When I took an introductory model theory course a few years ago, the instructor did not talk about ultrapowers, which seems fair to me, given contemporary model theorists' relative indifference to ultraproducts in general.

On the other hand, ultrapowers ares still sometimes used, e.g., to introduce to students applications of model theory, such as infinitesimals or modal logic (for the latter, see, e.g., Blackburn, de Rijke, Venema's Modal Logic). As far as I know, for these purposes you don't really need ultrapowers; you just need sufficiently saturated elementary extensions, which can easily be obtained by repeated use of compactness.

(A seemingly more recent applications are in operator algebra, but I'm not sure if you really need ultrapowers as opposed to any sufficiently saturated elementary extensions.)

I would imagine that whatever property you want in your elementary extension (e.g., $\kappa$-saturated or omitting a certain type), it is impossible or harder to achieve that with ultrapowers since the latter involves intricate infinite combinatorics.

So my question is: are there situations in which ultarpowers are more desirable than other kind of elementary extensions?

(Note: I understand that there are mainly set-theoretic/combinatorial interests in ultrapowers. I also vaguely understand the uses of ultraproducts; my question is specifically in regard to ultrapowers.)

  • $\begingroup$ I don't know if this is really compelling, but if $G$ is the linear graph on $\mathbb{N}$, and $\mathcal{U}$ is a nonprincipal ultrafilter on $\mathbb{N}$, then $G^\mathbb{N}/\mathcal{U} \equiv G$ and the lhs is not connected, while the rhs is connected. This shows that there is no theory of disconnected graphs or of connected graphs. It's really easier to prove it that way than with elementary extensions (I think) $\endgroup$ – Maxime Ramzi Oct 29 '17 at 21:55

One pleasant feature of ultrapowers is captured in the slogan "ultraproducts commute with reducts". That is, given two languages $L\subseteq L'$, a collection $(M_i)_{i\in I}$ of $L'$-structures, and an ultrafilter $U$ on $I$, we have $(\prod_I M_i/U)|_L = \prod_I (M_i|_L)/U$ as $L$-structures.

Reformulating this, suppose $M$ is an $L$-structure, and $N = M^I/U$ is an ultrapower of $M$ (in particular, $M\prec_L N$). If we expand $M$ to an $L'$-structure $M'$ by adding new relations or functions, then $N$ has a canonical expansion to an $L'$-structure $N' = (M')^I/U$, and we still have $M'\prec_{L'} N'$.

Moreover, this expansion is totally natural. For example, looking at $\mathbb{R}$ and an ultrapower $\mathbb{R}^*$ (the hyperreals), your favorite function $f\colon \mathbb{R}\to \mathbb{R}$ induces a function $f^*\colon \mathbb{R}^*\to \mathbb{R}^*$, defined by $f^*([a_i]) = [f(a_i)]$, where $[a_i]$ is the equivalence class of the sequence $(a_i)_{i\in I}$ under $U$-equivalence. And we have $(\mathbb{R},f)\prec (\mathbb{R}^*,f^*)$. The same goes for relations: if we add a predicate $Z$ to $\mathbb{R}$ picking out the integers, we have $(\mathbb{R},Z)\prec (\mathbb{R}^*,Z^*)$, where $[a_i]$ satisfies $Z^*$ just in case $U$-almost-all of the $a_i$ are integers.

Yet another way of putting this is that the canonical embedding of $M$ in its ultrapower $N$ doesn't just preserve first-order formulas, but also existential second order formulas: For any tuple $\overline{a}$ from $M$ and new relation symbols $R_1,\dots,R_k$, if $M\models \exists R_1,\dots,R_k\,\varphi(\overline{a})$, where $\varphi$ is a first-order formula in the language $L' = L\cup \{R_1,\dots,R_k\}$, then $N\models \exists R_1,\dots,R_k\,\varphi(\overline{a})$. We get this by expanding $M$ to $L'$ by relations $R_1,\dots,R_k$ witnessing the second-order existential quantifiers, expanding $N$ into an $L'$-structure in the canonical way, and then using $M\prec_{L'} N$. As Noah points out in the comments, the converse does not necessarily hold (the embedding $M\to N$ does not necessarily reflect truth of existential second-order formulas).

I believe that any saturated elementary extension of $M$ (i.e. saturated in its own cardinality) will share this property, and I also believe that it's possible for $M$ to have a $\kappa$-saturated elementary extension of size $\lambda>\kappa$ which fails this property. Unfortunately, I don't have proofs of these facts in mind (can anyone supply proofs/counterexamples?). But in any case, the expansion of an ultrapower is canonical and easy to explain, which is definitely an advantage (especially for people who want to know about nonstandard analysis but don't know much model theory).

  • $\begingroup$ +1. Quick addendum: although taking ultraproducts preserves $\Sigma^1_1$ formulas, it does not reflect them. E.g. there is a $\Sigma^1_1$ sentence in the language $\{<\}$ expressing ill-foundedness, but the ultrapower of a well-founded partial order can be ill-founded. We can ask whether there are ultrafilters with the property that the ultrapower maps they induce also reflect $\Sigma^1_1$ sentences. This leads straight into large cardinal territory; for example, $\mathcal{U}$ is a countably complete ultrafilter iff its ultrapower map is $\Sigma^1_1$-reflecting on countable structures. $\endgroup$ – Noah Schweber Oct 31 '17 at 20:25
  • $\begingroup$ Oh, good point! So what I wrote was wrong. Fixing it now. $\endgroup$ – Alex Kruckman Oct 31 '17 at 20:27

You pose a very nice idea in your question. You write,

"I would imagine that whatever property you want in your elementary extension (e.g., $\kappa$-saturated or omitting a certain type), is impossible or harder to achieve that with ultrapowers since the latter involves intricate infinite combinatorics."

There is an (partial) ordering on first order theories which exploits this very idea. This is known as Keisler's Order and the general idea goes something like this: If it is very hard (i.e. one needs very strong ultrafilters) to saturate models of a particular theory, then the theory is toward the top of Keisler order. If it is very easy to saturate models of this theory (using ultrapowers), then this theory is towards the bottom of the order.

Keisler's order illuminates important dividing lines. For instance, the first two classes in Keisler's Order is the class of stable theories. Research on Keisler's Order and simple theories is still ongoing, but there are already countably many classes living here (some which correspond to "amalgamation problems"). NIP unstable theories are in the maximal class.

  • $\begingroup$ This sounds very interesting. Do you have any references for further details? $\endgroup$ – Nagase Oct 30 '17 at 23:10
  • 1
    $\begingroup$ @Nagase: I believe it's in bad interest to reference one's self, but I wrote an introduction to the Keisler Order for people who are interested in it. Here is a link. There are some errors toward the end that I still need to fix... $\endgroup$ – Kyle Gannon Oct 30 '17 at 23:12
  • $\begingroup$ That was extremely helpful, thanks for sharing! $\endgroup$ – Nagase Oct 30 '17 at 23:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.