# How did Kadison obtain it here?

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

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 $$A=0$$.

• 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... Commented Dec 17, 2021 at 0:04
• 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 Commented Dec 17, 2021 at 0:11
• Yes. He says exactly that (with no proof, beause it's kind of obvious) in lines 4-6 in this proof. Commented Dec 17, 2021 at 0:40
• Thank you so much Commented Dec 17, 2021 at 1:38