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

  • $\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 $G_\mathbb{A}$


Do the measures 1) and 2) coincide?

Thanks in advance,


* Additional information *


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.

  • $\begingroup$ In your diagram going from $\mathbb{H}$ to $G_{\mathbb{Q}}Z_{\mathbb{R}}\backslash G_{\mathbb{A}}$ there is a right action of $GL_2(\mathbb{R})$ at every stage, and so the measure stays invariant under this (In fact, you have to remove $\Gamma^{\pm}\backslash G^1_{\mathbb{R}}/SO(2)$ but that parts unnecessary to have in there anyways). So the final measure is right invariant under $GL_2{\mathbb{R}}\times \hat{G_{\mathbb{Z}}}$. Since this group acts transitively on your final space, the claim follows. $\endgroup$
    – jacob
    Mar 6, 2014 at 17:29
  • $\begingroup$ Thanks. I have some questions still: 1) what is the right action on $\Gamma\backslash\mathbb{H}$? 2) When i remove what you said, how to get from $\Gamma\backslash \mathbb{H}$ to the adelic quotient? I have to put some homeomorphisms and removals of compact groups in between... the map $\tau \mapsto G_Q Z_R (1|g) \widehat{G_Z}$ for any $g$ such that $g.i=\tau$ is not a homeomorphism! $\endgroup$ Mar 6, 2014 at 17:34
  • $\begingroup$ Oops, you're right, that was nonsense. Theres only a left action. Still, what about this: $$\mathbb{H}\rightarrow G^1_{\mathbb{R}}\rightarrow Z_{\mathbb{R}}\backslash G_{\mathbb{R}}\times \hat{G_{\mathbb{Z}}}$$ This is a diagram through which you can transfer measures, you clearly end up with the haar measure on the right, and it seems compatible with your diagram. $\endgroup$
    – jacob
    Mar 6, 2014 at 18:02
  • $\begingroup$ Just to make sure I get the point: You are talking about the map $\tau \mapsto (Z_Rg, 1)$ for any $g$ such that $gi=\tau$, right? I do not really understand, this map is not well defined (i.e. it depends on choices!)... $\endgroup$ Mar 6, 2014 at 18:31

3 Answers 3


Yes, the two are the same, but I'd be inclined to separate the issues into first converting automorphic forms (with automorphy factors) on the domain $\mathfrak H$ into automorphic forms on the group $G=SL_2(\mathbb R)$ or $GL_2(\mathbb R)$, and look at the measures as a separate issue.

That is, for $f(\gamma\cdot z)=f(z)\cdot j(\gamma,z)$ for some cocycle $j$ that extends to a cocycle on $G\times \mathfrak H$, let $F_f(g)=f(g\cdot i)\cdot j(g,i)^{-1}$. This is now left $\Gamma$-invariant (no cycle) and right $K=SO(2,\mathbb R)$-equivariant by $g\to j(g,i)^{\pm 1}$. The Petersson inner product $\langle f_1,f_2\rangle$ is the integral of $f_1(z)\bar{f}_2(z)y^{2k}$, and the latter is $\Gamma$-invariant. Further, this is equal to $F_{f_1}(g)\cdot\bar{F}_{f_2}(g)$ for any $g\in G$ such that $g(i)=z$.

The measures on all the available physical spaces are uniquely determined from the Haar measures on the various groups $GL_2(\mathbb R)$ and $GL_2(\mathbb Q_p)$. In fact, if we follow Weil's prescription (as in "Adeles and Algebraic Groups") we see that the product of all the local Haar measures is uniquely determined up to a constant. (Cf. "Tamagawa measure".)

Whether or not we use that prescription, the only thing needed to see that the various measure determine each other is that very general property that for any topological group $G$ and closed subgroup $H$, $G/H$ admits a left $G$-invariant measure if and only if the modular function of $G$ restricted to $H$ is the modular function of $H$, and, if so, then $$ \int_G f(g)\,dg = \int_{G/H} \int_H f(gh)\;dh\;dg $$ for $f\in C^o_c(G)$.

In the case at hand, the left $G$-invariant measure on $\mathfrak H\approx G/K$ is thus determined by choice of total measure on $K$ and choice of Haar measure on $G$.

To subsequently make automorphic forms on the Lie group $G$ into automorphic forms on the adele group, some sort of Strong Approximation suffices, but we still need to specify the constant multiple of Haar measure we take on $SL_2(\mathbb Q_p)$, etc.

Sure, we can specify the Haar measure on $SL_2(\mathbb A)$ or $GL_2(\mathbb A)$ all at once, too, rather than factor-wise.

In any case, I think there is no need to work so hard as to talk about measures on quotients of spaces by groups. All that is needed is quotients $G/H$ (or $H\backslash G$).

Edit: left $\Gamma$-invariant right $K$-invariant functions on $G$ are in $\Gamma\backslash H\approx \Gamma\backslash G/K$, but/and the latter is conveniently the right $K$-fixed elements in functions on $\Gamma\backslash G$, e.g., $L^2(\Gamma\backslash G)^K$. Since $K$ is compact, all the useful relations hold... In particular, the measure on $\Gamma\backslash G$ is the natural right $G$-invariant one, and the right action of $G$ is unitary, etc. That is, removing the obstacle of right-quotient by $K$ is just a matter of viewpoint.

Ignoring the center for notational simplicity, strong approximation gives $G_k\cdot G_\infty \cdot K_f=G_{\mathbb A}$ for any compact-open in the finite component. Thus, $$(G_k\cap K_f)\backslash G_\infty \approx G_k\backslash G_{\mathbb A}/K_f \approx (G_k\backslash G_{\mathbb A})^{K_f}$$ That is, the automorphic forms of various levels are images by the projectors associated to $K_f$ on $L^2(G_k\backslash G_{\mathbb A})$.

  • $\begingroup$ Im sorry, I do not understand your answer. Let us do what you said: Then we end up with a function on SL$_2(R)$/SO(2). How to get from there to $G_QZ_R\backslash G_A$? In particular, we will have to get there while still keeping some kind of relation of the measures as we want the adelization to be an isometry. Secondly: If we just proceed to SL$_2(R)/SO(2)$, how to show that the map is an isometry here? The natural thing to view $f\overline{g}y^{2k}$ on is $\Gamma\backslash H$ and this is 'space/group' and not 'group/subgroup'! $\endgroup$ Mar 7, 2014 at 6:38
  • $\begingroup$ Side remark: The case 'space/group' is not harder in any sense than 'group/subgroup' as the proof is very much the same as for the case 'group/subgroup'... $\endgroup$ Mar 7, 2014 at 6:41
  • $\begingroup$ Notes on your edit: (1) What is the reason for $L^2(\Gamma \backslash G /K) \cong L^2(\Gamma \backslash G)^K$? This is precisely the assertion in the theorem stated in my post. [(a) or (b) depends on the direction from where you come] (2) So, what you are saying is that the second rightarrow in the diagram is a homeomorphism. How does that help? It is not at all clear that the measure on $\Gamma^\pm \backslash G^1_R$ is the natural quotient measure! All that one knows is that it arises as a measure from $\Gamma\backslash H$ as in the theorem in my post. $\endgroup$ Mar 8, 2014 at 5:59
  • $\begingroup$ @FabianWerner, apparently I do not understand your context. E.g., anything being unclear about the measure on $\Gamma\backslash G$ not being clearly the quotient measure, and this not being clearly the measure on $\Gamma\backslash \mathfrak H$, etc. Is it possible that you are not asking precisely the question you intend? E.g., one might ask why the measure on $G=P\cdot K$ is left Haar on $P$ and right Haar on $K$, since, as you note, in effect the left Haar on $P$ is what "more obviously" descends to $\mathfrak H$? $\endgroup$ Mar 8, 2014 at 13:43
  • $\begingroup$ The context is as stated above. Starting with a measure on the upper half plane $H$ , let it descend to $\Gamma \backslash H$. This measure has no invariance property as there is no action of, say GL$_2^+(R)$ on $\Gamma\backslash H$ (GL$_2^+(R)$ just acts on $H$ itself, not on the quotient)! Then we view this measure as a measure on $\Gamma^\pm \backslash G^1_R /SO(2)$ using that the spaces are homeomorphic. We let it ascend to a measure on $\Gamma^\pm \backslash G^1_R$. Question: is this the Haar measure of $G^1_R$ 'descended' to the quotient? This is not immediate! $\endgroup$ Mar 8, 2014 at 14:49

You simply have to use uniqueness.

Let $G$ be a locally compact group with closed subgroup $H$ such that the modular functions of $H$ and $G$ coincide on $G$, then there is up to normalization by a positive real only one $G$ right invariant measure on $H \backslash G$.

Now you have defined an $G$-equivariant isomorphic maps between the two spaces $i: X_1 \rightarrow X_2$, so you only have to evaluate the corresponding integrals for one specific $f$ on $X_2$ and $f \circ i$ on $X_1$, and check if they coincide.

The measure on the adelic group is often chosen to be the Tamagawa measure, which does not coincide with the classical measure on the upper halfplane, which is given in a different normalization. The measure of the compact subgroups is e.g. not normalized to be one in the adelic setting, check Gelbart-Jacquet Analytic theory of $GL(2)$ automorphic forms...

  • $\begingroup$ But this is precisely what the question is about: why is this pushforward measure [that originates in the long diagram described above] from the one on $H$ right invariant under the whole group $G_A$? Usually, the algebraic invariance is easy and the regularity is tricky, but here it is the other way around! What is the action of GL$_2(R)$ supposed to do on $\Gamma\backslash H$ ? $\endgroup$ Mar 7, 2014 at 16:43
  • $\begingroup$ Möbius transformations, where $z$ is equivalent to $-z$. Classically one uses upper plus lower halfplane. What is regularity? You mean normalization? $\endgroup$
    – Marc Palm
    Mar 7, 2014 at 18:12
  • $\begingroup$ The point is that the normal(=Tamagawa) measure does not coincide with the normal measure. The Tamagawa measure is a rational number, the measure on $X$ is a multiple of $1/\pi$. So no if you mean normal=Tamagawa. $\endgroup$
    – Marc Palm
    Mar 7, 2014 at 18:18
  • $\begingroup$ No, I meant regularity = outer/inner/... regularity of the measure. How to act an $\Gamma\backslash H$ by Moebius transformations? The action of $GL_2^+(R)$ is a left action and we have divided out something on the left that does not commute with $GL_2^+(R)$! $\endgroup$ Mar 7, 2014 at 18:36
  • $\begingroup$ You have a finite volume Radon measure. It is automatically regular. That's why I am confused. $\endgroup$
    – Marc Palm
    Mar 8, 2014 at 13:32

Ok, thanks to the discussion with Paul Garrett and Marc Palm I got the idea:

One starts with the Haar measure on $G_\mathbb{A}$. The one forms the quotient measure on $G_\mathbb{Q} Z_\mathbb{R} \backslash G_\mathbb{A}$. First of all, we form another quotient measure from this, namely on $G_\mathbb{Q} Z_\mathbb{R} \backslash G_\mathbb{A}/ K_\Gamma$ where $K_\Gamma = \prod_{p ~ \text{prime}} GL_2(\mathbf{Z_p}) \times \{id_\infty\}$. Although we divided out something on the right, we may still endow $G_\mathbb{Q} Z_\mathbb{R} \backslash G_\mathbb{A}/ K_\Gamma$ with a $G^1_\mathbb{R}$-right-action: Since the image of $G^1_\mathbb{R}$ in $G_\mathbb{A}$ and $K_\Gamma$ commute, $$(G_\mathbb{Q} Z_\mathbb{R} x K_\Gamma).g := G_\mathbb{Q} Z_\mathbb{R} x * (1|g) K_\Gamma$$ (where $*$ means the multiplication in $G_\mathbb{A}$) is well defined. The homeomorphism

$$\Gamma^\pm \backslash G^1_\mathbb{R} \to G_\mathbb{Q} Z_\mathbb{R} \backslash G_\mathbb{A}/ K_\Gamma$$

is clearly right-$G^1_\mathbb{R}$-invariant and hence, so is its inverse. Hence, if we push the measure from $G_\mathbb{Q} Z_\mathbb{R} \backslash G_\mathbb{A}/ K_\Gamma$ to $\Gamma^\pm \backslash G^1_\mathbb{R}$, this is right-$G^1_\mathbb{R}$-invariant. As measures with this (and some additional regularity) properties are unique up to a factor, this measure coincides with the natural quotient measure on $\Gamma^\pm \backslash G^1_\mathbb{R}$. So it suffices to show that the measure that we get from the 'chain'

$$ \Gamma^\pm \backslash G^1_\mathbb{R} \searrow (\Gamma^\pm \backslash G^1_\mathbb{R} ) / SO(2) \cong \Gamma \backslash \mathbb{H} $$

coincides with the natural quotient measure. Here we do not have an action that we can drag along but it goes like this:

By the same argument as above, the homeomorphism $\Gamma \backslash SL_2(\mathbb{R}) \cong \Gamma^\pm \backslash G^1_\mathbb{R}$ is right-$SL_2(\mathbb{R})$-invariant, so the push of the measure on $\Gamma^\pm \backslash G^1_\mathbb{R}$ to $\Gamma \backslash SL_2(\mathbb{R})$ is (up to constant) the natural quotient measure. So, including measures

$$(\Gamma^\pm \backslash G^1_\mathbb{R}) / SO(2) \cong (\Gamma \backslash SL_2(\mathbb{R}))/SO(2) $$

By a different calculation (or by definition, if you wish), the well known measure on $\mathbb{H}$ really is the push of the quotient measure of $SL_2(\mathbb{R})/SO(2)$. Hence, including measures

$$ \Gamma \backslash \mathbb{H} \cong \Gamma \backslash (SL_2(\mathbb{R}) / SO(2))$$

All in all, it suffices to see the following:

Given an LCH group $G$ and closed subgroups $A, B$ such that $B$ is compact and $A$ operates (continuously and) properly on $G/B$ from the left then the pushforward of the iterated quotient measure $A\backslash (G/B)$ along the homeomorphism $\omega^{-1} : A \backslash (G/B) \cong (A \backslash G)/B$ is precisely the natural (iterated) quotient measure of the right hand side. This follows from the fact that for every $f \in C_c(G \to \mathbb{C})$, the processes

$$\omega^* \circ ~ \text{symmetrization to the left by $A$} ~\circ ~\text{symmetrization to the right by $B$}~ (f)$$ and

$$\text{symmetrization to the left by $B$} ~\circ ~\text{symmetrization to the right by $A$}~ (f)$$ give precisely the same result. This in turn can be shown using the Theorem of Fubini ('$(a,b) \mapsto f(axb)$' is in $L^1(A \times B)$ as it is compactly supported).


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