restricted product puzzle

Stream: maths

[Kevin Buzzard]

It has never occurred to me to check various well-known facts about groups like $GL_2(\mathbb{A}_{\mathbb{Q}})$ before, and now I'm checking them I find that I don't quite understand what generality they're supposed to be true in.

Mathlib has restricted products. A quick primer: if $I$ is an index set, if $X_i$ are sets for $i\in I$ and if $Y_i$ are subsets then $\prod'_iX_i$ is the subset of the full product $\prod_iX_i$ consisting of $(x_i)$ such that $x_i\in Y_i$ for all but finitely many $i$. If the $X_i$ are groups/rings/$R$-modules and the $Y_i$ are subgroups/subrings/submodules then the restricted product is a group/ring/$R$-module.

But topology is subtler. If the $X_i$ are topological spaces then in the applications to number theory we don't want $\prod'_iX_i$ to have the subspace topology coming from the embedding into $\prod_iX_i$; this is too coarse. Anatole has formalised what we do want: the restricted product is the union of $\prod_{i\in S}X_i\times\prod_{i\notin S}Y_i$ as $S$ varies over finite subsets of $I$ and the topology we want on the restricted product is the finest one such that all the inclusions from these products (with the product topology) are continuous. Anatole has proved that a map from the restricted product is continuous iff its restriction to $\prod_{i\in S}X_i\times\prod_{i\notin S}Y_i$ is continuous for all finite $S$ (it's docs#RestrictedProduct.continuous_dom, note however that his $S$ is the complement of mine in $I$).

Now finally to my question. Does restricted product commute with finite products, or equivelently with binary products? More precisely, if $C_i\subseteq A_i$ and $D_i\subseteq B_i$ are topological spaces all indexed by $I$ then is $\prod'(A_i\times B_i)$ (with respect to $C_i\times D_i$) homeomorphic to $(\prod'A_i)\times(\prod'B_i)$? It's readily checked that there's an obvious bijection between the two sets, and that the map from the restricted product to the binary product is continuous.

But how does one check that a map from a binary product to a restricted product is continuous? Is this claim not actually true in general?

Why do I want to know this? Well the finite adeles $\mathbb{A}^\infty$ are a restricted product of $\mathbb{Q}_p$ with respect to the subrings $\Z_p$, and I want to check that $M_2(\mathbb{A^\infty})$ is a restricted product of $M_2(\mathbb{Q}_p)$, topologically. In this case I am lucky; the finite product (of 4 things not 2) is also a direct sum, and addition is continuous, and I can muddle my way through. But I am using algebraic structure which is not present in the general topological question I'm asking. Is this claim false in general? NB I only ever care about the case where the subset $Y_i$ is nonempty, compact and open, if this helps. But I'm having flashbacks to #maths > product of pushforward topologies @ 💬 -- it is all the same ingredients all over again...

[Calle Sönne]

Can this not be verified "by hand", by taking the preimage of an open in a basis of the restricted product of the binary product?

[Andrew Yang]

Is docs#RestrictedProduct.continuous_dom_prod_left (universal property for maps mapping from $X \times \prod' A_i$) useful here?

[Kevin Buzzard]

Yeah maybe but this was only proved under the extra assumption that the subsets were open (true in my case) and that the filter we're restrictedproducting over is the cofinite filter (true in my case).

[Kevin Buzzard]

Calle Sönne said:

Can this not be verified "by hand", by taking the preimage of an open in a basis of the restricted product of the binary product?

I am not sure I know a basis for the restricted product in this generality. I think Anatole basically discovered that things were pretty yucky topologcally iif you don't assume that the filter is the cofinite filter (which I assumed throughout this post because I don't know if anyone cares about a more general case) and that all the subspaces are open (which I didn't assume). Under those assumptions the maps $\prod_{i\in S}X_i\times\prod_{i\notin S}Y_i\to\prod'X_i$ are open embeddings and some kind of sanity is restored.

[Calle Sönne]

Okay so since your filter is so general, maybe this is the same as asking if filtered colimits commute with products in Top?

[Calle Sönne]

Someone asked this here but got no response: https://mathoverflow.net/questions/473788/when-do-filtered-colimits-commute-with-finite-products-in-top

[Calle Sönne]

Here is a better post: https://math.stackexchange.com/questions/1255678/do-finite-products-commute-with-colimits-in-the-category-of-spaces

[Kevin Buzzard]

Yeah maybe. A toy example: if the $X_i$ are pointed topological spaces and the $Y_i$ are the points, then $\prod'X_i$ is the "direct sum" of the spaces, i.e. the subset of the product consisting of elements which are equal to the base point at all but finitely many coordinates. Do you understand what the filtered colimit topology looks like on this object? A map from the space is continuous precisely when it's continuous when restricted to the finite products but I think that the intersection of this restricted product with an arbitrary product of open sets will be open (i.e. you can drop the hypothesis that the opens have to be $X_i$ for all but finitely many $i$)

[Kevin Buzzard]

Calle Sönne said:

Someone asked this here but got no response: https://mathoverflow.net/questions/473788/when-do-filtered-colimits-commute-with-finite-products-in-top

Note the comment in the question:

I managed to show that if the morphisms of the diagrams are open maps of topological spaces, then the colimit and product commute.

So again this is evidence that the restricted products really might not commute with binary products unless all the $Y_i$ are open (in which case Anatole proved that the morphisms are open maps).

Basically I was trying to work out the generality in which I should be proving API and perhaps this is it.

[Calle Sönne]

So will you be trying to formalize that filtered colimits with open transition maps commute with products in Top (if this is true)?

[Kevin Buzzard]

I think I'll just stick to what I need. If the subsets are all open (and the filter is cofinite) then the key result is docs#RestrictedProduct.continuous_dom_prod which gives a criterion for a map out of a binary product of restricted products to be continuous, and this is exactly the result I need to finish the argument that restricted product commutes with binary products.

Things are so ridiculously slow-going. Calle if I told you that $$GL_2(\mathbb{A}_{\mathbb{Q})$$ was obviously the restricted product of the $GL_2(\mathbb{Q}_v)$'s you would buy this, right? The argument above shows that things are more delicate than I had realised, but I need to prove that various compact open subgroups are compact open (or at least understand why this is the case) and I have a talk to give in 10 days so I will just go for the argument above. I still need to prove that the units of a restricted product of topological rings is homeomorphic to the restricted product of the units (to move from matrices to invertible matrices), but I suspect that it will be more of the same now we've realised that openness is the crucial input. Proving FLT is harder than you think!

[Ruben Van de Velde]

Kevin Buzzard said:

Proving FLT is harder than you think!

Some 350 years of mathematicians approve this message

[Calle Sönne]

Kevin Buzzard said:

I think I'll just stick to what I need. If the subsets are all open (and the filter is cofinite) then the key result is docs#RestrictedProduct.continuous_dom_prod which gives a criterion for a map out of a binary product of restricted products to be continuous, and this is exactly the result I need to finish the argument that restricted product commutes with binary products.

Things are so ridiculously slow-going. Calle if I told you that $$GL_2(\mathbb{A}_{\mathbb{Q})$$ was obviously the restricted product of the $GL_2(\mathbb{Q}_v)$'s you would buy this, right? The argument above shows that things are more delicate than I had realised, but I need to prove that various compact open subgroups are compact open (or at least understand why this is the case) and I have a talk to give in 10 days so I will just go for the argument above. I still need to prove that the units of a restricted product of topological rings is homeomorphic to the restricted product of the units (to move from matrices to invertible matrices), but I suspect that it will be more of the same now we've realised that openness is the crucial input. Proving FLT is harder than you think!

I think the reason I believe that $GL_2(\mathbb{A}_{\mathbb{Q}})$ is the restricted product of the $GL_2(\mathbb{Q}_v)$'s is because I tried to look into it once and found some long notes of Brian Conrad and that guy seems careful enough (so in the end I didn't even read them I think!)

[Anatole Dedecker]

Aha, I was on the road yesterday, but I've been thinking about this as well (you may notice the extra commit in FLT#561 where I tried proving the topological statement before realizing that it was non-trivial). This "commuting with products" is exactly the reason why the topological group/ring/... instances only exist in this particular setting.

In fact, on of the things that got me interested in this rstricted product business is that I was very surprised that taking an inductive limit topology was ever going to give you a topological group (so I assumed that people were implicitly taking that inductive limit among group topologies, as is done in e.g constructing the space of distributions). And indeed docs#RestrictedProduct.continuous_dom_prod only holds because you have a full neighborhood of the identity on which you know the topology (namely the product of the subgroups), hence you can make things work locally.

So while I'm not confident enough to argue that your claim is false in full generality, I certainly do not have high hopes for it.