Fibonacci B6 path projectors satisfy the braid relations

Starting with $(x_0,x_1)=(0,1)$, the fusion rule permits only $x_{j+1}=1$ when $x_j=0$, and permits $x_{j+1}\in \{0,1\}$ when $x_j=1$. Enumerating the five remaining transitions with endpoint $x_6=0$ or $1$ gives exactly the two path sets displayed in the Statement; hence the block dimensions are $5$ and $8$. Also $r^5=-1$ and $r\mathrm{\ne }-1$, so $r^4-r^3+r^2-r+1=0$. For $y=r+r^{-1}$, division by $r^2$ gives $y^2-y-1=0$; since $y=2\cos (3\pi \mathrm{/}5)<0$, $y=-{\varphi }^{-1}$. Thus $r^2+{\varphi }^{-1}r+1=0$.

In a fixed left-tree basis, $P_i$ is the projection onto pair channel $1$. The $F$-move gives the only nontrivial local block, on the two intermediate channels $(1,\tau )$: $\left(\begin{array}{cc}{\varphi }^{-2}& {\varphi }^{-3\mathrm{/}2}\\ {\varphi }^{-3\mathrm{/}2}& {\varphi }^{-1}\end{array}\right)\mathrm{.}$ Its square equals itself because ${\varphi }^{-4}+{\varphi }^{-3}={\varphi }^{-2}$, ${\varphi }^{-3\mathrm{/}2}({\varphi }^{-2}+{\varphi }^{-1})={\varphi }^{-3\mathrm{/}2}$, and ${\varphi }^{-3}+{\varphi }^{-2}={\varphi }^{-1}$; its entries are real symmetric. When the surrounding charge is the unit, the only allowed pair-channel block is $[1]$. Thus each $P_i=P_i^{\ast }=P_i^2$. Counting the diagonal entries of these blocks on the listed paths gives $\mathrm{tr}{P}_i=2$ on ${\mathcal{P}}_1$ and $3$ on ${\mathcal{P}}_{\tau }$, so these are the ranks.

On the nontrivial three-strand total-$\tau$ block, the adjacent projections are $P=\mathrm{\mid }{e}_1⟩⟨e_1\mathrm{\mid }$ and $Q=\mathrm{\mid }{f}_1⟩⟨f_1\mathrm{\mid }$, where $f_1=({\varphi }^{-1},{\varphi }^{-1\mathrm{/}2})$ is the first column of $F$. Therefore $PQP=\mathrm{\mid }⟨e_1,f_1⟩{\mathrm{\mid }}^{2}P={\varphi }^{-2}P$ and symmetrically $QPQ={\varphi }^{-2}Q$. In the one-dimensional total-$1$ block both adjacent projections are zero, so the same identities hold. Tensoring with spectator path factors leaves these identities unchanged. Operators on disjoint pairs commute by the interchange law for tensor products.

Put $q_0={\varphi }^{-2}$. For adjacent $P=P_i,Q=P_{i+1}$ and $s=1-r$, direct expansion using $P^2=P$, $Q^2=Q$, $PQP=q_{0}P$, and $QPQ=q_{0}Q$ gives $[rI+sP][rI+sQ][rI+sP]-[rI+sQ][rI+sP][rI+sQ]=s(r+q_0{s}^2)(P-Q)\mathrm{.}$ Because $r+r^{-1}=-{\varphi }^{-1}$, $(r+q_0(1-r{)}^2)\mathrm{/}r=1+{\varphi }^{-2}(r+r^{-1}-2)=1-{\varphi }^{-2}({\varphi }^{-1}+2)=0$; here ${\varphi }^{-1}+2={\varphi }^2$. Thus the adjacent braid relation holds.

Since $P_i$ is an orthogonal projection and $\mathrm{\mid }r\mathrm{\mid }=1$, $B_i$ acts by $1$ on $\mathrm{ran}{P}_i$ and by $r$ on its orthogonal complement, so it is unitary. The distant commutation and adjacent relation from steps 1.3 and 2.1 are the defining Artin relations, hence the matrices represent $B_6$. The rank table gives rank $2$ on the five-dimensional block and rank $3$ on the eight-dimensional block, so $\det B_i=r^{5-2}=r^3$ and $\det B_i=r^{8-3}=r^5$, respectively.