Let $R$ be the hyperfinite $II_1$-factor. It is well-known that it is the smallest $II_1$-factor, in the sense that every $II_1$-factor contains a copy of $R$.
Now, let $\omega$ be a free ultrafilter on the natural numbers and construct the tracial ultrapower $R^\omega$. It is well-known, that this is a quite big factor. Indeed
1) Its predual is not separable. Even, just to make a bit of self-advertisement, its unitary group contains an orthogonal copy of the free group on uncountable many generators (see Capraro V. and Paunescu L. Product of ultrafilters and applications to Connes Embedding Problem, to appear in J. Oper. Theory)
2) A well-known conjecture, Connes' Embedding Conjecture, states that every $II_1$-factor with separable predual should be embeddable into $R^\omega$. Namely, all known $II_1$-factors with separable predual are known to embed into $R^\omega$ (or we do not know if they embed)
On the other hand, $R^\omega$ is constructed starting from the smallest $II_1$-factor using a quite natural construction that should not increase to much. So my question is:
** Question: ** Is it true that any $II_1$-factor with non-separable predual contains a copy of (some) $R^\omega$?
Actually I don't think this is true, since for instance it is not clear how $R^\omega$ could be contained in the group factor of the free group on uncountable many generators. In some sense, the latter has very few relations and $R^\omega$ has lots of relations, but I don't understand quite well how this difference can be reflected in the group factors.
My interest in this problem comes from the following result by Taka Ozawa. He showed that there is indeed no way to find such a universal factor among the ones with separable predual. See N.Ozawa,* There is no separable universal $ II_1$-factor*, Proc. Amer. Math. Soc., vol. 132, iss. 2, pp. 487-490, 2004.
So, a way to reformulate my question is: Is Connes' embedding conjecture the best conjecture one can formulate?
Maybe, it's not the best conjecture, but it's the best one can hope for.
Thanks in advance for any comment.