Temperley–Lieb Towers, Spin Chains, and Scaling Interfaces

For any path $(a_0,\dots ,a_n,b)$ ending at $b$, its penultimate vertex $a$ is uniquely determined and satisfies $a\sim b$; deleting $b$ gives a unique prefix in ${\mathsf{P}}_{n,a}$. Conversely appending $b$ to any path in ${\mathsf{P}}_{n,a}$ with $a\sim b$ gives a path ending at $b$. Hence the set of paths ending at $b$ is the disjoint union of the appended path sets $\{pb:p\in {\mathsf{P}}_{n,a}\}$ over all such $a$. The basis vectors therefore split as the asserted direct sum, and $E_{p,q}^{(n,a)}$ acts on each appended block as the same matrix unit while acting by zero on the other predecessor blocks.

Take matrix units $X=E_{p,q}^{(n,a)}$ and $Y=E_{r,s}^{(n,c)}$. If $a\mathrm{\ne }c$, then $XY=0$; in the product ${\iota }_n(X){\iota }_n(Y)$ a nonzero matrix-unit product would require the middle path $qb$ to equal $rd$ for some $b\sim a$ and $d\sim c$, forcing both $q=r$ as length-$n$ paths and $a=c$, a contradiction. Thus the images multiply to zero.

If $a=c$, then $XY={\mathbf{1}}_{q=r}{E}_{p,s}^{(n,a)}$. On multiplying the two sums in [F2], the only nonzero terms have the same appended vertex $b$ and equal middle path $qb=rb$, which is equivalent to $q=r$. The product is consequently ${\mathbf{1}}_{q=r}{\sum }_{b\sim a}{E}_{pb,sb}^{(n+1,b)}={\iota }_n(XY)$. Adjoints are preserved because ${\iota }_n((E_{p,q}^{(n,a)}{)}^{\ast })={\sum }_{b\sim a}{E}_{qb,pb}^{(n+1,b)}={\iota }_n(E_{p,q}^{(n,a)}{)}^{\ast }$. Bilinearity proves multiplicativity and $\ast$-preservation on all of $K_n$.

The identity of $K_n$ is ${\sum }_{a,p}{E}_{p,p}^{(n,a)}$. Its image is the sum of the diagonal units indexed by every one-step extension $pb$ of every length-$n$ path $p$; by step 1.1 each length-$(n+1)$ path occurs exactly once. This is the identity of $K_{n+1}$. For injectivity, every vertex $a$ has at least one neighbor in $A_{\mathrm{\ell }-1}$, since $\mathrm{\ell }\ge 3$. For each source unit $E_{p,q}^{(n,a)}$ choose one neighbor $b$; the coordinate $E_{pb,qb}^{(n+1,b)}$ in its image determines the source unit uniquely by deleting the last vertex. Distinct source units therefore have disjoint image coordinates, so no nonzero linear combination maps to zero. Hence ${\iota }_n$ is a unital injective $\ast$-homomorphism and step 1.1 gives the multiplicity-one branching rule.

Multiplying the nonzero block gives $A^2=\left(\begin{array}{cc}{q}^{-2}+1& -(q+q^{-1})\\ -(q+q^{-1})& q^2+1\end{array}\right)=(q+q^{-1})A=\delta A$, because $q^{-2}+1=\delta q^{-1}$ and $q^2+1=\delta q$. The zero subspace remains zero, so $U(q{)}^2=\delta U(q)$, including when $\delta =0$.

On the three-spin subspace with one down spin, in the ordered basis $\downarrow \uparrow \uparrow ,\uparrow \downarrow \uparrow ,\uparrow \uparrow \downarrow$, the two embedded operators are $A_1=\left(\begin{array}{ccc}q& -1& 0\\ -1& q^{-1}& 0\\ 0& 0& 0\end{array}\right)$ and $A_2=\left(\begin{array}{ccc}0& 0& 0\\ 0& q& -1\\ 0& -1& q^{-1}\end{array}\right)$. Multiplication gives $A_1{A}_2=\left(\begin{array}{ccc}0& -q& 1\\ 0& 1& -q^{-1}\\ 0& 0& 0\end{array}\right)$ and $(A_1{A}_2)A_1=A_1$; likewise $A_2{A}_1=\left(\begin{array}{ccc}0& 0& 0\\ -q& 1& 0\\ 1& -q^{-1}& 0\end{array}\right)$ and $(A_2{A}_1)A_2=A_2$. The two-down-spin subspace has the same pair of matrices in reverse order, so both triple identities hold there as well. On the all-up and all-down vectors both operators vanish. These four invariant subspaces exhaust the three-spin space, proving both adjacent relations.

If $\mathrm{\mid }i-j\mathrm{\mid }>1$, the operators act on disjoint tensor factors. For elementary tensors, applying the two operators in either order changes separate tensor coordinates and yields the same vector; linearity gives $e_i{e}_j=e_j{e}_i$. This verifies all relations in [F2] for every chain length.

Negating the displayed formula for each $e_i$ changes its prefactor $-1\mathrm{/}2$ to $1\mathrm{/}2$. Separating the terms gives the bulk sum in the statement and the remaining term $\frac{i\sin \gamma }{2}{\sum }_{i=1}^{L-1}(Z_i-Z_{i+1})$.

In the endpoint sum, each interior $Z_j$, $2\le j\le L-1$, occurs once with coefficient $+1$ in $Z_j$ and once with coefficient $-1$ in $-Z_j$, hence cancels. Only $Z_1-Z_L$ remains. The original index range is $1\le i<L$, so it contributes exactly the open bonds and no wrap-around term.

Let $P=Q^{-1}{\sum }_{r=0}^{Q-1}{M}_Q^r=\mathrm{\mid }+⟩⟨+\mathrm{\mid }$, where $\mathrm{\mid }+⟩=Q^{-1\mathrm{/}2}{\sum }_a\mathrm{\mid }a⟩$, and let $D={\sum }_a\mathrm{\mid }aa⟩⟨aa\mathrm{\mid }$. The finite geometric sum gives $P^2=P$; orthogonality of the $R_Q$ eigenvalues gives $Q^{-1}{\sum }_r{R}_{Q,1}^r{R}_{Q,2}^{-r}=D$, since its action on $\mathrm{\mid }ab⟩$ is $Q^{-1}{\sum }_r{\omega }_Q^{r(a-b)}\mathrm{\mid }ab⟩$, equal to $1$ if $a=b$ and $0$ otherwise. Thus odd and even generators are $\sqrt{Q}{P}_j$ and $\sqrt{Q}{D}_{j,j+1}$. Their idempotence gives $e_i^2=\sqrt{Q}{e}_i$.

On three sites, $P_1{D}_{12}{P}_1=Q^{-1}{P}_1$ and $D_{12}{P}_2{D}_{12}=Q^{-1}{D}_{12}$. For the first identity, $P\mathrm{\mid }a⟩=Q^{-1\mathrm{/}2}\mathrm{\mid }+⟩$ and $⟨a\mathrm{\mid }P\mathrm{\mid }a⟩=Q^{-1}$, so summing $P\mathrm{\mid }a⟩⟨a\mathrm{\mid }P$ over the second-site labels yields $Q^{-1}{P}_1$. For the second, $P_2\mathrm{\mid }a⟩=Q^{-1}{\sum }_b\mathrm{\mid }b⟩$ and $D_{12}$ retains only the term $b=a$, giving coefficient $Q^{-1}$ on each $\mathrm{\mid }aa⟩$. Multiplying each identity by $(\sqrt{Q}{)}^3$ proves both adjacent triple relations.

Odd generators on distinct sites commute because they act on disjoint factors. Every even generator is a sum of products of the diagonal matrices $R_Q$ on its two sites, so any two even generators commute, including when their bonds overlap in one site. An odd and an even generator whose indices differ by more than one have disjoint supports, so commute. Finally $\sqrt{2}=2\cos (\pi \mathrm{/}4)$, $\sqrt{3}=2\cos (\pi \mathrm{/}6)$, and $2=2\cos 0$, verifying the three declared $q$ normalizations. These are exactly the relations in [F2].

On a basis diagram, appending a rightmost vertical strand leaves all existing matching pairs unchanged. Distinct diagrams remain distinct after appending it, so freeness in [F1] makes ${\iota }_n$ injective. Stacking two such appended diagrams creates no new closed component and is the appended version of their original stack, so ${\iota }_n$ preserves products and the unit. When $\delta$ is real, horizontal reflection also commutes with appending, so it preserves the involution. Right-closing the newly appended vertical strand creates exactly one circle, so $c_n({\iota }_n(D))=\delta D$ and $E_n{\iota }_n(D)=D$. Linearity gives $E_n{\iota }_n=\mathrm{i}\mathrm{d}$.

For basis diagrams $a,b$ on $n$ strands and $x$ on $n+1$ strands, stack the three diagrams and attach the right closure arc at the right boundary. Every path or cycle in this finite graph is unchanged if the left $n$-strand stacks are performed before the right closure: the closure arc meets only the two rightmost boundary endpoints, and the factors $a,b$ attach only to the left $n$ interfaces. The exterior matching is therefore the product $a{c}_n(x)b$, with the same number of removed cycles and the same powers of $\delta$. Thus $c_n({\iota }_n(a)x{\iota }_n(b))=a{c}_n(x)b$. Multiplication by ${\delta }^{-1}$ proves bimodularity on basis diagrams; bilinearity extends it.

Closing all remaining strands of $c_n(x)$ produces exactly the same closed matching graph as closing all $n+1$ strands of $x$. The normalized factor for $E_n$ is ${\delta }^{-1}$, so the total normalization changes from ${\delta }^{-n}$ to ${\delta }^{-(n+1)}$, exactly the normalization in [F3] for rank $n+1$. Hence ${\tau }_n(E_n(x))={\tau }_{n+1}(x)$ on basis diagrams and then on all $x$. The retraction identity also gives $({\iota }_n{E}_n{)}^2={\iota }_n{E}_n$ as endomorphisms of ${\mathrm{T}\mathrm{L}}_{n+1}$.

Right-closing ${\iota }_n(x)d_n$ connects the last cap and cup to the exterior closure arc; the result is the diagram $x$ on the first $n$ strands, with no new circle. Thus $c_n({\iota }_n(x)d_n)=x$, so $E_n({\iota }_n(x)d_n)={\delta }^{-1}x$. Trace preservation and [F4] give ${\tau }_{n+1}({\iota }_n(x)d_n)={\tau }_n(E_n({\iota }_n(x)d_n))={\delta }^{-1}{\tau }_n(x)$.

Suppose $x\ge 0$ in the finite-dimensional $C^{\ast }$-algebra ${\mathrm{T}\mathrm{L}}_{n+1}$ and the traces are faithful and positive. On each basis matching, reflecting before closing produces the same retained $(n,n)$ matching and the same closed components as closing before reflecting; the scalar ${\delta }^{-1}$ is real. Hence $E_n(x^{\ast })=E_n(x{)}^{\ast }$, so $E_n(x)$ is self-adjoint. For every $b\in {\mathrm{T}\mathrm{L}}_n$, trace preservation and bimodularity give ${\tau }_n(b^{\ast }{E}_n(x)b)={\tau }_{n+1}({\iota }_n(b{)}^{\ast }x{\iota }_n(b))\ge 0$. If $E_n(x)$ had a negative spectral subspace, its nonzero spectral projection $p$ would satisfy ${\tau }_n(p{E}_n(x)p)<0$, since $p{E}_n(x)p$ is strictly negative on that subspace and ${\tau }_n$ is faithful; this contradicts the same inequality with $b=p$. Thus $E_n(x)\ge 0$. It is unital, idempotent, bimodular and trace-preserving by the preceding steps, so it is a positive conditional expectation in the stated $C^{\ast }$-positive regime.

Let $x\in J_n$ and $a,b,y\in {\mathrm{T}\mathrm{L}}_n$. The defining radical condition gives ${\tau }_n((ya)x)=0$, so ${\tau }_n(yax)=0$. By cyclicity, ${\tau }_n(yxb)={\tau }_n((by)x)=0$, since $by\in {\mathrm{T}\mathrm{L}}_n$. Thus both $ax$ and $xb$ lie in $J_n$, proving that $J_n$ is a two-sided ideal.

If $x\in J_n$ and $z\in {\mathrm{T}\mathrm{L}}_{n+1}$, trace preservation and right bimodularity yield ${\tau }_{n+1}(z{\iota }_n(x))={\tau }_n(E_n(z{\iota }_n(x)))={\tau }_n(E_n(z)x)=0$, because $E_n(z)\in {\mathrm{T}\mathrm{L}}_n$. Hence ${\iota }_n(x)\in J_{n+1}$.

If $x\in J_{n+1}$ and $a\in {\mathrm{T}\mathrm{L}}_n$, then ${\tau }_n(a{E}_n(x))={\tau }_n(E_n({\iota }_n(a)x))$ by left bimodularity, and trace preservation makes this ${\tau }_{n+1}({\iota }_n(a)x)=0$ by the definition of $J_{n+1}$. Thus $E_n(x)\in J_n$. These ideal inclusions make both quotient maps independent of representatives, so the tower and right-closure maps descend.

The quantum-integer recurrence with real $\delta$ makes each $[r]$ real, and [F5] gives $[r]>0$ for $1\le r<\mathrm{\ell }$. Starting with $p_0=p_1=1$, induction in the JW recursion shows $p_r^{\ast }=p_r$ through $r=\mathrm{\ell }-1$: reflection fixes $P_r=p_r\otimes 1$ and $U_r$, while the coefficient $[r]\mathrm{/}[r+1]$ is real whenever this recursion is used. In particular the boundary projector $p_{\mathrm{\ell }-1}$ is self-adjoint. For $n<\mathrm{\ell }-1$, the generated ideal component $I_n$ is zero and is reflection-stable; for $n\ge \mathrm{\ell }-1$, its fixed embedded generator $p_{\mathrm{\ell }-1}\otimes 1$ is self-adjoint, so $I_n={\mathrm{T}\mathrm{L}}_n(p_{\mathrm{\ell }-1}\otimes 1){\mathrm{T}\mathrm{L}}_n$ is reflection-stable because $(x(p_{\mathrm{\ell }-1}\otimes 1)y{)}^{\ast }=y^{\ast }(p_{\mathrm{\ell }-1}\otimes 1)x^{\ast }$. By [F3], these are the endomorphism components of the generated negligible tensor ideal. Also [F7] gives ${\tau }_n(D^{\ast })=\overline{{\tau }_n(D)}$ on diagrams and hence on all elements. If $x\in J_n$, then for every $y$ cyclicity yields ${\tau }_n(y{x}^{\ast })=\overline{{\tau }_n(x{y}^{\ast })}=\overline{{\tau }_n(y^{\ast }x)}=0$; thus the trace radical is reflection-stable as well. Since [F3] identifies it with $I_n$, the quotient involution is well defined.

Fix a retained endpoint $1\le a\le k$, write $f_a=p_a\otimes 1$, and use the supplier's maps ${\alpha }_a=f_a(1_{a-1}\otimes \mathrm{c}\mathrm{u}\mathrm{p})p_{a-1}$ and ${\beta }_a=p_{a-1}(1_{a-1}\otimes \mathrm{c}\mathrm{a}\mathrm{p})f_a$. Horizontal reflection reverses composition and exchanges the cup and cap; step 1.1 fixes $f_a$ and $p_{a-1}$, so ${\alpha }_a^{\ast }={\beta }_a$. The supplier identity ${\beta }_a{\alpha }_a={\lambda }_a{p}_{a-1}$ therefore gives ${\alpha }_a^{\ast }{\alpha }_a={\lambda }_a{p}_{a-1}$. Set $v_a={\lambda }_a^{-1\mathrm{/}2}{\alpha }_a$. Then $v_a^{\ast }{v}_a=p_{a-1}$ and $v_a{v}_a^{\ast }={\lambda }_a^{-1}{\alpha }_a{\beta }_a$, the down projection in the JW split. For $a<k$, the complementary up projection is $p_{a+1}$; it is self-adjoint by step 1.1 and orthogonal to the down projection by the JW turnback relations in [F2]. At $a=k$, the quotient kills $p_{k+1}=p_{\mathrm{\ell }-1}$ before normalization, so only the down projection remains. The up inclusion and retraction are adjoints, as are $v_a$ and $v_a^{\ast }$. Composing these isometric branch maps along retained paths gives path maps $I_p$ with mutually orthogonal ranges and ${\sum }_p{I}_p{I}_p^{\ast }=1_n$. Hence $F_{p,q}=I_p{I}_q^{\ast }$ are complete matrix units and $F_{p,q}^{\ast }=F_{q,p}$. Their diagonal units equal the supplier's path projections (the reciprocal branch normalizations cancel), so ${\mathrm{Tr}}_n(F_{p,q})={\delta }_{p,q}[a+1]$ by [F2]. Thus the quotient with its inherited reflection is a direct sum of full complex matrix algebras with the usual adjoint, hence a finite-dimensional $C^{\ast }$-algebra. Since ${\tau }_n(F_{p,p})={\delta }^{-n}[a+1]>0$, its normalized closure trace is faithful positive.