indiscernible sequence?
Morley sequence?
Ramsey theorem?
Erdos-Rado theorem?
Ehrenfeucht-Fraïssé games (back-and-forth games)
Geometric stability theory is the principal part of the branch of model theory called geometric model theory?. It was introduced in works of Boris Zilber, Gregory Cherlin, Ehud Hrushovski, Anand Pillay, and others.
Geometric stability theory has largely to do with the model-theoretic classification of structures in terms of dimension-like quantities that can be axiomatized in terms of notions of combinatorial geometry such as matroid. Key guiding examples include vector spaces and algebraically closed fields, and many of the guiding concepts have an algebro-geometric flavor (e.g., Morley rank as a generalization of Krull dimension).
Such a structural geometric approach can make geometric stability theory an attractive key of entry into modern model theory for those mathematicians who are not already logicians.
Perhaps the key axiomatic notions of geometric stability theory (which at the outset don’t seem particularly wedded to logic) are those of pregeometry and geometry.
Let $X$ be a set. A pregeometry on $X$ is a closure operator (i.e., a monad $cl \colon P X \to P X$ on the power set), satisfying the following two conditions:
The monad $cl$ is finitary, i.e., $A \in X$ and $a \in cl(A)$, then there is a finite $A_0 \subseteq A$ such that $a \in cl(A_0)$.
(Exchange condition) If $A \in P X$, $a,b \in X$, and $a \in cl(A\cup\{b\})$, then $a \in cl(A)$ or $b \in cl(A \cup \{\a\})$. (Cf. matroid)
A geometry is a pregeometry such that $cl(\emptyset) = \emptyset$ and $cl(\{x\}) = \{x\}$ for all $x \in X$.
Let $X$ be a vector space, and let $cl$ be the monad on $P X$ whose algebras are vector subspaces of $X$. Clearly $cl$ is finitary (any subspace is the set-theoretic union of finite-dimensional subspaces), and the exchange condition is a classical fact about vector spaces related to the notion of independence. Thus $cl$ is a pregeometry.
Similarly, let $X$ be a projective space $\mathbb{P}V$, and let $cl$ be the monad on $P X$ whose algebras are projective subspaces. Then $cl$ is a geometry (the closure of a point is a point). Any pregeometry $cl$ gives rise to a geometry in a similar way, in the sense that a pregeometry $cl$ induces a geometry on the image of the function $X \to P X$, $x \mapsto cl(\{x\})$, as explained in Remark .
Let $X$ be an algebraically closed field; let $cl$ be the monad on $P X$ whose algebras are algebraically closed subfields. Then $cl$ is a pregeometry. That the exchange condition is satisfied is a classical result credited to Steinitz^{1}.
Given a pregeometry $(X, cl)$, a subset $A \in P X$ is independent if for all $a \in A$, $a \notin cl(A - \{a\})$. An independent set $A$ said to be a basis for $Y \in P X$ if $Y \subseteq cl(A)$. All bases of $Y$ have the same cardinality, called the dimension of $Y$.
Given a pregeometry $(X, cl)$ and a subset $Y \subseteq X$, there is a restriction pregeometry $cl^Y: P(Y) \to P(Y)$ defined by the formula
It is immediate that $A \subseteq Y$ is independent in $(X, cl)$ iff it is independent in $(Y, cl^Y)$, and that it is a basis for $Y$ in $(X, cl)$ iff it is a basis for $Y$ in $(Y, cl^Y)$. By this observation, to prove that any $Y \in P(X)$ has a well-defined dimension, we may assume without loss of generality that $Y = X$. That a pregeometry $X$ has well-defined dimension was proven here.
There is a standard way of getting a geometry from a pregeometry $(X, cl)$. First, if $cl(\emptyset) \neq \emptyset$, then replace $X$ by $X' \coloneqq X - cl(\emptyset)$, equipped with the restriction pregeometry of the previous remark. Then define an equivalence relation on $X'$ by $x \sim y$ if $cl(\{x\}) = cl(\{y\})$, and define a pregeometry on the quotient set $X'/\sim$ by
where $[b]$ denotes the equivalence class of $b$. This abstracts the process of taking a projectivization of a vector space.
We’ll suppose for now that $\mathbf{L}$ is a countable signature, so that the language it generates consists of countably many formulas. Let $\mathbf{M}$ be an $\mathbf{L}$-structure, with underlying set $M$.
For $A \subseteq M$, an element $b \in M$ is algebraic over $A$ if there is a formula $\phi(y, w_1, \ldots, w_n)$ and elements $a_1, \ldots, a_n \in A$ such that $\mathbf{M} \models \phi(b, a_1, \ldots, a_n)$ and
is finite. The algebraic closure of $A$, denoted $acl(A)$, is the set of elements of $M$ that are algebraic over $A$.
The algebraic closure $A \mapsto acl(A)$ defines a finitary closure operator on $M$.
That $acl$ is monotone (preserves order) is obvious. Also the fact that $acl$ is finitary is easy: given that $b \in acl(A)$ satisfies $\mathbf{M} \models \phi(b, a_1, \ldots, a_n)$ for $a_1, \ldots, a_n \in A$, then similarly $b \in acl(A_0)$ for $A_0 = \{a_1, \ldots, a_n\}$.
Taking $\phi(y, w)$ to be the equality predicate $y = w$, we see $A \subseteq acl(A)$.
For idempotence of $acl$, suppose $b_1, \ldots, b_k \in acl(A)$ and $\mathbf{M} \models \psi(c, b_1, \ldots, b_k)$ where $\{y \in M: \mathbf{M} \models \psi(y, b_1, \ldots, b_k)\}$ has exactly $m$ elements. Write down a formula $F_{m, \psi}(x_1, \ldots, x_k)$ that says $\{y \in M: \mathbf{M} \models \psi(y, x_1, \ldots, x_k)\}$ has at most $m$ elements (by adding some extra inequalities, we can make this “exactly $m$ elements”):
Now for $i = 1, \ldots, k$, let $\phi_i$ be a formula witnessing $b_i \in acl(A)$, i.e., $\mathbf{M} \models \phi_i(b_i, a_{i, 1}, \ldots, a_{i, n_i})$ where $a_{i, k} \in A$ and $\{x \in M: \mathbf{M} \models \phi_i(x, a_{i, 1}, \ldots, a_{i, n_i})\}$ has finitely many elements. Then
is a formula with parameters in $A$ that witnesses $c \in acl(A)$. This proves the idempotence of $acl$.
As before, this notion of algebraic closure $acl: P(M) \to P(M)$ can be restricted to a closure operator $acl^X: P(X) \to P(X)$ for a subset $X \subseteq M$, via the definition $acl^X(A) \coloneqq X \cap acl(A)$ for $A \subseteq X$. One often writes just $acl$ instead of $acl^X$, provided that $X$ is understood.
For an $\mathbf{L}$-structure $\mathbf{M}$, let $D \subseteq M^n$ be definable. $D$ is minimal if the only definable subsets of $D$ are finite or cofinite in $D$. Slightly abusing language, if $\phi(x_1, \ldots, x_n, a_1, \ldots, a_k)$ is a formula with parameters that defines $D$, we also say $\phi$ is minimal. We say $D$ (or $\phi$) is strongly minimal if it is minimal in any elementary extension of $\mathbf{M}$.
A theory $\mathbf{T}$ is strongly minimal if for any model $\mathbf{M}$ of $\mathbf{T}$, the underlying set $M$ (definable by the formula $x = x$) is strongly minimal.
Let $K$ be an algebraically closed field. (Elimination of quantifiers, Chevalley’s theorem, etc.) Conclusion: ACF is a strongly minimal theory.
Divisible torsionfree abelian groups.
Non-example of dense linear orders.
(Baldwin, Lachlan) The algebraic closure operator on a minimal set $X$ is a pregeometry.
Let $A \subseteq X$ and suppose $c \in acl(A \cup \{b\}) - acl(A)$ for $b, c \in X$; we want to show $b \in acl(A \cup \{c\})$. Thus, suppose $\phi(c, b)$ is a formula with parameters from $A$ such that $\mathbf{M} \models \phi(c, b)$ and $card(\{x \in X: \mathbf{M} \models \phi(x, b)\}) = n$, a finite number. As in the proof of Proposition , let $\psi(w)$ be a formula that says $card(\{x \in X: \mathbf{M} \models \phi(x, w)\}) = n$. This is a formula with parameters from $A$ and $\psi(b)$ is satisfied in $\mathbf{M}$.
If $\{y \in X: \mathbf{M} \models \phi(c, y) \wedge \psi(y)\}$ is finite, then since $b$ belongs to this set, $\phi(c, y) \wedge \psi(y)$ would witness $b \in acl(A \cup \{c\})$ and we would be done. So suppose otherwise. Then this set is cofinite in $X$, so that
for some finite $m$, and we can again write down a formula $\chi(x)$ with parameters from $A$ that says
If $\chi(x)$ defines a finite subset of $X$, then since $\chi(c)$ is satisfied, we would reach the conclusion that $c \in acl(A)$, a contradiction. Hence $\chi(x)$ defines a subset cofinite in $X$. We may therefore choose $n+1$ elements $a_1, \ldots, a_{n+1} \in X$ such that $\chi(a_i)$ is satisfied. By our supposition,
is cofinite for $i = 1, \ldots, n+1$. Hence $\bigcap_{i=1}^{n+1} B_i$ is inhabited, say by an element $b'$. We have at least $n+1$ elements $x \in X$ such that $\phi(x, b')$ is satisfied, namely $x = a_1, \ldots, a_{n+1}$. But now this contradicts the fact $\psi(b')$.
=–
Anand Pillay, Geometric stability theory, Oxford Logic Guides 32
slides from conference “Geometric model theory”, Oxford 2010: directory html
Misha Gavrilovich, Model theory of universal covering space of complex algebraic varieties, thesis, pdf
Boris Zilber, Elements of Geometric Stability Theory, 2003 (pdf)
Although oddly enough, as explained at the MacTutor biography page, what is called the Steinitz exchange condition was set out by Ernst Steinitz in a 1913 publication on convergent series and apparently not (?), as might be supposed, in his Algebraische Theorie der Körper (Crelle’s Journal, 1910). His 1913 lemma was for vector spaces. ↩
Last revised on February 17, 2014 at 15:59:34. See the history of this page for a list of all contributions to it.