# Categorical description of the restricted product (Adeles)

## Background on the Adèles

The Adèles $\mathbb{A}_K$ of a number field or function field $K$ are defined as a restricted product of the complete local fields $K_\nu$, where $\nu$ ranges over all places of $K$. The restricted product is usually defined as the subset of $\prod_\nu K_\nu$ given by

$\mathbb{A}_K := \prod_\nu' K_\nu := \{ (x_\nu)_\nu \in \prod_\nu K_\nu\ |\ \text{ all but finitely many } x_\nu \in \mathcal{O}_\nu\}$

where $\mathcal{O}_\nu$ is the ring of integers in $K_\nu$.

### Tensor product description

An alternative description, for the sake of concreteness given for the rationals $K=\mathbb{Q}$ can be made by using the tensor product:

$\mathbb{A}_\mathbb{Q} = \left(\left(\prod_p \mathbb{Z}_p\right) \otimes_\mathbb{Z} \mathbb{Q}\right) \times \mathbb{R}.$

This is the same, because $\mathbb{Z}_p \otimes \mathbb{Q} = \mathbb{Q}_p$ and the tensor product captures the finiteness condition. As there are always only finitely many infinite places, this description can be given for any number field as well (and of course function fields, since they don't have infinite places at all).

## The topology on the Adèles

The restricted product comes with a restricted product topology, which is not the subspace topology from the ordinary product (despite its name), but the topology whose subbasis sets are

$V_{\eta,U_\eta} := \{(x_\nu)_\nu \in \prod_\nu K_\nu\ |\ x_\nu \in \mathcal{O}_\nu \text{ for } \nu \neq \eta, \text{ and } x_\eta \in U_\eta\}$

with $\eta$ a place and $U_\eta \subseteq K_\eta$ any open subset. The subspace topology from the product differs from this by requiring only $x_\nu \in \mathcal{O}_\nu$ for all but finitely many places, which are not fixed uniformly for a subbasis set.

Given a subset $U$ of $\mathbb{A}_K$ which is open in the ordinary subspace topology from the ordinary product, for every place $\nu$ there might be an $x \in U$ such that $x_\nu \notin \mathcal{O}_\nu$. If instead $U$ is open in the restricted product topology, there is a fixed finite set of places $\{\nu_1,...,\nu_m\}$ such that for every $x \in U$ and every other place $\nu \neq \nu_i$ we have $x_\nu \in \mathcal{O}_\nu$.

Nice properties of this topology are: You get again a locally compact group with compact open subgroup $\prod_\nu \mathcal{O}_\nu$ and that the Haar measure on $\mathbb{A}_K$ gives the quotient $\mathbb{A}_K/K$ a finite measure (with $K$ embedded diagonally by the maps $K \to K_\nu$).

## The question: how to describe the Adèles categorically?

More specifically, I'd like to understand the restricted topology as well. The ordinary product is a limit, and as such it carries the initial topology. Any subspace carries the initial topology as well, but this gives the wrong topology, not the restricted product topology but the topology restricted from the product.

• Is it impossible to give a categorical description?
• Would it even be useful to have a categorical description?
• Does one have to apply a limit-colimit procedure or might a single limit or colimit suffice?
• There are some similarities with ultraproducts, which are classically not defined in a categorical way, but it is possible. The restricted product is somewhat dual to an ultraproduct. Could that help?
• Is there a good canonical way to topologize the tensor product of topological algebras over a topological ring? Would that solve my problem?
• Which (universal) properties do the Adèles satisfy?

(there was a section with my (non-working) ideas on this, which I removed after the answers came in.)

• You say that the restricted product topology is not the subspace topology despite its name, and I think that means you are parsing the term incorrectly: think about it as (restricted product) topology rather than restricted (product topology), since the adeles themselves are an example of a (restricted product) of topological groups/rings. There is a beautiful characterization of the adele ring of a global field, due to Iwasawa (On the rings of valuation vectors, Annals of Math 57 (1953), 331--3356): if $R$ is a locally compact topological ring containing a subfield $K$ that is (contd.) May 6, 2012 at 18:06
• discrete and co-compact in $R$ (meaning $K$ is discrete in the subspace topology and $R/K$ is compact in the quotient topology), $R$ is not discrete or compact, and the intersection of the closed maximal ideals in $R$ is $\{0\}$, then $K$ is a global field and $R$ is the adele ring of $K$. May 6, 2012 at 18:07
• There is a universal mapping property for adeles, due to Goldman and Sah (On a Special Class of Locally Compact Rings, J. Algebra 4 (1966), 71--95): if a locally compact ring $R$ is an algebra over a global field $K$, has no proper open ideal, and its closed maximal ideals have intersection $\{0\}$, then there is a unique $K$-algebra homomorphism ${\mathbf A}_K \rightarrow R$. So the adeles of $K$ are an initial object in a suitable category of locally compact $K$-algebras. May 6, 2012 at 18:43
• @KConrad: This is an answer. May 7, 2012 at 7:35
• @KConrad: Thank you! That is what I was after. May 9, 2012 at 16:37

I was puzzled by this question as well a while ago. Here is a suggestion, which worked for me:

1. The product and the co-product of categories are best defined by an universal mapping property.

2. The adeles $\mathbb{A}$ are the inductive limit of all $S$-adeles $$\mathbb{A}(S) = \mathbb{R} \times \prod\limits_{p \in S} \mathbb{Q}_p \times \prod\limits_{p \notin S} \mathbb{Z}_p$$ for a finite set of places. The universal property is described as follows: If you are given a map $$\phi : \mathbb{A} \rightarrow X$$ to some topological space $X$, then there exists for every large enough set $S$ a unique $\phi_S : \mathbb{A}(S) \rightarrow X$ such that $\phi_S = \phi$ on $\mathbb{A}(S)$.

Remark: For this to work, it is important that all but finitely many compact subrings ($\mathbb{Z}_p$ in the above context) are actually open subrings, since you have to choose the characteristic functions of this set at the places $p \notin S$ to construct $\phi_S$. So I doubt that the restricted product can be defined in the category of locally compact groups/rings/spaces in such a fashion. On the other hand, the restricted product can be defined for a family of pairs of topological spaces $(A_i \supset B_i)_i$ with almost all $B_i$ are open in $A_i$.

• I've seen Tate write the restricted product as $\rlap{\prod}\coprod$; i.e. a product combined with a coproduct, just as you decsribe it. May 6, 2012 at 19:07
• I like that symbol for the restricted product. How do you code for it in TeX or LateX ? May 7, 2012 at 2:30
• \mathrlap{coprod}\prod should do, see e.g. here... May 7, 2012 at 6:45
• tex.stackexchange.com/questions/21644/… May 7, 2012 at 6:45
• @Mrc Plm: Thank you! So the solution of my headache was to see the infinitely many conditions as a colimit over finitely many conditions. That could work in other settings, too. @Ryan Reich: Thanks for the suggestion, I like Tate's symbol. May 9, 2012 at 16:40

Personally I think that the restricted product description should be avoided. It is best to define $\widehat{\mathbb{Z}}$ to be the inverse limit of the system of all quotients $\mathbb{Z}/n$ (without gratuitously factoring $n$ as a product of primes) and then put $\mathbb{A}=(\mathbb{Q}\otimes\widehat{\mathbb{Z}})\times\mathbb{R}$. We can topologise this by giving $\mathbb{R}$ the usual topology, and $\mathbb{Q}\otimes\widehat{\mathbb{Z}}$ the topology for which the sets $q\otimes\widehat{\mathbb{Z}}$ form a basis of neighbourhoods of zero. Now the adeles for any number field $K$ can be defined as $\mathbb{A}\otimes K$. Any $\mathbb{Q}$-basis for $K$ identifies $\mathbb{A}\otimes K$ with $\mathbb{A}^d$ and thus gives a topology on $\mathbb{A}\otimes K$, which is easily seen to be independent of the choice of basis. The connection with primes/valuations for $K$ should be a theorem, not a definition.

• Thanks. To make that description work for me, there must be a natural topology on the tensor product of topological rings. What is it? May 9, 2012 at 21:06
• @Konrad: I edited the answer to include some comments about topologies. May 9, 2012 at 21:26
• In the spirit of the situation, rather than literally saying that the opens are $q\times \hat{\mathbb Z}$, perhaps saying to give $\mathbb Q$ the discrete topology would be nicer, and then "tensor product" as linearizing bilinear maps, etc. Similarly, stylistically, if we've talked about topologies on tensor products for that reason, then we can avoid talking about choosing bases for $K$, etc. Just quibbles... :) May 10, 2012 at 0:08
• Isn't the fact that your construction is the same is called strong approximation. The ring $\mathbb{Q} \otimes_{\mathbb{Z}} \widehat{\mathbb{Z}}$ seems not so explicit to my taste, but nice picture. May 10, 2012 at 8:20

First, the references cited by KConrad above are unduly neglected/obscure... while answering the question as it stands.

Also, as Mrc Plm's answer, quite literally the adeles are a colimit of products.

Third, indeed, one can (successfully) abstract the "restricted-product topology" (again, as KConrad notes, not "restricted product-topology", but the other grouping), BUT what is not clear is what the point of that would be, I think. That is, apart from discussion of adeles, ideles, adelizations of reductive groups over global fields, and their repn theory, there are not many examples of naturally-occurring "restricted products". Thus, although, if we persevere, we can abstract the notion, after having done so we do not have revelations about how these things were all around us but merely un-named. :)

That is, the notion of "restricted product" is reasonably perceivable as a bit of a let-down, since it explains little.

I find it an interesting rhetorical question "Do adeles occur in nature?", as opposed to "can we construct/axiomatize" them, or "are they useful?" A simpler analogue is the p-adic integers $\mathbb Z_p$ which, although eminently constructible as a completion of $\mathbb Z$, one could ask why make that metric, ... considering that it takes some thought to verify that it is a metric at all, and, then, why complete? That is, by now, to say that $\mathbb Z_p$ is the (projective) limit of $\mathbb Z/p^n$ is more persuasive to me of the occurring-in-nature aspect.

Similarly, the "solenoids" made by taking limits of $\mathbb R/N\cdot \mathbb Z$ have a limit which is arguably a natural object. When we already have $\mathbb Q_p$ in hand, we can exhibit an action of $\mathbb Q_p$ on this solenoid. Honest investigation of how close we can come to making the genuine product of p-adics act leads to the restriction appearing in the blunt definition of adeles. In various logical sequences, one finds that the solenoid is $\mathbb A/\mathbb Q$. In this setting, the compactness of the quotient is immediate, since (Tychonoff) limits of Hausdorff compacts are compact.

That is, one can "discover" much of the "restricted product" notion by trying to write the solenoid as a quotient by $\mathbb Q$ of something that has an action of $\mathbb R$ and all the $\mathbb Q_p$'s on it, etc.

That is, it is possible to give some sense of inevitability to these notions perhaps better than merely giving the definitions.

• It might be true that the restricted product doesn't appear much more often than to define adelic reductive groups, but for me the solution of my problem was quite enlightening: Here is a way to convert an element-dependent description into an element-free one. That is a technique I will surely want to use in other contexts. Thank you for the Solenoids-description, I wasn't aware of that either (it seems I'm an adelic newbie). May 9, 2012 at 16:43
• Yes, I agree that "mapping-property characterizations", rather than "set-theoretic constructions" (=element-wise descriptions) are much better in many situations! Absolutely! The present example may not be the best illustration, but, yes, still, we can do this. And Neal Strickland's answer, that starts with $\hat{\mathbb Z}$ as the limit of $\mathbb Z/n$ (noting, as Neal S. does, that we need not factor $n$), shows another way that "the adele ring" simply appears, rather than needing to "be defined into existence". May 9, 2012 at 18:40
• The restricted product appears quite a bit in nature. The function field analogy in the guise of the (geometric) Langlands correspondence shows that the restricted product that gives the adeles is just a special case of the restricted product of function algebras on all formal discs of an arithmetic/algebraic curve, with the restriction being along the inclusion of the functions on the unpunctured disk. These restricted product function algebras appear naturally as parts of the Cech cocycles for bundles over curves. See section 3.2 of ncatlab.org/nlab/show/restricted+product#Frenkel05 Jul 20, 2014 at 19:34

Let $P$ the pullbak (in the topological groups category) of the natural maps $\prod_{\nu\in N} K_\nu\to \prod_\nu K_\nu/\mathcal{O_\nu}$ and $\coprod_\nu K_\nu/\mathcal{O_\nu}\to \prod_\nu K_\nu/\mathcal{O_\nu}$, then $P$ as set is the restricted product, but has the subspace topology of the product $\prod_\nu K_\nu$. Now $\coprod_\nu K_\nu/\mathcal{O_\nu}$ is the (filtrant) colimit of $(\coprod_{\nu\in F} K_\nu)_{F\subset N finite}$

with coproiections $\coprod_{\nu\in F}K_\nu\to \coprod_\nu K_\nu\to \coprod_\nu K_\nu/\mathcal{O_\nu}$. And by pullback we have that the set $P$ is a (filtrant) colimits $(P_F\to P)_{F\subset N\ finite}$ and $P_F$ is the produt of the $K_\nu\ \nu\in F$ and the $\mathcal{O_\nu},\ \nu\not\in F$, and observe that $P_{F_1}\cap P_{F_2}=P_{F_1\cap F_2}$. Give on $P_F$ the topology inducted by the projection $P_F\to \prod_{\nu\in F} K_\nu$ (the codomain by the product topology).

If we take the colimit topology on $P$ we get the restricted topology: If $U$ is open on the colimit topology then each $U\cap P_F$ is open, then is open in the restricted topology (where the family of the $P_F$'s is a open covering), viceversa, let $U$ a unions of a family of $P_F$, is enought show that each $P_F$ is open in the colimit topology, i.e. that the intersection with any $P_{F'}$ is open in $P_{F'}$, but this follow from the observation above (and observe that any $\mathcal{O}\subset K$ is supposed open, see http://modular.math.washington.edu/129/ant/html/node82.html).

• Dear Buschi Sergio, while I appreciate your efforts on MO, I have had a hard time deciphering your answers. Given that you're most likely not a native english speaker (neither am I), I hope it's not too much to ask you to put a little more time into getting the spelling and grammar correct. You'll reach a much larger audience this way. May 23, 2012 at 15:05
• thank you, I'll do my best, the trouble is that I read English math articles further, but I learned only a rought personal technical slang, and don't realize very well the right English grammar. May 23, 2012 at 16:59