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Sep 6 at 16:55 comment added E. Papapetros Exactly, it is still open. Also, the very large class of those $II_{\infty}$ commutants with $II_1$ commutants, which are hyperreflexive, automatically gives us a very large class of $II_1$ factors satisfying Kadison's similarity problem.
Sep 6 at 16:46 comment added David Gao Hmm, that would be more challenging. There are plenty of II$_1$ factors that can be generated by finitely many projections, but whether this holds for all II$_1$ factors should still be open, if I’m not mistaken. But this should still mean your conjecture implies a very large class of II$_\infty$ factors with II$_1$ commutants are hyperreflexive, which makes this quite hard.
Sep 6 at 16:41 comment added E. Papapetros Dear @DavidGao, it is known by Lemma 1.3 on a paper of J. Kraus and David Larson entitle "Reflexivity and distance formulae". Also, any $II_1$ factor $M$ with $II_{\infty}$ commutant $M^\prime$ is hyperreflexive. The open case is if $M^\prime$ is hyperreflexive.
Sep 6 at 14:09 history edited Michael Hardy CC BY-SA 4.0
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Sep 6 at 12:31 comment added David Gao Also, if I’m not mistaken, any type II$_\infty$ factor can be generated by finitely many projections. $B(H)$ is hyperreflexive, so your claim would imply any II$_1$ factor with II$_\infty$ commutant is hyperreflexive, which, if I’m not mistaken, is still open according to your comments on one of your previous questions.
Sep 6 at 12:15 comment added David Gao What exactly is the known result regarding $\text{ker}(\pi)$ that you referred to? Any central compression of a hyperreflexive vNa is hyperreflexive? (Also, everything here is a vNa, you can define hyperreflexivity for just vNa’s instead of for general operator spaces.)
S Sep 6 at 9:45 review First questions
Sep 6 at 9:50
S Sep 6 at 9:45 history asked E. Papapetros CC BY-SA 4.0