# Question about the type decomposition of von Neumann algebras, Blackadar's notes.

this is a little bit of a dumb question, please be nice, I had some doubts about the type decomposition of von Neumann algebras.

I was reading Bruce Blackadar's "Operator algebras. Theory of C*-algebras and von Neumann algebras" and on page 246 he says that a von Neumann algebra $M$ can be decomposed as follows:

$M = Mz_I \oplus Mz_{II_1} \oplus Mz_{II_{\infty}} \oplus Mz_{III}$,

where $Mz_I = z_{I}Mz_{I}$, $Mz_{II_1} = z_{II_1}Mz_{II_1}$, $Mz_{II_\infty} = z_{II_\infty}Mz_{II_\infty}$ and $Mz_{III} = z_{III}Mz_{III}$.

Here if $\{ p_i\}$ is a family of abelian projections which is maximal with respect to the condition that the central carriers $z_{p_i}$ are pairwise orthogonal, then $z_I = \sum z_{p_i}$, with

$z_I = \bigvee \{p:$ p is a discrete projection$\} = \bigvee \{p:$ p is abelian$\}$,

and $z_{II} = 1 - z_{I} - z_{III}$ (there are other formulas for $z_{II_1}$ and $z_{II_\infty}$).

A von Neumann algebra is said to be of pure type $I$ if $z_{I} = 1$, with similar definitions for types $II_1$, $II_\infty$, and $III$.

I wanna make sure I'm understanding this right:

This means $Mz_{I}$, $Mz_{II_1}$, $Mz_{II_\infty}$ and $Mz_{III}$ are themselves von Neumann algebras of pure type?

That means that for example $Mz_{III}$, as a von Neumann algebra, has all the respective projections (let's call them $j_{I}$, $j_{II_1}$, $j_{II_{\infty}}$ and $j_{III}$), where $j_{III} = 1$ (which in this case $1 = z_{III}$ in $Mz_{III}$?), and $j_{I} = 0$, $j_{II_1} = 0$, $j_{II_{\infty}} = 0$?? Am I making sense here? If this is true or not true it is not obvious to me or just going over my head and I don't want there to be any room for confusion.

Yes. Those central projections exist for any von Neumann algebra. When the algebra is of a certain type, it means that the central projection corresponding to that type is $1$ and the ones corresponding to the other types are zero.