Finite Jones index gives bounded vector pull-down

Tracial standard form, normal functional vectors, and expectation produces $E_N$. Orthogonal projection gives $exe=E_N(x)e$ directly on the dense subspace $M$, and $Je=eJ$ since $L^2(N)$ is invariant under the conjugation $J$. If an element of $M^{\mathrm{\prime }}$ commutes with $e$, Tracial standard form, normal functional vectors, and expectation writes it as $R_a$, $a\in M$; applying the commutation equation to $1$ says $a=E_N(a)\in N$. Conversely right $N$ commutes with both $M$ and $e$. Bicommutants therefore give $W^{\ast }(M,e)=(R(N){)}^{\mathrm{\prime }}=B$. Ordinary Hilbert-module dimension and finite projection models makes this a finite-trace factor; it is diffuse because it has the diffuse corner $eBe=Ne$, or by its finite matrix corner model. To verify the corner assertion independently, restriction of $T\in eBe$ to $L^2(N)$ commutes with right $N$, so is left multiplication by some $n\in N$ by Tracial standard form, normal functional vectors, and expectation. Hence $T=ne$.

Let $V_x:L^2(N)\to H$ be $n\mapsto xn$. Then $V_x{V}_y^{\ast }=xe{y}^{\ast }$ and $V_y^{\ast }{V}_x=E_N(y^{\ast }x)$ as a left multiplier on $L^2(N)$. The rectangular trace calculation from Ordinary Hilbert-module dimension and finite projection models gives ${\mathrm{Tr}}_B(xey)=\tau (yx)$ by polarization. In particular ${\mathrm{Tr}}_B(e)=1$. Its normalized restriction ${\lambda }^{-1}{\mathrm{Tr}}_B{\mathrm{\mid }}_M$ is a tracial state, hence equals $\tau$ by Projection comparison and unique normalized traces in finite factors. Tracial standard form, normal functional vectors, and expectation applied to the finite tracial algebra $(B,{\lambda }^{-1}{\mathrm{Tr}}_B)$ gives the expectation $F$. For $a\in M$, $\lambda \tau (aF(e))={\mathrm{Tr}}_B(ae)=\tau (a)$. Nondegeneracy of the trace pairing gives $F(e)={\lambda }^{-1}1$, and bimodularity gives the displayed formula for $F(xey)$.

The -algebra $A_0=\mathrm{span}MeM$ is strongly dense in $B$. One way to see this with no frame assumption is that its WOT closure is a two-sided ideal of $B$ and contains $e$. Finite trace comparison gives finitely many orthogonal projections summing to $1$, each subequivalent to $e$: remove copies of $e$ while the remainder has trace greater than $1$, then embed the last remainder in $e$. If $v_j$ are the resulting partial isometries, $1={\sum }_j{v}_{j}e{v}_j^{\ast }$ belongs to this ideal, so the ideal is all of $B$. Applying [F8] to the nondegenerate C-closure of $A_0$ gives, for each $T\in B$, a net $a_i\in A_0$ converging strongly to $T$ with a common operator bound (apply it first to the C*-closure of $A_0$, then approximate each resulting bounded element in norm by $A_0$, allowing a fixed arbitrarily small bound slack). Writing $a_i=\sum x_{r}e{y}_r$ gives $a_{i}e=x_{i}e$, where $x_i=\sum x_r{E}_N(y_r)\in M$. The expectation formula shows $x_i=\lambda F(a_{i}e)$ and $\mathrm{\parallel }{x}_i\mathrm{\parallel }\le \lambda \mathrm{\parallel }{a}_i\mathrm{\parallel }$. A bounded subnet has an ultraweak limit $x\in M$; the matrix coefficients of left multiplication on $L^2(M)$ are normal, so $x_{i}e\to xe$ weakly. But $a_{i}e\to Te$ strongly, thus $Te=xe$. Applying $F$ to this identity gives $x=\lambda F(Te)$; uniqueness also follows by applying $xe$ to $1$. This proof does not silently pass an unbounded inverse through a weak limit.

If $\xi$ is right-$N$ bounded, its bounded map $V_{\xi }:L^2(N)\to H$ extends by zero on $L^2(N{)}^{\perp }$ to $T=V_{\xi }e\in B$. Pull-down gives $T=xe$, so $\xi =T1=x$. This proves equality of the right-$N$ bounded vectors and $M$.