Let $(M,\tau)$ be a tracial von Neumann algebra, i.e. 


* a unital subalgebra $M=M''\subset \mathbb{B}(H)$;
* a finite (faithful) trace $\tau: M\to \mathbb{C}$ (faithful means that $\tau(x^\*x)=0$ implies $x=0$; moreover $\tau(q)\leq 1$ for any projection $q\in M$).


Consider a projection $p\in M$, then its *central support* (denoted by $z(p)$) is defined as the smallest projection in the centre $Z(M)=M\cap M'$ greater than $p$. It is known that


* $z(p)=\vee\_{u\in U(M)} upu^\*$, where $U(M)$ are the unitaries of $M$ (i.e. elements $u$ such that $uu^\*=u^\*u=1$).


I read that under the above hypotheses one can prove that 
there exists a finite number of projections $\{p\_i\}\_{i=1}^n$ such that $p\_i\preceq p$, $p\_i\in M$ and $z(p)=\sum\_{i=1}^np\_i$. I think that the above characterization of $z(p)$ should be useful in the proof of this result but I have not been able to make a proof. Can anyone help me understanding why the above claim is true? Thank you in advance for the help.



Approximation of the central support
I don't think your claim is true. As an easy counter example, just take $p=e\_{1,1}\in M\_n(\mathbb{C})$. Its central support is $1$, but it is minimal, so $z(p)$ is never a sum of projections dominated by $p$.


Perhaps you meant that each $p\_i \preceq p$, in the sense that there is a partial isometry $u$ such that $uu^\* = p\_i$ and $u^\*u \leq p$. Even in this case, I don't think it's true. You can consider the finite von Neumann algebra 
$$
M = \bigoplus\_{n\geq 1} M\_n(\mathbb{C}),
$$
which has faithful normalized traces. Just take any strictly positive sequence $x=(x\_n)\in \ell^1$ with $\|x\|\_1 = 1$, and define $\tau = \sum\_{n\geq 1} x\_n \tau\_n$, where $\tau\_n$ is the normalized trace on $M\_n(\mathbb{C})$.


Now take a minimal projection $e\_n$ in each summand $M\_n(\mathbb{C})$, and let $p$ be the sum of the $e\_n$. Then any finite sum of $p\_i \preceq p$ will never equal $1=z(p)$. A finite sum of $N$ projections $p\_i \preceq p$ can only have $N$ orthogonal projections equivalent to each $e\_n$, and that won't be enough as $n\to \infty$.


Now if the center of your finite von Neumann algebra is finite dimensional, then you can find finitely many $p\_i \preceq p$ such that $\sum p\_i = z(p)$. Just note that $z(p)$ is a finite sum of minimal central projections (minimal in $Z(M)$). So for each minimal central projection $q\_n \in Z(M)$ with $q\_n \leq z(p)$, there is a non-zero $e\_n \leq p$ such that $e\_n \leq q\_n$. Now standard tricks will give you the result. (You'll need that $q\_n M$ is a finite factor for all $n$, so all projections in $q\_n M$ are comparable with respect to $\preceq$.)


