$\DeclareMathOperator\GL{GL}$This question might be beyond naive but I really can't find it nor figure it out. It's related to explicit idelic integration as opposed to just bounding its values.
Let $K$ be an algebraic number field whose class number is $h_K$ and number of roots of unity is $w_K$.
Consider $\GL(2, \mathbb{A})$ the general linear group over the adeles with some measure choices that give a Haar measure, $G(K)$ the counting measure, $Z_+$ the Lebesgue measure $dz/z$.
Let $G_{\infty}$ be the infinite part of $\GL(2, \mathbb{A})$.
The question is the following:
Let $g = \displaystyle\prod_{\mathfrak{q}}g_{\mathfrak{q}} \times g_{\infty}$ be a factorable function with $g\in C_c(Z_+\backslash G_{\infty})$. Then we have \begin{equation*} \displaystyle\int_{Z_+\backslash G(\mathbb{A})} g(a)da = h_K\displaystyle\prod_{\mathfrak{q}} \int_{\GL_2(K_{\mathfrak{q}})}g_{\mathfrak{q}}(x_{\mathfrak{q}})dx_{\mathfrak{q}} \times \int_{Z_+\backslash G_{\infty}}g_{\infty}(x_{\infty})dx_{\infty}, \end{equation*} where $h_K$ is the class number.
The difference being that on the left I am integrating the quotient measure, while on the right I first break up into parts, take the quotient on the second factor and then the product measure.
I have checked all over the place and I find that these sort of constants do appear (mostly $h_K/w_k$, when you further embed the units into the infinite part of the adeles via the log map (when you prove analytic class number formulas or when you do strong approximation theorems). However I don't find anything that states that you can nicely break up the integral once you do it in $Z_+\backslash G(\mathbb{A})$ (as opposed to $G(\mathbb{A})$, where you can of course do it) nor can justify it myself.
Does anyone know if the above is correct or what is the correction (or if its just totally not this way?)?
Thanks!
