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7 votes
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How to show isomorphism between an adele ring on a number field and its Pontryagin dual?

(This should probably just be a comment, but I don't have enough reputation yet). One can reduce this problem to proving that local fields are self-dual by theorem 5.4, which states that the ...
Daniel5803's user avatar
6 votes
Accepted

Can a compact group have an infinite sequence of closed subgroups?

Yes. For example we can take $G = \mathbb{Z}_p$ to be the $p$-adic integers and $G_n = p^n \mathbb{Z}_p$. Each of the successive quotients $G_n/G_{n+1}$ is the finite cyclic group $\mathbb{Z}/p$. In ...
Qiaochu Yuan's user avatar
6 votes
Accepted

Structure theorem for profinite abelian groups

No. For me it's easier to think about the Pontryagin dual question: recall that Pontryagin duality $A \mapsto \hat{A} \cong \text{Hom}(A, S^1)$ restricts to a contravariant equivalence of categories ...
Qiaochu Yuan's user avatar
4 votes
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Every simple topological group is either discrete or connected

In order to convert my comments to an answer. Q1. It is well-known and discussed numerous times on this site (see for instance here, here, here) that for every infinite field $F$ and $n\ge 2$, the ...
Moishe Kohan's user avatar
3 votes
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Generalization of the fact that for $f: S^1 \rightarrow \Bbb R$ continuous, $\exists x \in S^1 : f(x) = f(-x)$

Your argument is correct, here is a more approriate generalization, since you never really need the group operation but only the inversion: Let $X$ be a connected topological space equipped with a ...
Just a user's user avatar
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2 votes

A certain inverse limit

Yes. It is a consequence of Galois theory : Let $L|K$ be a finite Galois extension, of Galois group $G$. Then there is a decreasing bijection between distinguished subgroups of $G$, and subextensions $...
Enguerrand Moulinier's user avatar

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