Spin-half swaps realize the Temperley–Lieb algebra at loop weight two

On ${\mathbb{C}}^2\otimes {\mathbb{C}}^2$, direct application to the four basis vectors gives $s=\frac{1}{2}(I+{\sigma }_x\otimes {\sigma }_x+{\sigma }_y\otimes {\sigma }_y+{\sigma }_z\otimes {\sigma }_z)\mathrm{.}$ For $\mathrm{\mid }00⟩$ and $\mathrm{\mid }11⟩$, the sum of the three Pauli products is the vector itself; for $\mathrm{\mid }01⟩$ it is $2\mathrm{\mid }10⟩-\mathrm{\mid }01⟩$, and for $\mathrm{\mid }10⟩$ it is $2\mathrm{\mid }01⟩-\mathrm{\mid }10⟩$, so the displayed operator is the factor swap in all four cases. The symmetric three-dimensional subspace and antisymmetric line are an orthogonal decomposition; $s$ is $+I$ on the former and $-I$ on the latter, whose unit vector is $\omega$. Therefore $I-s=2P_{\omega }$, and $(I-s{)}^2=4P_{\omega }^2=4P_{\omega }=2(I-s)$. Tensoring with identities gives $e_i=2P_i$, self-adjointness, and $e_i^2=2e_i$ for every bond.

On three adjacent factors, $P_{12}$ has image spanned by $\mathrm{\mid }\omega {⟩}_{12}\otimes \mathrm{\mid }0⟩$ and $\mathrm{\mid }\omega {⟩}_{12}\otimes \mathrm{\mid }1⟩$. Applying $P_{23}$ and then $P_{12}$ to these two vectors gives, respectively, $P_{12}{P}_{23}{P}_{12}(\mathrm{\mid }\omega {⟩}_{12}\mathrm{\mid }0{⟩}_3)=\frac{1}{4}\mathrm{\mid }\omega {⟩}_{12}\mathrm{\mid }0{⟩}_3,P_{12}{P}_{23}{P}_{12}(\mathrm{\mid }\omega {⟩}_{12}\mathrm{\mid }1{⟩}_3)=\frac{1}{4}\mathrm{\mid }\omega {⟩}_{12}\mathrm{\mid }1{⟩}_3\mathrm{.}$ For example, the first intermediate vector is $P_{23}(\mathrm{\mid }\omega {⟩}_{12}\mathrm{\mid }0{⟩}_3)=(-\mathrm{\mid }001⟩+\mathrm{\mid }010⟩)\mathrm{/}(2\sqrt{2})$; $P_{12}$ kills $\mathrm{\mid }001⟩$ and sends $\mathrm{\mid }010⟩$ to $\mathrm{\mid }\omega {⟩}_{12}\mathrm{\mid }0{⟩}_3\mathrm{/}\sqrt{2}$. The second computation is the explicit calculation $P_{23}(\mathrm{\mid }\omega {⟩}_{12}\mathrm{\mid }1{⟩}_3)=(-\mathrm{\mid }101⟩+\mathrm{\mid }110⟩)\mathrm{/}(2\sqrt{2})$; $P_{12}$ kills $\mathrm{\mid }110⟩$ and sends $\mathrm{\mid }101⟩$ to $-\mathrm{\mid }\omega {⟩}_{12}\mathrm{\mid }1{⟩}_3\mathrm{/}\sqrt{2}$, again giving the coefficient $1\mathrm{/}4$. Since $P_{12}{P}_{23}{P}_{12}$ first projects into the displayed image, it follows on all eight basis vectors that $P_{12}{P}_{23}{P}_{12}=P_{12}\mathrm{/}4$. Reversing the three tensor factors exchanges $P_{12}$ and $P_{23}$ (the singlet line is unchanged by reversing its two entries), so also $P_{23}{P}_{12}{P}_{23}=P_{23}\mathrm{/}4$. Hence $e_i{e}_{i+1}{e}_i=8P_i{P}_{i+1}{P}_i=2P_i=e_i$ and the reversed relation holds. Operators on disjoint tensor factors commute, giving the distant relation.

The relations proved in steps 1.1 and 2.1 and [F3] give an algebra homomorphism from the presented quotient, since every defining relation has zero image. Each $e_i$ is self-adjoint by step 1.1, so the operator adjoint of a word reverses its factors and fixes its generator images; conjugating coefficients agrees with the algebra involution [F4]. Thus the representation preserves $\ast$.

The adjacent swap is the place action of the transposition $(i,i+1)$, since that transposition is its own inverse in the left-place convention. Therefore $e_i=I-\rho ((i,i+1))$. Applying the factor placement in [F5] shows that on each summand $S^{\lambda }\otimes M_{\lambda }$ it is $(I-{\rho }_{\lambda }((i,i+1)))\otimes I_{M_{\lambda }}$, as claimed.