Restriction of a link-state module has a two-step filtration

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.