In the book of Kadison-Ringrose vol II the authors claim the following:

Let $\mathscr{R}$ be a von Neumann algebra (acting on $\mathcal{H}$) of type $I$ with no infinite central summand, and let $P_n$ be a central projection in $\mathscr{R}$ such that $\mathscr{R} P_n$ is of type $I_n$. If $n>1$, then $\mathscr{R} P_n$ acting on $P_n \mathcal{H}$ is a von Neumann algebra without abelian central summand! Why?

Definition. 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.

I don't understand why doesn't $\mathscr{R} P_n$ have an abelian central summand under those conditions? From "Type Decomposition Theorem" since $\mathscr{R}$ is of type I and have no infinite central summand so here $n\in \Bbb{Z}$. Also since $\mathscr{R} P_n$ is of type $I_n$ so $P_n$ is a sum of $n$ many equivalent abelian projections $\{E_j\}$. I cannot see further. Any help is appereciated. Thanks.


1 Answer 1


This is a weird statement. Where exactly in KRII is it?

Without additional hypotheses, a type I$_n$ von Neumann algebra cannot have an abelian central summand if $n>2$.

Suppose that $R$ is type I$_n$ and $R=A\oplus M$, with $A$ abelian. By hypothesis there exist abelian projections $E_1,\ldots,E_n\in R$ and partial isometries $W_{kj}$ such that $W_{kj}^*W_{kj}=E_j$, $W_{kj}W_{kj}^*=E_k$. Let $P_A$ be the central projection corresponding to $A$. By definition $P_AE_k\in A$ and $P_AW_{kj}\in A$. So, using that $P_A$ is central and that $A$ is abelian, $$ P_AE_k=P_AW_{kj}W_{kj}^*=(P_AW_{kj})(P_AW_{kj}^*)=(P_AW_{kj}^*)(P_AW_{kj})=P_AW_{kj}^*W_{kj}=P_AE_j. $$ Then $$ P_AE_k=P_AE_kP_AE_j=P_AE_kE_j=0 $$ if $k\ne j$. This gives us $$ P_A=\sum_kP_AE_k=0, $$ and so $A=0$.

  • $\begingroup$ Thank you. This is not exactly a statement in the book rather it appears during an exercise in KRII. That exercise is actually Lemma 3.5 of this paper: jstor.org/stable/2374400?seq=7#metadata_info_tab_contents....In the 3rd line of that proof Kadison seem to use that argument which I posted in the question... $\endgroup$
    – sigma
    Commented Dec 17, 2021 at 0:04
  • $\begingroup$ But your answer shows that for any type $I_n$ algebra with ($n>1$) cannot have abelian central summand! Right?? And that's exactly Kadison had in mind in that Lemma 3.5 at 3rd line...!! Please point out if this is the case or I am missing something again! Thanks $\endgroup$
    – sigma
    Commented Dec 17, 2021 at 0:11
  • 1
    $\begingroup$ Yes. He says exactly that (with no proof, beause it's kind of obvious) in lines 4-6 in this proof. $\endgroup$ Commented Dec 17, 2021 at 0:40
  • $\begingroup$ Thank you so much $\endgroup$
    – sigma
    Commented Dec 17, 2021 at 1:38

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