MATRICES OF UNITARY MOMENTS ∗ KEN DYKEMA , KATE JUSCHENKO 9 Abstract. We investigatecertainmatricescomposedofmixed,second–ordermo- 0 ∗ ments of unitaries. The unitaries are taken from C –algebras with moments taken 0 with respect to traces, or,alternatively, from matrix algebras with the usual trace. 2 These sets are of interest in light of a theorem of E. Kirchberg about Connes’ n embedding problem. a J 5 1 1. Introduction ] A One fundamental question about operator algebras is Connes’ embedding problem, O which in its original formulation asks whether every II –factor embeds in the h. ultrapower Rω of the hyperfinite II –factor R. This is w1ell knowMn to be equivalent t 1 a to the question of whether all elements of II –factors possess matricial microstates, m 1 (which were introduced by Voiculescu [16] for free entropy), namely, whether such [ elements are approximable in –moments by matrices. Connes’ embedding problem ∗ 2 is known to be equivalent to a number of different problems, in large part due to v a remarkable paper [6] of Kirchberg. (See also the survey [10], and the papers [11], 8 8 [12], [13], [1], [14], [3], [7], [15], [5] for results with bearing on Connes’ embedding 2 problem.) 0 . In Proposition 4.6 of [6], Kirchberg proved that, in order to show that a finite von 1 Neumann algebra withfaithfultracialstateτ embeds inRω, it wouldbeenoughto 0 M 9 show that for all n, all unitary elements U ,...,U in and all ǫ > 0, there is k N 1 n 0 M ∈ andtherearek k unitarymatricesV ,...,V such that τ(U U ) tr (V V ) < ǫfor v: all i,j 1,...×,n , where tr is the1normalinzed trace o|n Mi∗(Cj )−. (Hke ai∗lsoj r|equired i ∈ { } k k X τ(U ) tr (V ) < ǫ, but this formally stronger condition is easily satisfied by taking i k i | − | r the n+ 1 unitaries U ,...,U ,U = I in finding k k unitaries V ,...,V , a 1 n n+1 1 n+1 M × so that τ(U U ) tr (V V ) < ǫ, and letting V = V V .) It is, therefore, of | i∗ j − k i∗ j | i n∗+1 i e e interest to consider the set of possible second–order mixed moments of unitaries in e e e e such ( ,τ) or, equivalently, of unitaries in C –algebraswith respect to tracial states. ∗ M (See also [12], where some similar sets were considered by F. R˘adulescu.) Definition 1.1. Let be the set of all n n matrices X of the form n G × X = τ(U U ) (1) i∗ j 1 i,j n (cid:0) (cid:1) ≤ ≤ as (U ,...,U ) runs over all n–tuples of unitaries in all C –algebras A possessing a 1 n ∗ faithful tracial state τ. 2000 Mathematics Subject Classification. 46L10, (15A48). Key words and phrases. Connes’ embedding problem, unitary moments, correlation matrices. ∗ Research supported in part by NSF grant DMS-0600814. 1 2 DYKEMA,JUSCHENKO Remark 1.2. The set–theoretic difficulties in the phrasing of Definition 1.1 can be evaded by insisting that A be represented on a given separable Hilbert space. Alternatively, let A = C U ,...,U denote the universal, unital, complex –algebra 1 n generated by unitary elehments U ,.i..,U . A linear functional φ on A is p∗ositive if 1 n φ(a a) 0 for all a A. By the usual Gelfand–Naimark–Segal construction, any ∗ suchpo≥sitivefunction∈alφgivesrisetoaHilbertspaceL2(A,φ)anda –representation π : A B(L2(A,φ)). Thus, the set equals the set of all matrice∗s X as in (1) as φ n τ runs→over all positive, tracial, unitalG, linear functionals τ on A. Definition 1.3. Let be the closure of the set n F tr (V V ) k N, V ,...,V , k i∗ j 1 i,j n | ∈ 1 n ∈ Uk (cid:8)(cid:0) (cid:1) ≤ ≤ (cid:9) where is the group of k k unitary matrices. k U × A correlation matrix is a complex, positive semidefinite matrix having all diagonal entries equal to 1. Let Θ be the set of all n n correlation matrices. Clearly, we n × have Θ . n n n F ⊆ G ⊆ Kirchberg’s result is that Connes’ embedding problem is equivalent to the problem of whether = holds for all n. n n F G Proposition 1.4. For each n, (i) and are invariant under conjugation with n n diagonal unitary matrices n n F G × and permutation matrices, (ii) and are compact, convex subsets of Θ , n n n F G (iii) and are closed under taking Schur products of matrices. n n F G Proof. Part (i) is clear. Note that Θ is a norm–bounded subset of M (C). That n n n F is closed is evident. That is closed follows from the description in Remark 1.2 and n the fact that a pointwise lGimit of positive traces on A is a positive trace. This proves compactness. Convexity of follows from by observing that if V is a k k unitary n F × and V is a k k unitary, then for arbitrary ℓ,ℓ N, ′ ′ ′ ′ × ∈ V V V V ′ ′ ⊕···⊕ ⊕ ⊕···⊕ ℓtimes ℓ′ times can be realized as a block–di|agon{azl (kℓ}+k|ℓ) ({kzℓ+k}ℓ) matrix whose normalized ′ ′ ′ ′ × trace is kℓ k ℓ ′ ′ trk(V)+ trk′(V′). kℓ+k ℓ kℓ+k ℓ ′ ′ ′ ′ Convexity of follows because a convex combination of positive traces on A is a n G positive trace. This proves (ii). Closedness of under taking Schur products follows by observing that if V and n F V are unitaries as above, then V V is a kk kk unitary whose normalized trace ′ ′ ′ ′ ⊗ × is trk(V)trk′(V′). For n, we observe that if U and respectively, U′, are unitaries in G C –algebras A and A having tracial states τ and τ , then the spatial tensor product ∗ ′ ′ C –algebra A A has tracial state τ τ that takes value τ(U)τ (U ) on the unitary ∗ ′ ′ ′ ′ U U . This ⊗proves (iii). ⊗ (cid:3) ′ ⊗ MATRICES OF UNITARY MOMENTS 3 Since it is important to decide whether we have = for all n, it is interesting n n F G to learn more about the sets . A first question is whether = Θ holds. In n n n F F Section 2, we show that this holds for n = 3 but fails for n 4. The proof relies on ≥ a characterization of extreme points of Θ , and it uses also the set of matrices of n n C moments of commuting unitaries. In Section 3 we prove M (R) Θ , and some n n n ∩ ⊆ F further results concerning . In Section 4, we show that has nonempty interior, n n C F as a subset of Θ . n 2. Extreme points of Θ and some consequences n The set Θ of n n correlation matrices is embedded in the affine space consisting n × of the self–adjoint complex matrices having all diagonal entries equal to 1; it is just the intersection of the set of positive, semidefinite matrices with this space. Every element of Θ is bounded in normby n (cf Remark 2.9), andΘ is a compact, convex n n space. Since, in the space of self–adjoint matrices, every positive definite matrix is the center of a ball consisting of positive matrices, it is clear that the boundary of Θ (for n 2) consists of singular matrices. n ≥ The extreme points of Θ and Θ M (R) have been studied in [2], [9], [4] and [8]. n n n ∩ In this section, we will use an easy characterization of the extreme points of Θ to n draw some conclusions about matrices of unitary moments. The papers cited above contain the facts about extreme points of Θ found below, and have results going n well beyond. However, for completeness and for use later in examples, we provide proofs, which are brief. We also introduce the subset of , consisting of matrices of moments of com- n n C F muting unitaries. This is a convenient place to recall the following standard fact. We include a proof for convenience. Lemma 2.1. The set of all X Θ of rank r is the set of all frame operators n ∈ X = F F of frames F = (f ,...,f ), consisting of n unit vectors f Cr, where ∗ 1 n j ∈ r = rank(X). If, in addition, X M (R), then the frame f ,...,f can be chosen n 1 n ∈ in Rr. Proof. Every frame operator F F as above clearly belongs to Θ and has rank r. ∗ n Recall that the support projection of a Hermitian matrix X is the projection onto the orthocomplement of the nullspace of X. Let P be the support projection of X and let λ λ > 0 be the nonzero eigenvalues of X with corresponding 1 r ≥ ··· ≥ orthonormal eigenvectors g ,...,g Cn. Let V : Cr P(Cn) be the isometry 1 r ∈ → definedbye g ,wheree ,...,e arethestandardbasisvectorsofCr. SoP = VV . i i 1 r ∗ 7→ Then X = F F, where F is the r n matrix ∗ × F = V X1/2 = diag(λ ,...,λ )1/2V . ∗ 1 r ∗ If f ,...,f Cr are the columns of F, then f = X = 1 and the linear span of 1 n i ii ∈ k k f ...,f is Cr. Thus, f ,...,f comprise a frame. 1 n 1 n If X is real, then the vectors g ,...,g can be chosen in Rn. Then V and X1/2 are 1 r real matrices and f ,...,f are in Rr. (cid:3) 1 n 4 DYKEMA,JUSCHENKO Lemma 2.2. Let X M (C) be a positive semidefinite matrix and let P be the n ∈ support projection of X. Then a Hermitian n n matrix Y has the property that × there is ǫ > 0 such that X +tY is positive semidefinite for all t ( ǫ,ǫ) if and only ∈ − if Y = PYP. Proof. If X = 0 then this is trivially true, so suppose X = 0. After conjugating with 6 a unitary, we may without loss of generality assume P = diag(1,...,1,0,...,0) with rank(X) = rank(P) = r. Then PXP, thought of as an r r matrix, is positive × definite. By continuity of the determinant, we see that if Y = PYP, then Y enjoys the property described above. Conversely, if Y = PYP, then we may choose two standard basis vectors e and i 6 e for i r < j, such that the compressions of X and Y to the subspace spanned by j ≤ e and e are given by the matrices i j x 0 a b X = , Y = (cid:18)0 0(cid:19) (cid:18)b c(cid:19) b b for some x > 0, a,c R and b C with c and b not both zero. But ∈ ∈ det(X +tY) = txc+t2(ac b 2). −| | If c = 0, then det(X+tY) < 0bfor abll nonzero t sufficiently small in magnitude and of 6 the appropriate sign, while if c = 0 then b = 0 and det(X+tY) < 0 for all t = 0. (cid:3) b b 6 6 Proposition 2.3. Let n N, let X Θ and let Pbbe thbe support projection of n ∈ ∈ X. A necessary and sufficient condition for X to be an extreme point of Θ is that n there be no nonzero Hermitian n n matrix Y having zero diagonal and satisfying × Y = PYP. Consequently, if X is an extreme point of Θ , then rank(X) √n. n ≤ Proof. X is an extreme point of Θ if and only if there is no nonzero Hermitian n n n × matrix Y such that X +tY Θ for all t R sufficiently small in magnitude. Now n ∈ ∈ use Lemma 2.2 and the fact that Θ consists of the positive semidefinite matrices n with all diagonal values equal to 1. For the final statement, if r = rank(X) then the set of Hermitian matrices with support projection under P is a real vector space of dimension r2, while the space of n n Hermitian matrices with zero diagonal has dimension n2 n. If r2 > n, then × − (cid:3) the intersection of these two spaces is nonzero. Proposition 2.4. Let X Θ . Suppose f ,...,f is a frame consisting of n unit n 1 n ∈ vectors in Cr, where r = rank(X), so that X = F F with F = (f ,...,f ) is the ∗ 1 n corresponding frame operator. (See Lemma 2.1.) Then X is an extreme point of Θ n if and only if the only r r self–adjoint matrix Z satisfying Zf ,f = 0 for all j j × h i j 1,...,n is the zero matrix. ∈ { } Proof. Since F is an r n matrix of rank r, the map M (C) M (C) given r s.a. n s.a. × → by Z F ZF is an injective linear map onto PM (C) P, where P is the support ∗ n s.a. 7→ projection of X. If Y = F ZF, then Y = Zf ,f . Thus, the condition for X to ∗ jj j j h i (cid:3) be extreme now follows from the characterization found in Proposition 2.3. MATRICES OF UNITARY MOMENTS 5 Proposition 2.5. Let n N and suppose X Θ satifies rank(X) = 1. Then X n ∈ ∈ is an extreme point of Θ and X . Moreover, using the notation introduced in n n ∈ F Remark 1.2, we have conv X Θ rank(X) = 1 = n { ∈ | } = τ(U U ) τ : A C a positive trace, τ(1) = 1, π (A) commutative { i∗ j 1 i,j n | → τ } (cid:0) (cid:1) ≤ ≤ (2) and this set is closed in Θ . n Notation 2.6. We let denote the set given in (2). Thus, we have . n n n C C ⊆ F Moreover, (cf Remark 1.2), is the set of matrices as in (1) where (U ,...,U ) run n 1 n C over all n–tuples of commuting unitarires in C –algebras A with faithful tracial state ∗ τ. Proof of Proposition 2.5. By Lemma 2.1, we have X = F F where F = (f ,...,f ) ∗ 1 n for complex numbers f with f = 1. Using Proposition 2.4, we see immediately j j | | that X is an extreme point of Θ . Thinking of each f as a 1 1 unitary, we have n j × X and, moreover, X = τ(U U ) , where τ : A C is the character ∈ Fn i∗ j 1 i,j n → defined by τ(U ) = f ; in fact, i(cid:0)t is appar(cid:1)en≤t t≤hat every character on A yields a rank i i one element of Θ . Since the set of traces τ on A having π (A) commutative is n τ convex, this implies the inclusion in (2). ⊆ That the left–hand–side of (2) is compact follows from Caratheodory’s theoem, because the rank one projections form a compact set. If τ : A C is a positive trace with τ(1) = 1 and π (A) commutative, then τ = ψ π fo→r a state ψ on the τ τ C –algebra completion of π (A). Since every state on a ◦unital, commutative C – ∗ τ ∗ algebra is in the closed convex hull of the characters of that C –algebra, τ is itself the ∗ limit in norm of a sequence of finite convex combinations of characters of A. Thus, X = τ(U U ) is the limit of a sequence of finite convex combinations of rank i∗ j 1 i,j n one e(cid:0)lements o(cid:1)f Θ≤ ,≤and we have in (2). (cid:3) n ⊇ Remark 2.7. We see immediately from (2) that is a closed convex set that is n C closed under conjugation with diagonal unitary matrices and permutation matrices; also, since the set of rank one elements of Θ is closed under taking Schur products, n so is the set . Furthermore, since lies in a vector space of real dimension n n C C m := n2 n, by Caratheodory’s theorem every element of is a convex combination n − C of not more than m+1 rank one elements of Θ . n An immediate application of Propositions 2.3 and 2.5 is the following. Corollary 2.8. The extreme points of Θ are precisely the rank one elements of Θ . 3 3 Moreover, we have = = = Θ . 3 3 3 3 C F G Remark 2.9. Let X and take A, τ and U ,...,U as in Definition 1.1 so n 1 n ∈ G that (1) holds, and assume without loss of generality that τ is faithful on A. If we identify Mn(A) with A⊗Mn(C), then we have X = n(τ ⊗idMn(C))(P), where P is 6 DYKEMA,JUSCHENKO the projection U 1∗ 1 U2∗ P = . (U U U ) n .. 1 2 ··· n   U   n∗ in M (A). If c = (c ,...,c )t Cn is such that Xc = 0, then this yields τ(Z Z) = 0, n 1 n ∗ ∈ where Z = c U + +c U . Since τ is a faithful, we have Z = 0. 1 1 n n ··· Proposition 2.10. Let n N. If X and rank(X) 2, then X . n n ∈ ∈ G ≤ ∈ C Proof. If rank(X) = 1, then this follows from Propostion 2.5, so assume rank(X) = 2. Let τ : A C be a positive, unital trace such that X = τ(U U ) and → i∗ j 1 i,j n let π : A B(L2(A,τ)) the the –representation as described(cid:0)in Rema(cid:1)rk≤ 1.≤2. Let τ σ : A π→(A) be the –represent∗ation defined by σ(U ) = π (U ) π (U ) for each τ i τ 1 ∗ τ i i 1→,...,n and let τ∗ = τ σ. Then τ is a positive, unital trace on A and the ′ ′ ∈ { } ◦ matrix τ′(Ui∗Uj) 1 i,j n is equal to X. Furthermore, πτ′(U1) = I. Consequently, we may wit(cid:0)hout loss(cid:1)of≤ge≤nerality assume π (U ) = I. τ 1 Let e ,...,e denote the standard basis vectors of Cn. Let i,j 2,...,n , 1 n ∈ { } with i = j. Since rank(X) = 2, there are c ,c ,c C with c = 0 such that 1 i j 1 6 ∈ 6 X(c e +c e +c e ) = 0. By Remark 2.9, we have π (c I +c U +c U ) = 0. We do 1 1 i i j j τ 1 i i j j not have c = c = 0, so assume c = 0. If c = 0, then π (U ) is a scalar multiple of i j i j τ i 6 the identity, while if c = 0, then π (U ) and π (U ) generate the same C –algebra, j τ i τ j ∗ 6 which is commutative. In either case, we have that the –algebras generated by π (U ) and π (U ) commute with each other. Therefore, π ∗(A) is commutative, and τ i τ j τ X . (cid:3) n ∈ C Corollary 2.11. = Θ . 4 4 G 6 Proof. Combining Proposition 2.10 and Proposition 2.5, we see that has no ex- 4 G treme points of rank 2. It will suffice to find an extreme point X of Θ with 4 rank(X) = 2. By Proposition 2.4, it will suffice to find four unit vectors f ,...,f 1 4 spanning C2 such that the only self–adjoint Z M (C) satisfying Zf ,f = 0 for 2 i i ∈ h i all i = 1,...,4 is the zero matrix. It is easily verified that the frame 1 0 1/√2 i/√2 f = , f = , f = , f = 1 (cid:18)0(cid:19) 2 (cid:18)1(cid:19) 3 (cid:18)1/√2(cid:19) 4 (cid:18)1/√2(cid:19) does the job, and, with F = (f ,f ,f ,f ), this yields the matrix 1 2 3 4 1 0 1 i √2 √2  0 1 1 1  X = F F = √2 √2 Θ . (3) ∗  1 1 1 1+i ∈ 4\G4 √2 √2 2    −i 1 1−i 1  √2 √2 2  (cid:3) Remark 2.12. We cannot have = for all n, because by an easy a modification n n C F of Kirchberg’s proof of Proposition 4.6 of [6], this would imply that M (C) can be 2 MATRICES OF UNITARY MOMENTS 7 faithfullyrepresented inacommutative vonNeumannalgebra. (This argumentshows that for some n there must be two–by–two unitaries V ,...,V such that the matrix 1 n tr (V V ) does not belong to .) In fact, in Proposition 3.6 we will show 2 i∗ j 1 i,j n Cn (cid:0) = . H(cid:1)ow≤ev≤er, we don’t know whether = holds or not for n = 4 or n = 5. 6 6 n n F 6 C F C 3. Real matrices The main result of this section is the following, which easily follows from the usual representation of the Clifford algebra. Theorem 3.1. For every n N, we have ∈ M (R) Θ . n n n ∩ ⊆ F We first recall the representation of the Clifford algebra. Let Λ be a linear map from a real Hilbert space H into the bounded, self–adjoint operators B(K) , for s.a. some complex Hilbert space K, satisfying Λ(x)Λ(y)+Λ(y)Λ(x) = 2 x,y I , (x,y H). (4) H h i ∈ The real algebra generated by range of Λ is uniquely determined by H and called the real Clifford algebra. Consider a real Hilbert space H of finite dimension r with its canonical basis e . i { } Let 1 0 0 1 1 0 U = , V = , I = . (cid:18) 0 1 (cid:19) (cid:18) 1 0 (cid:19) 2 (cid:18) 0 1 (cid:19) − ThentherealCliffordalgebraofH hasthefollowingrepresentationby2r 2r matrixes × Λ(x) = λiU⊗i−1 ⊗V ⊗I2⊗(n−i), X where x = λ e . It easy to check that the relation (4) is satisfied. Moreover if i i x = 1 thePn Λ(x) is symmetry, i.e. Λ(x)∗ = Λ(x) and Λ(x)2 = I. || || Proof of Theorem 3.1. Let r be the rank of X. By Lemma 2.1, there are unit vectors f ,...,f Rr such that X = f ,f for all i and j. Taking Λ as described above, 1 n i,j i j ∈ h i we get 2r 2r unitary matrices Λ(f ) (in fact, they are symmetries), and from (4) we i have tr(Λ×(f )Λ(f )) = f ,f . (cid:3) i j i j h i Below is the result for real matrices that is entirely analogous to Proposition 2.3. Proposition 3.2. Let n N, let X M (R) Θ and let P be the support projection n n ∈ ∈ ∩ of X. A necessaryand sufficientconditionfor X to be an extreme pointof M (R) Θ n n ∩ is that there be no nonzero Hermitian real n n matrix Y having zero diagonal and × satisfying Y = PYP. Consequently, if X is an extreme point of M (R) Θ and n n ∩ r = rank(X), then r(r +1)/2 n. ≤ Proof. This is just like the proof of Proposition 2.3, the only difference being that the dimension of PM (R) P for a projection P of rank r is r(r +1)/2. (cid:3) n s.a. Corollary 3.3. If n 5, then ≤ M (R) Θ . (5) n n n ∩ ⊆ C 8 DYKEMA,JUSCHENKO Proof. From Proposition 3.2, we see that every extreme point X of M (R) Θ for n n ∩ n 5 has rank r 2. But X , by Theorem 3.1, so using Proposition 2.10, n n ≤ ≤ ∈ F ⊆ G it follows that all extreme points of M (R) Θ lie in . Since is closed and n n n n ∩ C C (cid:3) convex (see Proposition 2.5), the inclusion (5) follows. Of course, we also have the result for real matrices (and real frames) that is anal- ogous to Proposition 2.4, which is stated below. The proof is the same. Proposition 3.4. Let X M (R) Θ . Suppose f ,...,f is a frame consisting of n n 1 n ∈ ∩ n unit vectors in Rr, where r = rank(X), so that X = F F with F = (f ,...,f ) ∗ 1 n is the corresponding frame operator. (See Lemma 2.1.) Then X is an extreme point of M (R) Θ if and only if the only real Hermitian r r matrix Z satisfying n n ∩ × Zf ,f = 0 for all j 1,...,n is the zero matrix. j j h i ∈ { } Although Corollary 3.3 shows that every element of M (R) Θ for n 5 is in the n n ∩ ≤ closed convex hull of the rank one operators in Θ , it is not true that every element n of M (R) Θ lies in the closed convex hull of rank one operators in M (R) Θ , n n n n ∩ ∩ even for n = 3, as the following example shows. Example 3.5. Consider the frame 1 0 1 1 f = , f = , f = 1 (cid:18)0(cid:19) 2 (cid:18)1(cid:19) 3 √2 (cid:18)1(cid:19) of three unit vectors in R2. It is easily verified that the only real Hermitian 2 2 × matrix Z such that Zf ,f = 0 for all i = 1,2,3 is the zero matrix. Thus, by i i h i Proposition 3.4, 1 0 1 √2 X =  0 1 1  √2  1 1 1  √2 √2  is a rank–two extreme point of M (R) Θ . However, an explicit decomposition as 3 3 ∩ a convex combination of rank one operators in Θ is 3 1 i 1+i 1 i 1 i − √2 − √2 1 1 X =  i 1 1−i+  i 1 1+i. 2 − √2 2 √2 1 i 1+i 1  1+i 1 i 1  − − √2 √2  √2 √2  Proposition 3.6. We have M (R) Θ . 6 6 6 ∩ 6⊆ C Thus, we have = . 6 6 F 6 C Proof. We construct an example of X (M (R) Θ ) . In fact, it will be a 6 6 6 ∈ ∩ \C rank–three extreme point of M (R) Θ . 6 6 ∩ MATRICES OF UNITARY MOMENTS 9 Consider the frame 1 0 0 f = 0, f = 1, f = 0, 1 2 3 0 0 1       1 0 1 1 1 1 f = 1, f = 1, f = 1 4 5 6 √2 √2 √3 0 1 1       of six unit vectors in R3. It is easily verified that the only real Hermitian 3 3 × matrix Z such that Zf ,f = 0 for all i 1,...,6 is the zero matrix. Thus, by i i h i ∈ { } Proposition 3.4, 1 0 0 1 0 1 √2 √3  0 1 0 1 1 1  √2 √2 √3  0 0 1 0 1 1   √2 √3    X =    1 1 0 1 1 2 √2 √2 2 q3    0 1 1 1 1 2  √2 √2 2 3  q   1 1 1 2 2 1  √3 √3 √3 3 3   q q  is a rank–three extreme point of M (R) Θ . The nullspace of X is spanned by the 6 6 ∩ vectors v = ( 1 , 1 ,0, 1,0,0)t 1 √2 √2 − v = (0, 1 , 1 ,0, 1,0)t 2 √2 √2 − v = ( 1 , 1 , 1 ,0,0, 1)t. 3 √3 √3 √3 − Suppose, to obtain a contradiction, that we have X . Then there is a com- 6 ∈ C mutative C –algebra A = C(Ω) with a faithful tracial state τ and there are unitaries ∗ I = U ,U ,...,U A such that X = τ(U U ) . Taking the vectors v , v 1 2 6 ∈ i∗ j 1 i,j 6 1 2 and v , above, by Remark 2.9 we have (cid:0) (cid:1) ≤ ≤ 3 1 U = (U +U ) (6) 4 1 2 √2 1 U = (U +U ) (7) 5 2 3 √2 1 U = (U +U +U ). (8) 6 1 2 3 √3 Fixing any ω Ω, we have that ζ := U (ω) is a point on the unit circle T, (1 j j j ∈ ≤ ≤ 6). From (6) and ζ = 1, we get ζ = iζ and similarly from (7) we get ζ = iζ . 4 1 2 3 2 | | ± ± However, from (8), we then have 1 2i 1 1+2i ζ − ζ , ζ , ζ , 6 2 2 2 ∈ { √3 √3 √3 } which contradicts ζ = ζ = 1. (cid:3) 6 2 | | | | 10 DYKEMA,JUSCHENKO 4. Nonempty interior In this section, we show that the interior of and, in fact, of , is nonempty, n n F C when considered as a subset of Θ . (Since = Θ for n = 1,2,3, this needs proving n n n C only for n 4.) ≥ Given X Θ , let n ∈ a = sup t [0,1] tX +(1 t)I X n { ∈ | − ∈ F } c = sup t [0,1] tX +(1 t)I . X n { ∈ | − ∈ C } Of course, c a . We now show that c is bounded below by a nonzero constant X X X ≤ that depends only on n. In particular, we have that the identity element lies in the interior of , when thisistaken asasubset oftheaffinespace of self–adjoint matrices n C having all diagonal entries equal to 1. Proposition 4.1. Let n N, n 3, and let X Θ . Then n ∈ ≥ ∈ 6 c . (9) X ≥ n2 n − Moreover, if λ is the smallest eigenvalue of X, then 0 6 c min( ,1). (10) X ≥ (n2 n)(1 λ ) 0 − − Proof. We have X = (x )n with x = 1 for all i = 1,...,n. Denote G = σ ij i,j=1 ii { ∈ S σ(1) < σ(2) < σ(3) . Then n | } n #G = (n 3)! (cid:18)3(cid:19) − Let U = (u ) be the permutation unitary matrix where u = δ . Then U XU = σ ij ij i,σ(i) ∗ (xσ−1(i)σ−1(j))i,j. Define the block-diagonal matrix 1 x x σ(1)σ(2) σ(1)σ(3) B =  x 1 x  I . σ σ(2)σ(1) σ(2)σ(3) n 3 ⊕ − x x 1 σ(3)σ(1) σ(3)σ(2)   Using Corollary 2.8 (and Remark 2.7), we easily see B . σ n ∈ C Let J = (σ(1),σ(2)),(σ(1),σ(3)),(σ(2),σ(3)) . Put X = U B U. Then σ σ ∗ σ { } 0, if (k,ℓ) (1,1),...,(n,n) J , σ 6∈ { }∪ (X ) = 1, if k = ℓ σ kℓ   x , if (k,ℓ) J . kℓ σ ∈  Since for any k < ℓ we have # σ G σ(1) = k,σ(2) = ℓ or σ(1) = k,σ(3) = ℓ or σ(2) = k,σ(3) = ℓ = { ∈ | } ((n ℓ)+(ℓ k 1)+(k 1))(n 3)! = (n 2)! − − − − − − it follows that matrix 1 X = X ′ σ #G X σ G ∈ has entries x = 1, and x = 6 x if k = ℓ. ′ii ′kℓ n2 n kℓ 6 −