Let $M$ be a $\mathrm{II}_{1}$ factor acting on $L^{2}(M, \tau)$ in standard form, let $\{e_{n}:n \in \mathbb{N}\}$ be fixed orthonormal basis of $L^{2}(M, \tau)$, does there exist sequence of unitaries $U_{n}$ in $M$ such that $U_{n}(e_{1})=e_{1};(U_{n}(e_{n})=e_{n+1}; n\neq 1)$??
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4$\begingroup$ Suppose your orthonormal basis is comprised of unitaries of M, and that e_1 is the identity. Can you come up with a concrete counterexample then? For example, look at the group element basis of a group factor. $\endgroup$– Jon BannonCommented May 5, 2019 at 8:28
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$\begingroup$ Yes,If you consider right shift operator $R(e_{n})=e_{n-1}$, then $U_{n}R=RU_{n}$, since $U_{n} \in L(G)$, but the equation does not hold for the element $e_{1}$, so contradiction. @Bannon what you can say for type III factors? $\endgroup$– user136400Commented May 5, 2019 at 9:01
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2$\begingroup$ I'll bet you can play the above game whenever you have a separating cyclic vector, since that vector will be like the identity element in the group. So as long as you have a faithful normal state on your vn alg, you should be able to find an on basis containing such a vector and get a counterexample, right? $\endgroup$– Jon BannonCommented May 5, 2019 at 10:23
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$\begingroup$ yes it seems!! but for semicyclic representations, there are infinitely many copies of type I subfactors sitting inside type III factors, but the above question survives type I class! Is the type I the only class where this question has a positive answer? $\endgroup$– user136400Commented May 5, 2019 at 11:46
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1$\begingroup$ To sharpen what I have been saying, it is the requirement of fixing the e_1 that breaks down, if you let e_1 be separating and cyclic. Beyond this case, though... $\endgroup$– Jon BannonCommented May 5, 2019 at 12:07
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