Is the intersection of a (possibly infinite) family $\{\mathcal M_i\}$ of type-I von-Neumann algebra factors (over the same Hilbert space $\mathcal H$) again a type-I von-Neumann algebra factor?
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2$\begingroup$ No. By the bicommutant theorem, you are asking if the (increasing) union closure of type I factors is again type I, which is not true. $\endgroup$– Narutaka OZAWACommented Mar 16, 2023 at 15:36
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$\begingroup$ @NarutakaOZAWA I don't understand. It seems to me that you are using something like $(A\cap B)' = A' \cup B'$ which is not true (not even modulo bicommutant-closure). Can you elaborate further? (Ideally, with an answer containing a proof?) $\endgroup$– Dominique UnruhCommented Mar 16, 2023 at 16:45
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1$\begingroup$ $(A\cap B)'=W^*(A'\cup B')$ in general, but we need only decreasing/increasing sequence here: $A=\mbox{cl}\bigcup A_n$ can be any type for finite-dimensional factors $A_1\subset A_2\subset\cdots$. Hence $A'=\bigcap A_n'$ can have any type even though $A_n'$ are type I factors. $\endgroup$– Narutaka OZAWACommented Mar 16, 2023 at 17:13
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$\begingroup$ @NarutakaOZAWA Oh, I think the counterexample for $(A\cap B)'\neq W^*(A'\cup B')$ I had in mind was wrong... (I assume $W^*(\dots)$ means the bicommutant here.) Do you have a reference or proof for $(A\cap B)' = W^*(A'\cup B')$? And for the fact that $\mathit{cl}\bigcup A_n$ can have any type for type-I $A_i$? (I assume $\mathit{cl}\dots$ also means bicommutant here?) $\endgroup$– Dominique UnruhCommented Mar 16, 2023 at 17:55
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$\begingroup$ I meant by "cl" the SOT closure. Hyperfinite factors arise as the SOT-closure of the increasing union of finite-dimensional factors. It's easy to see $(W^*(A'\cup B'))'=A\cap B$. Then the bicommutant theorem implies $W^*(A'\cup B')=(A\cap B)'$. $\endgroup$– Narutaka OZAWACommented Mar 16, 2023 at 17:59
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