I know that given C*-algebras $A, B$ with faithful states $\omega,\varpi$, the tensor product state $\omega\otimes\varpi$ on the minimal tensor product $A\otimes_{\text{min}}B$ is faithful.

I also believe I can prove that if $A,B$ equal the von Neumann algebras $B(\mathcal{H}), B(\mathcal{K})$ for some Hilbert spaces $\mathcal{H},\mathcal{K}$, then if $\omega, \varpi$ are additionally normal, the tensor product state $\omega\otimes\varpi$ is faithful (and normal) on the standard tensor product $A\otimes_{\text{vN}}B=B(\mathcal{H}\otimes\mathcal{K})$, essentially by representing $\omega,\varpi$ by trace-class operators and then reasoning in an appropriate eigenbasis of $\mathcal{H}\otimes\mathcal{K}$.

This sets up the following question: suppose $A,B$ are any von Neumann algebras and $\omega,\varpi$ faithful normal states. Is it true that $\omega\otimes\varpi$ is faithful on $A\otimes_{\text{vN}}B$? Unfortunately $\otimes_{\text{vN}}\neq\otimes_{\text{min}}$ for von Neumann algebas, so I cannot apply the first result I cited.


1 Answer 1


This is true. Consider the GNS representations $(\pi_\omega, H_\omega)$ and $(\pi_\varpi, H_\varpi)$. Both are normal faithful representations, so $A \otimes_{vN} B$ can be regarded as acting on $H_\omega \otimes H_\varpi$. Since $\omega$ is faithful, $\hat{1} \in H_\omega$ is separating for $A$, whence it is cyclic for $A’$. Similarly, $\hat{1} \in H_\varpi$ is cyclic for $B’$. Hence $\hat{1} \otimes \hat{1} \in H_\omega \otimes H_\varpi$ is cyclic for $(A \otimes_{vN} B)’ = A’ \otimes_{vN} B’$, so it is separating for $A \otimes_{vN} B$. As $\hat{1} \otimes \hat{1} \in H_\omega \otimes H_\varpi$ implements the state $\omega \otimes \varpi$, this is the same as saying $\omega \otimes \varpi$ is faithful.

  • $\begingroup$ Nice, my operator algebra is a bit rusty and I forgot about the faithful--cyclic dichotomy... $\endgroup$
    – J_P
    Oct 26, 2023 at 19:12
  • $\begingroup$ *separating--cyclic dichotomy $\endgroup$
    – J_P
    Oct 26, 2023 at 19:21

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