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For each $p\in\mathbb{P}$, the Prüfer group $\mathbb{Z}(p^{\infty})$ is a divisible abelian group which can be given the discrete topology or the topology it inherits via its identification as a subgroup of $\mathbb{Q}/\mathbb{Z}\subset\mathbb{R}/\mathbb{Z}$, where $\mathbb{R}/\mathbb{Z}$ is given the quotient topology relative to the Euclidean topology on $\mathbb{R}$ and the discrete subspace topology on $\mathbb{Z}$.

$\bigoplus\limits_{p\in\mathbb{P}}\mathbb{Z}(p^{\infty})$, which is algebraically isomorphic to the discrete abelian group $\mathbb{Q}/\mathbb{Z}$, can be given the discrete topology or the topology it inherits via its identification as a subgroup $\mathbb{Q}/\mathbb{Z}\subset\mathbb{R}/\mathbb{Z}$, with the subspace topology inherited from the quotient topology on $\mathbb{R}/\mathbb{Z}$. (All of this can be cast in terms of $S^1$ and cyclotomic units, but that is not relevant to the questions I will ask.)

The Pontryagin dual of the discrete abelian group $\mathbb{Z}(p^{\infty})$ is the $p$-adic integers $\mathbb{Z}_p$. Using the notation in the paper Topological Realizations of Absolute Galois Groups by Scholze and Kucharczyk: $\mathbb{Z}(p^{\infty})^ {\vee}\cong\mathbb{Z}_p$. The Pontryagin dual of the group $T\,\colon\!=\bigoplus\limits_{p\in\mathbb{P}}\mathbb{Z}(p^{\infty})\cong\mathbb{Q}/\mathbb{Z}$ with the discrete topology is thus topologically isomorphic to the profinite abelian group $\prod\limits_{p\in\mathbb{P}}\mathbb{Z}_p\cong\widehat{\mathbb{Z}}$.

Identifying $T$ with $\mathbb{Q}/\mathbb{Z}$ as a subspace of $\mathbb{R}/\mathbb{Z}$ under the quotient/Euclidean topology, $T$ is dense and non-locally compact. Because a continuous character extends uniquely to the closure of its domain, the set of characters for $T$ coinicides with the set of characters of $\mathbb{R}/\mathbb{Z}$, namely $\mathbb{Z}$. In this case, $T^\vee$ is $\mathbb{Z}$ under a topology courser than the discrete topology.

Similarly, $S\,\colon\!=\bigoplus\limits_{p\in\mathbb{P}}\mathbb{Z}/p\mathbb{Z}$ with topology inherited from the profinite abelian group $D\,\colon\!=\prod\limits_{p\in\mathbb{P}}\mathbb{Z}/p\mathbb{Z}$, is dense and non-locally compact, which we denote $(S,\tau)$.

Aside: $D$ is a commutative profinite ring with $\boldsymbol{1}=(1+2\mathbb{Z},1+3\mathbb{Z},1+5\mathbb{Z},\dots)$. Identify $\mathbb{Z}$ with the dense subgroup $\mathbb{Z}\boldsymbol{1}\subseteq D$. It always struck me as fascinating that $D$ has a dense torsion subgroup, $S$, as well as a dense torsion-free subgroup, $\mathbb{Z}\boldsymbol{1}$. Introducing $D$ via profinite theory, as in Ribes and Zalesskii, $S$ emerges organically as a dense subgroup, whereas defining $D$ via the $\mathfrak{a}$-adic topology as in Section 10 of Hewitt and Ross, $\mathbb{Z}\boldsymbol{1}$ emerges organically as a dense subgroup. If one toggles back-and-forth between these two presentations of the ring structure and associated topology of $D$, suffice it to say one learns some quite deep mathematics.

Let $(S,d)$ denote $S$ with the discrete topology. $D^\vee$ is topologically isomorphic to $(S,d)$ and $(S,\tau)^\vee$ is topologically isomorphic to $S$ with a topology courser than $d$.

We can also identify $S$ as a subspace of $\mathbb{R}/\mathbb{Z}$ under the quotient/Euclidean topoology, which we denote by $(S,\sigma)$. Note that any infinite subgroup of $S$, say $\bigoplus\limits_{p\in P}\mathbb{Z}/p\mathbb{Z}$ for some infinite set $P\subseteq\mathbb{P}$, is isomorphic to a dense subgroup of $\mathbb{R}/\mathbb{Z}$; this implies that in some sense $(S,\sigma)$ represents a minimally dense torsion subgroup of $\mathbb{R}/\mathbb{Z}$. In any case, we get that $(S,\sigma)^\vee$ is $\mathbb{Z}$ with a topology courser than $d$.

  1. Question: Are $(S,\tau)$ and $(S,\sigma)$ topologically isomorphic? Are $(S,\tau)^\vee$ and $(S,\sigma)^\vee$ topologically isomorphic?

Fuchs points out in Example 1 on page 105 of his Infinite Abelian Groups book (Volume I, 1970) that the discrete abelian group $E\,\colon\!=\prod\limits_{p\in\mathbb{P}}\mathbb{Z}(p^{\infty})$ is algebraically isomorphic to the discrete abelian group $\mathbb{R}/\mathbb{Z}\cong Q\oplus\mathbb{Q}/\mathbb{Z} \cong Q\oplus \bigoplus\limits_{p\in\mathbb{P}}\mathbb{Z}(p^{\infty})$ where $Q$ is a direct sum of a continuum of copies of $\mathbb{Q}$.

$T$ is dense under the subspace topology inherited via identification with $\mathbb{Q}/\mathbb{Z}\subset\mathbb{R}/\mathbb{Z}$, which is equivalent to the subspace topology $T$ inherits from its identification with $E$, where $E$ has topology induced via its algebraic isomorphism with $\mathbb{R}/\mathbb{Z}$.

  1. Question: What is the topology on $\prod\limits_{p\in\mathbb{P}}\mathbb{Z}(p^{\infty})$ induced by its algebraic isomorphism with $\mathbb{R}/\mathbb{Z}$?

Give each Prüfer group factor of $E$ the discrete topology; then, in particular, the unique copy of $\mathbb{Z}/p\mathbb{Z}$ in each respective Prüfer factor is open in that factor. Give $E$ the topology with open basis at $0$ consisting of sets of the form $\prod\limits_{p\in P}U_p \times \prod\limits_{p\notin P}\mathbb{Z}/p\mathbb{Z}$ where $P$ is a finite subset of $\mathbb{P}$ and $0\in U_p\subseteq\mathbb{Z}/p\mathbb{Z}$ for each $p\in P$. $D$ is an open subgroup of $E$ under this topology.

Let $\mathbb{Q}D$ denote the subgroup of $E$ consisting of elements $\boldsymbol{x}=(x_p)_{p\in\mathbb{P}}$ where $x_p\in\mathbb{Z}/p\mathbb{Z}$ for all but finitely many $p\in\mathbb{P}$. Then $\mathbb{Q}D$ under the topology inherited from $E$ is the restricted product topology relative to the open subgroups $\mathbb{Z}/p\mathbb{Z}$.

  1. Question: Is the topology on $E$ equivalent to the topology on $E$ from Question 2? Is the subspace topology on $D$ inherited from $E$ the profinite topology? Is $\mathbb{Q}D$ algebraically isomorphic to $E$?

Lastly,

  • The algebraically isomorphic copy of $\mathbb{Q}D$ in the solenoid $H\,\colon=\frac{D\times\mathbb{R}}{\mathbb{Z}(\boldsymbol{1},1)}\cong_{\rm t}\frac{\mathbb{Q}D\times\mathbb{R}}{X(\boldsymbol{1},1)}$ under the subspace topology is a $0$-dimensional, non-locally-compact, divisible, incomplete metric subgroup.
  • $H$ has a dense subgroup $X\cong\sum\limits_{p\in\mathbb{P}}\frac{1}{p}\mathbb{Z}$ algebraically isomorphic to the Pontryagin dual of $H$.
  • There are natural identfications $D\subseteq\mathbb{Q}D \subseteq H$ and $X\subseteq\mathbb{Q}D\subseteq H$, subject to the caveat that under the identifications the algebro-topological realizations of $\mathbb{Q}D$ and $X$ go from locally compact outside of $H$ to non-locally-compact as subgroups of $H$.
  • $\mathbb{Q}D$ is the subgroup of $H$ generated by all profinite subgroups.
  • $\mathbb{Q}D$ is the union of all subgroups $\Delta$ of $H$ containing $D$ for which $[\Delta\,\colon D]<\infty$.
  • The topology on $H$ is induced by a metric.
  • The metric on $H$ restricts to a non-Archimedean metric on $D$.
  • $H$ has WLOG total Haar measure 1.

All of the bullets above remain valid if $D$ is replaced by any Hausdorff quotient of $\widehat{\mathbb{Z}}$, say $K$, $\mathbb{Q}D$ is replaced by $\mathbb{Q}K$, $(\mathbb{Q}D\times\mathbb{R})/X(\boldsymbol{1},1)$ is replaced by $Z\,\colon=(\mathbb{Q}K\times\mathbb{R})/Y(\boldsymbol{1},1)$ where $Y$ is the Pontryagin dual of $Z$. For example, $\mathbb{A}/\mathbb{Q}\cong_{\rm t}(\mathbb{Q}\widehat{\mathbb{Z}}\times\mathbb{R})/\mathbb{Q}(\boldsymbol{1},1)$. In view of this background,

  1. Question: $H$ (resp.$Z$) has a Haar measure, so integration over the embedded copy of $\mathbb{Q}D$ (resp.$\mathbb{Q}K$) is possible even though $\mathbb{Q}D$ (resp.$\mathbb{Q}K$) is not locally compact. Would the analytical arguments of Tate's thesis be reproducible in this setting, with a non-locally-compact subspace topology, applying the Haar measure of the solenoid $H$ (resp.$Z$)?
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    $\begingroup$ Why not define the relevant metrics on $S$ and $\Bbb{Q/Z}$ ? There is the $d_1(s,s') = \min_n |\sum_k \frac{s(p_k)-s'(p_k)}{p_k}-n|$ one coming from $\Bbb{R/Z}$ and the $d_2(s,s') = \sum_{k, s(p_k) \ne s'(p_k)} 2^{-k}$ one coming from the restricted product topology. They are not isomorphic. $\Bbb{R/Z}$ and $\prod\limits_k\Bbb{Z[1/p_k]/Z}$ are completions of $\Bbb{Q/Z}$ for those different metrics, how do you obtain a concrete group isomorphism ? In term of the metric how do you make $S$ the least dense subgroup of $\Bbb{R/Z}, d_1$ ? $\endgroup$ – reuns Apr 8 at 5:23
  • $\begingroup$ reuns, FYI, it may take me a while to work through the details per your suggestions, but I will - just so you know I am not ignoring your comment. $\endgroup$ – Wayne Apr 8 at 8:51
  • $\begingroup$ reuns, define a subset of $E$ to be open if its complement can be partitioned into a countable collection of subsets $\{ X_i\}$ where each $\{ X_i\}$ consists of elements of $E$ for which the coordinates converge to the same limit in $\mathbb{R}/\mathbb{Z}$ (each $\mathbb{Z}(p^\infty)$ is identified as a subgroup of $\mathbb{R}/\mathbb{Z}$), and where the limit of the respective limits of each $\{ X_i\}$ exists in $\mathbb{R}/{\mathbb{Z}$. $\endgroup$ – Wayne Apr 9 at 0:32
  • $\begingroup$ reuns (cont.), The resulting topology on $E$ makes it homeomorphic, but not topologically isomorphic, to $\mathbb{R}/\mathbb{Z}$. $\endgroup$ – Wayne Apr 9 at 0:36
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    $\begingroup$ With $\Bbb{Q}D$ you mean those elements $a \in \prod_p \Bbb{Z[1/p]/Z}$ such that for some $n$ then $an \in \prod_p \Bbb{p^{-1}Z/Z}$. The equivalent in adeles is obtained form the completion of $ \Bbb{A_{Q,fin}/\hat{Z}}$ for the metric $|x| = \max_p |x_p|_p$. $\endgroup$ – reuns Apr 11 at 11:44
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$S$ with the product topology $\tau$ has the property that for every sequence $(u_p)$, $p$ ranging over primes, and $u_p\in S$ has order $p$, one has $\lim u_p=1$.

This is clearly false in the topology $\sigma$ induced by inclusion in the unit circle (just choose $u_p$ to have nonpositive real part).

So the topological groups $(S,\tau)$ and $(S,\sigma)$ are not isomorphic.

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