The annular central element constructs Temperley–Lieb reflection maps

It suffices to perform computations for left modules: send a right module to the left module with $a\cdot v=v\cdot a^{\ast }$, where $\ast$ is diagram reflection. This is an exact equivalence preserving Hom direction, induction and restriction. Fix a square root of $q$ and put $g_i=q^{1\mathrm{/}2}1-q^{-1\mathrm{/}2}{d}_i$. Expanding with [F1] gives $g_i^{-1}=q^{-1\mathrm{/}2}1-q^{1\mathrm{/}2}{d}_i$ and $g_i{g}_{i+1}{g}_i=g_{i+1}{g}_i{g}_{i+1}$. Let $T_n:A_{n+1}\to A_n$ close the last top endpoint to the last bottom endpoint externally on the right, removing any resulting loop with factor $\delta$. This is an $A_n$-bimodule map by the matching rule. Define $M_n=g_n\cdots g_2{g}_1^2{g}_2\cdots g_n$ and $F_n=T_n(M_n)$; set $F_0=\delta$. Every summand of $M_n$ has an even total number of square-root factors, so $F_n$ has coefficients in $\mathbb{Z}[q,q^{-1}]$, independent of the chosen square root.

For clarity give a basis proof of the induction sequence. In left-module notation put $h_r=d_r{d}_{r+1}\cdots d_n$ for $1\le r\le n$, and $h_{n+1}=1$. A state $d$ is $r$-admissible when it has no adjacent arc at an index $i\ge r$. The tensors $h_r\otimes d$ with $d$ $r$-admissible, including all $1\otimes d$, span $A_{n+1}{\otimes }_{A_n}{V}_{n,p}$. First, a diagram word is a sum of $h_{r}u$ with $u\in A_n$: reverse a spanning Jones normal form from [F1]. Its initial ascending block is either $h_r$ or contains no $d_n$, and every later block lies in $A_n$. The surjection from the presented algebra to the diagram algebra is supplied by the local relations, so this spanning assertion applies to diagram words. Next, if $d$ has its largest adjacent arc at $s\ge r$, write $d=d_{s}f$. For $n\ge 3$ this can be done without division: move the endpoint immediately outside that adjacent arc through the arc, so applying $d_s$ restores $d$ and creates no loop. If $s=1$ use site $3$; otherwise use site $s-1$. The outside site is either a defect or an endpoint of another arc, and reconnecting its former partner supplies the inverse state $f$. Now the relations give $h_r\otimes d=h_{s+2}\otimes d_r\cdots d_{s-1}d$. The latter state is $(s+2)$-admissible: the adjacent arc at $s$ has moved to $r$, all intervening endpoints shift by two, and the tail beyond $s+1$ still has no adjacent arc. Repetition, increasing $r$, proves spanning. At $n=1,2$ the same conclusion is checked on the at most two states, except $n=2,p=1,\delta =0$, when the needed factorization $d=d_{1}f$ fails.

The braid relations give $g_i{M}_n=M_n{g}_i$ for $i<n$. To see this directly, commute $g_i$ past all $g_j$ with $\mathrm{\mid }j-i\mathrm{\mid }>1$ in the descending and ascending strings; the remaining equality is $g_i{g}_{i+1}{g}_i=g_{i+1}{g}_i{g}_{i+1}$ on each side of the middle string. This proves the equality inductively in the length of the strings, starting with $g_1{g}_2{g}_1^2{g}_2=g_2{g}_1^2{g}_2{g}_1$. Bimodule linearity of $T_n$ gives $g_i{F}_n=F_n{g}_i$, hence $d_i{F}_n=F_n{d}_i$. Thus $F_n$ is central. Reversing the palindromic word $M_n$ and reflecting its closure leaves it unchanged, so $F_n^{\ast }=F_n$.

Compute $F_n$ on the state $z_p$ with adjacent arcs $(1,2),(3,4),\dots ,(2p-1,2p)$ and all remaining sites defects. At a cup, expansion of the two adjacent crossings uses the identities $d_i{g}_{i+1}{g}_i=-d_i{d}_{i+1}$ and $g_i{g}_{i+1}{d}_i=-d_{i+1}{d}_i$; for example the first is $q{d}_i-d_i{d}_{i+1}-\delta d_i+q^{-1}{d}_i{d}_{i+1}{d}_i=-d_i{d}_{i+1}$. The two minus signs on the two sides of each cup cancel. Thus each initial cup can be removed without changing the coefficient of $z_p$. For the $t=n-2p$ defect strands, expand the auxiliary strand successively from its first crossing. A smoothing connecting two defects vanishes in the standard quotient. There are two remaining continuations: following the auxiliary strand on its two sides gives respectively $q^t\delta -q^{t-1}=q^{t+1}$ and $q^{-t}\delta -q^{-t+1}=q^{-(t+1)}$. The factors arise by choosing the identity smoothing at each subsequent crossing until the final cap; that cap either closes a loop or chooses the second smoothing, giving the displayed two terms on each side. Hence the coefficient of $z_p$ is $f_{n,p}$. For generic $q$, [F3] makes the form nondegenerate. Capping $z$ against $y$ and then attaching $x$ gives the rank-one diagram operator $z\mapsto \varphi (y,z)x$, because lower-propagating terms vanish. Thus every nonzero vector generates the standard, which is simple. Centrality now makes $F_n$ scalar on the entire standard: over $\mathbb{C}$ a central operator has an eigenvalue, whose nonzero eigenspace is the simple module. Since the action matrices and $f_{n,p}$ are Laurent polynomial in $q$, this scalar identity extends to every $q\mathrm{\ne }0$.

On the submodule and quotient in [F2], $F_{n-1}$ acts by $f_{n-1,p}$ and $f_{n-1,p-1}$. Their difference is $-(q-q^{-1})(q^{n-2p+1}-q^{-(n-2p+1)})$. Under the noncritical and $q\mathrm{\ne }\pm 1$ hypotheses it is nonzero. The operator on the restriction satisfies $(F_{n-1}-f_{n-1,p})(F_{n-1}-f_{n-1,p-1})=0$, because each factor kills its respective subquotient. Its two spectral projections therefore give an invariant direct sum splitting the restriction sequence.

Append an adjacent arc at sites $n+1,n+2$ to $d$, obtaining $\bar{d}\in V_{n+2,p+1}$. The map $\mathrm{\Phi }(u\otimes d)=u\bar{d}$ into $V_{n+2,p+1}\downarrow$ is balanced and $A_{n+1}$-linear: elements of $A_n$ act only on the first $n$ sites. Its image of $h_r\otimes d$ has its rightmost adjacent arc at $r$; deleting that arc recovers $d$. Conversely every $(n+2,p+1)$-state has a rightmost adjacent arc; delete it at position $r$, and the remaining state is $r$-admissible. These two operations are inverse, so the spanning tensors in step 1.2 map bijectively to the link-state basis. They are therefore a basis and $\mathrm{\Phi }$ is an isomorphism, except in the stated exceptional case. Applying the restriction sequence [F2] to $V_{n+2,p+1}$ now gives the asserted induction sequence. Applying step 4.1 at rank $n+2$ splits it whenever $(n,p)$ is noncritical, since $n+2-2(p+1)+1=n-2p+1$.

Suppose $(n,p)$ is critical, $q\mathrm{\ne }\pm 1$, and $z_p$ is as in step 3.1. In the induced basis the coefficient of $h_{2p+1}\otimes z_p$ in $F_{n+1}(1\otimes z_p)$ is $(-1{)}^n({\delta }^2-4)\mathrm{\ne }0.$ Here is the local expansion computing it. First take $p=0$. To leave $h_1=d_1\cdots d_n$ after reduction, the first pair of crossings of the closed auxiliary strand has three contributing smoothings, with weights $-1,-1,\delta q^{-1}$; the fourth, of weight $q$, closes off the needed through path. This gives $-2+\delta q^{-1}$. At each of the next $n-1$ pairs exactly one continuation retains the required external matching, of weight $-1$: the other continuation either caps the auxiliary path prematurely or makes an additional bottom arc. At the final pair the corresponding three weights are $-1,-1,\delta q$, giving $-2+\delta q$. Their product is $(-1{)}^{n-1}(-2+\delta q^{-1})(-2+\delta q)=(-1{)}^n({\delta }^2-4)$. The required exterior matching of $h_1$ has top pair $(1,2)$, bottom pair $(n,n+1)$, and through paths from top $j+2$ to bottom $j$ for $1\le j<n$. Follow the auxiliary strand subject to these connections: a premature cap at an intermediate row would create another exterior arc instead of the required through path, so its continuation is excluded. For $p>0$, remove the first $p$ cups using the two local identities in step 3.1, on both sides of the auxiliary path. This leaves the same computation on $n-2p$ sites; $n-2p>0$ by criticality and $\kappa \ge 2$, and its parity equals that of $n$. The target basis tensor is $h_{2p+1}\otimes z_p=h_1\otimes z_p$, by successive $d_i{d}_{i+1}{d}_i=d_i$ reductions. One may do the cup removal first for $\delta \mathrm{\ne }0$ by inserting ${\delta }^{-1}{d}_{2j-1}$ at each initial cup. The resulting coefficient identity extends to $\delta =0$: the induced basis reduction in step 1.2 uses no division, so its matrices, and this coefficient, are Laurent polynomials in $q$. Finally ${\delta }^2-4=(q-q^{-1}{)}^2\mathrm{\ne }0$.

In the induction sequence at a critical $(n,p)$ the two standard constituents have the same eigenvalue $f=f_{n+1,p}=f_{n+1,p+1}$. The central endomorphism $F_{n+1}-f$ kills both constituents but is nonzero on the induction by step 6.1. Consequently it factors through the quotient $V_{n+1,p}$ into the submodule $V_{n+1,p+1}$, giving a nonzero map between these adjacent reflected standards. Explicitly, lift a quotient vector to the induction, apply $F_{n+1}-f$, and use the inverse of the submodule injection; the result is independent of the lift because this operator kills the submodule.

Every symmetric pair has the form $(n+j,p),(n+j,p+j)$ for a critical $(n,p)$ and $1\le j<\kappa$. For $j\ge 2$, induction and restriction at the noncritical labels give $\mathrm{Hom}(V_{n+j,p},V_{n+j,p+j})\cong \mathrm{Hom}(V_{n+j-1,p}\uparrow ,V_{n+j,p+j})\cong \mathrm{Hom}(V_{n+j-1,p},V_{n+j,p+j}\downarrow )\cong \mathrm{Hom}(V_{n+j-1,p},V_{n+j-1,p+j-1})\mathrm{.}$ In the first and last isomorphisms the extra standard summand has a different $F$-eigenvalue: substitute $q^{2(n-2p+1)}=1$ in step 3.1, and equality would force $j=1$ or $j=\kappa$. Thus its Hom is zero. The middle adjunction is explicit: $f\mapsto (v\mapsto f(1\otimes v))$, with inverse $g\mapsto (a\otimes v\mapsto ag(v))$. This works also when an extraneous summand is out of range. Reducing $j$ to $1$ and applying step 7.1 proves nonvanishing for every reflected pair. Returning along step 1.1 gives all statements for the repository's right modules.