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Consider the group of matrices $B$ generated by taking products of the matrices

\begin{equation} \rho_1 = \begin{pmatrix}\exp\left(\dfrac{-4\pi i}{5}\right) & 0\\ 0 & \exp\left(\dfrac{3\pi i}{5}\right)\end{pmatrix}\\ \rho_2 = \begin{pmatrix}\frac{1}{\phi}\exp\left(\dfrac{4\pi i}{5}\right) & \frac{1}{\sqrt{\phi}}\exp\left(-\dfrac{3\pi i}{5}\right)\\ \frac{1}{\sqrt{\phi}}\exp \left(-\dfrac{3\pi i}{5}\right)& -\frac{1}{\phi}\end{pmatrix}, \end{equation} where $\phi=\frac{1+\sqrt{5}}{2}$. Prove that $B$ is dense in $U(2)$.

This problem comes from the braiding of Fibonacci anyons in topological quantum computing and the above statement is equivalent to saying that braiding of Fibonacci anyons can produce any single qubit quantum gate.

I was wondering how one would typically go about proving such a density statement.

Edit: I was thinking of the following argument: $\rho_1$ and $\rho_2$ are non-commuting matrices and $\rho_1$ has order 10 and I think that $\rho_2$ has infinite order. Hence $B$ would be an infinite nonabelian subgroup of $U(2)$. The only non-abelian subgroups are $SU(2)$ and $U(2)$, since $\rho_1$ and $\rho_2$ are not in $SU(2)$, $B$ must be $U(2)$.

TShiong
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PhPanda
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  • I didn't understand what is meant by the set of matrices generated by products. Maybe you mean the subgroup generated by matrices $\rho_1$ and $\rho_2$, i.e. $B=\langle\rho_1,\rho_2\rangle$? – kabenyuk Aug 14 '23 at 13:22
  • Yes that is exactly what I meant, sorry for the confusing wording. – PhPanda Aug 14 '23 at 13:46
  • I changed a tag to make your post more visible. – Lee Mosher Aug 14 '23 at 14:37
  • Do you mean $SU(2)$ and $U(2)$ are the only non-abelian infinite closed subgroups? But that's not true either, e.g. the matrices $\pmatrix{e^{i\alpha} & 0\cr 0 & e^{i\beta}\cr}$ and $\pmatrix{0 & e^{i\gamma}\cr e^{i\delta} & 0\cr}$ form such a subgroup. – Robert Israel Aug 14 '23 at 14:48
  • You are right, Thank you to point out he flaw in my argument. – PhPanda Aug 14 '23 at 14:56

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