2
$\begingroup$

In the "Type decomposition" theorem 6.5.2 in Kadison-Ringrose, vol II, they proved that every von Neumann algebra can be decomposed as a direct sum of type $I_n$, type $II_1$, type $II_\infty$ and type $III$ parts (some of these summands may be zero). I understood this. My question is:

Does there exists a von Neumann algebra which is neither of type $I_n, II_1,II_\infty, \text{ nor } III$?

For example, consider $\mathscr{R}:=L^\infty(X, \mu)$ acting on $\mathcal{H}:=L^2(X, \mu)$. Then since $\mathscr{R}$ is abelian so every non-zero projection in it is abelian projection. Hence $\mathscr{R}$ cannot be of type $II$ nor $III$. Can I say from there that $L^\infty(X, \mu)$ is a type $I$ von Neumann algebra (without even finding a projeciton with central carrier $I$. In deed here $\chi_X$ is an abelian projection with central carrier $I=\chi_X$ proving $L^\infty$ or any abelian algebra is of type $I$.)?

$\endgroup$
2
  • $\begingroup$ I am not sure if I understand your question. Every von Neumann algebra is a direct sum of algebras of type I, II and III. So of course if it is a sum of (non-zero) type I and type II algebras for example, it is neither of type I, II or III. Is that what you mean? $\endgroup$
    – MaoWao
    Dec 16, 2021 at 7:54
  • $\begingroup$ @MaoWao Yes that is my point....So such algebras (i.e. direct sum of non zero types) are neither of the each type?? Why??? Is this some thing very trivial; follows from the definition ?? $\endgroup$
    – sigma
    Dec 16, 2021 at 9:25

1 Answer 1

3
$\begingroup$

In any abelian von Neumann algebra, each projection is its own central carrier. The only projection with central carrier I is I; that's enough to make it type I.

A von Neumann algebra can have all the types as central summands. For instance take $M$ and $N$ of your favourite (different types), and form $M\oplus N$.

If you look at the proof of Theorem 6.5.2 in KRII, the central projection $P_d$ is constructed in such a way that $(I-P_d)R$ has no abelian projections. That makes it unique.

In a similar way, $P_{c_1}$ is constructed as the largest finite central projection in $(I-P_d)R$. And so on.

The point is that you don't get to choose. A von Neumann algebra $R$ is a direct sum of von Neumann algebras of fixed types.

That a direct sum of algebras of distinct types is of neither type is trivial to check. For instance if $R=M\oplus N$ with $M$ type I and $N$ type II$_1$, then there are not abelian projections under $P_N$, so $R$ is not type I. And there are abelian projections under $P_M$, so $R$ is not type II.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .