Timeline for On $L^{1}(M',\tau')$
Current License: CC BY-SA 4.0
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| Oct 27, 2019 at 14:49 | comment | added | Matthew Daws | So, with the new follow-on question: we have that $M\subseteq B(H)$ and $\Omega_\tau$ is a cyclic and separating and tracial vector (some redundancy here). So, yes, canoncially $H \cong L^2(M,\omega_\tau) \cong L^2(M',\omega'_\tau)$. But we do not have equality: there is an isomorphism here, and in fact it's the "same" one which shows $L^1(M)$ is isomorphic to $L^1(M')$. Notice here I write $\omega'_\tau$ as this is a different functional, as it is on $M'$ and not $M$. | |
| Oct 27, 2019 at 12:16 | history | edited | user136400 | CC BY-SA 4.0 |
added 390 characters in body
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| Oct 27, 2019 at 11:30 | comment | added | Matthew Daws | They cannot be "equal" as, by definition, these are abstract Banach spaces, completions of different von Neumann algebras. However, $L^1(M,\tau)$ is isometrically isomorphic to $M_*$ the predual of $M$, and similarly for $M'$, and $M_*$ and $(M')_*$ are isomorphic via the pre-adjoint of the anti-homomorphism $M\rightarrow M'; x\mapsto Jx^*J$. | |
| Oct 27, 2019 at 11:11 | history | asked | user136400 | CC BY-SA 4.0 |