17
THE SEMIALGEBRAIC CASE
Today’s Goal Tarski-Seidenberg Theorem. For every n =1, 2, 3,...,every Lalg-definable subset of Rn can be defined by a quantifier-free Lalg-formula. n Thus every Lalg-definable subset of R is a finite boolean combination (i.e., finitely many intersections, unions, and comple- ments) of sets of the form
n {(x1,...,xn) ∈ R | p(x1,...,xn) > 0} where p(x1,...,xn)isapolynomial with coefficients in R. These are called the semialgebraic sets. A function f: A ⊂ Rn → Rm is semialgebraic if its graph is a semialgebraic subset of Rn × Rm. 18
As for the semilinear sets, every semialgebraic set can be written as a finite union of the intersection of finitely many sets defined by conditions of the form
p(x1,...,xn)=0
q(x1,...,xn) > 0 where p(x1,...,xn) and q(x1,...,xn) are polynomials with coefficients in R.
Proof Strategy • Prove a geometric structure theorem that shows that any semialgebraic set can be decomposed into finitely many semialgebraic generalized cylinders and graphs. • Deduce quantifier elimination from this. 19
Thom’s Lemma. Let p1(X),...,pk(X) be polynomials in the variable X with R coefficients in such that if pj(X) =0then pj(X) is included among p1,...,pk. Let S ⊂ R have the form
S = ∩jpj(X) ∗j 0 where ∗j is one of <, >,or=, then S is either empty, a point, or an open interval. Moreover, the (topological) closure of S is obtained by changing relaxing the sign conditions (changing < to ≤ and > to ≥.
Note There are 3k such possible sets, and these form a partition of R. 20
Some Standard Tricks • Identify the complex numbers C with R2 via
a + bi ←→ (a, b) √ where a, b ∈ R and i = −1. With this identification, multiplication of complex numbers is a semialgebraic function from R2 × R2 to R2. Also, Cn is identified with R2n • The collection of polynomials in the variable X with coefficients in R of degree not greater than n can be identified with Rn+1 via
n a0+a1X+···+anX ↔ (a0,a1,...,an). Similarly for polynomials with coeffi- cients in C. Addition, multiplication, differentiation of polynomials are semi- algebraic functions. 21
n R Rn+1 Let Bk ( ) denote (as a subset of ) the collection of polynomials in the variable X with real coefficients of degree not greater than n that have exactly k distinct complex roots. n R ⊂ n R Let Mk ( ) Bk ( )bethose polynomials of degree n with this property. ⊂ n R Lemma. Suppose that A Mk ( ) is connected. For each a¯ ∈ A let ra¯ be the number of distinct real roots of the polynomial pa¯(X) associated with a¯. Then
a. ra¯ = r is constant on A; b. There are continuous functions
f1,...,fr: A → R
such that for all a¯ ∈ A we have fi(¯a) (“Continuity of roots”) 22 n R n R Lemma. The subsets Bk ( ) and Mk ( ) of Rn+1 are semialgebraic. Idea • The polynomial p(X) has a repeated root if and only if p(X) and its derivative p (X)haveacommon factor. • This can be expressed by the condition that the determinant of a matrix constructed from the coefficients of p(X), the so-called resultant of p and p (also called the discriminant of p), has value 0. • This is a semialgebraic condition on the coefficients. • Then extend this idea to capture n R n R Bk ( ) and Mk ( ) semialgebraically. 23 Graphs and Cylinders The structure theorem shows that a semialgebraic set S ⊆ Rn can be partitioned into finitely many sets of two kinds, all of which are semialgebraic. Graphs Let A ⊂ Rk and f: A → R be continuous. The graph of f is the subset of Rk+1 given by Graph(f)={(¯x, y) ∈ Rk+1 | x¯ ∈ A and y = f(¯x)}. Generalized Cylinders Let A ⊂ Rk, and let f,g: A → R be continuous and satisfy f(¯x) k+1 (f,g)A = {(¯x, y) ∈ R | x¯ ∈ A and f(¯x) • If A is connected, then graphs and cylinders based on A are connected. Structure Theorem. Let S ⊂ Rn be semialgebraic. Then: In. S has finitely many connected compo- nents and each one is semialgebraic n−1 IIn. There is a finite partition P of R into connected semialgebraic sets such that for each A ∈Pthere is kA ∈ N A → R ∪ {±∞} and fi : A for i = 0, 1 ...,kA +1satisfying a. f A = −∞, f A = ∞, f A is 0 kA+1 i continuous for 1 ≤ i ≤ kA, and A A ≤ ≤ fi−1(¯x) The Proof • The proof is by induction on n, and I shall outline the induction step. • Most broadly, the argument is as follows: show In−1 ⇒ IIn and IIn ⇒ In. • IIn ⇒ In is evident; the crux is In−1 ⇒ IIn. • Split the coordinates of Rn as (x1,...,xn−1,t). • Using standard set theory, write S as the union of finitely many finite intersections of polynomial equalities and inequalities. • Extend the finite collection of polyno- mials in the given definition of S by in- cluding all iterated partial derivatives with respect to t. Let this expanded list of polynomials be q1,...,qr. 26 • For each subset S⊂{1,...,r}, form the polynomial QS (¯x, t)= qj(¯x, t). j∈S Viewx ¯ as parameter variables and consider the polynomial as QS,x¯(t), apolynomial in the variable t whose coefficients are polynomials inx ¯. • For each