Representations of Positive Polynomials: Theory, Practice ...vicki/talks/TempleNov2014_handout.pdf · R[X]2, from which one can deduce the positivity condition immediately. A theorem

Introduction and History psd ternary quartics PSD, not SOS polynomials Polya’s Theorem Finding SOS decompositions Rational sums of squares Rational certificates of positivity

Representations of Positive Polynomials: Theory,Practice, and Applications

Vicki Powers

Dept. of Mathematics and Computer ScienceEmory University, Atlanta, GA

Currently: National Science Foundation

Temple UniversityNovember 17, 2014

Vicki Powers Representations of Positive Polynomials: Theory, Practice, and Applications


In theory, theory andpractice are the same. But inpractice, they are different.

– Albert Einstein



SOS and Positive Polynomials

Let R[X ] := R[X1, . . . ,Xn]. f ∈ R[X ] is a sumof squares, or sos if f = g2

1 + · · ·+ g2k for

gi ∈ R[X ].

Let∑

R[X ]2 denote the sos polynomials.

For S ⊆ Rn, we write f ≥ 0 (f > 0) on S to mean f takes onlynonnegative (strictly positive) values on S . If f ≥ 0 on Rn, we sayf is positive semidefinite, or psd.

Clearly f sos implies f is psd – the square of a real number ispositive. This simple fact and generalizations of it underlie a largebody of theoretical and computational work on representations ofpositive polynomials.



Suppose S ⊆ Rn, f ∈ R[X ], and f ≥ 0 on S . By a (positivity)representation of f , or a certificate of positivity for f , on S wemean an algebraic expression for f , usually involving elements in∑

R[X ]2, from which one can deduce the positivity conditionimmediately. A theorem about the existence of such certificates isa “representation theorem”.

For example, I claim f is psd, i.e. f ≥ 0 on R2:

f (x , y) := x4 + 4y4 − 4xy2 + 2x2y − x2 + y2 − 2y + 1

The following representation for f proves the claim:

f = (x2 + y − 1)2 + (2y2 − x)2



This talk is about the theory and practice of such representations.

Theory: Results concerning the existence of a representation.

Practice: Results on computational and algorithmic issues, e.g.,finding and counting representations.

We will give some history and background and take a random walkthrough results and mention an application or two.



Our story begins in 1888, when the 26-year-oldDavid Hilbert published his seminal paper onsums of squares.

It was well-known at the time that psd impliessos in two cases:

In R[x ] (one variable), by theFundamental Theorem of Algebra, and

f ∈ R[X ] of degree 2, by Sylvester’s work.

Hilbert showed (the homogenized version of) the following:

every psd f of degree 4 in R[x , y ] is sos,

in all other cases there exist psd polynomials that are not sos.



In 1893, Hilbert proved that for n = 2 every psd polynomial inR[X ] can be written as a sum of squares of rational functions,equivalently, for such f there exists nonzero h ∈ R[X ] such thath2f ∈

∑R[X ]2.

Unable to prove that this holds for all psdpolynomials, it became the 17th problem onHilbert’s list of 23 problems he gave in hisaddress to the International Congress ofMathematicians in 1900.



E. Artin solve Hilbert’s 17th problem:

Theorem (Artin 1927)

Suppose f ∈ R[X ] is psd. Then there is a nonzero h ∈ R[X ] suchthat h2f is sos.

Note that an identity h2f = g21 + · · ·+ g2

k is a certificate ofpositivity for f on Rn.

Artin (along with Schreier) developed the theory of ordered fieldsin order to solve the 17th problem. And thus the subject of RealAlgebra was born (more or less).



PSD ternary quartics

The first result in Hilbert’s 1888 paper was that every psd ternaryquartic – homogenous polynomial in 3 variables of degree 4 – is asum of squares of three quadratic forms. Hilbert’s proof is not easyto understand and lacks detail in a number of key points.

In 1977, Choi and Lam gave a short elementary proof thatevery psd ternary quartic is a sum of squares of five quadraticforms.

In 2004, A. Pfister gave an elementary proof with fivereplaced by four and a constructive argument in the case thatthe t. q. has a non-trival real zero.

In 2012, Pfister and C. Scheiderer gave a complete proof ofHilberts Theorem, using only elementary tools such as thetheorems on implicit functions and symmetric functions.



Hilbert’s paper did not address constructivequestions:

Given a psd ternary quartic p, how canone find q. forms f , g , h so thatp = f 2 + g2 + h2?

In how many “fundamentally different”ways can this be done?

In 1929, A. Coble showed that every ternary quartic over C can bewritten as a sum of three squares of quadratic forms in 63inequivalent ways.

Work on quartic curves due to D. Plaumann, B. Sturmfels, andC. Vinzant from 2011 gives a new proof of the Coble result,yielding an algorithm for computing all representations in thesmooth case.



What about the computational questions above?

In 2000, in joint work with B. Reznick, we described methodsfor counting representations of a psd ternary quartic as a sumof 3 squares of q. forms. Experimentally, the Coble result (63representations over C) occured, with 8 representations as asum of 3 squares of q. forms over R.

In 2004, in joint work with Reznick, F. Sottile, andC. Scheiderer, we showed that if p = 0 is smooth, a psdternary quartic p has exactly 8 inequivalent representations asa sum of 3 squares of quadratic forms.

In 2010, Scheiderer extended this analysis to the singular case.



Hilbert did not give an explicit example of a psd, not sos,polynomial. The first published examples did not appear untilnearly 80 years later; the most famous example is due to Motzkin:

M(x , y , z) = x4y2 + x2y4 + z6 − 3x2y2z2

M(x , y , z) is psd by the arithmetic-geometric inequality.

Suppose M is sos, then we have

M =∑

i g2i ; gi = Aix

3+Bix2y+Cix

2z+Dixy2+Eixyz+...+Jiz

3

Since M has no x6 term Ai = 0 for all i . Similarly, there are no y3

terms in gi . Then since M has no x4z2 and no y4z2 terms, gi hasno x2z and no y2z terms. Similarly, the gi have no xz2, yz2 terms.

Thus we have M =∑

i (Bix2y + Dixy

2 + Eixyz + Jiz3)2, and the

coefficient of x2y2z2 is∑

i E2i ≥ 0, which is impossible.



Since then, many other examples and families of psd, not sos,polynomials have appeared, due to Robinson; Choi, Lam, andReznick; and others.

More recently, there have been attempts to understand Hilbert’sproof and make it accessible.

In a recent preprint, Reznick has isolated the underlying mechanismof Hilbert’s proof and shown that it applies to more generalsituations. He produced new families of psd forms that are not sos.



The Cayley-Bacharach Theorem saysthat if two cubics in the projective planemeet in nine different points, then everycubic that passes through any eight of thepoints passes through the ninth point.

Hilbert’s proof of the existence of a psd, not sos, form uses thistheorem. His “construction” starts from two cubics in generalposition; Reznick’s work simplifies Hilbert’s proof by starting withspecific cubics.

In 2012, G. Blekherman showed that the Cayley-Bacharachrelations are in some sense the fundamental reason that there arepsd polynomials that are not sos. In small cases, he gives acharacterization of the difference between psd and sos forms.



In a recent preprint, using ideas and results from convex algebraicgeometry, Blekherman, G. Smith, and M. Velasco have proven astriking theorem related to Hilbert’s 1888 paper.

They show that if X is a real irreducible nondegenerate projectivesubvariety where the real points are Zariski dense, then everynonnegative real quadratic form on X is a sum of squares of linearforms iff X is a variety of minimal degree. From this result they areable to deduce Hilbert’s result on psd, not sos, forms.

Further, they show that

the exceptional Veronese surface corresponds to theexceptional case of ternary quartics

thus giving a geometric explanation for the algebraic fact that apsd ternary quartic is a sum of squares of quadratic forms.



Polya’s Theorem for forms positive on a simplex

In 1928, Polya’s proved a representation theoremfor forms positive on the standard simplex∆n := {(x1, . . . , xn) ∈ Rn | xi ≥ 0,

∑i xi = 1}.

Theorem (Polya’s Theorem)

Suppose f ∈ R[X ] is homogeneous and is strictlypositive on ∆n, then for sufficiently large N, all ofthe coefficients of (X1 + · · ·+ Xn)N f are strictlypositive.

Notice that (∑

Xi )N f =

∑aαX

α with all aα > 0 is a certificate ofpositivity for f on ∆n.



Bounds for Polya’s Theorem and generalizations

In 2000, in joint work with Reznick, we gave an explicit boundon the exponent N in terms of the degree, the coefficients andthe minimum of the form on the simplex.

It is not too hard to show that there cannot exist a bound onN which depends only on the degree.

In 2012, in joint work with M. Castle and Reznick, we gave acomplete characterization of forms p such that p is Polyasemi-positive, i.e., for large enough N ∈ N,(X1 + · · ·+ Xn)Np has nonnegative coefficients, along with a

bound on the N needed.

If p has zeros on ∆n, then (X1 + · · ·+ Xn)Np will always havesome zero coefficients.



Application: Copositive programming

Copositive programming is linear programming over the cone ofcopositive matrices:

Cn = {n×n symmetric matrices M | x ·M ·xT ≥ 0, for all x ∈ Rn+}.

In 2001, De Klerk & Pasechnik showed thatthe stability number of a graph can be writtenas the solution to a copositive programmingproblem.

They approximated the copositive programming problem using asequence of cones which converge to Cn and using the degreebounds for Polya’s Theorem, gave results for approximating thestability number of a graph.



Finding SOS decompositions

For a given sos f ∈ R[X ], thanks to the magicof linear algebra, there are effective algorithmsfor writing f as a sum of squares.

Suppose deg f = 2d and

f =k∑

i=1

gi (x)2 (1)

The linear algebra version of (1) is

f (X ) = V · BBT · V T ,

V = monomials of deg ≤ d , and B is the matrix with i-th columnthe coefficients of gi .

A = B · BT is psd symmetric matrix, called a Gram matrix for fcorresponding to (1).



For example, p = x4 + y4 + 1 = (x2)2 + (y2)2 + 12.With V =

[x2 y2 xy x y 1

],

p = V ·

1 0 0 0 0 00 1 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 1

· VT

But also p = (x2 − y2)2 + 2(xy)2 + (1)2 =

V ·

1 −1 0 0 0 0−1 1 0 0 0 00 0 2 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 1

· VT



“f is sos” is equivalent to the existence of a Gram matrix for f ,i.e., a psd symmetric matrix A such that

f (X ) = V · A · V T , (2)

This was proven by Choi, Lam, and Reznick who also noted that:

Inequivalent sos representations of f correspond to differentGram matrices

If A is a Gram matrix for f then rank A is the number ofsquares in the sos representation of f corresponding to A.

Further, there is an easy algorithm for producing an explicitdecomposition f =

∑g2i from a Gram matrix for f .



Suppose f ∈ R[X ] is sos with deg f = 2d , and A is a Gram matrixfor f . Then A is an N × N matrix, where N =

(n+dd

).

The set of all symmetric N × N matrices A such that

f (X ) = V · A · V T (3)

is an affine subset L of the space of N × N symmetric matrices.

f is sos iff L ∩ PN 6= ∅, where PN is the convex cone of psdsymmetric N × N matrices over R.

Finding this intersection is a semidefinite program (SDP). Thereare good numerical algorithms – and software! – for solvingsemidefinite programs. There is even software for deciding if f issos: SOSTools (P. Parrilo et. al.)



Application: Global Optimization

Problem: Find inf{f (α) | α ∈ Rn} (if it exists).

Equivalent to finding the inf of {λ ∈ R | f − λ is psd }.

We can look for the smallest λ so that f − λ is sos:

fsos := min{λ | f − λ ∈∑

R2}. (4)

Then fsos is a lower bound for the minimum of f .

Of course, fsos = −∞ is possible: psd 6⇒ sos!

If fsos 6= −∞, then there is an a priori bound on the degree of thesquares that occur and finding fsos can be implemented as an SDP.



Representations on semialgebraic sets

In 1974, Stengle proved his celebrated Positivstellensatz – arepresentation theorem for polynomials on semialgebraic sets.Later it was discovered that in the 1964, Krivine proved a similartheorem that went largely unnoticed.

A semialgebraic set in Rn is a subset defined by finitely manypolynomial inequalities.

For G = {g1, . . . , gr} ⊆ R[X ], S(G ) denotes the basic closedsemialgebraic set generated by G :

S(G ) = {α ∈ Rn | g1(α) ≥ 0, . . . , gr (α) ≥ 0}.

Question: Suppose f ≥ 0 or f > 0 on S(G ), does there exist acertificate of positivity for f on S(G )?



Fix S = S(G ), there are two algebraic objects associated to G :

The quadratic module generated by G ,

Q(G ) := {s0 + s1g1 + · · ·+ srgr | each si ∈∑

R[X ]2}.

The preorder generated by G ,

P(G ) := {∑

ε∈{0,1}nsε g

ε11 . . . g εrr | sε ∈

∑R[X ]2}.

A representation of f in either Q(G ) or P(G ) yields a certificate ofpositivity for f on S .

Classic Positivstellensatz. (Stengle, Krivine) f > 0 on S iff thereexist h1, h2 ∈ P(G ) such that f h1 = 1 + h2.



In the 1990’s, connections with moment problems were discovered,which led to “denominator free” representation theorems due toSchmudgen and Putinar.

Fix a basic closed semialgebraic set S = S(G ), the correspondingquadratic module Q = Q(G ), and the corresponding preorderP = P(G ).

P (resp. Q) is archimedean if there is N ∈ N such thatN −

∑X 2i ∈ P (resp. Q).

Notice that N −∑

X 2i ∈ P =⇒ N −

∑X 2i ≥ 0 on S . Hence S

must be contained in the compact set {N −∑

X 2i ≥ 0} and

therefore is compact. And similarly for Q.

We can think of an explicit representation of N −∑

X 2i ∈ P or Q

as a “certificate of compactness” for S .



Two classic theorems in real algebraic geometry

We continue to assume S is our basic closed semialgebraic set andP the corresponding preorder.

Theorem

S 6= ∅ if and only if −1 6∈ P.

Theorem

S is compact if and only if P is archimedean.

This is algebraic geometry at its most beautiful – very naturalalgebraic properties correspond to very natural geometricproperties.

Quadratic modules are simpler computationally than preorders andare very useful. However, it is not true that if S is compact thenthe corresponding quadratic module is always archimedian.



Representations on compact semialgebraic sets

Theorem (Schmudgen, 1991)

If S is compact, then f > 0 on S implies f ∈ P.

Theorem (Putinar, 1993)

If Q is archimedean and f > 0 on S , then f ∈ Q.

This means that if S is compact, then there exists a certificate ofpositivity for every f > 0 on S . Notice that these hold *regardlessof the polynomials we choose to describe S*.

Schuudgen’s result was a corollary to his solution to themulti-dimensional Moment Problem for compact semialgebraicsets. The proof uses the classic Positivstellensatz tonon-contructively produce a certificate of compactness for S .



Optimizing on compact semialgebraic sets

Suppose S = S(g1, . . . , gk) is a compact semialgebraic set and Qthe quadratic module. Assume Q is archimedean, so Putinar’sTheorem applies.

For f ∈ R[X ], let f∗ denote the minimum of f on S (which alwaysexists, since S is compact). Let

fsos := min{λ | f − λ ∈ Q}.

As in the sos case, the problem of finding a representation of f − λin Q with a fixed upper bound on the degree of the sums ofsquares that occur can be implemented as an SDP.

Unfortunately, there cannot be any such bound in terms of deg fonly – any bound must depend on some measure of how close f isto having a zero on Q.



The Lasserre Method

Unlike the global optimization case, fsos always exists.

But we cannot implement finding fsos as a SDP directly.

Let Qa := {g ∈ Q | deg g ≤ a}.

J.-P. Lasserre developed a method for approximating f∗ byimplementing a series of SDP’s corresponding to finding

f asos := min{λ | f − λ ∈ Qa}.By Putinar’s Theorem, the values of f asos converge to f∗.

We can ensure that Q is archimedean by adding a certificate ofcompactness N −

∑X 2i to the set of generators.

Question: Given G = {g1, . . . , gk | ⊆ R[X ], is there an algorithm todecide if S(G ) is compact and/or compute N so thatN −

∑X 2i > 0 on S?



In 2002, M. Schweighofer gave a proof of Schmudgen’s theoremthat is algorithmic modulo one application of the classicPositivstellensatz. His proof reduces the theorem to Polya’sTheorem.

Using the bounds for Polya’s Theorem, Schweighofer obtainsdegree bounds for the sums of squares needed in a represenation,in terms of the coefficients and degrees of the “input data” andthe degree of the sums of squares needed in a certificate ofcompactness.

In 2007, Schweighofer and J. Nie gave a similar constructive proofof Putinar’s Theorem using Polya’s Theorem with bound.

Their results give information about the convergence rate ofLasserre’s method for optimization.



Question: Can the Polya-with-zeros theorem allow the constructiveproof of Schmudgen’s Theorem and the bound to be generalized toinclude polynomials in the preorder which have zeros on S? (OrPutinar’s Theorem)

One problem: The algorithm in the Schweighofer’s proof yieldsrepresentations in a particularly nice form. It is not hard to showthat not all elements in P have this nice form – there is somethingspecial about f which are strictly positive on S .

A corollary to a (deep!) result of C. Scheiderer says that any fwhich is nonnegative on the square{1− x ≥ 0, 1 + x ≥ 0, 1− y ≥ 0, 1 + y ≥ 0} is in the quadraticmodule generated by {1± x , 1± y}.

Open question: Does there exist an algorithmic/constructive proofof this?



Rational sums of squares

Consider the following example, due to C. Hillar:

Is f = 3− 12x − 6x3 + 18y2 + 3x6 + 12x3y − 6xy3 + 6x2y4 sos?

Our SDP program might say yes and return a certificate:

f = (x3 + 3.53y + .347xy2 − 1)2 + (x3 + .12y + 1.53xy2 − 1)2+

(x3 + 2.35y − 1.88xy2 − 1)2

Note that f− RHS has terms such as −.006xy2 which are nonzero.Is f really a sum of squares?

It turns out that our SDP program has approximated an sosdecomposition for f of the form

(x3+a2y+bxy2−1)2+(x3+b2y+cxy2−1)2+(x3+c2y+axy2−1)2,

where a, b, c are real roots of the cubic x3 − 3x + 1.Vicki Powers Representations of Positive Polynomials: Theory, Practice, and Applications


For many applications of certificates of positivity, we need exactcertificates, but SDP’s are numerical.

In theory, SDP problems can be solved purely algebraically, e.g.,using quantifier elimination. In practice, this is impossible for allbut trivial problems.

Work by J. Nie, K. Ranestand, and B. Sturmfels in 2010 showsthat optimal solutions of relatively small SDP’s can have minimumdefining polynomials of huge degree.

Much more promising are hybrid symbolic-numeric approaches.



Problem: Given a numerical (approximate) certificate f =∑

g2i

(via SDP software, say) find an exact certificate.

Peyrl and Parrilo gave an algorithm for solving this problem, insome cases. The idea: We want to find a symmetric psd matrix Awith rational entries so that

f = V · A · V T (5)

The SDP software will produce a psd matrix A which onlyapproximately satisfies (5). The idea is to project A onto the affinespace of solutions to (5) in such a way that the projection remainsin the cone of psd symmetric matrices.



The Peyrl-Parrilo method is (in theory!)guaranteed to work IF there exists a rationalsolution and the underlying SDP is strictlyfeasible, i.e., there is a solution with full rank.

In 2009, E. Kaltofen, B. Li, Z. Yang, and L. Zhi generalized thetechnique of Peyrl and Parrilo and found sos certificates certifyingrational lower bounds for some approximate polynomial g.c.d. andirreducibility problems.



Rational sums of squares decompositions

The above algorithms assume that there is a rational sos certificatefor f . But maybe one doesn’t exist, even if f has rationalcoefficients.

Suppose f ∈ Q[X ] is in∑

R[X ]2.

Sturmfels asked: Is f ∈∑

Q[X ]2?

A trivial, but illustrative example:

2x2 = (√

2x)2 ∈ R[x ]2

2x2 = x2 + x2 ∈∑

Q[x ]2



Recall the Hillar example

f = 3− 12x − 6x3 + 18y2 + 3x6 + 12x3y − 6xy3 + 6x2y4,

which is in∑

R[X ]2.

It turns out that f ∈∑

Q[X ]2:

f = (x3 + xy2 +3

2y − 1)2 + (x3 + 2y − 1)2+

(x3 − xy2 +5

2y − 1)2 + (2y − xy2)2 +

3

2y2 + 3x2y4

In both of these examples, the sos decompositions in∑

Q[X ]2 usemore squares than the ones in

∑R[X ]2.



Sturmfels asked: If f ∈ Q[X ] and f ∈∑

R[X ]2, is f ∈∑

Q[X ]2?

In 2009, Hillar showed that the answer to Sturmfel’s question is“yes” if f ∈

∑K 2, where K is a totally real extension of Q. He

gave bounds for the number of squares needed.

There is a simple proof of a slightly more general result with abetter bound given (independently) by Kaltofen, Scheiderer, andQuarez.

In 2012, Scheiderer showed that the answer is “no”, in general.



Back to the compact semialgebraic sets

What can we say about rational certificates of positivity in thecompact semialgebraic set case?

Fix G = {g1, . . . , gk} ⊆ R[X ] and let S = S(G ), the basic closedsemialgebraic set generated by G , Q = Q(G ), the quadraticmodule generated by G , and P = P(G ), the preorder generated byG .

Question: Suppose G ⊆ Q[X ], f ∈ Q[X ], and f > 0 on S . Doesthere exist a rational certificate of positivity for f on S?



Theorem

Suppose S = S(G ) is compact. Then for any f ∈ Q[X ] such thatf > 0 on S , there is a representation of f in the preordering P(G ),

f =∑

e∈{0,1}sσeg

e11 . . . g es

s ,

with all σe ∈∑

Q[X ]2.

The proof follows from an algebraic proof of Schmudgen’sTheorem, due to T. Wormann.



Theorem (2011)

Suppose G = {g1, . . . , gs} ⊆ Q[X ] and N −∑

X 2i ∈ Q for some

N ∈ N. Then given any f ∈ Q[X ] such that f > 0 on S , thereexist σ0 . . . σs , σ ∈

∑Q[X ]2 so that

f = σ0 + σ1g1 + · · ·+ σsgs + σ(N −∑

X 2i ).

This follows from the algorithmic proof of Putinar’s Theorem dueto Schweighofer. We follow the proof, making sure each steppreserves rationality.

Open question: If G ⊆ Q[X ] and N −∑

X 2i ∈ Q, can we find a

certificate of compactness N −∑

X 2i in Q with the sos’s in∑

Q[X ]2?



Other Topics

There are so many interesting topics that we haven’t mentioned!Here are some topics we missed:

Representations on noncompact semialgebraic sets

Invariant sums of squares

Convex real algebraic geometry

Positivity in a non-commutative setting, free positivity.

Applications to graph theoretic problems – max cut, maxclique and others.

Applications to control theory, e.g., Lapynov functions

Applications to operator algebras and functional analysis, e.g.,moment problems.

Linear matrix inequalities, hyperbolic polynomials, spectralcones



Thanks for listening!


Documents

Representations of Positive Polynomials: Theory, Practice ...vicki/talks/TempleNov2014_handout.pdf · R[X]2, from which one can deduce the positivity condition immediately. A theorem