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Let $\mathcal{H,K}$ be Hilbert spaces. Consider $S = \mathbb{C} I_{\mathcal{H}} \otimes \mathcal{B(K)} \subseteq \mathcal{B(H \otimes K)}$. Then, it is known that the commutant of $S$, $S' = \mathcal{B(H)} \otimes \mathbb{C} I_{\mathcal{K}}$.

Is there a natural generalisation in the following way:

Let $\mathcal{A}$ be a $C^*$-algebra and $\pi: \mathcal{A} \to \mathcal{B(K)}$ be an irreducible representation. Let $S = \mathbb{C} I_{\mathcal{H}} \otimes \pi(\mathcal{A}) \subseteq \mathcal{B(H \otimes K)}$. Then is it true that $S' = \mathcal{B(H)} \otimes \mathbb{C} I_{\mathcal{K}}$ ?

Note: $\pi(\mathcal{A})$ need not be the entire $\mathcal{B(K)}$ but $\pi(\mathcal{A})' = \mathbb{C} I_{\mathcal{K}}$ because $\pi$ is irreducible.

Anand O R
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Yes, this is true. We have $$S'= (\mathbb{C}I_{\mathcal{H}}\otimes \pi(\mathcal{A})'')' = B(\mathcal{H})\bar{\otimes} \pi(\mathcal{A})''' = B(\mathcal{H})\bar{\otimes} \pi(\mathcal{A})' = B(\mathcal{H})\otimes \mathbb{C}I_{\mathcal{K}}.$$

J. De Ro
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  • Can you elaborate more on the second equation? What property of the commutant have you used there? Are commutant of tensor product and strong closure of tensor product of commutant equal? – Anand O R Aug 22 '24 at 09:42
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    @AnandOR Yes, for (normal, unital) inclusions of von Neumann algebras $M\subseteq B(H)$ and $N\subseteq B(K)$, we have $(M\bar{\otimes} N)' = M'\bar{\otimes} N'$. This is a difficult result in von Neumann algebra theory. You can for example find a proof in Takesaki's first book. – J. De Ro Aug 22 '24 at 09:58