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Sep 25 at 8:58 comment added André Henriques @DavidGao Yes. The image of $1\in L^2(A,\phi)$ under the isomorphism $L^2(A,\phi)\cong L^2(A)$ is thing which I denoted by $\sqrt\phi\in L^2(A)$.
Sep 24 at 16:51 comment added David Gao In the $\sigma$-finite case, $L^2(A)$ can be identified with $L^2(A, \varphi)$ for some faithful $\varphi$ and similarly for $L^2(B)$. Then $1 \otimes 1 \in L^2(A) \boxtimes_Z L^2(B)$ induces a faithful state $\psi$ on $A \times_Z B$. Then $ab \in L^2(A \times_Z B) \to a \otimes b \in L^2(A) \boxtimes_Z L^2(B)$ should be an isomorphism.
Sep 24 at 15:32 history edited André Henriques CC BY-SA 4.0
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Sep 24 at 9:56 history asked André Henriques CC BY-SA 4.0