Denote by $\sim$ the "Murray-von 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 $u_i^*e_iu_i=f_i$?
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$\begingroup$ In an infinite von Neumann algebra the projection 1=e is MvN-equivalent 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
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$\begingroup$ @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
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$\begingroup$ @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 σ-finite case e~f if and only if tr(e)=tr(f), where tr denotes the canonical center-valued 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
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$\begingroup$ what about $e_1=f_1=0$? Moreover, if $e$ is a rank-one projection in $B(H)$, one of $e_i$'s must be 0. $\endgroup$ – C.Ding Jul 12 '18 at 1:08
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$\begingroup$ @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
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If two projections in a von Neumann algebra are related in the sense of the question then they are Murray-von 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 Murray-von Neumann equivalent.
Note: For an arbitrary C*-algebra, I don't think that the relation defined in the question is an equivalence relation.
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$\begingroup$ 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
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$\begingroup$ I see, it remained to show that, conversely, Murray-von 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
