Show that $\bar G=\left\{ \begin{pmatrix} e^{it} & 0 \\ 0 & e^{is} \\ \end{pmatrix}| t ,s \in \Bbb R\right\}$.

Question

Let $a \not \in \Bbb Q$ and let $G$ be the following subgroup of $GL(2, \Bbb C)$:

Let $$G=\left\{ \begin{pmatrix} e^{it} & 0 \\ 0 & e^{ita} \\ \end{pmatrix}| t \in \Bbb R\right\} $$

Show that $\bar G=\left\{ \begin{pmatrix} e^{it} & 0 \\ 0 & e^{is} \\ \end{pmatrix}| t,s \in \Bbb R\right\}$.

I know that I have to use the fact that $\{e^{2\pi ina}: n\in \Bbb Z\}$ is dense in $S^1$. But I can't put all the missing links correctly.

What I guess that if I put $t=2\pi n$ and vary $n \in \Bbb Z$ then $e^{it}=1$ and $e^{ita}$ would be dense in $S^1$ then taking closure I would get $$\left\{ \begin{pmatrix} 1 & 0 \\ 0 & e^{is} \\ \end{pmatrix}| s \in \Bbb R\right\}$$

Next is what? Please help

Answer 1

A hint in the way you were thinking,

you can break $$ \begin{pmatrix} e^{it} & 0 \\ 0 & e^{ita}\\ \end{pmatrix}= \begin{pmatrix} e^{it} & 0 \\ 0 & 1 \\ \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & e^{ita} \\ \end{pmatrix}$$. Then use two different sequences of matrices in $G$ converging these two matrices $\begin{pmatrix} e^{it} & 0 \\ 0 & 1 \\ \end{pmatrix}$ and $\begin{pmatrix} 1 & 0 \\ 0 & e^{is} \\ \end{pmatrix}$ separately and then take product of them. Although an answer is given.

Answer 2

Let $t' \in [0,2\pi)$ be fixed. Take now $t_n=t'+2 \pi n$. Then $e^{it} = e^{it'}$ and $e^{ita} = e^{it'a}e^{2 \pi i an}$. Now, you can use that $\{e^{2\pi a n} \, : \, n \in \mathbb{Z}\}$ is dense in $S^1$: For $s \in [0,2\pi)$ you can find a subsequene $n_k$ with $e^{2 \pi i an_k} \rightarrow e^{-it'a+is}$.