Let $M$ be a (not necessarily separable) von Neumann algebra. I am interested to understand the non separable case.

In the book [1, page 49], the authors says that $M$ is hyperfinite if we can write $M=\overline{\cup_\alpha M_\alpha}^{w^*}$ where $(M_\alpha)$ is a net of finite dimensional subalgebras directed by inclusion.

1) Is it equivalent to the definition of [2, page 97] ? This definition says that $M$ is approximately finite dimensional if for any $x_1,x_2,\ldots,x_n \in M$ and any $\sigma$-strong* neighborhood $V$ of $0$ in $M$ there exists a finite dimensional $*$-subalgebra $N$ such that $x_j \in N + V$ for any $j = 1,2,\ldots,n$.

2) In these definitions, is it to be assumed that $1_M \in M_\alpha$ and $1_M \in N$ (i.e. the subalgebras are unital subalgebras)? Is there any difference ?

[1] Sinclair, Smith, Finite von Neumann algebras and MASAS

[2] Takesaki, Theory of operator algebras 3

  • "Hyperfinite" is usually applied only to II$_1$-factors (and then there is a single one). Saying that $B(H)$ is hyperfinite, or that a type III-factor is hyperfinite sounds weird, since these algebras are properly infinite. So AFD is preferred.

  • If $M=\overline{\cup_\alpha M_\alpha}^{w^*}$ and $x_1,\ldots,x_n\in M$ and $V$ is a neighbourhood of zero, there is an alpha such that $x_1,\ldots,x_n\in M_\alpha+V$.

  • Conversely, if $M$ satisfies Takesaki's definition, let $\mathcal F$ be the net of finite subsets of $M_*$, ordered by inclusion. Note that a base of neighbourhoods of zero is given by $\{V_F\}_{F\in\mathcal F}$, where $$ V_F=\{x\in M:\ |f(x)|+|f(x^*)|<1:\ f\in F\}. $$ Let $\mathcal E$ be the net of finite subsets of $M$, ordered by inclusion. Form a net $$\mathcal A=\{F\times E:\ F\in \mathcal F,\ E\in\mathcal E\},$$ again ordered by inclusion. For any $\alpha=F\times E\in \mathcal A$, applying Takesaki's definition there exists a finite-dimensional $*$-subalgebra $N_\alpha$ of $M$ such that $E\subset N_\alpha+V_F$. If we let $M_\alpha=C^*(N_\alpha)$, this algebra is finite-dimensional because $N_\alpha$ is. If $\alpha\leq\beta$ then $M_\alpha\subset M_\beta$, because $E_\alpha\subset E_\beta$. Given $x\in M$ and a neighbourhood $V$ of zero, there exist $f_1,\ldots,f_n$ such that $0\in V_F\subset V$, where $F=\{f_1,\ldots,f_n\}$. By construction of the net, $x\in M_\alpha+V_\alpha$, where $\alpha=(F,\{x\})$. That is, there exists $x_0\in M_\alpha$ such that $x-x_0\in V_\alpha\subset V$. That is, $$ M=\overline{\bigcup_\alpha M_\alpha}^{\ \sigma-\text{strong}^*}. $$

  • The distinction between $\sigma$-weak and $\sigma$-strong$^*$ above is irrelevant, because the union $M_0=\bigcup_\alpha M_\alpha$ is a $*$-algebra (as the net is increasing). By [Takeksaki I, Theorem II.2.6] the $\sigma$-weak and $\sigma$-strong$^*$ continuous functionals are the same; then this result guarantees that the closures of convex sets are the same.

  • The unit of the subalgebras here is irrelevant. They can share the same unit as $M$ or not, it doesn't change the results.

  • $\begingroup$ Thank you for this beautiful answer. Why the unit of the subalgebras is here is irrelevant ? it is not clear for me. $\endgroup$ – Zouba Apr 10 '17 at 20:25
  • $\begingroup$ Because it is enough that the units converge to the unit of $M$. But they could be all different, and nothing changes in the arguments. $\endgroup$ – Martin Argerami Apr 11 '17 at 3:15

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