Let $M$ be a $\mathrm{II}_{1}$ factor with the faithful trace $\tau$ in standard form. Suppose $\tau'(x')=\tau(Jx'^{*}J)$. If we complete $M$ with $\|\cdot\|_{1}$ norm, i.e., $\|x\|_{1}=\tau(|x|)$, then the completion is $L^{1}(M,\tau)$. My qustion is if we complete $M'$ by $\|x'\|_{1}=\tau'(|x'|)$, is it true that $L^{1}(M',\tau')=L^{1}(M,\tau)$, or they isometrically isomorphic? Well this is clear that it is sometric isomorphism by the comment of Matthew. Now instead of that if we take the vector state $\omega_{\tau}=\langle x\Omega_{\tau},\Omega_{\tau}\rangle$. Complete $M$ and $M'$ with respect to this $\|\cdot\|_{1}$ norm coming from state, are they same? I am confused because this equality $L^{2}(M,\omega_{\tau})=L^{2}(M',\omega_{\tau})$. Thanks in advance!!
$\begingroup$
$\endgroup$
2
-
2$\begingroup$ 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 $M\rightarrow M'; x\mapsto Jx^*J$. $\endgroup$– Matthew DawsCommented Oct 27, 2019 at 11:30
-
1$\begingroup$ 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 $M$. $\endgroup$– Matthew DawsCommented Oct 27, 2019 at 14:49
Add a comment
|