On the three space property for $C(K)$-spaces

The reader is referred to the Preliminaries section for all unexplained terminology. This paper focuses on solving a problem that Castillo has repeatedly posed [11,12,9,2]: must every twisted sum of c 0 (ℵ) and c 0 (ℵ ′ ) be isomorphic to a C(K)-space? The problem stems from an idea of Cabello mentioned in [10] -that a twisted sum of two C(K)-spaces does not need to be a C(K)-space itself. The proof was sketched in [10] and, in all its details, in [2]. The argument uses a clever construction of Benyamini [4], a renorming ∥ • ∥ n of ℓ ∞ such that (ℓ ∞ , ∥ • ∥ n ) cannot be n-complemented in any C(K)-space. This eventually results in an exact sequence 0 c 0 X c 0 (ℓ ∞ /c 0 ) 0 in which X cannot be a complemented subspace of a C(K)-space.

Benyamini’s construction uses in an essential way properties of the compact space βω \ω. Its complexity suggests that such construction cannot be carried out with simpler compacta; for instance: Problem 1.1. Given any cardinal number ℵ, is every twisted sum space X in

isomorphic to a C(K)-space?

We present here a relatively consistent negative solution to Problem 1.1: Assuming p = c, a weak version of Martin’s axiom, we prove that every Eberlein compact space K of weight c admits an exact sequence 0 → c 0 → X → C(K) → 0 where X is not isomorphic to a C(K)-space. Recall that for any ℵ, the space c 0 (ℵ) is isomorphic to C(A(ℵ)), the space of continuous functions on the Eberlein compact space A(ℵ), the one-point compactification of a discrete set of size ℵ. Hence, our main result provides the negative answer to Problem 1.1 for ℵ = c. where the kernel of each arrow coincides with the image of the preceding one. We also say that X is a twisted sum of Y and Z. In this setting, the open mapping theorem yields that Y is isomorphic to a subspace of X and Z to the quotient X/Y . Two exact sequences 0 → Y → X i → Z → 0, i = 1, 2, are equivalent if there is an operator u :

Again, the open mapping theorem makes u an isomorphism. We say an exact sequence is trivial, or that it splits, if the operator p has a right-inverse r : Z → X. In other words, when it is equivalent to the twisted sum 0 → Y → Y ⊕ Z → Z → 0. When this happens, X must be of course isomorphic to Y ⊕ Z, but the converse is not true. Let us write Ext(Z, Y ) to represent the space of twisted sums of Y and Z, modulo equivalence. Hence Ext(Z, Y ) = 0 means that the only twisted sum of Y and Z is the trivial one. Two basic constructions regarding exact sequences are the pull-back and the push-out. Given a short exact sequence 0 → Y i → X p → Z → 0 and an operator T : Z ′ → Z, then there is a unique space P B, together with an operator P B → X, which makes commutative the following diagram:

Dually, if we take the previous exact sequence and an operator S : Y → Y ′ , there is a unique space P O and an operator X → P O such that the following diagram is commutative:

Explicitely, P O = (Y ′ ⊕X)/∆ where ∆ = {(i(y), -S(y)) : y ∈ Y }, with the quotient norm.

2.2. p = c. Recall that Martin’s axiom for σ-centered posets p = c amounts to saying that whenever B ⊆ P(ω) is a family with |B| < c such that B 1 ∩ . . . ∩ B n is infinite for every n and any B i ∈ B, then there is an infinite set A such that A ⊆ * B for every B ∈ B, see Fremlin [17]. Recall also that, in the topological setting, under p = c, every compact space K of weight < c is sequentially compact, see [17, 24A].

2.3. C-spaces. The symbol K always stands for a compact Hausdorff space. We say X is a C-space if it is isomorphic to a space of continuous real-valued functions C(K) for some K. For a given space C(K), we, as usual, identify the dual space C(K) * with the space M (K) of signed Radon measures on K of finite variation. M 1 (K) stands for the unit ball of M (K), equipped with the weak * topology inherited from C(K) * . For a measure µ ∈ M (K), |µ| denotes its variation.

Definition 2.1. Let X be a Banach space. Given any weak * compact F in B X * , we can consider the norm-one operator

Given 0 < c ≤ 1, we say F is c-norming if T is injective, has closed range, and has an inverse T -1 : T (X) → X such that ∥T -1 ∥ < 1/c. We say that F is free if T is onto.

Lemma 2.2. For a Banach space X, the following are equivalent:

(i) X is a C-space;

(ii) there is a free subset F of B X * which is c-norming for some 0 < c ≤ 1.

Proof. Assume T : X → C(K) is an isomorphism such that c • ∥x∥ ≤ ∥T x∥ ≤ ∥x∥ for every x ∈ X. This means that the norm |||x||| = sup k∈K |T x(k)| is equivalent to the norm in X. Hence write ∆ K = {δ k : k ∈ K} and consider F = T * (∆ K ). Such F is weak * compact, since T * is weak * continuous, and it is clearly c-norming and free. The converse is obvious. □

The main idea of our construction is to eliminate every possibility of B X * having a free c-norming subset, so that the resulting space X cannot be a C-space. The next simple lemma will enable us to recognize sets that are not free. Lemma 2.3. Let F ⊆ B X * be weak * compact and c-norming

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