Even infinite dimensional real Banach spaces
7 Even infinite dimensional real Banach spaces
Valentin Ferenczi∗ and Elói Medina Galego
Abstract
This article is a continuation of a paper of the first author [5] about com-
plex structures on real Banach spaces. We define a notion of even infinite
dimensional real Banach space, and prove that there exist even spaces, in-
cluding HI or unconditional examples from [5] and C(K) examples due to
Plebanek [13]. We extend results of [5] relating the set of complex structures
up to isomorphism on a real space to a group associated to inessential opera-
tors on that space, and give characterizations of even spaces in terms of this
group. We also generalize results of [5] about totally incomparable complex
structures to essentially incomparable complex structures, while showing
that the complex version of a space defined by S. Argyros and A. Manous-
sakis [2] provide examples of essentially incomparable complex structures
which are not totally incomparable. 1 2
1 Introduction
Any complex Banach space is also a real Banach space. Conversely, the linear
structure on a real Banach space X may be induced by a C-linear structure; the
corresponding complex Banach space is said to be a complex structure on X .
The existing theory of complex structure is up to isomorphism. In this setting,
a complex structure on a real Banach space X is a complex space which is R-
linearly isomorphic to X . Any complex structure up to isomorphism is associated
∗This article was written during a visit of the first author at the University of São Paulo financed
by the Centre National de la Recherche Scientifique.
1MSC numbers: 46B03, 47A53.
2Keywords: complex structures, inessential operators, even Banach spaces, spectral theory on
real spaces.
http://arxiv.org/abs/0704.1459v1
to an R-linear isomorphism I on X such that I2 = −Id. Conversely, for any such
isomorphism I , an associated complex structure may be defined by the law
∀λ, µ ∈ R, (λ+ iµ).x = (λId+ µI)(x),
and the equivalent norm
|||x||| = sup
‖cos θx+ sin θIx‖ .
Isomorphic theory of complex structure addresses questions of existence, uni-
queness, and when there is more than one complex structure, the possible structure
of the set of complex structures up to isomorphism (for example in terms of car-
dinality).
It is well-known that complex structures do not always exist (up to isomor-
phism) on a Banach space. The HI space of Gowers and Maurey [9] is a good
example of this, or more generally any space with the λId+S property (i.e. every
operators is a strictly singular perturbation of a multiple of the identity); and note
that since this property passes to hyperplanes, complex structures neither exist
on finite codimensional subspaces of such a space (relate this observation to the
forthcoming Question 17).
By the examples of [3] and [10] of complex spaces not isomorphic to their
conjugates, there exists real spaces with at least two complex structures up to iso-
morphism, and the examples of [3] and [1] (which are separable) actually admit
a continuum of complex structures. In [5] the first author showed that for each
n ≥ 2 there exists a space with exactly n complex structures. He also gave var-
ious examples of spaces with unique complex structure up to isomorphism and
different from the classical example of ℓ2, including a HI example and a space
with an unconditional basis.
A fundamental tool in [5] is an identification of isomorphism classes of com-
plex structures on a space X with conjugation classes in some group associated to
strictly singular operators on X . It remained open whether the associated map was
bijective. In this paper, we show that it is not bijective in general, but that there
actually exists a natural bijection between isomorphism classes of complex struc-
ture on a space X and on its hyperplanes with conjugation classes in the group
associated to strictly singular operators in X , Theorem 11. We also note that this
holds as well when one replaces strictly singular operators by any Fredholm ideal
in L(X), such as In(X) the ideal of inessential operators.
More precisely, it turns out to be fundamental to determine when a given oper-
ator of square −Id modulo inessential operators lifts to an operator of square −Id
on the space. While the answer is always positive in the complex case, Proposi-
tion 6, it turns out that on a real space, such an operator lifts either to an operator
on X of square −Id or, in a sense made precise in Lemma 9, to an operator on
an hyperplane of X of square −Id, Proposition 10; furthermore the two cases
are exclusive, Proposition 8. This implies that operators of square −Id modulo
inessential operators characterize complex structures on X and on its hyperplanes.
This leads us to define a notion of even and odd real Banach space extending
the classical notion for finite dimensional spaces. Even spaces are those spaces
which admit complex structure but whose hyperplanes do not. Odd spaces are the
hyperplanes of even spaces.
We provide characterizations of even and odd spaces in terms of the previously
mentioned groups associated to inessential operators on X and on its hyperplanes,
Corollary 12 or more precisely, in terms of lifting properties of operators of square
−Id modulo inessential operators, Proposition 13. We prove that there exist even
infinite dimensional Banach spaces, using various examples from [5], including
a HI and an unconditional example, Theorem 14. Moreover we use spaces con-
structed in [13] to give examples of even and odd spaces of the form C(K), The-
orem 18. We also show that the direct sum of essentially incomparable infinite
dimensional spaces is even whenever both spaces are even, Proposition 16.
Finally we extend and simplify the proof of some results of [5] about totally
incomparable complex structures by showing that essentially incomparable com-
plex structures are necessarily conjugate, Theorem 21 and Corollary 22. We also
show that the complex version of a space built by S. Argyros and A. Manoussakis
[2] provides examples of complex structures on a space which are essentially in-
comparable yet not totally incomparable, Proposition 25.
2 Parity of infinite dimensional spaces
It may be natural to think that a real infinite dimensional space of the form X⊕X
should be considered to be even. This seems to be restrictive however, as we
should consider as candidates for spaces with even dimension the spaces which
admit a complex structure, and there are spaces with complex structure which
are not isomorphic to a cartesian square (actually, not even decomposable, by
the examples of [5]). Another problem is that we would wish the hyperplanes
of a space with even infinite dimension not to share the same property. In other
words, parity should imply a structural difference between the whole space and its
hyperplanes. This suggests the following definition, which obviously generalizes
the case of finite dimensional spaces, and will be our guideline for this section.
Definition 1 A real Banach space is even if it admits a complex structure but its
hyperplanes do not admit a complex structure. It is odd if its hyperplanes are
even.
Equivalently a Banach space is odd if it does not have a complex structure but
its hyperplanes do, and clearly 2-codimensional subspaces of an even (resp. odd)
space are even (resp. odd).
The following crucial fact will be used repeatedly without explicit reference:
two complex structures XI and XJ are isomorphic if and only if I and J are
conjugate, i.e. there exists an isomorphism P on X such that J = PIP−1 (P
is then C-linear from XI onto XJ ). There is therefore a natural correspondance
between isomorphism classes of complex structure on X and conjugacy classes of
elements of square −Id in GL(X), and we shall sometimes identify the two sets.
Our first results are improvements of results from [5]. We recall that an op-
erator T : Y → Z is Fredholm if its kernel is finite dimensional and its image is
finite codimensional, in which case the Fredholm index of T is defined by
i(T ) = dim(Ker(T ))− dim(Z/TY ).
We shall use the easy facts that an operator T is Fredholm if and only if T 2 is
Fredholm (with i(T 2) = 2i(T )), and that a C-linear operator is Fredholm if and
only if it is Fredholm as an R-linear operator (and the corresponding indices are
related by iR(T ) = 2iC(T )).
A closed two-sided ideal U(X) in L(X) is a Fredholm ideal when an operator
T ∈ L(X) is Fredholm if and only if the corresponding class is invertible in
L(X)/U(X). It follows from well-known results in Fredholm theory that U(X)
is contained in the ideal In(X) of inessential operators, i.e. operators S such that
IdX−V S is a Fredholm operator for all operators V ∈ L(X), see for example [6].
Note that by continuity of the Fredholm index, IdX−V S is necessarily Fredholm
with index 0 when S is inessential, and V ∈ L(X).
Lemma 2 Let X be an infinite dimensional real Banach space. Let I ∈ L(X)
satisfy I2 = −Id, and let S be inessential such that (I + S)2 = −Id. Then I
and I + S are conjugate, or equivalently, the complex structures XI and XI+S
associated to I and I + S respectively are isomorphic.
Proof : The map 2I + S is immediately seen to be C-linear from XI into XI+S .
Furthermore, it is an inessential perturbation of an isomorphism, and therefore
Fredholm with index 0 as an R-linear operator on X . So it is also Fredholm
with index 0 as a C-linear map, which implies that XI and XI+S are C-linearly
isomorphic. �
Let X be a Banach space. The set I(X) denotes the set of operators on X
of square −Id. Let U(X) be a Fredholm ideal in L(X) and πU (or π) denote
the quotient map from L(X) onto L(X)/U(X). Let (L(X)/U(X))0 denote the
group πU(GL(X)), and Ĩ(X) denote the set of elements of (L(X)/U(X))0 whose
square is equal to −πU (Id).
Lemma 3 Let X be an infinite dimensional real Banach space and U(X) be a
Fredholm ideal in L(X). Then the quotient map πU induces an injective map π̃U
from the set of GL(X)-conjugation classes on I(X) (and therefore from the set of
isomorphism classes of complex structures on X) into the set of (L(X)/U(X))0-
conjugation classes on Ĩ(X).
Proof : For any operator T on X , we write T̃ = π(T ). Let I and T be operators
in I(X). If α is a C-linear isomorphism from XI onto XT , then the C-linearity
means that αI = Tα. Therefore α̃Ĩ = T̃ α̃, and Ĩ and T̃ satisfy a conjugation
relation. Conversely, if Ĩ = α̃−1T̃ α̃ for some α ∈ GL(X), then α−1Tα = I + S,
where S belongs to U(X) and is therefore inessential . Note that (I+S)2 = −Id,
and since Tα = α(I + S), α is a C-linear isomorphism from XI+S onto XT .
By Lemma 2, it follows that XI and XT are isomorphic. This proves that π̃ is
well-defined and injective. �
We shall now discuss when the above induced map is actually a bijection. This
is equivalent to saying that U(X) has the following lifting property.
Definition 4 Let X be an infinite dimensional Banach space. The Fredholm ideal
U(X) is said to have the lifting property if any α in (L(X)/U(X))0 satisfying
α2 = −1 is the image under πU of an operator T such that T
2 = −Id.
The following was essentially observed in [5].
Lemma 5 Let X be an infinite dimensional Banach space and let U(X) be a
Fredholm ideal in L(X). If U(X) admits a supplement in L(X) which is a sub-
algebra of L(X), then U(X) has the lifting property.
Proof : If H(X) is a subalgebra of L(X) which supplements S(X), then let
T ∈ L(X) be such that T̃ 2 = −Ĩd; we may assume that T (and therefore T 2)
belongs to H(X). Then since T 2 + Id is in U(X) ∩ H(X), T 2 must be equal to
−Id. Any class T̃ ∈ Ĩ(X) may therefore be lifted to an element of I(X). �
We shall now prove that although any Fredholm ideal in a complex space has
the lifting property, this is not necessarily true in the real case. The proof of
the complex case is essentially the same as the similar classical result concerning
projections (see e.g. [6]), and could be deduced directly from it using the fact that
an operator A satisfies A2 = −Id if and only if 1
(Id − iA) is a projection. We
shall however give a direct proof of this result for the sake of completeness.
Proposition 6 Let X be an infinite dimensional complex Banach space and let
U(X) be a Fredholm ideal in L(X). Then every element a ∈ L(X)/U(X) with
a2 = −1 is image under the quotient map of some A ∈ L(X) with A2 = −Id.
Proof : Recall that π : L(X) → L(X)/U(X) denote the quotient map. We
choose B ∈ L(X) such that π(B) = a. So π(B2) = −1 and therefore there exists
S ∈ U(X) ⊂ In(X) such that B2 = −Id+ S.
Since the spectrum σ(−Id+S) of −Id+S is countable and its possible limit
point is -1, it follows by the spectral mapping theorem ([4], Theorem VII.3.11)
that the spectrum σ(B) is also countable and its possible limit points are −i and i.
Take a simple closed curve Γ in C \ σ(B) such that i is enclosed by Γ and −i
is not enclosed by Γ. Define the operator
(λI − B)−1dλ.
By [4], Theorem VII.3.10, P is a projection. Moreover, according to the continu-
ity of π,
π(P ) =
(λI − a)−1dλ. (1)
On the other hand, it is easy to check that
(λI − a)−1 =
λI + a
λ2 + 1
Thus (1) implies that
π(P ) =
and hence putting A = 2iP − iI , we see that π(A) = a and A2 = −I . �
Such a proof is not possible in the real case essentially because there is no
formula with real coefficients linking projections and operators of square −Id.
Actually it is known that if X is real every element p belonging to the quotient
by a Fredholm ideal and satisfying p2 = p may be lifted to a projection (see
[6]), and the proof uses the complexification of X and a curve with well-chosen
symmetry so that the complex projection produced by the proof in the complex
case is induced by a real projection which will answer the question by the positive.
However there is no choice of curve such that the complex map A of square −Id
obtained in the above proof applied in the complexification of X is induced by a
real operator. Actually the result is simply false in the real case:
Theorem 7 Let X be a real infinite dimensional Banach space whose hyper-
planes admit a complex structure. Then no Fredholm ideal in L(X) has the lifting
property.
This theorem is a consequence of Proposition 8. We need to recall that the
complexification X̂ of a real Banach space X (see, for example, [12], page 81) is
defined as the space X̂ = {x + iy : x, y ∈ X}, which is the space X ⊕ X with
the canonical complex structure associated to J defined on X ⊕X by J(x, y) =
(−y, x). Let A,B ∈ L(X). Then
(A+ iB)(x+ iy) := Ax− By + (Ay +Bx)
defines an operator A+ iB ∈ L(X̂) that satisfies max{‖A‖, ‖B‖} ≤ ‖A+ iB‖ ≤
21/2(‖A‖+‖B‖). Conversely, given T ∈ L(X̂), if we put T (x+i0) := Ax+iBx,
then we obtain A,B ∈ L(X) such that T = A + iB. We write T̂ = T + i0 for
T ∈ L(X), and say that such the operator T̂ is induced by the real operator T .
Proposition 8 Let Y be an infinite dimensional real Banach space and J ∈ L(Y )
such that J2 = −Id. Let A be defined on X = R ⊕ Y by the matrix
Then A2 is the sum of −Id and of a rank 1 operator, but there is no inessential
operator S on X such that (A+ S)2 = −Id.
Proof : Assume (A + S)2 = −Id for some inessential operator S. Passing to the
complexification X̂ of X , we obtain (Â + Ŝ)2 = −Îd. The map from [0, 1] into
L(X̂) defined by Tµ = Â + µŜ is polynomial, moreover by spectral properties
of inessential operators and the spectral theorem, the spectrum Sp(Tµ) is, with
the possible exception of i and −i, a countable set of isolated points, which are
eigenvalues with associated spectral projections, denoted E(λ, Tµ) for each λ ∈
Sp(Tµ), of finite rank. Furthermore the complex operator Tµ is induced by the
real operator A+ µS, therefore Sp(Tµ) is symmetric with respect to the real line.
Let n(µ) =
λ∈R∩Sp(Tµ)
rk(E(λ, Tµ)) and let I1 = {µ ∈ [0, 1] : n(µ) is odd},
I0 = {µ ∈ [0, 1] : n(µ) is even}.
Observe that 0 ∈ I1, since Â is defined on X̂ = R̂⊕ Ŷ by the matrix
and therefore has unique real eigenvalue 1, with associated spectral projection of
dimension 1. On the other hand, since T (1)2 = (Â + Ŝ)2 = −Îd, T (1) does not
admit real eigenvalues and therefore 1 ∈ I0.
We now pick some µ1 ∈ I1. Let U be an open set containing the real line,
symmetric with respect to it, and such that U ∩ Sp(Tµ1) ⊂ R. Then the spectral
projection E(µ) := E(Tµ, U∩Sp(Tµ)) is an analytic ([4] Lemma VII.6.6) projec-
tion valued function defined for all µ such that |µ−µ1| < γ for some small enough
γ > 0, and for which E(µ1) has rank n(µ1). Therefore by [4] Lemma VII.6.8, the
dimension
λ∈U∩Sp(Tµ)
rk(E(λ, Tµ)) of the image of E(µ) is also n(µ1), for µ
in a small enough open set V around µ1. By symmetry of U and of Sp(Tµ) with
respect to the real line (with preservation of the ranks of the associated spectral
projections),
n(µ1) =
λ∈R∩Sp(Tµ)
rk(E(λ, Tµ)) + 2
λ∈Sp(Tµ),Im(λ)>0
rk(E(λ, Tµ)),
when µ ∈ V . So n(µ) =
λ∈R∩Sp(Tµ)
rk(E(λ, Tµ)) is odd whenever µ is in the
neighborhood V of µ1.
We have therefore proved that I1 is open. In the same way, I0 is also open (in
the special case when n(µ0) = 0, then E(µ0) = 0 and so, E(µ) = 0 and therefore
n(µ) = 0 in a neighborhood of µ0). In conclusion, the sets I0 and I1 are open,
non-empty, and partition [0, 1], a contradiction. �
The obstruction for the lifting property is therefore that a complex structure
on a hyperplane of a space X does not correspond to a complex structure on X ,
although it does induce elements of square −1 in (L(X)/S(X))0, as explicited
by the following result.
Lemma 9 Let Y be an infinite dimensional Banach space and X = Y ⊕ R.
Let U(X) be a Fredholm ideal. Then the map π′U (or π
′) from GL(Y ) into
(L(X)/U(X))0 defined by π
U(A) = πU(
) maps I(Y ) into Ĩ(X) and
induces an injection from the set of conjugation classes on I(Y ) (and therefore
from the set of isomorphism classes of complex structures on Y ) into the set of
conjugation classes on Ĩ(X).
Proof : First note that if the conjugation relation J = PKP−1 is satisfied in
GL(Y ) then
0 P−1
is satisfied in GL(X), which provides a conjugation relation in (L(X)/U(X))0.
Conversely if
˜(1 0
˜(1 0
P̃−1,
write
Then since P is Fredholm with index 0, P|Y is Fredholm with index −1 as an op-
erator of L(Y,X), and D = P|Y − b
∗ is also Fredholm with index −1 in L(Y,X),
and with index 0 in L(Y ). Therefore some finite rank perturbation D′ of D is an
isomorphism. Furthermore it is easy to deduce from the conjugation relation that
J = D′KD′
+ S, where S ∈ U(Y ). Therefore J − S = D′KD′−1, i.e. J − S
and K are conjugate, and by Lemma 3, it follows that Y J is isomorphic to Y K . �
The previous lemma does not mean that the sets of isomorphism classes of
complex structures on X and on hyperplanes of X are disjoint (only the corre-
spoding images by π̃, and π̃′ respectively, are). Actually the sets of isomorphism
classes on X and on its hyperplanes can either be equal, when X is isomorphic to
its hyperplanes, or disjoint, when it is not. And we have:
Proposition 10 Let Y be an infinite dimensional Banach space and let X = R⊕
Y . Let U(X) be a Fredholm ideal in L(X). Let A ∈ GL(X) and assume A2 =
−Id+S, S ∈ U(X) (i.e. Ã ∈ Ĩ(X)). Then there exists s ∈ U(X) such that either
(A+ s)2 = −Id or (A+ s)2 =
where J ∈ L(Y ) satisfies J2 = −Id.
Before the proof, let us observe that by Proposition 8, only one of the two
alternatives of the conclusion can hold for a given A such that Ã ∈ Ĩ(X).
Proof : Passing to the complexification X̂ of X , we have that Â2 = −Îd+Ŝ, and Ŝ
is inessential. Now let Γ be a rectangular curve with horizontal and vertical edges,
symmetric with respect to the horizontal axis, included in the open unit disk, and
such that Γ∩Sp(Ŝ) = ∅ and let U be the interior of the domain delimited by Γ, V
be the interior of the complement of this domain. Let P̂ be the spectral projection
associated to Sp(Ŝ)∩U ; since Γ is rectangular and symmetric with respect to the
real axis, it is classical and easy to see that P̂ is indeed induced by a real operator
P on X , see e.g. [7] where this principle is used. Let also Q̂ be the spectral
projection associated to Sp(Ŝ) ∩ V .
Then Ŝ = ŜP̂ + ŜQ̂. The operator ŜP̂ has spectral radius strictly smaller
than 1, therefore the series
n≥1 bn(ŜP̂ )
n converges to an operator ŝ, where
n≥1 bnz
n = −1 + (1 − z)−1/2 for all |z| < 1, and since the bn’s are reals, it
is indeed induced by a real operator s =
bn(SP )
n in U(X). We observe
(P̂ + ŝ)2 = P̂ (Îd+ ŝ)2 = P̂ (Îd− Ŝ)−1,
therefore
(ÂP̂ + Âŝ)2 = −P̂ .
Assume now that Q has even rank, then there exists a finite rank operator F on
QX such that F 2 = −Id|QX . Let then s
′ = FQ− AQ, therefore ÂQ̂+ ŝ′ = F̂ Q̂
and (ÂQ̂+ ŝ′)2 = −Q̂. If we then let v = s+ As′, we deduce that
(Â+ v̂)2 = (ÂP̂ + Âŝ+ ÂQ̂ + ŝ′)2 = −P̂ − Q̂ = −Îd,
and therefore (A+ v)2 = −Id, with v ∈ U(X).
If Q has odd rank then there exists a finite rank operator F on QX such that
, with j2 = −Id, in an appropriate decomposition of QX . Defining
v ∈ U(X) in the same way as above, we obtain that A+ v may be written
A + v =
corresponding to some decomposition R′⊕Y ′ of X where R′ is 1-dimensional. If
Y ′ = Y then we may clearly find some rank 1 perturbation f such that A+ v + f
may be written
A+ v + f =
corresponding to the original decomposition R ⊕ Y of X . If Y ′ 6= Y then we
consider the space Z = Y ′ ∩ Y ∩ JY , which is stable by J . If Z has codimension
3 then we may decompose Y = G⊕Z, where G has dimension 2, and by using a
operator k of square −IdG on G, we may find a rank 3 perturbation f such that
A + v + f =
0 J ′
in the original decompositionR⊕Y of X , and with J ′|Z = J|Z and J
|G = k, so that
J ′2 = −Id|Y . If finally Z has codimension 2 then easily some rank 2 perturbation
of A + v on X has square −Id, but because of the decomposition of A + v on
R′ ⊕ Y ′, this would contradict Proposition 8, so this case is not possible. �
We sum up the results of Proposition 8, Lemma 9 and Proposition 10 in the
next theorem:
Theorem 11 Let Y be an infinite dimensional real Banach space and let X =
Y ⊕ R. Let U(X) be a Fredholm ideal in L(X). Then there exists a partition
{Ĩ0(X), Ĩ1(X)} of Ĩ(X) such that πU induces a bijection from the set of con-
jugation classes on I(X) (and therefore from the set of isomorphism classes of
complex structures on X) onto the set of conjugation classes on Ĩ0(X) and such
that π′U induces a bijection from the set of conjugation classes on I(Y ) (and there-
fore from the set of isomorphism classes of complex structures on Y ) onto the set
of conjugation classes on Ĩ1(X).
Corollary 12 An infinite dimensional real Banach space is even if and only if
Ĩ(X) = Ĩ0(X) 6= ∅, and odd if and only if Ĩ(X) = Ĩ1(X) 6= ∅. When Ĩ(X) is a
singleton then X is either even or odd.
The next proposition sums up when a real Banach space has the lifting prop-
erty.
Proposition 13 Let X be an infinite dimensional real Banach space. Then the
following are equivalent:
• i) any Fredholm ideal in L(X) has the lifting property.
• ii) some Fredholm ideal in L(X) has the lifting property.
• iii) the hyperplanes of X do not admit complex structure.
• iv) for any Fredholm ideal U(X), the map πU induces a bijection from the
set of isomorphism classes of complex structures on X onto the set of con-
jugation classes on ĨU(X).
• v) for some Fredholm ideal U(X), the map πU induces a bijection from
the set of isomorphism classes of complex structures on X onto the set of
conjugation classes on ĨU(X).
Proof : It is clear by definition that iv) ⇔ i) and v) ⇔ ii). Then i) ⇒ ii) is
obvious, ii) ⇒ iii) due to Proposition 8, and iii) ⇒ i) by Proposition 10. �
We use this proposition to solve an open question from the first author, which
in our formulation asked whether there existed infinite dimensional even Banach
spaces. Recall that the space XGM is the real version of the HI space of Gowers
and Maurey [9], on which every operator is of the form λId+S, and therefore does
not admit complex structure. The space X(C) is a HI space constructed in [5] and
such that the algebra L(X(C)) may be decomposed as C⊕S(X(C)), and X(H),
also HI, is a quaternionic version of X(C). It is proved in [5] that X(H) admits
a unique complex structure, while X(C) admits exactly two complex structures.
Finally X(D2) is a space with an unconditional basis on which every operator is a
strictly singular perturbation of a 2-block diagonal operator and which also admits
a unique complex structure. We refer to [5] for details.
Theorem 14 The spaces X(C), X(H), X2nGM for n ∈ N, and X(D2) are even.
Proof : For X(C), X(H) and X2nGM this is due to the fact that the ideal of strictly
singular operators has the lifting property, because there is a natural subalgebra
supplementing the ideal of strictly singular operators in each case, see [5], so
Lemma 5 applies. For X(D2) note that it is proved in [5] that X(D2) admits a
complex structure and that there is a unique conjugation class in ĨS(X), therefore
the induced injection π̃S is necessarily surjective, i.e. v) is verified. �
Before giving some more examples, let us note two open problems about even
spaces. The first problem is quite simple to formulate.
Question 15 Is the direct sum of two even Banach spaces necessarily even?
We obtain a positive answer when the spaces are assumed to be essentially
incomparable. Recall that two infinite dimensional spaces Y and Z are essentially
incomparable when every operator from Y to Z is inessential [8]. More details
about this notion may be found in the last section of this article.
Proposition 16 The direct sum of two infinite dimensional even Banach spaces
which are essentially incomparable is even.
Proof : Let X and Y be infinite dimensional even, and essentially incomparable.
Clearly X⊕Y admits a complex structure. Assume X⊕(Y ⊕R) admits a complex
structure and look for a contradiction. Let T =
∈ L(X ⊕ (Y ⊕ R))
be such that T 2 = −Id. Since X and Y ⊕ R are essentially incomparable, S and
S ′ are inessential. Furthermore T 21 + S
′S = −IdX , i.e. T̃1 ∈ Ĩ(X). Since X is
even, there exists an inessential operator s1 on X such that T1+s1 = J1 with J
−IdX (Theorem 11). Likewise T
2 +SS
′ = −IdY⊕R therefore since Y ⊕R is odd,
there exists an inessential operator s2 on Y ⊕R such that T2+s2 =
, J2 ∈
L(Y ) with J22 = −IdY . Therefore there exists S0 =
s1 −S
−S s2
inessential on
X ⊕ Y ⊕ R such that
T + S0 =
J1 0 0
0 J2 0
0 0 1
Since T 2 = −Id this contradicts Proposition 8. �
For the second problem, we note that Banach spaces which are not isomorphic
to their hyperplanes may be classified in four categories: even spaces, odd spaces,
spaces such that neither the whole space neither hyperplanes admit a complex
structure, and spaces such that both the whole space and any hyperplane admit
complex structure. While we have just produced examples of the first and the
second category, and the space XGM belongs to the third, no examples are known
which belong to the fourth.
Question 17 Does there exist a real Banach space X which is not isomorphic to
its hyperplanes and such that both X and its hyperplanes admit complex struc-
ture?
We now use some spaces constructed by Plebanek [13] to giveC(K) examples
of even and odd Banach spaces. Similar C(K) spaces were first constructed by
P. Koszmider [11] under the Continuum Hypothesis. Let K be one of the two
infinite, separable, compact Hausdorff spaces defined in [13]. Every operator on
C(K) is of the form g.Id + S where g ∈ C(K) (therefore g.Id denotes the
multiplication by g) and S is strictly singular (or equivalently weakly compact).
The first space is connected, and we shall indicate where our proofs simplify due
to this additional property.
The space K ∪K will denote the space which is the topological union of two
copies of K (i.e. open sets are unions of open sets of each copy), while K ∪0 K
denotes the amalgamation of two copies of K in some point 0 (open sets are
unions of open sets of each copy either both containing 0 or neither containing 0).
Both are separable compact Hausdorff spaces.
Theorem 18 The space C(K∪K) is even and admits a unique complex structure,
and the space C(K ∪0 K) is odd.
Proof : The space C(K ∪K) identifies isomorphically with C(K) ⊕ C(K) and
C(K ∪0 K) identifies with the quotient C(K)
2/Y , where Y = {(f, g) : f(x0) =
g(x0)} for some fixed x0 ∈ K; therefore C(K∪0K) is isomorphic to a hyperplane
of C(K ∪K) and it is enough to prove that C(K) ⊕ C(K) is even with unique
complex structure.
Write X = C(K) and let Is(K) be the set of isolated points of K. Let
N ⊂ C(K) be the closed ideal of almost null functions, i.e. g ∈ N iff g vanishes
on K \ Is(K) and converges to 0 on Is(K) (i.e. for any ǫ > 0, |g(x)| ≤ ǫ for all
x ∈ Is(K) except a finite number of points). We observe the following fact: if an
operator on X of the form g.Id is strictly singular, then g belongs to N . Indeed if
g(x) 6= 0 for some non-isolated point x then |g| ≥ α > 0 on some infinite subset L
of K containing x, and the restriction of gId to the space of functions with support
included in L is an isomorphism. Likewise if |g| ≥ α > 0 on some infinite subset
of Isol(K) then the corresponding restriction of gId is an isomorphism.
In the case where K is connected, we have simply that g.Id is never strictly
singular unless g = 0.
Since X ⊕ X has a canonical complex structure, to prove that it is even with
unique complex structure, it is enough by Theorem 11 to prove that the group
G0 := (L(X ⊕ X)/S(X ⊕ X))0 has a unique conjugation class of elements of
square −Id.
We observe that the group GL2(C(K)) admits a unique class of conjugation
of elements of square −I , where I =
, i.e. that whenever M =
f1 f2
f3 f4
satisfies M2 = −I , then it is conjugate to the canonical element J =
Indeed from M2 = −I we deduce easily f1 = −f4 and f
1 + f2f3 = −1.
Note that f2 never takes the value 0. Let then P =
f1 f2
and let Q =
−f1/f2 1/f2
. It is routine to check that Q = P−1 and that QJP = M .
When K is connected then N = {0}, therefore L(X) ≃ C(K) ⊕ S(X) and
L(X ⊕ X) = M2(C(K)) ⊕ S(X ⊕ X). So S(X ⊕ X) has the lifting property
by Lemma 5, and G0 easily identifies with GL2(C(K)), and we therefore deduce
that there is a unique G0-conjugacy class of elements of square −Ĩd.
The general case is more complicated. Assume T̃ ∈ G0 satisfies T̃
2 = −Ĩd.
Then T 2+ Id is strictly singular, and up to a strictly singular perturbation we may
assume that T = M.Id with M ∈ M2(C(K)). Therefore (M
2 + I).Id is strictly
singular and M2 + I ∈ N . It follows that M2 = −I + n with ‖|n(x)‖| ≤ 1/2
except for x ∈ F , F a finite subset of Is(K) (here ‖|.‖| denotes the operator norm
on M2(R) ≃ L(R⊕2 R)). Note that M(x) and n(x) commute for all x ∈ K.
For x ∈ K \ F , let n′(x) = M(x)
k≥1 bkn(x)
k where (1 − z)−1/2 = 1 +
k≥1 bkz
k for |z| < 1, and let M ′(x) = (M + n′)(x). Then it is easy to check
that (M ′(x))2 = −IdR2 and therefore there exist P (x), Q(x) in M2(R) given
by formulas from the coefficients of M ′(x) which are explicited above in the
connected case, such that P (x)Q(x) = IdR2 and Q(x)jP (x) = M
′(x), where
For x ∈ F we let M ′(x) = (M + n′)(x) = j and P (x) = Q(x) = IdR2 . Note
that since the points of F are isolated and by uniform convergence of n′ and the
formulas giving P and Q, the matrices P , Q and n′ define elements of M2(C(K)).
Actually, since M ′(x) is invertible for all x and by the explicit formulas for the
inverses, we deduce that M ′ ∈ GL2(C(K)) and M
′.Id ∈ GL(X ⊕X). Likewise
P and Q belong to GL2(C(K)) and P.Id, Q.Id belong to GL(X ⊕X).
It is now enough to prove that S = n′.Id is a strictly singular operator on
X⊕X . Then the relation QJP = M ′ will imply a G0-conjugacy relation between
˜J.Id and ˜M ′.Id = ˜M.Id+ S̃ = T̃ , as desired.
For L ⊂ K, CL(K) denotes the space of functions of support included in L.
Let v denote the rank |F | projection onto CF (K) associated to the decomposition
C(K) = CF (K) ⊕ CK\F (K), and w = Id − v. Let V on X ⊕X be defined by
and let W =
It is easy to check that S =
k≥1 bk(n
k.Id)W + V (n′.Id), (just compute
S(f1, f2)(x) for all (f1, f2) ∈ X ⊕ X and x ∈ K). Since n
k.Id = (n.Id)k is
strictly singular for all k and V has finite rank, it follows that S is strictly singular.
To conclude this section we note an open question. Recall that there exist
spaces with exactly n complex structures, for any n ∈ N∗ [5], and spaces with
exactly 2ω complex structures [1].
Question 19 Does there exist a Banach space with exactly ω complex structures?
3 Essentially incomparable complex structures
We recall that two infinite dimensional spaces Y and Z are said to be essentially
incomparable if any bounded operator S ∈ L(Y, Z) is inessential, i.e., if IdY−V S
is a Fredholm operator (necessarily of index 0) for all operators V ∈ L(Z, Y ). Es-
sentially incomparable spaces were studied by M. González in [8]; it is clear that
the notion of essential incomparability generalizes the notion of total incompara-
bility. We also recall that Y and Z are projection totally incomparable if no infinite
dimensional complemented subspace of Y is isomorphic to a complemented sub-
space of Y . Essentially incomparable spaces are in particular projection totally
incomparable.
In this section we show how some results of [5] about totally incomparable
complex structures extend to essentially incomparable structures. Interestingly,
our more general proof turns out to be much simpler than the original one.
As was noted in [5], whenever T 2 = U2 = −Id, it follows that
(T + U)T = U(T + U),
which means that T + U is C-linear from XT into XU . The similar result holds
for T − U between XT and X−U .
Lemma 20 Let X be a real Banach space, T, U ∈ I(X). If T + U is inessential
from XT into XU , then XT is isomorphic to X−U .
Proof : Since T+U is inessential as an operator from XT into XU , and as T+U is
also linear from XU into XT , it follows by definition of inessential operators that
Id + λ(T + U)2 ∈ L(XT ) is Fredholm with index 0 whenever λ is real. Taking
λ = 1/4, we obtain that
4Id+ (T + U)2 = 2Id+ TU + UT = −(T − U)2.
Therefore (T −U)2 is Fredholm with index 0 as an operator on XT , and therefore
as an operator on X . It follows that T −U is Fredholm with index 0 as an operator
of L(X) and therefore as an operator of L(XT , X−U), hence XT and X−U are
C-linearly isomorphic. �
It was proved in [5] that two totally incomparable complex structures on a real
space must be conjugate and both saturated with HI subspaces. We show:
Theorem 21 Let X be a real Banach space with two essentially incomparable
complex structures. Then these complex structures are conjugate up to isomor-
phism and do not contain a complemented subspace with an unconditional basis.
Proof : Assume XT is essentially incomparable with XU . Then T +U is inessen-
tial from XT into XU and by Proposition 20, XU is isomorphic to X−T .
If Y is a C-linear complemented subspace of XT with an unconditional basis,
then Y is complemented in X−T and the coordinatewise conjugation map α as-
sociated to the unconditional basis is an isomorphism from Y onto Y . Therefore
XT and X−T are not projection totally incomparable, contradicting the essential
incomparability of XT with XU ≃ X−T . �
Note that Proposition 25 will prove that one cannot hope to improve Theorem
21 to obtain HI-saturated in its conclusion, as in the case of totally incomparable
complex structures.
Corollary 22 There cannot exist more than two mutually essentially incompara-
ble complex structures on a Banach space.
Recall that two Banach spaces are said to be nearly isomorphic (or sometimes
essentially isomorphic) if one is isomorphic to a finite-codimensional subspace of
the other. Equivalently this means that there exists a Fredholm operator acting
between them.
In the next proposition, we consider properties of complex structures which
are generalization of the λId + S-property. Note that each of these properties
implies that there do not exist non-trivial complemented subspaces. We first state
a lemma whose proof was given to us by M. González.
Lemma 23 Let X be an infinite dimensional real or complex Banach space such
that every operator is either Fredholm or inessential. Then every Fredholm oper-
ator on X has index 0 and L(X)/In(X) is a division algebra.
Proof : If in the above conditions, T were Fredholm with nonzero index, then
by the continuity of the index, K := sT + (1 − s)Id would be inessential for
some s with 0 < s < 1. Thus T := s−1
Id + 1
K; hence T has index 0, a
contradiction. Moreover, the only noninvertible element in L(X)/In(X) is 0;
hence L(X)/In(X) is a division algebra. �
Proposition 24 Let X be a real Banach space, and T ∈ I(X).
• i) If every operator on XT is either inessential or Fredholm, then either
XT is the only complex structure on X , or XT and X−T are the only two
complex structures on X and they are not nearly isomorphic.
• ii) If every operator on XT is either strictly singular or Fredholm, then
either XT is the only complex structure on X , or XT and X−T are the only
two complex structures on X and neither one embeds into the other.
Proof : i) Let U generate a complex structure on X . We use the relation
(T + U)2 + (T − U)2 = −4Id.
If (T −U)2, which is an operator on XT , is inessential, then (T +U)2 is Fredholm
with index 0 as a perturbation of −4Id, therefore T + U is Fredholm with index
0 and there exists an isomorphism from XT onto XU .
Therefore if there exists some U generating a complex structure non isomor-
phic to XT , then (T − U)2 is not inessential, and by the property of operators on
XT , (T − U)2 is Fredholm, with index 0 by Lemma 23, and therefore T − U as
well, so XU is isomorphic to X−T . We deduce that XT and X−T are the only
complex structures on X . To see that they are not nearly isomorphic, note that
if a map α is Fredholm from XT into X−T , then α2 is Fredholm on XT , hence
it is Fredholm with index 0, and α is Fredholm with index 0. Therefore XT is
isomorphic to X−T , contradicting our initial assumption.
ii) Applying i) we see that any complex structure on X is either isomorphic to
XT or X−T , and if α embeds XT into X−T , then α2 embeds XT into itself, hence
it is not strictly singular and so it is Fredholm, with index 0, and α is Fredholm
with index 0, which implies that XT is isomorphic to X−T . �
In [2] S. Argyros and A. Manoussakis constructed a real space which is un-
conditionnally saturated yet has the λId + S property. Although their result is
stated in the real case, no specific property of the reals is used in their definition
and proofs, and so the complex version XAM of their space satisfies the complex
version of the properties mentioned above. We observe:
Proposition 25 Every R-linear operator on the complex XAM is of the form
λId + S, λ ∈ C, S strictly singular. It follows that the complex XAM seen as
real admits exactly two complex structures, which are essentially incomparable
but not totally incomparable.
Proof : Denote XAM the complex version of the space of Argyros and Manous-
sakis, and let X be XAM seen as real. Lemma 4.17 from [2] in its complex ver-
sion states that every C-linear operator T on XAM satisfies lim d(Ten,Cen) = 0,
where (en)n is the canonical complex basis of XAM . A look at their proof shows
that actually only the R-linearity of T is required. Then if T is R-linear on X
we deduce easily that there exists λ ∈ C such that for any ǫ > 0, there exists M
an infinite subset of N such that (T − λId)|[en,n∈M ] is of norm at most ǫ. Here
[en, n ∈ M ] denotes the real subspace R-linearly generated by (en)n∈M .
Proposition 4.16 in [2] states that for any infinite subset M of N, any (yk) a
normalized block-sequence of (en), the distance d(S[en,n∈M ], S[yk,k∈N]) between
the respective unit spheres of the complex (i.e. C-linearly generated) block-
subspaces [en, n ∈ M ] and [yk, k ∈ N] is 0. A look at the proof shows that
one can obtain this using only the real block-subspaces which are R-linearly gen-
erated by (en)n∈M and (yk)k∈N (in particular note that [2] Lemma 4.12 used in the
proof only uses R-linear combinations of the (en)’s and the (yk)’s).
Combining the facts of the first and the second paragraph, we deduce that
when T is R-linear on X , there exists λ ∈ C such that for all ǫ > 0, for any
complex block-subspace (yk)k of (en)n, there exists a unit vector x in the R-linear
span of (yk)k such that (T − λId)x is of norm less than 2ǫ. We deduce easily that
every R-linear operator on X is of the form λId+ S, λ ∈ C, S strictly singular.
The complex structure properties now follow easily. By Proposition 24 ii) we
already know that either X has unique complex structure, or exactly two which
are XJ and X−J (where J ∈ L(X) is defined by Jx = ix). Note that whenever
α is C-linear from XJ into X−J then α(ix) = −iα(x) for all x ∈ X . Since
α = λ.Id + S, λ ∈ C and S R-strictly singular, we deduce that λ = 0 and
α = S. The operator S is in particular C-strictly singular; so L(XJ , X−J) =
S(XJ , X−J) and XJ and X−J are essentially incomparable. On the other hand
XJ is the complex version of XAM and so is unconditionally saturated, therefore
it is not HI saturated, and so not totally incomparable with its conjugate X−J , by
[5] Corollary 23. �
We end the paper with two open questions in the direction of further general-
izing the above results. Recall that two spaces are projection totally incomparable
if no infinite dimensional complemented subspace of one is isomorphic to a com-
plemented subspace of the other, and that essentially incomparable spaces are in
particular projection totally incomparable [8].
Question 26 If two complex structures on a real space X are projection totally
incomparable, must they be conjugate?
Question 27 Assume a complex space is projection totally incomparable with its
conjugate, is it necessarily essentially incomparable with it?
Acknowledgements The first author thanks B. Maurey for a useful discussion
which led the authors to Proposition 8. Both authors thank M. González for his
comments on a first version of this paper, and P. Koszmider for information about
the C(K) examples used in Theorem 18.
References
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ally saturated Banach space, Studia Math. 159 (2003), no. 1, 1–32.
[3] J. Bourgain, Real isomorphic complex Banach spaces need not be com-
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Valentin Ferenczi,
Institut de Mathématiques de Jussieu,
Université Pierre et Marie Curie - Paris 6,
Projet Analyse Fonctionnelle, Boı̂te 186,
4, place Jussieu, 75252 Paris Cedex 05,
France.
E-mail: ferenczi@ccr.jussieu.fr, ferenczi@ime.usp.br
Elói Medina Galego,
Departamento de Matemática,
Instituto de Matemática e Estatı́stica,
Universidade de São Paulo.
05311-970 São Paulo, SP,
Brasil.
E-mail: eloi@ime.usp.br.
	Introduction
	Parity of infinite dimensional spaces
	Essentially incomparable complex structures
