Suppose we have type $II_1$ factor $\mathcal{M}$ acting on separable Hilbert space $H$. Consider a faithful tracial state $f=tr$ (we know that such object exists) and produce $H_f$ as a Hilbert space obtained by the GNS construction from the state $f$. Does it follows that $H_f$ is separable Hilbert space? Going into details in GNS contruction suggest that this may fail to happen while $H_f$ is completion of $\mathcal{M}$-but we take a completion with respect to smaller norm.
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$\begingroup$ Are you assuming $f$ is normal? $\endgroup$– Yemon ChoiMay 29, 2012 at 0:14
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$\begingroup$ Any ${\rm II_1}$ factor has a unique trace, and it is indeed normal. $\endgroup$– Jesse PetersonMay 29, 2012 at 20:30
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Some hints:
- By replacing $H$ by $H\otimes\ell^2$, which doesn't change separability, you may suppose that every normal state $f$ is a vector state $x\mapsto (x\xi|\xi)$.
- Then $H_f$ is the completion of $M$ for $(x|y) = f(y^*x) = (y^*x\xi|\xi) = (x\xi|y\xi)$. So the map $x\mapsto x\xi$ extends to an isometry from $H_f$ into $H$. So $H_f$ is separable.
This doesn't use any special about $M$ or $f$.
answered May 29, 2012 at 8:08