Ordinary finite-index II1 inclusions give the full Jones tower and Temperley–Lieb relations

On dense vectors $x\in M$, trace preservation gives $⟨n,E_N(x){⟩}_{L^2}=⟨n,x{⟩}_{L^2}$ for $n\in N$, so $e_N$ is the Hilbert projection and $e_{N}x{e}_N=E_N(x)e_N$. By [F2], every right-module endomorphism has the finite expression $T={\sum }_{j}T(u_j)e_N{u}_j^{\ast }$, since in the finite Hilbert-space model $T$ is a matrix with entries in $N$, so $T(u_j)\in M$; its adjointable module bound also makes it bounded on $L^2(M)$. Hence the first algebra is precisely the finite corner $M_1=B\cong p{M}_k(N)p$.

The corner trace is ${\tau }_1={\lambda }^{-1}({\mathrm{Tr}}_k\otimes {\tau }_N)$, faithful and normal, and in the coordinate frame its unnormalized value is ${\sum }_j{\tau }_N(E_N(u_j^{\ast }T(u_j)))$. For $T=x{e}_{N}y$, the adjoint of the frame identity for $y^{\ast }$ gives ${\sum }_j{E}_N(y{u}_j)u_j^{\ast }=y$, whence $\lambda {\tau }_1(x{e}_{N}y)={\tau }_M(yx)$. By uniqueness [F5], ${\tau }_1{\mathrm{\mid }}_M={\tau }_M$. The expectation $F$ of [F3] is therefore for these normalized traces. Applying $F$ to $1={\sum }_j{u}_j{e}_N{u}_j^{\ast }$ gives ${\sum }_j{u}_j{u}_j^{\ast }=\lambda 1$.

Put $v_j=\sqrt{\lambda }{u}_j{e}_N\in M_1$. For $T=x{e}_{N}y$, [F3] gives ${\sum }_j{v}_{j}F(v_j^{\ast }T)={\sum }_j{u}_j{e}_N{E}_N(u_j^{\ast }x)y=x{e}_{N}y$. Every element of $M_1$ is a finite sum of such terms by step 1.1, so this is a Parseval frame on all of $M_1$. Its projection matrix $q_{ij}=F(v_i^{\ast }{v}_j)$ gives an onto $M$-valued inner-product isometry $T\mapsto (F(v_j^{\ast }T){)}_j$ from $M_1$ to $q{M}^k$, with inverse $(a_j)\mapsto {\sum }_j{v}_j{a}_j$. Its Hilbert-space completion is $q{L}^2(M{)}^k$. Moreover ${\sum }_j{\tau }_M(F(v_j^{\ast }{v}_j))=\lambda {\sum }_j{\tau }_1(e_N{u}_j^{\ast }{u}_j{e}_N)={\sum }_j{\tau }_M(u_j^{\ast }{u}_j)=\lambda$. Thus the ordinary index of $M\subseteq M_1$ is $\lambda$, and [F3]–[F5] apply again.

Suppose the initial inclusion is extremal. On $H=L^2(M)$, $M^{\mathrm{\prime }}\cap M_1=R(M)\cap R(N{)}^{\mathrm{\prime }}=\{R_a:a\in N^{\mathrm{\prime }}\cap M\}$ by [F4]. For such $a$, $E_N(a)\in Z(N)$ by bimodularity, so trace preservation gives $E_N(a)={\tau }_M(a)1$; hence for $c=R_a$ one has $e_{N}c{e}_N={\tau }_M(a)e_N$. Let $M_2={\mathrm{End}}_M{L}^2(M_1)$ with the trace from the frame in step 3.1. The right-action module trace is ${\tau }_2(R_c)={\lambda }^{-1}{\sum }_j{\tau }_1(v_j^{\ast }{v}_{j}c)={\sum }_j{\tau }_1(e_N{u}_j^{\ast }{u}_j{e}_{N}c)={\tau }_M(a){\sum }_j{\tau }_1(e_N{u}_j^{\ast }{u}_j{e}_N)={\tau }_M(a)\mathrm{.}$ In the middle equality one may insert a final $e_N$: if $A=e_{N}A{e}_N$, traciality gives ${\tau }_1(Ac)={\tau }_1(Ac{e}_N)$. The last sum is ${\lambda }^{-1}{\sum }_j{\tau }_M(u_j^{\ast }{u}_j)=1$. Initial extremality states ${\tau }_1(c)={\tau }_M(a)$, so ${\tau }_2(R_c)={\tau }_1(c)$ for every $c\in M^{\mathrm{\prime }}\cap M_1$, which is exactly extremality of $M\subseteq M_1$.

Iterate steps 1.1–3.1. Each new finite corner is a II1 factor by [F5], its trace restricts to the previous factor trace by uniqueness, and step 3.1 supplies a bounded finite frame of the same ordinary index $\lambda$. If extremality holds initially, step 4.1 propagates it. This constructs every finite stage of the asserted tower; no common finite frame length for all stages is required.

To verify both adjacent relations independently, work first on $L^2(M_1)$ and put $e=e_N$, $f=e_2$. The compression identity is $fTf=F(T)f$ for $T\in M_1$, so $fef=F(e)f={\lambda }^{-1}f$. For the reverse order evaluate on the dense vectors $T=xey\in M_1$: $efe(T)=eF(exey)={\lambda }^{-1}e{E}_N(x)y={\lambda }^{-1}{E}_N(x)ey={\lambda }^{-1}e(T)$. These vectors span $M_1$ by step 1.1, and all operators are bounded; hence $efe={\lambda }^{-1}e$. Apply the same two calculations at each adjacent inclusion from step 5.1. Every $e_i$ is a self-adjoint projection by definition.

For $j\ge i+2$, the projection $e_j$ commutes with $M_{j-2}$ by its bimodule projection formula, and $e_i\in M_i\subseteq M_{j-2}$; thus $e_i{e}_j=e_j{e}_i$. Setting $U_i=\delta e_i$, the square and both adjacent identities become the relations in [F6], since ${\delta }^2=\lambda$. Reflection is the $\ast$-operation on the real-parameter diagram algebra, so its presentation yields compatible unital $\ast$-representations ${\mathrm{T}\mathrm{L}}_n(\mathbb{C},\delta )\to M_{n-1}$.

The traces normalize the unit and restrict compatibly. For $x$ in the earlier algebra the Markov formula of step 2.1 at stage $i$ gives ${\tau }_i(x{U}_i)=\delta {\lambda }^{-1}{\tau }_{i-1}(x)={\delta }^{-1}{\tau }_{i-1}(x)$. To compare with normalized diagram closure, use [F7]: each element at rank $n+1$ is a sum of an earlier-rank element and terms $x{U}_{n}y$. Restriction fixes the first term, while cyclicity forces the value on $x{U}_{n}y$ to be ${\delta }^{-1}{\tau }_n(yx)$. Induction from rank zero therefore determines at most one compatible normalized Markov trace. Closure satisfies these same rules by [F7], so it is the pulled-back trace, which is positive because it comes from a positive trace on a II1 factor. Finally $\lambda \ge 1$ follows from ${\tau }_1(e_1)={\lambda }^{-1}\le 1$. If $\lambda =1$, faithfulness gives $e_1=1$, so $M=N$ and the subsequent inclusions and projections are identities; all the same formulas hold. No zero or empty frame occurs, since its rank operators sum to the identity. This proves all claims.