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Some papers I am reading talk about an "adelic" object $PGL(2, \mathbb{Q}) \backslash PGL(2, \mathbb{A})$ . This has sparked a lot of confusion since I don't know what such a quotient could mean.

A crude way of looking at the adéles is just as the product over primes:

$$ \mathbb{A}_\mathbb{Q} = \mathbb{R} \times \mathbb{Q}_{p_1} \times \dots \times \mathbb{Q}_{p_k} \dots $$

Naively one might assume this passes over to groups of fractional linear transformations. I believe the term is "strong approximation" though it doesn't make it any easier to understand.

$$ PGL(\mathbb{A}_\mathbb{Q}) = PGL(\mathbb{R}) \times PGL(\mathbb{Q}_{p_1}) \times \dots \times PGL(\mathbb{Q}_{p_k}) \dots $$

Actually even if we take just one part of that object the object is hard to understand, since $\mathbb{R}/\mathbb{Q}$ is already a nasty object:

$$PGL(2, \mathbb{Q}) \backslash PGL(2, \mathbb{R})$$

I don't really understand what the rationals are doing here. The only one I kind of understood is the identificaiton of the hyperbolic plane $\mathbb{H}^3 = PGL(2, \mathbb{R}) \backslash PGL(2, \mathbb{C})$. How to understand such a complicated group action?

It seems that for any two groups $H \subset G$ we could have $PGL(2, \mathbb{H} \backslash PGL(2, \mathbb{G})$.

OK. This object seems to be familiar to experts on automorphic forms - which I am definitely not:

Partial progress The issue of diagonal embedding $\mathbb{Q} \subset \mathbb{A}$ and the solenoid structure of $\mathbb{A}/\mathbb{Q}$ are two major points that I missed. The original question merely asked "What is $PGL(2, \mathbb{Q}) \backslash PGL(2, \mathbb{A})$?"

Although these points are in books, it would be great an outline of the "adèlic solenoid" structure of $PGL(2, \mathbb{Q}) \backslash PGL(2, \mathbb{A})$.

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    $\begingroup$ An easier question, which I think must be answered first, is: what is $\mathbb{Q}\backslash\mathbb{A}$? For answers, see e.g. this paper by A. Robert: retro.seals.ch/… You may also enjoy Weil's Basic number theory. $\endgroup$
    – Alain Valette
    Commented May 27, 2015 at 22:08
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    $\begingroup$ Of course you know that $\mathbb Q$ is embedded diagonally here, right? In this sense it is discrete because two 'real'ly close rationals have large denominators, so that there is some $\mathbb Q_p$ that sees them as being far apart. $\endgroup$
    – LSpice
    Commented May 27, 2015 at 22:58
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    $\begingroup$ Oh, also, the adèle (not adéle) ring $\mathbb A$ is significantly smaller than the full direct product $\mathbb R \times \prod_p \mathbb Q_p$, which would not be locally compact. One takes instead the subring of those tuples $t$ for which, for almost all (i.e., all but finitely many) $p$, we have $t_p \in \mathbb Z_p$. It may help in understanding the topology (or at least the discreteness of $\mathbb Q$) to see why this subring contains $\mathbb Q$. $\endgroup$
    – LSpice
    Commented May 27, 2015 at 23:01
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    $\begingroup$ The "solenoid structure" you are asking about is addressed in the last sentence of my response. Briefly, $\mathrm{PGL}_2(\mathbb{Q})\backslash\mathrm{PGL}_2(\mathbb{A})$ can be identified with the inverse limit of $\Gamma(N)\backslash\mathrm{PGL}_2(\mathbb{R})$, where $\Gamma(N)$ is the usual principal congruence subgroup modulo $N$. $\endgroup$
    – GH from MO
    Commented May 28, 2015 at 0:09
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    $\begingroup$ Before considering $G(\mathbb Q)\backslash G(\mathbb A)$, consider $G(\mathbb Z[\frac12])\backslash G(\mathbb R\times\mathbb Q_2)$. Also, before $G=PGL_2$, consider $G=\mathbb G_a,\mathbb G_m,SL_2$. $\endgroup$
    – Ben Wieland
    Commented May 28, 2015 at 1:07

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Yes, the quotient $\mathrm{PGL}_2(\mathbb{Q})\backslash\mathrm{PGL}_2(\mathbb{A})$ and its generalizations for other (reductive) algebraic groups is a complicated object, and this is to a large extent the reason why the theory of automorphic forms is a deep subject. The diagonal embedding of $\mathrm{PGL}_2(\mathbb{Q})$ into $\mathrm{PGL}_2(\mathbb{A})$ connects the quasi-factors $\mathrm{PGL}_2(\mathbb{Q}_v)$ in a subtle way, which otherwise would be completely independent. We would like to understand how much dependence is introduced and how much independence is lost by taking the quotient of $\mathrm{PGL}_2(\mathbb{A})$ by $\mathrm{PGL}_2(\mathbb{Q})$. This fits nicely in the general local-to-global philosophy of number theory.

By the way, one cannot just throw away some quasi-factors from $\mathrm{PGL}_2(\mathbb{A})$ and take a quotient by $\mathrm{PGL}_2(\mathbb{Q})$, because the latter is meant to be embedded diagonally into $\mathrm{PGL}_2(\mathbb{A})$, i.e. it appears in every quasi-factor. Hence $\mathrm{PGL}_2(\mathbb{Q})\backslash \mathrm{PGL}_2(\mathbb{R})$ or even $\mathrm{PGL}_2(\mathbb{Q})\backslash \prod_{v\neq 2}\mathrm{PGL}_2(\mathbb{Q}_v)$, say, have little to do with the true adelic quotient $\mathrm{PGL}_2(\mathbb{Q})\backslash\mathrm{PGL}_2(\mathbb{A})$.

At any rate, there are good introductions to adelic quotients. I recommend Chapter IV in Weil: Basic number theory, especially Section 2 there which explains why the adeles are separating the rationals much like the reals are separating the integers. Then one can read Sections 3.3 and 3.6 in Bump: Automorphic forms and representations, which explains in the setting of $\mathrm{PGL}_2$ the connection of the adelic quotient to classical congruence quotients.

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  • $\begingroup$ no I missed the part about diagonal embedding. Does it look like this? $$ \prod_{\{-1\} \cup \text{primes}}\mathrm{PGL}_2(\mathbb{Q})\backslash \mathrm{PGL}_2(\mathbb{Q}_p)$$ $\endgroup$
    – john mangual
    Commented May 27, 2015 at 23:29
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    $\begingroup$ @johnmangual: What you write is not the adelic quotient. You have to divide the restricted product $\mathrm{PGL}_2(\mathbb{A})$ instead of taking a restricted product of the local quotients (which are not even Hausdorff). Note also that $\mathbb{R}$ is usually denoted by $\mathbb{Q}_\infty$ instead of $\mathbb{Q}_{-1}$. $\endgroup$
    – GH from MO
    Commented May 27, 2015 at 23:32
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    $\begingroup$ @johnmangual: For comparison, $\mathbb{R}^2$ divided by $\{(x,x):\ x\in\mathbb{Z}\}$ is very different from $(\mathbb{R}/\mathbb{Z})^2$. In the first case you get a group isomorphic to $\mathbb{R}\times(\mathbb{R}/\mathbb{Z})$. $\endgroup$
    – GH from MO
    Commented May 27, 2015 at 23:40

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