Temperley–Lieb Towers, Spin Chains, and Scaling Interfaces: Examples and Counterexamples

Here $q^{-1}=(1-i)\mathrm{/}\sqrt{2}$ and $q=(1+i)\mathrm{/}\sqrt{2}$. The conjugate transpose of the mixed-spin block $\left(\begin{array}{cc}{q}^{-1}& -1\\ -1& q\end{array}\right)$ is $\left(\begin{array}{cc}q& -1\\ -1& q^{-1}\end{array}\right)$, which differs from it because $q\mathrm{\ne }{q}^{-1}$.

Since the other two basis vectors are annihilated by $U(q)$, the unequal mixed-spin blocks show $U(q{)}^{\ast }\mathrm{\ne }U(q)$. This single valid unit-circle parameter refutes universal self-adjointness in the declared basis convention.

The finite operator formulas in [F1] contain only adjacent generators within the listed open range. A periodic bond would require a separately defined operator acting on sites $L$ and $1$ (or the corresponding last and first Potts sites), and no such term occurs in either sum. Thus these formulas define open-chain models only.

The exact source labels and qualifications in [F2] make the lattice-to-Virasoro-mode correspondence conjectural/heuristic, despite the stated partial evidence. This item therefore makes no universal spectral, Bethe-completeness or boundary-algebra claim; that limit is an inference from the scope of the cited result, not a separate assertion by its authors.

For $x=[x_{ab}]\in M_k(P)$, $E_N(x)=k^{-1}{\sum }_a{x}_{aa}\otimes 1$. If $x\ge 0$, each diagonal compression $x_{aa}$ is positive, so this sum is positive. The formula makes $E_N$ unital, fixes $N$, is $N$-bimodular, and preserves the product trace; its finite sum of normal diagonal maps is normal. If $E_N(x)=0$ for $x\ge 0$, then ${\tau }_M(x)={\tau }_N(E_N(x))=0$, so faithfulness of ${\tau }_M$ gives $x=0$; hence $E_N$ is faithful. An element ${\sum }_{c,d}{p}_{cd}\otimes e_{cd}$ commuting with every $1\otimes e_{ab}$ must be $p\otimes 1$, and commuting also with $P\otimes 1$ forces $p\in Z(P)=\mathbb{C}1$; thus $M$ is a factor. It has no minimal projection: if $q\in M$ were minimal, some diagonal corner $r=(1\otimes e_{ii})q(1\otimes e_{ii})$ would be nonzero. Its support projection $s$ in the diffuse factor $P\otimes e_{ii}$ is nonzero. The polar decomposition of $q(1\otimes e_{ii})$ lies in $M$ by [F5] and gives a partial isometry from $s$ onto a nonzero subprojection of $q$. Since $P$ has no minimal projections, choose $0<t<s$; its image under that partial isometry is a nonzero proper subprojection of $q$, a contradiction. Thus $M$ is a II1 factor; $N\cong P$ is one as well.

For matrix units, $u_{ab}^{\ast }{u}_{cd}=k1\otimes e_{ba}{e}_{cd}=k{\delta }_{ac}(1\otimes e_{bd})$, so $E_N(u_{ab}^{\ast }{u}_{cd})={\delta }_{ac}{\delta }_{bd}1$. Thus the $k^2$ vectors are orthonormal as a right $N$-module frame. More generally, for $x={\sum }_{c,d}{x}_{cd}\otimes e_{cd}$, $E_N(u_{ab}^{\ast }x)=k^{-1\mathrm{/}2}{x}_{ab}\otimes 1$, and hence ${\sum }_{a,b}{u}_{ab}{E}_N(u_{ab}^{\ast }x)=x$. These coordinates identify $\mathcal{X}$ isometrically with $N^{k^2}$, so it is complete with projection model $p=1_{k^2}$; the inner-product traces sum to $k^2$, and [F2] gives index $k^2$.

An element ${\sum }_{c,d}{p}_{cd}\otimes e_{cd}$ commutes with every $p\otimes 1$ exactly when each $p_{cd}$ is central in the factor $P$, hence when it lies in $1\otimes M_k$. On the right action of this relative commutant, [F4] gives unnormalized module trace ${\sum }_{a,b}{\tau }_N(E_N(u_{ab}^{\ast }{u}_{ab}(1\otimes e_{cd})))=k{\mathbf{1}}_{c=d}$. Indeed $e_{ba}{e}_{ab}{e}_{cd}={\mathbf{1}}_{b=c}{e}_{bd}$, whose normalized matrix trace is ${\mathbf{1}}_{b=c=d}\mathrm{/}k$, the scalar factor $k$ in $u_{ab}$ cancels the normalized matrix trace $1\mathrm{/}k$, and the sum over $a$ has $k$ terms. Dividing by $\lambda =k^2$ gives ${\mathbf{1}}_{c=d}\mathrm{/}k={\mathrm{tr}}_k(e_{cd})$, so the right-action module trace and $\tau$ agree on $N^{\mathrm{\prime }}\cap M$; the inclusion is extremal.

Applying [F3] with $\lambda =k^2$ gives ${\tau }_1(x{e}_{N}y)=k^{-2}({\tau }_P\otimes {\mathrm{tr}}_k)(yx)$ for every $x,y\in M$. This is the claimed trace normalization.

Substituting $q=1$ in $\delta =q+q^{-1}$ gives $\delta =2$, and the mixed-spin block in [F1] becomes $B=\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$. In the one-down basis $(\downarrow \uparrow \uparrow ,\uparrow \downarrow \uparrow ,\uparrow \uparrow \downarrow )$, $e_1=\left(\begin{array}{ccc}1& -1& 0\\ -1& 1& 0\\ 0& 0& 0\end{array}\right)$ and $e_2=\left(\begin{array}{ccc}0& 0& 0\\ 0& 1& -1\\ 0& -1& 1\end{array}\right)$. Their sum is $K$. The two-down sector gives the same sum with $e_1,e_2$ interchanged, while the all-up and all-down vectors are annihilated.

On the one-down sector, direct multiplication gives $e_1{e}_2{e}_1=e_1$ and $e_2{e}_1{e}_2=e_2$; on the two-down sector the same two matrices are interchanged, so both identities still hold. The square identity follows from $B^2=2B$, and the distant commutation condition is vacuous for two generators. This agrees with [F2].

Negating the two embedded generators therefore gives $H_3=\mathrm{diag}(0,-K,-K,0)$ in the stated ordered number-of-down-spins blocks. Equivalently, [F3] gives $H_3=\frac{1}{2}(X_1{X}_2+Y_1{Y}_2+X_2{X}_3+Y_2{Y}_3+Z_1{Z}_2+Z_2{Z}_3-2I)$, with no bond $(3,1)$. Here $q=1$ gives $\cos \gamma =1$ and $\sin \gamma =0$, so the endpoint contribution in [F3] vanishes.

The geometric sum $Q^{-1}{\sum }_r{M}_Q^r$ is the rank-one projection onto $\mathrm{\mid }+⟩=Q^{-1\mathrm{/}2}{\sum }_a\mathrm{\mid }a⟩$, and $Q^{-1}{\sum }_r{R}_1^r{R}_2^{-r}={\sum }_a\mathrm{\mid }aa⟩⟨aa\mathrm{\mid }$. Multiplying these projections by $\sqrt{Q}$ gives the stated explicit generators. From the assigned $q$ values, $q+q^{-1}=2\cos (\pi \mathrm{/}4)=\sqrt{2}$, $2\cos (\pi \mathrm{/}6)=\sqrt{3}$, and $2\cos 0=2$.

There are $L$ odd indices $2j-1$ for $1\le j\le L$ and $L-1$ even indices $2j$ for $1\le j<L$. Their total is $2L-1$, with no additional term at the boundary. The relation verification is [F2], so each listed case has the declared open-chain Temperley–Lieb normalization.

Since $1_3={\iota }_2(1_2)$ and $d_1={\iota }_2(d_1)$, the identity $E_2{\iota }_2=\mathrm{i}\mathrm{d}$ gives the first two values. Also $d_2={\iota }_2(1_2)d_2$, so the right-generator Markov rule in [F2] gives $E_2(d_2)={\delta }^{-1}{1}_2$.

The tower inclusion sends $d_1$ to the first generator on three strands. Bimodularity and the last value therefore give $E_2(d_1{d}_2)=d_1{E}_2(d_2)={\delta }^{-1}{d}_1$ and $E_2(d_2{d}_1)=E_2(d_2)d_1={\delta }^{-1}{d}_1$. This lists all five basis images by [F1].

Trace preservation [F2] gives ${\tau }_2(E_2(D))={\tau }_3(D)$ for every basis element $D$ and hence for their linear combinations. Explicitly, the values are $1$ on $1_3$, ${\delta }^{-1}$ on $d_1,d_2$, and ${\delta }^{-2}$ on $d_1{d}_2,d_2{d}_1$, by [F3] and the displayed closures.