In the book S. Stratila, L. Zsido "LECTURES ON VON NEUMANN ALGEBRAS" authors prove the following proposition (4.10): If $e,f$ are abelian projections in von Neumann algebra $M$ and $z(e)\leq z(f)$, then $e\precsim f$, where $z(x)$ the central projection of $x\in M$. In the proof the authors by comparison theorem assume that $f\leq e$. Actually I don't understand how we use comparison theorem to assume that $f\leq e$. Thank's for any help.
About abelian projection in von Neumann algebras
operator-algebrasprojectionvon-neumann-algebras
Related Solutions
In your attempt for the easy inclusion, I'm not comfortable enough with tensor products to assess your exchange of tensor and intersection. But that's not needed. Elements of $Z_1\otimes Z_2$ commute with every elementary tensor in $M_1\otimes M_2$, so commute with all elements of the algebraic tensor product, and then commute with limits of these. This shows that $$ Z_1\otimes Z_2\subset Z. $$
For the reverse inclusion, what the authors do is to get, from the definition of centre, that $$\tag1 M_1\otimes M_2\subset Z'. $$ Then, since every element of $M_1'\otimes M_2'$ commutes with every elements of $M_1\otimes M_2$, $$\tag2 M_1'\otimes M_2'\subset (M_1\otimes M_2)'\subset Z', $$ the last inclusion being simply $Z\subset M_1\otimes M_2$. Then, $$\tag3 Z_1'\otimes Z_2'=(M_1\cap M_1')'\otimes (M_2\cap M_2')' =(M_1'\cup M_1)''\otimes (M_2'\cup M_2)''\subset Z'. $$ The last inclusion follows from $(1)$ and $(2)$, because if you have $a\otimes b$ with $a\in M_1'$, $b\in M_2$, then $$ a\otimes b=(a\otimes 1)(1\otimes b)\in Z' $$ since $a\otimes 1\in Z'$ by $(2)$ and $1\otimes b\in Z'$ by $(1)$. The other cases are treated similarly, to get $(M_1'\cup M_1)\otimes (M_2'\cup M_2)\subset Z'$. Taking sums and limits in the first coordinate we get $(M_1'\cup M_1)''\otimes (M_2'\cup M_2)\subset Z'$. And then taking sums and limits in the second coordinate, $(3)$ follows.
Finally, from $(3)$ one gets $$ Z\subset (Z_1'\otimes Z_2')'=Z_1''\otimes Z_2''=Z_1\otimes Z_2. $$
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).
Best Answer
By comparison, you have a central projection $p$ with $pe\preceq pf$ and $(1-p)f\prec (1-p)e$. What you want is to show that $1-p=0$. So they work with $(1-p)f$ and $(1-p)e$; they take a subprojection $f_0$ of $(1-p)e$ such that $(1-p)f\sim f_0$ and they consider the proper subprojection $f_0$ of $(1-p)e$. As equivalence of projections and multiplying by central projections preserve abelian projections, they still have that $f_0$ and $(1-p)e$ are abelian.
Then they label $f_0$ and $(1-p)e$ as $f$ and $e$.