A Banach space with a countable infinite number of complex structures

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by  Wilson Cuellar-Carrera

A BANACH SPACE WITH A COUNTABLE INFINITE NUMBER OF
COMPLEX STRUCTURES
W. CUELLAR CARRERA
4
1
0
2 Abstract. WegiveexamplesofrealBanachspaceswithexactlyinfinitecountablymany
n complex structuresand with ω1 many complex structures.
a
J
8 1. Introduction

A real Banach space X is said to admit a complex structure when there exists a linear
]
A operator I on X such that I2 = −Id. This turns X into a C-linear space by declaring a
F new law for the scalar multiplication:
.
h
(λ+iµ).x = λx+µI(x) (λ,µ ∈ R).
t
a
m Equipped with the equivalent norm
[
kxk = sup kcosθx+sinθIxk,

1 0≤θ≤2π
v
we obtain a complex Banach space which will be denoted by XI. The space XI is the
1
8 complex structure of X associated to the operator I, which is often referred itself as a
7 complex structure for X.
1
WhenthespaceX isalreadyacomplexBanach space, theoperatorIx = ixisacomplex
.
1
structureon X (i.e., X seen as areal space) which generates X. Recall that for acomplex
0 R
4 Banach space X its complex conjugate X is defined to be the space X equipped with the
1 new scalar multiplication λ.x = λx.
v: Two complex structures I and J on a real Banach space X are equivalent if there exists
i a real automorphism T on X such that TI = JT. This is equivalent to saying that the
X
spaces XI and XJ are C-linearly isomorphic. To see this, simply observe that the relation
r
a TI = JT actually means that the operator T is C-linear as defined from XI to XJ.
WenotethatacomplexstructureI onarealBanach spaceX isanautomorphismwhose
inverse is −I, which is itself another complex structure on X. In fact, the complex space
X−I is the complex conjugate space of XI. Clearly the spaces XI and X−I are always
R-linearly isometric. On the other hand, J. Bourgain [3] and N. Kalton [12] constructed
Thisarticleispartoftheauthor’sPh.DthesissupervisedbyProf. V.FerencziwithsupportofFAPESP
2010/17512-6
This paper was written during a visit of the author at the ICMAT-Madrid supervised by Prof. J. Lopez-
Abad and financed by FAPESP 2012/00631-8.
1
2 W. CUELLAR CARRERA
examples of complex Banach spaces not isomorphic to their corresponding complex con-
jugates, hence these spaces admit at least two different complex structures. Bourgain
example is an ℓ sum of finite dimensional spaces whose distance to their conjugates tends
2
to infinity. Kalton example is a twisted sum of two Hilbert spaces i.e., X has a closed
subspace E such that E and X/E are Hilbertian, while X itself is not isomorphic to a
Hilbert space.
Complex structures do not always exist on Banach spaces. The first example in the
literature was the James space, proved by J. Dieudonn´e [4]. Other examples of spaces
without complex structures are the uniformly convex space constructed by S. Szarek [15]
and the hereditary indecomposable space of W. T. Gowers and B. Maurey [8]. Gowers
[9, 10] also constructed a space with unconditional basis but without complex structures.
In general these spaces have few operators. For example, every operator on the Gowers-
Maureyspaceisastrictlysingularperturbationofamultipleoftheidentityandthisforbids
complex structures: suppose that T is an operator on this space such that T2 = −Id and
write T =λId+S with S a strictly singular operator. It follows that (λ2+1)Id is strictly
singular and of course this is impossible.
More examples of Banach spaces without complex structures were constructed by P.
Koszmider,M.Mart´ınandJ.Mer´ı[13,14]. Infact, theyintroducedthenotionofextremely
non-complex Banach space: A real Banach space X is extremely non-complex if every
bounded linear operator T : X → X satisfies the norm equality kId+T2k = 1+kTk2.
AmongtheirexamplesofextremelynoncomplexspacesareC(K)spaceswithfewoperators
(e.g. when every bounded linear operator T on C(K) is of the form T = gId+S where
g ∈ C(K) and S is a weakly compact operator on C(K)), a C(K) space containing a
complemented isomorphic copy of ℓ (thus having a richer space of operators than the
∞
first one mentioned) and an extremely non complex space not isomorphic to any C(K)
space.
Going back to the problem of uniqueness of complex structures, Kalton proved that
spaces whose complexification is a primary space have at most one complex structure [6].
In particular, the classical spaces c , ℓ (1 ≤ p ≤ ∞), L [0,1] (1 ≤ p ≤ ∞), and C[0,1]
0 p p
have a unique complex structure.
We have mentioned before examples of Banach spaces with at least two different com-
plex structures. In fact, V. Ferenczi [5] constructed a space X(C) such that the complex
structure X(C)J associated to some operator J and its conjugate are the only complex
structures on X(C) up to isomorphism. Furthermore, every R-linear operator T on X(C)
is of the form T = λId+µJ +S, where λ,µ are reals and S is strictly singular. Ferenczi
also proved that the space X(C)n has exactly n+1 complex structures for every positive
integern. Goingtotheextreme, R.Anisca[1]gave examplesofsubspacesofL (1 ≤ p < 2)
p
which admit continuum many non-isomorphic complex structures.
A BANACH SPACE WITH A COUNTABLE INFINITE NUMBER OF COMPLEX STRUCTURES 3
The question remains about finding examples of Banach spaces with exactly infinite
countablymanydifferentcomplexstructures. Afirstnaturalapproachtosolvethisproblem
is to construct an infinite sum of copies of X(C), and in order to control the number of
complex structures to take a regular sum, for instance, ℓ (X(C)). It follows that every
1
R-linear bounded operator T on ℓ (X(C)) is of the form T = λ(T)+S, where λ(T) is the
1
scalar part of T, i.e., an infinite matrix of operators on X(C) of the form λ Id+µ J,
i,j i,j
and S is an infinite matrix of strictly singular operators on X(C). It is easy to prove that
if T is a complex structure then λ(T) is also a complex structure. Recall from [5] that two
complex structures whose difference is strictly singular must be equivalent. Unfortunately,
theoperator S in therepresentation of T is notnecessarily strictly singular, andthismakes
very difficult to understand the complex structures on ℓ (X(C)).
1
It is necessary to consider a more “rigid” sum of copies of spaces like X(C). We found
this interesting property in the space X constructed by S. Argyros, J. Lopez-Abad and
ω1
S. Todorcevic [2]. Based on that construction we present a separable reflexive Banach
space X (C) with exactly infinite countably many different complex structures which
ω2
admits an infinite dimensional Schauder decomposition X (C) = X for which every
ω2 k k
R-linear operator T on X (C) can be written as T = D +S, where S is strictly singular,
ω2 T
L
D | = λ Id (λ ∈C) and (λ ) is a convergent sequence.
T Xk k Xk k k k
ThisconstructionalsoshowstheexistenceofcontinuummanyexamplesofBanachspaces
with the property of having exactly ω complex structures and the existence of a Banach
space with exactly ω complex structures.
1
2. Construction of the space X (C)
ω1
We construct a complex Banach space X (C) with a bimonotone transfinite Schauder
ω1
basis (e ) , such that every complex structure I on X (C) is of the form I = D+S,
α α<ω1 ω1
where D is a suitable diagonal operator and S is strictly singular.
By a bimonotone transfinite Schauder basis we mean that X (C) = span(e ) and
ω1 α α<ω1
suchthatforeveryintervalI ofω thenaturallydefinedmaponthelinearspan of(e )
1 α α<ω1
λ e 7→ λ e
α α α α
αX<ω1 Xα∈I
extends to a bounded projection P :X (C) → X = span (e ) with norm equal to 1.
I ω1 I C α α∈I
Basically X (C) corresponds to the complex version of the space X constructed in
ω1 ω1
[2] modifying the construction in a way that its R-linear operators have similar structural
properties to theoperators in the original space X (i.e. the operators are strictly singular
ω1
perturbation of a complex diagonal operator).
First we introduce the notation that will be used through all this paper.
2.1. Basic notation. Recall that ω and ω denotes the least infinite cardinal number
1
and the least uncountable cardinal number, respectively. Given ordinals γ,ξ we write
γ+ξ,γ·ξ,γξ for the usual arithmetic operations (see [11]). For an ordinal γ we denote by
4 W. CUELLAR CARRERA
Λ(γ) the set of limit ordinals < γ. Denote by c (ω ,C) the vector space of all functions
00 1
x : ω → C such that the set suppx = {α < ω : x(α) 6= 0} is finite and by (e ) its
1 1 α α<ω1
canonical Hamel basis. For a vector x ∈ c (ω ,C) ranx will denote the minimal interval
00 1
containingsuppx. GiventwosubsetsE ,E ofω wesaythatE < E ifmaxE < minE .
1 2 1 1 2 1 2
Then for x,y ∈ c (ω ,C) x < y means that suppx < suppy. For a vector x ∈ c (ω ,C)
00 1 00 1
and a subset E of ω we denote by Ex (or P x) the restriction of x on E or simply the
1 E
function xχ . Finally in some cases we shall denote elements of c (ω ,C) as f,g,h...
E 00 1
and its canonical Hamel basis as (e∗) meaning that we refer to these elements as being
α α<ω1
functionals in the norming set.
2.2. Definitionofthenormingset. ThespaceX (C)shallbedefinedasthecompletion
ω1
ofc (ω ,C)equippedwithanormgivenbyanormingsetK (C) ⊆ c (ω ,C). Thismeans
00 1 ω1 00 1
that the norm for every x ∈ c (ω ,C) is defined as sup{|φ(x)| = | φ(α)x(α)| : φ ∈
00 1 α<ω1
K (C)}. The norm of this space can also be defined inductively.
ω1 P
We start by fixing two fast increasing sequences (m ) and (n ) that are going to be used
j j
in the rest of this work. The sequences are defined recursively as follows:
1. m = 2 e m = m4;
1 j+1 j
2. n1 = 4 e nj+1 = (4nj)sj, where sj = log2m3j+1.
Let K (C) be the minimal subset of c (ω ,C) such that
ω1 00 1
1. It contains every e∗, α < ω . It satisfies that for every φ ∈ K (C) and for every
α 1 ω1
complex number θ = λ+iµ with λ and µ rationals and |θ|≤ 1, θφ∈ K (C). It is
ω1
closed under restriction to intervals of ω .
1
2. Forevery{φ , : i= 1,...,n }⊆ K (C)suchthatφ < ··· < φ ,thecombination
i 2j ω1 1 n2j
n2j
1
φ = φ ∈ K (C).
m i ω1
2j
i=1
X
In this case we say that φ is the result of an (m−1,n )-operation.
2j 2j
3. For every special sequence (φ ,...,φ ) (see the Definition 13), the combination
1 n2j+1
n2j+1
1
φ = φ ∈ K (C).
m i ω1
2j+1
i=1
X
In this case we say that φ is a special functional and that φ is the result of an
(m−1 ,n )-operation.
2j+1 2j+1
4. It is rationally convex.
Define a norm on c (ω ,C) by setting
00 1
kxk = sup φ(α)x(α) : φ∈ K (C) .
ω1
((cid:12) (cid:12) )
(cid:12)αX<ω1 (cid:12)
(cid:12) (cid:12)
The space Xω1(C) is defined as th(cid:12)(cid:12)e completion o(cid:12)(cid:12)f (c00(ω1,C),k.k).
A BANACH SPACE WITH A COUNTABLE INFINITE NUMBER OF COMPLEX STRUCTURES 5
This definition of the norming set K (C) is similar to the one in [2]. We add the
ω1
property of being closed under products with rational complex numbers of the unit ball.
This, together with property 2 above, guarantees the existence of some type of sequences
(like ℓn-averages and R.I.S see Appendix) in the same way they are constructed for X .
1 ω1
It follows that the norm is also defined by
kxk = sup φ(x) = φ(α)x(α) : φ∈ K (C), φ(x) ∈ R .
ω1
( )
αX<ω1
We also have the following implicit formula for the norm:
n2j
1
kxk = max kxk , supsup kE xk, E < E < ··· < E ∨
∞ m i 1 2 n2j
( j 2j )
i=1
X
n2j+1
1
sup φ (Ex) : (φ )n2j+1 is n - special, E interval .
(m2j+1 (cid:12) i (cid:12) i i=1 2j+1 )
(cid:12) Xi=1 (cid:12)
(cid:12) (cid:12)
It follows from the definition of the norming set that the canonical Hamel basis (e )
(cid:12) (cid:12) α α<ω1
is a transfinite bimono(cid:12)tone Schaude(cid:12)r basis of X (C). In fact, by Property 1 for every
ω1
interval I of ω the projection P has norm 1:
1 I
kP xk = sup |fP x| = sup |P fx|≤ kxk
I I I
f∈Kω1(C) f∈Kω1(C)
Moreover, we have that the basis (e ) is boundedly complete and shrinking, the proof
α α<ω1
is the obvious modification to the one for X (see [2, Proposition 4.13]). In consequence
ω1
X (C) is reflexive.
ω1
ω∗
Proposition 1. Kω1(C) = BX∗ω1(C).
ω∗
Proof. RecallthatthesetK (C)isbydefinitionrationalconvex. WenoticethatK (C)
ω1 ω1
ω∗ ω∗
is actually a convex set. Indeed let f,g ∈ K (C) and t ∈ (0,1). Suppose that f → f,
ω1 n
ω∗
g → g and t → t, where f ,g ∈ K (C) and t ∈ Q ∩(0,1) for every n ∈ N. Then
n n ω∗ n n ω1 n
tf +(1−t)g ∈ K (C) because
ω1
ω∗
t f +(1−t )g → tf +(1−t)g.
n n n n
Inthesamemannerwecan prove thatX∗ (C)is balanced i.e., λX∗ (C) ⊆ X∗ (C)forevery
ω1 ω1 ω1
ω∗
|λ| ≤ 1. To prove the Proposition suppose that there exists f ∈ BX∗ω1(C) \Kω1(C) . It
follows by a standard separation argument that there exists x ∈ X (C) such that
ω1
|f(x)| > sup{|g(x)| : g ∈ K (C)}
ω1
which is absurd. (cid:3)
6 W. CUELLAR CARRERA
3. Complex structures on X (C)
ω1
Let I ⊆ ω be an interval of ordinals, we denote by X (C) the closed subspace of X (C)
1 I ω1
generated by {e } . For every ordinal γ < ω we write X (C) = X (C). Notice that
α α∈I 1 γ [0,γ)
X (C) is a 1-complemented subspace of X (C): the restriction to coordinates in I is a
I ω1
projection of norm 1 onto X (C). We denote this projection by P and by PI = (Id−P )
I I I
thecorrespondingprojectiononto thecomplementspace(Id−P )X (C), whichwedenote
I ω1
by XI(C).
Atransfinitesequence(y ) iscalled ablocksequencewheny < y forallα < β < γ.
α α<γ α β
Given a block sequence (y ) a block subsequence of (y ) is a block sequence (x )
α α<γ α α<γ β β<ξ
in the span of (y ) . A real block subsequence of (y ) is a block subsequence in the
α α<γ α α<γ
real span of (y ) . A sequence (x ) is a block sequence of X (C) when it is a block
α α<γ n n∈N ω1
subsequence of (e ) .
α α<ω1
Theorem 2. Let T : X (C) → X (C) be a complex structure on X (C), that is, T is
ω1 ω1 ω1
a bounded R-linear operator such that T2 = −Id. Then there exists a bounded diagonal
operator D : X (C)→ X (C), which is another complex structure, such that T −D is
T ω1 ω1 T
strictly singular. Moreover D = k ǫ iP for some signs (ǫ )k and ordinal intervals
T j=1 j Ij j j=1
I < I < ... < I whose extremes are limit ordinals and such that ω = ∪k I .
1 2 k P 1 j=1 j
The strategy for the proof of Theorem 2 is the same than the one in [2, Theorem 5.32]
for the real case. However here we want to understand bounded R- linear operators in a
complex space. This forces us to justify that the ideas from [2] still work in our context.
The result is obtained using the following theorems that we explain with more details in
the Appendix.
Step I. There exists a family F of semi normalized block subsequences of (e ) , called
α α<ω1
R.I.S (Rapidly Increasing Sequences), such that every normalized block sequence (x )
n n∈N
of X (C) has a real block subsequence in F.
ω1
RecallthataBanachspaceX ishereditarilyindecomposable(orH.I)ifno(closed)subspace
of X can be written as the direct sum of infinite-dimensional subspaces. Equivalently, for
any two subspaces Y, Z of X and ǫ > 0, there exist y ∈ Y, z ∈ Z such that kyk = kzk = 1
and ky−zk < ǫ.
StepII.Foreverynormalizedblocksequence(x ) ofX (C),thesubspacespan (x )
n n∈N ω1 R n n∈N
of X (C) is a real H.I space.
ω1
Step III. Let (x ) be a R.I.S and T : span (x ) → X (C) be a bounded R-
n n∈N C n n∈N ω1
linear operator. Then lim d(Tx ,Cx )= 0.
n n
n→∞
The proof of Step I, II and III are given in the Appendix.
Step IV. Let (x ) be a R.I.S and T :span (x ) → X (C) be a bounded R-linear
n n∈N C n n∈N ω1
A BANACH SPACE WITH A COUNTABLE INFINITE NUMBER OF COMPLEX STRUCTURES 7
operator. Then the sequence λ : N −→ C defined by d(Tx ,Cx ) = kTx −λ (n)x k is
T n n n T n
convergent.
Proof of Step IV. First we note that the sequence (λ (n)) is bounded. Then consider
T n
(α ) and (β ) two strictly increasing sequences of positive integers and suppose that
n n n n
λ (α )−→ λ and λ (β ) −→ λ , when n −→ ∞. Going to a subsequence we can assume
T n 1 T n 2
that x < x < x for every n ∈ N.
αn βn αn+1
Fix ǫ > 0. Using the result of the Step III, we have that lim kTx −λ x k = 0. By
n→∞ αn 1 αn
passing to a subsequence if necessary, assume
ǫ
kTx −λ x k ≤ .
αn 1 αn 2n6
for every n ∈ N. Hence, for every w = a x ∈ span (x ) with kwk ≤ 1 we have
n n αn R αn n
P
kTw−λ wk ≤ |a |kTx −λ x k
1 n αn 1 αn
n
X
≤ ǫ/3.
because (e ) is a bimonotone transfinite basis. In the same way, we can assume that
α α<ω1
for every w ∈ span (x ) with kwk ≤ 1, kTw−λ wk ≤ ǫ/3. By Step II we have that
R βm m 2
span (x ) ∪ (x ) is real-H.I. Then there exist unit vectors w ∈ span (x ) and
R αn n βn n 1 R αn n
w ∈ span (x ) , such that kw −w k ≤ ǫkTk. Therefore,
2 R βm m 1 2 3
kλ w −λ w k ≤ kTw −λ w k+kTw −Tw k+kTw −λ w k ≤ ǫ.
1 1 2 2 1 1 1 1 2 2 2 2
By other side
kλ w −λ w k ≥ k(λ −λ )w k−kλ (w −w )k = |λ −λ |−|λ |ǫ.
1 1 2 2 1 2 1 2 1 2 1 2 2
In consequence, |λ −λ | ≤ (1+|λ |)ǫ. Since ǫ was arbitrary, it follows that λ = λ . (cid:3)
1 2 2 1 2
Let T : X (C) → X (C) be a bounded R-linear operator. There is a canonical way
ω1 ω1
to associate a bounded diagonal operator D (with respect to the basis (e ) ) such
T γ γ<ω1
that T − D is strictly singular: Fix α ∈ Λ(ω ) a limit ordinal, and (x ) , (y )
T 1 n n∈N n n∈N
two R.I.S such that sup maxsuppx = sup maxsupp y = α+ω. By a property of F
n n n n
we can mix the sequences (x ) , (y ) in order to form a new R.I.S (z ) such that
n n n n n n∈N
z ∈ {x } and z ∈ {y } for all k ∈ N (See Remark 16). Then it follows from
2k n n∈N 2k−1 n n∈N
Step IV that the sequences defined by the formulas d(Tx ,Cx ) = kTx −λ (n)x k and
n n n T n
d(Ty ,Cy ) = kTy −µ(n)y k are convergent, and by the mixing argument, they must
n n n n
have the same limit. Hence for each α ∈ Λ(ω ) there exists a unique complex number
1
ξ (α) such that
T
lim kTw −ξ (α)w k= 0
n T n
n→∞
for every (w ) R.I.S in X , where we write I to denote the ordinal interval [α,α+
n n∈N Iα α
ω). We proceed to define a diagonal linear operator D on the (linear) decomposition of
T
8 W. CUELLAR CARRERA
span(e )
α α<ω1
span(e ) = span(x )
α α<ω1 β β∈Iα
α∈MΛ(ω1)
by setting D (e ) =ξ (α)e when β ∈ I .
T β T β α
Observe in addition that this sequence (ξ (α)) is convergent. That is, for every
T α∈Λ(ω1)
strictly increasing sequence (α ) in Λ(ω ), the corresponding subsequence (ξ (α ))
n n∈N 1 T n n∈N
is convergent. In fact, for every n ∈ N, fix (yk) a R.I.S in X . Then we can take
n k∈N Iαn
(ynkn)n∈N a R.I.S such that kTynkn −ξT(αn +ω)ynknk < 1/n. It follows by Step IV there
exists λ ∈ C such that lim kTykn −λyknk = 0. This implies that lim ξ (α +ω)= λ.
n n n n T n
IngeneralthisoperatorD definesaboundedoperatoronX (C). Theproofisthesame
T ω1
that in [2, Proposition 5.31] and uses that certain James like space of a mixed Tsirelson
space is finitely interval representable in every normalized transfinite block sequence of
X (C). For the case of complex structures we have a simpler proof (see Proposition 6).
ω1
Proposition 3. Let A be a subset of ordinals contained in ω and X = span (e ) .
1 C α α∈A
Let T : X → X (C) be a bounded R-linear operator. Then T is strictly singular if and
ω1
only if for every (y ) R.I.S on X, lim Ty = 0.
n n∈N n n
Proof. The proposition is trivial when the set A is finite, then we assume that A is infinite.
Suppose that T is strictly singular. Let (y ) be a R.I.S on X such that lim Ty 6= 0,
n n∈N n n
then by Step IV there is λ 6= 0 with lim kTy −λy k = 0. Take 0 < ǫ < |λ|. By passing
n n n
to a subsequence if necessary, we assume that k(T −λId)| k< ǫ. This implies that
span(yn)n
T| is an isomorphism which is a contradiction.
span(yn)n
Conversely, suppose that for every (y ) R.I.S on X, lim Ty = 0. Assume that T
n n n n
is not strictly singular. Then there is a block sequence subspace Y = span(y ) of X
n n∈N
such that T restricted to Y is an isomorphism. By Step I we can assume that the sequence
(y ) is already a R.I.S on X. Then inf kTy k > 0. And we obtain a contradiction. (cid:3)
n n n n
Given Y ⊆ X (C) we denote by ι the canonical inclusion of Y into X (C).
ω1 Y ω1
Corollary 4. Let α ∈ Λ(ω ) and T : X (C) → X (C) be a bounded R-linear operator.
1 Iα ω1
Then there exists (unique) ξ (α) ∈ C such that T −ξ (α)ι is strictly singular.
T T XIα(C)
Proof. Let ξ (α) be the (unique) complex number such that limkTy −ξ (α)y k = 0 for
T n T n
every (y ) R.I.S on X (C). Thenby the previousProposition T−ξ (α)ι is strictly
singular.n n Iα T XIα(C) (cid:3)
Corollary 5. Let α ∈ Λ(ω ) and R : X (C) → XIα(C) be a bounded R-linear operator.
1 Iα
Then R is strictly singular.
Proof. By the previous result, ι R = λ ι + S with S strictly singular. Then
XIα(C) α XIα(C)
projecting by PIα we obtain R = PIα ◦ι R = PIαS which is strictly singular. (cid:3)
XIα(C)
Proposition 6. Let T be a complex structure on X (C). Then the linear operator D is
ω1 T
a bounded complex structure.
A BANACH SPACE WITH A COUNTABLE INFINITE NUMBER OF COMPLEX STRUCTURES 9
Proof. Let T be a complex structure on X (C) and D the corresponding diagonal oper-
ω1 T
ator defined above. Fix α ∈ Λ(ω ). We shall prove that ξ (α)2 = −1. In fact,
1 T
T ◦ι = P T ◦ι +PIαT ◦ι
XIα(C) Iα XIα(C) XIα(C)
= P T ◦ι +S
Iα XIα(C) 1
where S is strictly singular. This implies P T ◦ι = ξ (α)Id +S :X (C) →
1 Iα XIα(C) T XIα(C) 2 Iα
X (C) with S strictly singular. Now computing:
Iα 2
(P Tι )◦(P Tι ) = P T ◦P Tι
Iα XIα(C) Iα XIα(C) Iα Iα XIα(C)
= P T ◦(Id−PIα)Tι
Iα XIα(C)
= P T2ι −P TPIαTι
Iα XIα(C) Iα XIα(C)
= −Id +S
XIα(C) 3
where S is strictly singular because the underlined operator is strictly singular. Hence we
3
have that (ξ (α)2 +1)Id is strictly singular.Which allow us to conclude that ξ (α)2 =
T XIα T
−1. The continuity of D is then guaranteed by the convergence of (ξ (α)) . In
T T α∈Λ(ω1)
deed, we have that there exist ordinal intervals I < I < ... < I with ω = ∪k I and
1 2 k 1 j=1 j
such that D = k ǫ iP for some signs (ǫ )n .
T j=1 j Ij j j=1
(cid:3)
P
Remark 7. More generally, the proof of Proposition 6 actually shows that if T is a R-
linear bounded operator on X (C) such that T2+Id = S for some S strictly singular, then
ω1
D is bounded and D2 = −Id.
T T
Now we can conclude the proof of Theorem 2.
Proof of Theorem 2. Let T : X (C) → X (C) be a bounded R-linear operator which is
ω1 ω1
a complex structure and D be the diagonal bounded operator associated to it. It only
T
remainstoprovethatT−D isstrictlysingular. AndthisfollowsdirectlyfromProposition
T
3, because by definition lim (T −D )y = 0 for every (y ) R.I.S on X (C). (cid:3)
n T n n n ω1
We come back to the study of the complex structures on X (C). Denote by D the
ω1
family of complex structures D on X (C) as in Theorem 2, i. e., D = k ǫ iP
T ω1 T j=1 j Ij
where (ǫ )k are signs and I < I < ... < I are ordinal intervals whose extremes are
j j=1 1 2 k P
limit ordinals and such that ω = ∪k I . Notice that D has cardinality ω .
1 j=1 j 1
Recall that two spaces are said to be incomparable if neither of them embed into the
other.
Corollary 8. Thespace X (C)hasω manycomplexstructuresuptoisomorphism. More-
ω1 1
over any two non-isomorphic complex structures are incomparable.
Proof. LetJ beacomplexstructureonX (C). ByTheorem2wehavethatJ isequivalent
ω1
to one of the complex structures of the family D.
10 W. CUELLAR CARRERA
To complete the proof it is enough to show that given two different elements of D
they define non equivalent complex structures. Moreover, we prove that one structure
does not embed into the other. Fix J 6= K ∈ D. Then there exists an ordinal interval
I = [α,α+ω)suchthat, withoutlossofgenerality, J| = iId| andK| =−iId| .
α XIα XIα XIα XIα
Suppose that there exists T : X (C)J → X (C)K an isomorphic embedding. Then T
ω1 ω1
is in particular a R-linear operator such that TJ = KT. We write using Corollary 4,
T| = ξ (α)ι +S with S strictly singular. Then ξ (α)J| −ξ (α)K| = S
XIα T XIα(C) T XIα T XIα 1
where S is strictly singular. Im particular for each x ∈ X , S x = 2ξ (α)ix. It follows
1 Iα 1 T
from the fact that X is infinite dimensional that ξ (α) = 0. Hence T| = S but this a
contradiction becausIeαT is an isomorphic embeddingT. XIα (cid:3)
The next corollary offers uncountably many examples of Banach spaces with exactly
countably many complex structures.
Corollary 9. The space X (C) has ω complex structures up to isomorphism for every limit
γ
ordinal ω2 ≤ γ < ω .
1
Proof. LetJ beacomplexstructureonX (C). WeextendJ toacomplexstructuredefined
γ
in the whole space X (C) by setting T = JP +iPI, where I = [0,γ). It follows that
ω1 I
T = D +S for an strictly singularoperator S and adiagonal operator D like inTheorem
T T
2. Notice that D x = ix for every x ∈ XI, otherwise there would be a limit ordinal α
T
such that S| = 2iId| . Hence JP = D P + S. Which implies that J has the
XIα XIα I T I
form J = k ǫ iP +S where S is strictly singular on X (C), (ǫ )k are signs and
j=1 j Ij 1 1 ω1 j j=1
I < I < ... < I are ordinal intervals whose extremes are limit ordinals and such that
1 2 k
γ = ∪k IP. Now the rest of the proof is identical to the proof of the previous corollary.
j=1 j
In particular, all the non-isomorphic complex structures on X (C) are incomparable. (cid:3)
γ
We also have, using the same proof of the previous corollary, that for every increasing
sequence of limit ordinals A= (α ) , the space X = X (C), where I = [α ,α +
n n A n Iαn αn n n
ω), has exactly infinite countably many different complex structures. Hence there exists a
L
family, with the cardinality of the continuum, of Banach spaces such that every space in
it has exactly ω complex structures.
4. Question and Observations
Iseasy tocheck thatsubspacesof even codimensionof arealBanach space withcomplex
structurealsoadmitcomplexstructure. AninterestingpropertyofX (C)isthatanyofits
ω1
real hyperplanes (and thus every real subspace of odd codimension) do not admit complex
structure.
Proposition 10. The real hyperplanes of X (C) do not admit complex structure.
ω1
Proof. By the results of Ferenczi and E. Galego [7, Proposition 13] it is sufficient to prove
that the ideal of all R-linear strictly singular operators on X (C) has the lifting property,
ω1
that is, for any R-linear isomorphismon X (C)such that T2+Idis strictly singular, there
ω1
exists astrictly singularoperatorS suchthat(T−S)2 =−Id. Theproofnowfollows easily
from the Remark 7. (cid:3)