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In some calculations I am writing up, $\newcommand{\cR}{{\mathcal R}}$ I want to add - as a fairly throwaway remark - that any countable product (= $\ell^\infty$-direct sum) of matrix algebras can be embedded inside $\cR$, the hyperfinite ${\rm II}_1$. I think I have managed to hack out an explicit embedding, by realizing $\cR$ as an infinite tensor product of matrix algebras into which I embed successive blocks of my original product algebra, but the inductive construction seems both tedious and wasteful.

Assuming that I have not made a mistake, and that such product algebras do embed as von Neumann subalgebras of $\cR$, does anyone know of a reference for this, so that

(a) I don't have to tax the reader's patience by slogging through something tedious and routine

(b) I can refer the reader to a better construction than the one I currently have?

Moreover, my guess is that something slightly more general should hold: a finite Type ${\rm I}$ with separable predual should embed into $\cR$. Again, does anyone know of a reference, or a counter-example if I have made a mistake?

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up vote 11 down vote accepted

Murray and von Neumann showed that if $p \in \mathcal R$ is a non-zero projection then $p\mathcal R p \cong \mathcal R$, i.e., the fundamental group of $\mathcal R$ is all positive reals (You should be able to find this in most books that discuss the hyperfinite II$_1$ factor. Also, note that this is easy to see if $p$ has rational trace by viewing $\mathcal R$ as an infinite tensor product of matrices). Thus, if $\{ p_n \}_{n \in \mathbb N}$ is a partition of $1$ by non-zero projections then we obtain an embedding $$ \oplus_{n \in \mathbb N} \mathbb M_n(\mathbb C) \subset \oplus_{n \in \mathbb N} \mathcal R \cong \oplus_{n \in \mathbb N} p_n \mathcal R p_n \subset \mathcal R. $$

Note also that since $L^\infty([0, 1], \lambda)$ embeds into $\mathcal R$ (e.g., as tensor products of diagonal matrices), it follows that $\mathbb M_n(\mathbb C) \overline \otimes L^\infty([0, 1], \lambda) \subset \mathbb M_n(\mathbb C) \overline \otimes \mathcal R \cong \mathcal R$. Thus, since any separable abelian von Neumann algebra $A_n$ is a von Neumann subalgebra of $L^\infty([0, 1], \lambda)$ it follows from above that $\oplus_{n \in \mathbb N}( A_n \overline \otimes \mathbb M_n(\mathbb C) )$ embeds into $\mathcal R$. All separable finite type I von Neumann algebras are of this form where some of the $A_n \overline \otimes \mathbb M_n(\mathbb C)$ terms may be omitted (this can be found for instance in Takesaki, vol. 1, chapter 1). Thus all separable finite type I von Neumann algebras embed into $\mathcal R$.

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Thanks - that looks to be just the kind of argument I was hoping for. By the way, for the last part I stupidly forgot to add the word "finite", which I meant but bizarrely didn't type. As you say, obviously one can have infinite Type I which then cannot embed into any finite vN algebra. –  Yemon Choi Jan 17 '12 at 3:44
    
Ah, that makes much more sense. I'll edit my answer accordingly. –  Jesse Peterson Jan 17 '12 at 4:32
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