# Is a type $I_\infty$ von Neumann algebra “infinite”

operator-algebrasvon-neumann-algebras

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.

#### Best Answer

Yes.

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).