Let $f\_i$, $i=1,\dotsc,n$, be mutually orthogonal Abelian projections in a von Numann algebra, and let $e\leq\sum f\_i$. Is it true that there exist mutually orthogonal Abelian projections $e\_j$, $j=1,\dotsc,m$ such that $e=\sum e\_j$?



Subprojections of the sum of mutually orthogonal Abelian projections
This is true, more generally, in a C\*-algebra of real rank zero (i.e., such that the selfadjoint elements with finite spectrum are dense in the set of selfadjoints). Zhang proves in "A Riesz decomposition property and ideal structure of multiplier algebras" that the monoid of Murray-von Neumann equivalence classes of projections of a real rank zero C\*-algebra has the Riesz decomposition property. This means that if $e\leq \sum\_{i=1}^n f\_i$,
where $e$ is a projection and the $f\_i$s are pairwise orthogonal projections, then there exist partial isometries $v\_1,\ldots,v\_n$ such that
$e=\sum v\_i^\*v\_i$ and $v\_iv\_i^\*\leq f\_i$ for all $i$. Now, if $f$ is an abelian projection then so is any projection Murray-von Neumann subequivalent to it. So $e\_i=v\_i^\*v\_i$ is abelian for all $i$. 


