Effects of the fact that the idèles have a finer topology than the adèles

Question

The adelic ring (of, say, $\mathbb{Q})$ has a topology given by a basis of open sets of the form $$U_S \times \prod_{p \notin S} \mathbb{Z}_p$$ where $U_S$ is a finite product of open sets $U_v$ of $\mathbb{Q}_v$.

The group of idèles has a topology given by a basis of open sets of the form $$U_S \times \prod_{p \notin S} \mathbb{Z}_p^\times$$ where $U_S$ is a finite product of open sets $U_v$ of $\mathbb{Q}_v^\times$. I do not understand

• why is the idèle topology finer than the adèle topology? (is there an explicit set which is open for one but not for the other? we also need to show that all the open sets for one is open for the other)
• how important this distinction is in practice? (so we have examples or counter examples of what it would be that they have the same topology, to develop an intuition of the difference between both?)
• What is the relation with the adelic norm? I think that none of these is given by the adelic norm, what is the point of this norm then?

And if these topologies are so different, how can it be that we embed the idèles inside the adèles?

Answer 1

$\Bbb{A_Q^\times}$ is open for the ideles topology but not for the adeles topology:

let $e^p$ be the adele which is $1$ for all places but $p$ where it is $0$. Any $\Bbb{A_Q}$-open containing $1$ must contain some $e^p$ which is clearly not an idele.

The ideles topology is not very mysterious: a sequence $u_n\in \Bbb{A_Q^\times}$ converges for the $\Bbb{A_Q^\times}$ topoogy iff both $u_n,1/u_n$ converge in the $\Bbb{A_Q}$ topology.

Answer 2

This is more like a long comment, and I don't know if you'll find this abstraction helpful or unhelpful, but here is how I made my peace with this weird fact.

If $G$ is a quasi-projective algebraic group over $\mathbb{Q}$ and $A$ is a topological $\mathbb{Q}$-algebra, then it seems reasonable to believe that $G(A)$ should carry a natural topology defined as follows: pick any embedding of $G$ as a Zariski open subset of some projective space, topologize $\mathbb{P}^n(A)$ in the standard way (it's a quotient of a product of a finite number of copies of $A$), and give $G(A)$ the subspace topology.

Taking $G = \mathbb{G}_m$ and $A$ as the adèles gives a counterexample to this hypothesis. If we view $G$ as a Zariski open subset of the affine line and hence get an open embedding $G(\mathbb{A}_\mathbb{Q}) \to \mathbb{A}_\mathbb{Q}$, we get a different topology from what we get if we view $G$ as a closed subvariety of $SL_2$ via the map $$x \mapsto \begin{pmatrix} x & 0 \\ 0 & x^{-1} \end{pmatrix}.$$

Not only does this mean that the proposed means of giving $G(A)$ a topology is not well-defined in general, in this case the inversion map isn't even continuous for the first topology, which can't be right.

One can prove that if you work exclusively with affine algebraic groups, you can get a good definition independent of choices by taking any embedding which realizes $G$ as a Zariski closed subset of affine space, then giving it the subspace topology. This says in particular that you should topologize the idèles by thinking of them via the matrix embedding to $SL_2$, which you can check is equivalent to your topology.