About Schmüdgen's Theorem

Schmüdgen proved a Positivstellensatz about the representation of positive definite polynomials on a basic closed semialgebraic set using methods from functional analysis in connection with Stengle's Positivstellensatz. Later Wörmann gave a purely algebraic proof. The present note contains yet another proof that uses classical results by Kadison about the representation of partially ordered algebras and the Theorem of Krein-Milman. Positivity of polynomial functions is among the most classical topics in real algebraic geometry, but it is still a very active area of research. An excellent account of the current state-of-the-art (as of 2004) can be found in [Sch]. Another important source is [PD]. The present note describes a contribution by Schmüdgen [Sc] and gives a new and very elementary proof of his result. Suppose that A = R x1,K,xn [ ] is a finitely generated R-Algebra. Using a presentation R X1,K, Xn [ ] → R x1,K, xn [ ] as a factor ring of a polynomial ring, the algebra defines an algebraic set V ⊆ R n . The elements of A are considered as polynomial functions on V. Any finite subset S = s1,K,sk { } ⊆ A determines a cone in A, PS = a ∈A ∀ε ∈ 0,1 { } k ∃σε ∈ A 2 ∑ :a = σ ε ⋅ s1 ⋅K ⋅ sk k ε ∑       (where A ∑ denotes the set of sums of squares in A), and a basic closed semi-algebraic set in V, KS = x ∈V ∀ s ∈S:s x ( ) ≥ 0 { } . There is yet another cone in A, the set of polynomial functions that are positive semi-definite on KS : QS = a ∈A ∀ x ∈KS : a x ( ) ≥ 0 { } . Note that the set S determines the cone PS , but there are always infinitely many finite sets that determine the same cone. The semi-algebraic set KS depends only on the cone PS , and PS determines the cone QS . Main Problem What is the precise relationship between the cones PS and QS ? The inclusion PS ⊆ QS holds trivially; however, the cones are not equal in general. The first result in this direction was achieved by Hilbert ([Hi]) – he showed that there exist polynomials with two variables that are positive semi-definite on the entire plane, but cannot be represented as sums of *) The author acknowledges partial support from the European RTNetwork RAAG, Contract No. HPRN-CT2001-00271 These notes were presented in the seminar "Reelle algebraische Geometrie" in Regensburg in July 2003. Niels Schwartz About Schmüdgen's Theorem page 2 of 9 squares of polynomials. (In this case the algebra A is the polynomial ring in two variables over R, the algebraic set V is the plane and S is the empty set. Then the cone PS is the set of sums of squares in A, KS is the entire plane and QS is the set of polynomials that are positive semi-definite on the plane.) Even though Hilbert's answer was negative it was the beginning of a very fruitful development. There are many variations of the main problem, e.g.: • Is it possible to give conditions that ensure PS = QS ? • If PS ≠ QS : Is it possible to use the elements of the cone PS to give a description of the elements of QS ? Examples of results in this direction are: o Hilbert's 17th Problem asks whether every positive semi-definite polynomial in R X1,K, Xn [ ] can be written as a sum of squares of rational functions, i.e., as a sum of squares in R X1,K, Xn ( ) . The affirmative answer was given by Artin [A]. o Stengle's Positivstellensatz gives a characterization of those polynomial functions that are positive on a basic closed semi-algebraic set, cf. [St]; [KS], p. 141, Satz 2. • If PS ≠ QS : Hilbert proved his result in a non-constructive way. For a long time it was a challenge to name a specific polynomial that is positive semi-definite on the plane, but is not a sum of squares. The first explicit example was found by Motzkin [M]: X 4 ⋅ Y 2 + X ⋅Y 4 + 1− 3 ⋅ X ⋅Y 2 . Another such polynomial was determined by Robinson, cf. [R], pp. 258, 1 − X − Y 2 − X 4 + 3 ⋅ X ⋅Y 2 − Y 4 + X − X 4 ⋅ Y 2 − X ⋅Y 4 + Y 6 . More examples and an excellent account of these developments can be found in [R]. • How large is PS compared with QS ? Blekherman ([B]) showed: The space Fn,2k of homogeneous polynomials of degree 2k in n variables contains the cones Qn,2k of positive semi-definite forms and Pn,2k of sums of squares of forms of degree k. The intersections of the cones with a suitably chosen hyperplane in Fn,2k are compact convex sets. There are bounds for the volumes of these sets, which imply that the fraction of the sums of squares among the positive semi-definite forms tends to 0 as the number of variables goes to infinity. The focus of the present note is on Schmüdgen's contributions, which will be described below. A general hypothesis will be made that does not seriously restrict the generality of the results: It will be assumed throughout that the algebra A is reduced and that the semi-algebraic set KS is Zariski-dense in V. An equivalent condition is given in the following proposition. Note that, in general, both cones PS and QS are preorderings [KS], p. 140. They are partial orderings if PS ∩ −PS = 0 { } and QS ∩ −QS = 0 { } . Proposition With the notation introduced above, the following conditions are equivalent: (a) A is reduced and KS is Zariski-dense in V. (b) The preordering QS is a partial order for A. Proof For the proof of (a) ⇒ (b), suppose that a ∈QS ∩ −QS . Then a x ( ) = 0 for all x ∈KS . Since KS is Zariski-dense in V it follows that a x ( ) = 0 for all x ∈V , which implies that a is nilpotent. In the reduced ring A this means that a = 0 . Conversely, to prove (b) ⇒ (a) it must be shown that an Niels Schwartz About Schmüdgen's Theorem page 3 of 9 element a ∈A is 0 if it is nilpotent or if its restriction to KS is 0. If a is nilpotent then a x ( ) = 0 for all x ∈V , hence also for all x ∈KS . Thus in both cases a KS = 0 . It follows that a ∈QS ∩ −QS = 0 { } , i.e., a = 0 . Ω Clearly, if QS is a partial order then the same is true for PS . The converse is not true as the following simple example shows: Example The R-algebra A = R X [ ] X ( ) is totally ordered with positive cone P = α 0 +α1 ⋅ X + X 2 ( ) α 0 > 0 or α0 = 0 and α1 ≥ 0 { } . The algebraic set corresponding to A is V = 0 { } ⊆ R . If S = X + X 2 ( ) { } then PS = P , KS = V , and QS = α0 +α1 ⋅ X + X 2 ( ) α0 ≥ 0 { } . Thus, PS is a partial order, but QS is not. Ω From now on it will always be assumed that the equivalent conditions of the Proposition are satisfied. Schmüdgen has contributed to two problems that are quite different at a first glance, to the Moment Problem in functional analysis, and, with a new Positivstellensatz, to the study of the connections between the cones PS and QS . The following notation is needed to talk about the Moment Problem: The R-algebra A is a Rvector space of countable dimension. The dual space is denoted by A . The dual space contains the dual cones PS ∨ and QS ∨ . Evidently, the inclusion relation PS ∨ ⊇ QS ∨ is always true. The question about the connection between PS and QS has the following dual: • What is the exact relationship between PS ∨ and QS ∨ ? When is it true that PS ∨ = QS ∨ ? The following construction always yields elements of QS ∨ : Suppose that μ is a positive Borel measure on KS . Then the linear functional λμ : A → R:a → adμ ∫