Suppose $M$ is $II_{1}$ factor but need not be in standard form. Under what condition (on $M$ or Hilbert space) is the commutant $M'$ of $M$ again $II_{1}$ factor on the Hilbert space acted by $M$?
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5$\begingroup$ Exactly when the Hilbert space has finite von Neumann dimension over $M$. $\endgroup$– Mateusz WasilewskiCommented Apr 23, 2019 at 13:50
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1 Answer
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Look at V. Jones 2015 von Neumann notes https://math.vanderbilt.edu/jonesvf/VONNEUMANNALGEBRAS2015/VonNeumann2015.pdf
Theorem 10.2.1(1).
You shall learn the coupling constant first.
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$\begingroup$ how to show that coupling constant is finite if you don't know the exact size of trace, the only you know the existence!! $\endgroup$ Commented Apr 25, 2019 at 10:41
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$\begingroup$ 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. $\endgroup$ Commented Apr 28, 2019 at 4:33
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$\begingroup$ No not in $L^{2}(M)$? need not be in standrad form $\endgroup$ Commented Apr 28, 2019 at 6:22