Representations of Banach latticealgebras

H. G. Dales, Lancaster

24rd Banach algebra conference

Winnipeg, Manitoba, 11 July 2019

Joint work with Marcel de Jeu, Leiden

1

References

[DDLS] H. G. Dales, F. K. Dashiell, A. T.-M. Lau, andD. Strauss, Banach spaces of continuous functions asdual spaces, pp. 260, Canadian Mathematical SocietyBooks in Mathematics, Springer–Verlag, 2016.

[DJ] H. G. Dales and M. de Jeu, Lattice homomorphismsin harmonic analysis’ Positivity and NoncommutativeAnalysis: Festschrift in Honour of Ben de Pagter onthe Occasion of his 65th Birthday, Springer Birkhauser,2019, to appear.

[DLOT] H. G. Dales, N. J. Laustsen, T. Oikhberg, andV. Troitsky, Multi-norms and Banach lattices, Disserta-tiones Math., Volume 524 (2017), 1–115.

[GL] F. Ghahramani and A. T.-M. Lau, Multipliers andmodulus on Banach algebras associated to locally com-pact groups, J. Functional Analysis, 156 (1997), 476–497.

[K] D. Kok, Masters thesis, Leiden, 2016.

[W1] A. W. Wickstead, Two dimensional unital Rieszalgebras, their representations and norms, Positivity, 21(2017), 787–801.

[W2] A. W. Wickstead, Banach lattice algebras: somequestions, but very few answers, Positivity, 21 (2017),803–815.

[W3] A. W. Wickstead, Ordered Banach algebras andmulti-norms: some open problems, Positivity, 21 (2017),817–823.

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Banach algebras

An algebra is a linear, associative algebra overa field that is either C or R. A Banach algebrais complex or real.

For example, consider the following:

(1) C0(K) or C0(K,R) for a locally compactspace K, with uniform norm | · |K;

(2) L1(G) or L1(G,R) for a locally compactgroup G;

(3) M(G) or M(G,R) for a locally compactgroup G;

(4) B(E) for a (complex or real) Banach spaceE;

(5) `1(S, ω), where S is a semigroup and ω isa weight on S.

Given a real Banach algebra, there is a stan-dard complexification; see Rickart.

Vector lattices

A real linear space E is partially ordered ifthere is a partial order ≤ on E such that x+z ≥y+z when x ≥ y in E and z ∈ E and also αx ≥ 0when x ≥ 0 in E and α ≥ 0 in R. Set

E+ = {x ∈ E : x ≥ 0}.

The space E is a vector lattice if, further,any two elements x, y ∈ E have a sup calledx∨ y and an inf called x∧ y, and then we have|x| = x ∨ (−x), x+ = x ∨ 0, and x− = (−x) ∨ 0,so that

x = x+ − x−, |x| = x+ + x−.

The elements are disjoint if |x| ∧ |y| = 0, sothat x ⊥ y. In this case, |x+ y| = |x|+ |y|.

Let E be a vector lattice, with a subspace F .Then F is a sublattice if x∨y ∈ F when x, y ∈ F ,and F is an order ideal if, further, x ∈ F whenx ∈ E, y ∈ F , and |x| ≤ |y|.

3

Examples

A vector lattice is order complete (or Dedekind

complete) if every non-empty subset that isbounded above has a supremum. Each vectorlattice has an order completion.

(1) Take K compact. The space C(K,R) isa vector lattice. It is order complete iff K isextremely disconnected; see [DDLS]. For ex-ample, C(βN,R) is order complete.

(2) Let (X,µ) be a measure space (µ has valuesin [0,∞]), and take p with 1 ≤ p < ∞. ThenLp(X,µ,R) is an order complete vector lattice– and also for p =∞ when µ is σ-finite.

(3) Take a σ-algebra of subsets of a set X,and let M(X,R) be the linear space of all signedmeasures µ : X → R. This is an order completevector lattice. For µ ∈M(X,R), the usual totalvariation measure is |µ|.

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Side remark

The following is deducible from [DDLS].

Let K be a non-empty compact space. Then

the Dedekind completion of C(K) is C(GK),

where GK is the Gleason cover of K. It is

also equal to the Stone space of the complete

Boolean algebra, RO(K), which is the regular-

open algebra of K.

There are lots of interesting properties of GK.

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Regular maps

Let E and F be linear spaces. Then L(E,F ) is

the space of all linear maps from E to F .

Let E and F be vector lattices. A subset of

E is order bounded if it is contained in an

interval {x ∈ E : a ≤ x ≤ b} for some a, b,∈ E.

A linear map T : E → F is order bounded if

it maps order bounded sets into order bounded

sets. The set of these is the linear subspace

Lb(E,F ) of L(E,F ).

Take S, T ∈ L(E,F ). Then S ≥ T if

Sx ≥ Tx (x ∈ E). This gives a partial order on

Lb(E,F ). The regular operators are elements

of the subspace Lr(E,F ) of Lb(E,F ) that is

spanned by the positive maps, so each regular

operator has the form S − T , where S and T

are positive.

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Riesz–Kantorovitch

Let E and F be vector lattices. In general,

Lb(E,F ) and Lr(E,F ) are not vector lattices,

but we have:

Theorem Suppose that F is order complete.

Then Lb(E,F ) = Lr(E,F ) is also an order com-

plete vector lattice, and we have the Riesz–

Kantorovitch formulae, such as

|T | (x) = sup{|Ty| : |y| ≤ x}

and

(S∨T )(x) = sup{Sy+Tz : y, z ∈ E+, y+z = x}

for x ∈ E+. 2

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Lattice homomorphisms

Let E and F be vector lattices. A linear map

T : E → F is a lattice homomorphism if

T (x ∧ y) = Tx ∧ Ty for x, y ∈ E. Equivalently, we

require that |T (x)| = T (|x|) for x ∈ E.

These are positive, hence regular, linear oper-

ators.

The order dual space

Apply the above with F = R. Then order

bounded functionals are regular, and we ob-

tain another vector lattice called E∼; it is the

order dual space (and it is order complete).

The order adjoint of T : E → F is T∼ : F∼ → E∼

given by 〈T∼ϕ, x〉 = 〈ϕ, Tx〉 for x ∈ E and

ϕ ∈ F∼.

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Banach lattices

A Banach lattice is a (real) Banach space

(E, ‖ · ‖) such that ‖x‖ ≤ ‖y‖ for x, y ∈ E with

|x| ≤ |y|.

Examples Spaces C0(K,R), Lp(X,µ,R), and

M(X,R) with ‖µ‖ = |µ| (X). 2

‘Automatic continuity’ : Let E and F be Ba-

nach lattices. Then an order bounded linear

map T : E → F is automatically continuous,

and so we can write Bb(E,F ) and Br(E,F ).

Advertisement: There are many connections

with ‘multi-norms’: see [DLOT], for example.

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Key facts

(1) Let E be a Banach lattice. Then E∼ is

the same as E′, the Banach space dual, and

we obtain another Banach lattice.

(2) Let E and F be Banach lattices such that F

is order complete. Then Br(E,F ) is an order

complete Banach lattice with respect to the

regular norm ‖ · ‖r, defined by

‖T‖r = ‖ |T | ‖ (T ∈ Br(E,F )) .

2

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Complex Banach lattices

Let E be a (real) Banach lattice. Its complex-ified linear space EC has a modulus defined by

|x+ iy|C = sup{x cos θ + y sin θ : 0 ≤ θ ≤ 2π}

for x, y ∈ E. Then set ‖z‖C = ‖ |z|C ‖ (z ∈ EC)to obtain a norm on EC giving a complex Ba-nach space. This is a complex Banach lat-

tice.

Examples We obtain complex Banach latticesC0(K), Lp(X,µ), and M(X) in the usual way(with the usual norms). 2

Let EC and FC be such complex Banach lat-tices. An operator T : EC → FC has the formS1 + iS2, where S1, S2 ∈ B(E,F ), and T is or-

der bounded or regular if both S1 and S2

have these properties. Further T is a complex

lattice homomorphism if |Tz|C = T (|z|C) forz ∈ EC. We have ‖T‖r,C = ‖ |T |C ‖, etc.

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Banach lattice algebras

Definition Let A be a (real) Banach lattice

that is also a Banach algebra and such that

ab ∈ A+ whenever a, b ∈ A+. Then A is a

Banach lattice algebra = BLA.

The theory of BLAs is somewhat rudimentary,

but there are lots of examples, including in har-

monic analysis. For a discussion, see [W2].

Examples C0(K,R); Br(E) for each order com-

plete Banach lattice E. 2

Definition Let A and B be BLAs. A map

π : A→ B is a Banach lattice algebra homo-

morphism if it is an algebra homomorphism

and a lattice homomorphism.

Such maps are automatically continuous.

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Representations

Definition Let E be an order complete Banachlattice. A Banach lattice algebra represen-

tation of a BLA A on E is BLA homomorphismπ : A→ Br(E).

Definition Let A be a Banach algebra. Thenthe left regular representation of A is themap π : A→ B(A) defined by

π(a)(b) = ab (a, b ∈ A) .

More generally, take E to be a Banach left A-module and consider the representationπ : A → B(E) defined by π(a)(x) = a · x fora ∈ A and x ∈ E.

In the case where A is a BLA and E is a Ba-nach left A-module that is an order completeBanach lattice, we can consider whether themap π : A→ Br(E) is a BLA homomorphism.

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Order = Dedekind completions

Let E be a Banach lattice. Then E has an

obvious Dedekind completion that is a Banach

lattice. Now start from a BLA A. It would

be nice to extend the product on A to its

Dedekind completion to form an order com-

plete BLA, say B, and then look at BLA HMs

from A to Br(B).

Such an extension works in all the cases that

we are interested in, but the general process

seems to be somewhat murky.

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Complex Banach lattice algebras

Let A be a BLA. Then we have a complex

Banach lattice (AC, ‖ · ‖C), and this is also a

complex algebra for the obvious product.

In fact, (AC, ‖ · ‖C) is a complex Banach alge-

bra, and we have

|ab| ≤ |a| · |b| (a, b ∈ AC) .

(Not quite trivial.)

We also have complex Banach lattice alge-

bra homomorphisms and complex Banach

lattice algebra representations.

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Locally compact groups

Let G be a locally compact group, with left

Haar measure mG. Then we have the order

complete Banach lattice M(G,R) which is iso-

metrically lattice isomorphic to the dual Ba-

nach lattice C0(G,R)′. With respect to convo-

lution multiplication, M(G,R) is a BLA. Fur-

ther, the closed ideal L1(G,R) is also an order

complete BLA.

The process of complexification gives us the

usual M(G) and L1(G); they are complex Ba-

nach lattice algebras.

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Representations

The basic question that we consider is the

following. The left regular representation π of

M(G,R) takes its values in the algebra of reg-

ular operators Br(M(G,R)), and π is a contin-

uous Banach algebra HM. Since M(G,R) and

Br(M(G,R)) are BLAs, it is natural to ask if π

is also a Banach lattice homomorphism.

This is an open question in various lists, in-

cluding [W3].

We shall show that more than this is true.

More generally of course: Is there a similar

result for every order complete BLA A? What

about regular homomorphisms A → Br(E) for

suitable E?

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An easy result

Let K be an extremely disconnected compactspace. Then we claim that the left regular HMof C(K) is a lattice HM.

Define Lg(h) = gh (g, h ∈ C(K)), and takef ∈ C(K). Then L|f | is an upper bound for Lfand −Lf = L−f .

Suppose that T ∈ Br(C(K)) is also an upperbound. Take g ∈ C(K)+. For x ∈ K, we have

(Tg)(x) ≥ (fg)(x)

and

(Tg)(x) ≥ (L−fg)(x) = −(fg)(x) ,

and so (Tg)(x) ≥ (|f | g)(x). Hence Tg ≥ L|f |g.

This shows that T = L|f |, and so∣∣∣Lf ∣∣∣ = L|f |.

Theorem Suppose that K is extremely dis-connected. Then the left regular representa-tion π : C(K) → Br(C(K)) is a BLA homo-morphism. 2

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A generalization

For general compact K, with Gleason cover

GK, the space C(GK) is a Banach C(K)-module

and an order complete Banach lattice.

Theorem The left regular representation

π : C(K)→ Br(C(GK)) is a BLA HM. 2

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Earlier results

Results of Brainerd and Edwards (1963), Gilbert(1968), and Arendt (1981) show (in a rathercomplicated way):

Let G be a locally compact group. Then theleft representation π : M(G)→ Br(L1(G)) is anisometric embedding and a BLA homomorphism.There is a shorter proof in [GL], Theorem 3.1.This also true for the actions on Lp(G) when1 < p <∞ whenever G is amenable.

Set A = Br(E), where E is an order completeBanach lattice. Then the map A→ Br(A) is aBanach lattice homomorphism (Schep, 1989).

There are also strong partial results of Marceland Kok; see [K]. For example, the representa-tion of L1(G) on Br(Lp(G)) is a BLA HM evenwhen G is not amenable, but this does not ex-tend to a representation of M(G) on Br(Lp(G))or Br(M(G)); these remained open.

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Locally compact spaces

Now X is a locally compact space. For each

f ∈ C(X), its support is supp f , the closure of

the set {x ∈ X : f(x) 6= 0}. For a non-empty S

in X, we have

C(X,R, S) = {f ∈ C(X) : supp f ⊂ S},

and then

Cc(X,R) =⋃C(X,R,K)

for K compact in X. The space Cc(X,R) is

a vector lattice. Further, there is a natural

locally convex topology on Cc(X,R): a linear

map from Cc(X,R) into a LCS is continuous iff

its restriction to each C(X,R,K) is continuous.

(It is an inductive limit topology.)

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Radon measures

A Radon measure on X (in the sense of Bour-

baki) is a continuous linear functional on Cc(X,R);

the space is called M(X) in Bourbaki.

We can identify the space Cc(X,R)∼.

Fact The order dual space Cc(X,R)∼ of the

vector lattice Cc(X,R) is just M(X). 2

We can tie in M(X) with ‘regular Borel mea-

sures’, but the terminology differs in the liter-

ature.

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Supports

Let U be a non-empty, open subset of X. Anelement ϕ of Cc(X,R)∼ vanishes on U if〈ϕ, f〉 = 0 whenever f ∈ C(X,R, U). Let Vbe the union of the open sets in X on which ϕvanishes. Then the support of ϕ is X \ V ; callit suppϕ.

The subset of elements ϕ of Cc(X,R)∼ withsuppϕ ⊂ S is an order ideal in Cc(X,R)∼.

Technical remarks: (1) Two elements of Cc(X,R)∼

that have disjoint supports are disjoint elementsof Cc(X,R)∼.

(2) For a closed subspace Y of X, there isan embedding of Cc(Y,R)∼ in Cc(X,R)∼. (Notquite trivial.)

Definition Take ϕ ∈ Cc(X,R)∼. Then ϕ hasseparated supports if suppϕ+ and suppϕ−

are disjoint subsets of X.

(We can have suppϕ+ = suppϕ− = X.)

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Embedding familiar vector lattices in

Cc(X,R)∼

Let µ be a positive regular Borel measure on a

locally compact space X, and suppose that

g : X → R is Borel measurable. Then g is

locally integrable with respect to µ if∫K |g(x)| dµ < ∞ for each compact K ⊂ X.

The collection of these functions (usual iden-

tification) forms a vector lattice, called

L1,loc(X,µ,R), the locally integrable func-

tions.

Theorem For such a g ∈ L1,loc(X,µ,R), define

ϕg by

〈ϕg, f〉 =∫Xfg dµ (f ∈ Cc(X,R)) .

Then the map g 7→ ϕg is an injective lattice HM

from L1,loc(X,µ,R) into Cc(X,R)∼ =M(X). 2

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Embedding measures

Theorem Let X be a locally compact space.

For each µ ∈M(X), set

〈ϕµ, f〉 =∫Xf dµ (f ∈ Cc(X)) .

Then ϕµ ∈ Cc(X)∼, and the map

µ 7→ ϕµ, M(X)→ Cc(X)∼ ,

is an injective lattice HM. Further, the set of

ϕ ∈ M(X) with compact, separated support is

a dense subspace of the image.

It is not quite trivial that it is a lattice HM;

the proof uses the Riesz–Kantorovitch formu-

lae and Radon–Nikodym theorem. Then use

the Riesz representation theorem to show that

it is an injection.

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Other examples

Fact Take p with 1 ≤ p <∞. Then Lp(X,µ,R)

is a vector sublattice of L1,loc(X,µ,R) and so

Lp(X,µ,R) is identified with an order complete

Banach lattice in Cc(X,R)∼. Further, the func-

tions in Lp(X,µ,R) with compact, separated

support is a dense subspace of Lp(X,µ,R). 2

Fact Let G be a locally compact group, and

let S be a closed subspace that is a subsemi-

group. Suppose that ω is a continuous weight

on S. Then M(S, ω,R) is a vector sublattice of

L1,loc(X,µ,R) with similar properties. 2

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The main theorem

Theorem Let G be a locally compact group,and let X, Y , and Z be vector sublattices ofCc(G,R)∼ = M(G,R), with Z order complete.Suppose that ? : X × Y → Z is a bilinear mapsuch that x ? y ∈ Z+ (x ∈ X+, y ∈ Y +). Defineπ : X → Lr(Y, Z) by

π(x)(y) = x ? y (x ∈ X, y ∈ Y ),

and suppose that the following hold:

(1) supp (x ? y) ⊂ suppx · supp y for eachx ∈ X+ and y ∈ Y + with compact support;

(2) the elements in X with compact, separatedsupport are dense in X;

(3) the elements in Y + with compact supportare dense in Y +;

(4) χAy ∈ Y whenever y ∈ Y + has compactsupport and A is a Borel subset of supp y.

Then π is a lattice HM.27

Sketch of the proof

We have to show that |π(x)| = π(|x|) for each

x ∈ X.

Since the maps x 7→ |π(x)| and x 7→ π(|x|) are

both continuous, it is sufficient to prove this

when x ∈ X has separated, compact support.

Fix such an x = x+ − x−. Then there is a

relatively compact neighbourhood U of eG with

(suppx+)U ∩ (suppx−)U = ∅.

We need |π(x)| (y) = π(|x|)(y) for all y ∈ Y . It

is sufficient to suppose that y ∈ Y + and that

y has compact support.

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Sketch continued

First suppose, further, that supp y ⊂ Us for

some s ∈ G. Certainly |π(x)| (y) ≤ π(|x|)(y) be-

cause π is positive. Also |π(x)| (y) ≥ |π(x)(y)|.But we have supp (x+ ?y) ⊂ (suppx+) ·Us and

supp (x− ? y) ⊂ (suppx−) · Us, and so

|π(x)(y)| =∣∣∣x+ ? y − x− ? y

∣∣∣=

∣∣∣x+ ? y∣∣∣+

∣∣∣x− ? y∣∣∣ = π(|x|)(y) .

Thus |π(x)(y)| ≥ π(|x|)(y). So we win in this

special case.

Now suppose just that y ∈ Y + and that y has

compact support. Choose an open neighbour-

hood V of eG with V ⊂ U . Then supp y is con-

tained in a union of finitely many right trans-

lates of V .Then y is a finite sum of elements

of Y + each of which is supported in a right

translate of V . So we win by linearity. 2

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Consequences

Theorem Let G be a locally compact group.Then the left regular representationπ : M(G) → Br(M(G)) is an isometric Banachlattice algebra homomorphism.

Proof Put M(G,R) into Cc(G,R)∼ as earlier,with image X, and apply the theorem takingY = Z = X and ? to be the usual convolutionproduct. The complex case then follows. 2

Similarly, extending Arendt and Ghahramani–Lau.

Theorem Let G be a locally compact group,and take p with 1 ≤ p < ∞. Then the mapπ : M(G)→ Br(Lp(G)), where

π(µ)(g)(s) =∫Gg(t−1s) dµ(t) (s ∈ G)

for g ∈ Lp(G) and µ ∈ M(G), is an injectiveBanach lattice algebra homomorphism. 2

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Further results

By variations of the above, we obtain:

Theorem Let A be an order complete BLA.Then the left regular representation of A is aBanach lattice algebra homomorphism from A

into Br(A) in the following cases:

(1) A = M(G) or L1(G) for a locally compactgroup G;

(2) A = M(S, ω), where S is a closed semi-group of a locally compact group and ω is acontinuous weight on S (so that A is a Beurl-ing algebra);

(3) A = `1(S, ω), where S is a cancellativesemigroup and ω is a weight on S (such ex-amples could be radical BAs);

(4) ‘Twisted Orlicz algebras’ of Oztop andSamei. 2

31

Counter-examples

On the other hand Tony Wickstead [W1] has

shown that there are lots of finite-dimensional

mutually non-isomorphic, commutative BLAs

(with positive identity element) that have no

faithful, finite-dimensional Banach lattice

representations at all.

Some questions

Can we characterize the BLAs A for which the

previous theorem works? Maybe go for a rep-

resentation of A in Lr(E) for a more general

order complete Banach lattice E?

Does the above representation theorem lead

to some development of the theory of Banach

lattice algebras?

32