Symmetric Grothendieck Inequality

SYMMETRIC GROTHENDIECK INEQUALITY

SHMUEL FRIEDLAND AND LEK-HENG LIM

In memory of Joram Lindenstrauss, who introduced the ﬁrst author to the Grothendieck inequality.

Abstract. We establish an analogue of the Grothendieck inequality where the rectangular matrix is replaced by a symmetric (or Hermitian) matrix and the bilinear form by a quadratic form. We call this the symmetric Grothendieck inequality; despite its name, it is a generalization — the original Grothendieck inequality is a special case. While there are other proposals for such an inequality, ours differs in two important ways: (i) we have no additional requirement like positive semidefiniteness for the symmetric matrix; (ii) our symmetric Grothendieck constant is universal, i.e., independent of the matrix and its dimensions, like the original Grothendieck constant. A consequence of our symmetric Grothendieck inequality is a “conic Grothendieck inequality” for any family of cones of symmetric matrices: The original Grothendieck inequality is a special case; as is the Nesterov π/2-Theorem, which corresponds to the cones of positive semidefinite matrices; as well as the Goemans–Williamson inequality, which corresponds to the cones of weighted Laplacians. For yet other cones, e.g., of diagonally dominant matrices, we obtain new Grothendieck-like inequalities. With a slight extension, we obtain a unified framework that treats any Grothendieck-like inequality as an inequality between two norms within a family of “Grothendieck norms” restricted to an appropriate family of cones. This approach allows us to place on an equal footing the Goemans– Williamson inequality, Nesterov π/2-Theorem, Ben-Tal–Nemirovski–Roos 4/π-Theorem, generalized Grothendieck inequality, order-p Grothendieck inequality, rank-constrained positive semidefinite Grothendieck inequality, etc, and in turn allows us to simplify proofs, extend results from real to complex, obtain new bounds or establish sharpness of existing ones. Unsurprisingly, the symmetric Grothendieck inequality may also be applied to obtain uniform polynomial-time approximation bounds for various NP-hard combinatorial, integer, and nonconvex optimization problems.

1. Introduction

The Grothendieck inequality [24, 42] states that for k = R or C, there is a finite constant KG > 0 m×n such that for every d, m, n ∈ N with d ≥ m + n and every matrix B = (bij) ∈ k , m n m n X X X X (1) max bijhxi, yji ≤ KG max bijε¯iδj , kxik=kyj k=1 |εi|=|δj |=1 i=1 j=1 i=1 j=1 d where the maximum on the left is taken over all xi, yj in the unit sphere of k , and the maximum on the right is taken over all δi, εj in the unit circle of k, i = 1, . . . , m, j = 1, . . . , n. The original version in [24], while stated differently, is equivalent to the version in (1), which is due to [35]. The maxima on the left- and right-hand sides of (1) define matrix norms m n m n X X X X (2) kBkG := max bijhxi, yji and kBk∞,1 := max bijε¯iδj kxik=kyj k=1 |εi|=|δj |=1 i=1 j=1 i=1 j=1 respectively. So we have

(3) kBk∞,1 ≤ kBkG ≤ KGkBk∞,1,

2010 Mathematics Subject Classification. 47A07, 46B28, 68Q17, 81P45, 90C27, 05C50. Key words and phrases. Grothendieck inequality, Goemans–Williamson inequality, Nesterov π/2-Theorem, weighted Laplacians, diagonally dominant matrices. 1 2 SHMUEL FRIEDLAND AND LEK-HENG LIM where the first inequality is easy to see and the second inequality is the Grothendieck inequality (1). By the equivalence of norms in finite-dimensional spaces, an inequality of the form (3) must exist but the surprise here is that the constant KG is independent of the dimensions m, n! The existence of such a universal constant KG is the crux of the Grothendieck inequality and we expect any extension to also have such a property. The two norms in (2) as well as the value of the smallest possible KG in (3) depend on the k k k choice of the field k and we will denote them by k · kG, k · k∞,1, KG when we need to emphasize this 1 R C dependence. The sharpest constants KG and KG are known respectively as the real and complex Grothendieck constants. Their exact values are unknown but there are excellent bounds due to Davie [16, 17], Haagerup [25], and Krivine [33]:

R C (4) 1.67696 ≤ KG ≤ 1.78221, 1.33807 ≤ KG ≤ 1.40491; R and it has been shown [9] that Krivine’s upper bound for KG can be improved. That the Grothendieck constants have moderately small values has important implications [42]. The most direct one is that by virtue of (3), the G-norm, which can be computed in polynomial-time via semideﬁnite programming, gives us a good approximation of the (∞, 1)-norm, which is known to be NP-hard.

1.1. Symmetric Grothendieck inequality. The main goal of our article is to establish two n intimately related generalizations of (1). Let S be the vector space of n × n symmetric/Hermitian matrices over k. Then n n n n X X X X ¯ (5) kAkγ := max aijhxi, xji and kAkθ := max aijδiδj kxik=1 |δi|=1 i=1 j=1 i=1 j=1 n define seminorms on S , whereas n n n n X X X X ¯ (6) kAkΓ := max aijhxi, xji and kAkΘ := max aijδiδj kxik≤1 |δi|≤1 i=1 j=1 i=1 j=1 n define norms on S . The maximum in k · kγ (resp. k · kΓ) is taken over all x1, . . . , xn in the unit d sphere (resp. ball) of k while the maximum in k · kθ (resp. k · kΘ) is taken over all δ1, . . . , δn in the unit circle (resp. disk) of k, and where we have assumed d ≥ n. The seminorms in (5), the norms in (6), and those in (2) are all distinct, taking different values even on 2 × 2 diagonal matrices. For the norms in (2), whether we take maximum over the unit sphere or the unit ball makes no difference, i.e., one may replace “= 1” by “≤ 1” in the Grothendieck inequality (1). But here it makes a difference — we obtain two symmetric analogues of the Grothendieck inequality (3) by choosing “= 1” or “≤ 1” respectively and neither can be deduced from the other: In Section4, we will show that n n n n X X X X ¯ (7) max aijhxi, xji ≤ Kγ max aijδiδj , kxik=1 |δi|=1 i=1 j=1 i=1 j=1 n n n n X X X X ¯ (8) max aijhxi, xji ≤ KΓ max aijδiδj , kxik≤1 |δi|≤1 i=1 j=1 i=1 j=1 n hold for all A ∈ S with universal constants Kγ and KΓ independent of d and n satisfying ( sinh(π/2) ≈ 2.30130 over R, (9) KG ≤ KΓ ≤ Kγ ≤ 8/π − 1 ≈ 1.54648 over C.

1 This applies throughout our article, i.e., we add a superscript R or C whenever it is necessary to distinguish ﬁeld-dependent norms or constants. SYMMETRIC GROTHENDIECK INEQUALITY 3

We will refer to (7) and (8) as symmetric Grothendieck inequalities and the smallest constants Kγ and KΓ as symmetric Grothendieck constants. Should a distinction be necessary, we will call (7) the symmetric Grothendieck inequality for γ-seminorm and (8) as the symmetric Grothendieck inequality for Γ-norm, for reasons that will be clear later. These may also written in forms analogous to (3):

(10) kAkθ ≤ kAkγ ≤ KγkAkθ, kAkΘ ≤ kAkΓ ≤ KΓkAkΘ. While there are other inequalities in the literature that go by the same name, (7) and (8) are different, as we will explain next. Firstly, both symmetric Grothendieck inequalities (7) and (8) generalize the Grothendieck in- m×n equality: for any B ∈ k , setting 0 B (11) A = ∈ m+n B∗ 0 S recovers (1) from either (7) or (8). Secondly, we emphasize that the symmetric Grothendieck inequality is not a case of the Grothendieck inequality restricted to symmetric matrices — we will n see in Corollary 2.11 that k · kΘ 6= k · k∞,1 and k · kΓ 6= k · kG on S ; and obviously k · kθ 6= k · k∞,1 and k · kγ 6= k · kG as seminorms are not norms. Readers who think that they might have seen either (7) or (8) before (say, in [13, 29, 43]) should note that a key difference here is that the n matrix A ∈ S is not assumed to be positive semidefinite; in particular, the absolute values on both n sides of (8) cannot be dropped. If A ∈ S is in addition positive semidefinite, then we will see in Proposition 2.8 that

(12) kAkθ = kAkΘ = kAk∞,1 and kAkγ = kAkΓ = kAkG, and both the symmetric and original Grothendieck inequalities become weaker versions of the Nesterov π/2-Theorem [39, 43], sometimes also called the positive semidefinite Grothendieck inequality. Another key difference from other similarly-named inequalities is that Kγ and KΓ are truly universal constants independent of n, just like the original Grothendieck constant. Prior works. To the best of our knowledge, the symmetric Grothendieck inequality in (7) has never appeared before elsewhere. As for the version in (8), if an effective bound is not required, then one may deduce that it holds for some KΓ = O(1) from [3, Equation 2], as pointed out in [7, p. 17]. It has been pointed out to us that [45, Exercise 3.5.3] appears to give Kγ ≤ 2KG over R; 2 0 2 but the exercise is incorrect as stated: A = 0 −2 is a counterexample. We will have more to say about these after the proof of Theorem 4.3. The line of arguments introduced by Krivine [32] and Haagerup [25] to obtain bounds for KG, generalized and christened “Krivine schemes” in [37], unfortunately does not apply to Kγ and KΓ. 1.2. Conic Grothendieck inequalities. An immediate consequence of the symmetric Grothendieck inequalities is that they give various “Grothendieck inequalities for cones” by restricting (10) to n any family of cones C = {Cn ⊆ S : n ∈ N} and refining the constants. For example, we know from (10) that there is a constant KC ≤ Kγ so that

(13) kAkθ ≤ kAkγ ≤ KCkAkθ that holds for all A ∈ Cn; and we can often get a better bound or even the exact value of KC: The original Grothendieck inequality corresponds to the special case where the cones are subspaces of matrices of the form (11); but other well-known inequalities are also special cases of (13)— the Goemans–Williamson inequality [23] is the case where Cn is the cone of weighted Laplacians on n-vertex graphs; the Nesterov π/2-Theorem [39, 43] is the case where Cn is the cone of real symmetric positive semidefinite n × n matrices. For other family of cones, say, taking Cn to be 2 The author has corrected [45, Exercise 3.5.3] in https://www.math.uci.edu/~rvershyn/papers/HDP-book/ HDP-book.pdf by imposing zero diagonal/positive semidefiniteness. We make no such assumptions in (7) and (8). 4 SHMUEL FRIEDLAND AND LEK-HENG LIM the cone of diagonally dominant n × n matrices, we get new “Grothendieck-like inequalities”. The values of these conic Grothendieck constants (assuming R for simplicity),  n 1/αGW if Cn = {A ∈ S : A1 = 0, aij ≤ 0 for all i 6= j},   n kAk π/2 if Cn = S+, (14) K := sup max γ = C n n∈ A∈Cn⊆S kAkθ 0 B n m×(n−m) N KG if Cn = B∗ 0 ∈ S : B ∈ R ,   n Kγ if Cn = S , can sometimes be determined exactly (the Nesterov π/2-Theorem and Goemans–Williamson inequality) and other times be bounded (the original and symmetric Grothendieck inequalities); here αGW ≈ 0.878567 is the celebrated Goemans–Williamson constant. We also have complex analogues of (14) with different values for the respective constants. For example, over C, the constant π/2 in the Nesterov π/2-Theorem becomes the 4/π in the Ben-Tal–Nemirovski–Roos 4/π-Theorem [6]. There has been speculation [31] that the Goemans–Williamson inequality is somehow related to the original Grothendieck inequality (1) although we have not been able to find an unambiguous statement of their exact relation. Since a weighted Laplacian L is always positive semidefinite, we may invoke (12) and write kLk∞,1 ≤ kLkG ≤ KGkLk∞,1, n n×n and since the set of all weighted Laplacians L is a strictly smaller subset of R , the Grothendieck constant KG may be reduced to the (reciprocal of) Goemans–Williamson constant 1/αGW. This in our view spells the relationship between the two inequalities — Goemans–Williamson inequality n is Grothendieck inequality restricted to L . The crux of Goemans–Williamson inequality, namely, the existence of a universal constant independent of n, the size of the graph G, is characteristic of the Grothendieck inequality. 1.3. Grothendieck norms. One small novelty introduced in this article is that we approach the Grothendieck inequality and its extensions by viewing them as relations within a family of norms n m×n and seminorms. For any d, m, n ∈ N, A ∈ S , and B ∈ k , we define n n X X d kAkγ,d := max aijhxi, xji : xi ∈ k , kxik = 1 ,

i=1 j=1 n n X X d kAkΓ,d := max aijhxi, xji : xi ∈ k , kxik ≤ 1 ,

i=1 j=1 (15) m n X X d kBkG,d := max bijhxi, yji : xi, yj ∈ k , kxik = 1, kyjk = 1

i=1 j=1 m n X X d = max bijhxi, yji : xi, yj ∈ k , kxik ≤ 1, kyjk ≤ 1 .

i=1 j=1

We will show that k · kG,d and k · kΓ,d deﬁne norms and k · kγ,d a seminorm. We will refer to them collectively as Grothendieck norms; or the Grothendieck d-norms if we need to specify their rank d ∈ N; or more speciﬁcally as the (γ, d)-seminorm, (Γ, d)-norm, and (G, d)-norm respectively if we need to specify both their types and rank. The norms in (2) and (6) are special cases:

k · kG = k · kG,m+n, k · k∞,1 = k · kG,1, k · kΓ = k · kΓ,n, k · kΘ = k · kΓ,1. 2 In fact, we will see that k · kG,d = k · kG as soon as d(d + 1)/2 > m + n over R and d > m + n over C; similarly for k · kγ,d and k · kΓ,d. We will establish various characterizations of these norms; closed-form expressions for special matrices; and their relations across diﬀerent types, ranks, and base ﬁelds (R or C). SYMMETRIC GROTHENDIECK INEQUALITY 5

More generally, we prove that for each of these three types of norms/seminorms, there is a Grothendieck-like inequality relating any pair of d-norm and p-norm with d ≤ p:

kAkγ,p ≤ Kγ,d,pkAkγ,d, kAkΓ,p ≤ KΓ,d,pkAkΓ,d, kBkG,p ≤ KG,d,pkBkG,d n m×n for all A ∈ S and all B ∈ k . The original and symmetric Grothendieck inequalities (1), (7), and (8) then correspond to the limiting case d = 1, p → ∞. Other special cases include the order-p Grothendieck inequality [1, 19, 26, 33], when the pair of norms are k · kG,1 and k · kG,p; the generalized Grothendieck inequality [10], when the pair of norms are k · kG,d and k · kG,p, d ≤ p; and the rank-constrained positive semideﬁnite Grothendieck inequality [11, 12], when the pair of norms n are k · kγ,d and k · kγ, restricted to S+. In fact, when combined with our discussion in Section 1.2, we obtain an all-encompassing inequality

k k k (16) kAkγ,p ≤ KC,d,pkAkγ,d. Every inequality that we have mentioned up to this point may be obtained as a special case of (16) by choosing an appropriate family of cones C, appropriate values of d and p, and an appropriate ﬁeld k = R or C.

1.4. Polynomial-time approximations. Like the original Grothendieck inequality, the symmetric Grothendieck inequality has implications on the polynomial-time approximations of NP-hard quantities. We will study several such approximations in Section6, and will describe two here. For a bipartite graph with vertices in an independent set of size m and another of size n, its m+n m×n adjacency matrix A ∈ S takes the form in (11) with a weight matrix B ∈ R . The cut norm of B as deﬁned in [4] has important combinatorial properties but is NP-hard to compute. A consequence of the symmetric Grothendieck inequality is that 1 3 kLAkΓ ≤ kBkcut ≤ kLAkΓ, 8KΓ 8 where LA := diag(A1) − A is the Laplacian corresponding to A, thus aﬀording an approximation via the polynomial-time computable Γ-norm. n Another consequence of the symmetric Grothendieck inequality is that for any A ∈ S , the gap between two NP-hard combinatorial problems,

T T stretch(A) = maxx∈{−1,+1}n x Ax − minx∈{−1,+1}n x Ax , which was studied in [13], can be uniformly approximated by a polynomial-time computable spread(A) = maxX∈Gn tr(AX) − minX∈Gn tr(AX) , n where G is the convex set of n×n correlation matrices. More precisely, the symmetric Grothendieck inequality will imply that

stretch(A) ≤ spread(A) ≤ Kγ stretch(A), where Kγ is the symmetric Grothendieck constant corresponding to the (γ, d)-seminorms in (15).

Notations. We introduce some notations and conventions that will make an appearance later. n To avoid clutter, any statement about S would mean that it applies to both the space of real symmetric matrices and the space of complex Hermitian matrices alike. If a statement applies only n n to the former or the latter, we will write S (R) or S (C) respectively. This same convention will n apply to all other notations below. We remind the reader that S (C) is a vector space over R but n n n not over C; and that S × S → R,(A, B) 7→ tr(AB) deﬁnes a real inner product over both S (R) n and S (C) alike. 6 SHMUEL FRIEDLAND AND LEK-HENG LIM

n The following two subspaces of S will play important roles in our article: n n S= := {A ∈ S : a11 = ··· = ann}, n n S◦ := {A ∈ S : a11 = ··· = ann = 0}, i.e., the subspace of matrices with equal diagonal entries, and its subspace comprising those whose n n diagonal entries are all zeros. As usual, we write S+ and S++ for the cones of positive semidefinite and positive definite matrices respectively. Throughout this article a cone will always mean a convex cone. d For x, y ∈ k , we write ( xTy if k = R, hx, yik = ∗ x y if k = C, for the Euclidean and Hermitian inner products. Again, we will usually write hx, yi and only indicate field dependence when it is necessary. d For x1, . . . , xn ∈ k , we write   hx1, x1i hx1, x2i ... hx1, xni hx2, x1i hx2, x2i ... hx2, xni :   n G(x1, . . . , xn) =  . . .. .  ∈ S  . . . .  hxn, x1i hxn, x2i ... hxn, xni for the corresponding Gram matrix. Let n n d Gd := {G(x1, . . . , xn) ∈ S : kx1k = 1,..., kxnk = 1, x1, . . . , xn ∈ k } (17) n = {G ∈ S+ : g11 = ··· = gnn = 1, rank(G) ≤ d}

be the set of rank-d correlation matrices; and Ln n d d := {G(x1, . . . , xn) ∈ S : kx1k ≤ 1,..., kxnk ≤ 1, x1, . . . , xn ∈ k } (18) n = {G ∈ S+ : g11, . . . , gnn ∈ [0, 1], rank(G) ≤ d}

be the set of rank-d subcorrelation matrices. Clearly,

L L n n n n n Ln n n n

G1 ⊆ G2 ⊆ · · · ⊆ Gn ⊆ S= ∩ S+, 1 ⊆ 2 ⊆ · · · ⊆ n ⊆ S+.

L L n n Ln n n n For any d ≥ n, we have Gd = Gn and d = n, which we denote by G and respectively, i.e., n n

G = {G ∈ S+ : g11 = ··· = gnn = 1}, Ln n = {G ∈ S+ : g11, . . . , gnn ∈ [0, 1]}, are the convex sets of correlation matrices and subcorrelation matrices respectively. For the special case d = 1, the rank-one correlation and subcorrelation matrices may also be expressed as n ∗ n

G1 = {xx ∈ S : |x1| = 1,..., |xn| = 1}, Ln ∗ n 1 = {xx ∈ S : |x1| ≤ 1,..., |xn| ≤ 1}.

Note also that

L n n n n n Ln n n n span(G ) = S=, cone(G ) = S= ∩ S+, span( ) = S , cone( ) = S+.

n n n n n ReiteratingL the point we made in the first paragraph, we write S (k), S◦ (k), S=(k), S+(k), S++(k), n n G (k), (k) for k = R or C only when it is necessary to emphasize the field dependence. We write R+ = [0, ∞) for the nonnegative reals. The closed unit disk, denoted D, can either be real, D = [−1, 1], or complex, D = {z ∈ C : |z| ≤ 1}, and will be clearly specified. Likewise the unit circle, denoted T, can either be real, T = {−1, 1}, or complex, T = {z ∈ C : |z| = 1}, and m×n n n n will be clearly specified. Derivatives of these notations like R+ , S (R+), S◦ (R+), S=(D), etc, are self-explanatory. SYMMETRIC GROTHENDIECK INEQUALITY 7

n All vectors x ∈ k would be regarded as column vectors. Row vectors will always be denoted with T ∗ n transpose or adjoint, x or x . When enclosed in parentheses, a vector denoted (x1, . . . , xn) ∈ k would mean a column vector with entries x1, . . . , xn ∈ k. Matrices are denoted in uppercase letters, e.g., A, with its (i, j)th entry denoted in corresponding lowercase letters, e.g., aij. We deﬁne our sign function over k = R and C as ( z/|z| z 6= 0, (19) sgn z = 1 z = 0. The somewhat nonstandard convention sgn 0 := 1, ensuring that sgn always takes values in T, will simplify some proofs (e.g., Theorem 5.1) but makes no diﬀerence elsewhere (e.g., Lemma 3.2).

2. Grothendieck norms The perspective adopted in this article is that the Grothendieck inequality is a statement about the G-norm and the (∞, 1)-norm, and that the two symmetric Grothendieck inequalities (10) are analogous statements about the γ- and θ-seminorms, and the Γ- and Θ-norms respectively. More generally, all inequalities discussed in this article, including various generalizations [1, 10, 11, 12, 19, 26] and specializations [6, 23, 39, 43] of the Grothendieck inequality, will be viewed as relations between the Grothendieck d-norms for diﬀerent values of d. As such, we will provide a fairly extensive discussion of these norms in this section. We ﬁrst elaborate on a point raised in Section 1.1. An observation [22, Lemma 2.2] about the G-norm and the (∞, 1)-norm in (2) is that replacing “= 1” with “≤ 1” in the maxima leaves the norms unchanged,3 i.e., (2) may be rewritten as m n m n X X X X (20) kBkG = max bijhxi, yji and kBk∞,1 = max bijε¯iδj . kxik≤1, kyj k≤1 |εi|≤1, |δj |≤1 i=1 j=1 i=1 j=1 This is not the case for their symmetric counterparts in (5) and (6) — with “= 1” in the maxima, n n we obtain seminorms on S ; with “≤ 1” in the maxima, we obtain norms on S . Clearly,

(21) kAkθ ≤ kAkΘ, kAkγ ≤ kAkΓ, n and these inequalities are strict when A ∈ S is nonzero, diagonal, and traceless.

2.1. Basic properties. While there is an apparent dependence on d in the definitions of k · kG, k · kΓ, and k · kγ, our assumption that d ≥ m + n in the first case and d ≥ n in the latter two cases remove this dependence. However, it will aid our discussions here and in the rest of our article to study a family of norms/seminorms that interpolate between k · k∞,1 and k · kG, k · kΘ and k · kΓ, k · kθ and k · kγ — these are the Grothendieck d-norms that we introduced in Section 1.3. We will establish that they are indeed norms/seminorms in Lemma 2.3. We will see that results for Grothendieck d-norms are often no harder to establish for general d than for specific cases like d = 1, n, m + n. n Lemma 2.1. Let d, m, n ∈ N and the Grothendieck d-norms be as defined in (15). For any A ∈ S m×n and B ∈ k , we have kAkθ = kAkγ,1 ≤ kAkγ,2 ≤ · · · ≤ kAkγ,n = kAkγ,

(22) kAkΘ = kAkΓ,1 ≤ kAkΓ,2 ≤ · · · ≤ kAkΓ,n = kAkΓ,

kBk∞,1 = kBkG,1 ≤ kBkG,2 ≤ · · · ≤ kBkG,m+n = kBkG, and for all d ≥ n (ﬁrst two) or d ≥ m + n (third),

(23) kAkγ,d = kAkγ, kAkΓ,d = kAkΓ, kBkG,d = kBkG. 3 This applies more generally to k · kG,d, which is why it has two expressions in (15). 8 SHMUEL FRIEDLAND AND LEK-HENG LIM

Proof. The equalities and inequalities in (22) follow from the deﬁnitions in (15). The stabilization (23) is well-known for the (G, d)-norms and the same reason applies to the (Γ, d)-norms and (γ, d)- d d seminorms: When d ≥ n, any n vectors x1, . . . , xn ∈ k lie in an n-dimensional subspace of k and n maximizing over it gives the same value as maximizing over k . The inequalities (22) reﬁne

kAkθ ≤ kAkγ, kAkΘ ≤ kAkΓ, kBk∞,1 ≤ kBkG that we saw in (10) and (3). By virtue of (23), we may assume that d = n when we discuss k · kγ or k · kΓ, and d = m + n when we discuss k · kG. In fact, the stabilizations in (23) hold for much smaller values of d: We will see in Section 2.4 that over R, k · kγ,d = k · kγ and k · kΓ,d = k · kΓ as soon as d(d + 1)/2 > n and k · kG,d = k · kG as soon as d(d + 1)/2 > m + n; over C, these happen even sooner — d(d + 1)/2 may be replaced by d2. The (G, d)-norms have additional alternative expressions that will be useful later. The following hold over over R and C alike but we may of course drop the “Re” over R. Note however that there is no analogue of (24) for the (γ, d)-seminorms and (Γ, d)-norms, i.e., the absolute value cannot be replaced by the real part in their deﬁnitions in (15). m×n Proposition 2.2. Let d ∈ N. If B ∈ k , then m n X X d (24) kBkG,d = max Re bijhxi, yji : xi, yj ∈ k , kxik ≤ 1, kyjk ≤ 1 . i=1 j=1 n If A ∈ S , then n n 1 X X (25) kAk = max a (hx , y i + hy , x i): x , y ∈ d, kx k ≤ 1, ky k ≤ 1 . G,d 2 ij i j i j i j k i j i=1 j=1 In both (24) and (25), the “≤ 1” constraints may be replaced by “= 1”.

Proof. The equalities (24) are a simple consequence of the fact that any z ∈ C has |z| = Re tz for ∗ some t ∈ T, and ktxik = |t|kxik = kxik; ditto for R. If A = A , then tr AZ = tr AZ = tr (AZ)T = tr Z∗A∗ = tr Z∗A = tr AZ∗, and we get Z + Z∗ Re tr(AZ) = tr A , 2 from which (25) follows. n n Define the maps ∆, Ξ: S → S by    1  a11 n tr(A) a12 . . . a1n 1  a22   a21 tr(A) a2n  ∆(A) =   , Ξ(A) =  n   ..   . .. .   .   . . .  1 ann an1 an2 ... n tr(A) n for any A ∈ S . The maps ∆ and Ξ are respectively the orthogonal projections onto the subspace of n diagonal matrices and the subspace S=. Some immediate observations are that tr Ξ(A) = tr(A), ∗ ∗ ∗ n ∗ ∗ n that x Ax = x Ξ(A)x if xx ∈ G1 , and that ∆(xx ) = I iff xx ∈ G1 . n Lemma 2.3. Let A ∈ S and d ∈ N. Then (i) kAkγ,d = 0 if and only if Ξ(A) = 0; n n (ii) k · kγ,d defines a seminorm on S and a norm on S=; n (iii) k · kΓ,d defines a norm on S . SYMMETRIC GROTHENDIECK INEQUALITY 9

Proof. Nonnegativity, scaling, and triangle inequality are either obvious or straightforward to establish in all cases. This proof focuses on the definiteness. For n = 1, 2, the proofs of (i) and (ii) are easy. We will assume that n ≥ 3. Let T denote the unit circle in k = R or C and µ be the corresponding Haar measure. Thus iϑ T = {−1, 1} and µ the uniform counting measure if k = R; and T = {e ∈ C : ϑ ∈ [0, 2π)} and −1 dµ = (2π) dϑ if k = C. Our arguments below will apply to both fields. Consider the n-torus n n ∗ n ∗ n T := T × · · · × T = {x ∈ k : ∆(xx ) = I} = {x ∈ k : xx ∈ G1 }. n n×n Let µn be the product measure on T . For any A ∈ k , straightforward calculations give Z Z ∗ ∗ |tr(A)| = x Ax dµn ≤ |x Ax| dµn ≤ kAkθ. x∈Tn x∈Tn If kAkθ = 0, then tr(A) = 0 and so ∆ Ξ(A) = 0. We claim that aij = 0 for all i 6= j. Without loss of generality, we may assume i = n − 1 and j = n. We will show that an−1,n = 0. Let n n−2 x(s) = (s, t, 1) ∈ T where t ∈ T is fixed and s ∈ T . Again it is straightforward to see that Z 1 ∗ Re(tan−1,n) = x(s) Ax(s) dµn−2. 2 s∈Tn−2 Hence |Re(tan−1,n)| ≤ kAkθ = 0. Since t ∈ T is arbitrary, we have an−1,n = 0. Hence Ξ(A) = 0. The converse is obvious: If Ξ(A) = 0, then kAkθ = 0. In other words, if kAkγ,1 = kAkθ = 0, then A is a diagonal matrix with tr(A) = 0, and so kAkγ,d = 0 for all d ∈ N. On the other hand, if kAkγ,d = 0 for any d ∈ N, then kAkθ = 0 by (22). We have established (i). Since k · kγ,d vanishes exactly on ker(Ξ), the subspace of traceless diagonal matrices, it defines a n norm on the orthogonal complement im(Ξ) = S=. This establishes (ii). n For (iii), it is obvious by their definitions that for any A ∈ S ,

kAkγ,d ≤ kAkΓ,d.

So if kAkΓ,d = 0, then kAkγ,d = 0 and thus Ξ(A) = 0. So A is a traceless diagonal matrix. But T T it also follows from (22) that if kAkΓ,d = 0, then kAkΘ = 0, and so aii = eiAei = tr(Aeiei ) = 0, n i = 1, . . . , n. This shows that A = 0. Hence k · kΓ,d is a norm on S . n The proof of Lemma 2.3 in fact shows that k · kγ,d is a norm on a subspace W ⊆ S if and only if W does not contain a nonzero traceless diagonal matrix. Furthermore, any subspace W on which n k · kγ,d restricts to a norm can have dimension at most dim S=. We record this slightly stronger conclusion below. n n Corollary 2.4. The seminorm k · kγ,d on S restricts to a norm on any subspace W ⊆ S that does n not contain a nonzero diagonal matrix with zero trace. If k · kγ,d deﬁnes a norm on W ⊆ S , then ( n(n − 1)/2 + 1 k = R, dim W ≤ 2 n − n + 1 k = C. 2.2. Characterizations of and relations between Grothendieck norms. We will look at some alternative characterizations of the Grothendieck norms and relations between them that will be useful later. The most straightforward characterization is to simply rewrite the (Γ, d)-norms and (γ, d)- seminorms in Lemma 2.1 in terms of the correlation and subcorrelation matrices in (17) and (18). n

Proposition 2.5. Let A ∈ S and d ∈ N. Then n Ln (26) kAkγ,d = max{|tr(AG)| : G ∈ Gd }, kAkΓ,d = max{|tr(AG)| : G ∈ d }, 10 SHMUEL FRIEDLAND AND LEK-HENG LIM and

(27) kAkΓ,d = max{kDADkγ,d : D = diag(δ1, . . . , δn), δi ∈ [0, 1]}.

d d Proof. Let xi ∈ k with kxik ≤ 1. Write xi = δiti with ti ∈ k , ktik = 1, and δi ∈ [0, 1]. Then Pm,n Pm,n i,j=1 aijhxi, xji = i,j=1 δiaijδjhti, tji. n Corollary 2.6. Let d ∈ N. If A ∈ S , then

(28) kAkΓ,d ≤ kAkG,d ≤ 2kAkΓ,d. m×n If B ∈ k , then 1 0 B 1 0 B (29) kBkG,d = ∗ = ∗ . 2 B 0 γ,d 2 B 0 Γ,d d Proof. For xi, yi ∈ k , kxik ≤ 1, kyik ≤ 1, i = 1, . . . , n, set 1 1 u = (x + y ), v = (x − y ), i 2 i i i 2 i i n n×n and let Z = hxi, yji i,j=1 ∈ k . Note that kuik ≤ 1, kvik ≤ 1, and Z + Z∗ = G(u , . . . , u ) − G(v , . . . , v ). 2 1 n 1 n Now use (25) to deduce kAkG,d ≤ 2kAkΓ,d. m×n Let B ∈ k be given and set 0 B A = ∈ m+n. B∗ 0 S= d m+n Let x1, . . . , xm, y1, . . . , yn ∈ k be unit vectors and G = G(x1, . . . , xm, y1, . . . , yn) ∈ Gd . Observe that m n X X tr(AG) = 2 Re bijhxi, yji . i=1 j=1

Using the ﬁrst characterization of (G, d)-norms in (15), we see that kAkγ,d ≤ 2kBkG,d. Replacing yj with tyj for all j = 1, . . . , n, where t ∈ T is suitably chosen so that m n m n X X X X kBkG,d = bijhxi, yji = bijhxi, tyji,

i=1 j=1 i=1 j=1 we see that kAkγ,d = 2kBkG,d. Repeating the same arguments with the second characterization of (G, d)-norms in (15) gives kAkΓ,d = 2kBkG,d. The relations between the diﬀerent types of Grothendieck norms are best seen from their restric- n n n n tions to special subspaces and subcones of S , notably, S◦ , S+, and S (R+), which we will discuss in the following. n Lemma 2.7. Let A ∈ S◦ and d ∈ N. Then

kAkγ,d = kAkΓ,d. n Proof. Clearly kAkγ,d ≤ kAkΓ,d. For A ∈ S◦ , a11 = ··· = ann = 0 and so X tr(AG) = aijgji

i6=j Ln is independent of the diagonal entries of G. By the characterizations in (26), let G ∈ d be such n that kAkΓ,d = |tr(AG)|. We claim that G ∈ Gd , i.e., its diagonal entries are all 1’s. Suppose not; let G = G(x1, . . . , xn) have a maximal number of xi’s with kxik = 1 but that kxkk < 1 for some SYMMETRIC GROTHENDIECK INEQUALITY 11 k. Since kAkΓ,d = tr tAG(x1, . . . , xn) for t = ±1, it is a real linear function in xk when all other variables are ﬁxed, and hence attains its maximum when kxkk = 1, a contradiction.

n We now show that over S+, the three diﬀerent types of Grothendieck d-norms agree. As usual, n n for A, B ∈ S , we write A B iﬀ A − B ∈ S+. n Proposition 2.8. Let d ∈ N. If A ∈ S , then

(30) kAkγ,d ≤ kAkΓ,d ≤ kAkG,d. n If A ∈ S+, then

(31) kAkγ,d = kAkΓ,d = kAkG,d; n furthermore, for any B ∈ S with −A B A,

kBkγ,d ≤ kAkγ,d, kBkΓ,d ≤ kAkΓ,d, kBkG,d ≤ kAkG,d. Proof. The inequalities (30) are obvious from deﬁnitions. It remains to show that equality must n hold throughout if A ∈ S+. n n 2 ∗ d Any A ∈ S+ has a square root R ∈ S+, so A = R and R = R. Let x1, . . . , xn, y1, . . . , yn ∈ k d×n be of unit norms. Deﬁne the matrices X = [x1, . . . , xn], Y = [y1, . . . , yn] ∈ k . Observe that n ∗ hxi, yji i,j=1 = X Y . Hence n n X X ∗ ∗ ∗ ∗ ∗ aijhxi, yji = tr(AX Y ) = tr(Y AX ) = tr (RY ) (RX ) . i=1 j=1

∗ d×n Cauchy–Schwarz applied to the inner product hU, V i = tr(U V ) on k gives ∗ ∗ ∗ 2 ∗ ∗ ∗ 2 ∗ 2 tr (RY ) (RX ) ≤ tr(Y AY ) tr(XAX ) ≤ max tr(XAX ) , tr(Y AY ) .

Since kxik = kyjk = 1 for all i, j, the last inequality gives us kAkG,d = kAkγ,d and thus kAkG,d = kAkγ,d. n n Now suppose A ∈ S+ and B ∈ S with −A B A. By taking limits, it suﬃces to consider the n n −1 −1 case when A ∈ S++; in which case R ∈ S++ as well. Let C = R BR . Then −I C I, i.e., all eigenvalues of C lie in the interval [−1, 1]. So the last statement is equivalent to C2 I. d×n By Cauchy–Schwarz on k , |tr(BX∗Y )|2 = |tr(YBX∗)|2 = |tr(YRCRX∗)|2 ∗ ∗ ∗ 2 2 ∗ 2 ∗ = tr (RY ) (CRX ) ≤ tr(YR Y ) tr(XRC RX ).

2 ∗ 2 ∗ d×n As C I, it follows that tr(Z C Z) ≤ tr(Z Z) for any Z ∈ k . Hence |tr(BX∗Y )|2 ≤ maxtr(XAX∗)2, tr(Y AY ∗)2 .

This shows that kBkG,d ≤ kAkG,d. As kBkγ,d ≤ kBkΓ,d ≤ kBkG,d and kAkG,d = kAkΓ,d = kAkγ,d, the corresponding inequalities for k · kγ,d and k · kΓ,d follow. A consequence of (27) and Proposition 2.8 is the following characterization of the Grothendieck n norms on S+. Note that a departure from (26) is that we do not need to take absolute values of the trace.

n Corollary 2.9. Let A ∈ S+ and d ∈ N. Then n kAkγ,d = kAkΓ,d = kAkG,d = max{tr(AG): G ∈ Gd }. 12 SHMUEL FRIEDLAND AND LEK-HENG LIM

n n For the special case d = n, Gd = G is a closed convex set and Corollary 2.9 says that the n Grothendieck norms reduce to the support function [8, pp. 63 and 120] of G . The Grothendieck d-norms are diﬃcult to compute in general. For one they are not unitarily invariant; we suspect that they are NP-hard for a wide range of d’s — see Conjecture 2.18. To get a handle on these norms, it will often be helpful to examine the few simple cases where they have closed-form expressions. In cases (i) and (ii) below, note that the diagonal of a matrix in n n×n S (C) must have real values; in particular, a diagonal matrix in C does not necessarily belong n to S (C). Lemma 2.10. Let d ∈ N and k = R or C. n (i) If A = diag(a11, . . . , ann) ∈ S , then n n n X X X kAkγ,d = aii , kAkΓ,d = max max(aii, 0), − min(aii, 0) ;

i=1 i=1 i=1 n×n and if B = diag(b11, . . . , bnn) ∈ k , then n X kBkG,d = |bii|. i=1 n (ii) If A ∈ S is a symmetric tridiagonal matrix, then n n−1 X X kAkγ,d = aii + 2 |ai,i+1|.

i=1 i=1 n (iii) If A ∈ S (R+), then n n X X kAkγ,d = kAkΓ,d = kAkG,d = aij; i=1 j=1 m×n and if B ∈ R+ , then m n X X kBkG,d = bij. i=1 j=1

h A1 −B i m+n m n m×n (iv) If A = T ∈ ( ) has A1 ∈ ( +), A2 ∈ ( +), and B ∈ , then −B A2 S R S R S R R+ m+n m+n X X kAkγ,d = kAkΓ,d = kAkG,d = |aij|.

i=1 j=1 n n Ln Proof. For a diagonal A ∈ S , tr(AG) = tr(A) for any G ∈ Gd and so kAkγ,d = |tr(A)|. For G ∈ d , Pn tr(AG) = i=1 aiigii for gii ∈ [0, 1], giving us the required expression for kAkΓ,d. n Pn−1 n For a tridiagonal A ∈ S , tr(AG) = tr(A) + 2 i=1 Re(ai,i+1gi+1,i) for any G ∈ Gd . Hence Pn−1 n Pn−1 |tr(AG)| ≤ |tr(A)| + 2 i=1 |ai,i+1| for all G ∈ Gd and so kAkγ,d ≤ |tr(A)| + 2 i=1 |ai,i+1|. Replacing A by −A if tr(A) < 0, we may assume without loss of generality that tr(A) ≥ 0. Let ∗ ¯ ∗ δ1, . . . , δn ∈ T and D = diag(δ1, . . . , δn). As D AD = (δiaijδj), we have kD ADkγ,d = kAkγ,d. ∗ Replacing A by D AD if necessary, where D = diag(1, δ2, . . . , δn) is such that δ¯iai,i+1δi+1 ≥ 0, i = 1, . . . , n − 1, we may assume that the oﬀ-diagonal entries of A are nonnegative. Therefore 1T 1 Pn−1 Pn−1 A = tr(A) + 2 i=1 |ai,i+1|. Thus kAkγ,d ≥ |tr(A)| + 2 i=1 |ai,i+1|. Pm+n Pm+n For (iv), by their deﬁnitions in (15), kAkγ,d ≤ kAkΓ,d ≤ kAkG,d ≤ i=1 j=1 |aij|. Now m+n T observe that if we set x = (1m, −1n) ∈ R , then x Ax attains this upper bound. The equalities in (iii) are special cases of (iv), obtained either by setting m = 0 or setting A1 = A2 = 0. SYMMETRIC GROTHENDIECK INEQUALITY 13

For the tridiagonal case (ii), we are unable to obtain similar expressions for kAkΓ,d and kAkG,d. The last case (iv) is of interest since the weighted Laplacian of a bipartite graph takes this form with A1,A2 diagonal matrices. In all cases above, the norms happen to be independent of d, we know of course that this is not true in general. The diagonal case (i), while easy to prove, has an important implication — it shows that all three types of Grothendieck norms are distinct on n S . In particular, the symmetric Grothendieck inequalities in (10) cannot be obtained simply by n restricting the original Grothendieck inequality (3) to S . n Corollary 2.11. There exists A ∈ S such that

kAkγ,d < kAkΓ,d < kAkG,d. 1 0 Proof. Take A = 0 −1 . Then kAkγ,d = 0, kAkΓ,d = 1, kAkG,d = 2 by Lemma 2.10(i). 2.3. Real and complex Grothendieck norms. The value of a matrix norm in general depends on the choice of the base ﬁeld. This has been discussed for operator norms in [27]: Take the 1 −1 2×2 (∞, 1)-norm and B = 1 1 ∈ R for example, n 2 o R X kBk = max bijε¯iδj : εi, δj ∈ {−1, +1} = 2, ∞,1 i,j=1 n 2 o √ C X iϑ kBk = max bijε¯iδj : εi, δj ∈ {e : ϑ ∈ [0, 2π)} = 2 2. ∞,1 i,j=1 We will show that this is generally the case for the Grothendieck d-norms for smaller values of d but not when d is large enough. In particular, the values of the G-norm, Γ-norm, and γ-seminorm for a real matrix stay the same regardless of whether we take them over R or C. We begin by establishing a precise relation between the real and complex versions of the Grothendieck d-norms in (15).

n Proposition 2.12. Let d ∈ N and A ∈ S (R). Then C R C R (32) kAkγ,d = kAkγ,2d, kAkΓ,d = kAkΓ,2d. m×n Let d ∈ N and B ∈ R . Then C R (33) kBkG,d = kBkG,2d. d d 2 2 Proof. Let zi ∈ C , kzik = 1, i = 1, . . . , n. Write zi = ui + ivi with ui, vi ∈ R , kuik + kvik = 1, 2d n and set xi = (ui, vi) ∈ R with kxik = 1. Let A ∈ S (R). Then n n n n n n X X X X X X aijhzi, zjiC = aij Rehzi, zjiC = aijhxi, xjiR. i=1 j=1 i=1 j=1 i=1 j=1 2d Since x1, . . . , xd lie in a subspace of R of dimension at most d, it follows from the deﬁnitions in C R 2d (5) and (6) that kAkγ,d ≤ kAkγ,2d. Conversely, let xi ∈ R , kxik = 1, i = 1, . . . , n, be such that R Pn Pn 2d d kAkγ,2d = i=1 j=1 aijhxi, xjiR . Write xi = (ui, vi) ∈ R and set zi = ui + ivi ∈ C . Then n n R X X C kAk = aijhzi, zji ≤ kAk . γ,2d C γ,d i=1 j=1 Hence the ﬁrst equality in (32) holds; and the second equality can be similarly established. For (33), we start with the characterization of the G-norm in (20). Observe that m n C X X (34) kBkG = max Re bijhwi, zji , kw k=kz k=1 i j i=1 j=1 14 SHMUEL FRIEDLAND AND LEK-HENG LIM

d d where w1, . . . , wm, z1, . . . , zn ∈ C . Write wi = si + iti, zj = uj + ivj with si, ti, uj, vj ∈ R , and 2d xi = (si, ti), yj = (uj, tj) ∈ R , i = 1, . . . , m, j = 1, . . . , n. Note that T T Rehwi, zjiC = si uj + ti vj = hxi, yjiR. m×n Also if kwik = kzjk = 1, then kxik = kyjk = 1. Since B ∈ R , m n m n m n X X X X X X Re bijhwi, zjiC = bij Rehwi, zjiC = bijhxi, yjiR. i=1 j=1 i=1 j=1 i=1 j=1 Taking maximum as in (34), we get (33). If we apply (32) with d ≥ n and (33) with d ≥ m + n, then we have the following corollary by (23) in Lemma 2.1. n Corollary 2.13. Let A ∈ S (R). Then R C R C (35) kAkγ = kAkγ , kAkΓ = kAkΓ. m×n Let B ∈ R . Then R C (36) kBkG = kBkG. 1 −1 2×2 The analogue of (36) is false for the (∞, 1)-norm by the example B = 1 1 ∈ R that we saw 0 B R C R C earlier. By setting A = B∗ 0 and invoking (29), we get that kAkθ 6= kAkθ and kAkΘ 6= kAkΘ. So the analogue of (35) is likewise false for the θ-seminorm and the Θ-norm. However, the special case d = 1 in Proposition 2.12 gives us the following characterizations. n Corollary 2.14. Let A ∈ S (R). Then C R C R (37) kAkθ = kAkγ,2, kAkΘ = kAkΓ,2. m×n Let B ∈ R . Then C R (38) kBk∞,1 = kBkG,2. n n On the cone of positive semidefinite matrices S+(R) and the cone of nonnegative matrices S (R+), Corollary 2.9 and Lemma 2.10(iii) respectively hold true more generally across different fields. n Corollary 2.15. Let d ∈ N. If A ∈ S+(R), then R R R C C C (39) kAkγ,d = kAkΓ,d = kAkG,d = kAkγ,d = kAkΓ,d = kAkG,d. m×n If B ∈ R+ , then R C kBkG,d = kBkG,d. n If A ∈ S (R+), then not only does (39) hold, its value is the same for all d ∈ N. 2.4. Stabilization of Grothendieck d-norms. We show here that (23) can be improved — the stabilizations k · kγ,d = k · kγ, k · kΓ,d = k · kΓ, k · kG,d = k · kG occur for far smaller values of d than what is given in Lemma 2.1. This will require a more careful argument, beginning with an

observation about extreme points.

n Ln Lemma 2.16. For any d ∈ N, the extreme pointsL of the convex hulls conv(Gd ) and conv( d ), n n n viewed as subsets of S , are contained in Gp and p respectively, where √ −1 + 1 + 8n  + 1 k = R, p = 2  √ b nc + 1 k = C.

SYMMETRIC GROTHENDIECK INEQUALITY 15

L L

Ln n n

L L Proof. We begin withL k = R and d . Since d is compact, conv( d ) is compact, and so an extreme n n n point G of conv( d ) is contained in d . If d ≤ p, then G ∈ p and the lemma trivially holds. So suppose d > p. Let r := rank(G). If r ≤ p < d, then im(G) is isometric to an r-dimensional

p subspace W of R . So if G = GL(x1, . . . , xn), x1, . . . , xn ∈ im(G), then there exist y1, . . . , yn ∈ W such that G = G(y , . . . , y ) ∈ n. 1 n p √ n Now suppose r > p = b(−1 + 1 + 8n)/2c + 1. Since the subspace of matrices {A ∈ S : n im(A) ⊆ im(G)} has dimension r(r + 1)/2 > n and dim S◦ = n(n − 1)/2, their intersection must T contain a nonzero A, i.e., with im(A) ⊆ im(G) and a11 = ··· = ann = 0. Let G = VrΛrVr be a reduced eigenvalue decomposition with λ1 ≥ · · · ≥ λr > 0. Since im(A) ⊆ im(G), we must T r

have A = VrBrVr for some Br ∈ S . As Λr 0 it follows that there exists ε > 0 such that Λr + tBr 0 wheneverL |t| < ε. Hence G + tA 0 whenever |t| < ε. Since A has an all-zero

n diagonal, G + tA ∈ d . As A 6= 0, it follows that G+ := G + tA and G− := LG − tA are both not

1 n

L equal to G but yet G = 2 (G+ + G−). So GL is not an extreme point of conv( d ). The exact same n n n n proof applies with Gd and Gp in place of d and p . The arguments for k = C are similar with the following observation: If rank(G) = r > p = √ n 2 b nc + 1, then {A ∈ S (C) : im(A) ⊆ im(G)} has real dimension r > n; the real dimension of n n 2 S◦ (C) is n(n − 1), and that of S (C) is n . Since the maximum of a convex function over a compact convex set S is attained on its set of extreme points extr(S), we obtained the following stability result for Grothendieck d-norms.

Proposition 2.17. Let d, m, n ∈ N. n (i) If d(d + 1)/2 > n, then for all A ∈ S (R),

kAkγ,d = kAkγ,n = kAkγ, kAkΓ,d = kAkΓ,n = kAkΓ. m×n (ii) If d(d + 1)/2 > m + n, then for all B ∈ R ,

kBkG,d = kBkG,n = kBkG. 2 n (iii) If d > n, then for all A ∈ S (C),

kAkγ,d = kAkγ,n = kAkγ, kAkΓ,d = kAkΓ,n = kAkΓ. 2 m×n (iv) If d > m + n, then for all B ∈ C ,

kBkG,d = kBkG,n = kBkG.

Proof. We always have kAkγ,d ≤ kAkγ. By (26), n n kAkγ = max{|tr(AG)| : G ∈ G } = max{|tr(AG)| : G ∈ extr(G )} n ≤ max{|tr(AG)| : G ∈ Gd } = kAkγ,d, where the inequality follows from Lemma 2.16. The same argument applies to the (Γ, d)-norm. The equalities for the (G, d)-norm follows from (29).

The Grothendieck d-norms k · kγ,d, k · kΓ,d, k · kG,d are NP-hard over R when d = 1; see (86) for example. We suspect the following: Conjecture 2.18. The following norms are all NP-hard: R R n (i) k · kγ,d and k · kΓ,d on S (R) when d(d + 1)/2 ≤ n; R m×n (ii) k · kG,d on R when d(d + 1)/2 ≤ m + n; C C n 2 (iii) k · kγ,d and k · kΓ,d on S (C) when d ≤ n; C m×n 2 (iv) k · kG,d on C when d ≤ m + n.

16 SHMUEL FRIEDLAND AND LEK-HENG LIM Ln n

It is perhaps worthwhile to point out that d and Gd are nonconvex sets when d < n, thus

L computing k · kγ,d andL k · kΓ,d involve the nonconvex optimization problems in (26). On the other n n n n hand, when d ≥ n, d = and Gd = G are convex sets and the optimization problems in (26) are just√ semideﬁnite programs. An implication√ of Proposition 2.17 is that when d is in the range b(−1 + 1 + 8n)/2c + 1 ≤ d < n (for R) or b nc + 1 ≤ d < n (for C), the nonconvex optimization R problems in (26) are solvable in polynomial-time. For the case of k · kγ,d, d(d+1)/2 > n, this follows from [34, Theorem 2.1], a result that originally appeared in [5, 41].

3. Gaussian integrals of sign functions

k We will introduce a function ϕd : D → k, where D is the closed unit disc in k = R or C and d ∈ N, that will play a critical role in the proofs of all Grothendieck-like inequalities in this article. The function first appeared in [33] for d = 1 and k = R, in [25] for d = 1 and k = C, was generalized to arbitrary d ≥ 1 and k = R in [11, Equation 2], and will be further extended to arbitrary d ≥ 1 and R k = C in this section. For our later purposes, we will also need to establish a few facts about ϕd not found in [11]. d×n For any d, n ∈ N, the Gaussian function on k is defined by ( (2π)−dn/2 exp−kZk2/2 if = , k : k R Gd,n(Z) = −dn 2 π exp(−kZk ) if k = C, d×n where Z ∈ k and k · k is the Frobenius or Hilbert–Schmidt norm. The special case d = 1 will be denoted ( (2π)−n/2 exp−kzk2/2 if = , k k R Gn(z) = −n 2 π exp(−kzk ) if k = C, n n where z ∈ k . We will extend the definition of sign function in (19) to a vector variable z ∈ k by ( z/kzk z 6= 0, sgn z := e1 z = 0,

n k where e1 = (1, 0,..., 0) ∈ k . For each d ∈ N, we define the function ϕd : D → k by Z k k (40) ϕd(hu, vi) := sgn(Zu), sgn(Zv) Gd,n(Z) dZ, kd×n n defined for u, v ∈ k , kuk = kvk = 1. The integral depends on u, v only through their inner product hu, vi and does not depend on n as long as n ≥ 2, which explains the notation on the left of (40). To see this, observe that the integral is invariant under a change-of-variable Z 7→ ZQ by any n×n orthogonal/unitary Q ∈ k ; choosing Q so that Qu = (1, 0, 0,..., 0) and Qv = (x, y, 0,..., 0) where x = hu, vi, y = p1 − |hu, vi|2, the integral becomes Z Z z , hu, viz + p1 − |hu, vi|2z 1 1 2 k k (41) G (z1)G (z2) dz1dz2, p 2 n n kn kn kz1kkhu, viz1 + 1 − |hu, vi| z2k which evidently depends on u, v only through hu, vi. In the following, we will drop the superscript k when a statement holds for both R and C. It is straightforward to see from (41) that if kvk = 1, then ϕd(1) = ϕd(hv, vi) = 1; and if hu, vi = 0, then ϕd(0) = ϕd(hu, vi) = 0, as hz1, −z2i = −hz1, z2i. By (40), ϕd(−hu, vi) = ϕd(hu, −vi) = −ϕd(hu, vi); R in particular, ϕd is an odd function over R and so all its even degree Taylor coefficients are zero. In fact, we can say a lot more about these coefficients.4

4We acknowledge the precedence of [12], where the authors obtained closed-form expressions [12, Lemma 2.1] for R the coeﬃcients of ϕd , from which Lemma 3.1(i) follows. However, the statements in Lemma 3.1 are more appropriate for our needs and we require them for both R and C. Note also that (iii) is not the series in [12, Lemma 2.1]. SYMMETRIC GROTHENDIECK INEQUALITY 17

k Lemma 3.1. Let d ∈ N and ϕd be as deﬁned in (40) for k = R or C. R (i) Let bk,d be the kth Taylor coeﬃcient of ϕd . Then ∞ X bk,d ≥ 0, b2k,d = 0, bk,d = 1, k=0 for all k = 0, 1, 2,.... (ii) For all x ∈ C, |x| ≤ 1, ∞ ∞ R X 2k+1 C X 2k ϕd (x) = b2k+1,d x , ϕd (x) = b2k+1,2d x|x| . k=0 k=0 In particular, for all x ∈ R, |x| ≤ 1, C R ϕd (x) = ϕ2d(x). (iii) For all x ∈ R, |x| ≤ 1, ∞ 2k 2 2(1 − x2)d/2 X 2 Γ (d + 2k + 1)/2 Γ (2k + 3)/2 ϕR(x) = √ x2k+1. d π (2k + 1)!Γd/2Γ(d + 2k + 2)/2 k=0 (iv) For all x ∈ C, |x| ≤ 1, ∞ 2k 2 2(1 − |x|2)d X 2 Γ (2d + 2k + 1)/2 Γ (2k + 3)/2 ϕC(x) = √ x|x|2k. d π (2k + 1)!Γ(d)Γ(2d + 2k + 2)/2 k=0

n k Proof. Let d ∈ N and u1, . . . , ud ∈ k with ku1k = ··· = kudk = 1. Then the matrix ϕd(hui, uji) ∈ d S is positive semideﬁnite as d d Z d d X X k X X k a¯iϕd(hui, uji)aj = ai sgn(Zui), aj sgn(Zuj) Gd,n(Z) dZ ≥ 0 d×n i=1 j=1 k i=1 j=1 d for any a ∈ k . Let d ∈ N be ﬁxed. By the real and complex Schoenberg theorem [15], we must have power series of the forms: ∞ ∞ ∞ R X k C X X j k ϕd (x) = bkx , bk ≥ 0, and ϕd (x) = cjkx x¯ , cjk ≥ 0, k=0 j=0 k=0

k k R for all j, k ∈ N ∪ {0}. As ϕd(0) = 0, ϕd(1) = 1, and ϕd is odd, we obtain ∞ ∞ ∞ X X X b0 = 0, b2k = 0, c00 = 0, bk = 1, cjk = 1. k=0 j=0 k=0

R C This gives (i) and thus the expansion for ϕd in (ii). We will show that the double series for ϕd can C C be written as a series of the form in (ii). For any ζ ∈ C, |ζ| = 1, we have ϕd (ζx) = ζϕd (x) and thus ∞ ∞ C ¯ C ¯X X j k j ¯k ϕd (x) = ζϕd (ζx) = ζ cjkx x¯ ζ ζ . j=0 k=0 As usual, write the complex variable x = s + it with real variables s, t, and set ∂ 1 ∂ ∂ ∂ 1 ∂ ∂ := − i , := + i . ∂x 2 ∂s ∂t ∂x¯ 2 ∂s ∂t 18 SHMUEL FRIEDLAND AND LEK-HENG LIM

P∞ P∞ C Since j=0 k=0 cjk = 1, it follows that ϕd (s + it) is an analytic function of s, t in the unit disc |s|2 + |t|2 ≤ 1. As ∂x/∂x = ∂x/∂¯ x¯ = 1 and ∂x/∂x¯ = ∂x/∂x¯ = 0, ∂ j ∂ k ϕC(0) = j!k!c = j!k!c ζζ¯ jζ¯k ∂x ∂x¯ d jk jk for any j, k ∈ N ∪ {0} and |ζ| = 1. Hence cjk = 0 whenever j − k 6= 1 and we get that ∞ ∞ C X k X 2k ϕd (x) = x ck(xx¯) = x ck|x| , k=0 k=0

R where ck := ck+1,k. It remains to show that ck is exactly the kth Taylor coeﬃcient of ϕ2d. This C R will follow from the uniqueness of Taylor coeﬃcients and ϕd (x) = ϕ2d(x) for all x ∈ [−1, 1], which we will establish next. n Let u, v ∈ R . Then hu, vi = hv, ui. As in the discussion√ before (41), we may assume that n = 2 2 and u = (1, 0), v = (x, y) ∈ R , where x = hu, vi, y = 1 − x2 ∈ R. Then

1 Z Z hz , xz + yz i 2 2 C 1 1 2 C −kz1k −kz2k ϕd (x) = 2d e dz1dz2 π Cd Cd kz1kkxz1 + yz2k 1 Z Z Rehz , xz + yz i 2 2 1 1 2 C −kz1k −kz2k = 2d e dz1dz2 π Cd Cd kz1kkxz1 + yz2k 1 Z Z hw , xw + yw i 2 2 1 1 2 R −(kw1k +kw2k )/2 R = 2d e dw1dw2 = ϕ2d(x). (2π) R2d R2d kw1kkxw1 + yw2k d 2d The last step follows from replacing√zi = s√i + iti ∈ C by wi = (si, ti) ∈ R , i = 1, 2, and a change-of-variables (w1, w2) 7→ (w1/ 2, w2/ 2). 2 To√ prove (iii), we may again assume n = 2 and u = (1, 0), v = (x, y) ∈ R , where x = hu, vi and 2 d d y = 1 − x > 0. Let z1, z2 ∈ R and w = xz1 + yz2 ∈ R . Then x 1 x2 1 2x z = − z + w, kz k2 = kz k2 + kwk2 − hz , wi. 2 y 1 y 2 y2 1 y2 y2 1 −d With a change-of-variable z2 7→ w, dz2 = y dw,

1 Z Z D z xz + yz E 2 2 R 1 1 2 −(kz1k +kz2k )/2 ϕd (x) = d , e dz1dz2 (2π) Rd Rd kz1k kxz1 + yz2k 1 Z Z D z w E 2 2 2 2 1 −(kz1k +kwk )/2y (x/y )hz1,wi = d , e e dz1dw (2πy) Rd Rd kz1k kwk d y Z Z D z w E 2 2 1 −(kz1k +kwk )/2 xhz1,wi = d , e e dz1dw, (2π) Rd Rd kz1k kwk where the last expression follows from another change-of-variables (z1, z2) 7→ (z1/y, z2/y), and upon Taylor expanding exhz1,wi becomes

2 d/2 ∞ k k+1 (1 − x ) x Z Z hz , wi 2 2 X 1 −(kz1k +kwk )/2 = d e dz1dw. (2π) k! d d kz1kkwk k=0 R R k+1 k+1 Let Ik denote the double integral above. For even k, as h−z1, wi = −hz1, wi , we get Ik = 0. d For odd k, we introduce polar coordinates on each copy of R as in AppendixA, with kz1k = ρ1, kwk = ρ2, and ∞ ∞ Z Z Z Z 2 2 −1 −1 k+1 −(ρ1+ρ2)/2 d+k−1 d+k−1 Ik = hρ1 z1, ρ2 wi e ρ1 ρ2 dρ1dρ2dσ1dσ2. 0 0 kz1k=ρ1 kwk=ρ2 SYMMETRIC GROTHENDIECK INEQUALITY 19

d d/2 As the surface area of the (d − 1)-sphere in R is 2π /Γ(d/2), by (97) in Lemma A.1, we have Γ(d/2)2 Z Z 1 Γd/2Γ(k + 2)/2 −1 −1 k+1 √ d hρ1 z1, ρ2 wi dσ1dσ2 = ; 4π kz1k=ρ1 kwk=ρ2 π Γ (d + k + 1)/2 and a direct calculation gives ∞ ∞ ∞ 2 Z Z 2 2 Z 2 d+k−1 d+k−1 −ρ1/2 −ρ2/2 d+k−1 −ρ /2 d+k−2 2 ρ1 ρ2 e e dρ1dρ2 = ρ e dρ = 2 Γ((d + k)/2) . 0 0 0

Taken together, we obtain the value of Ik for odd k and thus (iii).

The Taylor coeﬃcients bk,d in Lemma 3.1(ii) may be obtained from Lemma 3.1(iii) by expanding (1 − x2)d/2 as a power series (d odd) or a polynomial (d even). The case d = 1 of (40) will be particularly important for us and requires special attention. For the rest of this article, we will write

R C k ϕR := ϕ1 , ϕC := ϕ1 , ϕk := ϕ1. In this case, the power series in Lemma 3.1(ii) have the following forms: ∞ 2 X (2k − 1)!! 2 (42) ϕ (x) = x2k+1 = arcsin x, R π (2k)!!(2k + 1) π k=0 ∞ 2 π X (2k − 1)!! 1 π 1 1 (43) ϕ (x) = x|x|2k = x F , ; 2; |x|2 . C 4 (2k)!! k + 1 4 2 1 2 2 k=0

As we noted earlier, ϕC was ﬁrst introduced by Haagerup [25], who also showed that it has an integral expressions on the complex unit disk |x| ≤ 1 given by Z π/2 cos2 t (44) ϕ (x) = x dt. C 2 2 1/2 0 (1 − |x| sin t)

The reason ϕk makes an appearance in the Grothendieck inequality is a result of the following identity (47) on the ‘Gaussian inner product of the sign of Hermitian inner products’ [28, 25]. The other two related identities (45) and (46) are due to Alon and Naor in [4] for k = R; we will ﬁll-in the proof for k = C here following their ideas. n Lemma 3.2. For any u, v ∈ k , Z k (45) hu, zihz, viGn(z) dz = hu, vi. kn If in addition kuk = kvk = 1, then ( Z p2/π hu, vi if = , k k R (46) hu, zi sgnhz, viGn(z) dz = p kn π/4 hu, vi if k = C, and Z k (47) sgnhu, zi sgnhz, viGn(z) dz = ϕk(hu, vi). kn Proof. Setting d = 1 in (40) gives us (47). The identities (45) and (46) for k = R appeared in [4, Equations 4.2 and 4.3]. For k = C, since Z 1 −kzk2 n ziz¯je dz = δij, π Cn 20 SHMUEL FRIEDLAND AND LEK-HENG LIM we obtain (45) via Z Z n 1 −kzk2 1 Xn Xn −kzk2 X n hu, zihz, vie dz = n uizi z¯jvj e dz = uivi = hu, vi. π n π n i=1 j=1 C C i=1 Let kuk = kvk = 1. We claim that Z √ 1 −kzk2 π (48) n hu, zi sgnhz, vie dz = hu, vi. π Cn 2 n×n Since both sides of (48) are invariant if u, v are replaced by Qu, Qv for any unitary Q ∈ C , we may assume that v = (1, 0,..., 0). Furthermore, since u = (u1, . . . , un) may be multiplied by ζ ∈ C, |ζ| = 1, so that ζu1 ≥ 0, we may assume that u1 ≥ 0. Hence n X iϑ iϑ hu, vi = u1, hu, zi = uizi, hz, vi =z ¯1 = re , sgnhz, vi = e i=1 n 2n for some r > 0, ϑ ∈ [0, 2π). Write z = x + iy where x, y ∈ R and set w = (x, y) ∈ R . Then the left side of (48) becomes Z 1 Xn −kxk2−kyk2 (49) uizi sgn(¯z1)e dxn dyn ··· dx1 dy1. n i=1 π R2n

By Fubini, we may integrate ﬁrst with respect to dxidyi to see that Z −kxk2−kyk2 zi sgn(¯z1)e dxn dyn ··· dx1 dy1 = 0, i = 2, . . . , n, R2n and so the integral in (49) simpliﬁes as Z 1 −kxk2−kyk2 n u1z1 sgn(¯z1)e dxn dyn ··· dx1 dy1 π R2n ∞ 2π √ u Z 2 u Z Z 2 π 1 −|z1| 1 2 −r = |z1|e dx1 dy1 = r e dr dt = u1, π R2 π 0 0 2 which gives the right side of (48) as u1 = u1 = hu, vi. For a function ϕ : S → S, we write ϕ◦k = ϕ ◦ · · · ◦ ϕ for the k-fold composition of ϕ with itself. We now establish an analog of [25, Lemma 3.5] for our use later.

Lemma 3.3. (i) The function ϕR is a homeomorphism of the closed interval [−1, 1]. Its ﬁxed points are {x ∈ [−1, 1] : |x| = 0 or 1}. If 0 < |x| < 1, then lim ϕ◦k(x) = 0. k→∞ R iϑ (ii) The function ϕC is a homeomorphism of the closed unit disk D = {re ∈ C : r ∈ [0, 1], ϑ ∈ [0, 2π)} that maps each line segment [0, eiϑ] := {reiϑ : r ∈ [0, 1]} to itself for any ϑ ∈ [0, 2π). Its ﬁxed points are {z ∈ D : |z| = 0 or 1} = T ∪ {0}. If 0 < |z| < 1, then lim ϕ◦k(z) = 0. k→∞ C SYMMETRIC GROTHENDIECK INEQUALITY 21

Proof. (i) Let ϕ = ϕR. By (44), it is clearly a homemorphism of [−1, 1] with ϕ(0) = 0, ϕ(1) = 1, and ϕ(−x) = −ϕ(x) for x ∈ [1, −1]. As ϕ0(x) > 0 and ϕ00(x) > 0 for x ∈ (0, 1), ϕ is strictly increasing and strictly convex on [0, 1]. Thus ϕ(x) < x for x ∈ (0, 1). Fix x ∈ (0, 1). The ◦k sequence yk = ϕ (x) is strictly decreasing and bounded below by 0. Let y = limk→∞ yk. ◦k Then ϕ(y) = y and so y = 0. Now fix x ∈ (−1, 0). Since ϕ(−x) = −ϕ(x), yk = ϕ (x) is a strictly increasing sequence converging to 0. (ii) Let ϕ = ϕC and let h be the restriction of ϕ to [0, 1]. By (44), h(0) = 0 and h(1) = 1. As h0(x) > 0 for x ∈ (0, 1), h is strictly increasing on [0, 1]. Thus h is a homeomorphism of [0, 1]. iθ Since ϕ(z) = sgn(z)h(|z|), it follows that ϕ is a homeomorphism of D taking each [0, e ] to itself. As h00(x) > 0 for x ∈ (0, 1), h is strictly convex on [0, 1]. Thus 0 < h(x) < x for x ∈ (0, 1) and so 0 and 1 are its unique fixed points on [0, 1]. Since ϕ(z) = sgn(z)h(|z|), it follows that the fixed points of ϕ are either the origin or on the unit circle. ◦k As in case (i), we have limk→∞ h (x) = 0 for any x ∈ (0, 1). Since ϕ(z) = sgn(z)h(|z|), we ◦k must also have limk→∞ ϕ (z) = 0 whenever |z| ∈ (0, 1).

4. Symmetric Grothendieck inequality with effective bounds In this section we will establish the two versions of symmetric Grothendieck inequalities mentioned earlier:

(50) kAkγ ≤ KγkAkθ, kAkΓ ≤ KΓkAkΘ over both R and C with explicit bounds for the constants. n n Lemma 4.1. Let k = R or C. Let D = {z ∈ k : |z| ≤ 1} and S=(D) = {A ∈ S= : aij ∈ D}. The map n n (51) Φk : S=(D) → S=(D), (aij) 7→ ϕk(aij) , n is a homeomorphism of S=(D). Its ﬁxed points are n {A ∈ S=(D): |aij| = 0 or 1 for all i, j = 1, . . . , n}.

Proof. For k = R or C, ϕk(z) = ϕk(¯z) and ϕk is a homeomorphism of D = {z ∈ k : |z| ≤ 1}, n it follows that Φk is a homeomorphism of S=(D). Since the fixed point of ϕk are the z’s with |z| ∈ {0, 1}, it follows that the fixed points of Φk are exactly the matrices with |aij| ∈ {0, 1} for all i, j = 1, . . . , d. n n We denote the Schur product of two matrices A, B ∈ S by A◦B ∈ S . Recall that this is simply the coordinatewise product, i.e., the (i, j)th entry of A ◦ B is aijbij, i, j = 1, . . . , n. We also write ◦k ◦k k A for the k-fold Schur product of A with itself, i.e., the (i, j)th entry of A is aij. n n Lemma 4.2. Let A ∈ S and G ∈ G . For any k ∈ N, set ◦(k+1) ◦k n (52) Mk = A ◦ Φk(G) ◦ Φk(G) ∈ S with Φk as defined in (51). Then for all k ∈ N,

(53) kMkkθ ≤ kAkθ. Proof. The proof will apply to both k = R and C alike — the only diﬀerence is that complex conjugation will have no eﬀect when k = R. We will write Φ = Φk and ϕ = ϕk below for notational simplicity. n n Let x1, . . . , xn ∈ k be unit vectors and G = G(x1, . . . , xn) ∈ G be their Gram matrix. Consider n the matrices A ◦ Φ(G), A ◦ Φ(G) ∈ S , whose (i, j)th entries are aijϕ(hxi, xji) and aijϕ(hxi, xji) respectively. 22 SHMUEL FRIEDLAND AND LEK-HENG LIM

n Let z ∈ k with hz, xji= 6 0 for all j = 1, . . . , n. By the deﬁnition of the θ-seminorm, n n n n X X X X aij sgnhxi, zi sgnhz, xji ≤ kAkθ, aijsgnhxi, zisgnhz, xji ≤ kAkθ.

i=1 j=1 i=1 j=1 By (47), these imply n n n n X X X X (54) aijϕ(hxi, xji) ≤ kAkθ, aijϕ(hxi, xji) ≤ kAkθ.

i=1 j=1 i=1 j=1

Consider a diagonal matrix D = diag(δ1, . . . , δn) with δ1, . . . , δn ∈ T. Clearly we always have ∗ ∗ kD ADkθ = kAkθ. So replacing A by D AD in (54) gives us n n n n X X ¯ X X ¯ aijϕ(hxi, xji)δiδj ≤ kAkθ, aijϕ(hxi, xji)δiδj ≤ kAkθ.

i=1 j=1 i=1 j=1

As δ1, . . . , δn ∈ T are arbitrary, it follows that

kA ◦ Φ(G)kθ ≤ kAkθ, kA ◦ Φ(G)kθ ≤ kAkθ, i.e., the operation of Schur multiplication by Φ(G) or Φ(G) does not increase the θ-seminorm. Hence (53) follows. We now deduce that the symmetric Grothendieck inequalities (7) and (8) hold with constants bounded by sinh(π/2) ≈ 2.30130 over R and 8/π − 1 ≈ 1.54648 over C respectively. n Theorem 4.3 (Symmetric Grothendieck inequalities). Let A ∈ S and d ≥ n. Then there exist constants Kγ,KΓ > 0 independent of d and n such that n n n n X X X X ¯ (55) max aijhxi, xji ≤ Kγ max aijδiδj kxik=1 |δi|=1 i=1 j=1 i=1 j=1 and n n n n X X X X ¯ (56) max aijhxi, xji ≤ KΓ max aijδiδj kxik≤1 |δi|≤1 i=1 j=1 i=1 j=1 d where x1, . . . , xn ∈ k . Furthermore, π 8 (57) KR ≤ KR ≤ sinh ,KC ≤ KC ≤ − 1. Γ γ 2 Γ γ π

Proof. Let k = R or C and ϕR and ϕC be as in (42) and (43). We will ﬁrst show that the inverse function of ϕk has a power series expansion of the form ∞ X (58) ϕ−1(z) = c z|z|2k, z ∈ , k 2k+1 k k=0 that is convergent when |z| ≤ 1 and that  π ∞ sinh if = , X  2 k R Bk := |c2k+1| = 8  − 1 if = . k=0 π k C Case I: = . By (44), ϕ (x) = (2/π) arcsin(x) and so ϕ−1(x) = sin(πx/2). Therefore B = k R R R R sinh(π/2). Note that ∞ X (−1)k π 2k+1 ϕ−1(x) = x|x|2k, R (2k + 1)! 2 k=0 SYMMETRIC GROTHENDIECK INEQUALITY 23 is convergent when |x| ≤ 1. Case II: k = C. It follows from Lemma 3.3(ii) that the power series of the inverse function of ϕC takes the form ∞ X (59) ϕ−1(z) = b z|z|2k C 2k+1 k=0 for some b2k+1 ∈ R, k = 0, 1, 2,.... Unlike the real case, we do not have closed-form expressions for these coeﬃcients. However, by (44), we do have that z 4 b1 = lim = ; z→0 ϕC(z) π and a result of Haagerup [25] (see also [22, Section 5]) shows that

b2k+1 ≤ 0 for all k ≥ 1. Therefore ∞ ∞ X X (60) 1 = ϕ−1(1) = b + b = b − |b |, C 1 2k+1 1 2k+1 k=1 k=1 and we obtain ∞ ∞ X X 8 B = |b | = b + |b | = 2b − 1 = − 1. C 2k+1 1 2k+1 1 π k=0 k=1 It also follows from (60) that the series in (59) converges when |z| ≤ 1. We may now prove (55) starting from the tautology hx , x i = ϕ−1ϕ (hx , x i). i j k k i j n n Let Mk ∈ S be as in (52) and 1 = (1,..., 1) ∈ R . Then by (58) and the observation that the 2k (i, j)th entry of Mk is aijhxi, xji|hxi, xji| , we get n n n n ∞ X X X X X a hx , x i = a ϕ−1ϕ (hx , x i) = c 1TM 1. ij i j ij k k i j 2k+1 k i=1 j=1 i=1 j=1 k=0 Applying Lemma 4.2, we get n n ∞ X X X T aijhxi, xji ≤ |c2k+1||1 Mk1| ≤ B kAkθ. k i=1 j=1 k=0

We now show that KΓ ≤ Kγ, from which (56) follows. The same argument works over both R and C. Let D = diag(δ1, . . . , δn) be a diagonal matrix with δi ∈ [0, 1], i = 1, . . . , n. Then by (27), kDADkθ ≤ kAkΘ. Since kDADkγ ≤ KγkDADkθ ≤ KγkAkΘ, by (27) again, we have that kAkΓ ≤ KγkAkΘ. Hence KΓ ≤ Kγ. Should a distinction be necessary, we will refer to (55) as the symmetric Grothendieck inequality for γ-seminorm and (56) as the symmetric Grothendieck inequality for Γ-norm. In case it is lost on the reader, the main point of Theorem 4.3 is in establishing the bounds in (57) for the symmetric Grothendieck constants. If we do not care about the size of the bound for KΓ, then it is straightforward to get KΓ ≤ 2KG from (28) and the original Grothendieck inequality (3):

(61) kAkΓ ≤ kAkG ≤ KGkAk∞,1 = KGkAkG,1 ≤ 2KGkAkΓ,1 = 2KGkAkΘ n for any A ∈ S . If a numerical value is not required, then one may deduce KΓ = O(1) from [3, Equation 2], as pointed out in [7, p. 17]. We emphasize that the preceding discussion only applies to the inequality (56) for the Γ-norm; as far as we can tell, the inequality (55) for the γ-seminorm cannot be deduced from such general arguments even if we do not care about the size of the bounds 24 SHMUEL FRIEDLAND AND LEK-HENG LIM for Kγ. In particular, as we mentioned at the end of Section 1.1,[45, Exercise 3.5.3] is erroneous n and does not yield kAkγ ≤ 2KGkAkθ for any A ∈ S from the original Grothendieck inequality. 2 n n The correction in its online version imposes the restriction that A ∈ S◦ or S+ — in the ﬁrst case, it reduces to (61) by Lemma 2.7 and in the second, it follows from Nesterov π/2-Theorem (69). The following proposition quantiﬁes the relations between the various Grothendieck constants k k k Kγ , KΓ, KG, and between the respective real and complex versions.

k k k Proposition 4.4. Let Kγ , KΓ, and KG be respectively the smallest constants such that (55), (56), and (1) hold. Then we have

k k k (62) KG ≤ KΓ ≤ Kγ , for k = R or C, and C R C R C R (63) Kγ ≤ Kγ ,KΓ ≤ KΓ ,KG ≤ KG .

0 B m+n m×n Proof. Let Wm,n = B∗ 0 ∈ S : B ∈ k . Then kAkγ kAkΓ KG = sup max = sup max . m,n∈N A∈Wm,n kAkθ m,n∈N A∈Wm,n kAkΘ m+n Since Wm,n ⊆ S , we get the ﬁrst inequality in (62). The second inequality was already established as part of Theorem 4.3. We next prove the ﬁrst inequality in (63); the remaining two may be similarly proved. Let n δ = (δ1, . . . , δn) ∈ C . We will write δi = αi + iβi, αi, βi ∈ R and 2n δb := (α1, −β1, . . . , αn, −βn) ∈ R .

h α β i 2×2 n×n Recall that a = α + iβ ∈ C may be represented as π(a) = −β α ∈ R . Let A ∈ C be Hermitian. Then 2n×2n Ab := π(aij) ∈ R is symmetric, and furthermore, δ∗Aδ = δbTAbδ.b √ 2 2 2 2 As the unit disk {(α,√ β) ∈ R : α + β ≤ 1} contains {(α, β) ∈ R : |α| = |β| = 1/ 2}, by choosing |αi| = |βi| = 1/ 2, i = 1, . . . , n, we deduce that

C R (64) 2kAkθ ≥ kAbkθ . n n Let x1, . . . , xn ∈ C . We write xi = ui + ivi, ui, vi ∈ R , and set ui −vi 2n y2i−1 := , y2i := ∈ R vi ui for i = 1, . . . , n. Then

n n 2n 2n X X 1 X X a hx , x i = π(a )hy , y i . ij i j C 2 ij i j R i=1 j=1 i=1 j=1

If kx1k = ··· = kxnk = 1, then ky1k = ··· = ky2nk = 1, and we obtain 1 1 kAkC ≤ kAbkR ≤ KRkAbkR ≤ KRkAkC, γ 2 γ 2 γ θ γ θ C R where the second and third inequalities are (??) and (64) respectively. Therefore Kγ ≤ Kγ . SYMMETRIC GROTHENDIECK INEQUALITY 25

There are slight generalizations of the symmetric and original Grothendieck inequalities to Grothendieck d-norms that follow from Lemma 2.1. In the following, the symmetric Grothendieck inequalities (50) and the original Grothendieck inequality (3) may be obtained by setting d = 1 and taking p → ∞. As usual, the constants Kγ,d,p, KΓ,d,p, KG,d,p below depend on the field but we will only indicate this dependence when comparing them over different fields. We will also write Kγ,d,∞ := limp→∞ Kγ,d,p and likewise for KΓ,d,∞, KG,d,∞ in the rest of this article.

Corollary 4.5. Let d, p, m, n ∈ N. If 1 ≤ d ≤ p ≤ n, then there exist ﬁnite constants Kγ,d,p,KΓ,d,p > 0 independent of n such that

kAkγ,p ≤ Kγ,d,pkAkγ,d, kAkΓ,p ≤ KΓ,d,pkAkΓ,d n for all A ∈ S . If 1 ≤ d ≤ p ≤ m + n, then there exists a ﬁnite constant KG,d,p > 0 independent of m, n, such that

(65) kBkG,p ≤ KG,d,pkBkG,d m×n for all B ∈ k . If these constants are chosen to be smallest possible, i.e., kAkγ,p kAkΓ,p kBkG,p Kγ,d,p := sup max ,KΓ,d,p := sup max ,KG,d,p := sup max , n n m×n n∈N A∈S kAkγ,d n∈N A∈S kAkΓ,d m,n∈N B∈k kBkG,d then for any 1 ≤ d ≤ p ≤ q ≤ ∞, we have R C Kγ,d,p ≤ Kγ,1,n ≤ Kγ,1,∞ = Kγ,Kγ,2d,2p ≤ Kγ,d,p,Kγ,d,q ≤ Kγ,d,pKγ,p,q; R C KΓ,d,p ≤ KΓ,1,n ≤ KΓ,1,∞ = KΓ,KΓ,2d,2p ≤ KΓ,d,p,KΓ,d,q ≤ KΓ,d,pKΓ,p,q; R C KG,d,p ≤ KG,1,m+n ≤ KG,1,∞ = KG,KG,2d,2p ≤ KG,d,p,KG,d,q ≤ KG,d,pKG,p,q. Furthermore, we have

KG,d,p ≤ KΓ,d,p ≤ Kγ,d,p and KΓ,d,p ≤ 2KG,d,p.

R Proof. All statements are obvious from deﬁnitions and earlier results. For example, Kγ,2d,2p ≤ C C C C R C R Kγ,d,p since kAkγ,p ≤ Kγ,d,pkAkγ,d and thus kAkγ,2p ≤ Kγ,d,pkAkγ,2d by (32). Such variants of the original Grothendieck inequality, i.e., for the (G, d)-norms, were ﬁrst studied by Krivine [33] and have appeared in many places [1, 10, 11, 12, 19, 26]. In [1, 19, 26, 33], KG,1,p is denoted KG(p) and called the order-p Grothendieck constant; in [12], the inequality (65) is called the generalized Grothendieck inequality and KG,d,p denoted KG(p 7→ d). We will have more to say about these inequalities in Section 5.4. In particular, Krivine [33, p. 17] showed that √ R (66) KG,1,2 = 2. m×n An interesting aside that follows from (38) and (66) is that for B ∈ R , we have √ C R R R R kBk∞,1 = kBkG,2 ≤ KG,1,2kBkG,1 = 2kBk∞,1, C R a sharp version of the bound kBk∞,1 ≤ 2kBk∞,1 in [27, Proposition 2.1].

5. Symmetric Grothendieck inequality for cones We will see that results such as the Nesterov π/2-Theorem [39, 43] and Goemans–Williamson inequality [23] are in fact just “symmetric Grothendieck inequality for positive semideﬁnite matrices” and “symmetric Grothendieck inequality for graph Laplacians.” To be more speciﬁc, let n C = {Cn ⊆ S : n ∈ N} be a family of convex cones. By a symmetric Grothendieck inequality for C, we mean an inequality of the form

(67) kAkγ ≤ KCkAkθ 26 SHMUEL FRIEDLAND AND LEK-HENG LIM where kAk (68) K := sup max γ . C n n∈N A∈Cn⊆S kAkθ Obviously, KC ≤ Kγ. In particular, KC is finite and does not depend on the dimension n; we will call it the conic Grothendieck constant for C. 0 B m+n m×n We have already seen that if we set Wm,n = B∗ 0 ∈ S : B ∈ k , a subspace and therefore trivially a convex cone, then we obtain the original Grothendieck constant KG. Another n immediate example is Cn = S (R+) — by Corollary 2.15, we see that KC = 1. We will discuss some nontrivial examples below. n In Section 5.1, we will see that for Cn = S+, the conic Grothendieck constants may be determined R C exactly: KC = π/2 and KC = 4/π. Furthermore, the inequalities over R and C are the well-known Nesterov π/2-Theorem and Ben-Tal–Nemirovski–Roos 4/π-Theorem respectively. n In Section 5.2, we will see that for Cn = L , the cone of weighted graph Laplacians on n- vertex graphs, we obtain the even better known inequality of Goemans–Williamson. The conic R R R Grothendieck constant in this case is KGW = 1/αGW where αGW ≈ 0.878567 is the Goemans– Williamson constant. We will prove a complex analogue. An immediate advantage of obtaining these inequalities as symmetric Grothendieck inequalities is that we obtain them over both R and C simultaneously. 5.1. Cones of positive semidefinite matrices. We will show that the Nesterov π/2-Theorem [39, Theorem 3.3], which actually appeared earlier in [43, Theorem 4], as well as its complex analogue, the Ben-Tal–Nemirovski–Roos 4/π-Theorem [6, Equation 57], will follow easily from the discussions in the last two sections. The sharpness of the Nesterov π/2-Theorem (69), i.e., that n π/2 is not just an upper bound but the exact value of the conic Grothendieck constant for S+(R), was established by Alon and Naor [4]. Our approach will yield 4/π as the exact value of the conic n Grothendieck constant for S+(C), showing that the Ben-Tal–Nemirovski–Roos 4/π-Theorem (70) is also sharp. While the key ideas for our sharpness proof are due to Alon and Naor [4], we take the opportunity to slightly refine their proof, using exact expressions established in Lemma A.1 in place of asymptotic estimates and providing a more careful argument in Lemma A.2, both to be found in the appendix. n Theorem 5.1. Let A ∈ S+(R). Then π (69) kAkR ≤ kAkR. γ 2 θ n Let A ∈ S+(C). Then 4 (70) kAkC ≤ kAkC. γ π θ These inequalities are sharp. In both inequalities, the seminorm k · kγ may be replaced by the norms k · kΓ or k · kG and the seminorm k · kθ may be replaced by the norms k · kΘ or k · k∞,1. Proof. Our proof works for k = R and C alike. We will indicate field dependence in our notations only when it is necessary to do so. n In terms of Schur product, for any A ∈ S=(D), the map Φ in (51) may be expressed as ∞ X ◦k Φ(A) = akA ◦ (A ◦ A) k=0 where, as in (42) and (43), 2 (2k − 1)!! π (2k − 1)!!2 1 aR = , aC = k π (2k)!!(2k + 1) k 4 (2k)!! k + 1 SYMMETRIC GROTHENDIECK INEQUALITY 27

n n for k = 0, 1, 2,.... Let G ∈ G . Then G ∈ S=(D) and

Φ(G) a0G,

◦k as ak > 0 and G ◦ (G ◦ G) 0 for all k ∈ N (Schur products of positive semidefinite matrices n n remain positive semidefinite). Hence for any A ∈ S+ and G ∈ G , kAkθ ≥ tr AΦ(G) ≥ a0 tr(AG), where the first inequality is by (54). The characterization of k · kγ in Corollary 2.9 yields

n kAkθ ≥ a0 max{tr AG : G ∈ G } = a0kAkγ.

R C It remains to observe that a0 = 2/π and a0 = π/4. That the θ- and γ-seminorms may be replaced by other Grothendieck norms is simply a consequence of (31). We now show that (69) and (70) are sharp. Let 0 < ε < 1/5n be arbitrary. Let m ∈ N and the n unit vectors x1, . . . , xm ∈ k be as in Lemma A.2. Set 1 G := G(x , . . . , x ) ∈ m,A := G ∈ m ∩ m. 1 m G m2 S= S+ Setting α = 2 in (97) of Lemma A.1 and in the lower bound in (98) of Lemma A.2, we obtain m m m 1 1 X 2 1 X X 2 − 5ε ≤ min |hxi, vi| ≤ |hxi, xji| = tr(AG), n kvk=1 m m2 i=1 j=1 i=1 and thus kAkγ ≥ 1/n − 5ε by (26). We next show that  1 Γ(n/2)2  + 3ε k = R, m 2 π 2 1 X  Γ (n + 1)/2 (71) kAkθ ≤ max |hxi, vi| ≤ kvk=1 m 2 i=1 π Γ(n)  + 3ε = .  4 Γ(n + 1/2)2 k C

The second inequality above comes from setting α =√ 1 in (98) of Lemma A.2 and replacing n by n − 1 in (97) of Lemma A.1, noting that Γ(3/2) = π/2. We will next show the ﬁrst inequality. m For t = (t1, . . . , tm) ∈ T , m m 2 ∗ 1 X X 1 Xm t At = hxi, xjit¯itj = t¯ixi . m2 m i=1 i=1 j=1

∗ m By deﬁnition, kAkθ is the maximum of t At over all t ∈ T and let this be attained at s = m Pm n m (s1, . . . , sm) ∈ T . Set x := i=1 s¯ixi ∈ C and v := x/kxk. Then for t ∈ T , m m DXm E DXm E X X t¯ixi, x = kxk t¯ixi, v ≤ kxk |t¯i||hxi, vi| = kxk |hxi, vi|, i=1 i=1 i=1 i=1 with equality if and only if ti = sgnhxi, vi, i = 1, . . . , m. On the other hand, Cauchy–Schwartz and the maximality of s yields

DXm E Xm Xm 2 t¯ixi, x ≤ t¯ixi s¯ixi ≤ kxk . i=1 i=1 i=1 28 SHMUEL FRIEDLAND AND LEK-HENG LIM

It follows that we must haves ¯i = sgnhxi, vi, i = 1, . . . , m. Therefore 1 2 D 1 Xm 1 E kAkθ = x = sgnhxi, vixi, kxkv m m i=1 m 1 1 Xm = x sgn(hxi, vi)hxi, vi m m i=1 1 Xm 2 1 Xm 2 = sgn(hxi, vi)hxi, vi = |hxi, vi| m i=1 m i=1 and we have the ﬁrst inequality in (71). Hence we get  1 1 Γ(n/2)2 −1 − 5ε + 3ε = ,  2 k R kAk  n π Γ (n + 1)/2 γ ≥ kAk θ  1 π Γ(n)2 −1  − 5ε + 3ε = , n 4 Γ(n + 1/2)2 k C and as ε > 0 is arbitrary, π Γ(n/2 + 1/2)2 π   p → k = R, kAk  2 Γ(n/2) n/2 2 γ ≥ kAk θ  4 Γ(n + 1/2)2 4  √ → = , π Γ(n) n π k C α as n → ∞, using the fact that limt→∞ Γ(t + α)/ Γ(t)t = 1 for any α ∈ C. Combined with (39) of Corollary 2.15 and (37) of Corollary 2.14, the inequality (70) shows that n for any A ∈ S+(R), we have 4 4 kAkR = kAkC ≤ kAkC = kAkR , γ γ π θ π γ,2 giving the n = 2 case in [11, Theorem 1]. Note however that (70) cannot be deduced from [11, Theorem 1]. We will have more to say about these inequalities in Proposition 5.6 where we extend Theorem 5.1 to an arbitrary pair of Grothendieck d- and p-norms.

5.2. Cones of weighted Laplacians. We begin with a precaution: All matrices considered in this section will be real but the norms can be taken over either R or C. See Section 2.3 for a discussion of real and complex Grothendieck norms. n n For any A ∈ S◦ = S◦ (R), the space of n × n real symmetric matrices with zero diagonal, we let n n LA ∈ S = S (R) be deﬁned by

(72) LA := diag(A1) − A, n n n where diag(x) ∈ S denotes the diagonal matrix whose diagonal is x ∈ R and 1 = (1,..., 1) ∈ R is the vector of all ones. n If in addition A ∈ S (R+), i.e., aij ≥ 0 for all i, j = 1, . . . , n, then LA is called a weighted Laplacian. Note that this implies LA has all off-diagonal entries nonpositive. It also follows from definition that LA1 = 0. In fact these last two conditions are enough to characterize the set of all weighted Laplacians: n n n n 1 L := {LA ∈ S : A ∈ S◦ (R+)} = {L ∈ S : L = 0, ìj ≤ 0 for all i 6= j}. n n n Clearly, L ⊆ S+ and L is a convex cone. n We will now establish the symmetric Grothendieck inequality for the cone L . The inequality in the case k = R was first discovered by Goemans and Williamson [23] and its sharpness established SYMMETRIC GROTHENDIECK INEQUALITY 29 by Feige and Schechtman [20, 21]. The corresponding inequality for k = C is new as far as we know. The constants in Theorem 5.2 have approximate values R C αGW ≈ 0.87856, αGW ≈ 0.93494, R C and so KGW ≈ 1.1382, KGW ≤ 1.0696.

Theorem 5.2. Let k = R or C and ϕk be as deﬁned in (42) or (43) respectively. Let 1 + ϕ (x) k k (73) αGW := inf . 0≤x≤1 1 + x n Then for any L ∈ L , k k k (74) kLkγ ≤ KGWkLkθ where the smallest possible constants 1 1 (75) KR = ,KC ≤ . GW R GW C αGW αGW

In (74), the seminorm k · kγ may be replaced by the norms k · kΓ or k · kG and the seminorm k · kθ may be replaced by the norms k · kΘ or k · k∞,1. Proof. We will show that for both k = R and C, 1 − Re ϕ (z) k k αGW = inf . |z|<1 1 − Re z The reader is reminded that excluding some or all boundary points makes no diﬀerence when taking inﬁmum over a region. Case I: k = R. For x ∈ [−1, 1], ϕR(x) = (2/π) arcsin x. If x ∈ [0, 1], then f(x) := 1 − ϕR(x) is a concave function with f(0) = 1 and f(1) = 0. Therefore the graph of f lies above the line 1 − x for all x ∈ [0, 1], i.e., f(x) ≥ 1 − x. Hence 1 − ϕ (x) inf R = 1. 0≤x<1 1 − x If x ∈ [−1, 0], let y = −x ∈ [0, 1], then 1 − ϕ (x) 1 + ϕ (y) R = R . 1 − x 1 + y Thus 1 + ϕ (y) 1 − ϕ (x) 1 − ϕ (x) inf R = inf R = inf R , 0≤y≤1 1 + y −1≤x≤0 1 − x −1

kLAkγ = kLAkΓ = kLAkG, kLAkθ = kLAkΘ = kLAk∞,1 n k k as LA ∈ S+. This establishes that KGW ≤ 1/αGW in (75); that equality holds in the case when k = R is a well-known result of Feige and Schechtman [20, 21]. SYMMETRIC GROTHENDIECK INEQUALITY 31

R To obtain the more familiar expression for αGW in [23], note that 1 + ϕ (x) 2 θ 2 θ R R αGW = inf = inf = inf . 0≤x≤1 1 + x π/2≤θ≤π π 1 − cos θ 0<θ≤π π 1 − cos θ Naturally, we suspect that equality also holds in (75) when k = C, i.e., 1 KC =? . GW C αGW While most of the construction in [20, 21] carries over to C, the ultimate diﬃculty is in obtaining C R the value of kLkθ for the weighted Laplacian L of the constructed graph — unlike kLkθ , which is C essentially maxcut, kLkθ has no combinatorial interpretation. n n Similar to our discussions at the end of Section 5.1, since L ⊆ S+(R), when combined with (39) of Corollary 2.15 and (37) of Corollary 2.14, the complex Goemans–Williamson inequality (74) n shows that for any L ∈ L , R C C C C R kLkγ = kLkγ ≤ KGWkLkθ = KGWkLkγ,2. We will have more to say about these inequalities and constants in Proposition 5.7 where Theo- rem 5.2 is extended to an arbitrary pair of Grothendieck d- and p-norms.

5.3. Cones of diagonally dominant matrices. As in the last section, all matrices considered in n n this section will be real but the norms can be taken over either R or C. Again, we write S = S (R) throughout the rest of this section. n n P Let Sdd := {A ∈ S : aii ≥ j6=i|aij|} be the cone of symmetric diagonally dominant matrices. Clearly, n n n L ⊆ Sdd ⊆ S+. n n The relation between the first two cones may be more precisely characterized. Let Sdd(R+) ⊆ Sdd be the subcone of diagonally dominant matrices with nonnegative entries. n n n Lemma 5.3. Every A ∈ Sdd has a unique decomposition A = H + L with H ∈ Sdd(R+), L ∈ L , n n n and hijìj = 0 whenever i 6= j, i.e., Sdd = Sdd(R+) + L . n Proof. Let B ∈ S◦ (R+), i.e., nonnegative symmetric with zero diagonal, be defined by ( −aij if aij < 0 and i 6= j, bij := 0 if aij ≥ 0 or i = j. n Then LB = diag(B1) − B ∈ L by (72). Let H := A − LB. Then hij ≥ 0 and hijìj = 0 for i 6= j. n P n Therefore H ∈ S (R+). Since aii ≥ j6=i hij + ìj, we also have H ∈ Sdd. Uniqueness follows since B is uniquely determined by A and thus so are LB and H. We now prove a Grothendieck inequality for diagonally dominant matrices that may be regarded as an inequality ‘in between’ the Nesterov π/2-Theorem and the Goemans–Willamson inequality over R, and ‘in between’ the Ben-Tal–Nemirovski–Roos 4/π-Theorem and the complex Goemans– Willamson inequality over C. k R C Proposition 5.4. Let k = R or C. Let αGW be as in (73) and a0 = 2/π, a0 = π/4. Then for any n A ∈ Sdd, 1 − ak (77) kAkk ≤ 1 + 0 kAkk. γ k θ αGW

In (77), the seminorm k · kγ may be replaced by the norms k · kΓ or k · kG and the seminorm k · kθ may be replaced by the norms k · kΘ or k · k∞,1. 32 SHMUEL FRIEDLAND AND LEK-HENG LIM

Proof. The same proof will work for both k = R and C and we drop k in our notations below. n We may assume that A 6= 0. Let G ∈ G be such that kAkγ = tr(AG), i.e., attaining the maximum in Corollary 2.9. Let A = H + L be the unique decomposition given by Lemma 5.3. Set t = tr(LG)/ tr(AG) ∈ [0, 1]. We claim that 1 (78) kAkγ ≤ kAkθ. max{a0 + (αGW − a0)t, 1 − t} Let ϕ be as in (42) or (43). Then kAkθ ≥ tr Aϕ(G) = tr Hϕ(G) + tr Lϕ(G) ≥ a0 tr(PG) + αGW tr(LG)

= a0 tr(AG) + (αGW − a0) tr(LG) = [a0 + (αGW − a0)t] tr(AG) = [a0 + (αGW − a0)t]kAkγ.

n T n T Since H ∈ S (R+), by Lemma 2.10(iii), we have kHkγ = 1 H1. Since L ∈ L , we have 1 L1 = 0. T T Thus kHkγ = 1 H1 = 1 A1 ≤ kAkθ and so

(1 − t)kAkγ = (1 − t) tr(AG) = tr(PG) ≤ kHkγ ≤ kAkθ, giving us (78). The required inequality (77) follows from (78) by minimizing max{a0 + (αGW − a0)t, 1 − t} over t ∈ [0, 1], observing that the minimum αGW/(1 − a0 + αGW) is achieved when a0 + (αGW − a0)t = 1 − t. 5.4. Mother of all Grothendieck inequalities. It is straightforward to combine Corollary 4.5 and (67) to obtain an (almost) all-encompassing inequality. Let d, p, n ∈ N with 1 ≤ d ≤ p ≤ n and n C = {Cn ⊆ S : n ∈ N} be a family of cones. Then

(79) kAkγ,p ≤ KC,d,pkAkγ,d where kAkγ,p (80) K := sup max ≤ K ≤ Kγ. C,d,p n C n∈N A∈Cn⊆S kAkγ,d Again the result holds over both k = R and C. We have the following special cases: (i) Grothendieck inequality [24, 35]:

0 B n m×(n−m) Cn = B∗ 0 ∈ S : B ∈ k , d = 1, p = n → ∞; (ii) order-p Grothendieck inequality [1, 19, 26, 33]:

0 B n m×(n−m) Cn = B∗ 0 ∈ S : B ∈ k , d = 1, p ≤ m; (iii) generalized Grothendieck inequality [10]:

0 B n m×(n−m) Cn = B∗ 0 ∈ S : B ∈ k , d < p ≤ m; (iv) symmetric Grothendieck inequality for γ-seminorm (55): n Cn = S , d = 1, p = n → ∞; (v) Nesterov π/2-Theorem [39, 43] and Ben-Tal–Nemirovski–Roos 4/π-Theorem [6]: n Cn = S+, d = 1, p = n → ∞; (vi) Goemans–Williamson inequality [23]: n Cn = L , d = 1, p = n → ∞; (vii) rank-constrained positive semideﬁnite Grothendieck inequality [11, 12]: n Cn = S+, d = 1, p ≤ n. SYMMETRIC GROTHENDIECK INEQUALITY 33

Missing from this list is the symmetric Grothendieck inequality for Γ-norm (56), which cannot be n obtained by restricting (79) to any subspace or cone of S . We may of course also define a Γ-norm n n version of (79) and (80) but since k · kΓ = k · kγ on both S◦ and S+, they give the same inequalities (i)–(vii) except for (iv), which becomes (56) instead. We have the following extension of Corollary 4.5 to any family of cones. n Lemma 5.5. Let d, p ∈ N with 1 ≤ d ≤ p ≤ ∞ and C = {Cn ⊆ S : n ∈ N} be a family of cones. Then C R KC,d,p ≤ KC,1,n ≤ KC,KC,1,∞ = KC,KC,d,p ≥ KC,2d,2p. Proof. The first inequality and the second limit follow from their definitions in (68) and (80). The C R n last inequality follows from (32), i.e., kAkγ,d = kAkγ,2d for any A ∈ S (R), and the obvious inclusion n n S (R) ⊆ S (C).

One may also obtain bounds for the constants KC,d,p for specific families of cones C such as the positive semidefinite cones and the cones of weighted Laplacians. n For the family of positive semidefinite cones C = {S+ : n ∈ N}, the following lower bound for R KC,d,p was established by Briët,Buhrman, and Toner in [10, Theorem 1] and its limiting expression R for KC,d,∞ by Briët,de Oliveira Filho, and Vallentin in [11, Theorem 2]. We will state the analogues C for C. Our lower bound for KC,d,p is a straightforward consequence of Lemma 5.5 and the lower R bound for KC,d,p in [10, Theorem 1]; but we will provide an alternative simple proof for the limiting k expression KC,d,∞ that applies to both k = R and C alike and that is nearly identical to our proof of Nesterov π/2-Theorem and Ben-Tal–Nemirovski–Roos 4/π-Theorem in Theorem 5.1. n Proposition 5.6. Let d, p ∈ N with 1 ≤ d ≤ p ≤ ∞ and C = {S+ : n ∈ N}. Then  dΓ(p + 1)/2Γ(d/2)2  = ,  k R p Γ(p/2)Γ (d + 1)/2 k (81) KC,d,p ≥ 2 dΓ (2p + 1)/2 Γ(d)  k = C. p Γ(p)Γ(2d + 1)/2 Furthermore, for any fixed d ∈ N,  dΓ(d/2)2  = ,  2 k R 2Γ (d + 1)/2 k (82) KC,d,∞ = 2  dΓ(d)  k = C. Γ(2d + 1)/22

Proof. The lower bound in (81) for R was established in [10, Theorem 1]. The lower bound for C C R thus follows from Lemma 5.5 since KC,d,p ≥ KC,2d,2p. The limiting expression for R in (82) was established in [11, Theorem 2] but we will provide an alternative simple proof that works for both R and C. Taking limits as p → ∞ in (81), we see that dΓ(d/2)2 dΓ(d)2 R C KC,d,∞ ≥ ,KC,d,∞ ≥ ; 2Γ(d + 1)/22 Γ(2d + 1)/22 and it remains to establish that ‘≥’ above may be replaced by ‘≤’. This will in turn follow if we n can show that for any A ∈ S+(R), dΓ(d/2)2 (83) kAkR ≤ kAkR ; γ 2Γ((d + 1)/2)2 γ,d 34 SHMUEL FRIEDLAND AND LEK-HENG LIM

n and for any A ∈ S+(C), dΓ(d)2 (84) kAkC ≤ kAkC . γ Γ((2d + 1)/2)2 γ,d n n Let k = R or C. As in the proof of Theorem 5.1, for any A ∈ S+(k) and G ∈ G (k), we have k k k Φd(G) := ϕd(gij) b1,dG, k k where b1,d is the first Taylor coefficient of ϕd as in Lemma 3.1. Using the characterization of k · kγ n k in Corollary 2.9, let G ∈ G (k) be such that tr(AG) = kAkγ. Then k k k k k kAkγ,d ≥ tr AΦd(G) ≥ b1,d tr(AG) = b1,dkAkγ. By Lemma 3.1(iii), 2 Γ((d + 1)/2)2Γ(3/2) 2 Γ((d + 1)/2)2 bR = √ = ; 1,d π Γ(d/2)Γ((d + 2)/2) d Γ(d/2)2 and so by Lemma 3.1(ii), 1 Γ((d + 1/2)2) bC = bR = . 1,d 1,2d d Γ(d)2 These give (83) and (84) respectively. R C For d = 1, we get KC,1,∞ = π/2 and KC,1,∞ = 4/π, so (83) generalizes the Nesterov π/2-Theorem and (84) generalizes the Ben-Tal–Nemirovski–Roos 4/π-Theorem. R n×n In [10, Theorem 1], the lower bounds for KC,d,p in (81) are stated for matrices in R as opposed n n×n n to S+(R) but note that their proof actually assumes the latter. Since k ⊇ S+, any lower bound for the latter is automatically one for the former. n The following discussion for the family of cones of weighted Laplacians C = {L : n ∈ N} is R essentially that of Briët, de Oliveira Filho, and Vallentin in [11, Section 6], where the constant αd is denoted v(d). Our two minor contributions here are to extend it to C and to relate the constants k over R and C. For d ∈ N, we will let KGW,d > 0 be sharpest constant so that k k k kLkγ ≤ KGW,dkLkγ,d n k k for all L ∈ L and all n ∈ N. Clearly, KGW,1 = KGW, where the latter is as defined in Theorem 5.2.

n k Proposition 5.7. Let C = {L : n ∈ N}. For each d ∈ N, let ϕd be as in Lemma 3.1, and 1 + ϕk(x) k d (85) αd := inf . 0≤x≤1 1 + x Then 1 Kk = Kk ≤ and KC = KR . C,d,∞ GW,d k GW,d GW,2d αd n n C R C R Proof. Since L ⊆ S (R), by Proposition 2.12, we have kLkγ,d = kLkγ,2d and thus KGW,d = KGW,2d. k By definition, we have KC,d,∞ = KGW,d, and it remains to establish 1/αd as an upper bound. By C R R R Lemma 3.1(ii), ϕd (x) = ϕ2d(x) for x ∈ [−1, 1], it suffices to show that KGW,d ≤ 1/αd and the corresponding result for C will follow. To avoid clutter, we drop the superscript R in the rest of this proof. The same proof that we gave for the d = 1 case in Theorem 5.2 applies here with minor modifi- cations and we will just outline the main steps. Note that f := 1 − ϕd is concave with f(0) = 1 and n n f(1) = 0 as in the d = 1 case. Let A ∈ S◦ (R+) and L = LA ∈ L . As in the proof of Theorem 5.2, n n X X kLkγ = aij(1 − hxi, xji) i=1 j=1 SYMMETRIC GROTHENDIECK INEQUALITY 35

n for some unit vectors x1, . . . , xn ∈ R ; and Z n n X X kLkγ,d ≥ `ij sgn(Zxi), sgn(Zxj) Gd,n(Z) dZ d×n R i=1 j=1 n n n n X X X X = aij 1 − ϕd(hxi, xji) ≥ αd aij(1 − hxi, xji) = αdkLAkγ. i=1 j=1 i=1 j=1 It also follows from Lemma 3.1(ii) and (85) that

C R αd = α2d. C C R In particular the complex Grothendieck constant in (73), αGW = α1 = α2 . In principle, one may R R use the explicit expression for ϕd (x) in Lemma 3.1(iii) to obtain the numerical value of αd . This was in fact done for d = 2 and 3 in [11, Section 6] using a diﬀerent expression.

6. Applications We will discuss some consequences of Theorem 4.3 to computer science and optimization. We will deduce some polynomial-time approximation bounds that we think are new.

6.1. Maxcut. Let G = (V,E) be an undirected graph on the vertex set V = {1, . . . , n}.A weighted n adjacency matrix of G is a matrix A ∈ S◦ where aij = 0 whenever {i, j} ∈/ E. In which case, LA as deﬁned in (72) is the corresponding weighted graph Laplacian. A cut of G is a partition of the vertex set V = S ∪ Sc into disjoint nonempty sets. Given a n weighted adjacency matrix A ∈ S◦ , the weight of the cut is

c X 1 T cut(S, S ) := aij = e LAeS i∈S, j∈Sc 4 S n where eS = (e1, . . . , en) ∈ {−1, 1} is deﬁned by ei = 1 if and only if i ∈ S. It follows from (76) that 1 c R (86) max|cut(S, S )| = kLAkθ . S(V 4 n n If the weights are nonnegative, i.e., A ∈ S◦ (R+), then LA ∈ S+ and we may drop the absolute value in the left-side of (86), which is called the maxcut of G weighted by A and is known to be NP-hard. The Goemans–Williamson inequality, i.e., combining (74) and (75) for k = R, R R R (87) αGWkLAkγ ≤ kLAkθ ,

R R yields a polynomial-time approximation to within a factor of αGW ≈ 0.87856 as both kLAkγ and the n unit vectors x1, . . . , xn ∈ R that attain its value as in (5) may be computed to arbitrary accuracy in polynomial-time. Incidentally, by Corollary 2.13, we have

C R C αGWkLAkγ ≤ kLAkθ ,

C C and so kLAkθ can be approximated to within a factor of αGW ≈ 0.93494 in polynomial-time. The C catch is that, unlike its real counterpart, kLAkθ does not give us the maxcut. n Goemans and Williamson [23] also showed that for a randomly chosen unit vector z ∈ R and c R R R Sz := {i ∈ V : hz, xii ≥ 0}, the expected value of cut(Sz,Sz) will be at least αGWkLAkγ ≤ kLAkθ . Consider a bipartite graph G = (V,E), i.e., V = V1 ∪ V2 and E ⊆ V1 × V2. Let V1 = {1, . . . , m} and V2 = {m + 1, . . . , m + n}. Then its weighted adjacency matrix takes the form 0 B A = ∈ m+n BT 0 S◦ 36 SHMUEL FRIEDLAND AND LEK-HENG LIM

m×n for some B ∈ R . Note that B is not required to be a nonnegative matrix. The cut norm of B is defined [4] as X kBkcut := max bij . I1⊆V1,I2⊆V2 i∈I1, j∈I2 m+n m×n Proposition 6.1. Let A ∈ S◦ and B ∈ R be as above. Then 1 3 (88) kL k ≤ kBk ≤ kL k . 8 A θ cut 8 A θ c Proof. For the partition S = I1 ∪ (V2 \ I2) and S = (V1 \ I1) ∪ I2, c X X cut(S, S ) = bij + bij. i∈I1, j∈I2 i∈V1\I1,j∈V2\I2 The equality (86) yields X X kLAkθ = 4 max bij + bij I1⊆V1,I2⊆V2 i∈I1, j∈I2 i∈V1\I1, j∈V2\I2 X X ≤ 4 max bij + bij ≤ 8kBkcut, I1⊆V1,I2⊆V2 i∈I1, j∈I2 i∈V1\I1,j∈V2\I2 i.e., the left inequality in (88). Setting I1 to be arbitrary and I2 = V1 or setting I1 = V2 and I2 to be arbitrary gives X X 1 max bij , bij ≤ kLAkθ. i∈I1, j∈V1 i∈V2, j∈I2 4 Hence X X X X 1 bij − bij = bij − bij ≤ kLAkθ. i∈I1, j∈V1\I2 i∈V2\I1, j∈I2 i∈I1, j∈V1 i∈V2, j∈I2 2 Setting I1 to be arbitrary and I2 = V2 \ I2, we get X 3 bij ≤ kLAkθ, i∈I1, j∈I2 8 which yields the right inequality in (88). Now note that the Goemans–Williamson inequality (87) does not apply to the cut norm as m×n m+n m+n B ∈ R is not required to be nonnegative; so A ∈ S◦ may have negative entries and LA ∈ S is no longer guaranteed to be positive semidefinite. This is where the symmetric Grothendieck inequality (55) can be useful, since it does not require positive semidefiniteness. When applied to Proposition 6.1, we obtain the following polynomial-time approximation for the cut norm. m+n m×n Corollary 6.2. Let A ∈ S◦ and B ∈ R be as above. Then 1 3 kLAkγ ≤ kBkcut ≤ kLAkγ. R 8Kγ 8 6.2. Nonconvex quadratic programming. Another consequence of the symmetric Grothendieck inequality is a new convex relaxation bound that we will describe below after providing some con- n text. Let A ∈ S (R) and consider the following ubiquitous quadratic programs in combinatorial optimization [2,3,4, 13, 23, 29, 30],

T T (89) q≤(A) := max x Ax, q=(A) := max x Ax, x∈[−1,1]n x∈{−1,1}n i.e., maximization of a quadratic function over the unit cube and the vertices of the unit cube respectively. Both problems are known to be NP-hard [18, 39] and it is customary to consider the following standard convex relaxations:

r≤(A) := max{tr(AX): xii ≤ 1, i = 1, . . . , n, X 0}, (90) r=(A) := max{tr(AX): xii = 1, i = 1, . . . , n, X 0}. SYMMETRIC GROTHENDIECK INEQUALITY 37

The problems in (90) are semideﬁnite programs whose solutions can be found by interior-point algorithms to any arbitrary precision ε > 0 in polynomial time [40]. Note that the identity matrix is within the sets of feasible matrices in the maximization problems (90). n n If we let A0 ∈ S◦ (R) be the matrix obtained from A ∈ S (R) by replacing its diagonal entries with zeros, then

q=(A) = q=(A0) + tr A = q≤(A0) + tr A, r=(A) = r=(A0) + tr A = r≤(A0) + tr A.

Nemirovski, Roos, and Terlaky [38] proved a fundamental inequality that relates r≤(A) and q≤(A), namely,

(91) r≤(A) ≤ 2 log 2n · q≤(A). When combined with (95) that we will establish later, we get

(92) r=(A) ≤ 2 log 2n · q=(A) if tr A ≥ 0. The nonnegative trace condition cannot be omitted, e.g., the inequality does not hold for A = −I. Megretski [36] improved the factor 2 log 2n in (91) and (92) to 2 log n for n > 60. Without n relying on [38], Charikar and Wirth [13] obtained an analog of (92) for A ∈ S◦ (R) and coined n the term “Grothendieck-type inequality” for (92). Given A ∈ S◦ (R), Alon et al. [3] defined a simple undirected graph G by the zero pattern of A, and thereby proved a variant of (92) where the factor 2 log 2n is replaced by Ωlog ω(G), with ω(G) the clique number of G. The result [45, R 2 0 Theorem 3.5.6] appears to show a much better r=(A) ≤ 2KG · q=(A), but this is false as A = 0 −2 is a counterexample. We note that [45, Theorem 3.5.6] has since been corrected in an online version2 n by requiring A ∈ S+. Nevertheless, with positive semidefiniteness, r=(A) = kAkγ, q=(A) = kAkθ, R and Nesterov π/2-Theorem (69) gives a better bound π/2 ≈ 1.57 than 2KG ≈ 3.56. All three estimates of the gap between r=(A) and q=(A) share the common feature that the upper bounds of their ratio either grow unbounded with the dimension of A (e.g., 2 log 2n or 2 log n) or at least depends on A (e.g., log ω(G)). In fact, these gaps are the smallest possible: Megretski [13, p. 2] showed that the best possible factor in (92) is Ω(log n) while Alon et al. [3] showed that it is Ωlog ω(G). With these in view, we find the following theorem somewhat surprising — unlike the bounds of Alon et al., Charikar and Wirth, Megretski, the bounds below are in terms of universal constants independent of n and A. n Theorem 6.3. For all n ∈ N, any A ∈ S (k), and both k = R and C, k max {q=(A), q=(−A)} ≤ max {r=(A), r=(−A)} ≤ Kγ max {q=(A), q=(−A)}, (93) k max {q≤(A), q≤(−A)} ≤ max {r≤(A), r≤(−A)} ≤ KΓ max {q≤(A), q≤(−A)}. k k These inequalities are sharp, i.e., Kγ and KΓ are the smallest possible constants for (93). While the objective function in (90) is always real-valued and thus (90) is well-defined over both R and C, we will have to rewrite (89) as ∗ ∗ q≤(A) := max x Ax, q=(A) := max x Ax, x∈Dn x∈Tn so that they apply to both R and C. Theorem 6.3 and all discussions below will hold for both R and C. By the next proposition, the inequalities in (93) are just restatements of the symmetric Grothendieck inequalities (55) and (56). n Proposition 6.4. Let A ∈ S . Then q=(A) ≤ q≤(A), r=(A) ≤ r≤(A), and kAkΘ = max {q≤(A), q≤(−A)}, kAkΓ = max {r≤(A), r≤(−A)}, (94) kAkθ = max {q=(A), q=(−A)}, kAkγ = max {r=(A), r=(−A)}.

If in addition the diagonal entries a11, . . . , ann ≥ 0, then

(95) q=(A) = q≤(A), r=(A) = r≤(A). 38 SHMUEL FRIEDLAND AND LEK-HENG LIM

Proof. The equalities in (94) follow from (26) and L

q (A) = max Ln tr(AX), r (A) = max n tr(AX), ≤ X∈ 1 ≤ X∈ q (A) = max n tr(AX), r (A) = max n tr(AX). = X∈G1 = X∈G

Let aii ≥ 0, i = 1, . . . , n. We will prove that q=(A) = q≤(A); similar arguments will apply to ∗ r=(A) = r≤(A); and the remaining two equalities in (95) then follow from (94). Let maxx∈Dn x Ax n be attained at x = (x1, . . . , xn) ∈ D , with a maximal number of coordinates xi ∈ T. We claim n that all x1, . . . , xn ∈ T, i.e., x ∈ T . Suppose to the contrary that |xi| < 1 for some i. T 2 Case I: k = R. Fixing all other coordinates except xi, the quadratic function x Ax = aiixi +··· is convex in the variable xi as aii ≥ 0. Thus its maximum is attained with xi ∈ T = {−1, 1}, a contradiction. 2 2 Case II: k = C. Fixing all other coordinates except xi and writing xi = ui +ivi with ui +vi < 1, ∗ 2 2 the quadratic function x Ax = aii(ui + vi ) + ··· is convex in (ui, vi) as aii ≥ 0. Thus its maximum iϑ is attained with xi ∈ T = {e ∈ C : ϑ ∈ [0, 2π)}, a contradiction.

n We will next deduce an analogue of Theorem 6.3 with sum in place of max. For any A ∈ S , we deﬁne the stretch of A as

∗ ∗ stretch(A) := maxx∈Tn x Ax − minx∈Tn x Ax = q=(A) + q=(−A), and the spread of A as spread(A) := maxX∈Gn tr(AX) − minX∈Gn tr(AX) = r=(A) + r=(−A).

The notion of stretch appeared in [13] but was not given a name. We next show that the stretch, despite being the diﬀerence of two NP-hard quantities, can be approximated up to a universal constant factor Kγ by the spread, which is polynomial-time computable to arbitrary precision.

n Theorem 6.5. Let A ∈ S . Then

stretch(A) ≤ spread(A) ≤ Kγ stretch(A).

Proof. The ﬁrst inequality is obvious. We will prove the second one. Let α ∈ R. Clearly,

q=(A + αI) = q=(A) + nα, q=(−A − αI) = q=(−A) − nα, spread(A + αI) = spread(A),

r=(A + αI) = r=(A) + nα, r=(−A − αI) = r=(−A) − nα, stretch(A + αI) = stretch(A). Now set α = q=(−A) − q=(A) /2n. Then q=(A + αI) = q=(−A − αI) = kA + αIkθ by (94). By (94) again and the symmetric Grothendieck inequality (55),

spread(A) = spread(A + αI) ≤ r=(A + αI) + r=(−A − αI) ≤ 2kA + αIkγ

≤ 2KγkA + αIkθ = Kγ stretch(A + αI) = Kγ stretch(A), as required.

Acknowledgment We thank Zehua Lai and Yury Makarychev for their helpful comments. The work of the ﬁrst author is partially supported by the Simons Collaboration Grant for Mathematicians. The work of the second author is partially supported by NSF IIS 1546413 and the Eckhardt Faculty Fund. SYMMETRIC GROTHENDIECK INEQUALITY 39

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Appendix A. Absolute value of inner product of random unit vectors When establishing that the sharpness of the Nesterov π/2-Theorem and its complex analogue in Theorem 5.1, we needed the ﬁrst and second moments of the absolute value of inner product of random vectors over R and C. To avoid distracting our readers, we have deferred these materials to this appendix. In the following, Γ and B will denote the Gamma and Beta functions. n n+1 We will write S (k) := {x ∈ k : kxk = 1} for the n-sphere over k = R or C. We begin our n+1 discussion with R. In terms of spherical coordinates on R given by (r, ϑ1, . . . , ϑn), the surface n area element of the real n-sphere S (R) is R n−1 n−2 dσn := sin (ϑ1) sin (ϑ2) ··· sin(ϑn−1) dϑ1 dϑ2 ··· dϑn

n+1 n R n (n+1)/2 and the volume element of R is r dr dσn . Recall that the surface area of S (R) is 2π /Γ((n+ n 1)/2. The spherical measure on S (R) is deﬁned as Γ((n + 1)/2) dωR := dσR, n 2π(n+1)/2 n n which is clearly a probability measure on S (R). The corresponding probability distribution, called n the uniform distribution on S (R), is invariant under the action of orthogonal matrices. n 2n When speaking of the above notions over C, we simply identify C with R and thus n 2n+1 C R C R S (C) = S (R), dσn = dσ2n+1, dωn = dω2n+1. iφ In the following, D will denote either the real unit disk [−1, 1] or the complex unit disk ρe ∈

C : ρ ∈ [0, 1], φ ∈ [0, 2π) as appropriate. The cases α = 2 and a less precise version of α = 1 of Lemma A.1 have appeared as [4, Equations 4.5 and 4.6] for R and [44, Lemma 3] for C. The version below holds for any α ∈ (0, ∞), not necessarily an integer.

Lemma A.1. Let Un,Vn be independent random vectors uniformly distributed on the n-sphere n n+1 n n S (k) = {x ∈ k : kxk = 1}. Let Tn : S (k) × S (k) → D be the random variable deﬁned by the SYMMETRIC GROTHENDIECK INEQUALITY 41 inner product Tn = hUn,Vni. Then the probability density function of Tn is given by  1 Γ(n + 1)/2 √ (1 − t2)(n−2)/2 = ,  π Γ(n/2) k R (96) fTn (t) = n  (1 − |t|2)n−1 = , π k C for all t ∈ D. For any α > 0,  1 Γ(n + 1)/2Γ(α + 1)/2 √ k = R,  π Γ(n + α + 1)/2 α  (97) E(|hUn,Vni| ) = Γ(α/2 + 1)Γ(n + 1)  = .  Γ(n + α/2 + 1) k C

Proof. For k = R,(96) may be found in [14]. It remains to prove (96) for k = C. We identify n+1 2n+2 2n+2 C = R and introduce the following coordinates for x = (x1, . . . , x2n+2) ∈ R :

p 2 2 p 2 2 x1 = r − ρ cos(ϑ1), x2n−1 = r − ρ sin(ϑ1) ··· sin(ϑ2n−2) cos(ϑ2n−1), p 2 2 p 2 2 x2 = r − ρ sin(ϑ1) cos(ϑ2), x2n = r − ρ sin(ϑ1) ··· sin(ϑ2n−2) sin(ϑ2n−1), p 2 2 x3 = r − ρ sin(ϑ1) sin(ϑ2) cos(ϑ3), x2n+1 = ρ cos φ, . . x2n+2 = ρ sin φ, where r = kxk, ρ ∈ [0, r], ϑ1, . . . , ϑ2n−1 ∈ [0, π], and ϑ2n−1, φ ∈ [0, 2π). This coordinate system 2n 3 2 combines the standard spherical coordinates on R and the Hopf coordinates on S ⊆ C . The Jacobian J = ∂(x1, . . . , x2n+2)/∂(r, ϑ1, . . . , ϑ2n−1, ρ, φ) has determinant 2 2 n−1 2n−2 2n−3 det J = r(r − ρ ) ρ sin (ϑ1) sin (ϑ2) ··· sin(ϑ2n−2) dϑ1 dϑ2 ··· dϑ2n−1; n 2n+1 and since r = 1 on S (C) = S (R), we get C 2 n−1 2n−2 2n−3 dσn = (1 − ρ ) ρ dρ sin (ϑ1) sin (ϑ2) ··· sin(ϑ2n−2) dϑ1 dϑ2 ··· dϑ2n−1 dφ. 2n+1 Integrating out ϑ1, . . . , ϑ2n−1 and dividing the result by the surface area of S (R) yields (n/π)(1− 2 n−1 iφ ρ ) ρ dρ dφ as the surface element of D = ρe ∈ C : ρ ∈ [0, 1], φ ∈ [0, 2π) , and thus 2 n−1 (n/π)(1 − ρ ) is the expression for fTn over C in polar coordinates. Over R,(97) follows from Z 1 α 1 Γ (n + 1)/2 α 2 (n−2)/2 E(|hUn,Vni| ) = √ |t| (1 − t ) dt π Γ(n/2) −1 1 Γ(n + 1)/2 Z 1 = √ s(α−1)/2(1 − s)(n−2)/2 ds π Γ(n/2) 0 1 Γ(n + 1)/2 α + 1 n 1 Γ(n + 1)/2Γ(α + 1)/2 = √ B , = √ ; π Γ(n/2) 2 2 π Γ(n + α + 1)/2 and over C, it follows from Z 1 Z 2π Z 1 α n α 2 n−1 α/2 n−1 E(|hUn,Vni| ) = r (1 − r ) rdrdφ = n s (1 − s) ds π 0 0 0 α Γ(α/2 + 1)Γ(n) = n B + 1, n = n . 2 Γ(n + α/2 + 1) 42 SHMUEL FRIEDLAND AND LEK-HENG LIM

√ In their proof that (69) is sharp, Alon and Naor [4] used the fact that over R, the random variable R n · Tn converges in distribution to the standard real Gaussian Zn . While we did not need this in our proof of the sharpness of (69) and (70), we would like to point out here that this asymptotic normality follows immediately from (96) for both R and C. Over R, direct calculation shows r 2 n/2−1 r √ 1 Γ(n/2 + 1/2) Z s 1 Z 2 −s /2 R P( n · Tn ≤ r) = √ p 1 − ds → √ e ds = P(Zn ≤ r) 2π Γ(n/2) n/2 −∞ n 2π −∞ √ √ as n → ∞. Over C, observe√ that r n · Tn has the same distribution as n · Tn for any |r| = 1, so C it suﬃces to show that | n · Tn| converges in distribution to |Zn |: Z √r Z 2π √ r n n 2 n−1 P(| n · Tn| ≤ r) = P |Tn| ≤ √ = (1 − ρ ) ρ dρ dθ n π 0 0 2 n r 2 = 1 − 1 − → 1 − e−r = (|ZC| ≤ r) n P n √ as n → ∞, for any r ≤ n. Our proof of sharpness in Theorem 5.1 will however rely on the following.

Lemma A.2. Let Un,Vn be independent random vectors uniformly distributed on the n-sphere n n+1 S (k) = {x ∈ k : kxk = 1}. Let α ≥ 1 and 0 < ε < 1. Then there exist m ∈ N and n x1, . . . , xm ∈ S (k) such that m α 1 X α E(|hUn,Vni| ) − (2α + 1)ε ≤ min |hxi, vi| v∈Sn(k) m (98) i=1 m 1 X α α ≤ max |hxi, vi| ≤ E(|hUn,Vni| ) + (2α + 1)ε. v∈Sn( ) m k i=1 n n k Proof. To avoid clutter, we write S = S (k) and dωn = dωn as the same proof works for k = R n n+1 and C alike. Let S be covered by closed ε-balls B1,...,Bp ⊆ k , where none of the balls are n n redundant, i.e., S ⊆ B1 ∪ · · · ∪ Bp but S * (B1 ∪ · · · ∪ Bp) \ Bk for any k. Let n [k−1 Fk := S ∩ Bk \ Bj , k = 1, . . . , p. j=1 n Then F1,...,Fp are disjoint Borel sets that partition S . We may assume that

Note that the orthogonal/unitary invariance of the spherical measure ωn implies that the conditional α expectation E(|hUn,Vni| | Un = u) does not depend on the value u and so Z α α α E(|hUn,Vni| ) = E(|hUn,Vni| | Un = u) = |hu, vi| dωn, v∈Sn giving us (98). n For m suﬃciently big, we expect m uniformly distributed points x1, . . . , xm ∈ S and a partition n S = F1 ∪ · · · ∪ Fp to approximately satisfy mk/m ≈ ωn(Fk), where mk is the number of xi’s falling on Fk. Our construction above relies on a more precise version of this observation. Department of Mathematics, Statistics and Computer Science, University of Illinois, Chicago, IL, 60607-7045. Email address: [email protected]

Computational and Applied Mathematics Initiative, Department of Statistics, University of Chicago, Chicago, IL 60637-1514. Email address: [email protected]