The positive-root Jones quotient is the type-A path tower with positive right closure

The quantum-integer recurrence with real $\delta$ makes each $[r]$ real, and [F5] gives $[r]>0$ for $1\le r<\mathrm{\ell }$. Starting with $p_0=p_1=1$, induction in the JW recursion shows $p_r^{\ast }=p_r$ through $r=\mathrm{\ell }-1$: reflection fixes $P_r=p_r\otimes 1$ and $U_r$, while the coefficient $[r]\mathrm{/}[r+1]$ is real whenever this recursion is used. In particular the boundary projector $p_{\mathrm{\ell }-1}$ is self-adjoint. For $n<\mathrm{\ell }-1$, the generated ideal component $I_n$ is zero and is reflection-stable; for $n\ge \mathrm{\ell }-1$, its fixed embedded generator $p_{\mathrm{\ell }-1}\otimes 1$ is self-adjoint, so $I_n={\mathrm{T}\mathrm{L}}_n(p_{\mathrm{\ell }-1}\otimes 1){\mathrm{T}\mathrm{L}}_n$ is reflection-stable because $(x(p_{\mathrm{\ell }-1}\otimes 1)y{)}^{\ast }=y^{\ast }(p_{\mathrm{\ell }-1}\otimes 1)x^{\ast }$. By [F3], these are the endomorphism components of the generated negligible tensor ideal. Also [F7] gives ${\tau }_n(D^{\ast })=\overline{{\tau }_n(D)}$ on diagrams and hence on all elements. If $x\in J_n$, then for every $y$ cyclicity yields ${\tau }_n(y{x}^{\ast })=\overline{{\tau }_n(x{y}^{\ast })}=\overline{{\tau }_n(y^{\ast }x)}=0$; thus the trace radical is reflection-stable as well. Since [F3] identifies it with $I_n$, the quotient involution is well defined.

Fix a retained endpoint $1\le a\le k$, write $f_a=p_a\otimes 1$, and use the supplier's maps ${\alpha }_a=f_a(1_{a-1}\otimes \mathrm{c}\mathrm{u}\mathrm{p})p_{a-1}$ and ${\beta }_a=p_{a-1}(1_{a-1}\otimes \mathrm{c}\mathrm{a}\mathrm{p})f_a$. Horizontal reflection reverses composition and exchanges the cup and cap; step 1.1 fixes $f_a$ and $p_{a-1}$, so ${\alpha }_a^{\ast }={\beta }_a$. The supplier identity ${\beta }_a{\alpha }_a={\lambda }_a{p}_{a-1}$ therefore gives ${\alpha }_a^{\ast }{\alpha }_a={\lambda }_a{p}_{a-1}$. Set $v_a={\lambda }_a^{-1\mathrm{/}2}{\alpha }_a$. Then $v_a^{\ast }{v}_a=p_{a-1}$ and $v_a{v}_a^{\ast }={\lambda }_a^{-1}{\alpha }_a{\beta }_a$, the down projection in the JW split. For $a<k$, the complementary up projection is $p_{a+1}$; it is self-adjoint by step 1.1 and orthogonal to the down projection by the JW turnback relations in [F2]. At $a=k$, the quotient kills $p_{k+1}=p_{\mathrm{\ell }-1}$ before normalization, so only the down projection remains. The up inclusion and retraction are adjoints, as are $v_a$ and $v_a^{\ast }$. Composing these isometric branch maps along retained paths gives path maps $I_p$ with mutually orthogonal ranges and ${\sum }_p{I}_p{I}_p^{\ast }=1_n$. Hence $F_{p,q}=I_p{I}_q^{\ast }$ are complete matrix units and $F_{p,q}^{\ast }=F_{q,p}$. Their diagonal units equal the supplier's path projections (the reciprocal branch normalizations cancel), so ${\mathrm{Tr}}_n(F_{p,q})={\delta }_{p,q}[a+1]$ by [F2]. Thus the quotient with its inherited reflection is a direct sum of full complex matrix algebras with the usual adjoint, hence a finite-dimensional $C^{\ast }$-algebra. Since ${\tau }_n(F_{p,p})={\delta }^{-n}[a+1]>0$, its normalized closure trace is faithful positive.

Tensoring an endpoint path map with a strand and resolving its endpoint identity by the same isometric up/down splitting gives $F_{p,q}\otimes 1={\sum }_c{F}_{pc,qc}$. The coefficients are exactly one because the inclusion and retraction at each common endpoint use reciprocal normalizations. At $a=k$ delete $p_{k+1}$ before splitting; only its down term remains, as in [F2]. Thus the actual quotient inclusions are exactly the prefix maps of [F4]; no compatibility is imposed as an additional hypothesis. Transporting the endpoint action along these matrix units and applying [F4] gives the asserted multiplicity-one branching, omitting zero path spaces and invalid boundary labels.

[F1] defines ${\bar{\iota }}_n([x])=[{\iota }_n(x)]$ and ${\bar{E}}_n([z])=[E_n(z)]$ independently of representatives. Bimodularity, trace preservation and ${\bar{E}}_n{\bar{\iota }}_n=\mathrm{i}\mathrm{d}$ pass to these quotients, so inclusion is injective. Its compatibility with reflection follows from real $\delta$. Closure also commutes with reflection: the two finite matching graphs obtained by reflecting before or after closing have the same components and the real factor ${\delta }^{-1}$. Thus ${\bar{E}}_n$ is star-preserving. For $z\ge 0$ and $b\in Q_n$, bimodularity gives ${\tau }_n(b^{\ast }{\bar{E}}_n(z)b)={\tau }_{n+1}({\bar{\iota }}_n(b{)}^{\ast }z{\bar{\iota }}_n(b))\ge 0$. If the self-adjoint ${\bar{E}}_n(z)$ had a negative spectral subspace in one of the finite matrix blocks from step 2.1, its nonzero spectral projection would contradict this inequality and faithfulness. Hence ${\bar{E}}_n$ is positive, unital, idempotent, bimodular and trace-preserving, the claimed conditional expectation.

Bimodularity with the endpoint central projections shows that ${\bar{E}}_n(F_{pc,qc})=0$ if the predecessor endpoints $a,b$ differ. If they agree, bimodularity with the individual predecessor matrix units shows that its value is a single scalar $t_{a,c}{F}_{p,q}$ independent of $p,q$: multiplying on the left and right by predecessor diagonal units isolates that matrix entry, and multiplying by predecessor off-diagonal units identifies their coefficients. Trace preservation on a diagonal unit now gives $t_{a,c}{\delta }^{-n}[a+1]={\delta }^{-(n+1)}[c+1]$, so $t_{a,c}=[c+1]\mathrm{/}(\delta [a+1])$. The coefficients are positive and sum to one over valid neighbours by [F5]; at the upper boundary the omitted term is $[\mathrm{\ell }]=0$, and at $a=0$ the nonexistent down term is $[0]=0$. For $\mathrm{\ell }=3$ every vertex has one retained neighbour and this coefficient is one. Rank zero and rank one have their unique scalar path algebra and the unital inclusion between them. This proves every assertion at the stated finite ranks.