Temperley–Lieb Link-State Modules and Cellularity

For basis diagrams $x:n\to t$ and $a,b:n\to n$, both $(x\cdot a)\cdot b$ and $x\cdot (ab)$ are represented by the same three-layer glued graph, with the same exterior paths and the same number of closed components. By [F1] their coefficients are the same power of $\delta$, and [F2] gives $(x\cdot a)\cdot b=x\cdot (ab)$. The identity diagram leaves $x$ unchanged. Bilinearity extends these equalities to $M_R(n,t)$, making it a right ${\mathrm{T}\mathrm{L}}_n(R,\delta )$-module before quotienting.

If $\mathrm{prop}(x)<t$, every propagating component of $x\circ a$ must pass through a distinct propagating component of $x$ at the glued $n$-boundary; stacking cannot create a new path from the top $t$-boundary of $x$ to the exterior bottom boundary of $a$ when no such path existed in $x$. Thus every basis term of $x\cdot a$ has fewer than $t$ propagating components. By [F3], $M_R^{<t}(n,t)$ is a right submodule.

The action therefore descends to the quotient $W_{n,t}(R,\delta )$, and the module identities from step 1.1 descend with it. Since the diagram basis of $M_R(n,t)$ is partitioned into diagrams with exactly $t$ propagating components and those with fewer, the surviving cosets form the stated link-state basis.

Encode each arc by an opening symbol at its left endpoint and a closing symbol at its right endpoint; encode each defect by an opening symbol. This gives a word of length $n$ with $a+t$ openings and $a$ closings. Every prefix has at least as many openings as closings, because each closing site in that prefix is paired with a distinct earlier opening site in the same prefix.

Conversely, scan any word with $a+t$ openings, $a$ closings and nonnegative prefix heights from left to right. Pair each closing symbol with the most recent unmatched opening. The prefix condition guarantees such an opening exists at every closing, and the final number of unmatched openings is $t$. The resulting arcs are nested or disjoint, so they form a noncrossing matching; the unmatched openings are its defects. This reverses step 1.1, giving a bijection between $T_{n,t}$ and those words.

There are $\left(\genfrac{}{}{0px}{}{n}{a}\right)$ words with $a$ closing positions, by [F2]. For $a\ge 1$, reflect openings and closings through the first prefix whose height is $-1$. This is a bijection from words that violate the prefix condition to all words with $a-1$ closings: the reflected word ends at height $t+2\ge 2$ and therefore has a first prefix of height $1$, whose reflection is the inverse. When $a=0$ there are no violating words, agreeing with the stated convention. Subtracting the bad count $\left(\genfrac{}{}{0px}{}{n}{a-1}\right)$ from the total gives the displayed formula.

For basis link states $x$ and $y$, the composite $y\circ x^{\ast }$ is a diagram from $t$ to $t$. If all $t$ boundary points propagate, planarity forces them to connect in their order, so after removing $c$ closed loops the result is ${\delta }^c{1}_t$ for a unique $c$. Otherwise it has fewer than $t$ propagating components. Thus the stated value is determined by the matching and loop count, and it extends uniquely to an $R$-bilinear form on the free link-state basis [F1, F2].

Reflection reverses the order of stacking, fixes each scalar $\delta$, and sends $y\circ x^{\ast }$ to $x\circ y^{\ast }$. Hence $y\circ x^{\ast }={\delta }^c{1}_t$ exactly when $x\circ y^{\ast }={\delta }^c{1}_t$ with the same $c$; the lower-propagating case is also preserved. Therefore ${\varphi }_{n,t}(x,y)={\varphi }_{n,t}(y,x)$ on basis states and, by bilinearity, on all module elements.

For basis states $x,y$ and a basis diagram $a\in {\mathrm{T}\mathrm{L}}_n$, reflection gives $(x\circ a{)}^{\ast }=a^{\ast }\circ x^{\ast }$. Both form values are therefore the coefficient of $1_t$ in the same associative composite $y\circ (x\circ a{)}^{\ast }=y\circ a^{\ast }\circ x^{\ast }=(y\circ a^{\ast })\circ x^{\ast }$. They have the same exterior matching and number of removed loops, so their coefficients of $1_t$ agree. If an intermediate term has fewer than $t$ propagating components, it remains lower-propagating after further stacking and contributes zero on both sides. This proves the invariance identity for basis states and basis diagrams; bilinearity proves it for all $x,y,a$.

Suppose $t>0$ and put $a=(n-t)\mathrm{/}2$. If $a=0$, take $x=y$ to be the all-defect state; gluing gives $t$ through paths and no loops, so ${\varphi }_{n,t}(x,y)=1_R\mathrm{\ne }0$. If $a>0$, let $x$ have adjacent arcs $(1,2),(3,4),\dots ,(2a-1,2a)$ and defects at $2a+1,\dots ,n$. Let $y$ have arcs $(2,3),(4,5),\dots ,(2a,2a+1)$, defects at $1$ and $2a+2,\dots ,n$. Both are link states. Their glued graph has one path alternating across sites $1,\dots ,2a+1$ and $t-1$ straight paths at the shared defects; it has no closed component. Hence ${\varphi }_{n,t}(x,y)=1_R\mathrm{\ne }0$.

Suppose $t=0$ and $n>0$, so $n=2a$ with $a\ge 1$. Every pair of link states glues to a finite graph in which every boundary site has degree two; each connected component is a cycle, and at least one cycle occurs. Thus every form value is divisible by $\delta$, so the form is identically zero when $\delta =0$. Conversely, let $x$ have adjacent arcs $(1,2),(3,4),\dots ,(n-1,n)$, and let $y$ have the outer arc $(1,n)$ together with the adjacent interior arcs $(2,3),(4,5),\dots ,(n-2,n-1)$. Their overlay is one cycle, so ${\varphi }_{n,0}(x,y)=\delta$. If $\delta \mathrm{\ne }0$, this value is nonzero and the form is not identically zero.

If $n=t=0$, there is one empty link state. Its reflection glues to the empty diagram with no loops, so ${\varphi }_{0,0}(x,x)=1_R\mathrm{\ne }0$. Together with the cases already treated, this exhausts all allowed $t$ and proves the equivalence.

Let $U$ be the span in $W_{n,t}$ of link states in which the rightmost bottom site $n$ is a defect. Removing that defect and its top endpoint gives a link state in $W_{n-1,t-1}$; conversely, adding a new rightmost bottom site joined to a new rightmost top defect gives the inverse map. These basis maps define an $R$-module isomorphism $W_{n-1,t-1}\to U$.

Under the inclusion in [F2], the rightmost strand of every diagram from ${\mathrm{T}\mathrm{L}}_{n-1}$ is vertical and does not meet the other $n-1$ sites. Stacking such a diagram below a link state in $U$ leaves the external site $n$ connected to its same defect. Thus $U$ is a ${\mathrm{T}\mathrm{L}}_{n-1}$-submodule, and the isomorphism in step 1.1 intertwines the restricted action.

The quotient by $U$ has basis the link states in which $n$ is paired to a site $m<n$. Cut the arc $(m,n)$ and turn $m$ into a defect. Noncrossing forces every defect to lie outside this arc, so this new defect is the rightmost one; the result is a link state with $n-1$ sites and $t+1$ defects. The inverse attaches a new site $n$ to the rightmost defect. These operations are inverse bijections of bases, so they give an $R$-module isomorphism of the quotient's underlying module with $W_{n-1,t+1}$.

Check the quotient map on a basis state whose rightmost site $n$ is paired to $m<n$. A generator $d_i$ with $i\mathrm{\notin }\{m-1,m\}$ does not meet the endpoint $m$; stacking and cutting therefore give the same exterior matching and the same loop count. If $i=m-1$, the site $m-1$ is either a defect or is paired to a site $k<m-1$ (it cannot be paired to a site inside $(m,n)$, since that would cross the arc $(m,n)$). In the paired case, stacking joins the distinct arcs $(k,m-1)$ and $(m,n)$ and replaces them by the bottom arc $(m-1,m)$ and the arc $(k,n)$; cutting either result at the arc ending at $n$ gives the same link state, with no loop removed. In the defect case, stacking makes $n$ a defect, so the result lies in $U$ and is zero in the quotient. Cutting first makes $m-1$ and $m$ two defects; stacking $d_{m-1}$ joins their two top endpoints, leaving fewer than $t+1$ propagating strands, so its class in $W_{n-1,t+1}$ is also zero. If $i=m$, then $m+1$ lies inside the arc $(m,n)$ and must be paired to some $k$ with $m+1<k<n$. Stacking joins the distinct arcs $(m,n)$ and $(m+1,k)$, replacing them by $(m,m+1)$ and $(k,n)$; cutting at the arc ending at $n$ before or after this operation gives the same state, again with no loop removed. These cases exhaust generators touching $m$; hence the quotient map intertwines every $d_i$. By [F3] it intertwines the whole ${\mathrm{T}\mathrm{L}}_{n-1}$-action. The basis descriptions in step 1.1 and step 2.2 give exactness at every term.

For the additional corollary, assume $R=\mathbb{C}$, $q\in {\mathbb{C}}^{\times }$, $\delta =q+q^{-1}$, and $q^{2(t+1)}\mathrm{\ne }1$. Under these hypotheses the short exact sequence above splits as a sequence of right ${\mathrm{T}\mathrm{L}}_{n-1}(\mathbb{C},\delta )$-modules. If $t=0$, then $W_{n-1,t-1}=0$; if $t=n$, then $W_{n-1,t+1}=0$. In either case exactness from step 3.1 makes the nonzero map in the sequence an isomorphism, so the sequence splits. Now suppose $0<t<n$, set $m=n-1$, and consider a nonzero summand $W_{m,s}$ with $s=t-1$ or $s=t+1$.

Choose $v\in \mathbb{C}$ with $v^2=q$ and, in ${\mathrm{T}\mathrm{L}}_{m+1}(\mathbb{C},\delta )$, put $g_i=v1-v^{-1}{d}_i$ for $1\le i\le m$. By [F4], multiplication gives $(v1-v^{-1}{d}_i)(v^{-1}1-v{d}_i)=1+(\delta -q-q^{-1})d_i=1$, and the reversed product is the same; hence $g_i^{-1}=v^{-1}1-v{d}_i$. For $1\le i<m$, expanding the adjacent products with both triple relations gives $g_i{g}_{i+1}{g}_i-g_{i+1}{g}_i{g}_{i+1}=-v^{-1}(q-\delta +q^{-1})(d_i-d_{i+1})=0$. For valid distant indices, [F4] likewise gives $g_i{g}_j=g_j{g}_i$. Define $M_m=g_m\cdots g_2{g}_1^2{g}_2\cdots g_m$. Let $T_m:{\mathrm{T}\mathrm{L}}_{m+1}\to {\mathrm{T}\mathrm{L}}_m$ close the last top and bottom points by an exterior arc and remove each resulting loop with factor $\delta$, and put $F_m=T_m(M_m)$. The product has $2m$ crossing factors, so $F_m$ is independent of the choice of square root $v$.

For $M_0=1$ and $M_j=g_j{M}_{j-1}{g}_j$, the braid and distant-commutation relations imply $g_i{M}_m=M_m{g}_i$ for $i<m$. Indeed, put $P=g_m\cdots g_{i+2}$ and $B_i=g_{i+1}{M}_i{g}_{i+1}=g_{i+1}{g}_i{M}_{i-1}{g}_i{g}_{i+1}$, so $M_m=P{B}_i{P}^{\mathrm{r}\mathrm{e}\mathrm{v}}$, where $P^{\mathrm{r}\mathrm{e}\mathrm{v}}=g_{i+2}\cdots g_m$. The element $g_i$ commutes with $P$ and $P^{\mathrm{r}\mathrm{e}\mathrm{v}}$, while $g_{i+1}$ commutes with $M_{i-1}$. Using $g_i{g}_{i+1}{g}_i=g_{i+1}{g}_i{g}_{i+1}$ on the left and right gives $g_i{B}_i=g_{i+1}{g}_i{M}_{i-1}{g}_{i+1}{g}_i{g}_{i+1}=B_i{g}_i$. Closing the last strand is a ${\mathrm{T}\mathrm{L}}_m$-bimodule map, since stacking on the first $m$ strands is disjoint from that closure. Thus $g_i{F}_m=F_m{g}_i$ for every $i<m$. Since $d_i=q1-v{g}_i$ and the $d_i$ generate ${\mathrm{T}\mathrm{L}}_m$ by [F3], $F_m$ is central.

For link states $x,y$ with $s$ defects, define $\psi (x,y)$ as the coefficient of $1_s$ in $y\circ x^{\ast }$, and set it to zero when this composite has fewer than $s$ propagating components. For link states $z,w$, associativity and the quotient action give $z\cdot (x^{\ast }\circ w)=\psi (x,z)w$: when $z\circ x^{\ast }$ has fewer than $s$ propagating components, the resulting term is zero in $W_{m,s}$. This pairing is nonzero if $s>0$: for no arcs, pair the all-defect state with itself; with $p=(m-s)\mathrm{/}2>0$, pair the states with arcs $(1,2),(3,4),\dots ,(2p-1,2p)$ and defects to their right, and with arcs $(2,3),(4,5),\dots ,(2p,2p+1)$ and a defect at site $1$, respectively. Their overlay has no loop and gives value $1$. If $s=0$, the case needed below has $t=1$ and $q^4\mathrm{\ne }1$, hence $\delta \mathrm{\ne }0$; pairing any perfect matching with itself gives ${\delta }^{m\mathrm{/}2}\mathrm{\ne }0$. Choose $x,y$ with $\psi (x,y)=1$. For every $w$, the element $x^{\ast }\circ w$ then satisfies $y\cdot (x^{\ast }\circ w)=w$. If $\mathrm{\Theta }$ is a module endomorphism, $\mathrm{\Theta }(w)=\mathrm{\Theta }(y)\cdot (x^{\ast }\circ w)=\psi (x,\mathrm{\Theta }(y))w$. Thus every module endomorphism of these summands is scalar.

The central operator given by right multiplication by $F_m$ is therefore scalar on each such $W_{m,s}$. To compute its scalar, use the state $z_{m,s}$ with $p=(m-s)\mathrm{/}2$ adjacent arcs followed by $s$ defects. Expand the defining crossings in $F_m$ and retain the coefficient of $z_{m,s}$ in the link-state quotient. The local identities $d_i{g}_{i+1}{g}_i=-d_i{d}_{i+1}$ and $g_i{g}_{i+1}{d}_i=-d_{i+1}{d}_i$, obtained from [F4] by $d_i{g}_{i+1}{g}_i=q{d}_i-d_i{d}_{i+1}-\delta d_i+q^{-1}{d}_i{d}_{i+1}{d}_i=-d_i{d}_{i+1}$ and $g_i{g}_{i+1}{d}_i=q{d}_i-d_i^2-d_{i+1}{d}_i+q^{-1}{d}_i{d}_{i+1}{d}_i=-d_{i+1}{d}_i$, for $1\le i<m$, remove the $p$ adjacent arcs. For $s\ge 1$, follow one orientation of the remaining closed strand from its outer end. At each of the first $s-1$ defect positions, only the through resolution on both arms survives: every alternative adds an arc and has fewer than $s$ propagating strands, so is zero in $W_{m,s}$. The initial half-crossing and these forced pairs contribute $q^{s-1\mathrm{/}2}$. At the terminal pair, the turn resolution closes one loop and contributes $q^{1\mathrm{/}2}\delta$, while the through resolution contributes $-q^{-1\mathrm{/}2}$. Thus this orientation contributes $q^{s-1\mathrm{/}2}(q^{1\mathrm{/}2}\delta -q^{-1\mathrm{/}2})=q^s\delta -q^{s-1}=q^{s+1}$. When $s=0$, there are no intervening defect positions; the two surviving terminal resolutions contribute $\delta$ and $-q^{-1}$, giving $\delta -q^{-1}=q$. For the second branch, first choose the cap-cup term $-v^{-1}{d}_i$ rather than the through term $v1$ at the outer crossing of $g_i=v1-v^{-1}{d}_i$. For $s\ge 1$, propagate this branch along the defect strings using the same cap-slide identities above: at each of the first $s-1$ defect positions, the surviving pair of smoothing coefficients is $(-v^{-1})(-v^{-1})=q^{-1}$; each alternative connects two defects and is zero in the link-state quotient. At the last crossing, the through smoothing contributes $v$ and the smoothing closing the remaining loop contributes $-v^{-1}\delta$. Including the initial coefficient, this branch therefore contributes $(-v^{-1})q^{-(s-1)}(v-v^{-1}\delta )=q^{-s}\delta -q^{-s+1}=q^{-s-1}$. For $s=0$, no defect string intervenes: the two terminal resolutions give $\delta -q=q^{-1}$. The surviving matchings in this second branch are reflected versions of the first branch's matchings, but the coefficients just computed come from its own smoothing choices; reflection is linear and does not invert the scalar $q$. Every discarded resolution adds an arc and is zero in the quotient. Hence $F_m$ acts on $W_{m,s}$ as $q^{s+1}+q^{-s-1}$.

For the submodule $W_{n-1,t-1}$, step 7.1 gives eigenvalue ${\lambda }_{-}=q^t+q^{-t}$; for the quotient $W_{n-1,t+1}$ it gives ${\lambda }_{+}=q^{t+2}+q^{-t-2}$. Both pairings used in step 6.2 are nonzero: the $t+1$ label has positive defects, and the $t-1$ label has positive defects unless $t=1$, in which case $\delta \mathrm{\ne }0$ follows from $q^4\mathrm{\ne }1$. Their difference is ${\lambda }_{+}-{\lambda }_{-}=(q-q^{-1})(q^{t+1}-q^{-t-1})$, which is nonzero by [A1].

Let $S$ be right multiplication by $F_{n-1}$ on the restricted middle module. It is a module endomorphism because $F_{n-1}$ is central. The polynomial operator $P=(S-{\lambda }_{+}I)\mathrm{/}({\lambda }_{-}-{\lambda }_{+})$ is the identity on the submodule and induces zero on the quotient. Therefore its image lies in the submodule and its restriction there is the identity, so $P$ is a retraction. The middle module is the direct sum of the submodule and $\ker P$, and projection to the quotient restricts to an isomorphism on $\ker P$. This proves the asserted splitting.

Every diagram $D:n\to n$ has a definite number $t$ of propagating strands, with $0\le t\le n$ and $n-t$ even. Cut each propagating strand at a horizontal midpoint. The upper half, reflected, is a link state $x:n\to t$; the lower half is a link state $y:n\to t$. Conversely, $x^{\ast }\circ y$ joins the $t$ corresponding defect endpoints in order and yields a diagram with exactly $t$ propagating strands. The cuts are determined by $D$, so these constructions are inverse. Thus the diagrams $C_{x,y}^t$ are in bijection with the basis diagrams of ${\mathrm{T}\mathrm{L}}_n$ and hence form an $R$-basis.

Define $\ast$ on every basis diagram by horizontal reflection and extend $R$-linearly. Reflecting a stack reverses its two factors and preserves each closed component, so $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$; reflecting twice restores the original pairing and $1_n^{\ast }=1_n$. Also $(C_{x,y}^t{)}^{\ast }=C_{y,x}^t$. Thus $\ast$ is the involutive anti-automorphism required by [F3].

For $t\in {\mathrm{\Lambda }}_n$, let $J_t$ be the span of diagrams with fewer than $t$ propagating strands. The stated order makes these exactly the basis elements indexed by labels strictly greater than ${\lambda }_t$, so $J_t=A^{⊳{\lambda }_t}$. Every propagating path in a stacked product must pass through a distinct propagating path in each factor, so a product has no more propagating strands than either factor; therefore $J_t$ is a two-sided ideal. For $a\in {\mathrm{T}\mathrm{L}}_n$, write the quotient of $y\circ a$ in $M_R(n,t)\mathrm{/}{M}_R^{<t}(n,t)$ as ${\sum }_v{r}_{v}v$ in its link-state basis. Then $C_{x,y}^{t}a=x^{\ast }\circ (y\circ a)\equiv {\sum }_v{r}_v{C}_{x,v}^t$ modulo $J_t$. The coefficients $r_v$ are those of $y\cdot a$ and are independent of $x$, so this is exactly the fixed-first-index cellular multiplication rule in [F3].

For link states $y,u$, the middle composite $y\circ u^{\ast }$ is a diagram $t\to t$. If it has all $t$ propagating strands, planarity makes it ${\delta }^c{1}_t$, and the full product is $C_{x,y}^t{C}_{u,v}^t=x^{\ast }\circ (y\circ u^{\ast })\circ v={\varphi }_{n,t}(u,y)C_{x,v}^t={\varphi }_{n,t}(y,u)C_{x,v}^t$, using symmetry of the form. If $y\circ u^{\ast }$ has fewer than $t$ propagating strands, the full product also has fewer than $t$ and lies in $J_t$, so the same congruence holds with coefficient zero. Hence ${\varphi }_{n,t}$ is the cellular cell form. Finally, the generic right cell-module action read from $C_{x,x}^{t}a$ has the same coefficients $r_v$ as $y\cdot a$ in step 2.2, so its basis and action identify with $W_{n,t}$.

By [F1], ${\mathrm{T}\mathrm{L}}_n(F,\delta )$ is a finite-dimensional unital cellular algebra with cell modules $W_{n,t}$ and cell forms ${\varphi }_{n,t}$. It is finite-dimensional because its diagram basis is a subset of the finite set of matchings of $2n$ boundary points.

Apply [F2] to this cellular datum. It gives absolute simplicity and pairwise distinction of every nonzero $D_{n,t}$, and says these quotients exhaust the simple right modules. It also gives $D_{n,t}\mathrm{\ne }0$ exactly when ${\varphi }_{n,t}$ is not identically zero and identifies $\mathrm{rad}({\varphi }_{n,t})$ as the unique maximal proper submodule in that case.

If ${\varphi }_{n,t}$ is identically zero, its radical is all of $W_{n,t}$, so $D_{n,t}=0$ by definition. The statement in [F2] makes no assertion about the actual head of $W_{n,t}$ in this case.