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Timeline for On commutant of $II_{1}$ factors

Current License: CC BY-SA 4.0

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Apr 28, 2019 at 6:22 comment added user136400 No not in $L^{2}(M)$? need not be in standrad form
Apr 28, 2019 at 4:33 comment added Quan Chen Do you mean the "Hilbert space acted by M" is $L^2(M)$, where the coupling constant is 1? In this case $M'\subset B(L^2(M))$ is a type $\text{II}_1$ factor.
Apr 25, 2019 at 10:41 comment added user136400 how to show that coupling constant is finite if you don't know the exact size of trace, the only you know the existence!!
Apr 25, 2019 at 4:10 vote accept user136400
Apr 25, 2019 at 3:58 history answered Quan Chen CC BY-SA 4.0