Grothendieck Spaces, Operators, and Beyond

Grothendieck spaces, operators, and beyond

Tomasz Kania

Academy of Sciences of the Czech Republic, Praha

Zimní škola z abstraktní analýzy Svratka, 13 leden 2019

joint work with K. Beanland & N. J. Laustsen

1 I closed under surjective linear images (hence complemented subspaces) –just apply Hahn–Banach! I not closed under taking subspaces (c0 ⊂ `∞), I closed under forming `p-sums with p ∈ (1, ∞) but not for p = ∞!

I call such spaces Grothendieck. This class is

I A separable space is G. if and only if it is reﬂexive–use Eberlein–Šmulian. s I Ergodic theory of op on G. spaces works rather well. So is it one of those ‘non-separable’ talks? Well...

Overview ∗ I Grothendieck’s motivation: measure-theoretic as `∞(Γ) comprises ﬁnitely additive measures on ℘(Γ); strengthening certain measure theoretic principles.

Background

Theorem (Grothendieck, 1953). Every weak* convergent sequence (fn) in the dual of `∞(Γ) converges with respect to the weak topology.

2 I closed under surjective linear images (hence complemented subspaces) –just apply Hahn–Banach! I not closed under taking subspaces (c0 ⊂ `∞), I closed under forming `p-sums with p ∈ (1, ∞) but not for p = ∞!

I call such spaces Grothendieck. This class is

I A separable space is G. if and only if it is reﬂexive–use Eberlein–Šmulian. s I Ergodic theory of op on G. spaces works rather well. So is it one of those ‘non-separable’ talks? Well...

Background

Theorem (Grothendieck, 1953). Every weak* convergent sequence (fn) in the dual of `∞(Γ) converges with respect to the weak topology.

Overview ∗ I Grothendieck’s motivation: measure-theoretic as `∞(Γ) comprises ﬁnitely additive measures on ℘(Γ); strengthening certain measure theoretic principles.

I A separable space is G. if and only if it is reﬂexive–use Eberlein–Šmulian. s I Ergodic theory of op on G. spaces works rather well. So is it one of those ‘non-separable’ talks? Well...

Background

Theorem (Grothendieck, 1953). Every weak* convergent sequence (fn) in the dual of `∞(Γ) converges with respect to the weak topology.

Overview ∗ I Grothendieck’s motivation: measure-theoretic as `∞(Γ) comprises ﬁnitely additive measures on ℘(Γ); strengthening certain measure theoretic principles.

I call such spaces Grothendieck. This class is

2 I not closed under taking subspaces (c0 ⊂ `∞), I closed under forming `p-sums with p ∈ (1, ∞) but not for p = ∞!

I A separable space is G. if and only if it is reﬂexive–use Eberlein–Šmulian. s I Ergodic theory of op on G. spaces works rather well. So is it one of those ‘non-separable’ talks? Well...

Background

Theorem (Grothendieck, 1953). Every weak* convergent sequence (fn) in the dual of `∞(Γ) converges with respect to the weak topology.

Overview ∗ I Grothendieck’s motivation: measure-theoretic as `∞(Γ) comprises ﬁnitely additive measures on ℘(Γ); strengthening certain measure theoretic principles.

I call such spaces Grothendieck. This class is

I closed under surjective linear images (hence complemented subspaces) –just apply Hahn–Banach!

2 I closed under forming `p-sums with p ∈ (1, ∞) but not for p = ∞!

I A separable space is G. if and only if it is reﬂexive–use Eberlein–Šmulian. s I Ergodic theory of op on G. spaces works rather well. So is it one of those ‘non-separable’ talks? Well...

Background

Theorem (Grothendieck, 1953). Every weak* convergent sequence (fn) in the dual of `∞(Γ) converges with respect to the weak topology.

Overview ∗ I Grothendieck’s motivation: measure-theoretic as `∞(Γ) comprises ﬁnitely additive measures on ℘(Γ); strengthening certain measure theoretic principles.

I call such spaces Grothendieck. This class is

I closed under surjective linear images (hence complemented subspaces) –just apply Hahn–Banach! I not closed under taking subspaces (c0 ⊂ `∞),

2 I A separable space is G. if and only if it is reﬂexive–use Eberlein–Šmulian. s I Ergodic theory of op on G. spaces works rather well. So is it one of those ‘non-separable’ talks? Well...

Background

Theorem (Grothendieck, 1953). Every weak* convergent sequence (fn) in the dual of `∞(Γ) converges with respect to the weak topology.

Overview ∗ I Grothendieck’s motivation: measure-theoretic as `∞(Γ) comprises ﬁnitely additive measures on ℘(Γ); strengthening certain measure theoretic principles.

I call such spaces Grothendieck. This class is

I closed under surjective linear images (hence complemented subspaces) –just apply Hahn–Banach! I not closed under taking subspaces (c0 ⊂ `∞), I closed under forming `p-sums with p ∈ (1, ∞) but not for p = ∞!

2 s I Ergodic theory of op on G. spaces works rather well. So is it one of those ‘non-separable’ talks? Well...

Background

Theorem (Grothendieck, 1953). Every weak* convergent sequence (fn) in the dual of `∞(Γ) converges with respect to the weak topology.

Overview ∗ I Grothendieck’s motivation: measure-theoretic as `∞(Γ) comprises ﬁnitely additive measures on ℘(Γ); strengthening certain measure theoretic principles.

I call such spaces Grothendieck. This class is

I A separable space is G. if and only if it is reﬂexive–use Eberlein–Šmulian.

2 So is it one of those ‘non-separable’ talks? Well...

Background

Theorem (Grothendieck, 1953). Every weak* convergent sequence (fn) in the dual of `∞(Γ) converges with respect to the weak topology.

Overview ∗ I Grothendieck’s motivation: measure-theoretic as `∞(Γ) comprises ﬁnitely additive measures on ℘(Γ); strengthening certain measure theoretic principles.

I call such spaces Grothendieck. This class is

I A separable space is G. if and only if it is reﬂexive–use Eberlein–Šmulian. s I Ergodic theory of op on G. spaces works rather well.

2 Background

Theorem (Grothendieck, 1953). Every weak* convergent sequence (fn) in the dual of `∞(Γ) converges with respect to the weak topology.

Overview ∗ I Grothendieck’s motivation: measure-theoretic as `∞(Γ) comprises ﬁnitely additive measures on ℘(Γ); strengthening certain measure theoretic principles.

I call such spaces Grothendieck. This class is

I A separable space is G. if and only if it is reﬂexive–use Eberlein–Šmulian. s I Ergodic theory of op on G. spaces works rather well. So is it one of those ‘non-separable’ talks? Well...

2 C(K) for K Stonean, σ-Stonean, F -space, all indecomposable spaces C(K) constructed by Koszmider, Plebanek, and others. No topological characterisation of when a C(K)-space is Grothendieck exists. If C(K) is Grothendieck, K does not have non-trivial convergent sequences.

∞ H (D), bdd analytic functions (Bourgain 1983) Dual spaces with Pełczyński’s property (V ). (X has prop. (V) whenever for any space Y every unconditionally converging operator T : X → Y is weakly compact.)

I C(K)-spaces have prop. (V). I C*-algebras have prop. (V) (Pﬁtzner, 1994); all von Neumann algebras are Grothendieck spaces B(H), the space of ops on a Hilbert space H, is Grothendieck.

This answered a question from Diestel–Uhl’s Vector measures:

Grothendieck spaces are rare animals

Some examples.

3 ∞ H (D), bdd analytic functions (Bourgain 1983) Dual spaces with Pełczyński’s property (V ). (X has prop. (V) whenever for any space Y every unconditionally converging operator T : X → Y is weakly compact.)

I C(K)-spaces have prop. (V). I C*-algebras have prop. (V) (Pﬁtzner, 1994); all von Neumann algebras are Grothendieck spaces B(H), the space of ops on a Hilbert space H, is Grothendieck.

This answered a question from Diestel–Uhl’s Vector measures:

Grothendieck spaces are rare animals

Some examples. C(K) for K Stonean, σ-Stonean, F -space, all indecomposable spaces C(K) constructed by Koszmider, Plebanek, and others. No topological characterisation of when a C(K)-space is Grothendieck exists. If C(K) is Grothendieck, K does not have non-trivial convergent sequences.

3 Dual spaces with Pełczyński’s property (V ). (X has prop. (V) whenever for any space Y every unconditionally converging operator T : X → Y is weakly compact.)

I C(K)-spaces have prop. (V). I C*-algebras have prop. (V) (Pﬁtzner, 1994); all von Neumann algebras are Grothendieck spaces B(H), the space of ops on a Hilbert space H, is Grothendieck.

This answered a question from Diestel–Uhl’s Vector measures:

Grothendieck spaces are rare animals

∞ H (D), bdd analytic functions (Bourgain 1983)

3 This answered a question from Diestel–Uhl’s Vector measures:

Grothendieck spaces are rare animals

I C(K)-spaces have prop. (V). I C*-algebras have prop. (V) (Pﬁtzner, 1994); all von Neumann algebras are Grothendieck spaces B(H), the space of ops on a Hilbert space H, is Grothendieck.

3 Finally, there is some evidence (Akemann [1967], [1968]) that the space L (H; H) of bounded linear operators on a Hilbert space is a Grothendieck space and that more generally the space L (X ; X ) is a Grothendieck space for any reﬂexive Banach space.

Grothendieck spaces are rare animals

I C(K)-spaces have prop. (V). I C*-algebras have prop. (V) (Pﬁtzner, 1994); all von Neumann algebras are Grothendieck spaces B(H), the space of ops on a Hilbert space H, is Grothendieck.

This answered a question from Diestel–Uhl’s Vector measures:

3 Grothendieck spaces are rare animals

I C(K)-spaces have prop. (V). I C*-algebras have prop. (V) (Pﬁtzner, 1994); all von Neumann algebras are Grothendieck spaces B(H), the space of ops on a Hilbert space H, is Grothendieck.

This answered a question from Diestel–Uhl’s Vector measures:

Finally, there is some evidence (Akemann [1967], [1968]) that the space L (H; H) of bounded linear operators on a Hilbert space is a Grothendieck space and that more generally the space L (X ; X ) is a Grothendieck space for any reﬂexive Banach space. 3 ∗ Note: B(Ep) is linearly isometric (via the adjoint map) to B(Ep ), the space of L n operators on n `∞)`p∗

L n Key ingredient (Johnson, 1972). The space E∞ = ( n `1)`∞ contains a complemented copy of `1, hence it is not Grothendieck.

Identify E∞ as a complemented subspace of B(Ep).

Remark. If B(X ) is Grothendieck then so are X and X ∗.

Question (Diestel): If X and X ∗ are Grothendieck, is X reﬂexive? I can prove it under property (V)...

Question: Is B(X ) ever reﬂexive? (X ∞-dim) see M. Ostrovskii’s post mathoverflow.net/q/232291/15129

Evidence was not good enough

L n Theorem (K. 2012). Let p ∈ (1, ∞). Set Ep = n `1)`p . Then the space of operators B(Ep) is not Grothendieck, even though Ep is reﬂexive.

4 L n Key ingredient (Johnson, 1972). The space E∞ = ( n `1)`∞ contains a complemented copy of `1, hence it is not Grothendieck.

Identify E∞ as a complemented subspace of B(Ep).

Remark. If B(X ) is Grothendieck then so are X and X ∗.

Question (Diestel): If X and X ∗ are Grothendieck, is X reﬂexive? I can prove it under property (V)...

Question: Is B(X ) ever reﬂexive? (X ∞-dim) see M. Ostrovskii’s post mathoverflow.net/q/232291/15129

Evidence was not good enough

L n Theorem (K. 2012). Let p ∈ (1, ∞). Set Ep = n `1)`p . Then the space of operators B(Ep) is not Grothendieck, even though Ep is reﬂexive. ∗ Note: B(Ep) is linearly isometric (via the adjoint map) to B(Ep ), the space of L n operators on n `∞)`p∗

4 Identify E∞ as a complemented subspace of B(Ep).

Remark. If B(X ) is Grothendieck then so are X and X ∗.

Question (Diestel): If X and X ∗ are Grothendieck, is X reﬂexive? I can prove it under property (V)...

Question: Is B(X ) ever reﬂexive? (X ∞-dim) see M. Ostrovskii’s post mathoverflow.net/q/232291/15129

Evidence was not good enough

L n Key ingredient (Johnson, 1972). The space E∞ = ( n `1)`∞ contains a complemented copy of `1, hence it is not Grothendieck.

4 Remark. If B(X ) is Grothendieck then so are X and X ∗.

Question (Diestel): If X and X ∗ are Grothendieck, is X reﬂexive? I can prove it under property (V)...

Question: Is B(X ) ever reﬂexive? (X ∞-dim) see M. Ostrovskii’s post mathoverflow.net/q/232291/15129

Evidence was not good enough

L n Key ingredient (Johnson, 1972). The space E∞ = ( n `1)`∞ contains a complemented copy of `1, hence it is not Grothendieck.

Identify E∞ as a complemented subspace of B(Ep).

4 Question: Is B(X ) ever reﬂexive? (X ∞-dim) see M. Ostrovskii’s post mathoverflow.net/q/232291/15129

Evidence was not good enough

L n Key ingredient (Johnson, 1972). The space E∞ = ( n `1)`∞ contains a complemented copy of `1, hence it is not Grothendieck.

Identify E∞ as a complemented subspace of B(Ep).

Remark. If B(X ) is Grothendieck then so are X and X ∗.

Question (Diestel): If X and X ∗ are Grothendieck, is X reﬂexive? I can prove it under property (V)...

4 Evidence was not good enough

L n Key ingredient (Johnson, 1972). The space E∞ = ( n `1)`∞ contains a complemented copy of `1, hence it is not Grothendieck.

Identify E∞ as a complemented subspace of B(Ep).

Remark. If B(X ) is Grothendieck then so are X and X ∗.

Question (Diestel): If X and X ∗ are Grothendieck, is X reﬂexive? I can prove it under property (V)...

Question: Is B(X ) ever reﬂexive? (X ∞-dim) see M. Ostrovskii’s post mathoverflow.net/q/232291/15129

4 I Common feature: the said spaces are reﬂexive but not superreﬂexive.

∞ I E a space with a normalised, 1-unconditional basis (en)n=1. ∞ The E-direct sum of a sequence (Xn)n=1 of spaces is given by ∞ M X Xn = (xn): xn ∈ Xn (n ∈ N) and kxnken converges in E . E n∈N n=1 P∞ This is a Banach space w.r.t k(xn)k = n=1 kxnken . ∞ s I Every uniformly bdd sequence (Un)n=1 of op , where Un ∈ B(Xn), induces L a ‘diagonal operator’ diag(Un) on Xn via n∈N E ∞ ∞ diag(Un)(xn)n=1 = (Unxn)n=1 k diag(Un)k = sup kUnk n

L Abstraction. Let X = Xn , where E is a Banach space with a n∈N E normalized, 1-unconditional basis and (Xn) is a sequence of Banach spaces. Then B(X ) contains a complemented subspace which is isometrically L isomorphic to Xn . n∈N `∞

Maybe this idea is not bad after all

Theorem (Beanland, K., Laustsen). Let X be the Tsirelson space or the pth Baernstein space (p ∈ (1, ∞)). Then B(X ) is not a Grothendieck space

5 ∞ I E a space with a normalised, 1-unconditional basis (en)n=1. ∞ The E-direct sum of a sequence (Xn)n=1 of spaces is given by ∞ M X Xn = (xn): xn ∈ Xn (n ∈ N) and kxnken converges in E . E n∈N n=1 P∞ This is a Banach space w.r.t k(xn)k = n=1 kxnken . ∞ s I Every uniformly bdd sequence (Un)n=1 of op , where Un ∈ B(Xn), induces L a ‘diagonal operator’ diag(Un) on Xn via n∈N E ∞ ∞ diag(Un)(xn)n=1 = (Unxn)n=1 k diag(Un)k = sup kUnk n

Maybe this idea is not bad after all

Theorem (Beanland, K., Laustsen). Let X be the Tsirelson space or the pth Baernstein space (p ∈ (1, ∞)). Then B(X ) is not a Grothendieck space

I Common feature: the said spaces are reﬂexive but not superreﬂexive.

5 ∞ s I Every uniformly bdd sequence (Un)n=1 of op , where Un ∈ B(Xn), induces L a ‘diagonal operator’ diag(Un) on Xn via n∈N E ∞ ∞ diag(Un)(xn)n=1 = (Unxn)n=1 k diag(Un)k = sup kUnk n

Maybe this idea is not bad after all

Theorem (Beanland, K., Laustsen). Let X be the Tsirelson space or the pth Baernstein space (p ∈ (1, ∞)). Then B(X ) is not a Grothendieck space

I Common feature: the said spaces are reﬂexive but not superreﬂexive.

5 L Abstraction. Let X = Xn , where E is a Banach space with a n∈N E normalized, 1-unconditional basis and (Xn) is a sequence of Banach spaces. Then B(X ) contains a complemented subspace which is isometrically L isomorphic to Xn . n∈N `∞

Maybe this idea is not bad after all

Theorem (Beanland, K., Laustsen). Let X be the Tsirelson space or the pth Baernstein space (p ∈ (1, ∞)). Then B(X ) is not a Grothendieck space

I Common feature: the said spaces are reﬂexive but not superreﬂexive.

5 Maybe this idea is not bad after all

Theorem (Beanland, K., Laustsen). Let X be the Tsirelson space or the pth Baernstein space (p ∈ (1, ∞)). Then B(X ) is not a Grothendieck space

I Common feature: the said spaces are reﬂexive but not superreﬂexive.

k 1 p X p X νp(x; N1,..., Nk ) = µ(x, Nj ) , where µ(x, Nj ) = |αn|.

j=1 n∈Nj

Deﬁnition. Bp is the completion of c00 w.r.t.

kxkBp = sup{νp(x; N1,..., Nk ): k ∈ N and N1 < N2 < ··· < Nk , |Nj | < min Nj }

B2 the ﬁrst example (1972) of a reﬂexive space without the Banach–Saks prop.

Caveat. Bp is not isomorphic to any E-sum of f.d. blocks of the basis mn isomorphic to `1 (mn → ∞) for any space E with a normalised 1-unconditional basis.

a word about the proof and some perspectives

Key observation. We observe that the (Johnson–Figiel) Tsirelson space X is L 54-isomorphic to an unconditional FDD ( Fn)T , where En is 2-isomorphic to `1(mn) and mn → ∞ as n → ∞.

6 For x ∈ c00 and Nj ⊂ N set

k 1 p X p X νp(x; N1,..., Nk ) = µ(x, Nj ) , where µ(x, Nj ) = |αn|.

j=1 n∈Nj

Deﬁnition. Bp is the completion of c00 w.r.t.

kxkBp = sup{νp(x; N1,..., Nk ): k ∈ N and N1 < N2 < ··· < Nk , |Nj | < min Nj }

B2 the ﬁrst example (1972) of a reﬂexive space without the Banach–Saks prop.

Caveat. Bp is not isomorphic to any E-sum of f.d. blocks of the basis mn isomorphic to `1 (mn → ∞) for any space E with a normalised 1-unconditional basis.

a word about the proof and some perspectives

Key observation. We observe that the (Johnson–Figiel) Tsirelson space X is L 54-isomorphic to an unconditional FDD ( Fn)T , where En is 2-isomorphic to `1(mn) and mn → ∞ as n → ∞.

Baernstein spaces.

6 Deﬁnition. Bp is the completion of c00 w.r.t.

kxkBp = sup{νp(x; N1,..., Nk ): k ∈ N and N1 < N2 < ··· < Nk , |Nj | < min Nj }

B2 the ﬁrst example (1972) of a reﬂexive space without the Banach–Saks prop.

Caveat. Bp is not isomorphic to any E-sum of f.d. blocks of the basis mn isomorphic to `1 (mn → ∞) for any space E with a normalised 1-unconditional basis.

a word about the proof and some perspectives

Key observation. We observe that the (Johnson–Figiel) Tsirelson space X is L 54-isomorphic to an unconditional FDD ( Fn)T , where En is 2-isomorphic to `1(mn) and mn → ∞ as n → ∞.

Baernstein spaces. For x ∈ c00 and Nj ⊂ N set

k 1 p X p X νp(x; N1,..., Nk ) = µ(x, Nj ) , where µ(x, Nj ) = |αn|.

j=1 n∈Nj

6 B2 the ﬁrst example (1972) of a reﬂexive space without the Banach–Saks prop.

Caveat. Bp is not isomorphic to any E-sum of f.d. blocks of the basis mn isomorphic to `1 (mn → ∞) for any space E with a normalised 1-unconditional basis.

a word about the proof and some perspectives

Key observation. We observe that the (Johnson–Figiel) Tsirelson space X is L 54-isomorphic to an unconditional FDD ( Fn)T , where En is 2-isomorphic to `1(mn) and mn → ∞ as n → ∞.

Baernstein spaces. For x ∈ c00 and Nj ⊂ N set

k 1 p X p X νp(x; N1,..., Nk ) = µ(x, Nj ) , where µ(x, Nj ) = |αn|.

j=1 n∈Nj

Deﬁnition. Bp is the completion of c00 w.r.t.

kxkBp = sup{νp(x; N1,..., Nk ): k ∈ N and N1 < N2 < ··· < Nk , |Nj | < min Nj }

6 Caveat. Bp is not isomorphic to any E-sum of f.d. blocks of the basis mn isomorphic to `1 (mn → ∞) for any space E with a normalised 1-unconditional basis.

a word about the proof and some perspectives

Key observation. We observe that the (Johnson–Figiel) Tsirelson space X is L 54-isomorphic to an unconditional FDD ( Fn)T , where En is 2-isomorphic to `1(mn) and mn → ∞ as n → ∞.

Baernstein spaces. For x ∈ c00 and Nj ⊂ N set

k 1 p X p X νp(x; N1,..., Nk ) = µ(x, Nj ) , where µ(x, Nj ) = |αn|.

j=1 n∈Nj

Deﬁnition. Bp is the completion of c00 w.r.t.

kxkBp = sup{νp(x; N1,..., Nk ): k ∈ N and N1 < N2 < ··· < Nk , |Nj | < min Nj }

B2 the ﬁrst example (1972) of a reﬂexive space without the Banach–Saks prop.

6 a word about the proof and some perspectives

Key observation. We observe that the (Johnson–Figiel) Tsirelson space X is L 54-isomorphic to an unconditional FDD ( Fn)T , where En is 2-isomorphic to `1(mn) and mn → ∞ as n → ∞.

Baernstein spaces. For x ∈ c00 and Nj ⊂ N set

k 1 p X p X νp(x; N1,..., Nk ) = µ(x, Nj ) , where µ(x, Nj ) = |αn|.

j=1 n∈Nj

Deﬁnition. Bp is the completion of c00 w.r.t.

kxkBp = sup{νp(x; N1,..., Nk ): k ∈ N and N1 < N2 < ··· < Nk , |Nj | < min Nj }

B2 the ﬁrst example (1972) of a reﬂexive space without the Banach–Saks prop.

Caveat. Bp is not isomorphic to any E-sum of f.d. blocks of the basis mn isomorphic to `1 (mn → ∞) for any space E with a normalised 1-unconditional basis.

6 Observation. 1. & 2. are incompatible with the Grothendieck property of B(X ). Indeed, 2. rules out reflexivity of B(X ), 1. forces separability of B(X ), and separable Grothendieck spaces are reflexive. Question. Is B(X ) Grothendieck for super-reflexive X ?

Embarrassingly, we do not know whether B(`p) is Grothendieck for ∼ p ∈ (1, 2) ∪ (2, ∞)... (btw. B(`p) = B(Lp) as Banach spaces). 1 n ∗∗ A distantly related question. Let En = C ([0, 1] ). Is En Grothendieck? (Yes, for n = 1.) If this is the case, this would have solved a recent question by n Candido, Cúth, and Doucha of whether Lip Z is Grothendieck.

Thank you very much!

Very few operators vs. Grothendieck

Argyros and Haydon built a space XAH such that

1. B(XAH) = C · IXAH ⊕ K (XAH)

2. XAH has a basis, ∗ ∼ 3. XAH = `1. Ultimate (?) problem in the area: is there a reﬂexive space with 1. & 2.?

7 Indeed, 2. rules out reflexivity of B(X ), 1. forces separability of B(X ), and separable Grothendieck spaces are reflexive. Question. Is B(X ) Grothendieck for super-reflexive X ?

Thank you very much!

Very few operators vs. Grothendieck

Argyros and Haydon built a space XAH such that

1. B(XAH) = C · IXAH ⊕ K (XAH)

2. XAH has a basis, ∗ ∼ 3. XAH = `1. Ultimate (?) problem in the area: is there a reﬂexive space with 1. & 2.? Observation. 1. & 2. are incompatible with the Grothendieck property of B(X ).

7 Question. Is B(X ) Grothendieck for super-reﬂexive X ?

Thank you very much!

Very few operators vs. Grothendieck

Argyros and Haydon built a space XAH such that

1. B(XAH) = C · IXAH ⊕ K (XAH)

2. XAH has a basis, ∗ ∼ 3. XAH = `1. Ultimate (?) problem in the area: is there a reflexive space with 1. & 2.? Observation. 1. & 2. are incompatible with the Grothendieck property of B(X ). Indeed, 2. rules out reflexivity of B(X ), 1. forces separability of B(X ), and separable Grothendieck spaces are reflexive.

7 Embarrassingly, we do not know whether B(`p) is Grothendieck for ∼ p ∈ (1, 2) ∪ (2, ∞)... (btw. B(`p) = B(Lp) as Banach spaces). 1 n ∗∗ A distantly related question. Let En = C ([0, 1] ). Is En Grothendieck? (Yes, for n = 1.) If this is the case, this would have solved a recent question by n Candido, Cúth, and Doucha of whether Lip Z is Grothendieck.

Thank you very much!

Very few operators vs. Grothendieck

Argyros and Haydon built a space XAH such that

1. B(XAH) = C · IXAH ⊕ K (XAH)

7 1 n ∗∗ A distantly related question. Let En = C ([0, 1] ). Is En Grothendieck? (Yes, for n = 1.) If this is the case, this would have solved a recent question by n Candido, Cúth, and Doucha of whether Lip Z is Grothendieck.

Thank you very much!

Very few operators vs. Grothendieck

Argyros and Haydon built a space XAH such that

1. B(XAH) = C · IXAH ⊕ K (XAH)

Embarrassingly, we do not know whether B(`p) is Grothendieck for ∼ p ∈ (1, 2) ∪ (2, ∞)... (btw. B(`p) = B(Lp) as Banach spaces).

7 Thank you very much!

Very few operators vs. Grothendieck

Argyros and Haydon built a space XAH such that

1. B(XAH) = C · IXAH ⊕ K (XAH)

7 Thank you very much!

Very few operators vs. Grothendieck

Argyros and Haydon built a space XAH such that

1. B(XAH) = C · IXAH ⊕ K (XAH)

7 Very few operators vs. Grothendieck

Argyros and Haydon built a space XAH such that

1. B(XAH) = C · IXAH ⊕ K (XAH)

Thank you very much!