Matrices of unitary moments

0.19 MB·

by  Ken Dykema

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
−