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