Vacuum projectors obey the reverse adjacent Temperley–Lieb relation

Choose a parenthesization grouping the three sites $a_i,a_{i+1},a_{i+2}$ and decompose the split-tree basis by their simple total charge $y$ and the spectator labels. If the first pair fuses to $\mathbf{1}$, then its fusion with the third $a$ has total charge $y=\mathbf{1}\otimes a\cong a$; the same is true if the last pair fuses to $\mathbf{1}$. On the $y=a$ block each vacuum-channel subspace has dimension $N_{aa}^{\mathbf{1}}{N}_{\mathbf{1}a}^a=1$, since $N_{aa}^{\mathbf{1}}=1$ and the tensor unit acts as the identity. On every $y\cong ̸a$ block both projectors vanish. Each fixed spectator label and multiplicity coordinate gives a copy of this local block, where $P_i$ and $P_{i+1}$ act as rank-one orthogonal projections onto unit vectors $e$ and $f$, by [F4].

By the F-symbol coordinate convention in [F3], the vacuum-to-vacuum entry of the three-site F-move is the coordinate of one unit basis vector in the other tree basis. Therefore $\mathrm{\mid }⟨e,f⟩\mathrm{\mid }=\mathrm{\mid }(F_a^{aaa}{)}_{\mathbf{1},\mathbf{1}}\mathrm{\mid }=\mathrm{\mid }{\kappa }_a\mathrm{/}{d}_a\mathrm{\mid }=1\mathrm{/}{d}_a$ by [F5] and [F6], so $\mathrm{\mid }⟨e,f⟩{\mathrm{\mid }}^2=d_a^{-2}$.

For unit vectors $e,f$, the linear-first convention in [F8] gives the rank-one projections $P_{e}x=⟨x,e⟩e$ and $P_{f}x=⟨x,f⟩f$. Hence $P_f{P}_e{P}_{f}x=⟨x,f⟩⟨f,e⟩⟨e,f⟩f=\mathrm{\mid }⟨e,f⟩{\mathrm{\mid }}^2{P}_{f}x$. Therefore $P_{i+1}{P}_i{P}_{i+1}=d_a^{-2}{P}_{i+1}$ on each $y=a$ spectator copy; on every other block both sides are zero. Tensoring with the spectator identities and transporting to the fixed path parenthesization preserves the relation by [F7].

Since ${\delta }_a={\kappa }_a{d}_a$ and ${\kappa }_a^2=1$, ${\delta }_a^2=d_a^2$. Thus $U_{i+1}{U}_i{U}_{i+1}={\delta }_a^3{P}_{i+1}{P}_i{P}_{i+1}={\delta }_a^3{d}_a^{-2}{P}_{i+1}={\delta }_a{P}_{i+1}=U_{i+1}$. The sign is retained in the final equality.