# sequence of projections tending to a projection in semifinite von Neumann algebra

Let $M$ be a semifinite von Neumann algebra with a faithful normal semifinite trace $\tau$. For any projection $P\in M$ with $\tau(P)=\infty$, we can always find an increasing net $P_\alpha$ with $\tau(P_\alpha)<\infty$ such that $P_\alpha \uparrow P$. Since $\tau(P)=\sup_\alpha\tau(P_\alpha)$, we can find a set $\{P_n\}$ of projections from $\{P_\alpha\}$ with $\tau(P_n) \ge n$.

Then, for every $P_\alpha$, we always can find an $n$ such that $\tau(P_n)\ge n > \tau(P_\alpha)$ (i.e. $P_n \ge P_\alpha$). So, $\{P_n\}$ is a subnet of $\{P_\alpha\}$. Therefore $P_n\uparrow_n P$.

Well, so, can I say that for every projection in a semifinite von Neumann algebra, we can always find an increasing sequence of $\tau$-finite projections tending to this projection? I have never seen something like that. I think probably I am wrong. Can anyone explain a little bit for me?

Now, I have realized that the problem is the net is not totally ordered！

• Your conclusion that $\tau(P_n) \ge \tau(P_{\alpha})$ implies $P_n \ge P_{\alpha}$ (I assume you mean Murray-von Neumann subequivalence here) is only true in a factor. In a general semifinite von Neumann algebra, you have a similar but weaker result: if $\tau(P) \ge \tau(Q)$ for every faithful normal semifinite trace $\tau$, then $P \ge Q$. Thus, the result for factors follows from the fact that semifinite factors have a unique (f.,n.,s.f.) trace (up to scalar multiple). However, I think your desired result (about sequence subnets) holds if the algebra is countably decomposable. Aug 15, 2017 at 2:26
• @J.Loreaux actually, since $P_n$ is from $\{P_\alpha\}$ and $P_\alpha$ increases, if $\tau(P_n)\ge \tau(P_\alpha)$, then $P_n\ge P_\alpha$. Right? Aug 15, 2017 at 2:33
• Well, I didn't read carefully enough the first time. Your issue is simpler. You definitely don't mean Murray--von Neumann subequivalence as I at first assumed; by $P \le Q$ you truly intend $P$ is a subprojection of $Q$. And now the issue is easy to see: $\tau(P) > \tau(Q)$ does not imply $P \ge Q$. Example: take $Q$ to be any finite projection in an infinite algebra. Then $Q^\perp$ is infinite. So $\tau(Q^\perp) = \infty > \tau(Q)$, but obviously $Q^\perp \not\ge Q$. Aug 16, 2017 at 13:54
• @J.Loreaux THX A LOT!! Aug 18, 2017 at 4:16

Let $H=\ell^2(\mathbb N)$, with $\{e_n\}$ the canonical basis. Take $M=B(H)$, with the canonical trace. Let $\mathcal A$ be the family of finite subsets of $\mathbb N$, ordered by inclusion. For each $\alpha\subset\mathcal A$, define $$P_\alpha=\sum_{n\in\alpha} E_n,$$ where $E_n$ is the rank-one projection onto the span of $e_n$. It is standard that $P_\alpha\nearrow I$.
Now you want to choose your sequence. Since you only choose by the size of the trace, say we choose $$P_n=P_{\{2,4,6,\ldots,2n\}}=\sum_{k=1}^n E_{2k}.$$ Then $\tau(P_n)=n$, and the sequence satisfies your condition. But it is not a subnet: for example $E_1=P_{\{1\}}$ is not below any $P_n$. And $$P_n\nearrow\sum_{k=1}^\infty E_{2k},$$ which is of course not the identity.