[proofplan] Let $P:=M\overline{\otimes}N\subseteq \mathcal{L}(H\otimes K)$. The proof uses the identity $Z(P)=P\cap P'$ and the commutant theorem for spatial tensor products, which gives $P'=M'\overline{\otimes}N'$. For an element in this intersection, normal slice maps show that all its right slices lie in $Z(M)$ and all its left slices lie in $Z(N)$; the slice-map characterization of spatial tensor products then forces the element to lie in $Z(M)\overline{\otimes}Z(N)$. The reverse inclusion follows by checking central elementary tensors and taking weak operator closures. [/proofplan]

[step:Rewrite the center using the spatial tensor product commutant theorem] Define \begin{align*} P:=M\overline{\otimes}N\subseteq \mathcal{L}(H\otimes K). \end{align*} By the definition of the center of a von Neumann algebra, \begin{align*} Z(P)=P\cap P'. \end{align*} The commutant theorem for spatial tensor products gives \begin{align*} P'=(M\overline{\otimes}N)'=M'\overline{\otimes}N'. \end{align*} Therefore \begin{align*} Z(P)=(M\overline{\otimes}N)\cap(M'\overline{\otimes}N'). \end{align*} [/step]

[step:Use normal slice maps to force central slices]Let $x\in Z(P)$. Let $\omega\in \mathcal{L}(K)_*$ be a normal linear functional on $\mathcal{L}(K)$, and let \begin{align*} \operatorname{id}_{\mathcal{L}(H)}\otimes\omega:\mathcal{L}(H\otimes K)\to\mathcal{L}(H) \end{align*} denote the corresponding normal right slice map. Since $x\in M\overline{\otimes}N$, the slice-map property gives \begin{align*} (\operatorname{id}_{\mathcal{L}(H)}\otimes\omega)(x)\in M. \end{align*} Since $x\in M'\overline{\otimes}N'$, the same slice-map property gives \begin{align*} (\operatorname{id}_{\mathcal{L}(H)}\otimes\omega)(x)\in M'. \end{align*} Thus \begin{align*} (\operatorname{id}_{\mathcal{L}(H)}\otimes\omega)(x)\in M\cap M'=Z(M). \end{align*} Similarly, for every normal linear functional $\varphi\in \mathcal{L}(H)_*$, with left slice map \begin{align*} \varphi\otimes\operatorname{id}_{\mathcal{L}(K)}:\mathcal{L}(H\otimes K)\to\mathcal{L}(K), \end{align*} we obtain \begin{align*} (\varphi\otimes\operatorname{id}_{\mathcal{L}(K)})(x)\in N\cap N'=Z(N). \end{align*}[/step]

[guided]We want to prove that an operator $x\in Z(P)$ actually belongs to the smaller [tensor product](/page/Tensor%20Product) $Z(M)\overline{\otimes}Z(N)$. The available way to detect where an element of a spatial tensor product lives is to apply normal slice maps. Fix a normal linear functional $\omega\in \mathcal{L}(K)_*$. The right slice map \begin{align*} \operatorname{id}_{\mathcal{L}(H)}\otimes\omega:\mathcal{L}(H\otimes K)\to\mathcal{L}(H) \end{align*} is the normal [linear map](/page/Linear%20Map) characterized on elementary tensors by \begin{align*} (\operatorname{id}_{\mathcal{L}(H)}\otimes\omega)(a\otimes b)=\omega(b)a \end{align*} for $a\in\mathcal{L}(H)$ and $b\in\mathcal{L}(K)$. Since $x\in Z(P)$, the previous step gives \begin{align*} x\in (M\overline{\otimes}N)\cap(M'\overline{\otimes}N'). \end{align*} Applying the slice-map property to the inclusion $x\in M\overline{\otimes}N$ gives \begin{align*} (\operatorname{id}_{\mathcal{L}(H)}\otimes\omega)(x)\in M. \end{align*} Applying the same property to the inclusion $x\in M'\overline{\otimes}N'$ gives \begin{align*} (\operatorname{id}_{\mathcal{L}(H)}\otimes\omega)(x)\in M'. \end{align*} Hence the right slice commutes with every element of $M$ and belongs to $M$ itself, so \begin{align*} (\operatorname{id}_{\mathcal{L}(H)}\otimes\omega)(x)\in M\cap M'=Z(M). \end{align*} The left slices are handled in the same way, but now we fix a normal linear functional $\varphi\in\mathcal{L}(H)_*$ and use the normal map \begin{align*} \varphi\otimes\operatorname{id}_{\mathcal{L}(K)}:\mathcal{L}(H\otimes K)\to\mathcal{L}(K). \end{align*} From $x\in M\overline{\otimes}N$ we get \begin{align*} (\varphi\otimes\operatorname{id}_{\mathcal{L}(K)})(x)\in N. \end{align*} From $x\in M'\overline{\otimes}N'$ we get \begin{align*} (\varphi\otimes\operatorname{id}_{\mathcal{L}(K)})(x)\in N'. \end{align*} Therefore \begin{align*} (\varphi\otimes\operatorname{id}_{\mathcal{L}(K)})(x)\in N\cap N'=Z(N). \end{align*} This is the point of using both descriptions of $x$: the first places each slice in the algebra, and the second places the same slice in the corresponding commutant.[/guided]

[step:Apply the slice-map characterization to obtain the forward inclusion] The slice-map characterization of von Neumann subalgebra tensor products says the following: if $A\subseteq M$ and $B\subseteq N$ are von Neumann subalgebras and $y\in M\overline{\otimes}N$ satisfies \begin{align*} (\operatorname{id}_{\mathcal{L}(H)}\otimes\omega)(y)\in A \end{align*} for every $\omega\in\mathcal{L}(K)_*$ and \begin{align*} (\varphi\otimes\operatorname{id}_{\mathcal{L}(K)})(y)\in B \end{align*} for every $\varphi\in\mathcal{L}(H)_*$, then $y\in A\overline{\otimes}B$. We apply this with $A:=Z(M)$, $B:=Z(N)$, and $y:=x$. The previous step verifies both slice hypotheses. Hence \begin{align*} x\in Z(M)\overline{\otimes}Z(N). \end{align*} Since $x\in Z(P)$ was arbitrary, \begin{align*} Z(P)\subseteq Z(M)\overline{\otimes}Z(N). \end{align*} [/step]

[step:Check central elementary tensors and close weakly] Let $z\in Z(M)$ and $w\in Z(N)$. For arbitrary $m\in M$ and $n\in N$, the elementary tensor product multiplication rule gives \begin{align*} (z\otimes w)(m\otimes n)=zm\otimes wn. \end{align*} Since $z\in Z(M)$ and $w\in Z(N)$, we have $zm=mz$ and $wn=nw$, hence \begin{align*} zm\otimes wn=mz\otimes nw=(m\otimes n)(z\otimes w). \end{align*} Thus every elementary tensor $z\otimes w$ with $z\in Z(M)$ and $w\in Z(N)$ commutes with every elementary tensor $m\otimes n$ with $m\in M$ and $n\in N$. The algebraic tensor product $M\odot N$ is weak-operator dense in $P=M\overline{\otimes}N$, and multiplication by a fixed bounded operator is weak-operator continuous on bounded sets against vector functionals. Since commutants are weak-operator closed, each $z\otimes w$ lies in $P'$. Also $z\otimes w\in P$. Therefore \begin{align*} z\otimes w\in P\cap P'=Z(P). \end{align*} Taking the weak operator closure of the linear span of such elementary tensors gives \begin{align*} Z(M)\overline{\otimes}Z(N)\subseteq Z(P). \end{align*} [/step]

[step:Conclude the center formula and the factor case] Combining the two inclusions obtained above gives \begin{align*} Z(M\overline{\otimes}N)=Z(M)\overline{\otimes}Z(N). \end{align*} If $M$ and $N$ are factors, then \begin{align*} Z(M)=\mathbb{C}I_H \end{align*} and \begin{align*} Z(N)=\mathbb{C}I_K. \end{align*} Therefore \begin{align*} Z(M\overline{\otimes}N)=(\mathbb{C}I_H)\overline{\otimes}(\mathbb{C}I_K)=\mathbb{C}I_{H\otimes K}. \end{align*} Hence $M\overline{\otimes}N$ has scalar center, so $M\overline{\otimes}N$ is a factor. [/step]