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)?

  • $\begingroup$ 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

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....

  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.