The sum-of-squares hierarchy on the sphere, and applications in quantum information theory

We consider the problem of maximizing a homogeneous polynomial on the unit sphere and its hierarchy of Sum-of-Squares (SOS) relaxations. Exploiting the polynomial kernel technique, we obtain a quadratic improvement of the known convergence rate by Reznick and Doherty&Wehner. Specifically, we show that the rate of convergence is no worse than $O(d^2/\ell^2)$ in the regime $\ell \geq \Omega(d)$ where $\ell$ is the level of the hierarchy and $d$ the dimension, solving a problem left open in the recent paper by de Klerk&Laurent (arXiv:1904.08828). Importantly, our analysis also works for matrix-valued polynomials on the sphere which has applications in quantum information for the Best Separable State problem. By exploiting the duality relation between sums of squares and the DPS hierarchy in quantum information theory, we show that our result generalizes to nonquadratic polynomials the convergence rates of Navascu\'es, Owari&Plenio.


Introduction
We consider in this paper a fundamental computational task, that of maximizing a multivariate polynomial p ∈ R[x] in d variables x = (x 1 , . . . , x d ) on the unit sphere: (1) where S d−1 = {x ∈ R d : x 2 1 + · · · + x 2 d = 1}. Optimization problems of the above form have applications in many areas. For example, computing the largest stable set of a graph is a special case of (1) for a suitable polynomial p of degree three, see [Nes03,DK08]. Computing the 2 → 4 norm of a matrix A corresponds to the maximization of the degree-four polynomial p(x) = Ax 4 4 on the sphere, see e.g., [BBH + 12] for more on this. In quantum information, the so-called Best Separable State problem very naturally relates to polynomial optimization on the sphere, as we explain later.
When p(x) is quadratic, problem (1) reduces to an eigenvalue problem which can be solved efficiently. However for general polynomials of degree greater than two, the problem is NP-hard as it contains as a special case the stable set problem [Nes03]. The sum-of-squares hierarchy is a hierarchy of semidefinite relaxations that approximate the value p max by a sequence of semidefinite programs of increasing size [Par00,Las01]. In this paper we study the approximation quality of this sequence of semidefinite relaxations.

Sum-of-squares hierarchy
The sum-of-squares hierarchy to approximate (1) is defined by The sequence (p ℓ ) ℓ∈N consists of monotone upper bounds on p max , i.e., for any ℓ we have p max ≤ p ℓ and p ℓ ≤ p ℓ−1 . For each ℓ, the value p ℓ can be computed by a semidefinite program of size d O(ℓ) , see e.g., [Par00,Las01].
A result of Reznick [Rez95] (see also [DW12]) shows that p ℓ → p max as ℓ → ∞. In fact Reznick shows, assuming p min = min x∈S d−1 p(x) = 0, that p ℓ /p max converges to 1 at the rate d/ℓ, for ℓ large enough. In this paper we show that the sum-of-squares hierarchy actually converges at the faster rate of (d/ℓ) 2 . More precisely, we prove the following Theorem 1 Assume p(x 1 , . . . , x d ) is a homogeneous polynomial of degree 2n in d variables with n ≤ d, and let p min denote the minimum of p on S d−1 . Then for any ℓ ≥ C n d 1 ≤ p ℓ − p min p max − p min ≤ 1 + (C n d/ℓ) 2 (2) for some constant C n that depends only on n.
In a recent paper, de Klerk and Laurent [dKL19] proved that a semidefinite hierarchy of lower bounds on p max converges at a rate of O(1/ℓ 2 ) and left open the question of whether the same is true for the hierarchy (p ℓ ) of upper bounds. Our Theorem 1 answers this question positively.

Matrix-valued polynomials
The proof technique we use in this paper actually allows us to get a significant generalization of Theorem 1, related to matrix-valued polynomials. Let S k be the space of real symmetric matrices of size k×k, and let S k [x] be the space of S k -valued polynomials in x = (x 1 , . . . , x d ). We will often use the lighter notation F ∈ S[x] when the size k is unimportant for the discussion. A polynomial F (x 1 , . . . , x d ) ∈ S[x] is positive if F (x) ≥ 0 for all x ∈ R d where the inequality is interpreted in the positive semidefinite sense. We say that F (x) ∈ S k [x] is a sum of squares if there exist polynomials U j (x) ∈ R k×k [x] such that F (x) = j U j (x)U j (x) T for all x ∈ R d . We say that F (x) is ℓ-sos on S d−1 if it agrees with a sum-of-squares polynomial on the sphere with deg U j ≤ ℓ. We are now ready to state our main theorem on sum of squares representations for matrix-valued polynomials.
Theorem 2 Assume F (x 1 , . . . , x d ) ∈ S[x] is a homogeneous matrix-valued polynomial of degree 2n in d variables with n ≤ d, such that F (x) is symmetric for all x. Assume furthermore that 0 ≤ F (x) ≤ I for all x ∈ S d−1 , where I is the identity matrix. There are constants C n and C ′ n that depend only on n such that for any ℓ ≥ C n d, F + C ′ n d ℓ 2 I is ℓ-sos on S d−1 .
Some remarks concerning the statement are in order: • Theorem 1 is a direct corollary of Theorem 2 where F (x) is the scalar polynomial given by F (x) = (p max − p)/(p max − p min ).
• A remarkable fact of Theorem 2 is that the result is totally independent on the size of the matrix F (x).
• Theorem 2 can be applied to get sum-of-squares certificates for scalar bihomogeneous polynomials on products of two spheres S k−1 × S d−1 . Indeed, one way to think about a matrix-valued polynomial F (x 1 , . . . , x d ) ∈ S k [x] is to consider the real-valued polynomial p(x, y) = y T F (x)y where x ∈ R d and y ∈ R k . This polynomial is bihomogeneous of degree (2n, 2) in the variables (x, y). One important application of this setting is in quantum information theory for the best separable state problem which we explain later in the paper.
• As stated, Theorem 2 is concerned only with levels ℓ ≥ Ω(d) of the sum-of-squares hierarchy. The main technical result we prove in this paper (Theorem 6 below) actually allows us to get a bound on the performance of the sum-of-squares hierarchy for all values of level ℓ, and not just the regime ℓ ≥ Ω(d).
The bounds we get however do not have closed-form expressions in general, and they depend on the eigenvalues of some generalized Toeplitz matrices. For small values of n (namely 2n = 2 and 2n = 4) our bounds can be computed efficiently though, as we explain later.
• For more details about the regime ℓ = o(d) of the sum-of-squares hierarchy, we refer the reader to the recent works [BGG + 17, BKS17] and references therein.

The Best Separable State problem in quantum information theory
The notion of entanglement plays a fundamental role in quantum mechanics. The set of separable states (i.e., non-entangled states) on the Hilbert space C d ⊗ C d is defined as the convex hull of all pure product states Here x † =x T is the conjugate transpose and is a convex subset of the set Herm(d 2 ) of Hermitian matrices of size d 2 × d 2 . A key computational task in quantum information theory is the so-called Best Separable State (BSS) problem: given M ∈ Herm(d 2 ), compute In that sense the BSS problem is very related to the polynomial optimization problem (1). The Doherty-Parrilo-Spedalieri (DPS) hierarchy [DPS04] is a hierarchy of semidefinite relaxations to the set of separable states, which is defined in terms of so-called state extensions (we recall the precise definitions later in the paper). It satisfies where DPS ℓ (d) is the ℓ'th level of the DPS hierarchy. It turns out that the DPS hierarchy can be interpreted, from the dual point of view, as a sum of squares hierarchy. This duality relation has been mentioned multiple times in the literature, however we could not find any formal and complete proof of this equivalence. In this paper we give a proof of this duality relation. To do this, we first need to specify the definition of sum of squares for Hermitian polynomials. We say that a Hermitian polynomial is a real sum of squares (rsos) if it can be written as a sum of squares of Hermitian polynomials. 2 To state the result it is more convenient to work in the conic setting and we denote the convex cones associated to Sep and DPS k by SEP and DPS k respectively (these convex cones simply correspond to dropping a trace normalization condition).
Theorem 3 (Duality DPS/sum-of-squares) Let SEP(d) be the convex cone of separable states on C d ⊗ C d , and let DPS ℓ (d) be the convex cone of quantum states corresponding to the ℓ'th level of the DPS hierarchy. Then we have: where K * denotes the dual cone to K and p M is the Hermitian polynomial of Equation (5).
Using this connection, our results on the convergence of the sum-of-squares hierarchy can be easily translated to bound the convergence rate of the DPS hierarchy. More precisely, since the polynomial p M of Equation (5) is bihomogeneous of degree (2, 2) (i.e., it is quadratic in x and y independently) we can get a bound on the rate of convergence of the DPS hierarchy from Theorem 2 where deg F = 2. The rate of convergence we get in this way actually coincides with the rate of convergence obtained by Navascues, Owari and Plenio [NOP09], who use a completely different (quantum-motivated) argument based on the primal definition of the DPS hierarchy using state extensions. From the sum-of-squares point of view, the theorem of Navascues et al. can thus be seen as a special case of Theorem 2 when deg F = 2. We conclude by stating the result on the convergence rate of the DPS hierarchy.
Theorem 4 (Convergence rate of DPS hierarchy, see also [NOP09]) Let M ∈ Herm(d 2 ) and assume that for any ℓ ≥ C ′ d, where C, C ′ > 0 are absolute constants.

Overview of proof
We give a brief overview of the proof of Theorem 2. We will focus on the case where F (x) is a scalar-valued polynomial for simplicity of exposition.
Given a univariate polynomial q(t) of degree ℓ consider the kernel K(x, y) = q( x, y ) 2 for (x, y) ∈ S d−1 × S d−1 . Define the integral transform, for h : where dσ is the rotation-invariant probability measure on S d−1 . If h ≥ 0 then the function Kh is ℓ-sos 3 on S d−1 , by construction of the kernel K(x, y). Let F (x) be a scalar-valued polynomial such that 0 ≤ F (x) ≤ 1 on S d−1 . Our goal is to find δ > 0 such thatF = F + δ is ℓ-sos. Assuming that the mapping K is invertible, we can always writeF = Kh with h = K −1F . If K is close to the identity (i.e., the kernel K(x, y) is close to a Dirac kernel δ(x, y)) then we expect that h ≈F , i.e., that h −F ∞ is small. SinceF ≥ δ, if we can guarantee that h −F ∞ ≤ δ it would follow that h ≥ 0, in which case the equationF = Kh = K(K −1F ) gives a degree-ℓ sum-of-squares representation ofF .
To make the argument above precise we need to measure how close the kernel K is to the identity. This is best done in the Fourier domain, where we analyze how close the Fourier coefficients of the the kernel K(x, y) are to 1. The Fourier coefficients of K(x, y) depend in a quadratic way on the coefficients in the expansion of q(t) in the basis of Gegenbauer polynomials. We show that there is a choice of q(t) such that the Fourier coefficients of K(x, y) converge to 1 at the rate d 2 ℓ 2 , as ℓ → ∞. The kernel we construct is the solution of an eigenvalue maximization for a generalized Toeplitz matrix, associated to the family of Gegenbauer polynomials. We use known results on the roots of such polynomials to obtain the desired rate of convergence.
The idea of proof here is similar to the approaches in Reznick [Rez95], and Doherty & Wehner [DW12], and Parrilo [Par13]. The work of Reznick uses the kernel K(x, y) = x, y 2ℓ /c for some normalizing constant c for which the Fourier coefficients can be computed explicitly. 4 The Fourier coefficients of this kernel happen to converge to 1 at a rate of d ℓ , which is slower than the kernels we construct.

Organization
In Section 2 we review some background material concerning Fourier decompositions on the sphere. The proof of Theorem 2 is in Section 3. Section 4 is devoted to the Best Separable State problem in quantum information theory.

Background
Spherical harmonics We review the basics of Fourier analysis on the sphere S d−1 . Any polynomial p of degree n on the sphere has a unique decomposition where each p i is a spherical harmonic of degree i. The decomposition (7) is known as the Fourier-Laplace decomposition of p. The space H d i is defined as the restriction on S d−1 of the set of homogeneous harmonic polynomials of degree i, i.e., Equivalently, the spaces H d i are also the irreducible subspaces of L 2 (S d−1 ) under the action of SO(d). For example H d 0 is the set of constant functions, H d 1 is the set of linear functions, and H d 2 is the set of traceless quadratic forms. The spaces H d i are mutually orthogonal with respect to the L 2 inner product f, g = f gdσ where dσ is the rotation-invariant probability measure on the sphere. Note that if p is an even polynomial (i.e., p(x) = p(−x)) then the only nonzero harmonic components of p are the ones of even order.
Integral transforms Consider a general SO(n)-invariant kernel K(x, y) = φ( x, y ) where φ is some univariate polynomial of degree L. The kernel K acts on functions f : S d−1 → R as follows To understand the action of K on arbitrary polynomials f , it is very convenient to decompose φ into the basis of Gegenbauer polynomials (also known as ultraspherical polynomials) (C k (t)) k∈N which are orthogonal polynomials on [−1, 1] with respect to the weight (1 − t 2 ) d−3 2 dt. Using appropriate normalization (which we adopt here) these polynomials satisfy the following important property: for any p i ∈ H d i . In other words, the kernel (x, y) → C k ( x, y ) is a reproducing kernel for H d k . Now going back to the kernel K(x, y) = φ( x, y ), if we expand φ = λ 0 C 0 + λ 1 C 1 + · · · + λ L C L , then it follows that for any polynomial p with Fourier expansion (7) we have The equation above tells us that the harmonic decomposition H 0 ⊕ H 1 ⊕ . . . diagonalizes K, with the Gegenbauer coefficients (λ i ) i=0,...,L being the eigenvalues. Equation (8) is also known as the Funk-Hecke formula.
The coefficients (λ i ) i=1,...,L in the expansion of φ in the basis of Gegenbauer polynomials are given by the following integral has unit norm with respect to w(t)dt.
We will use this simple property of the coefficients later in the proof.
A technical lemma The following lemma will be important for our proof later. It shows that the sup-norm of the harmonic components of a polynomial f can be bounded by a constant independent of the dimension d, times the sup-norm of f .

Proposition 5 For any integer n there exists a constant B 2n such that the following is true. For any homogeneous polynomial f with degree 2n and with decomposition into spherical harmonics
Proof The proof is in Appendix A.
The remarkable property in the previous proposition is that the constant B 2n is independent of the dimension d.
Remark 2 When f is a homogeneous polynomial of degree 2n such that 0 ≤ m ≤ f ≤ M on S d−1 , Proposition 5 gives us that f 2k ∞ ≤ B 2n M . However one can get a better bound by applying Proposition 5 instead to f − (m + M )/2; this gives f 2k ∞ ≤ (M − m)/2 for all k = 1, . . . , n.

Proof of Theorem 2
In this section we prove our main theorem, Theorem 2. We will actually prove a more general result giving bounds on the performance of the sum-of-squares hierarchy for all values of the level ℓ. (In Theorem 2 stated in the introduction, only the regime ℓ ≥ Ω(d) was presented.) For the statement of our theorem we need to introduce two quantities that play an important role in our analysis.
• The second quantity is the constant B 2n introduced in Proposition 5. It is the smallest constant such that for any homogeneous polynomial f of degree 2n, we have f 2k ∞ ≤ B 2n f ∞ for all k = 0, . . . , n, where f 2k are the 2k'th harmonic components of f . In other words, B 2n is an upper bound on the ∞ → ∞ operator norm of the linear map that projects a homogeneous polynomial of degree 2n onto its 2k'th harmonic component. Proposition 5 says that such an upper bound that only depends on n (i.e., independent of d) does exist. One can get explicit upper bounds on B 2n for small values of n. For example one can show that B 2 ≤ 2 and B 4 ≤ 9.
We are now ready to state our main theorem: Furthermore, the quantity ρ 2n (d, ℓ) satisfies the following: for any n ≤ d, there are constants Proof [Proof of first part of Theorem 6] We will start by proving the first part of the theorem. For clarity of exposition, we will assume that F is a scalar-valued polynomial, and we explain later why the argument also works for matrices. Let thus F be a homogeneous polynomial of degree 2n such that 0 ≤ F ≤ 1 on S d−1 . Let ) be the decomposition of F into spherical harmonics (since F is even, only harmonics of even order are nonzero). Given δ > 0 to be specified later, we will exhibit a sum-of-squares decomposition ofF = F + δ by writingF = KK −1F where K is an integral transform defined as where φ(t) = (q(t)) 2 is a univariate polynomial of degree 2ℓ. In order forF = KK −1F to be a valid sum-ofsquares decomposition ofF , we need that K −1F ≥ 0. The polynomial q(t) will be chosen so that K is close to a Dirac kernel; when combined withF ≥ δ > 0 we will be able to conclude that Let (λ i ) 0≤i≤2ℓ be the coefficients in the Gegenbauer expansion of φ, i.e., φ = λ 0 C 0 + λ 1 C 1 + · · · + λ 2ℓ C 2ℓ . By the Funk-Hecke formula we have Our analysis does not depend on the scaling of K so we will assume λ 0 = 1. Thus we get where in the last inequality we used Proposition 5 (see also Remark 2) together with the fact that 0 ≤ F ≤ 1. It thus follows that if then K −1 (F ) ≥ 0 and the equationF = KK −1 (F ) gives a valid sum-of-squares decomposition ofF = F +δ.
We have thus proved the first part of Theorem 1. It now remains to prove the second part of the theorem, which leads us to the analysis of the quantity ρ 2n (d, ℓ). Before doing so, we explain how the proof above applies in the case where F is a matrix-valued polynomial.
Matrix-valued polynomials Assume F ∈ S[x] homogeneous of degree 2n. We can decompose each entry of F into spherical harmonics to get F = F 0 + F 2 + · · · + F 2n . DefineF = F (x) + δI for a δ > 0 to be specified later. The steps in the argument above are identical, where · ∞ is defined as the maximum of F (x) over x ∈ S d−1 , where F (x) is the spectral norm of F (x), and the bound on F 2k ∞ follows from Proposition 17.
where U y (x) = q( x, y )H(y) 1/2 is a polynomial of degree ℓ in x. This is what we wanted.
We now proceed to the analysis of ρ 2n (d, ℓ).
Reformulating ρ 2n (d, ℓ) using generalized Toeplitz matrices It will be convenient to reformulate the optimization problem (10) in terms of certain suitable (generalized) Toeplitz matrices. We parametrize the degree-ℓ polynomial q(t) as where e 0 , . . . , e ℓ ∈ R. The presence of the term C i (1) is for convenience later. The Gegenbauer coefficients of φ(t) = (q(t)) 2 are then equal to (cf. Equation (9)) where for h : It can be easily checked that T [1] = I is the identity matrix (this follows from the fact that the polynomials have unit norm with respect to the weight function (1 − t 2 ) (d−3)/2 ), and so λ 0 = e T e = k e 2 k . It thus follows that ρ 2n (d, ℓ) can be formulated as: Case 2n = 2 Let us first analyze the case 2n = 2 which corresponds to quadratic polynomials. In this case the sum in (13) has simply one term. It is then not difficult to see that ρ 2 (d, ℓ) is given by where · denotes the spectral norm. Thus we see that ρ 2 (d, ℓ) can be computed efficiently by simply evaluating the spectral norm of T [C 2 /C 2 (1)]. The latter matrix can be formed explicitly using known formulas for the integrals of Gegenbauer polynomials (see e.g., [Hsu38]). Note that T [C 2 /C 2 (1)] is a banded matrix with bandwidth 3.
Case 2n = 4 We now turn to quartic polynomials. In this case ρ 4 (d, ℓ) takes the form Let R be the joint numerical range (also known as the field of values) of the matrices T From results about joint numerical ranges, it is known that R ⊂ R 2 is convex, see [Bri61] and also [PT07,Theorem 5.6]. It is not difficult to see then that R has a semidefinite representation, and that ρ 4 (d, ℓ) can be computed using semidefinite programming.
General degree 2n We now analyze the case of general degree 2n. To do this we formulate a proxy for the optimization problem that defines ρ 2n (d, ℓ) that is easier to analyze. Instead of minimizing 2n k=1 |λ −1 2k − 1| we will seek instead to minimize 2n k=1 (1 − λ 2k ). Since λ 2k ≤ λ 0 = 1, both problems seek to bring the λ 2k close to 1, but the latter problem is easier to analyze because it is linear in the λ 2k . Definẽ Since T is linear, i.e., . It thus remains to analyze λ max (T [h]). This is what we do next.
Proof We use the following standard result on orthogonal polynomials which gives the eigenvalues of T [f ] for any linear polynomial f . (The result below is stated in full generality for clarity, in our case the (p k ) is the family of normalized Gegenbauer polynomials.) Proposition 8 (Standard result on orthogonal polynomials) Let (p k ) k∈N be a family of orthogonal polynomials with respect to a weight function w(x) > 0. We assume the (p k ) are normalized, i.e., p 2 k w = 1. Given a linear polynomial f , define the (ℓ + 1) × (ℓ + 1) matrix Then the eigenvalues of d 2 ℓ 2 . It thus follows, using the fact that h(1) = 1 and h ′ (1) > 0, that where in the last inequality we used the exact value of h ′ (1) given by h ′ (1) = (n + 1)(3d + 4n − 4)/(3(d − 1)) and the fact that n ≤ d.
Our proof of Theorem 6 (i) is now almost complete. We just need to relateρ back to ρ. We use the following easy proposition.
Tightness Our analysis of ρ 2n (d, ℓ) in the regime ℓ ≥ Ω(d) can be shown to be tight. We show this in the case 2n = 2 below.
Theorem 10 (Tightness of convergence rate) There is an absolute constant C > 0 such that for ℓ ≥ Ω(d), Proof Given the expression for ρ 2 (d, ℓ) in (14), we need to produce an upper bound on T [C 2 /C 2 (1)] . Note that We now use the following property of generalized Toeplitz matrices constructed from sequences of orthogonal polynomials: If T ∞ denotes the semi-infinite version of (17), then for any polynomials f, g (this property follows immediately from the fact that the sequence of orthogonal polynomials (p k ) ∞ k=0 is an orthonormal basis of the space of polynomials, see e.g., [Bax71, Lemma 2.4]). In particular we have T ) 2 is positive semidefinite it then follows that

Relation to quantum state extendibility
Quantum entanglement is one of the key ingredients in quantum information processing. Certifying whether a given state is entangled or not is a hard computational task [Gur03] and considerable effort has been dedicated to this problem, e.g., [LBC + 00, HHH01]. Of particular interest is the hierarchy of tests known as the DPS hierarchy [DPS02,DPS04], applying semidefinite programs to verify quantum entanglement.
In this section, we explore the duality relation between the DPS hierarchy and sums of squares, and explain how our results from the previous section can be used to bound the convergence rate of the DPS hierarchy. We show that the result of Navascues et al. [NOP09] can be seen as the special case of our Theorem 6 when the polynomial F is quadratic.

Quantum extendible states
A quantum state is usually represented by a positive semidefinite operator normalized with unit trace. In this work, we mainly work with unnormalized quantum states and consider its convex cone. Given Hilbert spaces The convex cone of quantum separable states is denoted as SEP(H A ⊗ H B ) and it is strictly included in S(H A ⊗ H B ).
Positive partial transpose A well-known necessary condition for a state ρ AB to be in SEP is that it has a positive partial transpose (PPT). If we let T denote the transpose operation on Hermitian matrices of size d B × d B , then for ρ AB of the form (18) we have If we let PPT (H A ⊗ H B ) be the set of states with a positive partial transpose then we have the inclusions

SEP(H A ⊗ H B ) ⊂ PPT (H A ⊗ H B ) ⊂ S(H A ⊗ H B ).
A well-known result due to Woronowicz  Extendibility When the inclusion SEP = PPT is strict, one can find more accurate relaxations of SEP based on the notion of state extendibility. For simplicity of the following discussion, we introduce the notation [s 1 : s 2 ] := {s 1 , s 1 + 1, · · · , s 2 } and [s] := [1 : s] for short. Given a separable state expressed as Eq. (18) with 5 x i = y i = 1 we can consider its extension (on the B subsystem) as: The new system ρ AB In ( * ) we used the fact that y i = 1.
(d) Positive Partial Transpose: If we let T be the transpose map on Hermitian matrices of size for any s = 1, . . . , ℓ. For convenience later the state on the left of (22) will be denoted ρ The DPS hierarchy Define now the set DPS ℓ (H A ⊗ H B ) as s.t. conditions (20), (21), (22) are satisfied .
Note that DPS 1 = PPT . Also it is known that the hierarchy is complete in the sense that if ρ / ∈ SEP then there exists a ℓ ∈ N such that ρ / ∈ DPS ℓ [DPS02,DPS04].
Remark 3 (Extendibility without PPT conditions) One can also consider the weaker hierarchy where the Positive Partial Transpose constraints are dropped: (20) and (21) are satisfied . (24) It turns out that this weaker hierarchy EX T ℓ is already complete in the sense stated above. This is usually proven using de Finetti theorems [CKMR07,KM09].

Hermitian polynomials and sums of squares
In this section we leave the quantum world and introduce some terminology pertaining to Hermitian polynomials. A Hermitian polynomial p(z,z) is a polynomial with complex coefficients in the variables z = (z 1 , . . . , z n ) andz = (z 1 , . . . ,z n ) such that p(z,z) ∈ R for all z ∈ C n . The general form of a Hermitian polynomial is where the coefficients p uv satisfy p uv = p vu . We say that p(z) is nonnegative if p(z) ≥ 0 for all z ∈ C n .
Definition 11 (Hermitian polynomials and sums of squares) Let p(z,z) be a nonnegative Hermitian polynomial. We say that p(z,z) is a real sum-of-squares (rsos) if we can write p(z,z) = i g i (z,z) 2 where g i (z,z) are Hermitian polynomials. We say that p(z,z) is a complex sum-of-squares (csos) if we can write p(z,z) = i |q i (z)| 2 where q i (z) are (holomorphic) polynomial maps in z (i.e., q i are functions of z alone and notz).
Clearly if p(z,z) is csos then it is also rsos since |q(z)| 2 = Re[q(z)] 2 + Im[q(z)] 2 and Re[q(z)] and Im[q(z)] are both Hermitian polynomials. The converse however is not true. For example p(z,z) = (z +z) 2 is evidently rsos, however it is not csos [DP09,Example (a)]. Indeed the zero-set of a csos polynomial must be a complex variety, and the zero set of p(z,z) here is the imaginary axis. Note that a Hermitian polynomial p(z,z) is rsos iff the real polynomial P (a, b) = p(a + ib, a − ib) is a sum-of-squares (in the usual sense for real polynomials).

The duality relation
is nonnegative for all (x, y) ∈ C d A × C d B . We prove our first main result on the duality between the DPS hierarchy and sums of squares.
Theorem 12 (Duality between extendibility hierarchy and sums of squares) For M ∈ Herm(d A d B ), we let p M be the associated Hermitian polynomial in (25). Then we have: Proof Point (i) is immediate and follows from the definition of duality. Points (ii) and (iii) are proved in Appendix B.
The following diagram summarizes the situation.
In terms of the support functions The support function of the set DPS ℓ is defined as . The duality relation of Theorem 12 allows us to express h DPS ℓ (M ) in the following way: A somewhat more convenient formulation using matrix polynomials is as follows. For x ∈ C d , we letx ∈ R 2d be the vector of real and imaginary components of Then one can show the following equivalent formulation of h DPS ℓ (M ): This can be proved using the following lemma, which is a straightforward generalization of [dKLP05] to the matrix case.
Lemma 13 Let G(y 1 , . . . , y d ) be a homogeneous matrix-valued polynomial of even degree 2n, such that G(y) is symmetric for all y. Then G is ℓ-sos on S d−1 , if and only if, y 2(ℓ−n) G(y) is a sum of squares.

Convergence rate of the DPS hierarchy
Note that the state ρ A ⊗ I B /d B is clearly separable. In words, the result above says that if ρ AB in DPS ℓ , then by moving ρ AB in the direction ρ A ⊗ I B /d B by t = O(d 2 B /ℓ 2 ) results in a separable state. In terms of the Best Separable State problem, the result of [NOP09] has the following immediate implication. We show below how we can recover this result using our Theorem 6 from the previous section.
Proof We know from (26) that By assumption we have 0 ≤P M (ỹ) ≤ h Sep (M )I for allỹ ∈ S 2d B −1 . Our Theorem 6 from previous section tells us that for ℓ ≥ Cd B , which is what we wanted.

Conclusions
We have shown a quadratic improvement on the convergence rate of the SOS hierarchy on the sphere compared to the previous analysis of Reznick [Rez95] and Doherty & Wehner [DW12]. The proof technique also works for matrix-valued polynomials on the sphere and surprisingly, the bounds we get are independent of the dimension of the matrix polynomial. In the special case of quadratic matrix polynomials, we recover the same rate obtained by Navascues, Owari & Plenio [NOP09] using arguments from quantum information theory.
A Some technical results on polynomials on sphere We use the following lemma which appears in [Rez95]. Recall that the Laplacian of a twice differentiable function f : where ∆ is the Laplace operator and (n) m := n(n − 1) · · · (n − (m − 1)) is the falling factorial.
Proposition 5 (restatement) For any homogeneous polynomial f with degree 2n, denote its spherical harmonics decomposition as a constant that depends only on n (and independent of d). Also B 2 ≤ 2 and B 4 ≤ 9.
Proof For simplicity of exposition we prove first the cases 2n = 2 and 2n = 4, before considering the general case. The result is immediate when 2n = 2 with B 2 = 2 since the harmonic decomposition of a quadratic polynomial is The first nontrivial case is 2n = 4. The decomposition of a quartic polynomial on the sphere is f = f 0 + f 2 + f 4 . Clearly f 0 ∞ ≤ f ∞ . We thus focus on bounding f 2 ∞ . Since f is homogeneous note that f (x) can be written as f (x) = x 4 f 0 + x 2 f 2 (x) + f 4 (x) for all x ∈ R d . Using well-known identities concerning Laplacian one can check that ∆(f (x)) = 4(d + 2) x 2 f 0 + 2(d + 2)f 2 (x). (We used that ∆f 2k = 0 since the f 2k are harmonic, that ∆( x 2k ) = 2k(2k+d−2) and the identity x, ∇g(x) = 2kg(x) for any homogeneous polynomial g of degree 2k and x ∈ S d−1 .) It thus follows that f 2 (x) = 1 2d+4 ∆f (x) − 2 x 2 f 0 . By Lemma 16 we know that ∆f ∞ ≤ 12d f ∞ . It thus follows that We now proceed to prove the general case. Let f (x) be a homogeneous polynomial of degree 2n and let f = f 0 + f 2 + · · · + f 2n be its spherical harmonic decomposition. Note that f (x) has the following expression Thus by linearity, we have When restricted on the unit sphere, we have the inequalities where the second inequality follows from Lemma 16. For any 0 ≤ k ≤ n − m − 1, we have Moreover, for any m ≤ n we also have where the first inequality holds since each term in the falling factorial (n + (n − m) + d/2 − 1) m is no smaller than d/2. Thus we have By induction, we have the estimation Thus we have B 2n ≤ (2n)! [1 + (2n)!] n independent of d.
We can extend Proposition 5 to matrix-valued polynomials on the sphere.
Proof We can assume without loss of generality that F = 1. Note that F = max x∈S d−1 max y∈S k−1 |y T F (x)y|.
For any fixed y ∈ S k−1 define the real-valued polynomial f y (x) = y T F (x)y. By assumption on F we know that f y ∞ = max x∈S d−1 |y T F (x)y| ≤ 1. The spherical harmonic decomposition of f y is given by f y (x) = 2n k=0 y T F 2k (x)y since, for fixed y, y T F 2k (x)y is a linear combination of the entries of F 2k which are all in H d 2k . It thus follows from Proposition 5 that y T F 2k (·)y ∞ ≤ B 2n . This is true for all y ∈ S k−1 thus we get max y∈S k−1 max x∈S d−1 |y T F 2k (x)y| ≤ B 2n , i.e., F 2k ≤ B 2n as desired.
We will also need the following technical result about Gegenbauer polynomials.

B Duality relations DPS and SOS (Theorem 12)
In this section we prove that for any integer ℓ ≥ 1, we have the duality relation The key is the following lemma which gives a semidefinite programming characterization of the right-hand side of (35).

Lemma 19
For any M AB ∈ Herm(H A ⊗ H B ) and integer ℓ ≥ 1, then y 2(ℓ−1) p M is a rsos if and only if there exist positive semidefinite operators W s,AB [ℓ] ≥ 0, s = 0, 1, · · · , ℓ such that x ⊗ȳ ⊗s ⊗ y ⊗ℓ−s † W s,AB [ℓ] x ⊗ȳ ⊗s ⊗ y ⊗ℓ−s (36) The proof of the previous lemma is based on analyzing the biquadratic structure of p M to see which monomials can appear in a sum-of-squares decomposition of y 2(ℓ−1) p M . The proof is deferred to the end of this section.
Using Lemma 19, the proof of (35) follows from standard duality arguments which we know explain. First, we can dualize the semidefinite programming definition of DPS ℓ to get The variable W s,AB [ℓ] for s = 0 (resp. s = 1, . . . , ℓ) is the dual variable for the positivity constraint on ρ AB [ℓ] (resp. PPT constraint (22)).
Proof [Proof of Theorem 12] The proof consists of two directions. Assume M ∈ DPS * ℓ . Then there exists a Hermitian operator Y AB [ℓ] , and positive semidefinite operators W s,AB [ℓ] ≥ 0, s = 0, 1, · · · , ℓ such that Recalling that Π ℓ is the projector onto the symmetric subspace, we have Πy ⊗ℓ = y ⊗ℓ for any vector y. Thus Evaluating Eq. (38) on both sides at the state x ⊗ y ⊗ℓ , we have According to Proposition 19, we have y 2(ℓ−1) p M is a rsos. On the other hand, suppose y 2(ℓ−1) p M is a rsos. From Proposition 19, there exists positive semidefinite operators W s,AB [ℓ] ≥ 0, s = 0, 1, · · · , ℓ such that Eq. (40) holds. Since y ⊗ℓ forms a basis on the symmetric subspace of H B 1 ⊗ H B 2 ⊗ · · · ⊗ H B ℓ , it implies that the operators coincide when restricted on the symmetric subspace Sym(H ⊗ℓ ). That is, Take the Hermitian operator Then by the definition of Y AB [ℓ] , we have Π ℓ Y AB [ℓ] Π ℓ = 0 and which implies M AB 1 ∈ DPS * ℓ .
It remains to prove Lemma 19. To have an easier understanding of the result in Lemma 19, let us first have a look at the special case on the second level of the hierarchy, i.e, ℓ = 2. This will give us the key idea without loss of generality, and the higher level case is just a straightforward generalization.
Lemma 19 [special case ℓ = 2] For any Hermitian operator M AB 1 , we have that y 2 p M is rsos if and only if there exist positive semidefinite operators W 0,AB 1 B 2 , W 1,AB 1 B 2 , W 2,AB 1 B 2 ≥ 0, such that Proof If there exist operators W 0 , W 1 , W 2 ≥ 0 such that Eq. (45) holds, then y 2 p M can be shown to be rsos by using the spectral decompostion of W i . For the converse suppose y 2 p M is rsos. Then there exist polynomials f m (x,x, y,ȳ) such that y 2 p M (x,x, y,ȳ) = m f m (x,x, y,ȳ) 2 . Since the monomials of y 2 p M (x,x, y,ȳ) are all of the formsx i x mȳj y k y rȳr (they are degree 2 in (x,x) and degree 4 in (y,ȳ), then the possible monomials of f m (x,x, y,ȳ) can only be given by The existence of any other monomials in f m (x,x, y,ȳ), such as x i x j y k , will not be compatible with the monomials in y 2 p M (x,x, y,ȳ). Thus the most general form of f m (x,x, y,ȳ) can be written as the linear combinations, Since f m (x,x, y,ȳ) ∈ R, we havē a m,0 i,j,k = b m,0 i,j,k ,ā m,1 i,j,k = b m,1 i,j,k ,ā m,2 i,j,k = b m,2 i,j,k , ∀ i, j, k, m.
Proof Note that the second equality trivially holds due to the equation x † Zx = (x) † Z T (x). We will prove the first equality. If Eq. (53) holds for positive semidefinite operators W s,AB [ℓ] , then it is easy to check that y 2(ℓ−1) p M (x,x, y,ȳ) is a rsos by using the spectral decomposition of W s,AB [ℓ] .
On the other hand, if y 2(ℓ−1) p M (x,x, y,ȳ) is a rsos, by definition there exist Hermitian polynomials f m (x,x, y,ȳ) such that y 2(ℓ−1) p M (x,x, y,ȳ) = m f m (x,x, y,ȳ) 2 . In the following, we will compare the monomials on both sides of this equation and explicitly construct W s,AB [ℓ] from the coefficients of m f m (x,x, y,ȳ) 2 . We first note that the monomials of y 2(ℓ−1) p M (x,x, y,ȳ) are all of the form which is of degree 2 and 2ℓ with respect to x and y, respectively. Then the possible monomials of f m (x,x, y,ȳ) can only be of degree 1 and ℓ with respect to x and y, respectively. That is, the possible monomials are given by where we denote the term s 2 i=s 1 (·) = 1 if s 2 < s 1 . These monomials are basically formed by the ones with different number of complex conjugation over the symbol y. Therefore, the most general form of f m (x,x, y,ȳ) can be written as a linear combination of these monomials: Since f m (x,x, y,ȳ) ∈ R for all x, y, we know that the coefficients between the conjugate monomials have to be conjugate with each other. That is,ā m,s t,r [ℓ] = b m,s t,r [ℓ] holds for all m, s, t, r [ℓ] . Comparing the monomials of For any s ∈ [0 : ℓ], we construct the matrix W s whose elements are given by a m,s t,r [ℓ] .
Proof If there exists W AB [ℓ] ≥ 0 such that Eq. (61) holds, we can check that y 2(ℓ−1) p M (x,x, y,ȳ) is a csos by using the spectral decomposition of W . On the other hand, if y 2(ℓ−1) p M (x,x, y,ȳ) is a csos, by definition there exist polynomials f m (x, y) such that y 2(ℓ−1) p M (x,x, y,ȳ) = m |f m (x, y)| 2 . In the following, we will compare the monomials on both sides of this equation and explicitly construct W AB [ℓ] from the coefficients of m |f m (x, y)| 2 . We first note that the monomials of y 2(ℓ−1) p M (x,x, y,ȳ) are all of the form x txt ′ ℓ i=1 y r iȳr ′ i , which is of degree 2 and 2ℓ with respect to x and y, respectively. Then the possible monomials of f m (x, y) can only be of degree 1 and ℓ with respect to x and y, respectively. Furthermore, by definition f m (x, y) are polynomials with respect to x, y alone, thus the only possible monomial of f m (x, y) is x t ℓ i=1 y r i and we have the general form of f m (x, y) as f m (x, y) = t,r [ℓ] a m t,r [ℓ] x t ℓ i=1 y r i with coefficients a m t,r [ℓ] . Define the matrix W with elements a m t,r [ℓ] .
Then we have W ≥ 0, and which completes the proof.
Finally the result of EX T * ℓ = M : y 2(ℓ−1) p M is csos can be proved in a similar way by using Lemma 20.