Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to show that every Lindelöf topological group is isomorphic to a subgroup of the product of second countable topological groups. I received an answer using the fact that Lindelöf topological groups are $\omega$-narrow, but I want to show it by using the following theorem.

Theorem: Every Hausdorff topological group $G$ is topologically isomorphic to a subgroup of the group of isometries $Is(M)$ of some metric space $M$, where $Is(M)$ is taken with the topology of pointwise convergence.

Any help would be greatly appreciated!

share|cite|improve this question
Mathematical naming is not commutative. Algebra+topology$\neq$ algebraic topology... – Asaf Karagila Jun 2 '12 at 21:54
up vote 1 down vote accepted

Let $G$ be a Lindelöf group. By your theorem, you can assume that $G$ is a subgroup of $\operatorname{Iso}(M)$ for some metric space $M$. Now, consider the decomposition of $M$ into $G$-orbits, call them $M_a$. Since each orbit is an image of $G$, each $M_a$ is Lindelöf. Furthermore, since each $M_a$ is metrizable, they each have a countable base.

Now, for every $M_a$, since $G$ acts on $M_a$, there is a natural homomorphism of topological groups from $G$ to $\operatorname{Iso}(M)$. Then, the diagonal product of these homomorphisms gives you the embedding of $G$ into the product of second countable groups.

share|cite|improve this answer
I would never have thought of that. Thank you so much! – Maria Jun 7 '12 at 20:10

A nice reference for this sort of thing is the book Topological Groups and Related Structures by Arhangel'skii and Tkachenko.

A Hausdorff topological group $G$ is said to be $\omega$-narrow if for every open neighborhood $U$ of the identity $e$, there is a countable set $A$ such that $AU=G$.

Certainly every Lindelöf topological group is $\omega$-narrow; take $A\subset G$ to be a countable set such that $\{aU\}_{a\in A}$ is an open cover of $G$.

Guran's Theorem (3.4.23 in the book mentioned) states that a topological group is $\omega$-narrow iff it embeds as a topological subgroup of a product of second countable topological groups.

This result is more general than the one you are asking for and the proof can be found in the book. On the other hand, the proof here doesn't seem to use Uspenskij's theorem (that $G$ can be embedded in the isometry group of some metric space $M$, in particular the metric space of all bounded left uniformly continuous real-valued functions on $G$).

Perhaps for Lindelöf $G$, there is a simpler proof using Uspenskij's theorem and someone else can point the way to this. I am curious to know where it is said that such a proof is possible?

share|cite|improve this answer
That's the book I've been referencing for this question. I'm unsure how to show it directly using Uspenskij's theorem. – Maria Jun 2 '12 at 17:24
I see. You did not provide much about what you already knew in the question. You may want to specify the question to ask only for a proof with Uspenskij's Thm since you already know the answer to the question you wrote: "How can I show that every Lindelöf topological group is isomorphic to a subgroup of the product of second countable topological groups?" I am still curious why you want a proof with Uspenskij's Theorem and where it is said such a proof is possible. – Jeremy Brazas Jun 2 '12 at 19:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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