DESCRIPTIVE COMPLEXITY OF SOME ISOMORPHISM CLASSES OF BANACH SPACES.

GILLES GODEFROY - JEAN SAINT-RAYMOND

Abstract. We present a topological frame in which it is possible to estimate the complexity of some Borel families of separable Banach spaces. We use this frame for evaluating the complexity of the isomorphism class of the Hilbert space ℓ_{2}, of certain asymptotically Hilbertian spaces, and of the ℓ_{p} spaces under the condition 1 < p < ∞. The complexity appears to increase with the distance to the Hilbert space. Commented open problems conclude the article.

1. Introduction

The purpose of this work is to evaluate the descriptive complexity of some classes of real separable Banach spaces. Of course, the collection of separable Banach spaces is not a set, and we have first to code them with the elements of some Polish space. It turns out that when such a coding is performed, the equivalence relation of isometry between Banach spaces is not smooth (see [23] for a stronger result), and thus there is no hope of finding a coding which would map isometric spaces to the same code. Therefore we have to consider classes C of Banach spaces which are closed for the isometry relation: that is, if X ∈ C and Y is isometric to X, then Y ∈ C. It is clear that any meaningful property of Banach spaces will provide such a class C.

We recall that a topological space is Polish if it is homeomorphic to a separable metric complete space. We denote by Δ = 2^{ω} the Cantor set. The separable Banach space C(Δ) of real-valued continuous functions on Δ is isometrically universal for separable Banach spaces. We choose therefore to code the separable Banach spaces with the set SB of closed vector subspaces of C(Δ). For providing the set SB with a topology, we first equip the set F(C(Δ)) of non-empty closed subsets of C(Δ) with an admissible Polish topology, keeping in mind that such a topology is not canonical. However, the Borel σ-field is the same for all admissible topologies (it is called the Effros-Borel structure) and the Borel class of a subset of F(C(Δ)) is essentially independent of the choice of the admissible topology (see section 2). The subset SB of F(C(Δ)) is shown to be a Polish subset of F(C(Δ)) for any admissible topology (see Section 3). The frame is now ready for estimating the complexity of the isomorphism classes of some Banach spaces (see Section 4). After proving the existence of a continuous selection of a dense sequence in elements of SB, we show that the isomorphism class of the Hilbert space is a Borel set of class 2, that the isomorphism classes of certain near Hilbert spaces are Borel sets of class 4, and that the isomorphism classes of the spaces ℓ_{p} (1 < p < ∞, p ≠ 2) are Borel sets of class ω + 1. These results should be compared with the works of B. Bossard (see [3] and [6]), where the Effros-Borel structure is used and some isomorphism classes are shown to be non Borel (see also [9] for a recent progress). Our approach involves topologies on SB and this leads to distinguishing between different levels of complexity of Borel sets. Section 5 gathers some of the open problems which lie beyond the present work.

2010 Mathematics Subject Classification. Primary 46B20, Secondary 54E52.

Notation: The notation we use for Banach spaces is the usual one. In descriptive set theory, we follow the modern notation which can be found for instance in [20]. In particular, ω denotes the first infinite ordinal. Therefore, the set of natural numbers (positive integers), usually denoted by ℕ, is denoted ω in this article. Concerning Borel sets, we recall that for instance, F_{σ}-sets are Σ_{2}^{0}-sets, G_{δ}-sets are Π_{2}^{0}- sets, the notation Σ_{4}^{0} means F_{σδσ}, and Π_{ω+1}^{0} denotes the multiplicative class of order ω + 1.

2. Admissible topologies

When X is a compact metric space, there exists a canonical compact metrizable topology, sometimes called the Vietoris topology, on the set of closed subsets of X, generated by the sets E^{+}(U) = {F ∈ F(X) : F ∩ U ≠ ∅} and E^{−}(U) = {F ∈ F(X) : F ⊂ U}, where U is an arbitrary open subset of X. The Hausdorff distance induces this topology.

A Polish space is a separable, completely metrizable space. If X is Polish but not compact, the space F(X) equipped with the Vietoris topology is not separable any more, and there is no canonical way to equip F(X) with a Polish topology. Yet, let us say that a Polish topology τ on the set F(X) is admissible if it satisfies the following two conditions :

(i) For every open subset U of X, the set E^{+}(U) is τ-open.

(ii) Every τ-open subset W of F(X) is a countable union of sets of the form E^{+}(U) \ E^{+}(V), where U and V are open in X.

These conditions clearly imply that the Borel σ-field generated on F(X) by an admissible topology τ does not depend upon τ : it is the σ-field generated by the sets E^{+}(U), usually called the Effros-Borel σ-field. There exist several admissible topologies (see below). However, if τ and τ′ are two admissible topologies on F(X), it is easy to check that every τ-open set is Σ_{2}^{0} for τ′, and thus the identity map on F(X) is a first Baire class function in both directions. It follows that if ξ is any infinite ordinal, the Σ_{ξ}^{0}-sets (and Π_{ξ}^{0}-sets) are the same for all admissible topologies.

It follows easily from the above conditions that the relation x ∈ F is Π_{2}^{0} in X × F(X): indeed if (U_{n}) is a countable basis of the topology,

A = {(x, F) : ∀n x ∈ U_{n} ⟹ F ∩ U_{n} ≠ ∅} = ⋂_{n}(X × E^{+}(U_{n}) ∪ U_{n}^{c} × F(X))

It is therefore natural to request that admissible topologies also satisfy the condition

(iii) The set {(x, F) ∈ X × F(X) : x ∈ F } is closed in X × F(X).

We now turn to examples. If δ is a totally bounded distance on X, the completion X̂ of X is compact and any closed subset F of X can be identified with its closure \overline{F} ∈ F(X̂), hence we can identify F(X) with the set F^{∗}(X) = \{T ∈ F(X̂) : \overline{T ∩ X} = T }, which is a Π_{2}^{0}-subset of the compact set F(X̂), and thus a Polish space. We will check below (Lemma 2.1) that the restriction to F^{∗}(X) of the Vietoris topology of F(X̂) is admissible. The condition x ∈ F is equivalent to δ(x, \overline{F}) = 0, hence this condition is closed and (iii) is satisfied.

We can also equip F(X) with the Wijsman topology, which is somewhat canonically related with the original distance on X. We recall the definition : if (α_{j})_{j∈ω} is a fixed dense sequence in X, every non-empty closed subset F ∈ F(X) is mapped to the sequence (d(α_{j}, F))_{j∈ω} ∈ ℝ^{ω}. This map is one-to-one with values in ℝ^{ω} equipped with the product topology, and the Wijsman topology is defined as the topology on F(X) which turns this map into an homeomorphism. We will check below (Lemma 2.2) that F(X) is identified through this map with a Π_{2}^{0}-subset of ℝ^{ω}, hence a Polish space, and also that this Polish topology is admissible (Lemma 2.3). Since the function α ↦ d(α, F) is 1-Lipschitz for all F, the map (x, F) ↦ d(x, F) is continuous on X × F(X). Again, condition (iii) is satisfied and the Wijsman topology does not depend on the chosen sequence (α_{j})_{j∈ω}. Nevertheless it depends heavily on the distance d on X itself.

We now prove the technical lemmas which assert that the topologies defined above are admissible. Lemma 2.1. Let δ be a totally bounded distance on X, let X̂ be the completion of X and W an open subset of X̂. Then the set E^{−}(W) is a countable union of differences of sets E^{+}(U_{j}).

Proof. The set X̂ \ W is compact and U_{k} = {x ∈ X̂ : δ(x, W^{c}) < 2^{−k}} is open in X̂. Then if F is a closed subset of X̂ such that F ⊂ W, there exists k ∈ ω such that inf_{x∈F} δ(x, W^{c}) > 2^{−k}, and F ∉ E^{+}(U_{k}). Conversely, if F ∉ E^{+}(U_{k}) one has F ⊂ W. Hence

E^{−}(W) = ⋃_{k} F(X̂) \ E^{+}(U_{k}).  □

Thus Lemma 2.1 shows that δ induces an admissible topology on F(X). Indeed if U is open in X, we write U = Û ∩ X where Û is open in X̂. Then it is clear that E^{+}(U) = E^{+}(Û) ∩ F^{∗}(X). Hence condition (i) is satisfied, and then Lemma 2.1 shows that condition (ii) is satisfied as well. Let us now prove that the Wijsman topology is admissible as well. Lemma 2.2. Let (α_{j}) be a dense sequence in X, and denote (φ_{j}) the sequence of functions on F(F) defined by φ_{j}(F) = d(α_{j}, F). If Φ : F(X) → P = ℝ^{ω} is defined by Φ(F) = (φ_{j}(F))_{j}, then Φ is an homeomorphism between F(X) equipped with the Wijsman topology and the Π_{2}^{0}-subset Z of P defined by

{(t_{j})_{j∈ω} ∈ P : ∀j, k |t_{j} − t_{k}| ≤ d(α_{j}, α_{k})

and (∀n, j ∈ ω ∃k t_{k} < 2^{−n} and d(α_{j}, α_{k}) < t_{j} + 2^{−n})} Proof. It is clear that Φ(F) belongs to Z : since t_{j} = d(α_{j}, F), there exists x ∈ F such that d(α_{j}, x) < t_{j} + 2^{−1−n}, then k such that t_{k} = d(x, α_{k}) < 2^{−1−n}.

Conversely if (t_{j}) ∈ Z, the function α_{j} ↦ t_{j} uniquely extends to a 1-Lipschitz function χ from X to ℝ^{+}, and one lets F = χ^{−1}(0). We have Φ(F) = (t_{j}) : indeed, if x ∈ F, one has t_{j} = χ(α_{j}) − χ(x) ≤ d(x, α_{j}), hence t_{j} ≤ d(α_{j}, F). Moreover, if ε > 0 and j ∈ ω, one can construct a sequence k_{p} of integers such that k_{0} = j, t_{k_{p}} < ε2^{−p} and d(α_{k_{p}}, α_{k_{p+1}}) < t_{k_{p}} + ε2^{−p}. It follows that the sequence (α_{k_{p}}) is a Cauchy sequence which converges to some point x, which belongs to F since d(α_{k_{p}}, F) ≤ t_{k_{p}} → 0. Finally d(α_{j}, F) ≤ d(α_{j}, x) ≤ t_{j} +∑ ε2^{−p} = t_{j} +ε. This shows that d(α_{j}, F) ≤ t_{j} ≤ d(α_{j}, F) and this concludes the proof. □

Lemma 2.3. The Wijsman topology is admissible.

Proof. Let U be an open set in X. There exists J ⊂ ω and (r_{j})_{j∈J} such that U = ⋃_{j∈J} B(α_{j}, r_{j}). Then

E^{+}(U) = Φ^{−1}(⋃_{j∈J}{t ∈ Z : t_{j} < r_{j}})

and thus E^{+}(U) is open.

It remains to show that for all j ∈ ω and r ≥ 0 the set {F ∈ F(X) : d(α_{j}, F) > r} is a countable union of complements of open sets of type E^{+}(U). But if U_{k} = {x : d(x, α_{j}) < r + 2^{−k}}, one has

d(α_{j}, F) > r ⟺ ∃k d(α_{j}, F) ≥ r + 2^{−k} ⟺ ∃k F ∉ E^{+}(U_{k})

and this concludes the proof. □

3. The set SB of separable Banach spaces

Let Δ be the Cantor set 2^{ω} and C(Δ) the Banach space of continuous functions on Δ equipped with the usual supremum norm. It is well-known that every separable Banach space E is isometric to a subspace of C(Δ). We denote SB ⊂ F(C(Δ)) the set of closed vector subspaces of C(Δ).

Let τ be any admissible Polish topology on F(C(Δ)). Let (φ_{k}) be a dense sequence in C(Δ) with φ_{0} = 0. Then H ∈ F(C(Δ)) belongs to SB if and only if it satisfies the two conditions below:

(i) φ_{0} ∈ H
(ii) ∀p∀q∀n H ∩ B(φ_{p}, 2^{−n}) = ∅ or H ∩ B(φ_{q}, 2^{1−n}) = ∅

or H ∩ B(φ_{p} − \frac{φ_{q}}{2} , 2^{1−n}) ≠ ∅

Since for any admissible topology τ and any open subset U in C(Δ), the set {F ∈ SB : F ∩ U ≠ ∅} is τ-open, it follows that SB is a Π_{2}^{0} subset of F(C(Δ)) equipped with τ, hence a Polish space for the topology induced by τ.

In this work, SB is equipped with the topology induced by an admissible topology τ on F(C(Δ)). It turns out that the results we show do not depend upon the choice of this topology τ. Note that, as seen in Section 2, being a Π_{ω+1}^{0}-subset of F(C(Δ)) does not depend upon this choice.

4. Results

It is well-known that dense sequences within closed sets can be selected in a Borel way with respect to the Effros-Borel structure (see Theorem 12.13 in [20]). Our first theorem is a more precise result in the frame of Banach spaces, which follows from Michael’s selection theorem. It provides a handy continuous selection of dense sequences. In what follows, an admissible topology is chosen on the set F(C(Δ)), and the subset SB is equipped with the (Polish) restriction of this topology.

Theorem 4.1. There exists a sequence (f_{p})_{p∈ω} of continuous functions on SB with values in C(Δ) such that for all F ∈ SB, the set F is the closure in C(Δ) of the set {f_{p}(F) : p ∈ ω}.

Proof. Let (φ_{k}) be a dense sequence in C(Δ). We denote Ω_{k,n} = {φ ∈ C(Δ) : ∥φ − φ_{k}∥ < 2^{−n}} and V_{k,n} = {F ∈ SB : F ∩ Ω_{k,n} ≠ ∅}. Then, for all k and n, the set V_{k,n} is open in SB and on V_{k,n} the multivalued map φ_{k,n} : F ↦ \overline{F ∩ Ω_{k,n}} is l.s.c. and takes non-empty closed convex values : indeed, if U is open in C(Δ) and U ∩ φ_{k,n}(F) ≠ ∅, then F ∩ (U ∩ Ω_{k,n}) ≠ ∅ and for every F^{∗} in the open set {F^{∗} ∈ SB : F^{∗} ∩ (U ∩ Ω_{k,n}) ≠ ∅}, one has φ_{k,n}(F^{∗}) ∩ U ≠ ∅. It follows by Michael’s theorem that φ_{k,n} has a continuous selection ψ_{k,n} : V_{k,n} → C(Δ). Since V_{k,n} is open in SB, its characteristic function 1l_{V_{k,n}} is l.s.c. on SB and there exists an increasing sequence (χ_{k,n,m})_{m∈ω} of continuous functions from SB to [0, 1] such that 1l_{V_{k,n}} = sup_{m} χ_{k,n,m}. We set f_{k,n,m}(F) = χ_{k,n,m}(F) · ψ_{k,n}(F).

The function f_{k,n,m} : SB → C(Δ) is clearly continuous at each point of V_{k,n} and of SB \ \overline{V_{k,n}}. And if F ∈ ∂V_{k,n}, we have lim_{F^{∗}→F} χ_{k,n,m}(F^{∗}) = χ_{k,n,m}(F) = 0. Moreover ψ_{k,n} is bounded on V_{k,n} since ∥ψ_{k,n}(F)∥ ≤ ∥φ_{k}∥ + 2^{−n}. It follows that f_{k,n,m} tends to 0 in F ∈ ∂V_{k,n} and thus f_{k,n,m} is continuous.

Moreover, f_{k,n,m}(F) ∈ F since ψ_{k,n}(F) ∈ F if F ∈ V_{k,n} and since χ_{k,n,m}(F) ∈ ℝ, and f_{k,n,m}(F) = 0 ∈ F if F ∉ V_{k,n}. Finally, let a ∈ F. For ε > 0, there exist k and n such that a ∈ Ω_{k,n} ⊂ \overline{Ω_{k,n}} ⊂ B(a, ε). Hence F ∈ V_{k,n} and ψ_{k,n}(F) ∈ \overline{F ∩ Ω_{k,n}} ⊂ F ∩ B(a, ε). Then χ_{k,n,m}(F) → 1, hence f_{k,n,m}(F) → ψ_{k,n}(F) ∈ B(a, ε), which shows that B(a, ε) contains some f_{k,n,m}(F), hence that f_{k,n,m}(F) is dense in F. It suffices now to reorder the sequence (f_{k,n,m}) in a sequence (f_{p}) to conclude the proof. □

Corollary 4.2. The family Fin of finite-dimensional spaces is Σ_{2}^{0} in SB.

Proof. If dim(F) < ∞, there is an integer q such that F is spanned by the vectors f_{j}(F) for j ≤ q. Hence

Fin = ⋃_{q}{F : ∀k {f_{1}(F), f_{2}(F), ..., f_{q}(F), f_{k}(F)} is not free}

which shows the result, since the set {(x_{1}, x_{2}, …, x_{m}) free} is open in C(Δ)^{m}. □

Note that Corollary 4.2 means that the set SB_{∞} = SB\Fin of infinite-dimensional separable Banach spaces is a Π_{2}^{0} subset of SB, and thus is a Polish space for the topology induced by SB.

Theorem 4.3. The isomorphism class ⟨ℓ_{2}⟩ of the separable Hilbert space is D_{2}(Σ_{2}^{0}) in SB.

Proof. By Kwapien’s theorem, an infinite-dimensional separable Banach space E is isomorphic to ℓ_{2} if and only if there exists M ∈ ℝ such that for all finite sequences (x_{1}, ..., x_{n}) in E, one has

\frac{1}{M} · (∑_{j=1}^{n} ∥x_{j}∥^{2}) ≤ \frac{1}{2^{n}} · ∑_{ε∈\{−1,1\}^{n}} ∥∑ ε_{j}x_{j}∥^{2} ≤ M · (∑_{j=1}^{n} ∥x_{j}∥^{2})

Since the sequence f_{p}(F) is dense in F, a space F ∈ SB belongs to the isomorphism class of ℓ_{2} if and only if there is m ∈ ω such that for any finite subset J of ω, one has

\frac{1}{m} ∑_{p∈J} ∥f_{p}(F)∥^{2} ≤ 2^{−#J} ∑_{ε∈\{−1,1\}^{J}} ∥∑_{p∈J} ε_{p}f_{p}(F)∥^{2} ≤ m ∑_{p∈J} ∥f_{p}(F)∥^{2} .

For each m and each finite subset J of ω, the set of all F ∈ SB which satisfy the above inequalities is closed, since the f_{p}’s are continuous. The union E_{2} (over m) of the intersection (over J) of these sets is thus Σ_{2}^{0} in SB, and ⟨ℓ_{2}⟩ = E_{2}\Fin is D_{2}(Σ_{2}^{0}). The above proof shows that ⟨ℓ_{2}⟩ is a Σ_{2}^{0} subset of the Polish space SB_{∞}. □

Our next statement is an elementary perturbation lemma. Note that Auerbach’s lemma states that any n-dimensional subspace V of X has a basis (x_{1}, x_{2}, ..., x_{n}) which satisfies the conditions below (see Proposition 1.c.3 in [22]).

Lemma 4.4. Let X be a Banach space, E_{0} a subspace of X, (x_{1}, x_{2}, ..., x_{n}) vectors in X with norm 1, and (x_{1}^{∗}, x_{2}^{∗}, ..., x_{n}^{∗}) linear forms on X with norm 1. We assume

that ⟨x_{j}^{∗}, x_{k}⟩ = δ_{j}^{k} and that ∑_{j≤n} d(x_{j}, E_{0}) < \frac{1}{2}. Then there exists an isomorphism

S on X such that ∥S∥ . ∥S^{−1}∥ < 3 and such that x_{j} belongs to S(E_{0}) for all j ≤ n.

Proof. We pick some y_{j} ∈ E_{0} such that ∑_{j} ∥x_{j} − y_{j}∥ < \frac{1}{2} and we consider the finite rank operator T defined by

T u = ∑_{j}⟨x_{j}^{∗}, u⟩ · (x_{j} − y_{j})

Clearly ∥T ∥ ≤ ∑_{j} ∥x_{j}^{∗}∥ · ∥x_{j} − y_{j}∥ < \frac{1}{2} and (I − T)(x_{j}) = y_{j} ∈ E_{0}. The operator

I −T is invertible and S = (I −T)^{−1} satisfies ∥S∥ ≤ ∥∑_{n=0}^{∞} T^{n}∥ ≤ (1−∥T ∥)^{−1} < 2.

Moreover ∥S^{−1}∥ = ∥I − T ∥ ≤ 1+∥T ∥ < \frac{3}{2}, and this shows ∥S∥ . ∥S^{−1}∥ < 3. Finally x_{j} = S(I − T)(x_{j}) = Sy_{j} ∈ S(E_{0}). □

Proposition 4.5. Let 1 ≤ p ≤ ∞. The class L_{p} is Σ_{4}^{0} in SB.

Proof. Assume that F ∈ SB belongs to L_{p}. Then there is M ∈ ω such that for all n ∈ ω, the subspace of F spanned by the vectors (f_{p}(F))_{p≤n} is contained in an N-dimensional subspace of X which is M-isomorphic to ℓ_{p}^{N}. For J ⊂ ω finite, set E_{J} = span{f_{j} : j ∈ J}. Then F satisfies: ∃M ∀n ∃J such that

(i) ∀λ ∈ ℝ^{#J} \frac{1}{M + 1} ∥∑_{j∈J} λ_{j}f_{j}(F)∥ ≤ ∥λ∥_{p} ≤ (M + 1) ∥∑_{j∈J} λ_{j}f_{j}(F)∥

(ii)  ∀μ ∈ ℝ^{n} ∥∑_{p≤n} μ_{p}f_{p}(F)∥ = 1 ⟹ d(∑_{p≤n} μ_{p}f_{p}(F), E_{J}) ≤ \frac{1}{4n}

It is easily seen that the conditions (i) and (ii) are closed, and thus the above conditions define a Σ_{4}^{0} subset of SB.

Conversely, if F satisfies the above conditions and V is a d-dimensional subspace of F, there exists n ≥ p so large that the whole unit ball of V is within distance 1/4d of the space span{f_{p}(F) : p ≤ n}. Let J be a finite subset of ω satisfying (i) and (ii) for this n. By Lemma 4.4, there exists a subspace E′_{J} of F which is 3-isomorphic to E_{J} and contains V. The space E′_{J} is 3(M + 1)-isomorphic to ℓ_{p}^{#J}. This shows that F is L_{p}.

We have therefore shown that L_{p} is Σ_{4}^{0} in SB. □

The next result is crucially important for estimating the complexity of the isomorphism classes of reflexive ℓ_{p}’s. We use the language of games to prove it.

Proposition 4.6. There exists a Π_{ω+1}^{0} subset Ω of SB which contains ⟨ℓ_{p}⟩ for all p ∈]1, 2[ and such that if X ∈ Ω, then X does not contain an isomorphic copy of ℓ_{2}.

Proof. We choose a weak* dense sequence (μ_{j})_{j∈ω} in the unit ball of M(Δ) = C(Δ)^{∗}. For F ∈ SB, we define

Ω_{j,F} = {φ ∈ F : ∥φ∥ ≤ 1 and ∀k ≤ j, |⟨μ_{k}, φ⟩| < 2^{−j}}

Then V_{j} = Ω_{j,F} is a non-increasing sequence of weak neighborhoods of 0 in the unit ball F_{1} of F, and ⋂_{j} \overline{V_{j}} = \{0\}. So if F is reflexive the \overline{V_{j}}’s are weakly compact hence they form a basis of weak neighborhoods of 0 in F_{1}.

Denote W_{j,F} ⊂ ω the set of all ρ ∈ ω such that f_{ρ}(F) ∈ Ω_{j,F} and \frac{1}{2} < ∥f_{ρ}(F)∥ < 1.

Given M ≥ 1 and n, we consider the finite game G_{F}^{M,n} of length n on integers where Player I plays (j_{1}, j_{2}, ...j_{k}), Player II answers (ρ_{1}, ρ_{2}, ..., ρ_{k}) and where Player I wins if for some k ∈ [1, n], one has ρ_{k} ∉ W_{j_{k},F} or ∥∑_{i≤k} f_{ρ_{i}}(F)∥ ≥ M · \sqrt{k}.

For any run (j_{1}, ρ_{1}, j_{2}, ρ_{2}, …, j_{n}, ρ_{n}) in the game G_{F}^{M,n} the set of all F ∈ SB for which Player I wins is closed: indeed it is

⋃_{1≤k≤n}(\{F : f_{ρ_{k}}(F) ∉ \{φ : \frac{1}{2} < ∥φ∥ < 1\} ∩ Ω_{j_{k},F} \} ∪ \{F : ∥∑_{i≤k} f_{ρ_{i}}(F)∥ ≥ M · \sqrt{k}\})

It is easy to check by induction on (n−k) that the set E_{(j_{1},ρ_{1},j_{2},ρ_{2},...,j_{k},ρ_{k})}^{M,n} of all F ∈ SB for which Player I has a winning strategy starting from (j_{1}, ρ_{1}, j_{2}, ρ_{2}, …, j_{k}, ρ_{k}) in the game G_{F}^{M,n} is Σ_{2n−2k}^{0} if k < n (resp. Π_{1}^{0} if k = n): indeed

E_{(j_{1},ρ_{1},j_{2},ρ_{2},...,j_{k},ρ_{k})}^{M,n} = ⋃_{j} ⋂_{ρ} E_{(j_{1},ρ_{1},j_{2},ρ_{2},...,j_{k},ρ_{k},j,ρ)}^{M,n} .

Hence the set E^{M,n} of all F ∈ SB for which Player I has a winning strategy in G_{F}^{M,n} is Σ_{2n}^{0}.

We omit the simple proof of our next lemma. We refer to section 3 in [19] for much more along these lines.

Lemma 4.7. Pick ε ∈]0, \frac{1}{2}[, and let V and W be two finite-dimensional subspaces of ℓ_{p} (1 < p < ∞). We denote by π_{r} the canonical projection of ℓ_{p} onto the subspace generated by the r first basis vectors. We assume that there exists r ∈ ω such that ∥(I − π_{r})_{|V} ∥ < ε et ∥π_{r|W} ∥ < ε. Then for x ∈ V and y ∈ W, one has

(1 − 2ε)^{p}(∥x∥^{p} + ∥y∥^{p}) ≤ ∥x + y∥^{p} ≤ (1 + ε)^{p}(∥x∥^{p} + ∥y∥^{p})

We let now Ω = ⋂_{M∈ω} ⋃_{n} E^{M,n}. Since E^{M,n} is a Σ_{2n}^{0} Borel set, ⋃_{n} E^{M,n} is a Σ_{ω}^{0} set for all M, and the set Ω is a Π_{ω+1}^{0} Borel set. We check now that it fulfills the requirements of Proposition 4.6. This follows from the following two lemmas.

Lemma 4.8. If 1 < p < 2, the set ⟨ℓ_{p}⟩ is contained in Ω.

Proof. Indeed pick F ∈ ⟨ℓ_{p}⟩. There exists m ≥ 1 and an isomorphism T from ℓ_{p} on F such that for all x ∈ ℓ_{p}, one has (m^{−1}) ∥T x∥ ≤ ∥x∥ ≤ m ∥T x∥. Then Player I follows a strategy which ensures that for all k, one has

∥∑_{i≤k} T^{−1}f_{ρ_{i}}(F)∥^{p} ≥ (1 − ε)^{p} (∥∑_{i<k} T^{−1}f_{ρ_{i}}(F)∥^{p} + ∥T^{−1}f_{ρ_{k}}(f)∥^{p}) .

For doing so, one chooses a sequence (ε_{n}) such that ∏_{n=1}^{∞}(1 − 2ε_{n}) > 1 − ε. At each move, r is chosen so large that

∥(I − π_{r})(∑_{i<k} T^{−1}f_{ρ_{i}}(F))∥ < ε_{k} ∥∑_{i<k} T^{−1}f_{ρ_{i}}(F)∥

and then Player I plays j_{k} such that ∥π_{r}(f_{ρ}(F))∥ ≤ ε_{k} ∥f_{ρ}(F)∥ for ρ ∈ W_{j_{k},F}. This can be done since F is reflexive. If we denote s_{k}(F) = ∑_{i≤k} f_{ρ_{i}}(F) after II played ρ_{k}, we have now by induction (and Lemma 4.7) that

∥T^{−1}(s_{k}(F))∥^{p} ≥ ∏_{j≤k}(1 − 2ε_{j})^{p}.k.2^{−p}

since ∥T^{−1}f_{ρ_{j}}(F)∥ ≥ \frac{1}{2}. Hence ∥s_{k}(F)∥ ≥ \frac{1 − ε}{2m} · k^{1/p} and finally, ∥s_{k}(F)∥ ≥ M\sqrt{k}

if k^{1/p−1/2} ≥ \frac{2Mm}{1 − ε}. This shows that Player I wins the game if n ≥ (\frac{2Mm}{1 − ε})^{\frac{2p}{2−p}}, hence F ∈ Ω. □

Lemma 4.9. If F ∈ SB contains a subspace isomorphic to ℓ_{2}, then F ∉ Ω.

Proof. We have to show that there exists M such that for all n, F ∉ E^{M,n}.

There is a subspace G of F, an integer m ≥ 1 and an isomorphism T from ℓ_{2} onto G such that for x ∈ ℓ_{2}, (m)^{−1} ∥x∥_{2} ≤ ∥T x∥ ≤ m ∥x∥_{2}.

We denote (e_{n}) the canonical basis of ℓ_{2}, and we denote (ρ_{n}) a sequence of integers

such that ∥f_{ρ_{n}}(F)∥ < 1 and ∥f_{ρ_{n}}(F) − \frac{T e_{n}}{∥T e_{n}∥}∥ < 2^{−n}. If μ ∈ C(Δ)^{∗}, one has

∑_{n} |(μ∘T)(e_{n})|^{2} < +∞ and thus μ(f_{ρ_{n}}(F)) → 0. The strategy followed by Player II consists into choosing for each k ≥ 1 an integer n_{k} from that sequence such that ρ_{n_{k}}(F) ∈ W_{j_{k},F}.

For all k, if we let s_{k} = ∑_{j≤k} f_{ρ_{n_{j}}} (F), we have

∥s_{k} − ∑_{j≤k} \frac{T e_{n_{j}}}{∥T e_{n_{j}} ∥}∥ < 1

and

∥∑_{j≤k} \frac{T e_{n_{j}}}{∥T e_{n_{j}} ∥}∥^{2} = ∥T (∑_{j≤k} \frac{e_{n_{j}}}{∥T e_{n_{j}} ∥})∥^{2} ≤ m^{2} ∑_{j≤k} \frac{1}{∥T e_{n_{j}} ∥^{2}} ≤ m^{4} · k

hence ∥s_{k}∥ < 1 + m^{2} · \sqrt{k} ≤ (m^{2} + 1) · \sqrt{k}, and thus with this strategy Player II wins G_{F}^{M,n} for all n if M = m^{2} + 1. Therefore F ∉ Ω. This concludes the proof of Lemma 4.9 and of Proposition 4.6. □

Note that M and n could be incorporated to the game, if we ask Player II to choose M and Player I to answer n, before the j_{k}’s and ρ_{k}’s are played. Note also that the choice of j_{1} is irrelevant: the condition f_{ρ_{1}}(F) ∈ Ω_{j_{1},F} is not used in the proof that Player I has a winning strategy if F ∈ ⟨ℓ_{p}⟩.

Our next proposition states what happens on the other side of 2.

Proposition 4.10. There exists a Π_{ω+1}^{0} subset Ω′ of SB which contains ⟨ℓ_{p}⟩ for all p > 2 and such that if X ∈ Ω′, then X does not contain an isomorphic copy of ℓ_{2}.

Proof. The proof is completely similar to the proof of Proposition 4.6, but this time we consider a game H_{F}^{M,n} where the Player I chooses again weak neighborhoods of 0 and wins if

∥∑_{j≤k} f_{ρ_{j}}(F)∥ ≤ (M)^{−1}\sqrt{k}

for some k ≤ n. The same arguments show that Player II wins for some M and all n if F contains isomorphically ℓ_{2} by choosing a subsequence of a sequence in F equivalent to the canonical basis of ℓ_{2}, while Player I wins for all M and some n (which depends on M) by forcing Player II to play a sequence equivalent to the canonical basis of ℓ_{p}. □

We observe that the above proof shows that Ω contains all Banach spaces which have an asymptotically uniformly convex norm of power type p < 2. Similarly, the set Ω′ contains all Banach spaces which have an asymptotically uniformly smooth norm of power type p > 2.

We are now ready to show an improvement of the main result of [10]. This theorem provides an upper estimate of the Borel class of ⟨ℓ_{p}⟩, and thus answers Problem 3 from [10].

Theorem 4.11. Let 1 < p < ∞, with p ≠ 2. Then the isomorphism class ⟨ℓ_{p}⟩ is a Π_{ω+1}^{0} Borel set.

Proof. By the Propositions 4.6 and 4.10, the Π_{ω+1}^{0} Borel set S = Ω ∪ Ω′ contains ⟨ℓ_{p}⟩ for all 1 < p < ∞, with p ≠ 2, and if X ∈ S, then X does not contain an isomorphic copy of ℓ_{2}. Fix any such p. Then by Proposition 4.5, the set L_{p} ∩ S is also Π_{ω+1}^{0}. But it follows from a theorem of W.B. Johnson and E. Odell [15] that actually L_{p} ∩ S = ⟨ℓ_{p}⟩. Indeed by [15], a separable L_{p}-space which does not contain an isomorphic copy of ℓ_{2} is isomorphic to ℓ_{p}. □

We now investigate what happens with spaces which are “close to ℓ_{2}” without being necessarily isomorphic to ℓ_{2}. We use the terminology and results from [14]. We recall that a paving (E_{n}) of a separable Banach space X is an increasing sequence of finite-dimensional subspaces of X whose union is dense in X. Two spaces X and Y have a common paving if there exists a paving (E_{n}) of X and (F_{n}) of Y such that the Banach-Mazur distance between E_{n} and F_{n} is bounded uniformly in n. A space X is said to be determined by its pavings if any space Y which has a common paving with X is actually isomorphic to X. A separable space X is asymptotically Hilbertian if there exists K ≥ 1 such that for every n ≥ 1, there exists a finitecodimensional subspace X_{n} of X such that every n-dimensional subspace of X_{n} is K-isomorphic to ℓ_{2}^{n}. It is shown in [14] (Corollary 7.7) that if X is asymptotically Hilbertian and every infinite-dimensional subspace of X contains a complemented copy of ℓ_{2}, then X is determined by its pavings.

This result shows in particular that the ℓ_{2}-sums (∑ ⊕ℓ_{p_{n}}^{k_{n}})_{2} of ℓ_{p_{n}}^{k_{n}} spaces, with p_{n} → 2, are determined by their pavings. Such spaces are not isomorphic to ℓ_{2} when the dimension k_{n} increases to ∞ fast enough when compared with |p_{n} − 2|. For instance, the space (∑ ⊕ℓ_{2+(logn)^{−1}}^{2^{n}})_{2} is not isomorphic to ℓ_{2}. It is shown in [14] (Proposition 7.4) that if X is determined by its pavings, then X is near Hilbert, which means that X has type (2 − ε) and cotype (2 + ε) for every ε > 0.

Theorem 4.12. Let X be an asymptotically Hilbertian space, such that every infinite dimensional subspace of X contains a complemented subspace which is isomorphic to ℓ_{2}. Let Sub(X) ⊂ SB be the set of spaces which are isomorphic to a subspace of X. Let Isom_{X} ⊂ Sub(X)^{2} be the restriction to Sub(X) of the isomorphism relation. Then there is a Σ_{4}^{0} subset CP of SB^{2} such that CP ∩ Sub(X)^{2} = Isom_{X}. Therefore, if Z ∈ Sub(X) is any subspace of X, the isomorphism class ⟨Z⟩ of Z is Σ_{4}^{0} in SB.

Proof. It follows from [14] (see the above discussion) that any such X is determined by its pavings. Moreover, our assumptions on X are clearly hereditary, hence when restricted to Sub(X), the isomorphism relation coincide with the relation CP defined by

CP = {(Y, Z) ∈ SB^{2} : Y and Z have a common paving}

We observe now that CP is a Σ_{4}^{0} subset of SB^{2}. For checking this, we follow the approach from [11]. In what follows, d and N are positive integers, and s and t are finite sequences of integers of the same length |s| = |t|. We denote by E_{s,t,d,N} the set of all pairs ((a_{k})_{k∈ω}, (b_{k})_{k∈ω}) ∈ [C(Δ)^{ω}]^{2} which satisfy the following four conditions:

(i) the finite sequences (a_{s(j)})_{j<|s|} and (b_{t(j)})_{j<|t|} are linearly independent.

(ii) for all real numbers (λ_{j}) ∈ ℝ^{|s|}, one has

d^{−1} ∥∑_{j≤|t|} λ_{j}b_{t(j)}∥ ≤ ∥∑_{j≤|s|} λ_{j}a_{s(j)}∥ ≤ d ∥∑_{j≤|t|} λ_{j}b_{t(j)}∥

.

(iii) span[(a_{s(k)})_{k<|s|}] ∩ B(a_{j}, N^{−1}) ≠ ∅ for all j ≤ N.

(iv) span[(b_{t(k)})_{k<|t|}] ∩ B(b_{j}, N^{−1}) ≠ ∅ for all j ≤ N.

Note that in conditions (iii) and (iv) above, B(x, r) denotes the open ball of center x and radius r > 0. We claim that the set E_{s,t,d,N} is Σ_{2}^{0} in [C(Δ)^{ω}]^{2}. Indeed, the conditions (i), (iii) and (iv) define open sets, and the condition (ii) defines a closed set, hence E_{s,t,d,N} is the intersection of an open and a closed set. Therefore the set

B = ⋃_{d} ⋂_{N} ⋃_{(s,t),|s|=|t|} E_{s,t,d,N}

is a Σ_{4}^{0} subset of [C(Δ)^{ω}]^{2}. If now D = (f_{p})_{p∈ω} is the continuous function from SB to C(Δ)^{ω} provided by Theorem 4.1 , it is easy to check (see the proof of Lemma 6.1 in [11]) that

CP = [D × D]^{−1}(B)

and thus CP is a Σ_{4}^{0} subset of SB^{2}. Hence the set CP works. The last observation follows from the equation

⟨Z⟩ = {Y ∈ SB : (Z, Y) ∈ CP}.
□

We recall that a separable Banach space X is said to be ergodic [8] if the equivalence relation E_{0} Borel reduces to the the isomorphism relation on the set of subspaces of X. When this relation is Borel, the dichotomy shown in [12] applies, and X is ergodic as soon as this relation is not smooth. It has been shown by R. Anisca [1] that every asymptotically Hilbertian space which is not isomorphic to ℓ_{2} is ergodic, and on the other hand W. Cuellar proved that a non-ergodic Banach space is near Hilbert [5]. It is conjectured that every separable Banach space which is not isomorphic to ℓ_{2} is ergodic.

5. Open Problems

The descriptive complexity of isomorphic families of Banach spaces is far from being well understood, and many natural questions remain open. We select and comment below some of them.

Problem 5.1. Compute the exact Borel class of the isomorphic classes which are considered in the present work. In other words, the problem is to know if the theorems shown in this article are actually optimal. We conjecture for instance that Theorem 4.11 is optimal, which means in particular that if 1 < p < ∞ and p ≠ 2, then ⟨ℓ_{p}⟩ is not a Borel set of finite class. Such a proof could necessitate to construct continuous reductions of “complicated” subsets Z of 2^{ω} to ⟨ℓ_{p}⟩, hence continuous functions from 2^{ω} to SB taking values in ⟨ℓ_{p}⟩ iff their argument is in Z.

Problem 5.2. If φ is an Orlicz function such that ℓ^{φ} is reflexive, is the class ⟨ℓ^{φ}⟩ Borel ? Note that a positive answer would follow from Suslin’s separation theorem if the analogue of Johnson-Odell’s dichotomy [15] is shown for Orlicz spaces. Note that it follows from [4] that if X is reflexive, ⟨X⟩ is Borel if and only if ⟨X^{∗}⟩ is Borel.

Problem 5.3. Assume that X is a separable Banach space such that there exists a Borel subset I of SB^{2} such that for all Z ∈ Sub(X), the following equation holds: ⟨Z⟩ = {Y ∈ SB : (Z, Y) ∈ I}. Does it follow that X is near Hilbert ? Note that this assumption implies that if Y ∈ Sub(X) is isomorphic to a subspace of X, then ⟨Y ⟩ is Borel. Let us recall that if X ∈ SB, the set Sub(X) is in general not Borel. For instance, Sub(c_{0}) is not Borel, and actually ⟨c_{0}⟩ is not Borel [21].

Problem 5.4. Let 1 < p < ∞ and X = ℓ_{2} ⊕ ℓ_{p}. Is ⟨X⟩ Borel ? If it is so, what is the Borel class of ⟨X⟩ ? We strongly conjecture that ⟨X⟩ is Borel. Indeed we may assume (by [4]) that p > 2. Then [13] shows that if E ⊂ L_{p}, then E ∈ Sub(ℓ_{p}⊕ℓ_{2}) or ℓ_{p}(ℓ_{2}) ∈ Sub(E). This implies by Suslin’s separation theorem that L_{p} ∩ Sub(ℓ_{p} ⊕ ℓ_{2}) is Borel. A positive answer to Problem 5.5 would follow (again, by Suslin’s theorem) if one can show that L_{p}∩Sub(ℓ_{p}⊕ℓ_{2}) contains at most countably many isomorphism classes. It is shown in [16] that if X ∈ L_{p} ∩ Sub(ℓ_{p} ⊕ ℓ_{2}) has an unconditional basis, then X is isomorphic to one of the three spaces: ℓ_{p}, X_{p}, ℓ_{p} ⊕ ℓ_{2}. Hence a negative answer to Problem 5.4 would imply in particular the existence of a L_{p} space without unconditional basis. We refer to [7] for a recent work on “small” subspaces of L_{p}.

Problem 5.5. This last problem is rather a suggestion for further work. As checked in Section 3, the restriction of an admissible topology on F(C(Δ)) to the set SB is a Polish topology, and thus it makes sense to consider Baire category problems in SB. We can therefore investigate which properties (P) of separable Banach spaces are generic, in the sense that the subset of SB consisting of spaces which fail (P) is meager. A motivation for such questions is that it might lead to smooth existence proofs through Baire category arguments. It would nevertheless be extremely surprising that being meager for a subset of SB be independent on the chosen admissible topology.

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Sorbonne Universités, Université Pierre et Marie Curie (Univ Paris 06), CNRS Institut de Mathématique de Jussieu - Paris Rive Gauche Université Pierre et Marie Curie (Paris 6) Boîte 186 - 4 place Jussieu, F- 75252 Paris Cedex 05, FRANCE

E-mail address: gilles.godefroy@imj-prg.fr
E-mail address: jean.saint-raymond@imj-prg.fr