There is no largest proper operator ideal

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There is no largest proper operator ideal

Valentin Ferenczi, Universidade de S˜ao Paulo

Seminar in Operator Theory and Operator Algebras, University of Virginia, October 19th, 2021

Supported by Fapesp 2016/25574-8 and CNPq 303731/2019-2

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Outline

1 Background

2 A new class of proper ideals

3 More details on the proof

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Operator ideals in the sense of Pietsch

All notions from the 1979 Pietsch’s book “Operator ideals”

An operator idealUis a collection of subspaces U(X,Y) of L(X,Y) for all X,Y Banach spaces, such that

U(X,Y) contains F(X,Y) ={finite rank operators}

(for all spacesZ,W and all appropriate operatorsT,V) S ∈U(X,Y)⇒TSV∈U(Z,W)

In particular the classU(X) :=U(X,X) is a two-sided ideal of L(X) in the usual sense.

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Some classical operator ideals

F=ideal of finite rank operators K=ideal of compact operators S=ideal of strictly singular In=ideal ofinessential operators

F(F^closure (K(S(In

Inessential operators were defined by Kleinecke (63) whenX =Y, by Pietsch (78) in general.

Definition

S :X →Y is inessential iff ∀T :Y →X , IdX −TS is Fredholm When there exist Fredholm operators betweenX andY, then more intuitively,S is inessential iff∀T :X →Y

T Fredholm⇔ T +S Fredholm.

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Proper operator ideals

Definition (Pietsch)

A classU isproperif for any Banach space X , Id_X belongs toU ⇔ X is finite dimensional

Therefore an idealU is proper iff for anyinfinite-dimensional space X,U(X) is a proper ideal ofL(X).

Fact

The idealsF,K,S,In are proper.

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Two questions of Pietsch (79)

Question 1

IsInthe largest proper operador ideal?

Question 2

Does there exist a largest proper operador ideal?

Theorem

The answer is no to both. Actually:

In is not even a maximal proper ideal, i.e. there exists V proper ideal such that In(V

moreover, there exist two proper ideals In(V₁,V−1 with V₁+V−1 not proper.

We rely heavily on previous work by Aiena-Gonz´alez (00) and Gowers-Maurey (97) + notes by Maurey (96), and thank M.

Gonz´alez and A. Mart´ınez-Abej´on for useful conversations.

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Aiena-Gonz´ alez’s approach: improjective operators

Definition (Tarafdar 72)

An operator T :X →Y is projective if it induces an isomorphism between infinite dimensionalcomplemented subspaces of X and Y respectively; andimprojective otherwise.

Facts

Fredholm operators between ∞-dim spaces are projective.

Impis proper.

We have the inclusion In⊆Imp; actually for an ideal U, U is proper ⇔U⊆Imp.

Aiena-Gonz´alez (00) investigated whetherIn=Impor at least whetherImpis an ideal. In which caseImpwould be the largest proper ideal.

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Aiena-Gonz´ alez’s results: In ( Imp

IfX is an HI space then L(X)=Fredholm(X) ∪S(X)

So S(X) =In(X) =Imp(X) is the largest proper ideal of L(X). This is useless for their purpose.

Aiena-Gonz´alez note that if X is an indecomposable space, then

L(X)=Fredholm(X) ∪Imp(X)

So Aiena-Gonz´alez need an indecomposable, non HI space.

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Theorem (Gowers-Maurey 97)

There exists a Banach spaceXsuch that for any∞-dim subspace Y ofX, the following are equivalent

Y is finite codimensional Y is complemented in X Y is isomorphic to X

The spaceX is indecomposable and the isomorphism between X and its hyperplanes is provided by the Right Shift operatorR on the basis (e_n) ofX.

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Theorem (Aiena-Gonz´alez, last line of p. 477) We haveIn(X)(Imp(X)

Proof.

1 is in the essential spectrum ofR, therefore S1:=Id_X−R is not Fredholm. Therefore

(a) it is improjective

(b) but it is essential. Indeed 2Id_X−S1 =Id_X+R is not Fredholm, since −1 is also in the essential spectrum of R Note: for future use and wlog, replace S₁ by some compact perturbationT₁ taking value in someY with dimX/Y =∞.

IfImpwere an ideal, then it would be the largest proper ideal. But: Proposition (Aiena-Gonz´alez, second line of p. 478)

Impis not an ideal!

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Theorem (Aiena-Gonz´alez, last line of p. 477) We haveIn(X)(Imp(X)

Proof.

1 is in the essential spectrum ofR, therefore S1:=Id_X−R is not Fredholm. Therefore

(a) it is improjective

(b) but it is essential. Indeed 2Id_X−S1 =Id_X+R is not Fredholm, since −1 is also in the essential spectrum of R Note: for future use and wlog, replace S₁ by some compact perturbationT₁ taking value in someY with dimX/Y =∞.

IfImpwere an ideal, then it would be the largest proper ideal. But:

Proposition (Aiena-Gonz´alez, second line of p. 478) Impis not an ideal!

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Proposition (Aiena-Gonz´alez)

Impis not an ideal (actuallyImp(X) is not a linear subspace of L(X)).

But we can use their work byshrinking Impto an ideal:

Proposition

There exist a proper idealU₁⊆Imp such that T₁ ∈U₁.

SinceT₁ ∈/In, let us denote V₁:=In+U₁: Theorem

The idealInis not the largest proper ideal.

More precisely,In(V₁ and V₁ is a proper ideal.

But one can actually deduce more...

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Indeed Aiena-Gonz´alez observe the same as above holds for Id_X−λR,|λ|= 1 (and not only forλ= 1) and associated compact perturbationT_λ. I.e. T_λ is improjective∀λ.

HoweverT₁+T−1 = 2Id_X+compact which is projective.

This is how they deduce thatImp(X) is not a subspace ofL(X).

The case ofT−1 is “the same as” T₁ so summing up:

Proposition

There exist two proper idealsU1 andU−1 such that Ti ∈Ui, for i =−1,1.

ThereforeId_X∈U1(X) +U−1(X) and it follows Theorem

There exist two proper ideals whose sum is not proper. In particular there is no largest proper ideal.

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A new kind of proper ideal

Summing up everything boils down to proving:

Proposition

For i=−1,1, there exist a proper idealUisuch that Ti ∈Ui.

Definition

If X is an (∞-dim) Banach space,Op(X)is the class of operators factorizing through X .

This is an ideal as soon as, e.g.,X 'X².

(ifT =AB andT⁰=A⁰B⁰ thenT +T⁰ = A A⁰

B

B⁰

) However this is never proper ifX is infinite dimensional...

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Lemma

The classOp(X)∩Op(X⁰) is proper iff no ∞-dimensional complemented subspace of X is isomorphic to a complemented subspace of X⁰

(proof: IdZ ∈Op(X) if and only if IdZ =AB where

A∈L(X,Z),B ∈L(Z,X); this is equivalent to saying that Z embeds(by B) as a subspace ofX which is complemented (by the projectionBA). So Id_Z ∈Op(X)∩Op(X⁰) iff Z embeds

complementably into bothX andX⁰.)

So the properties of Gowers-Maurey’sXmean that: Proposition

The classOp(X)∩Op(Yi)is proper.

Furthermore recall thatT_i is defined on Xand takes values in

∞-codimensionalYi; so Ti ∈Op(X)∩Op(Yi).

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Lemma

The classOp(X)∩Op(X⁰) is proper iff no ∞-dimensional complemented subspace of X is isomorphic to a complemented subspace of X⁰

(proof: IdZ ∈Op(X) if and only if IdZ =AB where

A∈L(X,Z),B ∈L(Z,X); this is equivalent to saying that Z embeds(by B) as a subspace ofX which is complemented (by the projectionBA). So Id_Z ∈Op(X)∩Op(X⁰) iff Z embeds

complementably into bothX andX⁰.)

So the properties of Gowers-Maurey’sXmean that:

Proposition

The classOp(X)∩Op(Yi) is proper.

Furthermore recall thatT_i is defined on Xand takes values in

∞-codimensionalYi; so Ti ∈Op(X)∩Op(Yi).

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The problem is....

that X6'X² so we do not know whetherOp(X) andOp(Y_i) are ideals!

So weenlargeOp(X) to: Definition

Op^<ω(X) :=∪_n∈_NOp(Xⁿ) is theideal of operators factorizing

throughsome power of X ,

and we enlargeOp(X)∩Op(Y) toOp^<ω(X)∩Op^<ω(Y) hoping it is still proper. And indeed:

Proposition

If Y is infinite codimensional inX then the ideal Op^<ω(X)∩Op^<ω(Y) is proper.

This is enough: letU_i:=Op^<ω(X)∩Op^<ω(ImT_i)... then T_i ∈U_i.

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The problem is.... thatX6'X² so we do not know whetherOp(X) andOp(Y_i) are ideals!

So weenlargeOp(X) to: Definition

Proposition

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The problem is.... thatX6'X² so we do not know whetherOp(X) andOp(Y_i) are ideals!

So weenlargeOp(X) to:

Definition

Proposition

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So our main result follows from the technical result:

Theorem

If Y is infinite codimensional inX,and m,n∈N, then no infinite dimensional complemented subspace Z ofX^m embeds

complementably into Yⁿ.

(which involves extending the techniques of Gowers-Maurey to the multidimensional setting, as is presented at the end of this talk)

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A question

An abstractspace idealA (Pietsch) is a class of Banach spaces such that

F⊆A

E₁,E₂ ∈A⇒E₁⊕E₂ ∈A

F ∈AandE embeds complementably in F ⇒ E ∈A. Examples:

F,HILB:={hilbertian spaces},REFL,SEP, Space(U) :={X:Id_X∈U}, for any idealU

Pietsch proves that a space idealAis always Space(U) for some idealU , and asks (Problem 2.2.8) whether there is always a largestUsuch that A=Space(U). We just proved that the answer is no forA=F but we can also deal with other cases...

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A question

Indeed by considering ideal of operators of the form

A=Op^<ω(X⊕`2)∩Op^<ω(Y ⊕`2), and proving that

Space(A) =HILB, we obtain:

Proposition

There is no largest idealU such that Space(U) =HILB.

This extends to space ideals whose elements are “different” enough fromX.

So the casesREFL,SEP remain open.

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Details on the proof of the main technical result

Theorem

If Y is infinite codimensional inX, and m,n∈N, then no infinite dimensional complemented subspace ofX^m embeds into Yⁿ.

The proof (in the complex case) admits the following steps (1) Any ∞-dimensional complemented subspace of X^m is

isomorphic to X^p (for somep ≤m)

(2) “Any” embedding of Xinto Xⁿ is complemented...

(3) Use (2) to showX does not embed intoYⁿ (*) (*): if it did, then(Id_X +s)X⊆Y

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Sketch of (1)

By Gowers-Maurey,∃ projection and algebra homomorphismλ of L(X) onto some subalgebraA of operators (generated byR and the left shift L), such that

∀T, λ(T)−T is strictly singular

∃an isomorphism Ψ ofAonto the Wiener algebra A(T)⊆C(T) of continuous functions with absolutely summable Fourier series.

“Wlog” letP ∈Mm(A) be a projection onX^m. So Ψ(P) is an idempotent ofM_m(A(T))⊆M_m(C(T)).

Therefore for each θ, Ψ(P)(eⁱ^θ) is an idempotent ofM_m(C), with rank k(θ), which is constant =k by continuity.

Ifm= 1 we are done (GM), because then Ψ(P) = 0 or 1, and P = 0 orId_X.

But if say m= 2 andk = 1, the rank 1 (2,2)-matrix Ψ(P)(eⁱ^θ) may and will vary withθ.

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Sketch of (1)

Therefore for each θ, Ψ(P)(e^iθ) is an idempotent of Mm(C), with rank k(θ), which is constant =k by continuity.

But if say m= 2 andk = 1, the rank 1 (2,2)-matrix Ψ(P)(eⁱ^θ) may and will vary withθ.

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Sketch of (1)

Therefore for each θ, Ψ(P)(e^iθ) is an idempotent of Mm(C), with rank k(θ), which is constant =k by continuity.

But if say m= 2 andk = 1, the rank 1 (2,2)-matrix Ψ(P)(e^iθ) may and will vary with θ.

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Sketch of (1)

By Gowers-Maurey,∃ projection and algebra homomorphismλ of L(X) onto some algebra Aof operators (generated by R and the left shift L), such that

“Wlog” letP ∈M_m(A) be a projection onX^m. So Ψ(P) is an idempotent ofM_m(A(T))⊆M_m(C(T)).

Therefore for each θ, Ψ(P)e^iθ is an idempotent ofMm(C), with rank k(θ), which is constant =k by continuity.

K-theory (K1(C) ={0}) tells us that this rankk is the unique similarity invariant (a∼b ⇔ ∃c :b =cac⁻¹)for idempotents of M∞(C(T)). So Ψ(P)∼the canonical “rankk” idempotent of Mm(C(T)), i.e. Ik =

Id_k 0

0 0

, acting onC(T)^m

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A multidimensional version ofWiener’s Lemma tells us that this similarity occurs insideM_m(A(T)). So we can lift this similarity back to M_m(A)⊆L(X^m). (that working with A(T) instead of C(T) does not affect the use of K-theory took me some time to get convinced of, although this is probably obvious to K-theory specialists through the notion of “local Banach algebra”)

This means thatP is similar to the natural projectionP_k of X^m ontoX^k, which implies PX^m 'X^k.

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P. Aiena, M. Gonz´alez,Examples of improjective operators, Math. Z. 233 (2000), 471–479.

B. Blackadar,K-theory for operator algebras, MSRI Publications 5, Springer Verlag 1986.

V. Ferenczi,There is no largest proper operator ideal, arXiv:2005.07672 or my webpage.

W.T. Gowers, B. Maurey,Banach spaces with small spaces of operators. Math Ann. (307) (1997), 543–568.

B. Maurey,Operator theory and exotic Banach spaces. 1996.

Notes available on the author’s webpage.