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Let $\omega$ be a free ultrafilter on the natural numbers and $R$ be the hyperfinite $II_1$-factor (the definition of $R$ is recalled in the comments).

Question: Does there exist another free ultrafilter $\omega'$ such that $\left(R^{\omega}\right)^{\omega'}$ embeds into $R^\omega$?

Motivation. Some time ago, Jesse Peterson, after seeing the proof that the free group on uncountably many generators is hyperlinear (in the sense that its von Neumann algebra verifies Connes' embedding conjecture) for some particular ultrafilter, asked whether this is true for every ultrafilter. This question arose again today, asked by Martino Lupini. My question is a sort of more general statement about ultraproducts which would imply in particular the answer to Jesse's and Martino's question.

One way to reformulate the question is in terms of product of ultrafilters. Let $\omega$ and $\omega'$ be two free ultrafilters on the natural numbers. The product $\omega\times\omega'$ is the free ultrafilter on $\mathbb N\times\mathbb N$ defined by

$$ A\in\omega\times\omega' \qquad\text{iff}\qquad \lbrace k\in\mathbb N:\lbrace n\in\mathbb N:(k,n)\in A\rbrace\in\omega'\rbrace\in\omega. $$

It's standard to show that $\left(R^\omega\right)^{\omega'}\cong R^{\omega'\times\omega}$. Therefore

Question (reformulated): For every $\omega$, does there exist $\omega'$ such that $R^{\omega'\times\omega}$ embeds into $R^\omega$?

Update 12/06/2013: I was thinking that a weaker statement would already be enough.

Question 2: Fix a free ultrafilter $\omega$ on the natural numbers. Do there exist two free ultrafilters $\omega'$ and $\omega''$ such that $\left(R^{\omega'}\right)^{\omega''}$ embeds into $R^{\omega}$?

Thanks in advance for any help.

Valerio

P.s. Use of Continuum Hypothesis is forbidden! :)

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  • $\begingroup$ What you call the tensor product of ultrafilters is usually just called the product and is denoted $\omega\times\omega'$. And I think you mean to write $(R^{\omega'})^\omega\cong R^{\omega\otimes\omega'}$, which is a very general fact for any kind of structure $R$; that is, you seem to be iterating the ultrapowers in the wrong order. Could you clarify whether this issue affects your question? $\endgroup$
    – Joel David Hamkins
    Jun 11, 2013 at 23:14
  • $\begingroup$ I got the terminology from this paper dm.unipi.it/~dinasso/papers/19.pdf published in PAMS... Yes, you are right, the iteration is in the other way. I'm fixing it straight away. $\endgroup$
    – Valerio Capraro
    Jun 11, 2013 at 23:42
  • $\begingroup$ I see, in that paper, the authors write $\mu\times\nu$ for the filter generated by the rectangles, but I think this is unusual terminology (I have seen this product called the box product or the rectangular product). They do call $\mu\otimes\nu$ the "product ultrafilter (sometimes called tensor product)", and I think using product (and $\times$) for this is more common. But I suppose it is fine when one is clear about the meaning. Meanwhile, can you tell us what is the hyperfinite $II_1$ factor? $\endgroup$
    – Joel David Hamkins
    Jun 12, 2013 at 0:15
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    $\begingroup$ @Valerio: For the Definition 2.) Don't forget to take the closure. $\endgroup$
    – Owen Sizemore
    Jun 12, 2013 at 3:03
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    $\begingroup$ Also, although $(R^{\omega'})^\omega\cong R^{\omega\otimes\omega'}$ is true very geneerally for ultrapowers, the "ultraproduct" used by Banach space folks isn't the usual ultraproduct. Rather, it's obtained from the usual ultraproduct by discarding all elements whose norms are infinite and identifying any two elements whose distance is infinitesimal. So it's again a Banach space over the standard reals. I haven't checked the effect (if any) of this on $(R^{\omega'})^\omega\cong R^{\omega\otimes\omega'}$. $\endgroup$
    – Andreas Blass
    Jun 14, 2013 at 14:34

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