2
$\begingroup$

Why is there no closed subgroup of $\mathrm{GL}(2, \mathbb{C})$ with Lie algebra

$$\mathfrak{g} = \left\lbrace \begin{pmatrix} it & 0 \\ 0 & i\alpha t \end{pmatrix}: t\in\mathbb{R} \right\rbrace ?$$

Where $\alpha$ is a irrational number

$\endgroup$

2 Answers 2

4
$\begingroup$

There is if $\alpha\in\Bbb Q$.

If $\alpha\in\Bbb R\setminus\Bbb Q$ then the closure of the group generated by $\exp(A)$ for $A\in\frak g$ is the set of diagonal matrices $\pmatrix{a&0\\0&d}$ with $|a|=|d|=1$. The Lie algebra of this subgroup is two-dimensional.

$\endgroup$
1
$\begingroup$

Check that $\frak{g}$ is a real subalgebra of $gl(2, \Bbb C)$ and $dim_{\Bbb R} \frak{g}=1$ If $\alpha \not \in \Bbb Q$ Suppose $\exists G$ , matrix Lie group s.t $Lie(G)=\frak{g}$.

Now $H_0=\left\lbrace e^{X_t}= \begin{pmatrix} e^{it} & 1 \\ 1 & e^{i\alpha t} \end{pmatrix}: t\in\mathbb{R} \right\rbrace \subseteq G$

This implies that $\bar H_0 \subseteq \bar G=G $[As $G$ is a matrix lie group, it is closed in $GL(2, \Bbb C)$]

And $\bar H_0=\left\lbrace \begin{pmatrix} e^{it} & 1 \\ 1 & e^{is} \end{pmatrix}: s,t\in\mathbb{R}\right\rbrace$; $Lie(\bar H_0) \leq Lie(G)=\mathfrak g$ and $\mathfrak h_0=Lie(\bar H_0)=\left\lbrace \begin{pmatrix} e^{it} & 0 \\ 0 & e^{is} \end{pmatrix}: s,t\in\mathbb{R}\right\rbrace \subseteq \mathfrak g$ this implies $\dim_{\Bbb R} \mathfrak h_0=2$ a contradiction.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .