For the most part, we will be following Lou van den Dries’ book Tame Topology and O-minimal Structures for this seminar. We will supplement with material from reseach papers. If there are specific theorems about o-minimality you would like to see this term, let me know. Some possible topics include
- Speissegger’s theorem on Pfaffian closures of o-minimal structures
- The Peterzil-Starchenko o-minimal trichotomy
- The Pila-Wilkie counting theorem
- Proofs of o-minimality of specific structures (e.g. $\mathbb{R}_{an,\exp}$ or the expansion by Gevrey functions)
- The Peterzil-Starchenko o-minimal Chow theorem
Generally, we will work with structures in the usual logical sense. However, van den Dries offers an alternative definition closer in spirit to other mathematical constructions and borrowing from cylindrical algebra. Even though we will not be using his convention, it is useful to see how it relates to ours, especially as many mathematicians working in o-minimality use a van den Dries style definition of structure.
Definition: A structure $(R,\mathcal{S})$ consists of a nonempty set $R$ and a sequence $\mathcal{S} = (\mathcal{S}_n)_{n=0}^\infty$ such that
- each $\mathcal{S}_n$ is a Boolean subalgebra of the power set $\mathcal{P}(R^n)$,
- if $A \in \mathcal{S}_n$, then $R \times A \in \mathcal{S}_{n+1}$ and $A \times R \in \mathcal{S}_{n+1}$,
- for each $n$ and pair of numbers $i$ and $j$ with $1 \leq i, j \leq n$ the multidiagonal $\Delta_{i,j}^n := \{ (x_1, \ldots, x_n) \in R^n : x_i = x_j \}$ belongs to $\mathcal{S}_n$, and
- if $\varpi_{n}:R^{n+1} \to R^n$ is the projection $(x_1, \ldots, x_{n+1}) \mapsto (x_1, \ldots, x_n)$ and $A \in \mathcal{S}_{n+1}$, then $\varpi_{n}(A) \in \mathcal{S}_n$.
Proposition: For a structure $(R,\mathcal{S})$ there is a first-order language $\mathcal{L}$ and an $\mathcal{L}$-structure $\mathfrak{R}$ with universe $R$ so that the definable sets in $\mathfrak{R}$ are precisely the sets in $\mathcal{S}$.
Proof: Let $\mathcal{L}$ be the relational language that has an $n$-ary relation symbol $R_X$ for each $X \in \mathcal{S}_n$. Interpret $R_X^{\mathfrak{R}} := X \subseteq R^n$. Clearly, every set in $\mathcal{S}$ is definable in $\mathfrak{R}$. For the other inclusion we need the following claim.
Claim: For natural numbers $m$ and $n$, sequences $i_1, \ldots, i_n$ with $1 \leq i_1, \ldots, i_n \leq m$, and $X \in \mathcal{S}_n$, if we set $Y := \{ (a_1, \ldots, a_m) \in R^m : (a_{i_1}, \ldots, a_{i_n}) \in X \}$, then $Y \in \mathcal{S}_m$.
Proof of claim: We may express $Y$ as $Y = \varpi_m \varpi_{m+1} \cdots \varpi_{m+n-1} \left( (R^m \times X) \cap \bigcap_{j=1}^n \Delta_{i_j,m+j}^{n+m} \right)$ $\maltese$
With the claim in place, we may argue induction on the complexity of $\phi(x_1, \ldots, x_m) \in \mathcal{L}(x_1, \ldots, x_m)$ that $\phi(\mathfrak{R}) \in \mathcal{S}_m$. For $\phi$ atomic, $\phi$ is given by either $x_i = x_j$, which defines $\Delta_{i,j}^m$, or $R_X(x_{i_1}, \ldots, x_{i_n})$ fo some $X \in \mathcal{S}_n$ and sequence $i_1, \ldots, i_n$, in which $\phi$ defines the set $Y$ of the claim, which belongs to $\mathcal{S}_m$. In the inductive case that $\phi$ is a Boolean combination of simpler formulae we use the condition that $\mathcal{S}_m$ is a Boolean algebra. In the case that $\phi = \exists x_j \psi$ where $\psi$ is simpler, we first use the claim to reduce to the case that $j = n+1$, and then apply the final closure condion on $\mathcal{S}$ to see that the set defined by $\phi$, which is the projection of the set defined by $\psi$, belongs to $\mathcal{S}$. $\Box$.
Definition: An o-minimal structure is an $\mathcal{L}$-structure $(R,<,\ldots)$ for which $(R,<)$ is a dense linear order without endpoints and for which every $\mathcal{L}_R$-definable subset of $R$ is a finite union of points and intervals. Here an interval has the form $(a,b) := \{ x \in R : a < x < b \}$ where $a \in R \cup \{ -\infty \}$ and $b \in R \cup \{ \infty \}$. We give $R \cup \{ \pm \infty \}$ an order extending $<$ on $R$ by defining $-\infty < R < \infty$.
Remark: In the definition of o-minimality, it is important that $a$ and $b$ come from $R \cup \{ -\infty, \infty \}$. The structure $\mathfrak{Q} := (\mathfrak{Q},<, P)$ where $P^\mathfrak{Q} := (-\infty,\sqrt{2}) \cap \mathbb{Q}$ is not o-minimal.
Remark: Notice that we allow parameters when defining subsets of $R$. From now on, when we use the phrase “definable” without qualification we mean “definable with parameters”.
We will consider examples of o-minimal structures later in the term. From the quantifier elimination theorem for real closed fields, we know that if $(R,<,+,\cdot,0,1)$ is a real closed field, then it is an o-minimal structure. It follows that any reduct of a real closed field which still has the ordering is also o-minimal. We will offer a different proof of the o-minimality of real closed fields next week.
Lemma 1: O-minimal structures are definably complete: if $(R,<,\ldots)$ is o-minimal and $A \subseteq R$ is a nonempty definable subset, then $A$ has a supremum in $R \cup \{ \infty \}$.
Proof: By o-minimality, we may write $A = \bigcup_{i=1}^n \{ a_i \} \cup \bigcup_{j=1}^m (b_j,c_j)$ for some $a_i \in R$, $b_j \in R \cup \{ -\infty \}$, and $c_j \in R \cup \{ \infty \}$ with $b_j < c_j$ for $1 \leq i \leq n$ and $1 \leq j \leq m$. Set $s := \max (\{ a_i : i \leq n \} \cup \{c_j : j \leq m \})$, which exists as every finite nonempty subset of a totally ordered set has a maximum. It is easy to check that $s = \sup A$. $\Box$
Theorem: If $(G,<,\cdot,\ldots)$ is an o-minimal ordered group, then it is a divisible, ordered abelian group.
For the next few results, we fix $(G,<,\cdot,\ldots)$ an o-minimal ordered group.
Lemma 2: If $H \leq G$ is a definable subgroup, then $H$ is convex.
Proof: If $H$ is not convex, then there are elements $g \in G \smallsetminus H$ and $h \in H$ with $1 < g < h$. Multiplying by $h^n > 1$ for $n \in \mathbb{Z}_+$, we obtain a sequence of inequalities $1 < g < h < gh < h^2 < gh^2 < h^3 < \cdots < gh^n < h^{n+1} < \cdots$. For each $n \in \mathbb{N}$ we have $h^n \in H$ and $g h^n \in gH \subseteq G \smallsetminus H$. Hence, this sequence alternates between lying in $H$ and in its complemement. Such a situation violates o-minimality: as $H$ is definable, it is a finite union of points of intervals. There would have to be one such interval $(a,b)$ in this presentation for which $h^n \in (a,b)$ for all sufficiently large $n$, but $a <h^n < g h^n < h^{n+1} < b$ would yield $g h^n \in (a,b) \subseteq H$ which is not true. $\Box$
Lemma 3: If $H \leq G$ is a definable subgroup, then $H = \{ 1 \}$ or $H = G$.
Proof: By Lemma 1, $H$ is convex. By Lemma 2, $s := \sup H \in G \cup \{ \infty \}$. If $s = 1$, then $H = \{ 1 \}$ and if $s = \infty$, then $H = G$. Suppose that $1 < s < \infty$. We have that $s \notin H$ for if $s \in H$, then $s^2 \in H$ and $s < s^2$ showing that $s$ is not an upper bound of $H$. Note that because $1 < s$, we have that $H \neq \{ 1 \}$. So, there is some $h \in H$ with $h < 1$. Then $hs < s$ and both $s$ and $hs$ belong to $Hs$ which is disjoint from $H$. As $H$ is convex, there can be no element of $H$ between $hs$ and $s$. But then $hs$ is also an upper bound for $H$, showing that $s$ is not the supremum. $\Box$
Let us return to the proof of the theorem.
Proof of theorem: For any $g \in G$, the centralizer $C_G(g) := \{ x \in G : gx = xg \}$ is a definable subgroup of $G$. When $g = 1$, as is the case in every group, $C_G(g) = G$. When $g \neq 1$, then $g \in C_G(g) \smallsetminus \{ 1 \}$. Thus, by Lemma 3, $C_G(g) = G$. That is, every element of $G$ belongs to its center so that $G$ is abelian.
We will now write the group structure on $G$ additively. Fix some positive element $g \in G$. For each positive integers $n$, the set $nG = \{ nx : x \in G \} = \{ y \in G : (\exists x) nx = y \}$ is a definable subgroup of $G$ containing the nontrivial element $ng$. Thus, by Lemma 3 again, $nG = G$. As this identity holds for every $n$, we see that $G$ is divisible. $\Box$