The Lie algebra of the commutator subgroup

Question

If $G$ is a connected Lie group with Lie algebra $g$, then is its commutator subgroup $[G,G]$ a closed subgroup with Lie algebra $[g,g]$?

Answer 1

As far as I can tell, the mathoverflow thread linked by Qiaochu Yuan does not actually include an example of Lie group $G$ such that $[G,G]$ is not closed. The closest thing is the answer of Tom Goodwillie in which he exhibits closed subgroups $G_1,G_2 \subseteq G$ such that $[G_1,G_2]$ is not closed. It is not so difficult, however, to use the idea from that answer to get an example. As the basic ingredient, we take

$H$ the "Heisenberg-Weyl group". That is, $H=\mathbb{R}^2 \times \mathbb{T}$ with group operation \begin{align*} (x,y,z)(x',y',z') = (x+x',y+y',e^{ixy'}zz'). \end{align*} This group is a central extension of $\mathbb{R}^2$ by the circle group $\mathbb{T}$, where the central copy of $\mathbb{T}$ sits as $\{0\} \times \{0\} \times \mathbb{T}$. It is easy to check that \begin{align*}[g,h]=(0,0,e^{ixy'}) && \text{ when } && g=(x,y,z) && h=(x',y',z') \end{align*} and so $[H,H]$ is the copy of $\mathbb{T}$.

Now we make an alteration, in the spirit of Tom Goodwillie's answer:

Fix some irrational number $c$ and define $G$ to be the following closed subgroup of $H \times H$. $$ G = \{ \langle (x,y,z),(cx,y,z') \rangle: x,y \in \mathbb{R}, z,z' \in \mathbb{T} \}.$$ This group is a central extension of $\mathbb{R}^2$ by torus $\mathbb{T}^2$, where the central copy of $\mathbb{T}^2$ is all the elements of the form $\langle(0,0,z),(0,0,z')\rangle$. This time, we have \begin{align*}[g,h]=\langle(0,0,e^{ixy'}),(0,0,e^{icxy'}) \rangle \end{align*} when \begin{align*} g=\langle(x,y,z),(cx,y,z')\rangle && h=\langle(x',y',z''),(cx',y',z''')\rangle \end{align*} from which we can see that $[G,G]$ is the dense line $\{ (e^{it}, e^{ict}) : t \in \mathbb{R}\}$ in the copy of $\mathbb{T}^2$.