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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.

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1 Answer 1

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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 $\preceq$.)

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  • $\begingroup$ I am sorry, I used the wrong symbol: I should have used $\preceq$ instead of $\leq$. I'll correct my question. The property I was referring in my question was written in the paper AN INVITATION TO VON NEUMANN ALGEBRAS, Houdayer (page 18, fourth line from the bottom of the page) and if I understood correctly there aren't assumptions on the centre, so I don't know if I missed something in your answer and/or in the paper. Thanks for the help. $\endgroup$
    – John N.
    Nov 15, 2015 at 19:42
  • $\begingroup$ In this paper, Houdayer is approximating the central support, so his finite sum may not actually be equal to the central support. In my second example, it's possible to approximate the central support by a finite sum to whatever precision you like (in $\|\cdot\|_2$), but you will not be able get the central support on the nose with a finite sum. $\endgroup$ Nov 15, 2015 at 19:56
  • $\begingroup$ Thank you very much for the help. I didn't understand what Houdayer wrote. Thanks again for your time. $\endgroup$
    – John N.
    Nov 15, 2015 at 20:00

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