No doubt this proof can be modified to yield $\mathrm{vol}(\mathrm{SL}_2(\mathbb{A})/\mathrm{SL}_2(\mathbb{Q}))=1$ directly as well, but I don't know of any simple written account. Hopefully someone knows a good reference. In general, Langlands proved Weil's conjecture for simply connected Chevalley groups with the help of his general theory of Eisenstein series. 
A: I have only read Paul Garrett's paper. I find the use of Poisson summation rather mysterious, but it seems to me that it applies mutatis mutandis to the adelic case.
Very short version: the generic orbit of $SL(V)$ on $V$ is open. If we can compute volumes for $V$ and the stabilizer $N$, then we can do so for $SL(V)$.
It seems to me that most of the argument works for pretty general locally compact ring $A$ containing a lattice $B$. I see three steps:
1. We must choose Haar measures. First, scale Haar measure on $A$ so that $\mathop{vol}(A/B)=1$; this is important for Poisson summation. Choose invariant differential forms defined over $\mathbb Z$ for the three groups; these are unique up to sign. The differential forms lift Haar measure from $A$ to $G(A)$. Because the product of the differential forms on $V$ and $N$ is that of $SL(V)$, so are the Haar measures multiplicative. This gives the best normalization for understanding the large group in terms of the small groups, but if you care about other measures (eg, the measure on hyperbolic space), you may need a correction factor. For number rings, often a power of the discriminant is needed.
2. We need to understand the orbits of $SL(V)(R)$ on $V(R)$. In particular, if $R$ is a field, the action is transitive away from the origin. For any ring, $SL(V)(R)$ is transitive on the Zariski open $(V-0)(R)$. This is not true for general quotients. For example, $PGL_2$ is the quotient variety of $SL_2$, but $SL_2(\mathbb Q)$ does not surject onto $PGL_2(\mathbb Q)$. Thus for any representation of $SL_2$ that factors through $PGL_2$, such as the adjoint representation, $SL_2(\mathbb Q)$ is not transitive on the $\mathbb Q$-points of the algebraic orbit.
3. We need to understand the other orbits. They reflect the ideals of $R$. There is a tradeoff between the cases $\mathbb Z\subset\mathbb R$ and $\mathbb Q\subset\mathbb A$, in that we can't have both rings be fields. However, the complement for the locally compact ring has measure zero and is unimportant, while the counting measure on the discrete ring means that every point counts. This is an advantage of the adelic case and explains its simpler answer.
Finally, we have the argument. We take a Schwartz class function $f:V(A)\to\mathbb R$, and compute 
$$\int_{G(A)/G(B)}\sum_{x\in V(B)}f(gx)\,dg$$
One way to compute it is to break $V(B)$ into pieces according to the orbits of $SL(V)$. The zero orbit contributes $\mathop{vol}(G(A)/G(B))f(0)$. In the PID case, the other orbits are all scaled versions of the open orbit. Summing the scaling factors yields the zeta function of $B$ evaluated at the dimension of $V$. If $N$ is the stabilizer of $v$, then using point (1), the contribution of the open orbit is
$$\int_{G(A)/N(B)}f(gv)\,dg=\mathop{vol}(N(A)/N(B))\int_{(V-0)(A)}f(x)\,dx
=\mathop{vol}(N(A)/N(B))\hat f(0)$$ 
Thus the original integral is $\mathop{vol}(G(A)/G(B))f(0)+\zeta \mathop{vol}(N(A)/N(B))\hat f(0)$. Using Poisson summation for $V(B)\subset V(A)$ allows us to interchange $f$ with its Fourier transform and conclude that 
$\mathop{vol}(G(A)/G(B))=\zeta \mathop{vol}(N(A)/N(B))$. In the adelic case, the factor of $\zeta$ is always $1$ and this shows by induction that $\tau(SL(V))=1$. But the argument also explains where the factor of $\zeta(s)$ comes from in the case of $\mathbb Z\subset \mathbb R$ and extends to other number rings.
A: It is a pity you don't read French.  For those who do (or after you've learnt to read it), I would recommend Annexe B in the undergraduate text Éléments d’analyse et d’algèbre (et de théorie des nombres) by Pierre Colmez, and also his popular article Un autre monde est possible.
Addendum. The basic idea for computing the volume of $\mathrm{SL}_2(\mathbf{A})/\mathrm{SL}_2(\mathbf{Q})$ is that this space can 
be identified with 
$$
(\mathrm{SL}_2(\mathbf{R})/\mathrm{SL}_2(\mathbf{Z}))
\times \mathrm{SL}_2(\mathbf{Z}_2)
\times \mathrm{SL}_2(\mathbf{Z}_3)
\times \mathrm{SL}_2(\mathbf{Z}_5)
\times\cdots
$$
so the volume in question is the product of the volumes of the various factors.  The computation of a difficult integral shows that the volume of $\mathrm{SL}_2(\mathbf{R})/\mathrm{SL}_2(\mathbf{Z})$ is $\zeta(2)$, and an easy computation shows that the volume of 
$\mathrm{SL}_2(\mathbf{Z}_p)$ is $1-{1\over p^2}$, for every prime $p$.  Finally, the eulerian product $\zeta(2)=\prod_p(1-{1\over p^2})^{-1}$ allows you to conclude that the volume of   $\mathrm{SL}_2(\mathbf{A})/\mathrm{SL}_2(\mathbf{Q})$ is $1$.
