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If $a$ is an irrational number and $$G = \Big\lbrace \begin{pmatrix} e^{it}&0\\0 & e^{ita}:\end{pmatrix}:t\in\mathbb{R}\Big\rbrace$$ is a subgroup of $\operatorname{GL}(2;\mathbb{C})$:

  1. How can one prove that the closure of $G$ is $$\overline{G} = \Big\lbrace \begin{pmatrix} e^{i\theta}&0\\0 & e^{i\phi}\end{pmatrix}:\theta,\phi\in\mathbb{R}\Big\rbrace.$$

  2. Why $G$ can be seen as the "irrational line in a torus"? Physically, if I remember correctly and for example, you could have a dynamical system with two frequencies such that the sum of them is not an integer of $2\pi$. Then one "can cover densely" the torus with some mapping. Here, how is this related to $G$?

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    $\begingroup$ The torus is homeomorphic to the product of two circles $S^1 \times S^1$, and $G$ maps into that space. $\endgroup$
    – Jair Taylor
    Apr 6, 2020 at 16:16

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Jair Taylor's comment addresses $(2)$.

For $(1)$, suppose I want to show that a given choice of $\theta,\phi$ yields a point in the closure of $G$. Consider the sequence $t_n=\theta+2\pi n$. We have that for each $n$, $e^{it_n}=e^{i\theta}$. So it's enough to argue that $e^{i\phi}$ is in the closure of $$\{e^{i(\theta +2\pi n)a}:n\in\mathbb{N}\}=\{e^{i\theta a}e^{i2\pi n a}:n\in\mathbb{N}\}.$$ But this is basically just the fact that the orbit of any point (in this case, $e^{i\theta a}$) with respect to an irrational-multiple-of-$\pi$ rotation (in this case, $p\mapsto pe^{i2\pi a}$) is dense in $S^1$.

If you're unfamiliar with that fact, it may in turn be easier to prove it in the following form: that $\{\lfloor an+b \rfloor:n\in\mathbb{N}\}$ is dense in $[0,1]$ whenever $a$ is irrational (regardless of what $b$ is).

It may also help further clarify the situation to set $\theta=0$, so you're just looking at $\{e^{i(2\pi an)}:n\in\mathbb{N}\}$. In the previous italicized comment, this corresponds to setting $b=0$.

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    $\begingroup$ @user2820579 You understand the closure correctly but I don't follow your question. For simplicity let $$g_t=\begin{pmatrix} e^{it}&0\\0 & e^{ita}\end{pmatrix}.$$ For a given $\theta,\phi$ we build a sequence of $$g_{t_0}, g_{t_1}, g_{t_2}, g_{t_3},...$$ of elements of $G$ with $\lim_{n\rightarrow\infty}g_{t_n}=\begin{pmatrix} e^{i\theta}&0\\0 & e^{i\phi}\end{pmatrix}$. Specifically, we'll set $$t_i=\theta+2\pi n$$ so (after simplifying) we get $$g_{t_n}=\begin{pmatrix} e^{i\theta}&0\\0 & e^{i(\theta+2\pi n)a}\end{pmatrix}.$$ $\endgroup$
    – Noah Schweber
    Apr 6, 2020 at 22:14
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    $\begingroup$ Now we want to show that $\lim_{n\rightarrow\infty}g_{t_n}=\begin{pmatrix} e^{i\theta}&0\\0 & e^{i\phi}\end{pmatrix}$. Well, this amounts to checking the limits of each of the four coordinates. The point is that all but the bottom-right coordinate are already what we want - that is, we have $$\lim_{n\rightarrow\infty}g_{t_n}=\begin{pmatrix} e^{i\theta}&0\\0 & e^{i\phi}\end{pmatrix}\quad\iff\quad \lim_{n\rightarrow\infty}e^{i(\theta+2\pi n)a}=e^{i\phi}.$$ And that second limit is just the irrational rotation fact mentioned above! $\endgroup$
    – Noah Schweber
    Apr 6, 2020 at 22:20
  • $\begingroup$ (Sorry, in my initial comment a "$t_i$" should be replaced with "$t_n$." My bad.) $\endgroup$
    – Noah Schweber
    Apr 6, 2020 at 22:22

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