Let

$G_\mathbb{Q} = \text{GL}_2(\mathbb{Q})$

$\mathbb{A} = $ the adeles over $\mathbb{Q}$

$G_\mathbb{A} = \text{GL}_2(\mathbb{A})$

$Z_\mathbb{R} = \{(1,1,...,| \epsilon \cdot \text{id}) : \epsilon \in \mathbb{R}^\times\}$

There are essentially two ways to define a measure on

$$ G_\mathbb{Q} Z_\mathbb{R} \backslash G_{\mathbb{A}}$$

1) A measure that 'comes from' the one on the upper half plane making the map $\Phi$ an isometry from $S_k(\text{SL}_2(\mathbb{Z}))$ (see below for further description)

2) The natural quotient measure induced from the Haar measure on $G_\mathbb{A}$

Question:

## Do the measures 1) and 2) coincide?

Thanks in advance,

FW

*** Additional information ***

Motivation:

Let $f$ be a modular form for SL$_2(\mathbb{Z})$ then one can attach an interesting map $\Phi_f : G_\mathbb{A} \to \mathbb{C}$ to it. If $g = \iota(y) (1|g_\infty) (\kappa | 1)$ is a decomposition into $\iota(y) = (y,y,y,...|y), y \in \text{GL}_2(\mathbb{Q}), g_\infty \in \text{GL}_2(\mathbb{R}), \kappa \in \widehat{G_\mathbb{Z}} = \prod_{p ~\text{prime}}\text{GL}_2(\mathbf{Z_p})$ then $$\Phi_f(g) = f|_{g_\infty}(i)$$ is independent of the choosen decomposition and left invariant under $G_\mathbb{Q} Z_{\mathbb{R}}$. People use 1) to show that $f \mapsto \Phi_f$ is an isometry but they use 2) to verify that the right translation is unitary on the $L^2$ space but want, in fact, both to be true.

Attempts for solution:

It suffices to show that the measure 1) is, say, left invariant under $G_\mathbb{A}$. Clearly, it is left invariant under $G_\mathbb{Q} Z_\mathbb{R}$ and right invariant under $\widehat{G_\mathbb{Z}}$ (also left invariant under this?) but the GL$_2(\mathbb{R})$-part is missing completely.

Detailled description:

We have the following

THEOREM: Let $X$ be locally compact and hausdorffian (LCH for short). Let $G$ be an LCH group that acts on $X$ (say, from the right) such that the map $$\phi : X \times G \to X \times X, ~~~ \phi(x, g) = (xg, x)$$ is continuous and proper. Then

a) For every Radon measure (= a measure on the Borel-sigma-algebra satisfying certain regularity properties) $\mu$ on $X$ that satisfies $\mu(Ag) = \Delta_G(g) \mu(A)$ we get a unique measure $\nu$ on $X/G$ such that for all $f \in C_c(X\to\mathbb{C})$ (= compactly supported, continuous functions from $X$ to $\mathbb{C}$) $$\int_X f(x) d\mu(x) = \int_{X/G} f^b(C) d\nu(C)$$ where $f^b(xG) = f^{\text{sym}}(x)$ and $$f^\text{sym}(x) = \int_G f(xg) dg$$ here, $dg$ denotes a fixed choice of left Haar measure on $G$.

b) Conversely, for every Radon measure $\nu$ on the quotient $X/G$ we get a unique Radon measure $\mu$ on $X$ such that the formula above holds for all $f \in C_c(X \to \mathbb{C})$. This measure satisfies $\mu(Ag) = \Delta_G(g) \mu(A)$.

In case this is met, the quotient integral formula holds for all $f \in L^1(X)$.

If $X$ is an LCH group and $G$ is a closed subgroup then the action is continuous and proper and the condition about the 'invariance' translates into the well known one: $\Delta_X|_G = \Delta_G$.

Now 2) is just this quotient measure (one needs to show that GL$_2(\mathbb{Q})$, GL$_2(\mathbb{A})$ are unimodular here).

On 1)

The total process looks like

$$\mathbb{H} \searrow \Gamma\backslash\mathbb{H} \rightarrow \Gamma^{\pm} \backslash G^1_\mathbb{R} / \text{SO}(2) \nearrow \Gamma^{\pm} \backslash G^1_\mathbb{R} \rightarrow G_\mathbb{Q} Z_\mathbb{R} \backslash G_\mathbb{A} / \widehat{G_\mathbb{Z}} \nearrow G_\mathbb{Q} Z_\mathbb{R} \backslash G_\mathbb{A} $$ There is a measure $dxdy/y^2$ on $\mathbb{H} = \{x + iy \in \mathbb{C} ~|~ y > 0\}$ meaning that $\mu(A) = \int_\mathbb{R_{>0}} \int_{\mathbb{R}} \frac{1}{y^2} dx dy$.

We can use a) in the theorem to get a measure on $\Gamma\backslash\mathbb{H}$ where $\Gamma = \text{SL}_2(\mathbb{Z})$.

Using the homeomorphism $\Gamma \tau \mapsto \Gamma^{\pm} g\text{SO}(2)$ for any $g \in \text{SL}_2(\mathbb{R})$ with $g.i = \tau$ (here, $\Gamma^\pm = \Gamma \cup \begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix} \Gamma$) we can push this measure to $$\Gamma^{\pm} \backslash G^1_\mathbb{R} / \text{SO}(2)$$ where $G^1_\mathbb{R} = \{g \in \text{GL}_2(\mathbb{R}) : \det(g) = \pm 1\}$.

Using b) from the theorem above we get a measure on $\Gamma^\pm \backslash G^1_\mathbb{R}$.

Using the homeomorphism $\Gamma^\pm g \mapsto G_\mathbb{Q} Z_{\mathbb{R}} (1|g) \widehat{G_\mathbb{Z}}$ we can push ths measure to $$ G_\mathbb{Q} Z_\mathbb{R} \backslash G_\mathbb{A} / \widehat{G_\mathbb{Z}} $$

Using the theorem once more we finally get a measure on $G_\mathbb{Q} Z_\mathbb{R} \backslash G_\mathbb{A}$.

Since the groups we divided out on the right are compact, the symmetrization process $f \mapsto f^b$ is quite easy and does not do anything for functions that are right invariant under the compact group. If one follows the quotient integral formulas closely than indeed, $\Phi$ becomes an isometry.