REPRESENTATION OF CONTRACTIVELY COMPLEMENTED HILBERTIAN OPERATOR SPACES ON THE FOCK SPACE

MATTHEW NEAL AND BERNARD RUSSO

Abstract. The operator spaces H_{n}^{k} 1 ≤ k ≤ 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_{n}^{k}, 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_{n}^{k} are allowed as summands along with the Cartan factors ([7, Theorem 2]). It is pointed out in [7] that the space H_{n}^{k} 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_{n}^{1} 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_{n}^{k}. 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 ∩_{k=1}^{n}H_{n}^{k} is completely isometric 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_{n}^{k} and the other space lying in the kernel of the projection which maps onto X. Since any intersection of the spaces H_{n}^{k} 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.

Date: January 1, 1994 and, in revised form, June 22, 1994.

1991 Mathematics Subject Classification. Primary 46L07.

Key words and phrases. Hilbertian operator space, homogeneous operator space, contractive projection, creation operator, anti-symmetric Fock space, completely bounded Banach-Mazur distance.

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_{n}^{k} 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_{n}^{1} (Theorem 3). This answers a question we posed in [7] and shows, interestingly, that the points R_{n}, C_{n} and H_{n}^{k} 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(H, J) = {x ∈ B(H) : x^{t} = −x} (respectively S(H, J) = {x ∈ B(H) : x^{t} = −x}). 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 = \overline{sp}\{e_{i1} : i ≥ 1\} and the row Hilbert space R = \overline{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

dcb(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_{n}^{k} are homogeneous operator spaces. Although we use some multilinear algebra, our proof is direct and does not make use of the identification of H_{n}^{k} with a space of creation operators. In section 2 we establish the complete isometry of H_{n}^{k} 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_{n}^{k} to the column and row Hilbert spaces H_{n}^{1} and H_{n}^{n} and state some problems for further study.

1. Homogeneity of the spaces H_{n}^{k}

We begin by recalling from [7, Sections 6,7] the construction of the spaces H_{n}^{k}, 1 ≤ k ≤ n. Let I denote a subset of {1, 2, …, n} of cardinality |I| = k − 1. The number of such I is q := \binom{n}{k−1}. Let J denote a subset of \{1, 2, …, n\} of cardinality |J| = n − k. The number of such J is p := \binom{n}{n−k}. Unless otherwise noted, we shall assume that each I = {i_{1}, …, i_{k−1}} is such that i_{1} < ⋯ < i_{k−1}, and that the collection {I_{1}, …, I_{q}} of all such subsets is ordered lexicographically. Similarly, if J = {j_{1}, …, j_{n−k}}, then j_{1} < ⋯ < j_{n−k} and {J_{1}, …, J_{p}} is ordered lexicographically.

We shall use the notation e_{i} to denote the column vector with a 1 in the ith position and zeros elsewhere. Thus e_{1}, …, e_{n} denotes the canonical basis of column vectors for C^{n}, and for example e_{J_{1}}, …, e_{J_{p}} denotes the canonical basis of column vectors for C^{p}.

The space H_{n}^{k} is the linear span of matrices b_{i}^{n,k}, 1 ≤ i ≤ n, given by

b_{i}^{n,k} = ∑_{I∩J=∅,(I∪J)^{c}={i}} ϵ(I, i, J)e_{J,I},

where e_{J,I} = e_{J} ⊗ e_{I} = e_{J}e_{I}^{t} ∈ M_{p,q}(C) = B(C^{q}, C^{p}), and ϵ(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_{i}^{n,k} 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_{n}^{k} (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].

Lemma 1.1. Let X = [ξ_{ij}] be an n × m matrix. Let H ⊂ {1, …, n} and K ⊂ {1, …, m} both have cardinality r ≤ min{n, m}. Let X_{H,K} denote the corresponding r × r submatrix.

(i): If x_{i} = ∑_{j=1}^{n} ξ_{ji}e_{j} ∈ C^{n} for 1 ≤ p ≤ n, then x_{1}∧⋯∧x_{p} = ∑_{H} det X_{H,L}e_{H}, where the sum is over all H of cardinality p, L = {1, …, p}, and X is the n × p matrix [ξ_{ij}]. (Prop. 3.3, page 84 of [11])

1In [7], ϵ(I, i, J) is also denoted by ϵ(I, J). However, in this paper, ϵ(I, J) will denote the signature of the permutation taking (i_{1}, …, i_{k−1}, j_{1}, …, j_{n−k}) to (1, …, î, …, n)

(ii): If X is an n×n matrix and H ⊂ {1, …, n}, let H′ denote the complement of H. Then det X = ϵ(H, H′) ∑_{R} ϵ(R, R′) det X_{R,H} det X_{R′,H′}, where the sum is over all sets R having the same cardinality of H. (Prop. 3.4(1) page 87 of [11])

(iii): If H, K ⊂ {1, …, n} have cardinality r and n − r, and H ∩ K ≠ ∅, then ∑_{R} ϵ(R, R′) det X_{R,H} det X_{R′,K} = 0, where the sum is over all sets R having cardinality r. (Prop. 3.4(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 : C_{n} → H_{n}^{k} via ψ(e_{i}) = b_{i}^{n,k}, 1 ≤ i ≤ n. Let u = [ u_{1} ⋯ u_{n} ] be a unitary matrix so that u_{1}, …, u_{n} is an orthonormal basis for C_{n}. Then, with u_{i} = ∑_{j=1}^{n} u_{ji}e_{j}, we have

u = [ \begin{matrix}u_{11} & ⋯ & u_{1n} \\ ⋮ & ⋯ & ⋮ \\ u_{n1} & ⋯ & u_{nn} \end{matrix} ] ,

and

ψ(u_{i}) = ∑_{j=1}^{n} u_{ji}b_{j}^{n,k} = ∑_{j=1}^{n} u_{ji} ∑_{I∩J=∅,(I∪J)^{c}={i}} ϵ(I, i, J)e_{J,I}.

Lemma 1.2. The (J′, I′)-entry of the p × q matrix ψ(u_{i}) is given by

(1)  (ψ(u_{i}))_{J′,I′} = ∑_{I∩J=∅,(I∪J)^{c}=\{i\}} ϵ(I, i, J)\begin{matrix}det & \overline{u} & J′,J & det & \overline{u} & I′,I \\ det & \overline{u} \end{matrix} ,

where \overline{u} is the complex conjugate of u.

Proof. Let us first calculate the left side of (1):

(ψ(u_{i}))_{J′,I′} = e_{J′}^{t}ψ(u_{i})e_{I′}
= ∑_{j=1}^{n} u_{ji} ∑_{I∩J=∅,(I∪J)^{c}={j}} ϵ(I, j, J)e_{J′}^{t}e_{J}e_{I}^{t}e_{I′}

= \{ \begin{matrix}0 & J′ ∩ I′ ≠ ∅ \\ u_{li}ϵ(I′, l, J′) & J′ ∩ I′ = ∅, (I′ ∪ J′)^{c} = \{l\}. \end{matrix}

Before calculating the right side of (1), note that ϵ(I, i, J)ϵ(I, J) = (−1)^{i+k}; indeed,

ϵ(I, i, J)ϵ(I, J) = (−1)^{k−1}ϵ(i, I, J)ϵ(I, J)
= (−1)^{k−1}ϵ(I, J)ϵ(i, 1, 2, ⋯, î, ⋯, n − 1)ϵ(I, J)
= (−1)^{k−1}(−1)^{i−1} = (−1)^{k+i}.

Therefore, the right side of (1) is equal to

∑_{I∩J=∅,(I∪J)^{c}=\{i\}} ϵ(I, i, J)ϵ(I′, J′)ϵ(I′, J′)ϵ(I, J)ϵ(I, J)\begin{matrix}det & \overline{u} & J′,J & det & \overline{u} & I′,I \\ det & \overline{u} \end{matrix}

= (−1)^{k+i}ϵ(I′, J′) ∑_{I∩J=∅,(I∪J)^{c}=\{i\}} ϵ(I′, J′)ϵ(I, J)\begin{matrix}det & \overline{u} & J′,J & det & \overline{u} & I′,I \\ det & \overline{u} \end{matrix}

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

\frac{(−1)^{k−l}ϵ(I′, J′)(−1)^{i+l}}{det \overline{u}} ∑_{I∩J=∅,(I∪J)^{c}=\{i\}} ϵ(I′, J′)ϵ(I, J) det \overline{u}_{J′,J} det \overline{u}_{I′,I}.

Now by Lemma 1.1(ii), the above sum is the determinant of the (l, i)-minor of the matrix \overline{u}, call this \overline{M}_{li}. Thus, for J′ ∩ I′ = ∅, the right side of (1) is equal to

ϵ(I′, l, J′) [\frac{(−1)^{i+l} det \overline{M}_{li}}{det \overline{u}} ] = ϵ(I′, l, J′) × [ the (i, l)-entry of the inverse of \overline{u}]
= ϵ(I′, l, J′)u_{li}. □

Theorem 1. H_{n}^{k} is a homogeneous operator space.

Proof. Let α be a unitary operator on H_{n}^{k}. 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_{n}^{k} is the isometry defined by ψ(e_{i}) = b_{i}^{n,k}. 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 x ∈ H_{n}^{k}, where λ = det \overline{u}, w = [ ∧_{i∈I_{1}}u_{i} ⋯ ∧_{i∈I_{q}}u_{i} ], and

v = [ (∧_{j∈J_{1}}u_{j})^{t} ; ⋮ ; (∧_{j∈J_{p}}u_{j})^{t} ] .

(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_{i}^{n,k}. Thus it suffices to prove

(2)  vψ(u_{i})w = b_{i}^{n,k}/ det \overline{u}.

Let us first show that

(3)  ψ(u_{i}) = (det \overline{u})^{−1} ∑_{I∩J=∅,(I∪J)^{c}=\{i\}} ϵ(I, i, J) (∧_{j∈J} \overline{u}_{j}) (∧_{i∈I} \overline{u}_{i})^{t} .

By Lemma 1.2, the proof of (3) amounts to

(4)  [(∧_{j∈J}u_{j}) (∧_{i∈I}u_{i})^{t}]_{J′,I′} = det u_{J′,J} det u_{I′,I}.

The left side of (4) is given by e_{J′}^{t} (∧_{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_{J′}^{t} (∧_{j∈J}u_{j}) = ∑_{L} det u_{L,J}e_{J′}^{t}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} ⊗ \overline{e}_{I} = e_{J}(e_{I})^{t} = (∧_{j∈J}e_{j}) (∧_{i∈I}e_{i})^{t}, we may write

(5)  b_{i}^{n,k} = ∑_{I∩J=∅,(I∪J)^{c}={i}} ϵ(I, i, J) (∧_{j∈J}e_{j}) (∧_{i∈I}e_{i})^{t} .

By (3) and (5), it suffices to prove

v (∧_{j∈J} \overline{u}_{j}) (∧_{i∈I} \overline{u}_{i})^{t} w = (∧_{j∈J}e_{j}) (∧_{i∈I}e_{i})^{t} .

This is a simple calculation. Suppose for definiteness that J = J_{r} and I = I_{s}. Then v (∧_{j∈J} \overline{u}_{j}) = e_{J_{r}}, and (∧_{i∈I} \overline{u}_{i})^{t} w = e_{I_{s}}^{t}. □

Remark 1.3. In [7, page 2230], we defined an operator space construction denoted by Diag(H_{n}^{k_{1}}, …, H_{n}^{k_{m}}) which depended on a choice of orthonormal basis for each of the spaces H_{n}^{k_{j}}. Because of the homogeneity of the spaces H_{n}^{k} proved in Theorem 1, this space is independent of these choices up to complete isometry and is now seen to be the intersection H_{n}^{k_{1}} ∩ … ∩ H_{n}^{k_{m}} in the sense of operator space theory ([9, page 55]).

2. Anti-symmetric Fock spaces

Let C_{h}^{n,k} denote the wedge (or creation) operator from ∧^{k−1}C^{n} to ∧^{k}C^{n} given by

C_{h}^{n,k}(h_{1} ∧ ⋯ ∧ h_{k−1}) = h ∧ h_{1} ∧ ⋯ ∧ h_{k−1}.

Many properties of these classical operators on the full anti-symmetric Fock space are given in [5, Exercises 12.4.39-40].

As in section 1, let e_{1}, …, e_{n} be the usual column vector orthonormal basis for C^{n}, and let {e_{I_{1}}, …, e_{I_{q}}} and {e_{J_{1}}, …, e_{J_{p}}} be the column vector orthonormal bases for C^{q} and C^{p} respectively, and define the unitary operators U_{j}^{n} (j = k − 1 and j = n − k), W_{k}^{n}, V_{k}^{n} in the diagram below as follows:

• U_{k−1}^{n}(e_{I}) = e_{i_{1}} ∧ ⋯ ∧ e_{i_{k−1}}, where I = {i_{1} < ⋯ < i_{k−1}}.

• U_{n−k}^{n}(e_{J}) = e_{j_{1}} ∧ ⋯ ∧ e_{j_{n−k}}, where J = {j_{1} < ⋯ < j_{n−k}}.

• V_{k}^{n}(e_{i_{1}} ∧ ⋯ ∧ e_{i_{k}}) = e_{j_{1}} ∧ ⋯ ∧ e_{j_{n−k}}, where {j_{1} < ⋯ < j_{n−k}} is the complement of {i_{1} < ⋯ < i_{k}}.

• W_{k}^{n}(e_{j_{1}} ∧ ⋯ ∧ e_{j_{n−k}}) = ϵ(i, I)ϵ(I, i, J)e_{j_{1}} ∧ ⋯ ∧ 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).

C^{q}  ⟶^{b_{i}^{n,k}}  C^{p}
U_{k−1}^{n} ↓  ↓ U_{n−k}^{n}
∧^{k−1}C^{n}  ∧^{n−k}C^{n}
C_{e_{i}}^{n,k} ↓  ↓ W_{k}^{n}
∧^{k}C^{n}  ⟶^{V_{k}^{n}}  ∧^{n−k}C^{n}

Note that since b_{i}^{n,k} is a p × q matrix, it is viewed as an operator from C^{q} to C^{p}. In the definition of W_{k}^{n}, ϵ(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

ϵ(I, i, J) = ϵ(i_{1}, …, i_{k−1}, i, j_{1}, …, j_{n−k})
= (−1)^{k−1}ϵ(i, i_{1}, …, i_{k−1}, j_{1}, …, j_{n−k})
= (−1)^{k−1}ϵ(i, I)ϵ(i_{1}, …, i, …, i_{k−1}, j_{1}, …, j_{n−k}),

where i_{1} < ⋯ < i < ⋯ i_{k−1}. Similarly,

ϵ(I′, i′, J) = (−1)^{k−1}ϵ(i′, I′)ϵ(i′_{1}, …, i′, …, i′_{k−1}, j_{1}, …, j_{n−k}),

where i′_{1} < ⋯ < i′ < ⋯ i′_{k−1}. Hence, ϵ(i, I)ϵ(I, i, J) = ϵ(i′, I′)ϵ(I′, i′, J).

It is now a simple matter to check that W_{k}^{n}U_{n−k}^{n}b_{i}^{n,k} = V_{k}^{n}C_{e_{i}}^{n,k}U_{k−1}^{n}. Indeed, for any I′, W_{k}U_{n−k}^{n}b_{i}^{n,k}(e_{I′}) = W_{k}U_{n−k}^{n}ϵ(I′, i, J)(e_{J}) = W_{k}(ϵ(I′, i, J)e_{j_{1}} ∧ ⋯ ∧ e_{j_{n−k}} = ϵ(i, I′)e_{j_{1}} ∧ ⋯ ∧ e_{j_{n−k}}, and V_{k}^{n}C_{e_{i}}^{n,k}U_{k−1}^{n}(e_{I′}) = V_{k}^{n}(e_{i} ∧ e_{i′_{1}} ∧ ⋯ ∧ e_{i′_{k−1}}) = V_{k}^{n}(ϵ(i, I′)e_{i′_{1}} ∧ ⋯ ∧ e_{i} ∧ ⋯ ∧ e_{i′_{k−1}}) = ϵ(i, I′)e_{j_{1}} ∧ ⋯ ∧ e_{j_{n−k}}.

Hence, letting C^{n,k} denote the space sp{C_{e_{i}}^{n,k}} yields the following lemma.

Lemma 2.1. H_{n}^{k} 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_{n}^{k}. 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

(a): X = {pxq + (1 − p)x(1 − q) : x ∈ X};

(b): 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^{∗∗}.

Proof.

(a): Since P (A) is reflexive, X = P (A) = P^{∗∗}(A^{∗∗}) = pXq + (1 − p)X(1 − q),

the last equality following from [4, Prop. 4]; (b): By [7, Lemma 2.2]; (c): By Lemma 2.1 and [7, Prop. 2.6]; (d): For X this follows directly from [7, Corollary 2.8], and for pXq it follows

directly from [7, Theorem 3(b) and Corollary 7.3]. ^{2} (e): For x ∈ P (A), P^{∗∗}(pxq) ∈ P^{∗∗}(A^{∗∗}) = P (A), say P^{∗∗}(pxq) = y ∈ P (A) and by [4, Prop. 4], y = pyq + (1 − p)y(1 − q). Thus p(P^{∗∗}(pxq))q = pyq. Now it follows from [4, Prop. 1], that p(P^{∗∗}z)q = z for all z ∈ pP^{∗∗}(A^{∗∗})q. With z = pxq, we have pxq = pyq and by (b) x = y, proving that P^{∗∗}(pxq) = x. Moreover, P^{∗∗}((1 − p)x(1 − q)) = P^{∗∗}(x − pxq) = x − P^{∗∗}(pxq) = x − y = 0. □

2The authors wish to take this opportunity to point out the following correction to [7, Lemma 7.2 and Corollary 7.3]. The term \binom{n−1}{k−1}^{1/2} should be replaced by \binom{n−1}{k−1} in the statements of Lemma 7.2 and Corollary 7.3, and in the proof of Lemma 7.2. Accordingly, in the proof of Corollary 7.3, m^{1/2} should be replaced by m.

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.

C_{h}^{n,k∗}C_{h}^{n,k}(h_{1} ∧ ⋯ ∧ h_{k−1}) = (h|h)h_{1} ∧ ⋯ ∧ h_{k−1} − (h_{1}|h)h ∧ h_{2} ∧ ⋯ ∧ h_{k−1}

+ ⋯ ± (h_{k−1}|h)h ∧ h_{1} ∧ ⋯ ∧ h_{k−2}. (6)

and

C_{h}^{n,k}C_{h}^{n,k∗}(h_{1} ∧ ⋯ ∧ h_{k}) = ∑_{j=1}^{k} (h_{j}|h)h_{1} ∧ ⋯ ∧ h′_{j} ∧ ⋯ ∧ h_{k} (h′_{j} = h).

In particular, C_{h}^{n,1}C_{h}^{n,1∗} = h ⊗ \overline{h}, for h ∈ C^{n}.

Lemma 2.3. (a): tr(C_{h}^{n,k∗}C_{h}^{n,k}) = \binom{n−1}{k−1}∥h∥^{2}. In particular, C_{h}^{n,1∗}C_{h}^{n,1} =

∥h∥^{2}. (b): Let the (repeated) eigenvalues of ∑_{i=1}^{m} C_{h_{i}}^{n,1}C_{h_{i}}^{n,1∗} be λ_{1} ≥ ⋯ ≥ λ_{n}. Then the eigenvalues of ∑_{i=1}^{m} C_{h_{i}}^{n,k}C_{h_{i}}^{n,k∗} are precisely the sums of k eigen-

values of ∑_{i=1}^{m} C_{h_{i}}^{n,1}C_{h_{i}}^{n,1∗}.

Proof. In the first place, we have

tr(C_{h}^{n,k}C_{h}^{n,k∗}) = tr(C_{h}^{n,k∗}C_{h}^{n,k})
= ∑_{i_{1},...,i_{k−1}} (C_{h}^{n,k∗}C_{h}^{n,k}(e_{i_{1}} ∧ ⋯ ∧ e_{i_{k−1}})|e_{i_{1}} ∧ ⋯ ∧ e_{i_{k−1}})
= ∑_{i_{1},...,i_{k−1}} (h ∧ e_{i_{1}} ∧ ⋯ ∧ e_{i_{k−1}}|h ∧ e_{i_{1}} ∧ ⋯ ∧ e_{i_{k−1}})

= ∑_{i_{1},...,i_{k−1}} det [ \begin{matrix}(h|h) & (h|e_{i_{1}}) & ⋯ & (h|e_{i_{k−1}}) \\ (e_{i_{1}}|h) & 1 & ⋯ & 0 \\ ⋯ & ⋯ & ⋯ & ⋯ \\ (e_{i_{k−1}}|h) & 0 & ⋯ & 1 \end{matrix} ]

= ∑_{i_{1},...,i_{k−1}} (∥h∥^{2} − ∑_{j=1}^{k−1} |(h|e_{i_{j}})|^{2})

= \binom{n}{k − 1}∥h∥^{2} − ∑_{i_{1},...,i_{k−1}} ∑_{j=1}^{k−1} |(h|e_{i_{j}})|^{2}

= \binom{n}{k − 1}∥h∥^{2} − ∑_{l=1}^{n} \binom{n − 1}{k − 2}|(h|e_{l})|^{2}

= (\binom{n}{k − 1} − \binom{n − 1}{k − 2}) ∥h∥^{2} = \binom{n − 1}{k − 1}∥h∥^{2}.

To prove the second statement, let ξ_{1}, …, ξ_{n} be an orthonormal basis of C^{n} consisting of eigenvectors of ∑_{i=1}^{m} C_{h_{i}}^{n,1}C_{h_{i}}^{n,1∗} so that ∑_{i} (ξ_{k}|h_{i})h_{i} = λ_{k}ξ_{k}. Then

∑_{i=1}^{m} C_{h_{i}}^{n,k}C_{h_{i}}^{n,k∗}(ξ_{i_{1}} ∧ ⋯ ∧ ξ_{i_{k}}) = ∑_{i} ∑_{j=1}^{k} (ξ_{i_{j}}|h_{i})ξ_{i_{1}} ∧ ⋯ ∧ h_{i} ∧ ⋯ ∧ ξ_{i_{k}}

= ∑_{j} ξ_{i_{1}} ∧ ⋯ ∧ [∑_{i} (ξ_{i_{j}}|h_{i})h_{i}] ∧ ⋯ ξ_{i_{k}}

= [∑_{j} λ_{i_{j}} ] ξ_{i_{1}} ∧ ⋯ ∧ ξ_{i_{k}}.

Conversely, if ξ = ∑ α_{i_{1},...,i_{k}}ξ_{i_{1}} ∧ ⋯ ∧ ξ_{i_{k}} is an eigenvector of ∑_{i=1}^{m} C_{h_{i}}^{n,k}C_{h_{i}}^{n,k∗}, with eigenvalue λ, then

λ ∑ α_{i_{1},...,i_{k}}ξ_{i_{1}} ∧ ⋯ ∧ ξ_{i_{k}} =
= ∑_{i=1}^{m} C_{h_{i}}^{n,k}C_{h_{i}}^{n,k∗} ∑ α_{i_{1},...,i_{k}}ξ_{i_{1}} ∧ ⋯ ∧ ξ_{i_{k}}
= ∑ α_{i_{1},...,i_{k}} ∑_{i} ∑_{j} (ξ_{i_{j}}|h_{i})ξ_{i_{1}} ∧ ⋯ ∧ h_{i} ∧ ⋯ ∧ ξ_{i_{k}}

= ∑ α_{i_{1},...,i_{k}} ∑_{j} ξ_{i_{1}} ∧ ⋯ ∧ [∑_{i=1}^{m} C_{h_{i}}^{n,1}C_{h_{i}}^{n,1∗}ξ_{i_{j}}] ∧ ⋯ ∧ ξ_{i_{k}}

= ∑ α_{i_{1},...,i_{k}} [∑_{j} λ_{i_{j}} ] ξ_{i_{1}} ∧ ⋯ ∧ ξ_{i_{k}}.

From this, the second statement follows. □

3. Completely bounded Banach-Mazur distance

Recall that the completely bounded Banach-Mazur distance between two (completely isomorphic) operator spaces E, F is defined by

dcb(E, F) = inf{∥u∥cb∥u^{−1}∥cb : u : E → F complete isomorphism }.

We shall explicitly compute dcb(H_{n}^{k}, H_{n}^{1}). By Lemma 2.1, we can identify H_{n}^{k} with C^{n,k}. For a fixed k, let ψ : H_{n}^{1} → H_{n}^{k} be the isometry given by ψ(C_{e_{i}}^{n,1}) = C_{e_{i}}^{n,k}. Recall from [6, Prop. 4] that, since ψ is a mapping from the column Hilbert space H_{n}^{1}, ∥ψ∥cb = ∥ψ∥row-cb, where ∥ψ∥row-cb := sup{∥(ψ(h_{1}), …, ψ(h_{m}))∥} where the supremum is extended over all m ≥ 1 and all row vectors with ∥(h_{1}, …, h_{m})∥ ≤ 1. The norm ∥ψ∥col-cb is defined analogously and by [6, Prop. 2], ∥ψ^{−1}∥cb = ∥ψ^{−1}∥col-cb.

Lemma 3.1. ∥ψ∥row-cb = \sqrt{k} and ∥ψ^{−1}∥col-cb = \sqrt{\frac{n}{n−k+1}}.

Proof. Let A = (C_{h_{1}}^{n,1}, …, C_{h_{m}}^{n,1}) and B = (C_{h_{1}}^{n,k}, …, C_{h_{m}}^{n,k}). We show first that ∥B∥ ≤ \sqrt{k} ∥A∥. We have AA^{∗} = ∑_{i=1}^{m} C_{h_{i}}^{n,1}C_{h_{i}}^{n,1∗} = ∑_{i=1}^{m} h_{i} ⊗ \overline{h_{i}} and BB^{∗} = ∑_{i=1}^{m} C_{h_{i}}^{n,k}C_{h_{i}}^{n,k∗}.

Let the (repeated) eigenvalues of AA^{∗} be λ_{1} ≥ ⋯ ≥ λ_{n}. By Lemma 2.3(b), we have

∥B∥^{2} = ∥BB^{∗}∥ = λ_{1} + ⋯ + λ_{k} ≤ kλ_{1} = k∥AA^{∗}∥ = k∥A∥^{2}.

Taking m = n and h_{j} = e_{j}, we have AA^{∗} = I and BB^{∗} = kI, proving the first statement of the Lemma.

Let D = (C_{h_{1}}^{n,1}, …, C_{h_{m}}^{n,1})^{t} and C = (C_{h_{1}}^{n,k}, …, C_{h_{m}}^{n,k})^{t}. We show next that ∥D∥ ≤

\sqrt{\frac{n}{n−k+1}}∥C∥. By Lemma 2.3(a), we have D^{∗}D = ∑_{i=1}^{m} C_{h_{i}}^{n,1∗}C_{h_{i}}^{n,1} = ∑_{i=1}^{m} ∥h_{i}∥^{2}

and C^{∗}C = ∑_{i=1}^{m} C_{h_{i}}^{n,k∗}C_{h_{i}}^{n,k}. Since C^{∗}C is a square matrix of size \binom{n}{k−1}, again by Lemma 2.3(a),

∥C^{∗}C∥\binom{n}{k − 1} ≥ tr(C^{∗}C) = \binom{n − 1}{k − 1} ∑_{i=1}^{m} ∥h_{i}∥^{2}.

Therefore,

\frac{∥D∥^{2}}{∥C∥^{2}} = \frac{∑_{i=1}^{m} ∥h_{i}∥^{2}}{∥C^{∗}C∥} ≤ \frac{∑_{i=1}^{m} ∥h_{i}∥^{2}}{\begin{matrix}( & {n−1 \atop k−1} & ) \\ ( & {n \atop k−1} & ) \end{matrix} ∑_{i=1}^{m} ∥h_{i}∥^{2}} = \frac{n}{n − k + 1}.

Taking m = n and h_{i} = e_{i}, we have D^{∗}D = n. By (6), C_{e_{i}}^{n,k∗}C_{e_{i}}^{n,k}(e_{i_{1}} ∧ ⋯ ∧ e_{i_{k−1}}) = 0 if i ∈ {i_{1}, …, i_{k−1}} and equal to e_{i_{1}} ∧ ⋯ ∧ e_{i_{k−1}} otherwise. Hence C^{∗}C = (n − k + 1)I, proving the second statement. □

Theorem 3. dcb(H_{n}^{k}, H_{n}^{1}) = \sqrt{\frac{kn}{n−k+1}}, for 1 ≤ k ≤ n.

Proof. By [12, Theorem 3.1], and the first paragraph of its proof, dcb(H_{n}^{k}, H_{n}^{1}) = ∥ψ∥cb∥ψ^{−1}∥cb. Now apply Lemma 3.1 and the remarks just preceding it. □

Not surprisingly, we obtain the result published first by Mathes ([6, Prop. 7], [9, p. 21]).

Corollary 3.2. dcb(R_{n}, C_{n}) = n.

Symmetry considerations in Theorem 3 suggest dcb(H_{n}^{k}, H_{n}^{n}) = dcb(H_{n}^{n−k+1}, H_{n}^{1}), which can be proved by exactly the same methods. Hence we obtain the following, which is the answer to Problem 1 in [7].

Corollary 3.3. dcb(H_{n}^{k}, H_{n}^{n}) = \sqrt{\frac{(n−k+1)n}{k}}, for 1 ≤ k ≤ n.

Remark 3.4. It is curious to note that by Theorem 3 and its two corollaries, dcb(H_{n}^{k}, H_{n}^{n})dcb(H_{n}^{k}, H_{n}^{1}) = dcb(H_{n}^{n}, H_{n}^{1}), so that in the metric space of all operator spaces of dimension n ([9, page 335]), the three points H_{n}^{k}, H_{n}^{1}, H_{n}^{n} form a degenerate triangle.

Since H_{n}^{k} is distinct from row or column Hilbert space, new ideas will be needed to solve the following problem.

Problem 1. Find dcb(H_{n}^{k_{1}}, H_{n}^{k_{2}}) 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_{n}^{1} 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].

REPRESENTATION OF HILBERTIAN OPERATOR SPACES ON THE FOCK SPACE 11

Problem 2. Find the operator space dual of H_{n}^{k}.

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?

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Department of Mathematics, Denison University, Granville, Ohio 43023

E-mail address: nealm@denison.edu

Department of Mathematics, University of California, Irvine, California 92697-3875

E-mail address: brusso@math.uci.edu