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Let $M$ be a semifinite von Neumann algebra equipped with a semifinite normal faithful trace $\rho$. Let $p\in M$ be a nonzero projection in $M$.

If $\rho(p)=\infty$, can we deduce that $p$ is not a finite projection?

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If $ρ(p)=∞$, can we deduce that $p$ is not a finite projection?

No, this can fail. For instance let $M=\ell^\infty(\mathbb N)$ with the fns trace $$ \rho(x)=\sum_n\frac1n\,x_n. $$ We have $\rho(1)=\infty$, while every projection is finite since the algebra is abelian.

On a factor, on the other hand, the assertion is true. If $p$ is finite, then the semifinitness guarantees that there exists $q\leq p$ with $\rho(q)<\infty$. Let $\{q_1,\ldots,q_n\}$ be a maximal family of pairwise orthogonal projections such that $q_k\sim q$ and $q_k\leq p$ for all $k $. The family is necessarily finite for otherwise $p$ would be infinite. We have that $p-\sum_kq_k\prec q$. Then $$ \rho(p)=\rho\bigg(p-\sum_kq_k\bigg)+\sum_k\rho(q_k)\leq(n+1)\rho(q)<\infty. $$

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  • $\begingroup$ I mean the projection is nonzero. $\endgroup$ Sep 7, 2023 at 16:21
  • $\begingroup$ That depends on how you define "trace". $\endgroup$ Sep 7, 2023 at 17:36
  • $\begingroup$ That's exactly what my answer says. $\endgroup$ Sep 8, 2023 at 6:00
  • $\begingroup$ Sorry, I don't know what you mean "the other definition of trace". $\endgroup$ Sep 8, 2023 at 15:41

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