Higher degree S-lemma and the stability of quadratic modules

0.77 MB·

by  Philipp Jukic

Higher degree S-lemma and the stability of
quadratic modules
Philipp Jukic
Abstract
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0 In this work we will investigate a certain generalization of the so called S-lemma in
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higher degrees. The importance of this generalization is, that it is closely related to

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a Hilbert’s 1888 theorem about tenary quartics. In fact, if such a generalization exits,
J
then one can state a Hilbert-like theorem, where positivity is only demanded on some
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2 semi-algebraic set. We will show that such a generalization is not possible, at least

] not without additional conditions. To prove this, we will use and generalize certain
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A tools developed in [Ne]. In fact, these new tools will allow us to conclude that this
. generalization of the S-lemma is not possible because of geometric reasons. Furthermore,
h
t
a we are able to establish a link between geometric reasons and algebraic reasons. This
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will be accomplished within the framework of quadratic modules.
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Contents
0 Introduction 1
1 The S-lemma 4
1.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Proof of the S-lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2 Higher degree S-lemma 13
2.1 Counterexample . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.2 Formulating a higher degree S-lemma . . . . . . . . . . . . . . . . . . . . 14
2.3 S4-conjecture in two variables . . . . . . . . . . . . . . . . . . . . . . . . 15
2.4 The S4-conjecture: A counterexample . . . . . . . . . . . . . . . . . . . . 21
2.5 Geometric analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.6 A generalization of the counterexample . . . . . . . . . . . . . . . . . . . 27
3 Quadratic modules and stability 31
3.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3.2 Stability and tentacles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
3.3 Tentacles and the S4-conjecture . . . . . . . . . . . . . . . . . . . . . . . 39
3.4 A non-geometric counterexample . . . . . . . . . . . . . . . . . . . . . . 47
3.5 Final thoughts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
0 Introduction
First of all, let us talk about the motivation of this article. In 1888 Hilbert showed in
his work [Hi] that a ternary quartic f, that is a 4-form in three variables, can be written
as a sum of three squares of quadratic forms if and only if f is non-negative on R3.
The question is: Can we find a Hilbert-like theorem in a more general setting? What
does a more general setting mean in this context? Instead of considering non-negative
ternary quartics, we consider a ternary quartic that needs to be non-negative on a semi-
algebraic set S ⊆ R3. Furthermore, the semi-algebraic set S should also satisfy the
following two conditions: First, there exists a quadratic form g in three variables such
that S = S(g) := {x ∈ R3 : g(x) ≥ 0}. Second, the set S has a non-empty interior.
Of course, if S (cid:54)= R3 then f can, in general, not be written as a sum of three squares
of quadratic forms. In this case we need a sort of a correcting term. This correcting
term should also satisfy some conditions. First, we demand that this term is of the form
−tg, where t is a non-negative quadratic form. Second, f −tg should be a non-negative
ternary quartic. Thus a generalization of Hilbert’s theorem could look like the follow-
ing: Let g be a quadratic form such that there exists a point x(cid:48) ∈ R3 with g(x(cid:48)) > 0. A
ternaryquarticf isnon-negativeonthesetS(g)ifandonlyifthereexistsanon-negative
quadratic form t such that f −tg can be written as a sum of three squares of quadratic
forms.
The interpretation of this statement is simple. If g is non-negative, then this state-
ment is equal to Hilbert’s statement. If g is not non-negative, then −tg measures ’how
far away’ f is from being a sum of three squares of quadratic forms.
Let us illustrate this statement by considering the two polynomials g = x2 − x2 and
1 2
f = x4 − x4. It is easy to see that S(g) has a non-empty interior and that f is non-
1 2
negative on S(g). Furthermore, we have f −2x2g = x4−2x2x2+x4 = (x2 −x2)2. Thus
2 1 1 2 2 1 2
f −2x2g is a sum of three squares of quadratic forms: One quadratic form is given by
2
x2 −x2, the other two are 0.
1 2
1
We are looking to clarify the following question: Can such a generalization of Hilbert’s
theorem be made? It turns out that this question is closely related to the so called S-
lemma resp. to a certain generalization of the S-lemma. Hence the first Chapter is all
about the introduction and the proof of the S-lemma. The machinery presented in this
chapter relies heavily on the work of [PT] and [Bar]. The results in the first chapter are
all well known. Therefore there is nothing new in this part of the article. In this chapter
and throughout the whole article no fancy knowledge will be required. One should be
familiar with basic linear algebra, convex geometry, and real algebraic geometry. In the
second Chapter we will formulate a generalization of the S-lemma. For the sake of
simplicity we will refer to the generalization as the S4-conjecture. The importance of
this S4-conjecture is the following: If the conjecture is true, then the generalization of
Hilbert’s theorem is possible. If it is not true, then such a generalization is impossible.
However, it turns out that it is impossible because we can find a counterexample for the
S4-conjecture. Although we can find a counterexample, we will still refer to this men-
tioned generalization as the S4-conjecture. Next, we do some geometric investigations
and finally generalize the counterexample to higher degrees. In the third and last
chapter we use and generalize the machinery developed in [Ne] to further investigate
the counterexample. It turns out that the tools presented in [Ne] are quite suitable in
analyzing the S4-conjecture. In fact, by using these new methods we will see that the
conjecture fails because of geometric reasons. Since [Ne] connects geometric properties
and algebraic properties, we will see that there is an interesting link between the S4-
conjecture and the stability of quadratic modules. Finally, this article will be concluded
by presenting some new questions that should serve as a motivation for further studies.
2
We will use the following notation throughout this article:
• R, C, Z, N,N : The real, complex, integer, natural numbers and the natural numbers
0
with 0.
• Rn, Cn, Zn: The 0 ≤ n dimensional vector spaces Rn resp. Cn and the free Z-module
Zn.
• (cid:104)·,·(cid:105): The standard scalar product in Rn.
• K[x ,...,x ]: The polynomial ring over a field K in n ≥ 1 variables. Polynomial
1 n
variables will always be denoted by upright letters x,y,β,λ etc.
• K[x ,...,x ] : The set of all polynomials f ∈ K[x ,...,x ] with deg(f) ≤ d.
1 n d 1 n
• A polynomial f ∈ R[x ,...,x ] is called non-negative if ∀x ∈ Rn : f(x) ≥ 0. Negative,
1 n
positive and non-positive polynomials are defined in the same manner.
• The homogenization of a polynomial f ∈ K[x ,...,x ] will be denoted by f. The
1 n
dehomogenization of a homogeneous polynomial g with g˜.
• An, Pn: The n-dimensional affine space and the n-dimensional projective space.
• V(f ,...,f ): For polynomials f ,...,f ∈ K[x ,...,x ] the set V(f ,...,f ) is defined
1 s 1 s 1 n 1 s
to be the set of all solutions x ∈ Kn (K denotes the algebraic closure of K) of the
polynomial equalities f (x) = 0,...,f (x) = 0. If K = R then we will fix C as the
1 s
algebraic closure of R. If f ,...,f are homogeneous, then we can interpret V(f ,...,f )
1 s 1 s
as the set of all solutions in the projective space Pn.
• Let V be a variety defined over a field K and L|K an algebraic extension of K. The
L-rational points of V are denoted by V(L).
• S(f ,...,f ): The basic closed semi-algebraic set
1 s
S(f ,...,f ) = {x ∈ Rn : f (x) ≥ 0,...,f (x) ≥ 0}
1 s 1 s
defined by the polynomials f ,...,f ∈ R[x ,...,x ].
1 s 1 n
• GL , O : The general linear group over R and its orthogonal subgroup over R.
n n
• Let X be a topological space and A a subset. The interior of A is denoted with int(A)
and the closure with A.
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1 The S-lemma
In this chapter we will formulate and prove the so called S-lemma. Before doing this,
however, it shall be noted that the S-lemma has many variations in the literature. While
all versions are in fact equivalent, we will use a version that is closer to real algebraic
geometry. Thus the original statement of the S-lemma made by Yakubovich, that can
be found in the work of Polik and Terlaky [PT], will not be used. In the sense of real
algebraic geometry the S-lemma is formulated in the following way:
Theorem 1.0.1. S-lemma: Let f,g be polynomials in R[x ,...,x ] . If there exists a
1 n 2
point x(cid:48) ∈ Rn with g(x(cid:48)) > 0, then the following statements are equivalent:
(a) The inclusion S(g) ⊆ S(f) holds.
(b) There exists a non-negative real number t such that f(x)−tg(x) ≥ 0 for all x ∈ Rn
The aim of this chapter is to provide a proof for Theorem 1.0.1. Simultaneously, it
should serve as an introduction in what is to come later. Before we are ready to prove
Theorem 1.0.1, we need some preparatory results, which will be bundled together in the
following section.
1.1 Preliminaries
First of all, it is worth mentioning that one could prove Theorem 1.0.1 directly, with-
out any notable machinery. One such proof can be found in [PT, pp. 376-378]. The
disadvantage is, however, that it needs quite a lot of computations. As already pointed
out, we will use a different approach. For the proof of theorem 1.0.1 we will need the
following definitions, lemmas, and propositions:
Definition 1.1.1. Let f = (cid:80)n (cid:80)n a x x be a quadratic form in R[x ,...,x ],
i=1 j=1 ij i j 1 n
where all coefficients a of f lie in R. The matrix that corresponds to f is defined to be
ij
(cid:0) (cid:1)
the symmetric matrix A = 1(a +a ) . If f is an arbitrary form of degree
f 2 ij ji 1≤i≤n,1≤j≤n
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d ≥ 0 in R[x ,...,x ], then f is said to be positive semi-definite resp. positive definite
1 n
if ∀x ∈ Rn : f(x) ≥ 0 resp. ∀x ∈ Rn\{0} : f(x) > 0.
Remark 1.1.2. A quadratic form f is positive (semi-) definite if and only if the corre-
sponding matrix A is positive (semi-) definite.
f
In the following we will assume that the coefficients of a quadratic form in R[x ,...,x ]
1 n
lie in R.
Definition 1.1.3. Let P be the set of all forms of even degree d > 0 in R[x ,...,x ].
d,n 1 n
With P+ we denote the subset of P that consist of all positive semi-definite forms in
d,n d,n
P .
d,n
Definition 1.1.4. Let V be finite dimensional R-vector space. A (convex) cone C ⊆ V
is a subset of V that satisfies the following two conditions:
• The set C is not empty.
• For any real number λ ≥ 0 and any element g ∈ C we have λg ∈ C.
We say that a cone C ⊆ V is pointed, if the identity C ∩−C = {0} holds.
Remark 1.1.5. One can easily see that P is a finite dimensional R-vector space. To
2,n
be more precise, there is a vector-space isomorphism P −∼→ Rn(n+1). For f,g ∈ P
2,n 2 2,n
the dot product on P is defined by (cid:104)f,g(cid:105) = tr(A A ), which is just the pullback of
2,n f g
the dot product in Rn(n+1). Thus P is an euclidean space. The same is also true for
2 2,n
P , where d ≥ 0.
d,n
Finally, it should be noted that this vector space P has a more or less surprising
2,n
upcoming in algebraic geometry: See [Sha I, Example 3, p. 44] about determinantal
varieties.
Definition 1.1.6. Let V be a finite dimensional real vector space and C ⊆ V a convex
subset. A convex subset F of C is called a face of C if the following statement holds:
Suppose u and v are two points in C. If there exists a λ ∈ (0,1) such that λu+(1−λ)v ∈
F, then u and v lie already in F.
A face F of C is called proper if ∅ (cid:36) F (cid:36) C holds. If there is a point u ∈ C such
that {u} is a face of C, then the point u is called an extremal point. With ex(C) we
will denote the set of all extremal points of C.
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Definition 1.1.7. Let V be a finite dimensional real vector space and C ⊆ V a convex
subset. Let H be a hyper plane given by H = {x ∈ V : (cid:96)(x) = 0}, where (cid:96) : V → R is a
linear form. Set H = {x ∈ V : (cid:96)(x) ≥ 0} and H = {x ∈ V : (cid:96)(x) ≤ 0}. A face F of C
+ −
is called exposed if there exists a linear form (cid:96) : V → R such that C is contained in H
+
or H and F = C ∩H.
−
Definition 1.1.8. Let V be a finite dimensional real vector space and S an arbitrary
subset of V. The affine hull aff(S) of S is defined by
(cid:40) (cid:41)
n n
(cid:88) (cid:88)
aff(S) = λ s : s ,...,s ∈ S,λ ,...,λ ∈ R, λ = 1,n ∈ N .
i i 1 n 1 n i
i=1 i=1
Definition 1.1.9. Let V be a finite dimensional real vector space. The dimension
of a convex set C ⊆ V is defined to be the dimension of its affine hull. In short
dim(C) = dim(aff(C)).
Definition 1.1.10. Let V be a finite dimensional real vector space and C a convex
subset of V. Let B (x) denote the open ball in x with radius ε > 0. The relative interior
ε
relint(C) of C in V is defined by
relint(C) = {x ∈ C : ∃ε > 0 : B (x)∩aff(C) ⊆ C}.
ε
Lemma 1.1.11. (a): For every x ∈ Rn\{0} the symmetric n×n-matrix xxT matrix is
positive semi-definite of rank 1.
(b): Let A be a positive semi-definite matrix. The rank of A is the smallest natural
number r such that A can be written as A = (cid:80)r x xT for some x ,...,x ∈ Rn.
i=1 i i 1 r
Proof: (a): Trivial.
(b): Firstofall,notethatstatement(b)isindependentwithrespecttotransformations
STAS, where S ∈ O . Indeed, set D = STAS and assume that A = (cid:80)r x xT, where r
n i=1 i i
is minimal. Then we have D = ST (cid:80)r x xTS = (cid:80)r STx xTS = (cid:80)r STx (STx )T.
i=1 i i i=1 i i i=1 i i
It is clear that r is also the minimal length of the sum for D: Otherwise, A could be
written as a sum of smaller length, which would be a contradiction. Choose S ∈ O such
n
that D is a diagonal matrix. The diagonal of D consists of the eigenvalues of A, which
are all non-negative. Thus it is easy to see that D can be written as a sum (cid:80)r x xT
i=1 i i
for some x ,...,x ∈ Rn and r = rk(A). It remains to verify that r is minimal. But
1 n
this follows from rk(cid:0)(cid:80)r x xT(cid:1) ≤ rrk(x xT) = r.
i=1 i i i i
(cid:31)
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Proposition 1.1.12. Let d ≥ 0 be an even number.
(a) The set P+ is a closed cone in P .
d,n d,n
(b) The cone P+ is pointed.
d,n
(c) Let L ⊆ Rn be a subspace and F := (cid:8)f ∈ P+ : ∀x ∈ L : f(x) = 0(cid:9). The set F
L 2,n L
is an exposed face with dim(F ) = r(r+1) and r = n − dim(L). If f ∈ P+ and
L 2 2,n
L = ker(A ), then f is in relint(F ).
f L
Proof: (a): We will show that P := P \P+ is open. Take an element f ∈ P.
d,n d,n
Since f ∈ P, there exists a point x ∈ Rn such that f(x) < 0. Consider the evaluation
homomorphism ev : P → R,p (cid:55)→ p(x). Furthermore, P is, as already stated in
x d,n d,n
Remark 1.1.5, an euclidean space. Thus ev is continuous. Let U ⊆ R be an open
x
neighborhood of f(x) such that all elements of U are negative real numbers. The set
U(cid:48) = ev−1(U) is an open neighborhood of f that satisfies U(cid:48) ⊆ P. Thus we proved that
x
P is open resp. that P+ is closed. The second assertion that P+ is a cone is trivial.
d,n d,n
(b): Trivial.
(c): In the following we will just omit the trivial parts of the proof.1 Let us begin
with the easiest part, verifying that dim(F ) = r(r+1), where r = n − dim(L). This
L 2
can be done by proving aff(F ) ∼= Rr(r+1). Let L⊥ be the orthogonal complement
L 2
of L in Rn. Without loss of generality we can identify L with Rn−r and L⊥ with
Rr. Consider the cone P+. It is easy to see that F can be identified with P+.
2,r L 2,r
Furthermore, P+ has a non-empty interior. A well known result in convex geometry
2,r
(cid:0) (cid:1)
states that a cone with non-empty interior is full. This means that aff P+ = P .
2,r 2,r
Thus aff(cid:0)P+(cid:1) ∼= Rr(r+1). Identifying aff(F ) with aff(cid:0)P+(cid:1) proves the assertion. Next,
2,r 2 L 2,r
we show that F is an exposed face. A quadratic form h ∈ P+ and a quadratic
L 1 2,n−r
form h ∈ P+ give rise to a quadratic form h ∈ P+ in an obvious manner. In fact, the
2 2,r 2,n
(cid:32) (cid:33)
A 0
corresponding matrix A of h is given by A = h1 . Fix h = x2+···+x2
h h 1 1 n−r
0 A
h2
(cid:32) (cid:33) (cid:32) (cid:33)
A 0 0 0
and define A˜ = h1 , A˜ = . Then we can identify F with
h1 0 0 h2 0 A L
h2
(cid:110) (cid:111)
h ∈ P+ : ∃h ∈ P+ : A = A˜ . It is easy to see that F consists of all quadratic
2,n 2 2,r h h2 L
(cid:16) (cid:17)
forms h ∈ P+ that satisfy tr A A˜ = 0. Thus it is convenient to consider the linear
2,n h h1
(cid:16) (cid:17)
form (cid:96) : P → R,p (cid:55)→ tr A A˜ and the hyper plane H = {p ∈ P : (cid:96)(p) = 0}.
2,n p h1 2,n
1Pay attention to statements that begin with ’It is easy to see’.
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So far, we know that F = P+ ∩ H. Finally, we just have to deal with the inclusion
L 2,n
P+ ⊆ H . Let h = (cid:80) a x x be a quadratic form in P+ . Since h is non-negative,
2,n + i,j ij i j 2,n
the coefficients a must be non-negative for all i = 1,...,n. Thus the diagonal of A
ii h
consists of non-negative real numbers. This implies (cid:96)(h) ≥ 0. Altogether we proved that
F is an exposed face. Let us deal with the last statement in (c). Suppose f ∈ P+ and
L 2,n
L = ker(A ). As before set r = n−dim(L). Again we identify F with P+ and interpret
f L 2,r
f as a quadratic form in P+. Then f does only vanish at the origin in Rr. Hence f lies
2,r
in the interior of the cone P+ resp. in relint(F ), which proves the assertion.
2,r L
(cid:31)
Lemma 1.1.13. Let C ⊂ Rn be a non-empty closed convex set which does not contain
any straight line. Then ex(C) is a non-empty set.
Proof: See [Bar, Lemma 3.5, p. 53].
(cid:31)
Because the next result is very important, it will be proven, although there exists a
suitable reference.
Proposition 1.1.14. Let L be an affine subspace of P such that S = L∩P+ is not
2,n 2,n
empty. Suppose the inequality codim (L) < (r+2)(r+1) holds for some r ∈ N . Then
P2,n 2 0
there exists a quadratic form f ∈ S such that the rank of A is bounded by r.
f
Proof[Bar, Proposition 13.1, p. 83]: According to Proposition 1.1.12 the cone P+
2,n
is pointed and closed. This means that there is no way that the cone P+ contains
2,n
a straight line. If P+ does not contain such a line so does not the subset S of P+ .
2,n 2,n
By using Lemma 1.1.13 we get ex(S) (cid:54)= ∅. Choose an arbitrary f ∈ S and let A be
f
its corresponding matrix of rank m. Consider W = ker(A ) and the exposed face F
f W
(Proposition 1.1.12). We want to show that f is an element of the set relint(L∩F ).
W
Since f is an element of relint(F ) and relint(L), it is enough to verify the inclusion
W
relint(F )∩relint(L) ⊂ relint(L∩F ). Take a point g ∈ relint(F )∩relint(L). There
W W W
exist ε ,ε > 0 such that B (g)∩aff(F ) ⊂ F and B (g)∩aff(L) ⊂ L. By setting
1 2 ε1 W W ε2
ε = min{ε ,ε } we get B (g) ∩ aff(L) ∩ aff(F ) ⊂ L ∩ F . Since aff(L ∩ F ) ⊂
1 2 ε W W W
aff(L)∩aff(F ), we have B (g)∩aff(L∩F ) ⊂ L∩F which implies the assertion.
W ε W W
We know that f lies in both sets, relint(L∩F ) and ex(L∩F ). This can only work
W W
if dim(L∩F ) = 0 holds: Suppose dim(L∩F ) > 0. For every ε > 0 we can find two
W W
different points δ ,δ ∈ B (f) such that δ ,δ (cid:54)= f and δ ,δ ,f ∈ aff(L∩F ). Choose
1 2 ε 1 2 1 2 W
ε > 0 such that B (f)∩aff(L∩F ) ⊂ L∩F holds. Now, f is some point on the line
ε W W
segment that connects the two points δ and δ . But both points lie in L ∩ F . This
1 2 W
contradicts the fact that f is an extremal point of L∩F .
W
8