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Closure of a Subgroup of GL(2,C)

Ruun

New member
Feb 1, 2012
10
Let \(\displaystyle GL(2;\mathbb{C})\) be the complex 2x2 invertible matrices group. Let \(\displaystyle a\) be an irrational number and \(\displaystyle G\) be the following subgroup

\(\displaystyle G=\Big\{ \begin{pmatrix}e^{it} & 0 \\
0 & e^{iat}
\end{pmatrix} \Big| t \in \mathbb{R} \Big\}\)

I have to show that the closure of the set \(\displaystyle G\) is


\(\displaystyle \bar{G}=\Big\{ \begin{pmatrix}e^{it} & 0 \\
0 & e^{is}
\end{pmatrix} \Big| t \in \mathbb{R}, s \in \mathbb{R} \Big\}\)

I don't know even how to start. I'm afraid my topolgy knowledge needs serious improvement, but I didn't think it was necessary since I picked up a Group Theory book (Lie Groups, Lie Algebras and their representations: An elementary introduction by Brian C Hall). It sad because the books looks fascinating​
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Let \(\displaystyle GL(2;\mathbb{C})\) be the complex 2x2 invertible matrices group. Let \(\displaystyle a\) be an irrational number and \(\displaystyle G\) be the following subgroup

\(\displaystyle G=\Big\{ \begin{pmatrix}e^{it} & 0 \\
0 & e^{iat}
\end{pmatrix} \Big| t \in \mathbb{R} \Big\}\)

I have to show that the closure of the set \(\displaystyle G\) is


\(\displaystyle \bar{G}=\Big\{ \begin{pmatrix}e^{it} & 0 \\
0 & e^{is}
\end{pmatrix} \Big| t \in \mathbb{R}, s \in \mathbb{R} \Big\}\)

I don't know even how to start. I'm afraid my topolgy knowledge needs serious improvement, but I didn't think it was necessary since I picked up a Group Theory book (Lie Groups, Lie Algebras and their representations: An elementary introduction by Brian C Hall). It sad because the books looks fascinating​
Obviously $G \subseteq \bar{G}$, so you need to show that every element in $\bar{G}$ can be approximated by an element of $G$. The elements $\begin{bmatrix}e^{it} & 0 \\ 0 & e^{iat}\end{bmatrix}$ and $\begin{bmatrix}e^{it} & 0 \\ 0 & e^{is}\end{bmatrix}$ have the same entry $e^{it}$ in the top left corner, and the value of this element is unchanged if we replace $t$ by $t+2n\pi$. Also, the value of $e^{is}$ is unchanged if we replace $s$ by $s+2m\pi$. So can you choose $n$ and $m$ in such a way that $a(t+2n\pi)$ is close to $s+2m\pi$?

Hint: To do that, use Hurwitz's theorem.
 
Last edited:

Ruun

New member
Feb 1, 2012
10
So you use that the closure of a set is itself plus its boundary? Let me explain it myself to see if I'm understanding this: Because \(\displaystyle G \subseteq \bar{G}\) we only need to compute the boundary part.

Now we want to see if the number \(\displaystyle \frac{a(t+2\pi n)}{s+2\pi m } << 1\) but I can't put that expression in the form \(\displaystyle \left| \xi-\frac{p}{q} \right|\) to use the theorem
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
So you use that the closure of a set is itself plus its boundary? Let me explain it myself to see if I'm understanding this: Because \(\displaystyle G \subseteq \bar{G}\) we only need to compute the boundary part.
One way to show that a set $B$ is the closure of a subset $A$ is to check that (i) every point of $B$ is a limit of a sequence in $A$ (and is therefore in either $A$ or the boundary of $A$), and (ii) $B$ is closed. In my previous comment I suggested that you should prove (i) but I neglected to say that you should also check that the set $\bar{G}$ is closed.

Now we want to see if the number \(\displaystyle \frac{a(t+2\pi n)}{s+2\pi m } << 1\) but I can't put that expression in the form \(\displaystyle \left| \xi-\frac{p}{q} \right|\) to use the theorem
Look at the difference rather than the quotient. You want to find $m,\,n$ such that $a(t+2\pi n) \approx s+2\pi m$, or in other words $s-at \approx 2\pi(an-m).$ Now notice that if $\left|a-\frac mn\right| < \frac1{n^2}$ then $|an-m| < \frac1n.$
 
Last edited:

Ruun

New member
Feb 1, 2012
10
I will think about your post and reply here later, thank you very much.
 

Ruun

New member
Feb 1, 2012
10
I understand your proof and the closure of \(\displaystyle \bar{G}\) follows quite easily just by multiplying the matrices, however I'm not quite convinced on the solution of this exercise. I'm quite grateful on your help, but I think I need to study some prerequisites of this book and in mathematics in general, say set theory, topology and even some number theory. My math knowledge comes from physics, so I'm not very familiar to the level of rigor required, this is somethin I want to put some effort into too.

I wanted to study group theory because in my general relativity course we were studing the mathematical properties of the Lorentz Group, and I found all of that fascinating.