The Ocneanu Markov trace exists and is unique

Uniqueness. Suppose $({\mathrm{tr}}_n)$ is a Markov trace. The base is $H(1)=\mathrm{\Lambda }$, where ${\mathrm{tr}}_1(\lambda )=\lambda$ by (M1) and $\mathrm{\Lambda }$-linearity. For $n\ge 2$, [F2] at level $n-1$ gives each $y\in H(n)$ the unique expansion $y={\sum }_{i=0}^{n-1}{x}_i{T}_{w^{(i)}}$ with $x_i\in H(n-1)$ and $T_{w^{(i)}}=T_{n-1}\cdots T_{n-i}$. By (M2), ${\mathrm{tr}}_n(x_0)={\mathrm{tr}}_{n-1}(x_0)$; by (M3) and the two-sided form of (M4), ${\mathrm{tr}}_n(x_i{T}_{w^{(i)}})={\mathrm{tr}}_n(T_{w^{(i)}}{x}_i)=z{\mathrm{tr}}_{n-1}(T_{n-2}\cdots T_{n-i}{x}_i)$ for $i\ge 1$, an element of $H(n-1)$ on which ${\mathrm{tr}}_{n-1}$ is already defined. Hence ${\mathrm{tr}}_n$ is determined by ${\mathrm{tr}}_{n-1}$; induction gives uniqueness and (c).

Recursive construction. Define ${\mathrm{tr}}_1:\mathrm{\Lambda }\to \mathrm{\Lambda }$ by ${\mathrm{tr}}_1(a)=a$. Suppose ${\mathrm{tr}}_n$ is defined. The bimodule isomorphism in [F2] is induced by ${\mu }_n(x\otimes y)=x{T}_{n}y$ and gives $H(n+1)=H(n)\oplus \mathrm{im}({\mu }_n)$. The $\mathrm{\Lambda }$-linear map ${\tau }_n:H(n){\otimes }_{H(n-1)}H(n)\to \mathrm{\Lambda }$, $x\otimes y\mapsto {\mathrm{tr}}_n(xy)$, is balanced: $(xh)\otimes y=x\otimes (hy)$ for $h\in H(n-1)$, and both tensors map to ${\mathrm{tr}}_n(xhy)$. Define ${\mathrm{tr}}_{n+1}(a+{\mu }_n(\xi )):={\mathrm{tr}}_n(a)+z{\tau }_n(\xi ),a\in H(n),\xi \in H(n){\otimes }_{H(n-1)}H(n)\mathrm{.}$ The direct-sum decomposition and the isomorphism ${\mu }_n$ make this definition well defined and $\mathrm{\Lambda }$-linear. Restriction to $H(n)$ gives (M2), and ${\mathrm{tr}}_{n+1}(1)={\mathrm{tr}}_n(1)$ gives (M1). For each $i<n$, repeated restriction gives ${\mathrm{tr}}_n(T_i)={\mathrm{tr}}_{i+1}(T_i)=z$ by the recursion at level $i$, proving (a). By construction, ${\mathrm{tr}}_{n+1}(x{T}_{n}y)=z{\mathrm{tr}}_n(xy)(x,y\in H(n));$ this is the two-sided recursion, and $y=1$ gives (M4). Iterating it along the left basis of [F2] gives the recursive formula in (c).

Cyclicity: reduction. We prove cyclicity by induction. The base $H(1)=\mathrm{\Lambda }$ is commutative, and $H(2)$ is generated over $\mathrm{\Lambda }$ by the single element $T_1$, so its trace is cyclic. For $n\ge 2$, assume ${\mathrm{tr}}_n$ is cyclic and consider $H(n+1)=H(n)\oplus I_n$, where $I_n$ is the $H(n)$-sub-bimodule spanned by $x{T}_{n}y$, as in [F2]. If $a,b\in H(n)$, cyclicity is the induction hypothesis. If $a\in H(n)$ and $b=x{T}_{n}y\in I_n$, then the construction in step 1.2 gives ${\mathrm{tr}}_{n+1}(ab)=z{\mathrm{tr}}_n(axy)$ and ${\mathrm{tr}}_{n+1}(ba)=z{\mathrm{tr}}_n(xya)$, equal by induction; linearity handles sums in $I_n$. Thus it remains the case $a=x{T}_{n}y$, $b=u{T}_{n}v$ with $x,y,u,v\in H(n)$. Applying the already proved one-in-$H(n)$ case to the outer factors reduces ${\mathrm{tr}}_{n+1}(ab)={\mathrm{tr}}_{n+1}(T_{n}X{T}_{n}Y)$ and ${\mathrm{tr}}_{n+1}(ba)={\mathrm{tr}}_{n+1}(T_{n}Y{T}_{n}X)$, where $X=yu$ and $Y=vx$. By that same case, this is equivalent to ${\mathrm{tr}}_{n+1}(T_{n}X{T}_{n}Y)={\mathrm{tr}}_{n+1}(X{T}_{n}Y{T}_n)(X,Y\in H(n))\mathrm{.}$ It remains to prove this identity.

The final cases. Use [F2] at level $n-1$ to write $H(n)=H(n-1)\oplus H(n-1)T_{n-1}H(n-1)$; the balance here is over $H(n-2)$, since $T_{n-1}$ commutes with $H(n-2)$. If $X,Y\in H(n-1)$, then $T_n$ commutes with both and the desired identity follows from the quadratic relation for $T_n$. For $X=x^{\mathrm{\prime }}{T}_{n-1}{x}^{\mathrm{\prime }\mathrm{\prime }}$ with $x^{\mathrm{\prime }},x^{\mathrm{\prime }\mathrm{\prime }}\in H(n-1)$ and $Y\in H(n-1)$, commuting $T_n$ past $x^{\mathrm{\prime }},x^{\mathrm{\prime }\mathrm{\prime }},Y$ and applying the braid relation gives ${\mathrm{tr}}_{n+1}(T_{n}X{T}_{n}Y)=z{\mathrm{tr}}_n(x^{\mathrm{\prime }}{T}_{n-1}^2{x}^{\mathrm{\prime }\mathrm{\prime }}Y)\mathrm{.}$ On the other side, commute $T_n$ past $x^{\mathrm{\prime }\mathrm{\prime }}$ and $Y$, expand $T_n^2$, and use the two-sided recursion from step 1.2 to obtain ${\mathrm{tr}}_{n+1}(X{T}_{n}Y{T}_n)=(v-1)z{\mathrm{tr}}_n(x^{\mathrm{\prime }}{T}_{n-1}{x}^{\mathrm{\prime }\mathrm{\prime }}Y)+v{\mathrm{tr}}_n(x^{\mathrm{\prime }}{T}_{n-1}{x}^{\mathrm{\prime }\mathrm{\prime }}Y)\mathrm{.}$ The level-$n$ recursion gives ${\mathrm{tr}}_n(x^{\mathrm{\prime }}{T}_{n-1}{x}^{\mathrm{\prime }\mathrm{\prime }}Y)=z{\mathrm{tr}}_{n-1}(x^{\mathrm{\prime }}{x}^{\mathrm{\prime }\mathrm{\prime }}Y)$, while restriction gives ${\mathrm{tr}}_n(x^{\mathrm{\prime }}{x}^{\mathrm{\prime }\mathrm{\prime }}Y)={\mathrm{tr}}_{n-1}(x^{\mathrm{\prime }}{x}^{\mathrm{\prime }\mathrm{\prime }}Y)$. Expanding $T_{n-1}^2$ in the first display therefore yields the same expression as the right side. If $X\in H(n-1)$ and $Y\in H(n-1)T_{n-1}H(n-1)$, write $Y=y^{\mathrm{\prime }}{T}_{n-1}{y}^{\mathrm{\prime }\mathrm{\prime }}$ and put $a:=X{y}^{\mathrm{\prime }}\in H(n-1)$. Since $T_n$ commutes with $X,y^{\mathrm{\prime }},y^{\mathrm{\prime }\mathrm{\prime }}$, the quadratic relation and two-sided recursion give $\begin{array}{rl}{\displaystyle {\mathrm{tr}}_{n+1}(T_{n}X{T}_{n}Y)}& {\displaystyle =(v-1)z{\mathrm{tr}}_n(a{T}_{n-1}{y}^{\mathrm{\prime }\mathrm{\prime }})+v{\mathrm{tr}}_n(a{T}_{n-1}{y}^{\mathrm{\prime }\mathrm{\prime }}),}\\ {\displaystyle {\mathrm{tr}}_{n+1}(X{T}_{n}Y{T}_n)}& {\displaystyle =z(v-1){\mathrm{tr}}_n(a{T}_{n-1}{y}^{\mathrm{\prime }\mathrm{\prime }})+zv{\mathrm{tr}}_n(a{y}^{\mathrm{\prime }\mathrm{\prime }})\mathrm{.}}\end{array}$ By (M2) and the two-sided recursion at level $n$, ${\mathrm{tr}}_n(a{y}^{\mathrm{\prime }\mathrm{\prime }})={\mathrm{tr}}_{n-1}(a{y}^{\mathrm{\prime }\mathrm{\prime }})$ and ${\mathrm{tr}}_n(a{T}_{n-1}{y}^{\mathrm{\prime }\mathrm{\prime }})=z{\mathrm{tr}}_{n-1}(a{y}^{\mathrm{\prime }\mathrm{\prime }})$; hence these expressions agree. Finally let $X=x^{\mathrm{\prime }}{T}_{n-1}{x}^{\mathrm{\prime }\mathrm{\prime }}$ and $Y=y^{\mathrm{\prime }}{T}_{n-1}{y}^{\mathrm{\prime }\mathrm{\prime }}$ with all four coefficients in $H(n-1)$. Braid, commutation, and the two-sided recursion give $\begin{array}{rl}{\displaystyle {\mathrm{tr}}_{n+1}(T_{n}X{T}_{n}Y)}& {\displaystyle =z{\mathrm{tr}}_n(x^{\mathrm{\prime }}{T}_{n-1}^2{x}^{\mathrm{\prime }\mathrm{\prime }}{y}^{\mathrm{\prime }}{T}_{n-1}{y}^{\mathrm{\prime }\mathrm{\prime }}),}\\ {\displaystyle {\mathrm{tr}}_{n+1}(X{T}_{n}Y{T}_n)}& {\displaystyle =z{\mathrm{tr}}_n(x^{\mathrm{\prime }}{T}_{n-1}{x}^{\mathrm{\prime }\mathrm{\prime }}{y}^{\mathrm{\prime }}{T}_{n-1}^2{y}^{\mathrm{\prime }\mathrm{\prime }})\mathrm{.}}\end{array}$ After expanding the squared generators, the terms with coefficient $z(v-1)$ agree. The remaining terms agree because the level-$n$ recursion and the induction hypotheses that ${\mathrm{tr}}_n$ and ${\mathrm{tr}}_{n-1}$ are cyclic give ${\mathrm{tr}}_n(x^{\mathrm{\prime }}{x}^{\mathrm{\prime }\mathrm{\prime }}{y}^{\mathrm{\prime }}{T}_{n-1}{y}^{\mathrm{\prime }\mathrm{\prime }})=z{\mathrm{tr}}_{n-1}(x^{\mathrm{\prime }}{x}^{\mathrm{\prime }\mathrm{\prime }}{y}^{\mathrm{\prime }}{y}^{\mathrm{\prime }\mathrm{\prime }}),{\mathrm{tr}}_n(x^{\mathrm{\prime }}{T}_{n-1}{x}^{\mathrm{\prime }\mathrm{\prime }}{y}^{\mathrm{\prime }}{y}^{\mathrm{\prime }\mathrm{\prime }})=z{\mathrm{tr}}_{n-1}(x^{\mathrm{\prime }}{x}^{\mathrm{\prime }\mathrm{\prime }}{y}^{\mathrm{\prime }}{y}^{\mathrm{\prime }\mathrm{\prime }})\mathrm{.}$ Thus the central identity holds in every case, (M3) follows, and the induction is complete.