Order isomophisms between Riesz spaces

The first aim of this paper is to give a description of the (not necessarily linear) order isomorphisms \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C(X)\rightarrow C(Y)$$\end{document}C(X)→C(Y) where X, Y are compact Hausdorff spaces. For a simple case, suppose X is metrizable and T is such an order isomorphism. By a theorem of Kaplansky, T induces a homeomorphism \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\tau :X\rightarrow Y$$\end{document}τ:X→Y. We prove the existence of a homeomorphism \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$X\times \mathbb {R}\rightarrow Y\times \mathbb {R}$$\end{document}X×R→Y×R that maps the graph of any \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f\in C(X)$$\end{document}f∈C(X) onto the graph of Tf. For nonmetrizable spaces the result is similar, although slightly more complicated. Secondly, we let X and Y be compact and extremally disconnected. The theory of the first part extends directly to order isomorphisms \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C^{\infty }(X)\rightarrow C^{\infty }(Y)$$\end{document}C∞(X)→C∞(Y). (Here \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C^{\infty }(X)$$\end{document}C∞(X) is the space of all continuous functions \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$X\rightarrow [-\infty ,\infty ]$$\end{document}X→[-∞,∞] that are finite on a dense set.) The third part of the paper considers order isomorphisms T between arbitrary Archimedean Riesz spaces E and F. We prove that such a T extends uniquely to an order isomorphism between their universal completions. (In the absence of linearity this is not obvious.) It follows, that there exist an extremally disconnected compact Hausdorff space X, Riesz isomorphisms \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\hat{}$$\end{document}^ of E and F onto order dense Riesz subspaces of \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C^{\infty }(X)$$\end{document}C∞(X) and an order isomorphism \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$S:C^{\infty }(X)\rightarrow C^{\infty }(X)$$\end{document}S:C∞(X)→C∞(X) such that \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\hat{Tf}=S\hat{f}$$\end{document}Tf^=Sf^ (\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f\in E$$\end{document}f∈E).


Introduction
Two Riesz spaces may be order isomorphic without being Riesz isomorphic, an example being formed by l 1 and l 2 . However, order isomorphic Riesz spaces necessarily share certain properties that at first sight seem to depend on the Riesz space structure and not only on the ordering. For instance, if two Archimedean Riesz spaces are order isomorphic, then their universal completions (in the sense of [1]) are order and even Riesz isomorphic, although the definition of the universal completion involves the vector space structure.
In this paper we consider order isomorphisms between Riesz spaces. We start with the Riesz spaces C(X ), C(Y ) where X and Y are compact Hausdorff spaces. A homeomorphism π : Y → X induces a linear order isomorphism f f •π of C(X ) onto C(Y ). Conversely, I. Kaplansky proves in [2] that X and Y are homeomorphic as soon as there exists an order isomorphism C(X ) → C(Y ), linear or not. In Sect. 2, we investigate the order isomorphisms C(X ) → C(Y ).
For X and Y distinct from [0, 1] the sailing may be less smooth. In general, an order isomorphism C(X ) → C(Y ) can be described in terms of a homeomorphism X 0 × R → Y 0 × R where X 0 and Y 0 are obtained from X and Y by removing certain meagre sets.
In Sect. 3, X and Y are extremally disconnected and we prove analogues of the above results for order isomorphisms C ∞ (X ) → C ∞ (Y ). The purpose of that lies in Sect. 4, where we look at an order isomorphism T between arbitrary Archimedean Riesz spaces E and F. We show that such a T extends uniquely to an order isomorphism between the universal completions of E and F. These universal completions are Riesz isomorphic to some C ∞ (X ) and we can apply the results of Sect. 3.
For s ∈ R we denote by s the constant function with value s (on any given set).
In the reverse direction we have: Theorem 2 (Kaplansky [2]) Let X and Y be compact Hausdorff spaces such that C(X ) and C(Y ) are order isomorphic. Then X and Y are homeomorphic.
Proof (Our proof is Kaplansky's in disguise but it yields some extra information) (1) Let a ∈ X , g ∈ C(X ) and T : C(X ) → C(Y ) be an isomorphism. Claim: There The latter inequality implies that f ≥ g: a contradiction.
(2) Let a ∈ X , g, h ∈ C(X ), b, c ∈ Y , and assume: (3) It follows from (1) and (2) that, for a given a ∈ X , there is a unique b ∈ Y with (4) This enables us to define a map σ : X → Y by: Similarly, there is a τ : Y → X , determined by: It follows that τ b lies in the closure of A. (6) For a ∈ X , by taking A := {a} and b := σ a one obtains τ σ a = a. Thus τ • σ is the identity map of X . Similarly σ • τ is the identity map of Y . Letting A be an arbitrary closed subset of X and observing that σ = τ −1 one sees that τ is continuous. So is σ .

Observation 3
In (3) of the above proof it is shown that for every x ∈ X , there is precisely one y ∈ Y with Definition 4 Let X and Y be compact Hausdorff spaces. A homeomorphism σ : X → Y is said to be associated with an order isomorphism T : By the above proof of Theorem 2, for every order isomorphism In the remainder of this section X and Y are compact Hausdorff spaces, T is an order isomorphism C(X ) → C(Y ), and σ is the associated homeomorphism X → Y .
. Now by the first part, A has empty interior.

Definition 6
Let σ : X → Y be the homeomorphism associated with T . Define the subset X 0 of X by It follows from Lemma 5 that X \X 0 is meagre (= first category). In particular, X 0 is dense in X .
Lemma 7 Let X 0 be as above. Let x ∈ X 0 , f, g ∈ C(X ). Then The second part of the lemma follows. .
does not intersect the dense set X 0 and therefore must be empty, i.e., The rest of the corollary follows. [3], Exercise 22.10).
Corollary 10 Assume T 0 = 0 and let σ : X → Y be associated with T .
Proof (1) Take X 0 as above. Let W be {x ∈ X : f (x) = 0}. By Lemma 7 and by the assumption Theorem 11 (Functoriality) Suppose Z is a third compact Hausdorff space and S is an order isomorphism C(Y ) → C(Z ). Let σ : X → Y and ω : Y → Z be the homeomorphisms associated with T and S, respectively. Then ω • σ is the homeomorphism We see that ω • σ and φ coincide on the dense subset X 0 of X . Then ω • σ = φ.

Definition 12
We say that T acts horizontally if T s = s for all s ∈ R and (in case X = Y ) that T acts vertically if the homeomorphism associated with T is the identity map of X . If σ : X → Y is a homeomorphism, then the isomorphism σ * (see Definition 4) acts horizontally. If X = Y and u ∈ C(X ), then the translation map f f + u is an order isomorphism C(X ) → C(X ) that acts vertically.
Comment 13 (1) Kaplansky's original proof of Theorem 2 is quite roundabout. The proof we give yields a more direct connection between the order isomorphism T and the homeomorphism τ . Our Corollary 8 (2) is essentially the characterisation of τ obtained by Geuze in his thesis [4]. (2) Kaplanksy's Theorem is more general than our Theorem 2, considering spaces of continuous functions with values in an arbitrary chain instead of R (in which case the spaces X and Y have to satisfy natural regularity conditions). See also [4] for further generalisations.

Vertically acting order isomorphisms
In this section, X, Y, T are as above.
Proof Apply Theorem 11 together with the fact that σ −1 is associated with σ * (see Definition 4).

Corollary 15 If T acts horizontally and σ is the associated homeomorphism, then
Corollary 16 (for X = Y ) The only order isomorphism C(X ) → C(X ) that acts both horizontally and vertically is the identity map.
Theorem 17 Let X = Y , T 0 = 0. Then the following are equivalent: (3) and by Corollary 10 (2) We proceed to prove the representation theorem announced in the Introduction.
Theorem 18 Let X = Y and let T be acting vertically. Take X 0 as in Definition 6.
Then F is a homeomorphism and for every f ∈ C(X ) we have Putting it differently: if for g ∈ C(X ) we define g := {(x, y) ∈ X 0 × R : y = g(x)} (so that g is dense in the graph of g), then F maps f onto T f ( f ∈ C(X )).
Proof ( . We prove to be continuous: Let r ∈ R. We show that the set A : we make an open W ⊆ X 0 and an s ∈ R for which: We have (x 0 , s 0 ) > r . By In the same way one shows that for every r ∈ R the set {(x, s) ∈ X 0 × R : (x, s) < r } is open. Thus, is continuous, and so is F.
(4) With the order automorphism T −1 we associate a continuous G : Then In particular, for all ( , s), Was the introduction of X 0 necessary? Or is X 0 an artefact of our reasoning and may we replace "X 0 " by "X " in Eq. (1)? We consider these questions in the following paragraphs.

Lemma 19 Let X be metrizable. Then X
; it suffices to deduce a contradiction.
As x ∈ X \X 0 and X 0 is dense there is a sequence (x n ) n∈N in X 0 that converges to x such that x n = x m = x a soon as n = m. Let Z be the closed set {x, x 1 , x 2 , . . .}.

By Tietze's Extension Theorem there is an
Example 20 An example to show that X \X 0 may be non-empty.
For X we take the Stone-Čech compactification of N = {1, 2, 3, . . .}. We view N as a subset of X . There is an order (even: Riesz) isomorphism l ∞ → C(X ) sending For n ∈ N we make a strictly increasing bijection ϕ n : R → R by The formula (T 0 a) n := ϕ n (a n ) (a ∈ l ∞ , n ∈ N) defines an order isomorphism T 0 : l ∞ → l ∞ with T 0 t ∈ c 0 for all t ∈ [0, 1). This T 0 determines an order isomorphism T :

Comment 21
The introduction of X 0 in Theorem 18 is not redundant. Indeed, suppose for every X and every vertically acting order isomorphism T : Now let X, T be as in Example 20. Take x in X \N and f = t for some t in (0, 1). Then

Contradiction.
At last we want to prove what we have promised in the introduction.

Theorem 22 Let X, Y be compact Hausdorff spaces, T an order isomorphism C(X ) → C(Y )
, and σ the associated homeomorphism X → Y . Form X 0 as in Definition 6 and put Y 0 = σ (X 0 ) ( so that X \X 0 and Y \Y 0 are meagre, and X 0 = X, Y 0 = Y if X and Y are metrizable). Define F : For f ∈ C(X ), set f := {(x, f (x)) : x ∈ X 0 }. Similarly, make g for g ∈ C(Y ).Then F is a homeomorphism and maps f onto T f ( f ∈ C(X )).
Proof Define σ * as in Definition 1 and σ * −1 as its inverse, set S = (σ * • T ) and note that T = σ * −1 • (σ * • T ) = σ * −1 • S and also that Lemma 14 proves that S is acting vertically. Define B : First of all note that F = B • G. Secondly note that since σ is a homeomorphism we know that B is continuous. Thirdly we see that G(x, (T s)(σ (x))) = G(x, (Ss)(x)) which enables us to fall back on Theorem 18 for the proof that G is continuous and that G maps f onto S f ( f ∈ C(X )). We conclude that F = B • G is continuous and Theorem 22 is an amplification of known representation theorems such as Theorem 1(b) in [5]. This paper gives further representations of lattice homomorphisms C(X ) → C(Y ) that are not required to be isomorphisms, and for isomorphisms that are norm-continuous. It also considers order isomorphisms L ∞ (μ) → L ∞ (ν) where μ and ν are measures.
A related paper is [6], dealing with order isomorphisms between sufficiently large subspaces of C(X ) and C(Y ) for metrizable X and Y . (2) Let Z be, indeed, extremally disconnected.
By C R (Z ) we denote the collection of all continuous functions Z → R. As an ordered set, C R (Z ) is isomorphic to It follows that C R (Z ) is a complete lattice and is completely distributive (i.e., if we denote the set of all continuous f : Z → R for which the closed set {z : | f (z)| = ∞} has empty interior. For f, g ∈ C ∞ (Z ) there is a unique f + g in C ∞ (Z ) with the property that Under the addition "+", C ∞ (Z ) is a Riesz space that contains C(Z ) as an order dense Riesz ideal. For details we refer to page 189 of [1].

Theorem 24 Let X and Y be extremally disconnected compact Hausdorff spaces and let T be an order isomorphism C ∞ (X ) → C ∞ (Y ). Then X and Y are homeomorphic.
There exists a (unique) homeomorphism σ : X → Y satisfying: Proof Copy the proof of Theorem 2, just replacing "C" by "C ∞ " and, in (2), inserting the condition g(a), h(a) = −∞. This does not affect the rest of the proof.

Comment 25
For later use we file away this obvious consequence: Let X and Y be extremally disconnected compact Hausdorff spaces. If C ∞ (X ) and C ∞ (Y ) are order isomorphic, then they are Riesz isomorphic.

Definition 26
We define "associated" as in Definition 4, "acting horizontally" and "acting vertically" as in Definition 12.
The natural analogues of Theorem 11 and Lemma 14 and Theorem 17 are valid.

Definition 27
We define the set X 0 verbatim as in Definition 6. One extra observation: If x ∈ X 0 and f ∈ C ∞ (X ), then Theorem 28 Let X be extremally disconnected and let T be a vertically acting order isomorphism C ∞ (X ) → C ∞ (X ). Take X 0 as in Definition 27. Define F : Then F is a homeomorphism and for every f ∈ C ∞ (X ) we have Proof (Essentially rephrasing the proof of Theorem 18) (1) Equation (3) follows from Lemma 7 and the observation in Definition 27.
(2) Take x ∈ X 0 and consider the function φ : R → R defined by: From (2) of the proof of Theorem 18 one sees that the restriction of φ is a strictly increasing bijection R → R. Consequently φ itself is a strictly increasing bijection R → R, hence is continuous.
(3) From here onwards, follow the proof of Theorem 18, replacing by : X 0 ×R → R defined by: In this section we consider an order isomorphism T between two Riesz spaces, E and F. We show that there exists an extremally disconnected compact Hausdorff space X such that C ∞ (X ) is a universal completion of E and of F, and such that, if we identify E and F with suitable Riesz subspaces of C ∞ (X ), the map T extends uniquely to a vertically acting order isomorphism C ∞ (X ) → C ∞ (X ).

Observations 30
The following is preparation for the proof of Lemma 32. Here E and F are Riesz spaces, T is an order isomorphism E → F with T 0 = 0.
(1) The formula defines an order isomorphism T * : (4) Let f, g ∈ E. Then For a proof, assume f ⊥ g. Then ( f + g) Hence, the elements T f As a consequence of (4) we obtain.
In the same way, the order isomorphism T −1 : F → E generates an increasing which is the same as For h ∈ C ∞ (X ) + we obtain so A Ah = h by the order denseness of E. Thus, A A is the identity map of C ∞ (X ) + . By symmetry, A A is the identity map on C ∞ (Y ) + : The maps A and A are order isomorphisms and each other's inverses. (6) Just as T engenders A, so T * (see Observations 30) leads to an order isomorphism B : Led by Observations 30, we define T : Observe that T is an extension of T . We conclude our proof by showing that T is an order isomorphism. Firstly, T is increasing: if h ≤ j in C ∞ (X ), then h + ≤ j + and h − ≥ j − , whence T h ≤ T j. Secondly, for all h ∈ C ∞ (X ), Ah + lies in the band generated by h + (look at the definition) and, similarly Bh − lies in the band generated by h − . Hence Ah + ⊥ Bh − , and Thirdly, the formula Sj := A −1 j + −B −1 j − defines a map S : C ∞ (Y ) → C ∞ (X ) with properties similar to those of T . In particular Then for all h ∈ C ∞ (X ) and ST is the identity map of C ∞ (X ). Also, T S is the identity map of C ∞ (Y ). It follows that T and S are order isomorphisms.
In terms of abstract Riesz space theory, the lemma says: Theorem 33 An order isomorphism between two Archimedean Riesz spaces extends uniquely to an order isomorphism between their universal completions.

Corollary 34 Let E be a universally complete Riesz space. If a Riesz space F is order isomorphic to E, then F is even Riesz isomorphic to E.
Proof As E is Dedekind complete, so is F. In particular, F is Archimedean. Let F u be a universal completion of F. Choose an order isomorphism T : E → F. By Theorem 33, this T "extends" to an order isomorphism E → F u . Then F u = F and F is universally complete. Now apply Definition 29(1) and Theorem 24.
Theorem 35 Let E and F be Archimedean Riesz spaces and T an order isomorphism E → F. Then there exist an extremally disconnected compact Hausdorff space X , order dense Riesz subspacesÊ andF of C ∞ (X ), Riesz isomorphisms R E : E →Ê and R F : F →F, and a vertically acting order isomorphismT : C ∞ (X ) → C ∞ (X ) such that R F T =T R E : Proof Choose extremally disconnected compact Hausdorff spaces X and Y such that C ∞ (X ) and C ∞ (Y ) are universal completions of E and F, respectively. Choose a Riesz isomorphism R E of E onto an order dense Riesz subspaceÊ of C ∞ (X ), and a Riesz isomorphism A of F onto an order dense Riesz subspace A(F) of C ∞ (Y ).
We have an order isomorphism S :Ê → A(F) given by S R E = AT . By Lemma 32 this S extends to an order isomorphism S : C ∞ (X ) → C ∞ (Y ). Let σ : X → Y be the homeomorphism associated with S. Then σ induces a Riesz isomorphism σ * : C ∞ (Y ) → C ∞ (X ).