A matrix amplification gives an explicit finite-index II1 inclusion

For $x=[x_{ab}]\in M_k(P)$, $E_N(x)=k^{-1}{\sum }_a{x}_{aa}\otimes 1$. If $x\ge 0$, each diagonal compression $x_{aa}$ is positive, so this sum is positive. The formula makes $E_N$ unital, fixes $N$, is $N$-bimodular, and preserves the product trace; its finite sum of normal diagonal maps is normal. If $E_N(x)=0$ for $x\ge 0$, then ${\tau }_M(x)={\tau }_N(E_N(x))=0$, so faithfulness of ${\tau }_M$ gives $x=0$; hence $E_N$ is faithful. An element ${\sum }_{c,d}{p}_{cd}\otimes e_{cd}$ commuting with every $1\otimes e_{ab}$ must be $p\otimes 1$, and commuting also with $P\otimes 1$ forces $p\in Z(P)=\mathbb{C}1$; thus $M$ is a factor. It has no minimal projection: if $q\in M$ were minimal, some diagonal corner $r=(1\otimes e_{ii})q(1\otimes e_{ii})$ would be nonzero. Its support projection $s$ in the diffuse factor $P\otimes e_{ii}$ is nonzero. The polar decomposition of $q(1\otimes e_{ii})$ lies in $M$ by [F5] and gives a partial isometry from $s$ onto a nonzero subprojection of $q$. Since $P$ has no minimal projections, choose $0<t<s$; its image under that partial isometry is a nonzero proper subprojection of $q$, a contradiction. Thus $M$ is a II1 factor; $N\cong P$ is one as well.

For matrix units, $u_{ab}^{\ast }{u}_{cd}=k1\otimes e_{ba}{e}_{cd}=k{\delta }_{ac}(1\otimes e_{bd})$, so $E_N(u_{ab}^{\ast }{u}_{cd})={\delta }_{ac}{\delta }_{bd}1$. Thus the $k^2$ vectors are orthonormal as a right $N$-module frame. More generally, for $x={\sum }_{c,d}{x}_{cd}\otimes e_{cd}$, $E_N(u_{ab}^{\ast }x)=k^{-1\mathrm{/}2}{x}_{ab}\otimes 1$, and hence ${\sum }_{a,b}{u}_{ab}{E}_N(u_{ab}^{\ast }x)=x$. These coordinates identify $\mathcal{X}$ isometrically with $N^{k^2}$, so it is complete with projection model $p=1_{k^2}$; the inner-product traces sum to $k^2$, and [F2] gives index $k^2$.

An element ${\sum }_{c,d}{p}_{cd}\otimes e_{cd}$ commutes with every $p\otimes 1$ exactly when each $p_{cd}$ is central in the factor $P$, hence when it lies in $1\otimes M_k$. On the right action of this relative commutant, [F4] gives unnormalized module trace ${\sum }_{a,b}{\tau }_N(E_N(u_{ab}^{\ast }{u}_{ab}(1\otimes e_{cd})))=k{\mathbf{1}}_{c=d}$. Indeed $e_{ba}{e}_{ab}{e}_{cd}={\mathbf{1}}_{b=c}{e}_{bd}$, whose normalized matrix trace is ${\mathbf{1}}_{b=c=d}\mathrm{/}k$, the scalar factor $k$ in $u_{ab}$ cancels the normalized matrix trace $1\mathrm{/}k$, and the sum over $a$ has $k$ terms. Dividing by $\lambda =k^2$ gives ${\mathbf{1}}_{c=d}\mathrm{/}k={\mathrm{tr}}_k(e_{cd})$, so the right-action module trace and $\tau$ agree on $N^{\mathrm{\prime }}\cap M$; the inclusion is extremal.

Applying [F3] with $\lambda =k^2$ gives ${\tau }_1(x{e}_{N}y)=k^{-2}({\tau }_P\otimes {\mathrm{tr}}_k)(yx)$ for every $x,y\in M$. This is the claimed trace normalization.