I'm studying von Neumann algebras type decomposition and I already noticed two "different" definitions of, for instance, type I :
(i) a vN algebra is said to be of type I, if it has an abelian projection with central carrier Id (Id stands for identity)
(ii) a vN algebra is said to be of type I, if there is an abelian projection $A$ such that for every central projection $Q$ we have that $Q\geq A$
Are these conditions equivalent ? I tried to prove $(i)\Rightarrow (ii)$, but I stopped here : Let $A$ be the abelian projection with central carrier $C_{A}=Id$ and let $P$ be a central projection. Since it is a central projection, $QC_{A}=C_{QA}=Q$ and $C_{Q}=Q$. Thus, we have that $C_{QA}=C_{Q}$. On the other hand, we have that $A\leq C_{A}=Id$, so that $Q\geq QA$. If we suppose that $Q> QA$ can I conclude, under these conditions, that $C_{Q}>C_{QA}$ ? If so, we easily conclude that $Q\geq A$.
I didnt try on the other direction. Any suggestions ? Thank you
It is not true that (i)$\implies$ (ii). Consider for instance the von Neumann algebra $L^\infty[0,1]$. Since it is abelian, every projection is abelian, and in particular the central carrier of the identity is the identity. But it has no minimal projections, so condition (ii) cannot be satisfied: given any abelian projection $A$ (that is, a characteristic function of some non-null Borel set), we can write it as the sum of two central projections (corresponding to a division of the Borel set into two non-null parts, obtained by intersection with appropriately chosen intervals).
The idea in (ii) seems to work well for direct sums of full matrix algebras, but not for diffuse algebras as the previous paragraph shows.
It is true that (ii)$\implies$(i), though. Because the central carrier of such $A$ ought to be the identity. Indeed, if $C_A$ is not the identity, then $I-C_A$ is a central projection with $I-C_A\not\geq A$.