17
$\begingroup$

My PhD was on so called "pure" model theory, and my advisor was not very much interested in applications of model theory to algebra. Now I feel the need to fill in the gap, and I'd like to educate myself on applied model theory. One of the questions I always asked myself is about the study of groups definable (or interpretable) in some structures : it seems that model theorists are very fond of that, since a long time ago (at least since the 70's to my knowledge, and even quite recently with works about groups definable in NIP structures by Pillay). Why is it so ? I guess that the origin of all this is the fact that groups definable in ACF are precisely algebraic groups. This is indeed a nice link between model theory and algebraic geometry (by the way, does this link has brought something interesting and new to algebraic geometry, or has it always been just a slightly different point of view ?). But if it is so, why going on to study extensively groups definable on such exotic kind of theories as NIP or simple for example ? Is it only because something can be said about those groups and that groups are prestigious objects within mathematics, or are there deeper reasons ?

$\endgroup$
18
  • 3
    $\begingroup$ I'm definitely not an expert; but isn't it simply because model theory is about definable things, and so the only groups that sit inside structures that you can talk about are the definable/interpretable ones ? $\endgroup$ Feb 10, 2019 at 14:57
  • 13
    $\begingroup$ Oh you were asking about why "definable groups", I thought you were asking about why "definable groups"; my bad $\endgroup$ Feb 10, 2019 at 15:36
  • 3
    $\begingroup$ I think that groups being prestigious is a significant reason (which of course is related to good reasons to being prestigious). And also it is easier to continue in a well-studied topic (model theory of groups) than opening a new topic. $\endgroup$
    – YCor
    Feb 10, 2019 at 17:03
  • 3
    $\begingroup$ @MattF. "Definable" means "definable by a formula of first-order logic". Very very few topological manifolds can be realized as definable sets relative to the theory of the real field. On the other hand, you could add enough to the language to make all sorts of crazy topological manifolds definable, but the resulting theory would be an extremely "wild" theory that general theorems of model theory wouldn't apply to. As a middle-ground, you can look at an o-minimal expansion of the real field, e.g. by the exponential function and analytic functions restricted to compact domains... $\endgroup$ Feb 10, 2019 at 17:21
  • 4
    $\begingroup$ @MattF. The claim that every topological $4$-manifold is homeomorphic to a semi-algebraic set seems unlikely to me, but I don't know enough topology to assess it. Even if it's true, the category of topological $4$-manifolds definitely won't be equivalent to the semi-algebraic category - there just aren't enough semi-algebraically definable functions. But all that aside, I'm not sure what your point is here. How is this discussion relevant to the question? $\endgroup$ Feb 10, 2019 at 17:50

2 Answers 2

13
$\begingroup$

Here are a few reasons. These are just my perspectives - some people may disagree, and I'm sure there are good reasons to study definable groups that I'm missing.

  1. Mathematicians in general are fond of groups. There is a general theme in mathematics (with different motivations in different contexts) of developing a theory of group objects in any category of interest. For example, algebraic groups are group objects in the category of algebraic varieties, Lie groups are group objects in the category of differentiable manifolds, topological groups are group objects in the category of topological spaces, etc. Model theorists study the category of definable sets relative to a first-order theory, and group objects in this category are definable groups. [Aside: I'm confused by your comment "model theory is not mainly about definable things in my opinion". If it's not about definable things, what is it about?] Moreover, it sometimes happens that definable groups relative to a theory $T$ correspond to groups of classical interest in mathematics, e.g. as you mention in the question, definable groups relative to the theory of algebraically closed fields are essentially the same as algebraic groups.

  2. Stable groups and their generalizations. One of the most successful applications of Shelah's general stability theory has been the theory of stable groups (i.e. groups definable in stable theories). Model theorists love to generalize results, so there's a natural motivation to try to prove theorems analogous to theorems about stable groups in a wide variety of more general model theoretic contexts, or in individual (unstable) theories of interest. Extra motivation comes from the fact that the theory of stable groups, together with the kinds of connections between definable groups and groups in algebraic geometry mentioned in the previous point, led to Hrushovski's impressive applications of model theory to Mordell-Lang and related problems. This is just one example of applications of the theory of definable groups to other areas of mathematics. For others, you could look at Hrushovski's work on approximate groups, or recent applications to regularity lemmas in groups - both of these make heavy use of theorems about definable groups outside the stable context which were inspired by theorems about stable groups.

  3. Binding groups and internality. Getting more technical here, internality is a key concept in geometric stability theory. Roughly speaking, if one (type-)definable set $X$ is internal to another (type-)definable set $Y$, then the automorphism group of $X$ over $Y$ in the monster model is realizable as a definable group, called the binding group. The classic example is that an $n$-dimensional vector space $V$ over $k$ is internal to the field $k$, with binding group $\text{GL}_n(k)$. If you can understand what kinds of groups are definable in your theory, then you can understand what kinds of internality relations are possible relative to your theory, which can lead to powerful structural results.


Edit: In the comments, you ask "Why not study definable lattices, or rings, or whatever?" In fact, model theorists do study such things. Classifying definable equivalence relations (elimination of imaginaries) is extremely important - it's one of the first things you want to do when you start studying a theory. It's also very important to know whether your theory has any definable orders. And the question of the definability of a field in certain structures is the central question of the Zilber trichotomy and Zariski geometries, which has been a central motivating force in modern model theory.

My point is that definable groups get more attention than other kinds of structures (like lattices, for example) for reasons including those I outlined above. But a huge variety of instances of the general question of interpretability of certain theories in other theories come up everywhere in model theory.

$\endgroup$
13
  • 1
    $\begingroup$ About your aside: from my perspective, model theory is about invariant things. When you go outside of stability theory (well, maybe also outside of $\omega$-categorical), definable things go near-extinct fast. They are still important, but often they don't really get to the heart of things. $\endgroup$
    – tomasz
    Feb 10, 2019 at 19:13
  • $\begingroup$ @tomasz Yeah, that's a totally reasonable point of view. I think you can come at it from either side: If you're primarily interested in the invariant category, you should still care about definability, because definable / type-definable / $\bigvee$-definable / Borel-definable sets have better properties (like compactness!). On the other hand, to me, the definable category is the main object of interest, and one is naturally led to enlarge the category by invariant gadgets (this is essentially what Stone duality teaches us) in order to prove things about definable sets. $\endgroup$ Feb 10, 2019 at 19:53
  • $\begingroup$ I guess I'm just too attached to the compactness theorem to agree that model theory is primarily about invariant things. $\endgroup$ Feb 10, 2019 at 19:53
  • 1
    $\begingroup$ Isn't model theory also in a large part about classifyng models of a first order theory as Shelah did ? this program involves of course definability, but as a tool more than a mean. $\endgroup$
    – huurd
    Feb 10, 2019 at 20:18
  • 1
    $\begingroup$ @huurd: That's what I mean. About classifying models --- I don't feel that it is quite so central in modern model theory, even if it has been historically important. Classifying theories seems to be a much more popular theme, I think. $\endgroup$
    – tomasz
    Feb 11, 2019 at 11:24
5
$\begingroup$

Alex's answer is very good, but there is another reason that is worth mentioning. One of the main goals of model theory is to known when first order theories of interest are interpretable in other theories of interest and when they are not. Many structures of interest expand a group, so it is important to understand interpretable groups. For the same reason it is important to understand interpretable fields, and work on interpretable fields often builds on work on interpretable groups.

Here's a nice example: Suppose that $\mathcal{R}$ is an o-minimal expansion of a real closed field $R$. It is a theorem of Otero, Peterzil, and Pillay that any infinite field interpretable in $\mathcal{R}$ is definably isomorphic to either $R$ or $R[\sqrt{-1}]$. (O-minimal expansions of fields eliminate imaginaries, so here "interpretable" is equivalent to "definable".) It easily follows that if $\mathcal{S}$ is another o-minimal expansion of a real closed field, and $\mathcal{S}$ is interpretable in $\mathcal{R}$, then $\mathcal{S}$ is isomorphic to a reduct of $\mathcal{R}$.

This shows that there are no non-trivial interpretations between o-minimal expansions of ordered fields. So for example $(\mathbb{R},+,\cdot,\exp)$ is not interpretable in $(\mathbb{R},+,\cdot)$. The proof of their theorem uses the theory of definable groups in o-minimal structures.

$\endgroup$
10
  • $\begingroup$ Can you mention some open questions on interpretability, especially for algebraic or geometric theories? Or are most of the questions in the area on arithmetics and set theories? $\endgroup$
    – user44143
    Oct 4, 2020 at 1:34
  • $\begingroup$ @MattF Sure, although a different person might give you a very different list. It's an open question to give a characterization of theories interpretable in the theory of an infinite set with equality, in other words to characterize the theories that are interpretable in the theory of any infinite structure. This has come up on mathoverflow mathoverflow.net/questions/231791/…. $\endgroup$ Oct 4, 2020 at 1:46
  • 1
    $\begingroup$ It is also an open question to show that any infinite field interpretable in $\mathbb{Q}_p$ is definably isomorphic to a finite extension of $\mathbb{Q}_p$. This is known for definable fields, but for example $(\mathbb{Z},+)$ is interpretable in $\mathbb{Q}_p$ as the value group, but not definable. One can ask similar questions for other valued fields like $\mathbb{C}((t))$. The only case where this has been answered to my knowledge is work of Hrushovksi and Rideau and on algebraically closed valued fields. $\endgroup$ Oct 4, 2020 at 1:47
  • 1
    $\begingroup$ Thanks, those are great -- and worthy of a survey article if you want to write it ! And I like monadic second-order theories; you can see my question here: mathoverflow.net/questions/332867/… $\endgroup$
    – user44143
    Oct 4, 2020 at 2:09
  • 1
    $\begingroup$ Yeah, I don't think there has been any work on the monadic second order theory of the reals for a long time, people where probably scared off by the difficulty of the work of Shelah and Guervich. I believe it's an open question whether the monadic second order theory of $(\mathbb{R},<)$ is decidable under ZF + AD, this should be very close to the question of whether the monadic second order theory of $(\mathbb{R},<)$ is decidable when set quantifiers are restricted to Borel sets. $\endgroup$ Oct 4, 2020 at 22:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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