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! :)