A NON-EXTENDABLE BOUNDED LINEAR MAP BETWEEN C^{∗}-ALGEBRAS

NARUTAKA OZAWA

Abstract. We present an example of a C^{∗}-subalgebra A of 𝔹(H) and a bounded linear map from A to 𝔹(K) which does not admit any bounded linear extension. This generalizes the result of Robertson and gives the answer to a problem raised by Pisier. Using the same idea, we compute the exactness constants of some Q-spaces. This solves a problem raised by Oikhberg. We also construct a Q-space which is not locally reflexive.

1. A non-extendable bounded linear map

Definition. Let X be a closed subspace of a Banach space Y. We say X is complemented in Y if there is a bounded linear projection from Y onto X. We say X is weakly complemented in Y if there is a bounded linear map T from Y to X^{∗∗} such that T |_{X} = ι_{X}, where ι_{X} is the canonical inclusion map of X into X^{∗∗}.

Let A be a C^{∗}-algebra. If there is a faithful ∗-representation A ⊂ 𝔹(H) such that A is (weakly) complemented in 𝔹(H), then it follows from the injectivity of 𝔹(H) that for any faithful ∗-representation A ⊂ 𝔹(K), A is (weakly) complemented in 𝔹(K). A C^{∗}-algebra A has the WEP [La] if for every faithful ∗-representation A ⊂ 𝔹(H), there is a complete contraction T: 𝔹(H) → A^{∗∗} such that T |_{A} = ι_{A}.

Lemma 1. Let A be a C^{∗}-algebra and let A ⊂ 𝔹(H) be the universal representation, i.e, \overline{A}^{ultraweak} = A^{∗∗}. If A is weakly complemented in 𝔹(H) and is locally reflexive, then A^{∗∗} is complemented in 𝔹(H).

Proof. Let T: 𝔹(H) → A^{∗∗} be a bounded linear map such that T |_{A} = id_{A}. Let I be a set of all pairs i = (E, F) consisting of finite dimensional subspaces E of A^{∗∗} and F of 𝔹(H)_{∗}. Then, I is directedly ordered by inclusions. Fix i = (E, F) in I. Since A is locally reflexive, there is a map S_{i}: E → A with ∥S_{i}∥_{cb} ≤ 1 + \frac{1}{dim(E)} such that ⟨S_{i}(e), f⟩ = ⟨e, f⟩ for all e ∈ E and f ∈ F. Since 𝔹(H) is injective, we can extend S_{i} to S̄_{i}: 𝔹(H) → 𝔹(H) with ∥S̄_{i}∥_{cb} ≤ ∥S_{i}∥_{cb}. Define T_{i}: 𝔹(H) → A^{∗∗} by T_{i} = T ∘ S̄_{i}. Then, we have lim sup ∥T_{i}∥ ≤ ∥T ∥ and lim⟨T_{i}(e), f⟩ = ⟨e, f⟩ for all e ∈ A^{∗∗} and f ∈ 𝔹(H)_{∗}. Let T̃: 𝔹(H) → A^{∗∗} be a cluster point of the net {T_{i}}_{i} in the point-ultraweak topology. Then, T̃ is the desired bounded linear projection.

We use next results due to De Cannière and Haagerup and due to Kirchberg. Let C_{λ}^{∗}(𝔽_{n}) be the reduced group C^{∗}-algebra of the free group 𝔽_{n} with n generators (n ≥ 2). Then, by [DCH], C_{λ}^{∗}(𝔽_{n}) has the complete metric approximation property. Thus, C_{λ}^{∗}(𝔽_{n}) is exact and a fortiori is locally reflexive [K1]. By [K2], there are a C^{∗}-algebra B with the WEP and a surjective ∗-homomorphism π from B onto C_{λ}^{∗}(𝔽_{n}). Since C_{λ}^{∗}(𝔽_{n}) has the metric approximation property, there is a contractive linear lifting φ: C_{λ}^{∗}(𝔽_{n}) → B, i.e., π ∘ φ = id_{C_{λ}^{∗}(𝔽_{n})}. (There is even a unital k-positive lifting for each k ∈ ℕ, see [RS].) We need one more ingredient due to Haagerup and Pisier. Let V N(𝔽_{n}) be the group von Neumann algebra of the free group 𝔽_{n} with n generators (n ≥ 2) and let V N(𝔽_{n}) ⊂ 𝔹(K) be any faithful ∗-representation. Then, V N(𝔽_{n}) is not complemented in 𝔹(K) (Corollary 4.9 in [HP]).

Lemma 2. Let C_{λ}^{∗}(𝔽_{n}) ⊂ 𝔹(H) be any faithful ∗-representation. Then C_{λ}^{∗}(𝔽_{n}) is not weakly complemented in 𝔹(H).

Proof. If C_{λ}^{∗}(𝔽_{n}) is weakly complemented in 𝔹(H), then by Lemma 1 and the preceding remarks, C_{λ}^{∗}(𝔽_{n})^{∗∗} is complemented in 𝔹(K) for any faithful ∗-representation C_{λ}^{∗}(𝔽_{n})^{∗∗} ⊂ 𝔹(K). Since the group von Neumann algebra V N(𝔽_{n}) is complemented in C_{λ}^{∗}(𝔽_{n})^{∗∗}, a fortiori it is complemented in 𝔹(K). This contradicts Corollary 4.9 in [HP].

Date: April 16, 2000.

The author is supported by JSPS.

Theorem 1. Let C_{λ}^{∗}(𝔽_{n}) ⊂ 𝔹(H) be a faithful ∗-representation. Let B ⊂ 𝔹(K) be a C^{∗}-subalgebra with the WEP and π be a surjective ∗-homomorphism from B onto C_{λ}^{∗}(𝔽_{n}). If φ: C_{λ}^{∗}(𝔽_{n}) → B is a bounded linear lifting of π, then there is no bounded linear extension φ̄: 𝔹(H) → 𝔹(K) of φ.

Proof. Suppose that there is a bounded linear extension φ̄: 𝔹(H) → 𝔹(K) of φ. Since B has the WEP, there is a bounded linear map ψ: 𝔹(K) → B^{∗∗} such that ψ|_{B} = ι_{B}. Let us define T: 𝔹(H) → C_{λ}^{∗}(𝔽_{n})^{∗∗} by T = π^{∗∗} ∘ ψ ∘ φ̄. Then, T is a bounded linear map and T |_{C_{λ}^{∗}(𝔽_{n})} = ι_{C_{λ}^{∗}(𝔽_{n})}. This contradicts Lemma 2.

Remark. Since C_{λ}^{∗}(𝔽_{n}) is weakly complemented in V N(𝔽_{n}) (see the proof of Lemma 7.6 in [K2]), there is a bounded linear map from V N(𝔽_{n}) to 𝔹(K) without bounded linear extension to 𝔹(ℓ_{2}(𝔽_{n})).

Problem 1. Let A is a C^{∗}-subalgebra of 𝔹(H) and assume that any bounded linear map from A to 𝔹(H) extends to a bounded linear map on 𝔹(H). Is A weakly complemented in 𝔹(H)?

2. Computing exactness constants of some Q-spaces

We will compute the exactness constant of some Q-spaces. A Q-space is a quotient operator space of a minimal operator space. By the duality between maximal and minimal operator spaces, a dual operator space of a subspace of a maximal operator space is a Q-space. See [B] and [BP] for details.

Let E be a finite dimensional operator space. For any C^{∗}-algebra B and any closed two-sided ideal J in B, there is a canonical isomorphism T_{E}: (E ⊗ B)/(E ⊗ J) → E ⊗ (B/J), where ⊗ means the minimal tensor product. Let C be a constant. We say E is C-exact if ∥T_{E}^{−1}∥ ≤ C for all choices of B and J. By the canonical isometric identification E ⊗ X = CB(E^{∗}, X) for an operator space X (see [BP, ER1]), E is C-exact if any complete contraction from the dual operator space E^{∗} to any quotient C^{∗}-algebra B/J has a lifting with cb-norm ≤ C. For an infinite dimensional operator space X, we say X is C-exact if every finite dimensional operator subspace of X is C-exact and we say X is exact if it is C-exact for some constant C. The exactness constant ex(X) of X is defined by ex(X) = inf{C : X is C-exact}. See [P2] for details.

Define an operator space E_{n} ⊂ 𝕄_{n} ⊕ 𝕄_{n} ⊂ 𝕄_{2n} by E_{n} = span{e_{k1} ⊕ e_{1k} : k = 1, 2, …, n}, where {e_{jk}} is a standard matrix unit in 𝕄_{n}. By Proposition 1.3 in [HP], there are two maps w: E_{n} → C_{λ}^{∗}(𝔽_{n}) and v: C_{λ}^{∗}(𝔽_{n}) → E_{n} such that v ∘ w = id_{E_{n}} and ∥v∥_{cb} ≤ 2, ∥w∥_{cb} ≤ 1. By Lemma 4.2 in [HP], any projection P from 𝕄_{2n} onto E_{n} has cb-norm ≥ \frac{1}{2}(1 + \sqrt{n}). By Smith’s lemma (Theorem 2.1 in [S]), we have ∥P ∥_{cb} = ∥P ∥_{2n} for any map P: 𝕄_{2n} → E_{n}. Hence, by a standard averaging argument (see [RS, HP]), we have ∥P ∥ ≥ \frac{1}{2}(1 + \sqrt{n}) for any projection P from 𝕄_{2n}(𝕄_{2n}) onto 𝕄_{2n}(E_{n}). On the other hand, by Remark 4.3 in [HP], there is a projection Q from 𝕄_{2n}(𝕄_{2n}) onto 𝕄_{2n}(E_{n}) with ∥Q∥ = \frac{1}{2}(1 + \sqrt{n}). See [HP] for details.

Theorem 2. We equip 𝕄_{2n}(E_{n}) with a new operator space structure induced by the canonical embedding into MAX(𝕄_{2n}(𝕄_{2n})) and denote the resultant operator space by F_{n}, i.e., F_{n} = 𝕄_{2n}(E_{n}) as a Banach space and F_{n} ⊂ MAX(𝕄_{2n}(𝕄_{2n})) as an operator space. Then, F_{n}^{∗} is a 4n^{3}-dimensional Q-space such that

\frac{1}{4}(1 + \sqrt{n}) ≤ ex(F_{n}^{∗}) ≤ ∥id: MIN(F_{n}^{∗}) → F_{n}^{∗}∥_{cb} ≤ \frac{1}{2}(1 + \sqrt{n}).

Proof. We note that the formal identity J: F_{n} → 𝕄_{2n}(E_{n}) is completely contractive. Let v and w be as in the preceding remarks and let w̃: F_{n} → 𝕄_{2n}(C_{λ}^{∗}(𝔽_{n})) be a complete contraction defined by w̃ = (id_{𝕄_{2n}} ⊗ w) ∘ J. By [K2], there are a C^{∗}-algebra B with the WEP and a surjective ∗-homomorphism π from B onto 𝕄_{2n}(C_{λ}^{∗}(𝔽_{n})). Suppose that F_{n}^{∗} is C-exact. By definition, there is a lifting φ: F_{n} → B of w̃ with ∥φ∥_{cb} ≤ C. Since B has the WEP, φ extends to φ̄: MAX(𝕄_{2n}(𝕄_{2n})) → B^{∗∗} with ∥φ̄∥_{cb} ≤ C. Let us define P: 𝕄_{2n}(𝕄_{2n}) → 𝕄_{2n}(E_{n}) by P = (id_{𝕄_{2n}} ⊗ v)^{∗∗} ∘ π^{∗∗} ∘ φ̄. Then, P is a projection with ∥P ∥ ≤ 2C. Thus, we have C ≥ \frac{1}{4}(1+\sqrt{n}). This proves the first inequality. Since ex(MIN(F_{n}^{∗})) = 1, we have the second inequality.

Next, let Q be a projection as the preceding remarks. Then, we have

∥id: F_{n} → MAX(F_{n})∥_{cb} ≤ ∥Q: MAX(𝕄_{2n}(𝕄_{2n})) → MAX(F_{n})∥_{cb} = ∥Q∥ = \frac{1}{2}(1 + \sqrt{n}).

Taking the dual of this identity map, we obtain the third inequality.

Since an ℓ_{∞}-sum of Q-spaces is also a Q-space, we obtain the next corollary.

Corollary. There is a Q-space which is not exact.

Problem 2. What is the asymptotic behavior of the constant

sup{ex(E) : E an n-dimensional Q-space}

as n tends to infinity?

3. A Q-space which is not locally reflexive

Definition. Let C ≥ 1 be a constant. We say an operator space X is C-locally reflexive if for any finite dimensional subspaces E ⊂ X^{∗∗} and F ⊂ X^{∗} and any ε > 0, there is a map φ: E → X with ∥φ∥_{cb} < C + ε such that ⟨φ(e), f⟩ = ⟨e, f⟩ for all e ∈ E and f ∈ F. We say an operator space X is locally reflexive if X is C-locally reflexive for some constant C.

We note that a C^{∗}-algebra is 1-locally reflexive if it is locally reflexive and that a subspace of a C-locally reflexive operator space is also C-locally reflexive. See [ER2] for details. Now, let us construct a Q-space which is not locally reflexive. First, we need a lemma due to Oikhberg. Let us recall that the cb version of the Banach-Mazur distance between two completely isomorphic operator spaces X and Y is defined by

d_{cb}(X, Y) = inf{∥φ∥_{cb}∥φ^{−1}∥_{cb} : φ a completely bounded isomorphism from X onto Y }.

Put d_{cb}(X, Y) = ∞ if X and Y are not completely isomorphic.

Lemma 3 (Lemma 3.4 in [O]). For every C′ > 0, there is a finite dimensional subspace F ⊂ MAX(𝔹(H)) such that d_{cb}(F, G) > C′ for all n and G ⊂ MAX(𝕄_{n}).

Taking the dual of the inclusion F ⊂ MAX(𝔹(H)), we obtain a complete metric surjection q: MIN(S_{1}) → E, where E = F^{∗}. By the above lemma and a small perturbation argument, we obtain

Lemma 4. For every C′ > 0, there is a finite rank complete metric surjection q: MIN(S_{1}) → E such that d_{cb}(E, F/(ker q ∩ F)) > C′ for all finite dimensional subspace F ⊂ MIN(S_{1}).

Now, we prove

Lemma 5. For every C > 0 there is a Q-space which is not C-locally reflexive.

Proof. Fix C′ > C and take a finite rank complete metric surjection q: MIN(S_{1}) → E as in Lemma 4. Let \{F_{n}\} be an increasing sequence of finite dimensional subspaces of MIN(S_{1}) such that \overline{⋃ F_{n}} = MIN(S_{1}) and q(F_{1}) = E. Let E_{n} = F_{n}/(ker q ∩ F_{n}) and let φ_{n}: E_{n} → E be the complete contraction induced by q|_{F_{n}}: F_{n} → E. By Lemma 4, we have ∥φ_{n}^{−1}∥_{cb} > C′ for all n. On the other hand, it can be seen that lim_{n→∞}∥φ_{n}^{−1}∥_{k} = 1 for all k ∈ ℕ. Finally, let X be an operator space defined by

X = {(x_{n}) ∈ (∏ E_{n})_{ℓ_{∞}} : lim_{n→∞} φ_{n}(x_{n}) exists in E }.

Since all E_{n}’s are Q-spaces, X is also a Q-space. We will show that X is not C-locally reflexive. There is a natural map φ: X → E defined by φ((x_{n})) = lim φ_{n}(x_{n}). For each n, define ψ_{n}: E → X by

ψ_{n}(x) = (0, …, 0, φ_{n}^{−1}(x), φ_{n+1}^{−1}(x), …).

Since φ_{m}^{−1} ∘ φ_{n}: E_{n} → E_{m} is completely contractive for all m ≥ n, we have

lim_{n→∞} ∥ψ_{n}∥_{k} = lim_{n→∞} ∥φ_{n}^{−1}∥_{k} = 1

for all k ∈ ℕ. Let ψ: E → X^{∗∗} be a cluster point of the sequence {ψ_{n}}_{n} in the point-weak^{∗} topology. Then, by the previous argument, we have ∥ψ∥_{cb} ≤ 1. Since φ ∘ ψ_{n} = id_{E} for all n, we have φ^{∗∗} ∘ ψ = id_{E}. Now, suppose that X is C-locally reflexive. Since φ: X → E is finite rank, applying the local reflexivity to the complete contraction ψ: E → X^{∗∗}, we obtain a map θ: E → X with ∥θ∥_{cb} < C′ such that φ ∘ θ = id_{E}. Let θ_{n}: E → E_{n} be the “n-th coordinate” of θ. Then, by the definition of φ, we have

lim_{n→∞} φ_{n} ∘ θ_{n}(x) = x

for all x ∈ E. Since E is finite dimensional, we have

lim sup_{n→∞} ∥φ_{n}^{−1}∥_{cb} ≤ lim sup_{n→∞}(∥θ_{n}∥_{cb} + dim(E)∥φ_{n}^{−1} − θ_{n}∥) ≤ ∥θ∥_{cb} < C′.

This contradicts the choice of E.

Theorem 3. There is a Q-space which is not locally reflexive.

Proof. For each n, there is a Q-space X_{n} which is not n-locally reflexive by Lemma 5. Define a Q-space Y by Y = (⨁ X_{n})_{c_{0}}. It is easy to see that Y is not locally reflexive.

Problem 3. In the proof of Lemma 5, it can be seen that ex(X) = sup ex(E_{n}). Can we control this value?

4. Crude representability and local reflexivity

Let (⨁ X_{n})_{ℓ_{1}} be the ℓ_{1}-direct sum of a sequence {X_{n}} of operator spaces. We equip (⨁ X_{n})_{ℓ_{1}} with the natural operator space structure (see p34–36 in [P3]). Then, (⨁ X_{n})_{ℓ_{1}} is an operator space with the following properties. If φ_{n}: X_{n} → 𝔹(H) is a complete contraction for all n, then

φ: (⨁ X_{n})_{ℓ_{1}} ∋ (x_{n}) ↦ ∑ φ_{n}(x_{n}) ∈ 𝔹(H)

is a complete contraction. We have a completely isometric identity (⨁ X_{n})_{ℓ_{1}}^{∗} = (∏ X_{n}^{∗})_{ℓ_{∞}}. And if Y_{n} ⊂ X_{n} for all n, then we have (⨁ Y_{n})_{ℓ_{1}} ⊂ (⨁ X_{n})_{ℓ_{1}} completely isometrically. When X_{n} = X for all n, we simply denote (⨁ X_{n})_{ℓ_{1}} by ℓ_{1}(X).

Lemma 6. Let X be a separable operator space and let {E_{n}} be an increasing sequence of finite dimensional subspaces of X such that \overline{⋃ E_{n}} = X. If Y = (⨁ E_{n})_{ℓ_{1}} is C-locally reflexive, then so is X.

Proof. We follow the construction due to Johnson [Jo]. Let q: Y → X be a complete metric surjection defined by q((x_{n})) = ∑_{n=1}^{∞} x_{n}. Fix a free ultrafilter U on ℕ and define a map r: Y^{∗} → X^{∗} by

⟨r(f), x⟩ = lim_{U}⟨f_{n}, x⟩

for all f = (f_{n}) ∈ Y^{∗} = (∏ E_{n}^{∗})_{ℓ_{∞}} and x ∈ ⋃ E_{n}. Then, r is a well defined complete contraction and r ∘q^{∗} = id_{X^{∗}}. To prove that X is C-locally reflexive, we give ourselves finite dimensional subspaces E ⊂ X^{∗∗} and F ⊂ X^{∗} and ε > 0. Let Ẽ = r^{∗}(E) ⊂ Y^{∗∗} and let F̃ = q^{∗}(F) ⊂ Y^{∗}. Since Y is C-locally reflexive, there is a map φ: Ẽ → Y with ∥φ∥_{cb} ≤ C + ε such that ⟨φ(ẽ), f̃⟩ = ⟨ẽ, f̃⟩ for all ẽ ∈ Ẽ and f̃ ∈ F̃. Now, define ψ: E → X by ψ = q ∘ φ ∘ (r^{∗}|_{E}). Then, we have ∥ψ∥_{cb} ≤ C + ε and

⟨ψ(e), f⟩ = ⟨φ(r^{∗}(e)), q^{∗}(f)⟩
= ⟨r^{∗}(e), q^{∗}(f)⟩
= ⟨e, r ∘ q^{∗}(f)⟩
= ⟨e, f⟩

for all e ∈ E and f ∈ F. This completes the proof.

Lemma 7 (Theorem 4.3 in [EOR]). An operator space X is locally reflexive if every separable subspace of X is locally reflexive.

Proof. The “isometric” version of this lemma has been proved in [EOR]. Observe that the assumption implies that there is a constant C so that every separable subspace of X is C-locally reflexive. Now the proof of C-local reflexivity of X is almost same as the proof of Theorem 4.3 in [EOR].

Let Z and X be operator spaces. We say X is crudely representable in Z if there is a constant C such that for any finite dimensional subspace E of X, there is a subspace F of Z with d_{cb}(F, E) < C.

Theorem 4. Let Z be an operator space such that Z contains a completely isomorphic copy of ℓ_{1}(Z). Assume that Z is locally reflexive. If X is an operator space which is crudely representable in Z, then X is locally reflexive.

Proof. By Lemma 7, we may assume that X is separable. Take an increasing sequence {E_{n}} of finite dimensional subspaces of X with \overline{⋃ E_{n}} = X. Since X is crudely representable in Z, Y = (⨁ E_{n})_{ℓ_{1}} can be embedded into ℓ_{1}(Z) completely isomorphically. Since a subspace of locally reflexive operator space is also locally reflexive, by Lemma 6, we are done.

Remark. In [Ju], Junge has proved that the operator space S_{1} of trace class operators satisfies the assumption of Theorem 4 and the consequence is already known [EJR]. There is a locally reflexive operator space X such that ℓ_{1}(X) is not locally reflexive. Indeed, if V is a separable operator space which is not locally reflexive and \{E_{n}\} is an increasing sequence of finite dimensional subspaces with \overline{⋃ E_{n}} = V, then X = (⨁ E_{n})_{c_{0}} is locally reflexive, but ℓ_{1}(X) is not locally reflexive. This answers a question raised by Le Merdy (private communication). There is an “ℓ_{∞}-version” of Theorem 4 (use Lusky’s construction [Lu] at Lemma 6), but it seems that for only few operator spaces X, ℓ_{∞}(X) is locally reflexive.

Problem 4. Are exact operator spaces necessarily locally reflexive?

It has been shown in Corollary 4.8 in [EOR] that every 1-exact operator space is 1-locally reflexive.

Acknowledgment. The author would like to thank Professor Gilles Pisier for introducing him to the problem and a stimulating conversation.

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Texas A&M University, College Station, TX77843, USA and University of Tokyo, Komaba, 153-8914, Japan

   E-mail address: ozawa@math.tamu.edu