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Let $\pi_1$ and $\pi_2$ be two representations of a unital $C^*$-algebra $A$ on Hilbert spaces $H_1$ and $H_2$ respectively. Assume that $\pi_1$ and $\pi_2$ are unitarily equivalent. So there exists a unitary $U\in B(H_1,H_2)$ such that $U\pi_1(a)U^*=\pi_2(a)$. I want to prove that the von Neumann algebra generated by $\pi_1$ and $\pi_2$ are isomorphic. That is I need to prove that $\pi_1(A)''$ and $\pi_2(A)''$ are isomorphic.

I tried to define a map as follows. Let $T\in \pi_1(A)''$ and map this to $UTU^*$. I should prove that $UTU^* \in \pi_2(A)''$.

Case 1: Let $T\in \pi_1(A)$. Then $T=\pi_1(a)$ so that $UTU^*=U\pi_1(a)U^*=\pi_2(a) \in \pi_2(A)\subseteq \pi_2(A)''$.

Case 2: Let $T \in \overline{\pi_1(A)}$ where the closure is wrt SOT and $T$ does not belong to $\pi_1(A)$. Then there exists a net $\pi_1(a_i)$ converges to $T$ in SOT. Then $U\pi_1(a_i)U^*$ converges to $UTU^*$ in SOT. But $U\pi_1(a_i)U^*=\pi_2(a_i)$ and this proves that $UTU^* \in \overline{\pi_2(A)}=\pi_2(A)''$.

What more details I need to check that it is a von Neumann algebra isomorphism? Is my approach correct?

budi
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  • Sure thats one way to do it. Just don't forget to explain why this is an isomorphism (so far it just defines a map $\pi_1(A)''\to\pi_2(A)''$). – s.harp Jan 26 '21 at 23:41
  • injectivity and surjectivity is clear. Only thing is to check that it is an algebra morphism.It is clear that if $S \mapsto USU^$ and $T \mapsto UTU^$, then $ST\mapsto USTU^=(USU^)(UTU^*)$ and hence an algebra morphism. Is it enough now? – budi Jan 27 '21 at 06:36
  • Surjectivity is the only point that I would say requires an argument. Algebra morphism, compatibility with $*$ and injectivity are completely elementary here. – s.harp Jan 27 '21 at 11:44

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