Approximation of the central support

Question

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.

Answer 1

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 $\preceq$.)