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It is known that any closed linear subspace of $\mathbb R^\omega$ is topologically isomorphic to $\mathbb R^n$ for some $n\in\omega$.

Problem 1. Is each closed subgroup of $\mathbb Z^\omega$ (or better $\mathbb R^\omega$) topologically isomorphic to a Tychonoff product of LCA (= locally compact Abelian topological) groups?

If Problem 1 has affirmative answer, then we can ask

Problem 2. Is there a (desirably simple) example of a closed subgroup $H$ of a Tychonoff product $\prod_{n\in\omega}G_n$ of LCA groups, which is not topologically isomorphic to a Tychonoff product of LCA groups? What will be the answer if $H$ is compactly generated (i.e., contains a dense subgroup generated by some compact set)?

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  • $\begingroup$ In question 2, do you want each $G_n$ to be connected? If not, then surely any profinite group is a closed subgroup of a Tychonoff product of finite groups? $\endgroup$ – IJL May 24 '18 at 10:41
  • $\begingroup$ @IJL Closed subgroups of products of finite groups are compact, so are products of locally compact groups (with only one non-trivial factor). $\endgroup$ – Taras Banakh May 24 '18 at 10:47
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Concerning Question 1 for $\mathbb{Z}$:

Let $G$ be a closed subgroup of $\mathbb{Z}^\omega$ and let $G_i$ be the projection of $G$ to the first $i$ factors. We have compatible projections between the $G_i$'s and so we get a map $G\rightarrow \lim_i G_i$. This limit can also be thought of as a subgroup of $\mathbb{Z^\omega}$ (a model is given by compatible sequences). Actually it is precisely the closure of $G$; it consists of all sequences $g$ such that for each $n$ there is a $g_n\in G$ such that $g$ and $g_n$ agree on the first $n$ coordinates.

So we just have to look at such inverse limits. The structure maps $G_i\rightarrow G_{i-1}$ are surjections of finitely generated, free abelian groups and the rank increases by at most one. Assuming that we already have a basis for chosen for $G_{i-1}$, we can choose a basis of $G_i$ that maps the first $rk(G_{i-1})$-many basis vectors to the given basis. If the rank increases, the additional basis vector should get send to zero.

So we see that this inverse limit has a very special form and that is isomorphic to $\mathbb{Z}^n$ in the case that $rk(G_i)$ stabilizes to $n\in \mathbb{N}$ or to $\mathbb{Z}^\omega$ otherwise.


EDIT: added some ideas for Question 1 for $\mathbb{R}$:

In this situation things are more complicated. Again let $G\subset \mathbb{R}^\omega$ be a subgroup, let $pr_i:\mathbb{R}^\omega\rightarrow \mathbb{R}^n$ be the projection on the first $n$ coordinates

First $pr_n(G)$ is not automatically a closed subgroup of $\mathbb{R}^n$ (even if $G$ is closed). For example, you can embed $\mathbb{Z}^2$ into $\mathbb{R}^2$ (in a rotated way) such that the projection on each factor gives a nonclosed subgroup.

Lemma: The following subsets of $\mathbb{R}^\omega$ are the same:

  1. The closure $\overline{G}$ of $G$;
  2. The limit $\lim_n \overline{pr_n(G)}$;
  3. The intersection $\bigcap_n pr_n^{-1}(\overline{pr_n(G)})$.

Proof: The case (2) $\Leftrightarrow$ (3) is again the idea that elements in the inverse limit are given by compatible sequences.

So lets have a look at (1) $\Leftrightarrow$ (3). The set $\bigcap_n pr_n^{-1}(\overline{pr_n(G)})$ is a closed set containing $G$ and thus it also contains the closure of $G$. Conversely, given a point $g$ in the intersection, we have to construct a sequence of elements in $G$ that converge to that point. For a given $n$, we can choose a sequence $(g^n_m)_m$ of group elements such that $pr_n(g)=\lim_m pr_n(g^n_m)$.

Now we can build the desired sequence $(a_n)_n$ by a diagonalization argument: A local basis at $g$ is given by $U_n =pr_n^{-1}(B_{1/n}(pr_n(g)))$ for $n\in \mathbb{N}$; here $B_{1/n}$ denotes the $1/n$-ball in $\mathbb{R}^n$.

By definition of the $g^n_*$'s we can find for a given $n$ an index $m_n$ such that $g^n_{m_n}\in U_n$; now choose $a_n= g^n_{m_n}$. Since the $U_i's$ are nested, we know that $a_k\in U_n$ for $k>n$. Thus our sequence really converges to $g$ and this completes the proof of this lemma.

Now with this lemma we can try to argue as above; any closed subgroup of $\mathbb{R}^n$ is isomorphic to $\mathbb{R}^k\times \mathbb{Z}^l$ for some numbers $k$ and $l$. Sadly the structure maps in the inverse system (for $\lim_n \overline{pr_n(G)}$) need not be surjective anymore (the example with $\mathbb{Z}^2$ sitting inside $\mathbb{R}^2$ rotated). So it remains to examine what these structure maps look like....

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  • $\begingroup$ Very good, thank you! And what about the general case? For my purposes it would be sufficient to consider closed subgroups containing a dense subgroup, generated by some compact set. In this case we can apply the fundamental theorem on finitely generated groups and prove something similar? $\endgroup$ – Taras Banakh May 24 '18 at 11:40
  • $\begingroup$ I added some ideas about the case of the real numbers. I wonder how bad the inverse systems that appear there could be. $\endgroup$ – HenrikRüping May 31 '18 at 17:57

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