Open Potts clock-shift Temperley–Lieb generators

Let $Q\in \{2,3,4\}$, $L\ge 1$, ${\omega }_Q=e^{2\pi i\mathrm{/}Q}$, and $\delta =\sqrt{Q}$. On ${\mathbb{C}}^Q$ with basis $\mathrm{\mid }0⟩,\dots ,\mathrm{\mid }Q-1⟩$, define the clock and shift matrices by $R_Q\mathrm{\mid }a⟩={\omega }_Q^a\mathrm{\mid }a⟩,M_Q\mathrm{\mid }a⟩=\mathrm{\mid }a-1\text{\hspace{0.17em}⁣}(\mathrm{m}\mathrm{o}\mathrm{d}Q)⟩\mathrm{.}$ Thus the source matrices are $R_2=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),M_2=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right);R_3=\left(\begin{array}{ccc}1& 0& 0\\ 0& e^{2\pi i\mathrm{/}3}& 0\\ 0& 0& e^{4\pi i\mathrm{/}3}\end{array}\right),M_3=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right)\mathrm{.}$ $R_4=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& i& 0& 0\\ 0& 0& -1& 0\\ 0& 0& 0& -i\end{array}\right),M_4=\left(\begin{array}{cccc}0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ 1& 0& 0& 0\end{array}\right)\mathrm{.}$ On ${\mathcal{H}}_{Q,L}=({\mathbb{C}}^Q{)}^{\otimes L}$, put, for $1\le j\le L$ and $1\le j<L$ respectively, $e_{2j-1}=\frac{1}{\sqrt{Q}}{\sum }_{r=0}^{Q-1}{M}_{Q,j}^r,e_{2j}=\frac{1}{\sqrt{Q}}{\sum }_{r=0}^{Q-1}{R}_{Q,j}^r{R}_{Q,j+1}^{-r}\mathrm{.}$ where the subscript denotes the tensor factor on which the matrix acts. The Potts loop normalization is $\delta =q+q^{-1}$ with $q=\{\begin{array}{ll}{e}^{i\pi \mathrm{/}4},& Q=2,\\ e^{i\pi \mathrm{/}6},& Q=3,\\ 1,& Q=4.\end{array}$ The free/open-chain Hamiltonian convention is $H_{Q,L}=-{\sum }_{i=1}^{2L-1}{e}_i$; there is no generator joining the last site back to the first. The following result proves that these operators satisfy the stated Temperley–Lieb relations.

The odd generators act on one site and the even generators on neighboring pairs. For $Q=3$, the source uses $R_j{R}_{j+1}^2+R_j^2{R}_{j+1}$; for $Q=4$ it uses the powers $1,2,3$ with inverse powers on the adjacent site. The displayed summation convention gives those formulas because $R_Q^Q=I$.