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I'm studying for my topology comp and I'm at a bit of a loss on this question. (My experience with Algebra is very limited and my experience with topological groups in particular is almost non-existent.)

Let $G$ be a topological group and $f: R \to G$ be an injective homomorphism of topological groups. (i.e. a continuous injective homomorphism). Is it true that $f(R) < G$ is necessarily a closed subgroup?

If you could provide a prod in the right direction I'd be quite grateful!

Thank you.

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It is not true that $f(R)$ is necessarily closed under the induced topology.

The classic counterexample here is that of "shredding the torus". In particular, we take $G = S^1 \times S^1$ (where $S^1$ denotes the unit circle in $\Bbb C$ under multiplication), and take $R = \Bbb R$. For some $a \in \Bbb R \setminus \Bbb Q$, define $f$ by $$ f(t) = (e^{it},e^{ait}) $$ Show that $f$ is not surjective, but $\overline{f(R)} = G$.

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  • $\begingroup$ This looks fantastic. I've never seen this example before! Thanks! Also, there is a second part to the question. If H<G is a subgroup that is isomorphic to R as topological groups, is H necessarily closed in G? $\endgroup$ – Cros Jul 7 '15 at 23:58
  • $\begingroup$ @Cros if I understand the question correctly, then I think the same example disproves the second question. That is, take $H = f(R)$. $\endgroup$ – Ben Grossmann Jul 8 '15 at 0:00
  • $\begingroup$ I didn't take the time to realize f was also an isomorphism. It seems pretty clear now though, sorry to waste your time. $\endgroup$ – Cros Jul 8 '15 at 0:04
  • $\begingroup$ No problem! Note that every injective homomorphism is an isomorphism with its image. I'm not sure if this counts as an isomorphism of topological groups, however... so I'm actually not 100% about that answer. $\endgroup$ – Ben Grossmann Jul 8 '15 at 0:06
  • $\begingroup$ In particular, $f^{-1}: f(R) \to R$ is not a continuous map in this case $\endgroup$ – Ben Grossmann Jul 8 '15 at 0:06
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No. Say $T=\mathbb R/\mathbb Z$ is the torus, and consider the map $f:\mathbb R\to T^2$ given by $f(t)=(t,\pi t)$. The image is dense.

Edit: Someone asks why the image is dense. This is a classical fact, with a fun proof by Fourier analysis.

Notation. As often, we identify a function defined on $\Bbb T^2$ with a doubly periodic function on $\Bbb R^2$.

Define $$\Lambda \phi=\lim_{A\to\infty}\int_{-A}^A\phi(f(t))\,dt,$$for $\phi$ such that the limit exists. If $\phi(t_1,t_2)=1$ then $\Lambda\phi=1$. If $$\phi(t)=e^{int_1+imt_2}$$with $(n,m)\in\Bbb Z^2$ and $(n,m)\ne(0,0)$ then you can work out the integral explicitly and you see that $$\Lambda\phi=0.$$So $$\Lambda\phi=\left(\frac{1}{2\pi}\int_0^{2\pi}\right)^2\phi(t_1,t_2)\,dt_1dt_2$$if $\phi$ is a trigonometric polynomial. Since the trigonometric polynomials are dense in $C(\Bbb T^2)$ this shows that

$\Lambda\phi=\left(\frac{1}{2\pi}\int_0^{2\pi}\right)^2\phi(t_1,t_2)\,dt_1dt_2$ for every $\phi\in C(\Bbb T ^2)$.

And hence the image of $f$ is dense: If otoh $V\ne\emptyset$ is open and $V\cap f(\Bbb R)=\emptyset$ then taking $\phi\ge0$ supported in $V$ gives a contradiction (because then $\Lambda\phi=0$ while $\left(\frac{1}{2\pi}\int_0^{2\pi}\right)^2\phi(t_1,t_2)\,dt_1dt_2>0$).

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  • $\begingroup$ Why is the image dense? $\endgroup$ – math112358 Jan 9 '20 at 0:52
  • $\begingroup$ @math112358 See edit. $\endgroup$ – David C. Ullrich Jan 9 '20 at 12:04
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For your second question: if $G$ is a topological group and $H$ is a subgroup which is topologically isomorphic to $\mathbb{R}$ (when given the subspace topology) then $H$ is closed in $G$. In fact, it's enough to assume $H$ is locally compact (when given the subspace topology).

I don't remember how the proof goes, but I'm fairly certain this is true.

EDIT: I found the proof. It's in Topological Groups and Related Structures by Arhangel’skii and Tkachenko, Proposition 1.4.19 (page 30). It goes like this:

Suppose $H$ is a locally compact subgroup of a (topological) group $G$. Denote by $K$ the closure of $H$; $K$ is also a subgroup of $G$ (Corollary 1.4.14 in the same book). Thus $H$ is a dense locally compact subspace of $K$, and one can show this means it's open in $K$ (he refers to a point-set topology book for this fact; it's called General Topology and was written by Engelking. It was also answered here). This means $H$ is an open subgroups of $K$, which means it's a closed subgroup of $K$ (since the complement of $H$ in $K$ is the union of its left cosets in $K$, all of which are open since multiplication is an open map). Thus $H$ is closed in $K$ which is closed in $G$, which means $H$ is closed in $G$ (and hence $H=K$).

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