Functional Analysis – Semi-Finite Trace on a von Neumann Algebra: Equivalent Definitions

functional-analysisoperator-algebrasoperator-theorytracevon-neumann-algebras

Let $(N,\tau)$ be a semi-finite von Neumann algebra. This means that $\tau$ is a normal, faithful and semi-finite trace.

Normality means that $\tau(x) = \sup_i \tau(x_i)$ if $x \in N_+$ is the limit of an increasing net $x_i$ in $N_+$.

With $\tau$ one associates the following sets:

$$
N_\tau^+ := \{x \in N_+ : \tau(x) < \infty \}
$$
and
$$
\mathscr L_\tau^1(N) := \{ x \in N : \tau(\lvert x \rvert) < \infty \}.
$$
The latter set is in fact a $*$-ideal and equals the complex linear span of $N_\tau^+$.

I read a few definitions of semi-finiteness and I wonder if these are all equivalent. Lets recall a few definitions:

  1. Every $0 \neq x \in N_+$ majorizes some $0 \neq y \in N_\tau^+$. (Takesaki: Theory of Operator Algebras I, p. 309)
  2. $\tau(x) = \sup \{ \tau(y) : y \leq x, \ y \in N_\tau^+ \}$ for $x \in N_+$. (Dixmier: von Neumann algebras p. 93)
  3. The $\sigma$-weak (=ultra weak) closure of $N_\tau^+$ equals $N_+$. (A. M. Bimkchentaev: On a Property of $L^p$ Spaces on Semifinite von Neumann Algebras)
  4. For all $x \in N_+$ there exists an increasing net $x_\alpha$ in $N_\tau^+$ with strong (SOT) limit $x$. (Edward Nelson: Notes on Non-commutative Integration)

I hope that all these definitions are equivalent for a normal tracial weight $\tau$ on a von Neumann algebra. By tracial I mean that $\tau(x^*x) = \tau(xx^*)$ for all $x \in N$. I am also wondering if the normality may be dropped to see the equivalence.

Maybe the following result is helpful (Haagerup): A weight on a von Neumann algebra is normal iff it is ultra weakly lower semi-continuous.

Best Answer

$1\iff4$ : Assume $1$. Given $x\in N_+$, there exists nonzero $y\in N_\tau^+$ with $y\leq x$. Now use Zorn to find a maximal ordered family $\{y_j\}\subset N_\tau^+$ with $y_j\leq x$ for all $j$. As the net is bounded, it has a sup, say $y=\lim_{sot}y_j$. Then $y=x$, because otherwise a nonzero element of $N+\tau^+$ below $y-x$ contradicts the maximality. The converse is trivial, just take some $x_\alpha$.

$3\iff 4$ : note that $4$ implies $3$ directly. For the converse, note that $N_\tau^+$ is convex, and so its wot closure agrees with its sot closure. And a $\sigma$-weak limit is in particular a wot limit, so $3$ implies $4$.

$4\implies 2$ : this follows directly by definition of normality.

$2\implies 1$ : if $x\in N_+$ does not majorize nonzero elements of $N_\tau^+$, then by $2$ we have $\tau(x)=0$. Then faithfulness implies $x=0$.

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