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This is tangentially related to this recent question of mine.

Given nonprincipal ultrafilters $\mathcal{U},\mathcal{V}$ on sets $X,Y$, say $\mathcal{U}\sqsubseteq\mathcal{V}$ iff every finite signature $\mathcal{U}$-ultraproduct is isomorphic to some finite signature $\mathcal{V}$-ultraproduct; that is, for every finite signature $\Sigma$ and every $X$-sequence of $\Sigma$-structures $(\mathfrak{A}_x)_{x\in X}$ there is some $Y$-sequence of $\Sigma$-structures $(\mathfrak{B}_y)_{y\in Y}$ such that $\prod_{x\in X}\mathfrak{A}_x/\mathcal{U}\cong\prod_{y\in Y}\mathfrak{B}_y/\mathcal{V}$.

I'm broadly interested in anything that can be said about $\sqsubseteq$, but at a glance the case of nonprincipal ultrafilters on $\omega$ already seems interesting:

What can we say (in $\mathsf{ZFC}$ + large cardinals) about the preorder of nonprincipal ultrafilters on $\omega$ under $\sqsubseteq$?

Of course if $\mathsf{CH}$ holds then any two nonprincipal ultrafilters on $\omega$ have the same class of possible ultraproducts of countable structures. However, since I'm not requiring the $\mathfrak{A}/\mathfrak{B}$s to be countable, this doesn't seem to help directly.

[Removed silly remark about Rudin-Keisler reducibility.]

Noah Schweber
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  • I don't think it is true that $\sqsubseteq$ extends Rudin-Keisler reducibility. For example, if $U$ is a nonprincipal ultrafilter, then I believe $U\otimes U\sqsubseteq U$ and $U\not\sqsubseteq U\otimes U$. For countably complete ultrafilters it is consistent that the order is equal to the reverse of the Rudin-Frolik order. In fact this is a consequence of the Ultrapower Axiom. It is also consistent to have $\sqsubseteq$-equivalent normal ultrafilters that are not equal. – Gabe Goldberg Oct 31 '21 at 03:29
  • @GabeGoldberg Huh, not sure what I was thinking - certainly my claim was not justified! Can you explain why $U\not\sqsubseteq U\otimes U$ (I also don't see how UA implies that $\sqsubseteq$ coincides with the reverse Rudin-Frolik order or how it is consistent to have distinct $\sqsubseteq$-equivalent normal ultrafilters, but I suspect the explanations of those points may be more lengthy)? (I've also added a forgotten index.) – Noah Schweber Oct 31 '21 at 03:48
  • Do you mean the set (?) of ultraproducts, or the set of isomorphism classes of products? – YCor Oct 31 '21 at 07:58
  • If $M_W = \text{Ult}(V,W)$ is isomorphic to a subclass of $M_U$, then $W \sqsubseteq U$: the structures in $M_W$ in a finite signature are exactly the $W$-products up to isomorphism. But $M_{U\otimes U} \cong (M_{j_U(U)})^{M_U}$, so $U\otimes U \sqsubseteq U$. At least in the countably complete case, $U\not\sqsubseteq U\otimes U$: then $W\sqsubseteq U$ iff $M_W\subseteq M_U$. That $M_W\subseteq M_U$ implies $U\leq_{\text{RF}} W$ under UA is Corollary 5.4.21. For the normal ultrafilters, see Theorem 3.1. – Gabe Goldberg Oct 31 '21 at 16:23
  • @YCor The latter. I've edited to clarify. – Noah Schweber Oct 31 '21 at 16:27
  • Why is the “of course” statement true? – Gro-Tsen Nov 03 '21 at 19:24

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