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Asking for elementary proof.
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Given two type I (von-Neumann algebra) factors $\mathcal M,\mathcal N$, is there a smallest type I factor containing both $\mathcal M,\mathcal N$?

Notes:

  • $\mathcal M,\mathcal N$ are over the same Hilbert space, of course.
  • Obviously, a type I factor containing both exists (namely the set of all bounded operators). But I want the smallest. (I.e., it needs to be contained in all other type I factors containing both.)
  • A standard approach would be to take the intersection of all type I factors containing both. However, it is not obvious that this is a type I factor. (The intersection of type I factors is not necessarily a type I factor, see the comments here.)
  • Taking the von-Neumann algebra generated by $\mathcal M\cup\mathcal N$ (i.e., $(\mathcal M\cup\mathcal N)''$) also does not work, because that is not a type I factor in general (see the comments here).
  • If the answer would also cover the case where there are infinitely many factors (not just the two $\mathcal M,\mathcal N$) would be a bonus.
  • The more elementary the proof of existence, the better. (I would need to formalize it in a computer-aided theorem prover, so the more "known facts" are used, the more of those I need to formalize first.)

Given two type I (von-Neumann algebra) factors $\mathcal M,\mathcal N$, is there a smallest type I factor containing both $\mathcal M,\mathcal N$?

Notes:

  • $\mathcal M,\mathcal N$ are over the same Hilbert space, of course.
  • Obviously, a type I factor containing both exists (namely the set of all bounded operators). But I want the smallest. (I.e., it needs to be contained in all other type I factors containing both.)
  • A standard approach would be to take the intersection of all type I factors containing both. However, it is not obvious that this is a type I factor. (The intersection of type I factors is not necessarily a type I factor, see the comments here.)
  • Taking the von-Neumann algebra generated by $\mathcal M\cup\mathcal N$ (i.e., $(\mathcal M\cup\mathcal N)''$) also does not work, because that is not a type I factor in general (see the comments here).
  • If the answer would also cover the case where there are infinitely many factors (not just the two $\mathcal M,\mathcal N$) would be a bonus.

Given two type I (von-Neumann algebra) factors $\mathcal M,\mathcal N$, is there a smallest type I factor containing both $\mathcal M,\mathcal N$?

Notes:

  • $\mathcal M,\mathcal N$ are over the same Hilbert space, of course.
  • Obviously, a type I factor containing both exists (namely the set of all bounded operators). But I want the smallest. (I.e., it needs to be contained in all other type I factors containing both.)
  • A standard approach would be to take the intersection of all type I factors containing both. However, it is not obvious that this is a type I factor. (The intersection of type I factors is not necessarily a type I factor, see the comments here.)
  • Taking the von-Neumann algebra generated by $\mathcal M\cup\mathcal N$ (i.e., $(\mathcal M\cup\mathcal N)''$) also does not work, because that is not a type I factor in general (see the comments here).
  • If the answer would also cover the case where there are infinitely many factors (not just the two $\mathcal M,\mathcal N$) would be a bonus.
  • The more elementary the proof of existence, the better. (I would need to formalize it in a computer-aided theorem prover, so the more "known facts" are used, the more of those I need to formalize first.)
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Least upper bound of type I factors

Given two type I (von-Neumann algebra) factors $\mathcal M,\mathcal N$, is there a smallest type I factor containing both $\mathcal M,\mathcal N$?

Notes:

  • $\mathcal M,\mathcal N$ are over the same Hilbert space, of course.
  • Obviously, a type I factor containing both exists (namely the set of all bounded operators). But I want the smallest. (I.e., it needs to be contained in all other type I factors containing both.)
  • A standard approach would be to take the intersection of all type I factors containing both. However, it is not obvious that this is a type I factor. (The intersection of type I factors is not necessarily a type I factor, see the comments here.)
  • Taking the von-Neumann algebra generated by $\mathcal M\cup\mathcal N$ (i.e., $(\mathcal M\cup\mathcal N)''$) also does not work, because that is not a type I factor in general (see the comments here).
  • If the answer would also cover the case where there are infinitely many factors (not just the two $\mathcal M,\mathcal N$) would be a bonus.