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Consider the von Neumann subalgebra of $M\otimes M$ by $ B= \mathrm{vN} \{T\otimes T: T\in M\}$. What is the commutant of B?

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We need $M \subseteq B(H)$ in order for the commutant to make sense. So $B \subseteq B(H\otimes H)$. The commutant of $B$ is the von Neumann algebra $C$ generated by $M' \otimes M'$ and the flip unitary $u$ acting on $H \otimes H$. It's clear that $B$ is contained in $C'$; conversely, if $x \in C'$ then $\phi(x) = x$ where $\phi$ is the flip automorphism of $B(H\otimes H)$ (= conjugation by $u$), which implies that $x$ lies in the symmetric part of $B(H)\otimes B(H)$ (see the argument here, replacing norm limits by bounded weak* limits), and also, commuting with $M'\otimes M' \subseteq C$ implies that $x \in M\otimes M$. Thus $x$ lies in $B =$ the symmetric part of $M\otimes M$. This shows that $B = C'$.

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    $\begingroup$ Assume $M$ is in standard form sitting inside $B(L^2(M,\tau))$, sorry for not mentioning $\endgroup$ – user136400 Apr 15 at 13:37
  • $\begingroup$ Does there exist canonical trace on $C$ if M is tracial? $\endgroup$ – user136400 Apr 15 at 14:16
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    $\begingroup$ I think so, if $H = L^2(M)$ then $H\otimes H = L^2(M\otimes M)$ and there is a canonical trace on $(M\otimes M)' = M' \otimes M'$. I guess every element of $C$ has the form $x + uy$ where $x,y \in M'\otimes M'$, and then you just define $\tau(x + uy) = \tau(x)$. $\endgroup$ – Nik Weaver Apr 15 at 14:41
  • $\begingroup$ But that is not faithful trace!! $\endgroup$ – user136400 Apr 16 at 4:36
  • $\begingroup$ You sure about that? Note that every positive element has $y=0$. $\endgroup$ – Nik Weaver Apr 16 at 11:36

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