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$T$ is a set of some prime numbers. $S$ is the multiplictive set generated by $T$. How to compute the Pontryagin dual of $S^{-1}Z$ ($Z$ is integal ring and $S^{-1}Z$ is localization of $Z$ at $S$). If $T$ is an empty set, then the Pontryagin dual of $S^{-1}Z=Z$ is $R/Z$. If $T$ is the set of all prime numbers, the Pontryagin dual of $S^{-1}Z=Q$ is naturally isomorphic to $A_Q/Q$, where $A_Q$ is the adele ring of rational number $Q$.

Question: I guess the Pontryagin dual of $S^{-1}Z$ is isomorphic to $(R \times\prod_{p\in T}Q_{p})/(S^{-1}Z)$. Is it right?

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Are you sure about $\mathbf{Q}$? I think its dual is the solenoid. – Timo Keller Dec 21 '10 at 14:54
What is a solenoid in this context? – Martin Brandenburg Dec 21 '10 at 15:06
$A_Q/Q$ is isomorphic to $(R\times\hat{Z})/Z$, hence the name "solenoid" : you take $[0,1]\times\hat{Z}$ and glue $0\times\hat{Z}$ to $1\times\hat{Z}$ using addition by $1$ in $\hat{Z}$. And the guess is right, except for the fact that the product over $p\in T$ must be restricted if $T$ is infinite : only finitely many components should be in $Q_p\setminus Z_p$. – BS. Dec 21 '10 at 15:47
up vote 1 down vote accepted

Let $\mathbf A_T$ denote the restricted direct product $\mathbf R\times \prod'_{p\in T}\mathbf Q_p$ (relative to the subgroups $\mathbf Z_p$, $p\in T$).

The OP asked whether the Pontryagin dual of $S^{-1}\mathbf Z$ is isomorphic to $\mathbf A_T/(S^{-1}\mathbf Z)$. Let's take it up in two steps:

Step 1: $\mathbf Q_p/\mathbf Z_p$ is canonically isomorphic (as an abstract group) to the subgroup of $\mathbf R/\mathbf Z$ consisting of elements whose order is a power of $p$ (mapping $p^{-n}\in \mathbf Q_p$ to $p^{-n}\in \mathbf R/\mathbf Z$ determines this embedding). This gives rise to an element $\psi_p\in \hat{\mathbf Q}_p$ which vanishes on $\mathbf Z_p$ but not on $p^{-1}\mathbf Z_p$. Also define $\psi_\infty$ to be the quotient map $\mathbf R\to \mathbf R/\mathbf Q$ composed with the inversion map $x\mapsto -x$.

Step 2: The formula $(a_p)\mapsto ((x_p)\mapsto \exp(2\pi i \sum_{p\in T\cup \{\infty\}} \psi_p(a_p x_p)))$ defines an isomorphism $\mathbf A_T\to \widehat{\mathbf A_T}$. Under this isomorphism $S^{-1}\mathbf Z$ maps to precisely those characters which vanish on $S^{-1}\mathbf Z$, establishing the isomorphism $S^{-1}\mathbf Z=\widehat{\mathbf A_T/S^{-1}\mathbf Z}$ (this can be seen with the help of partial fraction expansions of rational numbers).

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The guess of the OP is not what you say (the product is only over $p\in T$). Apart from taking the restricted product, the guess is right. – BS. Dec 22 '10 at 9:12
BS, you are right; I misread it. – Amritanshu Prasad Dec 22 '10 at 9:30
I edited my answer; essentially the same technique gives the OP's guess. – Amritanshu Prasad Dec 22 '10 at 9:49

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