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Algebraic Geometry Lecture 30 – Semi-algebraic geometry part 2
Lee Butler
- A crash course in model theory
Now for the model theory. Model theory is essentially a branch of logic that formalises the lan- guage used in mathematics, kind of like how category theory formalises the objects in mathematics. But you can actually prove stuff with model theory.
Consider the following innocuous statement:
for every x greater than zero there is a y such that y is the square root of x.
Normally you wouldn’t bat an eyelid at this statement, but there’s a lot going on here. Before the statement makes any sense we need to have a notion of “greater than” and know what “zero” is. We also need to know about multiplication, since y being the square root of x really just means y · y = x. Finally we need to know what set we’re working in. As you’ve probably already figured out, the statement is true in the real numbers or the real algebraic numbers, but not in the rational numbers (since 2 doesn’t have a square root, say), nor in the complex numbers since there’s no concept of an order on C by theorem 4.1.
If we strip away the flesh of many mathematical statements we’re left with a very basic language to use:
(1) The logical symbols, =, ∃, ∀, ∨ (or), ∧ (and), ¬ (not), ⇒, and ⇔ and variables x, y, z,... or v 0 , v 1 , v 2 ,.. .. We don’t need all of these as some can be expressed in terms of the others, but they’re useful abbreviations. (2) A specific language for a given context. This consists of a set of function symbols of given arity, a set of relation symbols of given arity, and a set of constant symbols. The language of rings, for example, is Lr = {+, ·, 0 , 1 }. Here + and · are binary functions, there are no relations, and 0 and 1 are constants. The language of ordered rings is Lor = {+, ·, <, 0 , 1 }, which includes the binary relation <.
As they stand, the two languages mentioned above don’t mean anything. We could just as easily write Lr = {f 1 , f 2 , c 1 , c 2 } for binary functions f 1 , f 2 and constants c 1 , c 2. To give a language L context we need an L-structure M. This is a set M and an interpretation for each symbol in L. The interpretation of an n-ary function symbol f ∈ L is a function f M^ : M n^ → M , the interpretation of an n-ary relation symbol R ∈ L is a subset RM^ ⊂ M n, and the interpretation of a constant symbol c ∈ L is an element cM^ ∈ M. We usually don’t distinguish between the symbols in L and their interpretation in M, though.
As an example consider the Lr -structure M = (R, +, ·, 0 , 1). So this is the set of real numbers equipped with addition and multiplication and special constants zero and one.
The other basic idea in model theory is that of an L-theory. This is just a well-formed set of formulae using the logical symbols and the language L, and where every variable in each formula is bound either by ∃ or by ∀. So a typical Lor -theory might be
T = {∀x (0 < x → (∃y y · y = x)) , ∀x∃y x < y , ∃x (0 < x ∧ ∀y ¬(y · y = x))}. 1
The first formula says that every positive number has a square root. The second says that for every number, there’s a bigger number. The final one says that there’s a positive number with no square root. This is a valid Lr -theory, but the first and last formulae contradict each other, so whatever context we work in, we can’t satisfy all the formulae in T. Suppose we drop the last formula, though, to get a new theory T ′. The two formulae are then satisfied if we work in the Lr -structure M = (R, +, ·, 0 , 1), so we say that M is a model for T ′^ and write M |= T ′.
Consider two L-structures M = (M, L) and N = (N, L), and suppose that M ⊆ N and that the inclusion map i respects the interpretation of the symbols in L, so i(f M(a)) = f N^ (i(a)) for every a ∈ M , and so on. Suppose moreover that given any formula φ in the language L we have M |= φ if and only if N |= φ. Then we say M is an elementary substructure of N and write M ≺ N.
Given an L-theory T we say T is model-complete if, for any models M and N of T , if M ⊆ N then M ≺ N.
- Definable sets and quantifier elimination
We said that every formula in a theory had to have all its variables bound by quantifiers, such as in ∀x∃y x < y. That’s so the formula is either true or false in any given model. If we don’t bind all the variables it’s not so cut-and-dry. Consider the formula φ(x, y) given by
∃z(z 6 = 0 ∧ y = x + z · z).
This formula doesn’t bind x or y, so there’s no sense asking if it’s true in a given structure since it depends on your choice of x and y. Instead this formula defines a set in a given structure:
{(x, y) ∈ R^2 : ∃z(z 6 = 0 ∧ y = x + z · z)} = {(x, y) ∈ R^2 : x < y} {(x, y) ∈ N^20 : ∃z(z 6 = 0 ∧ y = x + z · z)} = {(x, y) ∈ N^20 : y − x is a nonzero square}.
In general we say a set X ⊂ M n^ is definable if there is a formula ψ(x, y) in L such that X = {x ∈ M n^ : M |= ψ(x, b)}
for some b ∈ M m.
Some examples using the language Lr of rings include:
- In the structure (Z, +, ·, 0 , 1) the set {(m, n) ∈ Z^2 : m < n} is definable using Lagrange’s four squares theorem.
- If F is a field and we consider the structure (F [X], +, ·, 0 , 1) then F is definable in this structure – it’s the set of units.
- More surprisingly, we can define C in the structure (C(X), +, ·, 0 , 1) using arguments in- volving elliptic curves.
- Zp is definable in (Qp, +, ·, 0 , 1) using Hensel’s lemma.
- One of the great results in model theory in the twentieth century was a result by Julia Robinson who showed that the integers are definable in (Q, +, ·, 0 , 1). To define them let φ(x, y, z) be the formula ∃a∃b∃c xyz^2 + 2 + yc^2 = a^2 + xy^2 and let ψ(x) be the formula ∀y∀z ([φ(y, z, 0) ∧ (∀w (φ(y, z, w) → φ(y, z, w + 1)))] → φ(y, z, x)). Then ψ(x) define Z.
- Hilbert’s seventeenth problem
Definition 6.1. Let F be a real closed field and f (X) ∈ F (X 1 ,... , Xn) be a rational function. We say that f is positive semidefinite if f (a) > 0 for all a ∈ F n.
Hilbert asked if real positive semidefinite polynomials could always be written as sums of squares of rational functions, similarly to how positive integers can be written as sums of squares. The answer is: yes, yes they can.
Theorem 6.2 (Hilbert’s seventeenth problem). If f is a positive semidefinite rational function over a real closed field F , then f is a sum of squares of rational functions.
Proof. Let f (X 1 ,... , Xn) be a positive semidefinite rational function over F that isn’t a sum of squares. So by theorem 3.1 there’s an ordering of F (X) such that f (X) is negative. Let R be the real closure of F (X) extending this order, which we have by corollary 3.4. Then
R |= ∃v f (v) < 0
since the variable now ranges over R and X ∈ R, and we have f (X) < 0 in R. But RCF is model-complete so anything true in an extension structure is true in a substructure, so
F |= ∃v f (v) < 0 ,
where now the variable ranges over F n. This contradicts f being positive semidefinite. Thus no such f can exist.
