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In order to calculate Tamagawa numbers, I need to justify that for a nice (say Schwartz-Bruhat) function, the following identity holds :

$$\int_{\mathbf{A}^2} f(x)dx = \int_{SL_2(\mathbf{A})/SL_2(K)} \left( \sum_{a \in K^2 - 0} f(ua) \right) du$$

where $\mathbf{A}$ are the adeles on $K$, and $du$ is a Tamagawa measure on $SL_2(\mathbf{A})/SL_2(K)$. It seems to be easy if we know the Tamagawa numbers, but obviously it would be vicious...

The natural decomposition I want to star from is that of the transitive action of $SL_2(\mathbf{A})$ on $\mathbf{A}^2 - 0$, hence $\mathbf{A}^2 - 0 \cong SL_2(\mathbf{A}) / \mathbf{A}$, $N(\mathbf{A}) \cong \mathbf{A}$ being the stabilizer of $(1,0)$. But then, trying to decompose I cannot go further, for instance (dropping the integrands) and neglecting 0 (should I ?) :

$$\int_{A^2} = \int_{A^2/K^2} \sum_{K^2}$$

and then

$$\int_{A^2/K^2} = \int_{(SL_2(A)/A)/(SL_2(K)/K)}$$

is there so an isomorphism theorem I don't see how to use efficiently here ?

Thanks in advance for any help or idea, I'll dig further any clue. And sorry if it is indeed a trivial affair.

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    $\begingroup$ This is treated in A. Weil's book on 'Adeles and Algebraic Groups'. In particular, see section(s up to) 3.4. $\endgroup$
    – Daniel m3
    Commented Nov 25, 2014 at 10:46
  • $\begingroup$ Works well indeed, thanks for the reference ! $\endgroup$
    – Desiderius Severus
    Commented Nov 25, 2014 at 10:56
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    $\begingroup$ Note that ${\rm{SL}}_2(\mathbf{A})$ does not act transitively on $\mathbf{A}^2 - 0$, for the same reason that ${\rm{SL}}_2(\mathbf{Z}_p)$ does not act transitively on $\mathbf{Z}_p^2 - 0$. Explicitly, if $a \in \mathbf{A}$ is an adele whose $p$th component is contained in $p\mathbf{Z}_p$ for infinitely many primes $p$ then $(a,a)$ is not part of an $\mathbf{A}$-basis of $\mathbf{A}^2$. The same happens for $(b,b)$ for any adele $b$ which has some vanishing component and some non-vanishing component. $\endgroup$
    – user74230
    Commented Nov 25, 2014 at 11:06
  • $\begingroup$ For $A:=\mathbf{A}$, your error is that $A^2-0$ is bigger than $X=U(A)$ for the complement $U$ of 0 in the affine $K$-plane. Zariski-stratification doesn't commute with passage to adelic points. (For the affine $K$-line and its complement of 0 this says $A^{\times}$ is a proper subset of $A-0$; it even has measure 0, but the 1-dimensional case is deceptive.) The SL$_2(A)$-action on $X$ is a topological quotient map (apply I,3.6 of Oesterle's paper on Tamagawa numbers to the orbit map), but must be attentive to measures. Relating $G(A)/H(A)$ to $(G/H)(A)$ for topology and measure requires care. $\endgroup$
    – user74230
    Commented Nov 25, 2014 at 11:50
  • $\begingroup$ @user74230 Tanks for the warning, indeed it seems trickier than in the classical case... But then, it seems a problem to conclude with Weil's results, mainly lemma 2.4.2 which only express the integral over $SL_2(A)/A$. Any idea of what the orbits look like ? $\endgroup$
    – Desiderius Severus
    Commented Nov 25, 2014 at 13:26

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