Timeline for On existence of sequence of unitaries in $II_{1}$ factor $M$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 5, 2019 at 12:07 | comment | added | Jon Bannon | 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... | |
| May 5, 2019 at 11:46 | comment | added | user136400 | 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? | |
| May 5, 2019 at 10:23 | comment | added | Jon Bannon | 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? | |
| May 5, 2019 at 9:01 | comment | added | user136400 | 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? | |
| May 5, 2019 at 8:28 | comment | added | Jon Bannon | 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. | |
| May 5, 2019 at 8:21 | history | asked | user136400 | CC BY-SA 4.0 |