1
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ 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 May 27 '15 at 22:08
  • 2
    $\begingroup$ 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 May 27 '15 at 22:58
  • 2
    $\begingroup$ 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 May 27 '15 at 23:01
  • 1
    $\begingroup$ 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 May 28 '15 at 0:09
  • 2
    $\begingroup$ 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 May 28 '15 at 1:07
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 May 27 '15 at 23:29
  • 1
    $\begingroup$ @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 May 27 '15 at 23:32
  • 2
    $\begingroup$ @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 May 27 '15 at 23:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.