Hecke–bracket trace dictionary and local descent

Use source coordinates $(u,v)$ with their standard positive orientation and positive projection coordinates $(t,x)$ on the braid box. For $h>0$, the map $(t,x)=(v,-2hu)$ has determinant $2h>0$. It sends the source over direction $(1,1)$ to $(1,-2h)$ and the source under direction $(-1,1)$ to $(1,2h)$, whose determinant is $4h>0$. The source left/right pairs have constant $u$, hence constant braid position $x$ and give the through pairing $I$; the source top/bottom pairs have constant $v$, hence constant braid time $t$ and give the cup-cap $e_i$. The displayed page coordinates $(x,t)$ reverse the positive projection orientation; their visual quarter-turn is not the orientation comparison. Thus its two smoothings are $I$ with weight $A$ and $e_i$ with weight $A^{-1}$. For the opposite crossing the pairings exchange. Hence the local operators are $C_i=AI+A^{-1}{e}_i,C_i^{-}=A^{-1}I+A{e}_i\mathrm{.}$ The relations [F5] and [F6] show these operators satisfy the Artin relations and are mutual inverses; distant generators commute by [F5].

Define ${\rho }_n({\sigma }_i)=C_i$ and $S_i:=-A^{-1}{C}_i=-I-q{e}_i$. With $t=q^2$ and $\delta =-(q+q^{-1})$, direct multiplication gives $S_i^2=I+(2q+q^2\delta )e_i=I+(q-q^3)e_i=(t-1)S_i+t\mathrm{.}$ The braid and distant-commutation relations for $S_i$ follow from [F5] and step 1.1. Under the coefficient map $\psi :\mathrm{\Lambda }\to R_A$ given by $v\mapsto t$ and $z\mapsto z_0$, the assignment $T_i\mapsto S_i$ therefore defines a compatible homomorphism ${\rho }_H$ from the specialized Hecke tower to the Temperley--Lieb diagram algebras. For a braid word, this map sends ${\pi }_n(\beta )$ to the product of the corresponding $S_i^{\pm 1}$.

The element $A^{-4}+1$ is a unit in $R_A$ and $\delta =-A^2(A^{-4}+1)$ is therefore a unit. Set ${\tau }_n={\delta }^{-n}{\mathrm{Tr}}_n$. By [F7], ${\tau }_n(1)=1$, ${\tau }_{n+1}(x\otimes 1)={\tau }_n(x)$, and ${\tau }_n$ is cyclic. Since $S_n=-I-q{e}_n$, the last closure rule of [F7] gives ${\tau }_{n+1}((x\otimes 1)S_n)=-{\tau }_n(x)-q{\delta }^{-1}{\tau }_n(x)=-\frac{1}{q^2+1}{\tau }_n(x)=z_0{\tau }_n(x)\mathrm{.}$ Thus $\tau \circ {\rho }_H$ satisfies all specialized Markov-trace rules, where ${\rho }_H(T_i)=S_i$.

The decomposition in [F10] is a direct sum of modules spanned by disjoint subsets of the Hecke standard basis. Under $\psi$, that basis remains a basis, so for $H_{\psi }(n):=H(n){\otimes }_{\mathrm{\Lambda },\psi }{R}_A$ the specialized tower still has $H_{\psi }(n+1)=H_{\psi }(n)\oplus H_{\psi }(n)T_n{H}_{\psi }(n)$. Any specialized Markov trace is therefore determined by its restriction to $H_{\psi }(n)$: restriction determines the first summand, while cyclicity and the Markov rule force ${\mathrm{tr}}_{n+1}(x{T}_{n}y)=z_0{\mathrm{tr}}_n(yx)$ on every product in the second summand. Starting at $H(1)$, induction proves uniqueness, without dividing by any specialized parameter. The scalar extension of the Ocneanu trace exists by [F8], so the trace in step 3.1 equals that specialized trace.

Expand each crossing of $\beta$ by $C_i$ or $C_i^{-}$ and close the resulting Temperley–Lieb diagrams. The unnormalized trace counts each closed state circle with a factor $\delta$, whereas the bracket state sum counts it with ${\delta }^{\mathrm{\mid }S\mathrm{\mid }-1}$; hence $⟨\widehat{\beta }⟩={\delta }^{-1}{\mathrm{Tr}}_n({\rho }_n(\beta ))={\delta }^{n-1}{\tau }_n({\rho }_n(\beta ))\mathrm{.}$ Since $C_i=-A{S}_i$ and $C_i^{-}=(-A{)}^{-1}{S}_i^{-1}$, multiplying along an Artin word gives ${\rho }_n(\beta )=(-A{)}^{e(\beta )}{\rho }_H({\pi }_n(\beta ))$. Step 4.1 now gives the displayed trace identity.

The closure of the identity in $B_1$ has one circle; the state sum gives ${\delta }^{1-1}=1$, and its trace expression gives ${\delta }^0(-A{)}^0{\mathrm{tr}}_1(1)=1$. For positive stabilization, the Markov trace factor is $z_0=-1\mathrm{/}(q^2+1)$, so the bracket formula changes by $(-A)\delta z_0=-A^3\mathrm{.}$ For negative stabilization, [F9] and [F8] give trace factor $z_{-}:=t^{-1}(z_0+1-t)=-\frac{q^2}{q^2+1};$ the bracket formula changes by $\delta (-A{)}^{-1}{z}_{-}=-A^{-3}\mathrm{.}$ This proves the unknot value and both stabilization normalizations.