1
$\begingroup$

If $\mathcal M,\mathcal N$ are type-I von-Neumann algebra factors (over the same Hilbert space $\mathcal H$), is $\mathcal M\cap\mathcal N$ a type-I von-Neumann algebra factor?

Notes:

  • An elementary proof would be ideal, but I also take a textbook reference.
  • The comments after my earlier question indicate that this is not the case for an intersection of infinitely many factors, but they crucially rely on the "infinite" part of it.
  • I know that the intersection of two von-Neumann algebras is a von-Neumann algebra.
  • I don't know if the intersection of two factors is a factor.
Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ Again, by the bicommutant theorem, you are asking what kind of von Neumann algebra is generated by finitely many type I factors. It needs not be a factor and can be quite arbitrary. The tracial free product of $M_3(\mathbb{C})$ with itself is a factor of type II_1. The proof of factoriality is not so easy (see projecteuclid.org/journals/duke-mathematical-journal/volume-69/…), but by construction it is infinite dimensional and has a tracial state, hence not a type I factor. $\endgroup$
    – Narutaka OZAWA
    Commented Mar 17, 2023 at 11:49
  • $\begingroup$ Unfortunately, I lack the expertise to understand the linked paper (or even to find the result about $M_3(\mathbb C)$ in it), or to see that the tracial product you mention is indeed the closed union of finitely many type-I factors... $\endgroup$
    – Dominique Unruh
    Commented Mar 17, 2023 at 12:37
  • 2
    $\begingroup$ It's not closed union. It is $W^*(A\cup B)$ with $A\cong M_3(\mathbb{C})\cong B$, and $W^*(A\cup B)' = A' \cap B'$ is the intersection of type I factors $A'$ and $B'$. Since the type (I, II, III) is preserved under the commutant operation, the type of $A'\cap B'$ is same as that of $W^*(A\cup B)$, which is II under the tracial free product. $\endgroup$
    – Narutaka OZAWA
    Commented Mar 17, 2023 at 13:11

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .