 # Size-Degree Trade-Offs for Sums-of-Squares and Positivstellensatz Proofs

We show that if a system of degree-k polynomial inequalities on n Boolean variables has a Sums-of-Squares (SOS) proof of unsatisfiability with at most s many monomials, then it also has one whose degree is of the order of the square root of n s plus k. A similar statement holds for the more general Positivstellensatz (PS) proofs. This establishes size-degree trade-offs for SOS and PS that match their analogues for weaker proof systems such as Resolution, Polynomial Calculus, and the proof systems for the LP and SDP hierarchies of Lovász and Schrijver. As a corollary to this, and to the known degree lower bounds, we get optimal integrality gaps for exponential size SOS proofs for sparse random instances of the standard NP-hard constraint optimization problems. We also get exponential size SOS lower bounds for Tseitin and Knapsack formulas. The proof of our main result relies on a zero-gap duality theorem for pre-ordered vector spaces that admit an order unit, whose specialization to PS and SOS may be of independent interest.

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## 1 Introduction

A key result in the theory of semi-algebraic geometry is the Positivstellensatz [29, 18], whose weak form gives a version of the Nullstellensatz for semi-algebraic sets: A system of polynomial equations and polynomial inequalities on commuting variables has no solution over reals if and only if

 −1=s∅+∑J⊆[ℓ]J≠∅sJ∏j∈Jqj+∑j∈[m]tjpj, (1)

where the are sums of squares of polynomials, and the are arbitrary polynomials. Based on this, Grigoriev and Vorobjov  defined the Positivstellensatz (PS) proof system for certifying the unsatisfiability of systems of polynomial inequalities, and initiated the study of its proof complexity.

For most cases of interest, the statement of the Positivstellensatz stays true even if the first sum in (1) ranges only over singleton sets . This special case of PS yields a proof system called Sums-of-Squares (SOS). Starting with the work in , SOS has received a good deal of attention in recent years for its applications in algorithms and complexity theory. For the former, through the connection with the hierarchies of SDP relaxations [19, 24, 23, 6]. For the latter, through the lower bounds on the sizes of SDP lifts of combinatorial polytopes [9, 21, 20]. We refer the reader to the introduction of 

for a discussion on the history of these proof systems and their relevance for combinatorial optimization.

In this paper we concentrate on the proof complexity of PS and SOS when their variables range over the Boolean hypercube, i.e., the variables come in pairs of twin variables and , and are restricted through the axioms , and . This case is most relevant in combinatorial contexts. It is also the starting point for an obvious link with the traditional proof systems for propositional logic, such as Resolution, through the realization that monomials represent Boolean disjunctions, i.e., clauses. In return, this link brings concepts and methods from the area of propositional proof complexity to the study of PS and SOS proofs.

A question that is suggested by this link is whether, in analogy with several other propositional proof systems in the literature, the monomial size of a proof can be traded for its degree. For a proof as in (1), the monomial size of the proof is the number of monomials in an explicit sum-of-monomials representation of the summands in the right-hand side. The degree of the proof is the maximum of the degrees of those summands. These are the two most natural measures of complexity for PS proofs. The importance of the question whether size can be traded for degree stems from the fact that the complexity of PS and SOS proofs is relatively well understood when it is measured by degree, but rather poorly understood when it is measured by monomial size. If size could be traded for degree, then strong lower bounds on degree would transfer to strong lower bounds on monomial size. The converse has long been known by elementary linear algebra.

In this paper we answer the size-degree trade-off question for PS and SOS. We show that if a system of degree- polynomial inequalities on pairs of twin variables has a PS proof of unsatisfiability with no more than many monomials in total, then it also has one of degree , where is a bound on the number of inequalities that are multiplied together in (1). By taking , this yields a size-degree trade-off for SOS as a special case.

Our result matches its analogues for weaker proof systems that were considered before. Building on the work of  and , a size-width trade-off theorem was established for Resolution: a proof with many clauses can be converted into one in which all clauses have size , where is the size of the largest initial clause . The same type of trade-off was later established for monomial size and degree for the Polynomial Calculus (PC) in , and for proof length and rank for LS and LS , i.e., the proof systems that come out of the Lovász-Schrijver LP and SDP hierarchies . To date, the question for PS and SOS had remained open, and is answered here111Besides the proofs of the trade-off results for LS and LS, the conference version of  claims the result for the stronger Sherali-Adams and Lasserre/SOS proof systems, but the claim is made without proof. The very last section of the journal version  includes a sketch of a proof that, unfortunately, is wrong. See the forthcoming discussion for a comparison that clarifies how our proof is based on, and generalizes, the one for LS and LS in ..

Our proof of the trade-off lemma for PS follows the standard pattern of such previous proofs with one new key ingredient. Suppose is a system of equations and inequalities that has a size refutation. Going back to the main idea from , the argument for getting a degree refutation goes in four steps: (1) find a variable that appears in many large monomials, (2) set it to a value to kill all monomials where it appears, (3) induct on the number of variables to get refutations of and which, if is small enough, are of degrees and , respectively, and (4) compose these refutations together to get a degree refutation of . The main difficulty in making this work for PS is step (4), for two reasons.

The first difficulty is that, unlike Resolution and the other proof systems, whose proofs are deductive, the proofs of PS are formal identities, also known as static. This means that, for PS, the reasoning it takes to refute from the degree refutation of and the degree refutation of needs to be witnessed through a single polynomial identity, without exceeding the bound on the degree. This is challenging because the general simulation of a deductive proof by a static one incurs a degree loss. The second difficulty comes from the fact that, for establishing this identity, one needs to use a duality theorem that is not obviously available for degree-bounded PS proofs. What is needed is a zero-gap duality theorem for PS proofs of non-negativity that, in addition, holds at each fixed degree of proofs. For SOS, the desired zero-gap duals are provided by the levels of the Lasserre hierarchy. This was established in  under the sole assumption that the inequalities include a ball contraint for some . In the Boolean hypercube case, this can be assumed without loss of generality. For PS, we are not aware of any published result that establishes what we need, so we provide our own proof. At any rate, one of our contributions is the observation that a zero-gap duality theorem for PS-degree is all that is needed in order to complete step (4) in the proof of the trade-off lemma. We reached this conclusion from trying to generalize the proofs for LS and LS from  to SOS. In those proofs, the corresponding zero-gap duality theorems are required only for the very special case where and for deriving linear inequalities from linear constraints. The fact that these hold goes back to the work of Lovász and Schrijver .

In the end, the zero-gap duality theorem for PS-degree turned out to follow from very general results in the theory of ordered vector spaces. Using a result from  that whenever a pre-ordered vector space has an order-unit a zero-gap duality holds, we are able to establish the following general fact: for any convex cone of polynomials of at most some even degree , if the cone contains all squares of degree and is closed under equality modulo the ideal generated by the , then a zero-gap duality holds. The conditions obviously hold for PS-degree and SOS-degree in the Boolean hypercube case, and we have what we want.

In Section 5 we list some of the applications of the size-degree trade-off for PS that follow from known degree lower bounds. Among these we include exponential size SOS lower bounds for Tseitin formulas, Knapsack formulas, and optimal integrality gaps for sparse random instances of MAX-3-XOR and MAX-3-SAT. Except for Knapsack formulas, for which size lower bounds follow from an easy random restriction argument applied to the degree lower bounds in [11, 13], these size lower bounds for SOS appear to be new.

## 2 Preliminaries

For a natural number we use the notation for the set . We write and for the sets of non-negative and positive reals, respectively and for the set of natural numbers. The natural logarithm is denoted , and denotes base exponentiation.

### 2.1 Polynomials and the Boolean ideal

Let and be two disjoint sets of variables. Each is called a pair of twin variables, where is the basic variable and is its twin. We consider polynomials over the ring of polynomials with real coefficients and commuting variables , which we write simply as . The intention is that all the variables range over the Boolean domain , and that . Accordingly, let be the Boolean ideal, i.e., the ideal of polynomials generated by the following set of Boolean axioms on the pairs of twin variables:

 Bn={x2i−xi:i∈[n]}∪{¯x2i−¯xi:i∈[n]}∪{xi+¯xi−1:i∈[n]} (2)

We write if is in .

A monomial is a product of variables. A term is the product of a non-zero real and a monomial. A polynomial is a sum of terms. For , we write for the monomial , so polynomials take the form for some finite . The monomial size of a polynomial is the number of terms, and is denoted . A sum-of-squares polynomial is a polynomial of the form , where each is a polynomial in . For a polynomial we write for its degree. We think of as an infinite dimensional vector space, and we write for the subspace of polynomials of degree at most .

### 2.2 Sums-of-Squares proofs

Let be an indexed set of polynomials. We think of the polynomials as inequality constraints, and of the polynomials as equality constraints:

 q1≥0,…,qℓ≥0,p1=0,…,pm=0. (3)

Let be another polynomial. A Sums-of-Squares (SOS) proof of from is a formal identity of the form

 p=s0+∑j∈[ℓ]sjqj+∑j∈[m]tjpj+∑q∈Bnuqq, (4)

where and are sums of squares of polynomials, for , and and all are arbitrary polynomials. The proof is of monomial size at most  if

 k0∑i=1size(ri,0)+∑j∈[ℓ]kj∑i=1size(ri,j)+∑j∈[m]size(tj)≤s.

This definition of size corresponds to the size of an explicit SOS proof. Accordingly, the monomials of the ’s and the ’s are called the explicit monomials of the proof. The proof is of degree at most if and for each , and for each .

Note that the number of monomials in the polynomials are not considered in the definition we have chosen for monomial size of a proof. Our aim is to prove lower bounds in the number of monomials in a proof, and, of course, lower bounds for our more liberal definition of monomial size imply lower bounds on the number of all monomials in the proof.

Similarly, the degrees of the polynomials do not play a role in the definition of the degree of a proof. However, in this case we can always assume that the degree of the products does not surpass in a proof of degree . This follows from the fact that is a Gröbner basis for with respect to any monomial ordering – one can see this quite easily using Buchberger’s Criterion (see e.g. ). In other words, this means that lower bounds for the restricted definition of degree imply lower bounds also for our liberal definition of degree.

### 2.3 Positivstellensatz proofs

This will be an extension of SOS. Let be an indexed set of polynomials interpreted as in (3). A Positivstellensatz proof (PS) of from is a formal identity of the form

 p=s0+∑J∈JsJ∏j∈Jqj+∑j∈[m]tjpj+∑q∈Bnuqq, (5)

where is a collection of non-empty subsets of , each is a sum-of-squares polynomial, , and each and is an arbitrary polynomial. The proof has monomial size at most  if

 k0∑i=1size(ri,0)+∑J∈JkJ∑i=1size(ri,J)+∑j∈[m]size(tj)≤s.

The proof is of degree at most  if and for each , and for each . The explicit monomials of the proof are the monomials of the ’s and the ’s. It should be noted that PS applied to a that contains at most inequality constraint (i.e., ) is literally equivalent to SOS.

As in SOS proofs, the definitions of monomial size and degree of a proof do not take into account the polynomials. Likewise, the monomials in the products do not contribute to the definition of monomial size. As above, this liberal definition plays in favour of lower bound results in the case of monomial size. For degree, ignoring the ’s does not really matter, again, because is a Gröbner basis for .

## 3 Duality

In this section we start by recalling some basic facts about ordered vector spaces from . We prove the results for pre-ordered vector spaces rather than ordered ones since the polynomial spaces we will apply the results to carry a natural pre-order.

### 3.1 Vector spaces with order unit

A pre-ordered vector space is a pair , where is a real vector space and is a pre-order that respects vector addition and multiplication by a non-negative scalar, i.e. the following hold for all and :

• and only if ;

• only if .

Pre-ordered vector spaces arise naturally from convex cones of real vector spaces. If is a convex cone, then the relation defined by if satisfies the above requirements. An element is an order unit for if for any there is some such that .

For the rest of this section let be a pre-ordered vector space with an order unit .

###### Lemma 1.

The following hold.

• ;

• For every and with , if , then .

• For every there is such that ;

• If , then for every .

###### Proof.

(i) There is some such that , i.e. , and so . (ii) Now and so . Thus , i.e. . (iii) Let be such that and let be such that , and let . Now (iv) Suppose and let be such that . Now also and so , i.e. . ∎

Let be a subspace of . A linear functional is positive if implies for all . Equivalently, is positive if it is order-preserving, i.e., if implies for all . A positive linear functional on is a pseudo-expectation if . We denote the set of all pseudo-expectations of by .

Suppose contains the order unit and let . By Lemma 1.(iii) the following two sets are non-empty:

 DUp ={v∈U:p≥v}, UUp ={v∈U:v≥p}.

Thus, if is any positive linear functional that is defined on , then and are real numbers and .

###### Lemma 2.

Let be a subspace of containing the order unit , and let be a positive linear functional on . Then for any and for any satisfying there is a positive linear functional that is defined on , that extends , and such that .

###### Proof.

Every element of can be written uniquely in form , where and . Define by

 L′(ap+v)=aγ+L(v).

It is easy to check that is linear map. We show that is positive by considering a few cases.

Case (i) . If and , then and . Case (ii) . Suppose that and . Then , and so , i.e. . Case (iii) . Suppose that and . Then , and so . Hence , and so . ∎

Now we can prove the general duality theorem for pre-ordered vector spaces that admit an order unit. For a more general version of this result, see .

###### Theorem 1.

For any it holds that

 sup{r∈R:p≥re}=inf{E(p):E∈E(V)}.

Moreover, if the common value is a real number, then there is a pseudoexpectation achieving this value, i.e., is well-defined.

###### Proof.

The inequality from left to right is clear. For the inequality from right to left we distinguish two cases: whether or not. If , then , since , so . On the other hand by Lemma 1.(iv), so the claim follows. If , then implies , so the map defined by for all is a positive linear functional on . Since is non-empty, is finite, say . Thus, to prove the theorem, it suffices to show that there is some pseudoexpectation extending such that .

Consider the subspace , which contains , the map , and the corresponding upper and lower bounds and , which obviously satisfy . If , say , then since implies , and . If , then by Lemma 2, there is a positive linear functional extending on such that . Now consider the set of all positive linear functionals that are defined on some subspace and satisfy and . By the argument above . On the other hand is closed under unions of chains and so, by Zorn’s lemma, there is some maximal .

Now the domain of is the whole of , since otherwise we could extend by using Lemma 2, contradicting the maximality of . Hence is the pseudoexpectation we are after. ∎

### 3.2 Order units for semi-algebraic proof systems

We give a sufficient condition on convex cones of polynomials for its associated pre-ordered vector space to have the constant polynomial as an order unit. We say is closed under mod if whenever and , then , for any . Note that the claim is made only for spaces of polynomials of at most some even degree.

###### Lemma 3.

Let be a convex cone that contains for any polynomial of degree at most , and that is closed under mod . For every polynomial of degree at most , there is such that .

###### Proof.

We prove, by induction on , that for every term with there is some such that . The claim follows then immediately. In the following, write , where and is a monomial, and . For constant terms the claim is trivial, so assume .

If , we have two cases. If , then and the degrees are at most since ; thus follows since contains all squares and is closed under mod . If on the other hand , then and the degrees are at most again; thus .

Assume then that and let and be monomials such that and and . Now if , then

 as1/2+as+as2/2≡(√a/2s1+√a/2s2)2modIn

and the degrees are at most . Now, by induction assumption applied to and , there exist such that and . Hence . On the other hand, if , then

 −as1/2+as+−as2/2≡(√−a/2s1−√−a/2s2)2modIn

and the degrees are at most . Again the induction hypothesis applied to and gives such that and , and so . ∎

Next we apply Lemma 3 to the concrete semi-algebraic proof systems from Section 2. Let be an indexed set of polynomials interpreted as constraints as in (3). A cut-off function for is a function that satisfies for each , and for each . A PS proof as in (5) has degree mod at most if and for each , and for each . It has product-width at most if each has cardinality at most .

Let be the convex cone of all polynomials in that have a PS proof from of degree mod at most , and product-width at most . We will write if , and denote by the set of pseudo-expectations over the pre-ordered vector space determined by .

It is obvious from the definitions that is a convex cone that contains all squares of polynomials of degree at most , and that it is closed under mod . Therefore, Lemma 3 and gives:

###### Corollary 1.

Let be a positive integer, let be an indexed set of polynomials, let be a cut-off function for , let be a positive integer, and let be a polynomial of degree at most . Then

 sup{r∈R:Q⊢cw,2dp≥r}=inf{E(p):E∈Ecw,2d(Q)}.

Moreover, if the common value is a real number, then there is a pseudoexpectation achieving this value; i.e., is well-defined.

Note that the definitions of degree for SOS and PS proofs as defined in Section 2 are special cases of the definitions above, for appropriate choices of and . Thus, Corollary 1 gives Duality Theorems for them.

In this section we prove the following.

###### Theorem 2.

For every two natural numbers and , every indexed set of polynomials of degree at most with pairs of twin variables, and every two positive integers and , if there is a PS refutation from of monomial size at most and product-width at most , then there is a PS refutation from of degree at most .

An immediate consequence is a degree criterion for size lower bounds:

###### Corollary 2.

Let be an indexed set of polynomials of degree at most with pairs of twin variables, and let be a positive integer. If is the minimum degree and is the minimum monomial size of PS refutations from of product-width at most , and , then .

The proof of Theorem 2 will follow the standard structure of proofs for degree-reduction lemmas for other proof systems, except for some complications in the unrestricting lemmas. These difficulties come from the fact that PS proofs are static, which complicates matters, and force us to apply the Duality Theorem for the notion of degree modulo cut-off functions that we introduced in Section 3.

### 4.1 Unrestricting lemmas

For all the lemmas in this section, fix three positive integers , and , an indexed set of polynomials on the pairs of twin variables, and a cut-off function for .

###### Lemma 4.

Let and be polynomials of degree at most . If , then for any .

###### Proof.

The assumption that implies that both and belong to . Hence for any . ∎

###### Lemma 5.

Let be one of the variables and let be a monomial of degree at most . Then implies for any .

###### Proof.

First we prove the lemma when has degree at most . By linearity we have . Moreover, both terms on the right-hand side are non-negative by Lemma 4 since and , and all degrees are at most . Hence and . It follows that if , then .

Now we prove the lemma when has degree at most . Let and be two monomials of degree at most such that . By the previous paragraph, . By linearity also . Now observe that and all the degrees are at most . Therefore by Lemma 4. Also and therefore . ∎

For a polynomial on the pairs of twin variables, an index, and a Boolean value, we denote by the polynomial that results from assigning to and to in . We extend the notation to indexed sets of such polynomials through . Note that and are polynomials on pairs of twin variables, and their degree is at most that of and , respectively.

In the following, we say that the cut-off function is strict if for each and for each . Observe that if is a strict cut-off function for , then it is also a strict cut-off function for .

###### Lemma 6.

Let , let and be the extensions of with the polynomials and , respectively, let be the extension of that maps to , and assume that is strict. The following hold:

1. The function is a strict cut-off function for both and ,

2. If , then .

3. If , then .

###### Proof.

() is obvious, given the assumption that is strict. We prove (i); the proof of (ii) is symmetric. Suppose that , say:

 −1=s0+∑J∈JsJ∏j∈Jqj[i/0]+∑j∈[m]tjpj[i/0]+∑q∈Bntqq[i/0]. (6)

For , write , let and and note that

 qj[i/0]=qj+∑α∈Jjaj,α(xα/xαii)(−xαii)+∑α∈Kjaj,α(xα/¯xαn+ii)(1−¯xαn+ii).

Therefore where

 tj=∑α∈Kjaj,α(xα/¯xαn+ii)−∑α∈Jjaj,α(xα/xαii).

Note that since for and for . Now

 sJ∏j∈Jqj[i/0] ≡sJ∏j∈J(qj+tjxi)modIn ≡sJ∑T⊆J∏j∈Tqj∏j∈J∖T(tjxi)modIn ≡sJ∏j∈Jqj+(∑T⊆JT≠JsJ∏j∈Tqj∏j∈J∖Ttj)ximodIn.

Moreover, and, for every , we have:

 deg(sJ∏j∈Tqj∏j∈J∖Ttj) ≤deg(sJ)+∑j∈Tdeg(qj)+∑j∈J∖Tdeg(tj) ≤2d−c(J)+∑j∈Jdeg(qj)−1 ≤2d−2.

The second inequality follows from the facts that and for all . The third inequality follows from strictness. Hence, . A similar and easier argument with and in place of and shows that . This gives proofs for all terms in the right-hand side of (6), and the proof of the lemma is complete. ∎

###### Lemma 7.

Let , let and be the extensions of with the polynomials and , respectively, let be the extension of that maps to , and assume is strict. The following hold:

• The function is a strict cut-off function for both and .

• If , then for any .

• If , then for any .

###### Proof.

() is obvious. We prove (i); the proof of (ii) is symmetric. Suppose towards a contradiction that there is such that . We want to show that is also in . This contradicts the assumption that . Let

 s0+∑J∈JsJ∏j∈Jqj+∑j∈[m]tjpj+tm+1xi+∑q∈Bntqq (7)

be a proof from of degree mod at most and product-width at most . First note that . Therefore, Lemma 5 applies to all the monomials of , so . The rest of (7) will get a non-negative value through , since by assumption is in and is restricted to . Thus, is in . ∎

###### Lemma 8.

Let and assume that and is strict. The following hold:

• If and , then .

• If and , then

###### Proof.

First note that , and , so

 ⊢cw,2d−¯xixi≥0. (8)

Next we prove (i); the proof of (ii) is entirely analogous.

Assume . By Lemmas 6 and 7 and we have for any . By the Duality Theorem, the infimum of over is achieved, so there exist such that for any . By the Duality Theorem again, this means that