I was studying the "Type decomposition" of von Neumann algebras (in Kadison-Ringrose; Vol II). And I feel some trouble to properly grasp the definitions of different "types". The authors have defined:

Type $I_n$: A von Neumann algebra $\mathscr{R}$ is said to be of type $I$ if it has an abelian projection with central carrier $I$- of type $I_n$ if $I$ is the sum of $n$ equivalent abelian projections.

My question is : from this definition can we determine whether $\mathscr{R}$ is finite or infinite? I mean is it true that the identity in a type $I_\infty$ von Neumann algebra is "infinite"?? Is $I$ in a type $I_n$ von Neumann algebra for $n<\infty$ finite?

To show $I$ is infinite (in the first question) we have to prove $I\sim E<I$.


1 Answer 1



If $R$ is type I$_\infty$, then $I=\sum_{n=1}^\infty E_n$, with each $E_n$ abelian, and pairwise equivalent. Then $I\sim E$, where $E=\sum_{n=2}^\infty E_n$ is a proper subprojection of $I$. So $I$ is infinite.

If $R$ is of type I$_n$, then we have $I=\sum_{k=1}^nE_k$, with the $E_k$ abelian and pairwise equivalent. Abelian projections are finite (Proposition 6.4.2). So $I$ is finite (Theorem 6.3.8).

  • $\begingroup$ Thank you I got it.. $\endgroup$
    – sigma
    Dec 17, 2021 at 1:36
  • 1
    $\begingroup$ Glad I could help. We don't get that many von Neumann algebra questions here :) $\endgroup$ Dec 17, 2021 at 1:40

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