[proofplan] We construct the isomorphism by first identifying the [Hilbert space](/page/Hilbert%20Space) $H$ with $e_{i_0i_0}H\otimes \ell^2(I)$ using the complete family of orthogonal projections $(e_{ii})_{i\in I}$. Under this unitary identification, each matrix unit $e_{ij}$ becomes $1\otimes E_{ij}$. We then show that every $x\in M$ has matrix coefficients in the corner $N$, so its conjugate lies in $N\overline{\otimes}\mathcal{L}(\ell^2(I))$, and conversely every elementary tensor over $N$ is obtained from an element of $M$. Strong convergence of finite diagonal compressions upgrades the finite matrix calculation to the full von Neumann algebra. [/proofplan] [step:Build the Hilbert space identification from the complete diagonal projections] Let $K:=e_{i_0i_0}H$. For each finite subset $F\subset I$, define \begin{align*} p_F:=\sum_{i\in F}e_{ii}\in M. \end{align*} The projections $(p_F)_F$, indexed by finite subsets of $I$ ordered by inclusion, form an increasing net. Since $\bigvee_{i\in I}e_{ii}=1$, the projection-lattice description in [citetheorem:9267] gives \begin{align*} p_F\to 1 \end{align*} strongly. Define a map \begin{align*} U:H&\to K\otimes \ell^2(I) \end{align*} by the Hilbert-space direct-sum formula \begin{align*} U\xi:=\sum_{i\in I}e_{i_0i}\xi\otimes \delta_i, \end{align*} where $(\delta_i)_{i\in I}$ is the standard [orthonormal basis](/page/Orthonormal%20Basis) of $\ell^2(I)$. This sum is well-defined because \begin{align*} \sum_{i\in F}\|e_{i_0i}\xi\|_H^2 =\sum_{i\in F}(e_{ii}\xi,\xi)_H =(p_F\xi,\xi)_H \le \|\xi\|_H^2 \end{align*} for every finite $F\subset I$. Moreover, taking the supremum over finite $F$ and using $p_F\to 1$ strongly gives \begin{align*} \|U\xi\|_{K\otimes \ell^2(I)}^2 =\sup_F\sum_{i\in F}\|e_{i_0i}\xi\|_H^2 =\|\xi\|_H^2. \end{align*} Thus $U$ is an isometry. The range of $U$ contains every elementary vector $\eta\otimes \delta_j$ with $\eta\in K$ and $j\in I$, because if $\eta\in K=e_{i_0i_0}H$, then \begin{align*} U(e_{ji_0}\eta)=\eta\otimes \delta_j. \end{align*} Hence $U$ has dense range. Since $U$ is an isometry, its range is closed, and therefore $U$ is unitary. [guided] The purpose of this step is to turn the abstract matrix units into the standard matrix units on a tensor-product Hilbert space. The diagonal projections $e_{ii}$ split $H$ into mutually orthogonal pieces, and the partial isometries $e_{i_0i}$ transport the $i$-th piece back to the fixed reference corner $K=e_{i_0i_0}H$. For each finite subset $F\subset I$, define \begin{align*} p_F:=\sum_{i\in F}e_{ii}. \end{align*} The matrix-unit relations imply that the projections $e_{ii}$ are pairwise orthogonal. Since their supremum is $1$, the net $(p_F)_F$ over finite subsets of $I$ increases strongly to $1$ by [citetheorem:9267]. Now define \begin{align*} U:H&\to K\otimes \ell^2(I) \end{align*} by \begin{align*} U\xi:=\sum_{i\in I}e_{i_0i}\xi\otimes \delta_i. \end{align*} This is not a formal infinite sum without justification: in the Hilbert direct sum, it suffices to check that the finite partial square norms are bounded. For finite $F\subset I$, \begin{align*} \sum_{i\in F}\|e_{i_0i}\xi\|_H^2 =\sum_{i\in F}(e_{ii}\xi,\xi)_H =(p_F\xi,\xi)_H \le \|\xi\|_H^2. \end{align*} Therefore the family $(e_{i_0i}\xi)_{i\in I}$ is square-summable in the Hilbert direct sum. Taking the supremum over all finite $F$ and using strong convergence $p_F\xi\to \xi$, we obtain \begin{align*} \|U\xi\|_{K\otimes \ell^2(I)}^2 =\sup_F(p_F\xi,\xi)_H =\|\xi\|_H^2. \end{align*} So $U$ is an isometry. It remains to see that $U$ is onto. Let $\eta\in K$ and $j\in I$. Since $\eta=e_{i_0i_0}\eta$, the matrix-unit relations give \begin{align*} U(e_{ji_0}\eta)=\eta\otimes \delta_j. \end{align*} Thus every elementary tensor $\eta\otimes\delta_j$ lies in the range of $U$. These elementary tensors span a dense subspace of $K\otimes \ell^2(I)$, and the range of an isometry is closed. Hence $U$ is unitary. [/guided] [/step] [step:Identify the given matrix units with the standard matrix units] Let $(E_{ij})_{i,j\in I}$ denote the standard matrix units on $\ell^2(I)$, defined by \begin{align*} E_{ij}\delta_k=\delta_{jk}\delta_i. \end{align*} For $\xi\in H$ and $a,b\in I$, the $i$-th component of $Ue_{ab}\xi$ is \begin{align*} e_{i_0i}e_{ab}\xi=\delta_{ia}e_{i_0b}\xi. \end{align*} The $i$-th component of $(1_{\mathcal{L}(K)}\otimes E_{ab})U\xi$ is the same vector. Hence \begin{align*} Ue_{ab}U^*=1_{\mathcal{L}(K)}\otimes E_{ab} \end{align*} for all $a,b\in I$. [/step] [step:Show that conjugating $M$ gives operators with coefficients in the corner $N$] For $x\in M$ and $i,j\in I$, define the matrix coefficient \begin{align*} x_{ij}:=e_{i_0i}xe_{ji_0}\in e_{i_0i_0}Me_{i_0i_0}=N. \end{align*} For $\eta\in K$ and $j\in I$, using $U^*(\eta\otimes \delta_j)=e_{ji_0}\eta$, we compute \begin{align*} (UxU^*)(\eta\otimes \delta_j) =Uxe_{ji_0}\eta. \end{align*} Its $i$-th component is \begin{align*} e_{i_0i}xe_{ji_0}\eta=x_{ij}\eta. \end{align*} Thus $UxU^*$ is represented by the operator matrix $(x_{ij})_{i,j\in I}$ with entries in $N$. For each finite $F\subset I$, define \begin{align*} q_F:=\sum_{i\in F}E_{ii}\in \mathcal{L}(\ell^2(I)). \end{align*} Then \begin{align*} (1\otimes q_F)UxU^*(1\otimes q_F) =\sum_{i,j\in F}x_{ij}\otimes E_{ij}\in N\overline{\otimes}\mathcal{L}(\ell^2(I)). \end{align*} Since $q_F\to 1$ strongly on $\ell^2(I)$, the projections $1\otimes q_F$ converge strongly to $1$ on $K\otimes \ell^2(I)$. Therefore the finite compressions above converge strongly to $UxU^*$. The von Neumann algebra $N\overline{\otimes}\mathcal{L}(\ell^2(I))$ is strongly closed, so \begin{align*} UxU^*\in N\overline{\otimes}\mathcal{L}(\ell^2(I)). \end{align*} [/step] [step:Show that every elementary tensor over $N$ comes from $M$] Let $n\in N$ and $i,j\in I$. Since $n=e_{i_0i_0}ne_{i_0i_0}$, the element \begin{align*} y_{ij}(n):=e_{ii_0}ne_{i_0j} \end{align*} belongs to $M$. For $\eta\in K$ and $k\in I$, \begin{align*} Uy_{ij}(n)U^*(\eta\otimes \delta_k) =Uy_{ij}(n)e_{ki_0}\eta. \end{align*} Using the matrix-unit relation $e_{i_0j}e_{ki_0}=\delta_{jk}e_{i_0i_0}$, this becomes \begin{align*} Uy_{ij}(n)e_{ki_0}\eta =\delta_{jk}Ue_{ii_0}n\eta =\delta_{jk}n\eta\otimes \delta_i. \end{align*} Hence \begin{align*} Uy_{ij}(n)U^*=n\otimes E_{ij}. \end{align*} The von Neumann algebra generated by all elementary tensors $n\otimes E_{ij}$ is $N\overline{\otimes}\mathcal{L}(\ell^2(I))$. Therefore \begin{align*} N\overline{\otimes}\mathcal{L}(\ell^2(I))\subseteq UMU^*. \end{align*} Together with the previous step, this gives \begin{align*} UMU^*=N\overline{\otimes}\mathcal{L}(\ell^2(I)). \end{align*} [/step] [step:Extract the normal isomorphism and the generation statement] Define \begin{align*} \Phi:M&\to N\overline{\otimes}\mathcal{L}(\ell^2(I)) \end{align*} \begin{align*} x&\mapsto UxU^*. \end{align*} Conjugation by a unitary is a normal faithful $*$-isomorphism from $M$ onto $UMU^*$, and the preceding step identifies $UMU^*$ with $N\overline{\otimes}\mathcal{L}(\ell^2(I))$. Therefore $\Phi$ is a normal $*$-isomorphism. For $n=e_{i_0i_0}xe_{i_0i_0}\in N$, the computation in the previous step gives \begin{align*} \Phi(e_{ii_0}ne_{i_0j})=n\otimes E_{ij}. \end{align*} In particular, \begin{align*} \Phi(e_{ij})=1_N\otimes E_{ij}. \end{align*} Equivalently, the inverse isomorphism sends \begin{align*} n\otimes E_{ij}\mapsto e_{ii_0}ne_{i_0j}. \end{align*} Finally, for $x\in M$ and finite $F\subset I$, \begin{align*} p_Fxp_F=\sum_{i,j\in F}e_{ii}xe_{jj} =\sum_{i,j\in F}e_{ii_0}(e_{i_0i}xe_{ji_0})e_{i_0j}. \end{align*} Each coefficient $e_{i_0i}xe_{ji_0}$ belongs to $N$, so $p_Fxp_F$ belongs to the von Neumann algebra generated by $N$ and the matrix units. Since $p_F\to 1$ strongly, we have $p_Fxp_F\to x$ strongly. A von Neumann algebra is strongly closed, so $x$ belongs to the von Neumann algebra generated by $N$ and the matrix units. Hence $M$ is generated by $N$ and $(e_{ij})_{i,j\in I}$, completing the proof. [/step]