Denote by $\sim$ the "Murrayvon Neumann" equivalence in the projection lattice of a von Neumann algebra. It is well known that in a finite von Neumann algebra the equivalence of projections can be obtained by unitary elements, i.e. if $e\sim f$ in a finite von Neumann algebra then there exists an unitary element $u$ such that $u^*eu=f$. What can we say in general cases; let $e\sim f$, if there exists projections $e_i,f_i$ with $e_1e_2=0=f_1f_2$, $e=e_1+e_2$, $f=f_1+f_2$ along with unitaries $u_i$ such that $u_i^*e_iu_i=f_i$?

$\begingroup$ In an infinite von Neumann algebra the projection 1=e is MvNequivalent to many other projections f≠1. But u*eu=u*1u=u*u=1≠f, so there is no such u. $\endgroup$ – Dmitri Pavlov Jul 5 '18 at 18:08

$\begingroup$ @DmitriPavlov Does this immediately show that you cannot write $e$ as the sum of two orthogonal projects, and the same for $1$, with those subprojections unitarily equivalent? $\endgroup$ – Matthew Daws Jul 7 '18 at 6:55

$\begingroup$ @MatthewDaws: You can always do it if you are willing to subdivide. For two projections e,f to be unitarily equivalent it is necessary and sufficient that e~f and 1−e~1−f. Furthermore, in the σfinite case e~f if and only if tr(e)=tr(f), where tr denotes the canonical centervalued trace. Thus it suffices to show that if tr(e)=tr(f), we can find e_i,f_i such that tr(e_i)=tr(f_i), tr(1−e_i)=tr(1−f_i), e=e_1+e_2, f=f_1+f_2. It suffices to deal with the case of an infinite factor and tr(e)=tr(f)=∞. In this case we take e_i and f_i such that tr(e_i)=tr(f_i)=∞. $\endgroup$ – Dmitri Pavlov Jul 7 '18 at 17:12

$\begingroup$ what about $e_1=f_1=0$? Moreover, if $e$ is a rankone projection in $B(H)$, one of $e_i$'s must be 0. $\endgroup$ – C.Ding Jul 12 '18 at 1:08

$\begingroup$ @C.Ding: What about them? I require tr(e_i)=tr(f_i)=∞, but tr(0)=0. $\endgroup$ – Dmitri Pavlov Jul 19 '18 at 14:23
If two projections in a von Neumann algebra are related in the sense of the question then they are Murrayvon Neumann equivalent. In Theorem 4.1 of "Equivalence in operator algebras", by Kadison and Pedersen, it is shown more generally that if $e,f$ are projections in a von Neumann algebra such that $e=\sum_i x_i^*x_i$ and $f=\sum_i x_ix_i^*$ for some $x_i$ (the sums are allowed to be infinite) then $e$ and $f$ are Murrayvon Neumann equivalent.
Note: For an arbitrary C*algebra, I don't think that the relation defined in the question is an equivalence relation.

$\begingroup$ But, this does not answer my question, please note the comments by Matthew and Dimitri above. $\endgroup$ – Meisam Soleimani Malekan Aug 6 '18 at 14:04

$\begingroup$ I see, it remained to show that, conversely, Murrayvon Neumann equivalence implies the relation of the question.. and a proof for this is sketched in Dmitri's comment (using direct integrals, though arguably this can be avoided). $\endgroup$ – Leonel Robert Aug 7 '18 at 15:45