Representation of contractively complemented Hilbertian operator spaces and the Fock space

The operator spaces $H_n^k$ $1\le k\le n$, generalizing the row and column Hilbert spaces, and arising in the authors' previous study of contractively complemented subspaces of $C^*$-algebras, are shown to be homogeneous and completely isometric to a space of creation operators on a subspace of the anti-symmetric Fock space. The completely bounded Banach-Mazur distance from $H_n^k$ to row or column space is explicitly calculated.


Introduction and Preliminaries
A well-known result of Friedman and Russo ([4, Theorem 2]) states that if a subspace X of a C * -algebra A is the range of a contractive projection on A, then X is isometric to a JC * -triple, that is, a norm closed subspace of B(H, K) stable under the triple product ab * c + cb * a. If X is atomic (in particular, finite-dimensional), then it is isometric to a direct sum of Cartan factors of types 1 to 4.
The authors showed in [7] that this latter result fails, as it stands, in the category of operator spaces. In that paper, we defined a family of n-dimensional Hilbertian operator spaces H k n , 1 ≤ k ≤ n, generalizing the row and column Hilbert spaces R n and C n and showed that in the above result, if X is atomic, the word "isometric" can be replaced by "completely semi-isometric," provided the spaces H k n are allowed as summands along with the Cartan factors ([7, Theorem 2]). It is pointed out in [7] that the space H k n is contractively complemented in some B(K), and for 1 < k < n, is not completely (semi-)isometric to either of the Cartan factors B(C, C n ) = H 1 n or B(C n , C) = H n n , and that these spaces appeared in a slightly different form and context in [1]. It is also shown in [7,Theorem 3] that finite dimensional JC *triples which are contractively complemented in a C * -algebra can be classified up to complete isometry.
In this paper, we study the operator space structure of the spaces H k n . Besides being a generalization of the row and column Hilbert spaces, as shown in Lemma 2.1 below, they are completely isometric to the span of creation operators on a subspace of the anti-symmetric Fock space. Thus they are related to the operator space denoted by Φ n in [9, section 9.3], which is the span of the creation operators on the full anti-symmetric Fock space. Φ n is the unique operator space which is completely isometric to the span of n operators satisfying the canonical anticommutation relations (CAR), [9,Theorem 9.3.1], and ∩ n k=1 H k n is completely isometric 1 to Φ n . We show in Theorem 2 below that all finite dimensional Hilbertian operator spaces X which are contractively complemented in some C * -algebra are completely isometric to the diagonal of two spaces, one space being an intersection of some of the spaces H k n and the other space lying in the kernel of the projection which maps onto X. Since any intersection of the spaces H k n is also completely isometric to a space of creation operators on a subspace of the full anti-symmetric Fock space, Theorem 2 can be interpreted as saying that every contractively complemented Hilbertian operator space is, up to complete isometry, essentially a space of creation operators. This result is analogous to the result of Robertson, [10], which states that every completely contractively complemented Hilbertian operator space is completely isometric to either row or column space.
The operator space structures of the row and column Hilbert spaces R n and C n have been well studied, and in particular it is known that they are homogeneous, dual to each other in the operator space sense, and have completely bounded Banach-Mazur distance n between them. We show here that H k n is homogeneous (Theorem 1) and we give an explicit formula for the completely bounded Banach-Mazur distance from it to R n = H n n and C n = H 1 n (Theorem 3). This answers a question we posed in [7] and shows, interestingly, that the points R n , C n and H k n lie on a straight line in the metric space of all operator spaces of dimension n.
Recall that a Cartan factor of type 1 is B(H, K) for complex Hilbert spaces H and K. To define the Cartan factors of types 2 and 3, fix a conjugation J on a complex Hilbert space H, that is, a conjugate-linear isometry of order 2, and for x ∈ B(H), let x t = Jx * J. A Cartan factor of type 2 (respectively of type 3) is A Cartan factor of type 4 is the spin factor (cf. [7,Subsection 3.1]).
An operator space is a subspace X of B(H), the space of bounded linear operators on a complex Hilbert space. Its operator space structure is given by the sequence of norms on the set of matrices M n (X) with entries from X, determined by the identification M n (X) ⊂ M n (B(H)) = B(H ⊕ H ⊕ · · · ⊕ H). See [9] for the general theory of operator spaces, which is now extensive and covered in several other monographs, for example [3], [8], and the forthcoming [2]. Let us just recall that a linear mapping ϕ : X → Y between two operator spaces is completely bounded if the induced mappings ϕ n : M n (X) → M n (Y ) defined by ϕ n ([x ij ]) = [ϕ(x ij )] satisfy ϕ cb := sup n ϕ n < ∞. A completely bounded map is a completely bounded isomorphism if its inverse exists and is completely bounded. Two operator spaces are completely isometric if there is a linear isomorphism T between them with T cb = T −1 cb = 1. We call T a complete isometry in this case.
In the matrix representation for B(ℓ 2 ) consider the column Hilbert space C = sp{e i1 : i ≥ 1} and the row Hilbert space R = sp{e 1j : j ≥ 1} and their finite dimensional versions C n = sp{e i1 : 1 ≤ i ≤ n} and R n = sp{e 1j : 1 ≤ j ≤ n}.
Here of course e ij is the operator defined by the matrix with a 1 in the (i, j)entry and zeros elsewhere. Although R and C are Banach isometric, they are not completely isomorphic; and R n and C n , while completely isomorphic, are not completely isometric.
An operator space is said to be homogeneous if every bounded linear map on it is completely bounded with the norm and completely bounded norm coinciding (see [9, 9.2]) and it is Hilbertian if it is isometric to a Hilbert space. A linear map of one operator space into another is said to be a complete semi-isometry if it is isometric and completely contractive. The completely bounded Banach-Mazur distance between two (completely isomorphic) operator spaces E, F is defined by d cb (E, F ) = inf{ u cb u −1 cb : u : E → F complete isomorphism }. This paper is organized as follows. In section 1, we show that the spaces H k n are homogeneous operator spaces. Although we use some multilinear algebra, our proof is direct and does not make use of the identification of H k n with a space of creation operators. In section 2 we establish the complete isometry of H k n with a space of creation operators and use it to describe the fine structure of the range of a contractive projection on a C * -algebra in case said range is isometric to a Hilbert space. We also establish some spectral properties of creation operators. In section 3 we compute explicitly the completely bounded Banach-Mazur distance from the space H k n to the column and row Hilbert spaces H 1 n and H n n and state some problems for further study.

Homogeneity of the spaces H k n
We begin by recalling from [7,Sections 6,7] the construction of the spaces H k n , We shall use the notation e i to denote the column vector with a 1 in the i th position and zeros elsewhere. Thus e 1 , . . . , e n denotes the canonical basis of column vectors for C n , and for example e J1 , . . . , e Jp denotes the canonical basis of column vectors for C p .
The space H k n is the linear span of matrices b n,k i , 1 ≤ i ≤ n, given by b n,k where e J,I = e J ⊗ e I = e J e t I ∈ M p,q (C) = B(C q , C p ), and ǫ(I, i, J) is the signature of the permutation taking (i 1 , . . . , i k−1 , i, j 1 , . . . , j n−k ) to (1, . . . , n). 1 Since the b n,k i are the image under a triple isomorphism (actually ternary isomorphism) of a rectangular grid in a JW * -triple of rank one, they form an orthonormal basis for H k n (cf. the beginning of subsection 5.3 and the beginning of section 7 of [7]).
In the rest of this section, we shall use the following lemma about determinants, whose proof can be found, for example, in [11].    (2), page 87 of [11]) Let e 1 , . . . , e n be the canonical basis for the column Hilbert space C n = M n,1 (C) = B(C, C n ) and define an isometry ψ : ] be a unitary matrix so that u 1 , . . . , u n is an orthonormal basis for C n . Then, with u i = n j=1 u ji e j , we have where u is the complex conjugate of u.
Proof. Let us first calculate the left side of (1): Before calculating the right side of (1), note that ǫ(I, i, J)ǫ(I, J) = (−1) i+k ; indeed, Therefore, the right side of (1) is equal to According to Lemma 1.1(iii), the above sum is 0 if J ′ ∩ I ′ = ∅. Otherwise, if J ′ ∩ I ′ = ∅ so that (I ′ ∪ J ′ ) c = {l}, the right side of (1) equals Now by Lemma 1.1(ii), the above sum is the determinant of the (l, i)-minor of the matrix u, call this M li . Thus, for J ′ ∩ I ′ = ∅, the right side of (1) is equal to Theorem 1. H k n is a homogeneous operator space. Proof. Let α be a unitary operator on H k n . To prove the theorem, it suffices, by [9, Prop. 9.2.1], to show that α is a complete isometry. We shall show that α(x) = λvxw for suitable unitary matrices v and w, and λ ∈ C, with |λ| = 1, which will complete the proof.
Recall that ψ : C n → H k n is the isometry defined by ψ(e i ) = b n,k i . Let ψ −1 αψ have matrix u −1 on C n with respect to the basis e 1 , . . . , e n . As in Lemma 1.2, let u 1 , . . . , u n be the columns of u. We shall show that α(x) = λvxw holds for every (The fact that v and w are unitary matrices follows from the definition of the inner product on ∧ r C n : (x 1 ∧ · · · ∧ x r |y 1 ∧ · · · ∧ y r ) = det[(x i |y j )].) In the first place, ψ −1 αψ(u i ) = u −1 (u i ) = e i , so that αψ(u i ) = ψ(e i ) = b n,k i . Thus it suffices to prove (2) vψ(u i )w = b n,k i / det u. Let us first show that By Lemma 1.2, the proof of (3) amounts to The left side of (4) is given by e t J ′ (∧ j∈J u j ) (∧ i∈I u i ) t e I ′ . By Lemma 1.1(i), ∧ j∈J u j = L det u L,J e L , where L runs over the subsets of cardinality n−k. Hence e t J ′ (∧ j∈J u j ) = L det u L,J e t J ′ e L = det u J ′ ,J . Similarly, (∧ i∈I u i ) t e I ′ = det u I ′ ,I , which proves (4) and hence (3).
We now use (3) to prove (2). Note that since e J,I = e J ⊗ e I = e J (e I ) t = (∧ j∈J e j ) (∧ i∈I e i ) t , we may write By (3) and (5), it suffices to prove This is a simple calculation. Suppose for definiteness that J = J r and I = I s . Then v (∧ j∈J u j ) = e Jr , and (∧ i∈I u i ) t w = e t Is .

Anti-symmetric Fock spaces
Let C n,k h denote the wedge (or creation) operator from ∧ k−1 C n to ∧ k C n given by As in section 1, let e 1 , . . . , e n be the usual column vector orthonormal basis for C n , and let {e I1 , . . . , e Iq } and {e J1 , . . . , e Jp } be the column vector orthonormal bases for C q and C p respectively, and define the unitary operators U n j (j = k − 1 and j = n − k), W n k , V n k in the diagram below as follows: • U n k−1 (e I ) = e i1 ∧ · · · ∧ e i k−1 , where I = {i 1 < · · · < i k−1 }. • U n n−k (e J ) = e j1 ∧ · · · ∧ e j n−k , where J = {j 1 < · · · < j n−k }. • V n k (e i1 ∧ · · · ∧ e i k ) = e j1 ∧ · · · ∧ e j n−k , where {j 1 < · · · < j n−k } is the complement of {i 1 < · · · < i k }. • W n k (e j1 ∧ · · · ∧ e j n−k ) = ǫ(i, I)ǫ(I, i, J)e j1 ∧ · · · ∧ e j n−k for any i and I such that I ∩ J = ∅ and (I ∪ J) c = {i} (which is independent of the choice of i or I).
Note that since b n,k i is a p × q matrix, it is viewed as an operator from C q to C p . In the definition of W n k , ǫ(i, I) is the signature of the permutation (i, i 1 , . . . , i k−1 ) → (i 1 , . . . , i, . . . , i k−1 ). To prove the non-dependence on i, suppose i, i ′ ∈ J. Then where i 1 < · · · < i < · · · i k−1 . Similarly, Hence, letting C n,k denote the space sp{C n,k ei } yields the following lemma.
Lemma 2.1. H k n is completely isometric to C n,k . By [7, Theorem 2, Corollary 2.8], every atomic contractively complemented subspace X of a C * -algebra is isometrically completely contractive to a direct sum of Cartan factors of types 1 to 4 and some of the spaces H k n . The following theorem gives more detailed information on what can be said up to complete isometry in the case of an Hilbertian X.
Recall that a linear map of one operator space into another is said to be a complete semi-isometry if it is isometric and completely contractive.
Theorem 2. Let X be the range of a contractive projection P on a C * -algebra A, and suppose that X is isometric to a Hilbert space. Then there exist projections p, q ∈ A * * such that The map E 0 x = pxq is a complete semi-isometry of X onto pXq; (c): If X is finite-dimensional, then pXq is completely isometric to an intersection of the spaces C n,k . If X is infinite-dimensional, then pXq is completely semi-isometric to either row or column Hilbert space; (d): Both X and pXq are completely isometric to the range of a contractive projection on B(K) for an appropriate Hilbert space K. (e): P * * (pxq) = x and P * * ((1 − p)x(1 − q)) = 0, for x ∈ P (A). Hence P * * : pXq → X is the inverse of E 0 and (1 − p)X(1 − q) ⊂ ker P * * .
Remark 2.2. For any subset S ⊂ {1, . . . , n}, ∩ k∈S C n,k is exactly the space of creation operators on the direct sum ⊕ k∈S ∧ k−1 C n . So Theorem 2 says that all finitedimensional contractively complemented Hilbertian operator spaces are essentially a space of creation operators in the anti-symmetric Fock space. In particular, if S = {1, . . . , n}, this is the space Φ n discussed in [9, section 9.3].
The following two properties of the wedge operators follow easily from [5, Exercises 12.4.39-40]. They will be used, together with Lemma 2.3, in section 3. and In particular, C n,1 hi C n,1 * hi be λ 1 ≥ · · · ≥ λ n . Then the eigenvalues of m i=1 C n,k hi C n,k * hi are precisely the sums of k eigenvalues of m i=1 C n,1 hi C n,1 * hi . Proof. In the first place, we have To prove the second statement, let ξ 1 , . . . , ξ n be an orthonormal basis of C n consisting of eigenvectors of From this, the second statement follows.
Let D = (C n,1 h1 , . . . , C n,1 hm ) t and C = (C n,k h1 , . . . , C n,k hm ) t . We show next that D ≤ n n−k+1 C . By Lemma 2.3(a), we have Since C * C is a square matrix of size n k−1 , again by Lemma 2.3(a), Therefore, Taking m = n and h i = e i , we have D * D = n. By (6), C n,k * ei C n,k ei (e i1 ∧ · · · ∧ e i k−1 ) = 0 if i ∈ {i 1 , . . . , i k−1 } and equal to e i1 ∧ · · · ∧ e i k−1 otherwise. Hence C * C = (n − k + 1)I, proving the second statement.  , H 1 n ), which can be proved by exactly the same methods. Hence we obtain the following, which is the answer to Problem 1 in [7].  Since H k n is distinct from row or column Hilbert space, new ideas will be needed to solve the following problem. Problem 1. Find d cb (H k1 n , H k2 n ) for 1 < k 1 < k 2 < n. We have already mentioned in the introduction that C * n = R n and R * n = C n in the category of operator spaces. Hence H 1 n and H n n are operator space duals of each other. The following problem is therefore of interest and its solution would certainly lead to insight into Pisier's question on the operator space dual of Φ n , [9, page 175].
Problem 2. Find the operator space dual of H k n . Note that Theorem 2 does not say anything about the infinite-dimensional case up to complete isometry, although in this case, the space is completely semiisometric to R or C.
Problem 3. Are all infinite-dimensional Hilbertian contractively complemented operator spaces completely isometric to a space of creation operators on a subspace of the anti-symmetric Fock space?
Problem 4. What is the completely bounded Banach-Mazur distance between two infinite-dimensional Hilbertian contractively complemented operator spaces?