# Is the von Neumann algebra of type I under the given hypothesis?

Suppose $$\mathscr{R}$$ is a von Neumann algebra acting on $$\mathcal{H}$$ and $$\{E_a\}$$ is a family of abelian projections in $$\mathscr{R}$$ such that $$I= \sum E_a$$.

Assume this is given. Now this means that "non-zero abelian" projection exists in the algebra $$\mathscr{R}$$. So it cannot be of type $$II$$ or of type $$III$$ (since by definition in those algebras there is no abelian projections). Then I have two question in mind:

1. Can I now say that $$\mathscr{R}$$ is of Type I (definition: there exists an abelian projection with central carrier $$I$$)?
2. If Q1 is true. then how to construct an abelian projection, $$E$$ with $$C_E=I$$, from the initial $$\{E_a\}$$?

There is no explicit way of doing this. The problem is that you don't know the relations between the distinct $$E_a$$.
Applying the Type Decomposition Theorem, there exists a central projection $$P$$ such that $$RP$$ is type I, and $$R(I-P)$$ has no abelian subprojections. Since $$E_a(I-P)$$ is abelian for all $$a$$, we get that $$E_a(I-P)=0$$ for all $$a$$; thus $$I-P=\sum_a E_a(I-P)=0$$, and $$P=I$$.
Explicitly constructing an abelian projection $$E$$ with $$C_E$$ is not doable with the data you have. Consider for instance $$R=M_2(\mathbb C^2)$$, and $$E_1=\begin{bmatrix} (1,1)&(0,0)\\ (0,0)&(0,0)\end{bmatrix}, \quad E_2=\begin{bmatrix} (0,0)&(0,0)\\ (0,0)&(1,0)\end{bmatrix}, \quad E_3=\begin{bmatrix} (0,0)&(0,0)\\ (0,0)&(0,1)\end{bmatrix}.$$ Examples of abelian projections $$E$$ with $$C_E=I$$ are $$E_1$$, and $$E_2+E_3$$. If instead we have $$E_1=\begin{bmatrix} (1,0)&(0,0)\\ (0,0)&(0,0)\end{bmatrix}, \quad E_2=\begin{bmatrix} (0,1)&(0,0)\\ (0,0)&(0,0)\end{bmatrix}, \quad E_3=\begin{bmatrix} (0,0)&(0,0)\\ (0,0)&(1,0)\end{bmatrix}, \quad E_4=\begin{bmatrix} (0,0)&(0,0)\\ (0,0)&(0,1)\end{bmatrix}.$$ Now the examples become $$E_1+E_2$$ and $$E_3+E_4$$. An orthogonal family of abelian projections is not enough information to explicitly find who the abelian projection with central carrier 1 will be.
• I haven't thought about it too much, but I guess that every projection $E_a$ has central carrier $I$ if the decomposition is minimal in the sense that the sum $E_a+E_b$ is never abelian. What do you think? Dec 16, 2021 at 21:46
• Yes, sounds right. An easy situation would be if when you group the abelian projections by equivalence classes, the corresponding central carriers are orthogonal. In that case, taking one projection per class gives you the abelian projection with central carrier $I$. Dec 16, 2021 at 21:55