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Let ${P_i}$ be a family of mutually orthogonal projections of a semifinite von Neumann algebra $M$ and $\tau$ be the semifinite faithful normal trace on $M$. If $\tau(\sum P_i) =\infty$, can we find a sequence of projections $\{Q_n\}$ from $\{P_i\}$ such that $\tau(\sum Q_n) =\infty$?

My idea is:

Without loss of generality, we may assume that $\tau(P_i)<\infty$ for every $i$. Define $\alpha_i := \tau(\sum_{j\le i} P_j)$. Then, since $\tau$ is normal, we have $\sup_i\alpha_i = \sup_i\tau(\sum_{j\le i} P_j)=\tau( \sup_i \sum_{j\le i} P_j) =\tau( \sum P_j) =\infty $. Since $\tau(P_i) >0$ for every $i$ ($\tau$ is faithful), the elements in $\{\alpha_i <\infty\}$ are distinct and therefore $\{\alpha_i <\infty\}$ consists of countably many elements. But I don't know how to show the supremum of $\{\alpha_i <\infty\}$ is infty.

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2 Answers 2

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Since $\tau$ is normal, we have $\tau(\sum P_i)=\sum \tau(P_i)$. So this becomes a question about numbers.

If $\{\alpha_j\}\subset[0,\infty)$ satisfies $\sum_j\alpha_j=\infty$, this means by definition that $$ \sup\{\sum_{j\in F}\alpha_j:\ F \text{ finite}\}=\infty $$ So we can find finite subsets $F_n$ such that $\sum_{F_n}\alpha\geq n$. Let $E=\bigcup_n F_n$. Then $E$ is countable and, for each $n$, $$ \sum_E\alpha_j\geq\sum_{F_n}\alpha_j\geq n, $$ so $\sum_E\alpha_j=\infty$.

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  • $\begingroup$ Thank you for your answer. Your answer is simpler. $\endgroup$
    – user92646
    Mar 30, 2017 at 4:15
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Assume by contradiction that there is no such a subset $\{P_n\}_{n=1}^\infty$ of $\{P_i\}_i$ consisting of countably many projections such that $\tau(\sum_{n=1}^\infty P_n )=\infty$. Then, there is a finite constant $C>0$ such that, for every subset $\{P_n\}_{n=1}^\infty$ of $\{P_i\}_i$ consisting of countably many projections, $\tau(\vee_{n=1}^\infty P_n) \le C$ (Otherwise, for every $k\in \mathbb{N}$, there is a subset $S_k:=\{P_{kn}\}_{n=1}^\infty$ such that $\tau(\vee _{n=1}^\infty P_{kn})>k $. Let $S=\vee_{k=1}^\infty S_k$. Then, $S$ consists of countably many projections from $\{P_i\}_i$ and $\tau(\vee_{P\in S} P ) \ge \tau(\vee_{P\in S_k } P ) >k$ for every $k$. That is, $S$ is the set we need.).

Define $\bar{S}:=\{\mbox{$S$ consisting of countably many projections from $\{P_i\}_i$} \}$ and $ M:= \{ \tau(\vee_{P\in S} P )\mid S\in \bar{S} \} $. Then, every element from $M$ has a upper bound $C$ and therefore the supremum $\bar{C}$ of $M$ is less than $C<\infty$. For every $n$, we can find an $S_n\in \bar{S}$ such that $\tau(\vee_{P\in S_n} P ) \ge \bar{C} -\frac{1}{n}$. Then, $\vee_{n=1} ^\infty S_n \in \bar{S}$ with $\tau(\vee_{P\in \vee_{n=1} ^\infty S_n } P ) \ge \tau(\vee_{P\in S_n } P ) \ge \bar{C}-\frac1n $ for every $n$. Hence, $\tau(\vee_{P\in \vee_{n=1} ^\infty S_n } P ) \ge \bar{C}$. However, since $\tau$ is faithful, $\tau(P_i)>0$ for every $P_i \in \{P_i\}$. We only need to take an $Q \in \{P_i\} \setminus \vee_{n=1} ^\infty S_n $ (If it is empty, then $\tau(\sum P_i) =C <\infty$, which is a contradiction.). Then, $\{Q\} \vee ( \vee_{n=1} ^\infty S_n) \in \bar{S} $ consists of countably many projections with $ \tau(\{Q\} \vee (\vee_{n=1} ^\infty S_n ) ) = \tau(Q) +\tau( \vee_{n=1} ^\infty S_n ) > \bar{C} $, which is a contradiction to that $\bar{C}<\infty $ is the supremum of $M$.

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