Ultrapower of an ultrapower of von Neumann algebras Let $M$ be a $\mathrm{II}_1$ factor, fix a ultrafilter $\omega$, we know the ultrapower $M^{\omega}$, is again $\mathrm{II}_1$ factor. The question is that what is the ultrapower of $M^{\omega}$, i.e., $(M^{\omega})^{\omega}$? Is it again isomorphic to $M^{\omega}$?
 A: This is an expanded version of my deleted answer and comment. Since the question does not include the definition of ultrapower and since it is not standard, I am using the definition from Ando and Haagerup, https://arxiv.org/abs/1212.5457 .  That is if $M$ is a vN $II_1$ factor, and $\omega$ is an non-principal ultrafilter over natural numbers, then $M^\omega$ is obtained as follows. First take  the standard ultrapower $\Pi M/\omega$. Then take the subalgebra $A$ from that algebra consisting of sequences $(x_i)$ with $tr(x_ix_i^*)$  bounded. $A$ contains the ideal $I$ consisting of all sequences $x_i$ with $\lim tr(x_ix_i^*)=0$ (all limits are $\omega$-limits). Then $M^\omega=A/I$. With that definition $M^\omega$ is the vN analog of ultralimits (NOT ultrapowers) of metric spaces. This weird definition of "ultrapower" is consistent with the definition of ulttraproducts of Banach algebras. Probably goes back to Banach. So perhaps it is not so weird after all.
As in the paper by Kramer, Shelah, Tent, Thomas [KSTT] (there is only one paper by these 4 authors, easily found in the arXiv) one can turn every vN $II_1$ factor  into an algebraic structure with countable signature ("countable" is important) trearing trace as they treat the distance function. Then $M^\omega$ is elementarily definable in $\Pi M/\omega$. Furthermore, assuming the Continuum Hypothesis $\Pi(\Pi M/\omega)/\omega$ is isomorphic to $\Pi M/\omega$ (the proof is exactly the same as in [KSTT] using the fact that an ultraproduct is always a saturated structure). Since $M^\omega$ is elementarily definable in $\Pi M/\omega$ and $(M^\omega)^\omega$ is definable by the same first order formulas in $\Pi(\Pi M/\omega)/\omega$, we get the result. If we do not assume CH, I do not know the answer. But I refer to [KSTT] again, it may be that the answer is "no" or even "totally no" (meaning that the isomorphism almost never occurs). 
