Three-Fibonacci-anyon qubit matrices

Put $u={\varphi }^{-1}$ and $v={\varphi }^{-1\mathrm{/}2}$. Since ${\varphi }^2=\varphi +1$, $u^2+u=1$ and $v^2=u$. Matrix multiplication then gives $F^2=\left(\begin{array}{cc}{u}^2+v^2& uv-uv\\ uv-uv& u^2+v^2\end{array}\right)=I$. Both diagonal entries of $B_1$ have modulus one, so $B_1$ is unitary; $F$ is unitary because it is real and $F^{\ast }F=F^2=I$. The data in [F1] give $r=R_{\tau }\mathrm{/}{R}_1=e^{-3\pi i\mathrm{/}5}$. Also $r^5=-1$ and $r\mathrm{\ne }-1$, so division of $r^5+1=0$ by $r+1$ gives $r^4-r^3+r^2-r+1=0$. With $y=r+r^{-1}$, division by $r^2$ gives $y^2-y-1=0$; since $y=2\cos (3\pi \mathrm{/}5)<0$, it follows that $y=-u$ and $r^2+ur+1=0$. Hence $B_2=F{B}_{1}F$ is unitary.

Write $B_2=\left(\begin{array}{cc}p& q\\ q& t\end{array}\right)$, where $p=u^2+ur$, $q=uv(1-r)$, and $t=u+u^{2}r$. Then $B_1{B}_2{B}_1=\left(\begin{array}{cc}p& qr\\ qr& r^{2}t\end{array}\right),B_2{B}_1{B}_2=\left(\begin{array}{cc}{p}^2+q^{2}r& q(p+rt)\\ q(p+rt)& q^2+t^{2}r\end{array}\right)\mathrm{.}$ The relations $r^2=-ur-1$ and $u^3=2u-1$ give $p+rt=r$; expanding the remaining diagonal entries and using $u^2+u=1$, $v^2=u$, and $r^2+ur+1=0$ gives $p^2+q^{2}r=p$ and $q^2+t^{2}r=-p$. Thus both matrices reduce to $\left(\begin{array}{cc}{u}^2+ur& uvr(1-r)\\ uvr(1-r)& -(u^2+ur)\end{array}\right)\mathrm{.}$ The Artin relation follows entry by entry.

The off-diagonal entry of $B_1{B}_2-B_2{B}_1$ is $q(1-r)=uv(1-r{)}^2\mathrm{\ne }0$, because $u,v\mathrm{\ne }0$ and $r\mathrm{\ne }1$. Thus the generators do not commute. They are not two copies of a single diagonal phase gate in the same fusion basis.