Structure of the quotient algebra ofcompact-by-approximable operators for Banach

spaces failing the approximation property

Hans-Olav Tylli

University of Helsinki(joint with Henrik Wirzenius)

Banach Algebras 2019, WinnipegJuly 15, 2019

Basic references

? [B76] G.F. Bachelis: A factorization theorem for compactoperators, Illinois J. Math. 20 (1976), 626–629.? [D13] H.G. Dales: A Banach algebra related to a Banach spacewithout AP, (unpublished notes, 2013).? [TW19] H-O. Tylli & H. Wirzenius: The quotient algebra ofcompact-by-approximable operators on Banach spaces failing theapproximation property, J. Aust. Math. Soc. (2019).(arXiv:1811.09402 [math.FA])? [TW2] H-O. Tylli & H. Wirzenius: Algebraic structure of thequotient algebra of compact-by-approximable operators (inpreparation).? [W92] G. Willis: The Compact Approximation Property does notimply the Approximation Property, Studia Math. 103 (1992),99–108.

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Setting

? Let X ,Y be Banach spaces, and

L(X ,Y ) = {bounded linear operators X → Y }K(X ,Y ) = {compact operators X → Y }A(X ,Y ) = {approximable operators X → Y } = F(X ,Y ),

where F(X ,Y ) = {bounded finite rank operators X → Y }.? For X = Y write K(X ) := K(X ,X ), A(X ) := A(X ,X ) etc.

? K(X ) and A(X ) are closed two-sided ideals of L(X ) in theoperator norm, so AX := K(X )/A(X ) is a Banach algebra(the ”compact-by-approximable” quotient algebra) with

||T +A(X )|| = dist(T ,A(X )) = infA∈A(X )

||T − A||

whenever X is a real or complex Banach space.

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Setting

? Let X ,Y be Banach spaces, and

L(X ,Y ) = {bounded linear operators X → Y }K(X ,Y ) = {compact operators X → Y }A(X ,Y ) = {approximable operators X → Y } = F(X ,Y ),

where F(X ,Y ) = {bounded finite rank operators X → Y }.? For X = Y write K(X ) := K(X ,X ), A(X ) := A(X ,X ) etc.

? K(X ) and A(X ) are closed two-sided ideals of L(X ) in theoperator norm, so AX := K(X )/A(X ) is a Banach algebra(the ”compact-by-approximable” quotient algebra) with

||T +A(X )|| = dist(T ,A(X )) = infA∈A(X )

||T − A||

whenever X is a real or complex Banach space.

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Approximation properties 1

? AX = K(X )/A(X ) is a (non-unital) radical Banach algebra. (Forreal scalars radicality is here interpreted as

σR(T +A(X )) = {0}, T ∈ K(X ),

which follows from classical Riesz-Fredholm theory). The structureof AX is very poorly understood, in part because of connection toapproximation properties.

Recall: Banach space X has the approximation property (AP) if forall ε > 0 and all compact subsets K ⊂ X there is a finite rankoperator T ∈ F(X ) such that

supx∈K||x − Tx || < ε

- All classical Banach spaces have AP: spaces with a Schauderbasis, Lp-spaces, C (K )-spaces etc.

- Enflo (1973): there is a Banach space X ⊂ c0 without AP.Davie (1973) & Szankowski (1978): for 1 ≤ p <∞, p 6= 2,there is a closed subspace X ⊂ `p failing AP.

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Approximation properties 1

? AX = K(X )/A(X ) is a (non-unital) radical Banach algebra. (Forreal scalars radicality is here interpreted as

σR(T +A(X )) = {0}, T ∈ K(X ),

which follows from classical Riesz-Fredholm theory). The structureof AX is very poorly understood, in part because of connection toapproximation properties.

Recall: Banach space X has the approximation property (AP) if forall ε > 0 and all compact subsets K ⊂ X there is a finite rankoperator T ∈ F(X ) such that

supx∈K||x − Tx || < ε

- All classical Banach spaces have AP: spaces with a Schauderbasis, Lp-spaces, C (K )-spaces etc.

- Enflo (1973): there is a Banach space X ⊂ c0 without AP.Davie (1973) & Szankowski (1978): for 1 ≤ p <∞, p 6= 2,there is a closed subspace X ⊂ `p failing AP.

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Approximation properties 2

? L(`2) does not have AP (Szankowski; 1981), and neither doesthe Calkin algebra L(`2)/K(`2) (Godefroy & Saphar; 1989)

? Relevance for quotient algebra AX : if X has AP, thenK(X ) = A(X ), so AX = {0}.? The converse is a notorious open question: if K(X ) = A(X ),does X have AP? (Lindenstrauss & Tzafriri, 1977)? Dales (2013): collection of results and open questions on AX .

- Is there a X without AP such that 0 < dim(AX ) <∞? (Stillopen)

- Is there a X such that AX contains infinitely many non-trivialclosed ideals? (Still open)

- Further questions about factorisation, amenability of K(X )etc.

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Approximation properties 2

? L(`2) does not have AP (Szankowski; 1981), and neither doesthe Calkin algebra L(`2)/K(`2) (Godefroy & Saphar; 1989)

? Relevance for quotient algebra AX : if X has AP, thenK(X ) = A(X ), so AX = {0}.? The converse is a notorious open question: if K(X ) = A(X ),does X have AP? (Lindenstrauss & Tzafriri, 1977)? Dales (2013): collection of results and open questions on AX .

- Is there a X without AP such that 0 < dim(AX ) <∞? (Stillopen)

- Is there a X such that AX contains infinitely many non-trivialclosed ideals? (Still open)

- Further questions about factorisation, amenability of K(X )etc.

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Warm-up: examples where AX 6= {0}? Characterisation of the AP (Grothendieck): X has AP if and onlyif K(Y ,X ) = A(Y ,X ) for every Banach space Y .This easily provides examples of Banach spaces Z with anon-trivial quotient algebra AZ :

Example Suppose X is a Banach space failing the AP, and pick Yand T ∈ K(Y ,X ) \ A(Y ,X ). Let Z = Y ⊕ X and define

T̃ : Z → Z , T̃ (y , x) = (0,Ty).

Then T̃ ∈ K(Z ) \ A(Z ) and AZ 6= {0}.? Drawback: above Y = (Y0, | · |), where Y0 ⊂ X linear subspaceand | · | complete norm in Y0. (Hence Y preserves little of X .)

? Bachelis [B76]: Suppose that E is a Banach space having thebounded AP, E ⊕ E ≈ E and E contains a closed subspace Xfailing the AP. Then there is a closed subspace Z ⊂ E such thatAZ 6= {0}. (This applies e.g. to E = `p for 1 ≤ p <∞ and p 6= 2,or E = c0. Generalises Alexander (1974) for 2 < p <∞.)

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Warm-up: examples where AX 6= {0}? Characterisation of the AP (Grothendieck): X has AP if and onlyif K(Y ,X ) = A(Y ,X ) for every Banach space Y .This easily provides examples of Banach spaces Z with anon-trivial quotient algebra AZ :

Example Suppose X is a Banach space failing the AP, and pick Yand T ∈ K(Y ,X ) \ A(Y ,X ). Let Z = Y ⊕ X and define

T̃ : Z → Z , T̃ (y , x) = (0,Ty).

Then T̃ ∈ K(Z ) \ A(Z ) and AZ 6= {0}.? Drawback: above Y = (Y0, | · |), where Y0 ⊂ X linear subspaceand | · | complete norm in Y0. (Hence Y preserves little of X .)

? Bachelis [B76]: Suppose that E is a Banach space having thebounded AP, E ⊕ E ≈ E and E contains a closed subspace Xfailing the AP. Then there is a closed subspace Z ⊂ E such thatAZ 6= {0}. (This applies e.g. to E = `p for 1 ≤ p <∞ and p 6= 2,or E = c0. Generalises Alexander (1974) for 2 < p <∞.)

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Examples with large AX , I

? Motivation: c0 embeds isometrically into K(`p) = A(`p) for1 ≤ p <∞ by the map a 7→ Ta for a = (ak) ∈ c0, where

Ta(xk) = (akxk), (xk) ∈ `p.

Infinite-dimensional analogue for AX ?

Proposition [TW19] Let X1,X2, . . . be sequence of Banach spaceswith AXn 6= {0} for all n ∈ N. Then there is a linear isomorphicembedding c0 → AX , where X = (⊕n∈NXn)p for 1 ≤ p ≤ ∞.(Interpret as direct c0-sum for p =∞.)

Proof (idea): For n ∈ N, pick Tn ∈ K(Xn) \ A(Xn) with ||Tn|| < 2and ||Tn +A(Xn)|| = 1. Let Pn : X → Xn, (xk) 7→ xn andJn : Xn → X , xn 7→ (0, 0, . . . , 0, xn, 0, 0 . . .) be the canonical maps.Then one may define a linear isomorphic embedding c0 → AX by

θ : (ak) 7→∞∑k=1

akJkTkPk +A(X ).

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Examples with large AX , I

? Motivation: c0 embeds isometrically into K(`p) = A(`p) for1 ≤ p <∞ by the map a 7→ Ta for a = (ak) ∈ c0, where

Ta(xk) = (akxk), (xk) ∈ `p.

Infinite-dimensional analogue for AX ?

Proposition [TW19] Let X1,X2, . . . be sequence of Banach spaceswith AXn 6= {0} for all n ∈ N. Then there is a linear isomorphicembedding c0 → AX , where X = (⊕n∈NXn)p for 1 ≤ p ≤ ∞.(Interpret as direct c0-sum for p =∞.)

Proof (idea): For n ∈ N, pick Tn ∈ K(Xn) \ A(Xn) with ||Tn|| < 2and ||Tn +A(Xn)|| = 1. Let Pn : X → Xn, (xk) 7→ xn andJn : Xn → X , xn 7→ (0, 0, . . . , 0, xn, 0, 0 . . .) be the canonical maps.Then one may define a linear isomorphic embedding c0 → AX by

θ : (ak) 7→∞∑k=1

akJkTkPk +A(X ).

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Examples with large AX , II

Corollary. Let 1 < p <∞, p 6= 2. Then there is a closed subspaceX ⊂ `p such that AX contains a complemented copy of c0.

Comment. ”Most” Banach spaces Z contains a closed subspaceisomorphic to `p (for some 1 ≤ p <∞) or to c0.

Proof (idea): By Bachelis [B76] there is a closed subspace Y ⊂ `psuch that AY 6= {0}. Put X = (⊕NY )p ⊂ (⊕N`p)p ≈ `p. Byabove Proposition, AX contains a copy of c0. Moreover, AX isseparable, so c0 is complemented in AX by Sobzcyk’s theorem.

Here also fact: if X ∗ is separable, then K (X ) (as well as AX ) isseparable. Reason: K := (BX∗ ,w∗)× (BX∗∗ ,w∗) is compactmetrisable, and isometric embedding χ : K(X )→ C (K ), where

χ(T )(x∗, x∗∗) = 〈T ∗x∗, x∗∗〉, T ∈ K(X ), (x∗, x∗∗) ∈ K .

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Examples with large AX , II

Corollary. Let 1 < p <∞, p 6= 2. Then there is a closed subspaceX ⊂ `p such that AX contains a complemented copy of c0.

Comment. ”Most” Banach spaces Z contains a closed subspaceisomorphic to `p (for some 1 ≤ p <∞) or to c0.

Proof (idea): By Bachelis [B76] there is a closed subspace Y ⊂ `psuch that AY 6= {0}. Put X = (⊕NY )p ⊂ (⊕N`p)p ≈ `p. Byabove Proposition, AX contains a copy of c0. Moreover, AX isseparable, so c0 is complemented in AX by Sobzcyk’s theorem.

Here also fact: if X ∗ is separable, then K (X ) (as well as AX ) isseparable. Reason: K := (BX∗ ,w∗)× (BX∗∗ ,w∗) is compactmetrisable, and isometric embedding χ : K(X )→ C (K ), where

χ(T )(x∗, x∗∗) = 〈T ∗x∗, x∗∗〉, T ∈ K(X ), (x∗, x∗∗) ∈ K .

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Banach algebra contents

Question: Banach algebraic relevance of linear embedding θ?

- Above θ : c0 → AX is never an algebra homomorphism (sinceAX radical B-algebra, but c0 not so)

- θ(c0) is a commuting set, and generates a closed commutativesubalgebra A[θ(c0)] of AX . But there is closed subspaceX ⊂ `p for p 6= 2, so that AX is non-commutative (whenceA[θ(c0)] 6= AX ).

- In the Corollary one may choose the embedding θ so that(i) θ(a) · θ(b) = 0 for all a, b ∈ c0,alternatively(ii) for each n ∈ N there are a(k) ∈ c0 for k = 1, . . . , n suchthat θ(a(1)) · · · θ(a(n)) 6= 0. (This means that A[θ(c0)], aswell as AX , is not nilpotent. Part (ii) is based on [B76].)

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Banach algebra contents

Question: Banach algebraic relevance of linear embedding θ?

- Above θ : c0 → AX is never an algebra homomorphism (sinceAX radical B-algebra, but c0 not so)

- θ(c0) is a commuting set, and generates a closed commutativesubalgebra A[θ(c0)] of AX . But there is closed subspaceX ⊂ `p for p 6= 2, so that AX is non-commutative (whenceA[θ(c0)] 6= AX ).

- In the Corollary one may choose the embedding θ so that(i) θ(a) · θ(b) = 0 for all a, b ∈ c0,alternatively(ii) for each n ∈ N there are a(k) ∈ c0 for k = 1, . . . , n suchthat θ(a(1)) · · · θ(a(n)) 6= 0. (This means that A[θ(c0)], aswell as AX , is not nilpotent. Part (ii) is based on [B76].)

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Compact approximation property

Recall: X has the compact approximation property (CAP) if for allε > 0 and all compact subsets K ⊂ X there is a compact operatorT ∈ K(X ) such that

supx∈K||x − Tx || < ε

Moreover, X has the bounded compact approximation property(BCAP) if there is constant M <∞ such that T ∈ K(X ) abovecan always be chosen with ||T || ≤ M.

This points to a relevant class of spaces!

Proposition (Dales, [TW19]) If X has BCAP but fails AP, thendim(AX ) =∞. Moreover, there is T ∈ K(X ) such thatT n /∈ A(X ) for every n ∈ N.

(Recall that ||T n +A(X )||1/n → 0 as n→∞ by radicality of AX .)

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Compact approximation property

Recall: X has the compact approximation property (CAP) if for allε > 0 and all compact subsets K ⊂ X there is a compact operatorT ∈ K(X ) such that

supx∈K||x − Tx || < ε

Moreover, X has the bounded compact approximation property(BCAP) if there is constant M <∞ such that T ∈ K(X ) abovecan always be chosen with ||T || ≤ M.

This points to a relevant class of spaces!

Proposition (Dales, [TW19]) If X has BCAP but fails AP, thendim(AX ) =∞. Moreover, there is T ∈ K(X ) such thatT n /∈ A(X ) for every n ∈ N.

(Recall that ||T n +A(X )||1/n → 0 as n→∞ by radicality of AX .)

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Spaces with BCAP that fail AP

Question: Do such Banach spaces X exist?

Theorem. (Willis [W92]) There are Banach spaces Z having theBCAP (with constant M = 1) that fail AP.

Outline of the construction:

Let X be a Banach space failing AP. Then there is c > 0 and acompact K = conv{xn | n ∈ N} ⊂ X , where ||xn|| → 0, such that

supx∈K||Tx − x || ≥ c for all T ∈ F(X ).

For all 0 < s < 1, let Us = absconv{ xn||xn||s

| n ∈ N} and let

Ys = span Us endowed with the complete norm

|y |s = inf{λ > 0 | y ∈ λUs}.

Let Z = span{yχ(s,t) | 0 < s < t < 1, y ∈ Ys} with the norm

||f || =∫ 10 |f (s)|sds. Main step: the completion Z of Z has BCAP

but fails AP.10 / 14

Spaces with BCAP that fail AP

Question: Do such Banach spaces X exist?

Theorem. (Willis [W92]) There are Banach spaces Z having theBCAP (with constant M = 1) that fail AP.

Outline of the construction:

Let X be a Banach space failing AP. Then there is c > 0 and acompact K = conv{xn | n ∈ N} ⊂ X , where ||xn|| → 0, such that

supx∈K||Tx − x || ≥ c for all T ∈ F(X ).

For all 0 < s < 1, let Us = absconv{ xn||xn||s

| n ∈ N} and let

Ys = span Us endowed with the complete norm

|y |s = inf{λ > 0 | y ∈ λUs}.

Let Z = span{yχ(s,t) | 0 < s < t < 1, y ∈ Ys} with the norm

||f || =∫ 10 |f (s)|sds. Main step: the completion Z of Z has BCAP

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AZ for Willis spaces Z , I

Theorem. [TW19] Let Z be a Willis space. Then there is a linearisomorphic embedding θ : c0 → AZ .Idea of proof: Fix 0 < s < t < 1. The following diagramcommutes up to a constant (which depends on s, t).

Ys X

Z

Z0 = P0Z

Rs(y) = y

Ts(y) = yχ(s,1) J(f ) =∫ 10 f (s)ds

P0(f ) = f χ(s,t) J0(f ) = f

Here Rs /∈ A(Ys ,X ) andTs ∈ K(Ys ,Z ). ThusP0Ts ∈ K(Ys ,Z0) \A(Ys ,Z0).

Hence the closedcomplemented subspace Z0

has (B)CAP but fails AP.Consequently AZ0 6= {0}.

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AZ for Willis spaces Z , I

Theorem. [TW19] Let Z be a Willis space. Then there is a linearisomorphic embedding θ : c0 → AZ .Idea of proof: Fix 0 < s < t < 1. The following diagramcommutes up to a constant (which depends on s, t).

Ys X

Z

Z0 = P0Z

Rs(y) = y

Ts(y) = yχ(s,1) J(f ) =∫ 10 f (s)ds

P0(f ) = f χ(s,t) J0(f ) = f

Here Rs /∈ A(Ys ,X ) andTs ∈ K(Ys ,Z ). ThusP0Ts ∈ K(Ys ,Z0) \A(Ys ,Z0).

Hence the closedcomplemented subspace Z0

has (B)CAP but fails AP.Consequently AZ0 6= {0}.

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AZ for Willis spaces Z , I

Theorem. [TW19] Let Z be a Willis space. Then there is a linearisomorphic embedding θ : c0 → AZ .Idea of proof: Fix 0 < s < t < 1. The following diagramcommutes up to a constant (which depends on s, t).

Ys X

Z

Z0 = P0Z

Rs(y) = y

Ts(y) = yχ(s,1) J(f ) =∫ 10 f (s)ds

P0(f ) = f χ(s,t) J0(f ) = f

Here Rs /∈ A(Ys ,X ) andTs ∈ K(Ys ,Z ). ThusP0Ts ∈ K(Ys ,Z0) \A(Ys ,Z0).

Hence the closedcomplemented subspace Z0

has (B)CAP but fails AP.Consequently AZ0 6= {0}.

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AZ for Willis spaces Z , II

Fix interlacing sequences (sn) and (tn) of (0, 1) such thatsn < tn < sn+1 for all n and sn → 1 as n→∞. Let Zn ⊂ Z be theclosed subspace consisting of functions in Z supported in(sn, tn) ⊂ (0, 1). By above argument AZn 6= {0} for all n ∈ N. Bythe general Proposition A(⊕nZn)1 contains an isomorphic copy ofc0. Deduce that c0 also embeds into AZ by the following fact.

Proposition. Let X ,Y be Banach spaces and suppose thatAX 6= {0}. Then AX is algebra isomorphic to a subalgebra ofAX⊕Y .

Comment: Theorem holds also for related separable reflexive Willisspaces Zp with 1 < p <∞ and p 6= 2, where Zp quotient of closedsubspace of Lp.

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AZ for Willis spaces Z , II

Fix interlacing sequences (sn) and (tn) of (0, 1) such thatsn < tn < sn+1 for all n and sn → 1 as n→∞. Let Zn ⊂ Z be theclosed subspace consisting of functions in Z supported in(sn, tn) ⊂ (0, 1). By above argument AZn 6= {0} for all n ∈ N. Bythe general Proposition A(⊕nZn)1 contains an isomorphic copy ofc0. Deduce that c0 also embeds into AZ by the following fact.

Proposition. Let X ,Y be Banach spaces and suppose thatAX 6= {0}. Then AX is algebra isomorphic to a subalgebra ofAX⊕Y .

Comment: Theorem holds also for related separable reflexive Willisspaces Zp with 1 < p <∞ and p 6= 2, where Zp quotient of closedsubspace of Lp.

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Further big quotient algebras

Let (Gm) be sequence of finite dimensional spaces dense in theBanach-Mazur distance in the class of finite dimensional Banachspaces. Let Cp = (⊕m∈NGm)`p for 1 ≤ p <∞. C1 is Johnson’scomplementably universal conjugate space.

Proposition. [TW19] AC∗1

contains a linear isomorphic copy of c0(but AC1 = {0} since C1 has AP).

Johnson (1972) & Figiel (1973): every compact T ∈ K(X ,Y )factors compactly through a subspace of Cp, that is, there is aclosed subspace M ⊂ Cp and compact operators A ∈ K(X ,M),B ∈ K(M,Y ) so that T = B ◦ A. Let 1 ≤ p <∞ and

I = {W ⊂ Cp |W closed infinite dimensional subspace}.

Consider ZpFJ = (⊕W∈IW )`p , which is a universal compact

factorisation space.

Proposition. [TW19] There is an uncountable index set Γ suchthat AZp

FJcontains a isomorphic copy of non-separable space c0(Γ).

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Further big quotient algebras

Let (Gm) be sequence of finite dimensional spaces dense in theBanach-Mazur distance in the class of finite dimensional Banachspaces. Let Cp = (⊕m∈NGm)`p for 1 ≤ p <∞. C1 is Johnson’scomplementably universal conjugate space.

Proposition. [TW19] AC∗1

contains a linear isomorphic copy of c0(but AC1 = {0} since C1 has AP).

Johnson (1972) & Figiel (1973): every compact T ∈ K(X ,Y )factors compactly through a subspace of Cp, that is, there is aclosed subspace M ⊂ Cp and compact operators A ∈ K(X ,M),B ∈ K(M,Y ) so that T = B ◦ A. Let 1 ≤ p <∞ and

I = {W ⊂ Cp |W closed infinite dimensional subspace}.

Consider ZpFJ = (⊕W∈IW )`p , which is a universal compact

factorisation space.

Proposition. [TW19] There is an uncountable index set Γ suchthat AZp

FJcontains a isomorphic copy of non-separable space c0(Γ).

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Concluding remarks and problems

Conclusions: size of A(X ) ⊂ K(X ) is very subtle for operatorsX → X , where X is given Banach space failing AP.

Problems: (i) how embed other spaces Z into AX ?(ii) how preserve Banach algebra properties?

For direct sums the following variant becomes relevant:

(iii) Given X ,Y such that X ∗ and Y both fail the AP, is it alwaystrue that

A(X ,Y ) K(X ,Y )?

Somewhat surprisingly this is not so! For example, if X has type 2and Y has cotype 2, then K(X ,Y ) = A(X ,Y ) (Kwapien, Maurey,John).This matters e.g. for structure of closed ideals of AX (Cf. nexttalk by Henrik Wirzenius.)

THANKS!

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Concluding remarks and problems

Conclusions: size of A(X ) ⊂ K(X ) is very subtle for operatorsX → X , where X is given Banach space failing AP.

Problems: (i) how embed other spaces Z into AX ?(ii) how preserve Banach algebra properties?

For direct sums the following variant becomes relevant:

(iii) Given X ,Y such that X ∗ and Y both fail the AP, is it alwaystrue that

A(X ,Y ) K(X ,Y )?

Somewhat surprisingly this is not so! For example, if X has type 2and Y has cotype 2, then K(X ,Y ) = A(X ,Y ) (Kwapien, Maurey,John).This matters e.g. for structure of closed ideals of AX (Cf. nexttalk by Henrik Wirzenius.)

THANKS!

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