# Prüfer Groups and Product Topologies

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.$$Z$$)?
• 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 $\Bbb{R/Z}, d_1$ ? – reuns Apr 8 at 5:23
• 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. – Wayne Apr 8 at 8:51
• 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 $\mathbb{R}/{\mathbb{Z}$. – Wayne Apr 9 at 0:32
• reuns (cont.), The resulting topology on $E$ makes it homeomorphic, but not topologically isomorphic, to $\mathbb{R}/\mathbb{Z}$. – Wayne Apr 9 at 0:36
• 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 $|x| = \max_p |x_p|_p$. – reuns Apr 11 at 11:44

$$S$$ 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.