Nonseparable $C(K)$-spaces can be twisted when $K$ is a finite height compact

0.16 MB·

by  Jesús M. F. Castillo

NONSEPARABLE C(K)-SPACES CAN BE TWISTED
WHEN K IS A FINITE HEIGHT COMPACT
6
1 JESU´SM.F.CASTILLO
0
2

n
a
Abstract. We show that, given a nonmetrizable compact space K having ω-derived set empty,
J
there always exist nontrivial exact sequences 0 → c0 → E → C(K) → 0. This partially solves a
8 problemposedinseveralpapers: IsExt(C(K),c0)6=0forK anonmetrizablecompact set?

]
A 1. Introduction
F
A general problem in Banach space theory is to determine, given two Banach spaces Y and Z,
.
h the existence and properties of Banach spaces X containing Y such that X/Y = Z. The space X is
t called a twisted sum of Y and Z. When only the trivial situation X =Y ⊕Z is possible one obtains
a
m an interesting structure result asserting that every copy of Y in a space X such that X/Y = Z
must be complemented. When some X 6= Y ⊕Z as above exists, one obtains a usually exotic and
[
interesting space with unexpected properties; perhaps the perfect example could be the Kalton-Peck

1 Z space [20] which has no unconditional basis while containing an uncomplemented copy of ℓ such
2 2
v
that Z /ℓ =ℓ . Recall that a short exact sequence of Banach spaces is a diagram
7 2 2 2
3 (1) i q
0 −−−−→ Y −−−−→ X −−−−→ Z −−−−→ 0
0
2 where Y, X and Z are Banach spaces and the arrows are operators in such a way that the kernel of
0 each arrow coincides with the image of the preceding one. By the open mapping theorem i embeds
. Y as a closed subspace of X and Z is isomorphic to the quotient X/i(Y). Two exact sequences
1
0 0→Y →X →Z →0and0→Y →X1 →Z →0aresaidtobeequivalent ifthereexistsanoperator
6 T :X −→X making commutative the diagram
1
1
0 −−−−→ Y −−−−→ X −−−−→ Z −−−−→ 0
:
v
Xi k T k

y
r 0 −−−−→ Y −−−−→ X1 −−−−→ Z −−−−→ 0.
a
The sequence (1) is said to be trivial (or that it splits) if i(Y) is complemented in X (i.e., if it is
equivalenttothe sequence0→Y →Y ⊕Z →Z →0). We writeExt(Z,Y)=0toindicatethatevery
sequenceofthe form(1)splits. Summing allup, thegeneralproblemistodetermine whenthere exist
nontrivial twisted sums of Y and Z, or else, when Ext(Z,Y)6=0.
WepassnowtoconsiderthespecificcaseinwhichbothY andZ arespacesofcontinuousfunctions
oncompactspaces. ThetwistingofseparableC(K)-spaceswastreatedand,toalargeextent,solvedin
[7]. Nevertheless, the problem of constructing nontrivialtwisted sums of large C(K)-spaces is mostly
unsolved. The following problem was posed andconsidered in [7, 8, 10]: Is it true that for every non-
metrizable compactK onehas Ext(C(K),c )6=0? In this paper we provethatExt(C(K),C(S))6=0
0
Thanksareduetoademandingrefereewhopushedtheauthortolookfornicerargumentsand,intheend,toproduce
abetter paper. AdditionalthanksareduetoDanielTauskwhomadeafewaccurateremarksaboutthecontent ofthe
paper.
ThisresearchwassupportedbyprojectMTM2013-45643-C2-1-P,Spain.
1
2 JESU´SM.F.CASTILLO
foranytwocompactspacesK,S withω-derivedsetempty andK nonmetrizablewhich,inparticular,
covers the case above for C(S)=c .
0
2. Main result, and its consequences
Inwhatfollows,thecardinalofasetΓwillbedenoted|Γ|. Wewillwritec (Γ)orc (|Γ|)depending
0 0
if one needs to stress the set or only the cardinal.
We begin with a couple of observations: First of all, that if Γ is an uncountable set then
Ext(c (Γ),c ) 6= 0. To explicitly construct an example it is enough to assume that |Γ| = ℵ , pick an
0 0 1
isomorphic embedding ϕ:c (ℵ )→ℓ /c and consider the commutative diagram
0 1 ∞ 0
p
0 −−−−→ c −−−−→ ℓ −−−−→ ℓ /c −−−−→ 0
0 ∞ ∞ 0
ϕ
(cid:13) x x
(cid:13)  
0 −−−−→ c(cid:13) −−−−→ p−1(ϕ(c (ℵ ))) −−−pϕ−→ c (ℵ ) −−−−→ 0
0 0 1 0 1
wherep (x)=ϕ−1p(x). Indeed,toseethatthelowersequencedoesnotsplit,oneonlyhastoobserve
ϕ
that in any commutative diagram
0 −−−−→ c −−−−→ ℓ −−−−→ ℓ /c −−−−→ 0
0 ∞ ∞ 0
(2) (cid:13) xϕ′ xϕ
(cid:13)  
(cid:13)  q 
0 −−−−→ c −−−−→ X −−−−→ c (Γ) −−−−→ 0.
0 0
in which ϕ is an isomorphic embedding then also ϕ′ is an isomorphic embedding, which implies that
the quotient map q cannot be an isomorphism onto any non separable subspace of c (Γ) since non-
0
separablec (Γ′) spacesarenotsubspacesof ℓ . The lowersequence indiagram(2)canhoweversplit
0 ∞
when ϕ is just an operator. Moreover,the results in [12] imply that in any exact sequence
q
0 −−−−→ c −−−−→ X −−−−→ c (J) −−−−→ 0
0 0
inwhich|J|largeenoughtheoperatorq becomesanisomorphismoncopiesofc (I)forlargeI;while,
0
on the other hand, in an exact sequence
q
0 −−−−→ Y −−−−→ X −−−−→ c (J) −−−−→ 0,
0
in which Ext(X,c )=0 then q cannot be an isomorphism onto any non-separable subspace of c (J):
0 0
otherwise,ifq becomesanisomorphismonsomesubspacec (J′),thismustbecomplementedinc (J)
0 0
by [17], and therefore q−1(c (J′)) must be complemented in X, which prevents Ext(X,c ) = 0. It
0 0
is still open to decide whether every twisted sum of two c (Γ) must be a C(K)-space (see [13] for
0
related information). We prove two preparatory lemmata of independent interest. The first one is a
reformulation of [11, Lemma 1]. Recall that given any exact sequence 0 → A → B → B/A → 0 and
anoperatorφ:A→Y there exists a superspaceX ofY suchthatX/Y =B/A (see [9, Prop. 1.3.a]),
thus forming a commutative diagram
0 −−−−→ A −−−−→ B −−−−→ B/A −−−−→ 0
(3) φ  (cid:13)
  (cid:13)
y y (cid:13)
0 −−−−→ Y −−−−→ X −−−−→ B/A −−−−→ 0.
The following lemma shows that the converse is somehow true:
NONSEPARABLE C(K)-SPACES CAN BE TWISTED WHEN K IS A FINITE HEIGHT COMPACT 3
Lemma 1. Let 0 → A → B → C → 0 be an exact sequence and let E be a Banach space. If
Ext(B,E) = 0 then for every exact sequence 0 → E → X → C → 0 there is an operator φ : A → E
so that there is a commutative diagram
0 −−−−→ A −−−−→ B −−−−→ C −−−−→ 0
(4) φ  (cid:13)
  (cid:13)
y y (cid:13)
0 −−−−→ E −−−−→ X −−−−→ C −−−−→ 0.
This result can be seen as a consequence of the homology sequence (see [5]), whose part relevant
for us is the existence of an exact sequence:
... −−−−→ L(A,E) −−ω−E−→ Ext(C,E) −−−−→ Ext(B,E) −−−−→ ··· .
Thus, if Ext(B,E) = 0 then the map ω : L(A,E) → Ext(C,E) is surjective. This map ω sends
E E
operators φ:A→E into the lower sequence in diagram (3).
Lemma 2. Let
(5) 0 −−−−→ Y −−−−→ X −−−−→ c (c) −−−−→ 0
0
be an exact sequence. If |Y∗|≤c then Ext(X,c )6=0.
0
Proof. Applying the homology sequence as above to sequence (5) with target E = c one gets that
0
the map
ω :L(Y,c ) −−−−→ Ext(c (c),c )
Y 0 0 0
is surjective when Ext(X,c ) = 0, from where |Ext(c (c),c )| ≤ |L(Y,c )|. Let us show that, under
0 0 0 0
the assumption |Y∗| ≤ c, this cannot be so: On one hand one has |L(Y,c )| ≤ |L(ℓ ,Y∗)|, which is
0 1
the set of bounded sequences of the unit ball of Y∗. Since there are |(Y∗)N| countable subsets of Y∗,
each of them admitting c bounded sequences one gets,
|L(Y,c )|≤|R×(|Y∗|ℵ0)ℵ0|=cℵ0 =c.
0
On the other hand, Marciszewski and Pol show in [21, 7.4] that there exist 2c non-equivalent exact
sequences 0→c →♠→c (c)→0; i.e., |Ext(c (c),c )|≥2c. (cid:3)
0 0 0 0
The reason behind the involved proof to come is that the straightforward proof one would obtain
from the argumentabovedoes not holds in generalsince the inequality ℵℵ0 <2ℵ does not necessarily
hold for all cardinals. Indeed, see [18, Thm. 5.15], under GCH, it occurs that when ℵ has cofinality
greater than ℵ0 then ℵℵ0 =ℵ; but for ℵ having cofinality ℵ0 then ℵℵ0 =2ℵ.
Moreover,the Continuum Hipothesis, in what follows [CH], is necessary to make c the first case to
consider. Indeed, the argument of Marciscewski-Pol could not work even for ℵ outside CH since it
1
depends on cardinal arithmetics: for c<2ℵ1 the same proof in [21] yields that there are 2ℵ1 different
exact sequences 0 → c0 → X → c0(ℵ1) → 0. However, if 2ℵ1 = c then the method in [21] does not
decide. We are indebted to W. Marciszewki who informed us from these facts.
Lemma 2 can be improved to:
Lemma 3. Let 0 → Y → X → Z → 0 be an exact sequence. Let B be a Banach space for which
|L(Y,B)|<|Ext(Z,B)|. Then |Ext(X,B)|≥|Ext(Z,B)|.
Proof. It follows from the exactness of the sequence
L(Y,B) −−ω−B−→ Ext(Z,B) −−−γ−→ Ext(X,B)
4 JESU´SM.F.CASTILLO
andthestandardfactorizationExt(Z,B)/kerγ =Imγthat|Ext(X,B)|≥|Imγ|=|Ext(Z,B)/ kerγ|=
|Ext(Z,B)/ Im ω |. On the other hand one has |Im ω | ≤ |L(Y,B)| < |Ext(Z,B)|, and thus
B B
|Ext(Z,B)/Imω |=|Ext(Z,B)|. (cid:3)
B
Tocontinuewiththepreparationfortheproofweneednowaresultthatcanbeconsideredfolklore
but which, to the best of our knowledge, cannot be found in the literature:
Lemma 4. Let Y be a subspace of X. There is a subspace Y• ⊂Y with density character densY• ≤
densX/Y and a subspace X• ⊂X such that X•/Y• =X/Y.
Proof. Let(z ) beadensesubsetoftheunitballofX/Y havingminimalsize. Bytheopenmapping
γ γ
theorem there is some constant C > 0 such that one can pick for each z an element x ∈ X with
γ γ
norm kx k≤Ckz k so that x +Y =z . Let L:X/Y →X be a linear (not necessarily continuous)
γ γ γ γ
selection for the quotient map q :X →X/Y. Thus x −Lz ∈Y and the closed subspace they span
γ γ
Y• = [x −Lz ] has density character smaller or equal than that of Z. Let X• = [Y•+[x ]] be the
γ γ γ
closed subspace of X spanned by Y• and all the x . It is clear that q : X• →X/Y is onto since the
γ
points xγ +Y =zγ are in the image of the ball of radius C of X•. Moreover,kerq|X• =Y•: indeed,
since X• =[Y•+[x ]]=[x −Lz ,x ]=[x ,Lz ], pick x∈X• of the form x= λ x + λ Lz
γ γ γ γ γ γ γ γ µ µ
P P
so that qx=0; which means λ z =− λ z . Thus
γ γ µ µ
P P
x= λ x + λ Lz = λ x +L λ z = λ x −L λ z = λ (x −Lz )
X γ γ X µ µ X γ γ (cid:16)X µ µ(cid:17) X γ γ (cid:16)X γ γ(cid:17) X γ γ γ
and thus x∈Y•. (cid:3)
Lemma 4 can be reformulated as: given an exact sequence 0 → Y → X → Z → 0 there are
subspaces Y• ⊂Y and X• ⊂X with densY• ≤densZ that form a commutative diagram
0 −−−−→ Y• −−−−→ X• −−−−→ Z −−−−→ 0
  (cid:13)
  (cid:13)
y y (cid:13)
0 −−−−→ Y −−−−→ X −−−−→ Z −−−−→ 0.
In particular, given an exact sequence 0 → c (Γ) → X → c (c) → 0, since for every subspace
0 0
E ⊂c (Γ) withdensE ≤ℵ≤|Γ| there is a copyofc (ℵ)for whichE ⊂c (ℵ)⊂c (Γ), itis clearthat
0 0 0 0
there is a commutative diagram
0 −−−−→ c (c) −−−−→ X• −−−−→ c (c) −−−−→ 0
0 0
i i• (cid:13)
  (cid:13)
0 −−−−→ c y(Γ) −−−−→ Xy −−−−→ c (cid:13)(c) −−−−→ 0,
0 0
where i,i• are the canonical inclusions. We are ready for the action:
Basic case: |J|=c:
Lemma 5. Given an exact sequence 0→c (Γ)→X →c (c)→0 then Ext(X,c )6=0.
0 0 0
Proof. If Ext(X,c ) = 0 then the map ω : L(c (Γ),c ) → Ext(c (c),c ) is surjective and thus, via
0 c0 0 0 0 0
compositionwithi,thereisalsoasurjectivemapL(c (c),c )→Ext(c (c),c ),whichisimpossible. (cid:3)
0 0 0 0
Reduction to the case |J|=c:
Lemma 6. Under CH, given an exact sequence 0 → c (Γ) → X → c (J) → 0 then Ext(X,c ) 6= 0
0 0 0
when X is nonseparable.
NONSEPARABLE C(K)-SPACES CAN BE TWISTED WHEN K IS A FINITE HEIGHT COMPACT 5
Proof. If|J|≤ℵ thenthesequencesplits(thespacesc (Γ)areseparablyinjective,whichmeansthat
0 0
they are complemented in every superspace X so that X/c (Γ) is separable (see, e.g., [3]). In which
0
case, X is isomorphic to c (Γ). Being X nonseparable, Γ must be uncountable and Ext(X,c ) 6= 0.
0 0
If |J| = c, we are in the case of Lemma 5. If |J| > c, take a set of size c in J and consider the
decomposition c (J)=c (c)⊕c (H) with canonical embedding u:c (c) →c (J). Take an arbitrary
0 0 0 0 0
exactsequence0→c →A→c (c)→0andmultiplyontherightbyc (H)togettheexactsequence
0 0 0
0→c →A⊕c (H)→c (c)⊕c (H)→0. Form then the commutative diagram
0 0 0 0
0 −−−−→ c (c) −−−−→ q−1(u(c (c)))• −−−−→ c (c) −−−−→ 0
0 0 0
i i• (cid:13)
  (cid:13)
0 −−−−→ c y(Γ) −−−−→ q−1(u(yc (c))) −−−qu−→ c(cid:13)(c −−−−→ 0
0 0 0
(cid:13) u′ u
(cid:13)  
(cid:13) y q y
0 −−−−→ c (Γ) −−−−→ X −−−−→ c (J) −−−−→ 0
0 0
φ φ′ (cid:13)
  (cid:13)
0 −−−−→ cy −−−−→ A⊕yc (H) −−−−→ c (c)⊕(cid:13)c (H) −−−−→ 0.
0 0 0 0
where the operator φ exist because Ext(X,c ) = 0. If p : A⊕c (H) → A denotes the canonical
0 A 0
projection onto A then there is a commutative diagram
0 −−−−→ c (c) −−−−→ q−1(u(c (c)))• −−−−→ c (c) −−−−→ 0
0 0 0
φi pAφ′u′i• (cid:13)
  (cid:13)
0 −−−−→ cy −−−−→ Ay −−−−→ c (cid:13)(c) −−−−→ 0.
0 0
This again implies that there is a surjective map L(c (c),c )→Ext(c (c),c ), which we have already
0 0 0 0
shown it is impossible. (cid:3)
We are now ready to show:
Proposition 1. Let X be a sequence of Banach spaces in which X is a nonseparable twisted sum
n 0
of c (Γ ) and c (J ), and X is a twisted sum of c (Γ ) and X for all n ≥ 1. Under CH,
0 0 0 0 n 0 n n−1
Ext(X ,c )6=0 for all n≥0.
n 0
Proof. We proceed (formally) inductively on n. That Ext(X ,c )6=0 has been already shown.
0 0
We show now that also Ext(X ,c )6=0. By the argumentspreceding, there is no loss of generality
1 0
assuming that |J | ≥ c, in which case there is also an exact sequence 0 → c (Γ′) → X → c (c) →
0 0 0 0 0
6 JESU´SM.F.CASTILLO
and then X and X are connected as in the commutative diagram
1 0
0 0
x x
 
c(c) = c(c)
0 0
p
x x
 
 q 
(6) 0 −−−−→ c0(Γ1) −−−−→ X1 −−−−→ X0 −−−−→ 0
(cid:13) x xu
(cid:13)  
(cid:13)  
0 −−−−→ c (Γ ) −−−−→ q−1(u(c (Γ′))) −−−−→ c (Γ′) −−−−→ 0
0 1 0 0 0 0
qu
x x
 
 
0 0.
Now, reasoning as in Lemma 5 and Lemma 6 we can assume that |Γ′| = c, and thus that also
0
|Γ |≤c; in which case |q−1(u(c (Γ′)))∗|≤c and Lemma 2 applies to yield Ext(X ,c )6=0.
1 0 0 1 0
The result for X can be obtained via the same method: call P = q−1(u(c (Γ′))) from diagram
2 1 0 0
(6). We get the diagram
0 0
x x
 
c(c) = c(c)
0 0
p
x x
 
 q 
(7) 0 −−−−→ c0(Γ2) −−−−→ X2 −−−−→ X1 −−−−→ 0
(cid:13) x xu
(cid:13)  
(cid:13)  
0 −−−−→ c (Γ ) −−−−→ q−1(u(P )) −−−−→ P −−−−→ 0
0 2 1 1
qu
x x
 
 
0 0.
After the appropriate size reduction we can assume that |P∗| ≤c and |Γ |≤ c which implies that
1 2
|q−1(u(c (Γ )))∗|≤c and thus Lemma applies to yield Ext(X ,c )6=0.
0 1 2 0
Call now P =q−1(u(P )) for n=1,2,... and iterate the argument. (cid:3)
n+1 n
Letus recallthatgivenacompactsetK,the derivedsetK(1) is the setofits accumulationpoints.
If one calls K(n+1) =(K(n))′, the ω-derived set is Kω =∩ K(n).
n
Theorem 1. [CH] Given two compact spaces K,S with ω-derived set empty and K nonmetrizable
then Ext(C(K),C(S))6=0.
Proof. The result is clear after recalling a few well-known facts:
• ThatC(S) spacesinwhichS isa compactwithSω =∅containc complementedasitfollows
0
from the existence of a convergentsequence in S.
NONSEPARABLE C(K)-SPACES CAN BE TWISTED WHEN K IS A FINITE HEIGHT COMPACT 7
• If I is the set of isolated points of K the restriction operator C(K) → C(K(1)) induces a
K
short exact sequence
0 −−−−→ c (I ) −−−−→ C(K) −−−−→ C(K(1)) −−−−→ 0.
0 K
• That,afterafinitenumberN+2ofderivationsofK,onearrivestoafinitecompact,obtaining
thus an exact sequence
0 −−−−→ c (I ) −−−−→ C(K(N)) −−−−→ c (J) −−−−→ 0.
0 K(N) 0
We are therefore in the hypotheses of Proposition 1 and thus Ext(C(K),c ) 6= 0 and therefore
0
Ext(C(K),C(S))6=0. (cid:3)
Godefroy,KaltonandLancienshowin[16,Thm. 2.2]thataC(K)spaceisLipschitzisomorphicto
c (N) if and only if it is linearly isomorphic to c . They also show that the result is no longer true in
0 0
the nonseparable case, showing that a C(K) space is Lipschitz isomorphic to some c (I) if and only
0
if the ω derived set of K is empty. Therefore,
Corollary 1. [CH]. If C(K) and C(S) are Lipschitz isomorphic to some (probably different) spaces
c (Γ) then Ext(C(K),C(S))6=0.
0
For the sake of completeness, let us gather in an omnibus theorem all known results about the
problem considered. Recall that a compact space K is said to have countable chain condition (ccc in
short) when every family of pairwise disjoint open sets is countable. It is well known [25, Thm. 4.5]
that K has ccc if and only if C(K) does not contain c (I) for I uncountable.
0
Proposition 2. One has Ext(C(S),c ) 6= 0 whenever C(S) contains a complemented non-separable
0
C(K) space for which at least one of the following conditions hold:
(1) K is an Eberlein compact.
(2) K is a Valdivia compact without ccc.
(3) [CH] K has finite height.
(4) C(K) admits a continuous injection into C(N∗) but not into ℓ .
∞
(5) K is an ordinal space.
(6) C(K) contains ℓ .
∞
Proof. The proof of (1) appears in [10]: According to Reif [24], since the space C(K) is WCG it
admits a Marku˜seviˇc basis (x ,f ) ∈ Γ, for which it can be assumed that (f ) is bounded. It is
γ γ γ γ
possible to define a dense-range operator T : C(K) → c (Γ) as T(f) = (f (f)). On the other hand,
0 γ
q
in any nontrivial exact sequences 0 → c → X → c (Γ) → 0 the space X cannot be WCG (Weakly
0 0
Compactly Generated) since c is complemented in WCG spaces. Thus, one has a commutative
0
diagram
q
0 −−−−→ c −−−−→ X −−−−→ c (Γ) −−−−→ 0
0 0
(cid:13) tx xT
(cid:13)  
(cid:13)  
0 −−−−→ c −−−−→ P −−−−→ C(K) −−−−→ 0
0 T
π
where P ={(f,g)∈X ⊕ C(K):q(f)=T(g)}, t(f,g)=f and π(f,g)=g (indeed, when T is not
T ∞
an embedding one cannot expect the middle operator t to be an embedding; i.e., one cannot expect
the lower twisted sum space to be a subspace of the upper twisted sum space). The “dense range”
versionof the 3-lemma implies that the operator t has dense range; hence, the lower sequence cannot
split since otherwise P ≃c ⊕C(K) would be WCG, as well as X.
T 0
(2)followsfrom[1, Thm. 1.2]: if K is a Valdivia compactand c (I)⊂C(K) thenthere is a subset
0
J ⊂I such that |J|=|I| and c (J) is complemented in C(K).
0
8 JESU´SM.F.CASTILLO
Assertion (3) is contained in the main results of the paper.
Assertion (4) follows from the commutative diagram
0 −−−−→ c −−−−→ ℓ −−−q−→ C(N∗) −−−−→ 0
0 ∞
(cid:13) tx xT
(cid:13)  
0 −−−−→ c(cid:13) −−−−→ q−1(ϕ(C(K))) −−−qT−→ C(K) −−−−→ 0
0
in which T is the continuous injection claimed in he hypothesis. If the lower sequence splits, t
|C(K)
would be a continuous injection C(K)→ℓ .
∞
Instances of the situation just described appear when, say, K contains a dense set of weight at
most ℵ but C(K) is not a subspace of ℓ , as it follows from Paroviˇcenko’s theorem [4], which
1 ∞
implies the existence of a continuous surjection ϕ : N∗ → K, which in turn provides the injection
ϕ◦ : C(K) → C(N∗). See also [27, Prop. 0.1]. Or else, C(K) spaces with non weak*-separable dual,
but admitting continuous injections into C(N∗).
(5) The case of uncountable ordinal spaces can be reduced to K =[0,ω ], and this is contained in
1
(4).
(6) is obvious since Ext(ℓ ,c )6=0 (see [6]) and ℓ will be complemented in C(K). (cid:3)
∞ 0 ∞
In[7,p. 4539-4540]weclaimedthattheargumentin(1)canbeextendedtocoverthecaseofCorson
compacta. WehavebeeninformedbyDanielTauskthatsuchresultdoesnotimmediatelyfollowswith
the same arguments: on one side, consider any nontrivial exact sequence 0 → c →X →c (I) → 0.
0 0
ThispreventsBX∗ frombeingaValdiviacompact. IfweassumeK isaCorsoncompactthenfrom[2,
Thm. 1.7] it follows that there is an injective operator T : C(K)→c (I), or a dense-range operator
0
T : C(K) → c (I) if one uses either Markushevich basis for C(K) or increasing PRI [26], argument
0
also valid for K Valdivia. Thus, one gets a commutative diagram
q
0 −−−−→ c −−−−→ X −−−−→ c (Γ) −−−−→ 0
0 0
(cid:13) tx xT
(cid:13)  
(cid:13)  
0 −−−−→ c −−−−→ P −−−−→ C(K) −−−−→ 0.
0 T
ButsomeargumentisrequirednowtoconcludethatnocontinuousliftingC(K)→X existsinorderto
make the lower sequence nontrivial. If T has been chosen having dense range, then t∗ :X∗ →C(K)∗
is an imbedding; thus, if BC(K)∗ is a Corson compact then also BX∗ would be a Corson compact,
which is a contradiction. However, it is consistent with ZFC the existence of Corson compacta K so
that the unit ball of C(K)∗ is not Corson (see [22, Thm. 5.9], or else [2, Thm.3.5, Ex.3.10]). The
situation for Valdivia compacta is not much better: while now K Valdivia implies that B is
C(K)∗
Valdivia [19, Thm. 5.2], it is no longer true that closed subspaces of Valdivia compacta are Valdivia.
Therefore, it is worth to state the problem:
Problem. Assume that K is a (nonmetrizable) Corson or Valdivia compact. Does one have
Ext(C(K),c )6=0?
0
Added in proof. In the meantime Correa and Tausk have shown in [15] that under CH all Corson
compact admit a nontrivial twisting with c and that the same holds for “most” Valvidia compact.
0
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Departamentode Matema´ticas,Universidadde Extremadura,Avenida de Elvas, 06011-Badajoz,Spain
E-mail address: [email protected]