Derivations from the disc algebra into natural modules Yemon Choi - - PowerPoint PPT Presentation

derivations from the disc algebra into natural modules
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Derivations from the disc algebra into natural modules Yemon Choi - - PowerPoint PPT Presentation

Derivations from the disc algebra into natural modules Yemon Choi University of Saskatchewan Banach Algebras and Applications 2013 Chalmers University, 4th August 2013 0 / 16 S ETTING THE SCENE D = the open unit disc in C ; T = D the unit


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Derivations from the disc algebra into natural modules

Yemon Choi University of Saskatchewan Banach Algebras and Applications 2013 Chalmers University, 4th August 2013

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SETTING THE SCENE

D = the open unit disc in C; T = ∂D the unit circle. O(D) = the algebra of holomorphic functions D → C.

The disc algebra

A(D) = {f ∈ C(D) and f|D ∈ O(D)} ∼ = {f ∈ C(T) : f(n) = 0 for all n < 0} Fundamental example in Banach algebra theory. We know many things about A(D), but not so much about its continuous Hochschild cohomology. This talk explores a small corner.

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WHY CARE ABOUT A(D)?

Completion in a natural norm of the polynomial ring C[z], which has a basic role in commutative algebra/algebraic geometry. Has connections with operator theory (e.g. von Neumann’s inequality). Serves as a prototype/motivation for more exotic Banach algebras of functions with analytic structure. Indirect motivation: the paper

  • M. C. WHITE, Injective modules over uniform algebras.
  • Proc. LMS 73 (1996) 155–184.

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DERIVATIONS

Let A be an algebra and M an A-bimodule. A derivation is a linear map D : A → M satisfying D(ab) = a · D(b) + D(a) · b for all a, b ∈ A. (In this talk, all derivations etc are tacitly norm-continuous.) Der(A, M) = space of derivations A → M. In this talk, only consider symmetric bimodules (a · m = m · a).

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If D ∈ Der(A(D), M) an easy induction gives D(Zn) = nZn−1 · D(Z) (n ≥ 1) so by linearity D(f) = f ′ · D(Z) for every f ∈ C[z].

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If D ∈ Der(A(D), M) an easy induction gives D(Zn) = nZn−1 · D(Z) (n ≥ 1) so by linearity D(f) = f ′ · D(Z) for every f ∈ C[z].

Easy result

Let M be a C(T)-bimodule and D : A(D) → M a derivation. Then D = 0. Hence: Der(A(D), C(T)) = 0; and Der(A(D), A(D)) = 0.

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DERIVATIONS INTO C(T)/A(D)

Notation

Define P+ : L1(T) → O(D) and P− : L1(T) → Conj O(D) by (P+ f)(z) = ∑

n≥1

  • f(n)zn

, (P− f)(z) = ∑

n≥1

  • f(−n)zn

For k ∈ L1(T) let ∂k be the formal/distributional derivative of k:

  • ∂k(n) = in

k(n) (n ∈ Z) Hp denotes Hardy space on the disc/circle (1 ≤ p ≤ ∞).

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EXPLICIT DESCRIPTION OF Der(A(D), C(T)/A(D))

The following theorem paraphrases results of ALEKSANDROV–PELLER, IMRN 1996.

Theorem

Let k ∈ C(T). Then the following are equivalent:

1 f ′k + A(D)C(T)/A(D) f∞ for all f ∈ C[z]; 2

∂k ∈ (H1)∗, i.e. sup

  • T ∂k(z)h(z) |dz|
  • : h ∈ C[z], h1 ≤ 1
  • < ∞

With a bit more work: Der(A(D), C(T)/A(D)) ∼ = P− C(T) via Dk ↔ ∂ P− k

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MORE REMARKS ON THIS RESULT

Can present the proof without using real-variable techniques. Instead, use an isomorphism between P− L∞(T) and a space of multipliers on some Hilbert space of analytic functions. Most proofs of the theorem use, implicitly or explicitly, factorization theorems for derivatives of functions in Hp. DAVIDSON–PAULSEN, JRAM 1997 had an operator-theoretic viewpoint, and gave a proof that gives more.

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MY ORIGINAL MOTIVATION

Interested in the second Hochschild cohomology group H2(A(D), A(D)) = Z2(A(D), A(D)) B2(A(D), A(D)) Long known (e.g. JOHNSON, MAMS 1972) that H2(A(D), A(D)) = 0, in contrast with what one gets for cohomology of C[z]. Seems that there is nothing more in the literature on H2(A(D), A(D)). . .

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THEOREMS IN THE REAR VIEW MIRROR

APPEAR CLOSER THAN THEY ARE Theorem (C., unpublished, circa 2002)

H2(A(D), A(D)) is a Banach space, and there is a bounded linear map ψ : H2(A(D), A(D)) → Der(A(D), C(T)/A(D)) which is injective with dense range.

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THEOREMS IN THE REAR VIEW MIRROR

APPEAR CLOSER THAN THEY ARE Theorem (C., unpublished, circa 2002)

H2(A(D), A(D)) is a Banach space, and there is a bounded linear map ψ : H2(A(D), A(D)) → Der(A(D), C(T)/A(D)) which is injective with dense range.

Question

Is the map ψ surjective? Answers will be gratefully received!

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OUTLINE OF THE PROOF

By general theory (WHITE, PLMS 1996) or direct averaging arguments, Hn(A(D), L∞(T)) = 0 for all n ≥ 1. In the case n = 2 we can actually show H2(A(D), C(T)) = 0 (borrowing ideas from JOHNSON, MAMS 1972) and prove that B2(A(D), A(D)) is closed in Z2(A(D), A(D)). Now use Z2(A(D), A(D)) ֒ → Z2(A(D), C(T)) = B2(A(D), C(T)) and Der(A(D), C(T)) = 0 to define a continuous linear map H2(A(D), A(D)) → Der(A(D), C(T)/A(D)). [“Excision argument”] Finally, check this map is injective with dense range.

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AN A(D)-MODULE OF MULTIPLIERS

Let Λ be the measure on D given by dΛ(z) := 4 log 1 |z|dx dy; and let C = {h ∈ O(D) : hkL2(Λ) kL2(T) for all k ∈ H2}. Note that if f ∈ H∞ and h ∈ C then fh ∈ C.

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AN A(D)-MODULE OF MULTIPLIERS

Let Λ be the measure on D given by dΛ(z) := 4 log 1 |z|dx dy; and let C = {h ∈ O(D) : hkL2(Λ) kL2(T) for all k ∈ H2}. Note that if f ∈ H∞ and h ∈ C then fh ∈ C. Equip C with the natural norm, and let Ω be the closure of C[z] inside C. This is a Banach A(D)-module.

Theorem (Paraphrase of known result)

Equip P+ L∞(T) with the quotient norm induced from L∞(T). Then f → f ′ is a continuous bijection from P+ L∞(T) onto C.

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Remark

This result can be proved by concatenating two hard results: P+ L∞(T) ∼ = BMOA ∼ = C where the second isomorphism is given by differentiation of functions. But there are proofs which just use complex analysis and Green’s function identities.

Corollary

f ′C f∞ for all f ∈ C[z]. Thus f → f ′ gives a non-zero, bounded derivation dΩ : A(D) → Ω.

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NON-COMPACT DERIVATIONS?

HEATH, PhD thesis 2008 studied when various derivations on commutative Banach algebras are compact. To my knowledge, the only recorded example of a non-compact derivation from a uniform algebra to a symmetric module is a road-runner-ish construction of J. FEINSTEIN.

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NON-COMPACT DERIVATIONS?

HEATH, PhD thesis 2008 studied when various derivations on commutative Banach algebras are compact. To my knowledge, the only recorded example of a non-compact derivation from a uniform algebra to a symmetric module is a road-runner-ish construction of J. FEINSTEIN.

Observation (C.)

The derivation dΩ : A(D) → Ω is not even weakly compact. Idea of the proof. Fix f ∈ H∞(D), then look at (fn) ⊂ A(D) given by fn(z) = f(rnz), where rn ր 1. If dΩ were weakly compact we’d end up with f ′ ∈ Ω. Now choose f with hindsight so this can’t happen.

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A UNIVERSAL PROPERTY

Proposition (MORRIS, PhD thesis 1993)

If D : A(D) → M is a derivation and f ∈ A(D) then D(f) ≤ 2e fP+ L∞

Corollary (Universal derivation)

Given D : A(D) → M, there is a bounded linear map Ω → M sending h to h · D(Z), and a factorization of D through dΩ : A(D) → Ω.

Remark

Any unital CBA will have a “universal symmetric module for derivations”: see RUNDE, GMJ 1992. The point is that for A(D) we have an explicit description of a universal module.

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EXPLICIT DESCRIPTION OF Der(A(D), A(D)∗)

Theorem (C.–HEATH, PAMS 2011)

Let h ∈ O(D), h(0) = 0, and suppose h′ ∈ H1. Then Dh(f)(g) =

  • T f ′(z)g(z)h(z)|dz|

defines a bounded derivation Dh : A(D) → A(D)∗, with Dh h′1. Moreover, every D ∈ Der(A(D), A(D)∗) has the form Dh for a unique h as above.

Consequences

Every Dh is compact. Every Dh is 2-summing, and we can write down an explicit Pietsch control measure in terms of h.

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REMAINING PROBLEMS

Determining H2(A(D), A(D)) explicitly

Does every derivation D : A(D) → C(T)/A(D) admit a bounded linear lifting A(D) → C(T)? Can we use dilation techniques applied to (Foguel–)Hankel operators?

Finding the universal derivation for other R(X)

Some partial results for X a circular domain (C. + HEATH, unpublished) but nothing yet for infinitely connected domains.

Function algebras on polydiscs or Euclidean balls

Which algebras do we choose as “suitable” generalizations of A(D)?

Noncommutative versions

What are the right questions for the NC disc algebra?

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