ON THE MAXIMAL G-COMPACTIFICATION OF PRODUCTS OF TWO G-SPACES

Let G be any Hausdorff topological group and let βGX denote the maximal G-compactification of a G-Tychonoff space X (i.e., a Tychonoff G-space possessing a G-compactification). Recall that a completely regular Hausdorff topological space is called pseudocompact if every continuous function f : X →R is bounded. In this paper, we prove that if X and Y are two G-Tychonoff spaces such that the product X ×Y is pseudocompact, then βG(X ×Y)= βGX ×βGX (see Theorem 2.2). This is a G-equivariant version of the well-known result of Glicksberg [16], which for G a locally compact group was proved earlier by de Vries in [10]. Note that even in the case of a locally compact acting group G, our proof is shorter than that of [10, Theorem 4.1]. It follows from Proposition 2.7 that the equality βG(X ×Y)= βGX ×βGX does not imply, in general, the pseudocompactness of X ×Y even if X and Y both are infinite (cf. [16, Theorem 1]). Theorem 2.10 says that if a pseudocompact group G acts continuously on a pseudocompact space X , then βGX = βX . Let us introduce some terminology we will use in the paper. Throughout the paper, all topological spaces are assumed to be Tychonoff (i.e., completely regular and Hausdorff). The letter “G” will always denote a Hausdorff (and hence, completely regular) topological group unless otherwise stated. For the basic ideas and facts of the theory of G-spaces or topological transformation groups, we refer the reader to [5, 7, 11]. However, we recall below some more special notions and facts we need in the paper.


Introduction
Let G be any Hausdorff topological group and let β G X denote the maximal G-compactification of a G-Tychonoff space X (i.e., a Tychonoff G-space possessing a G-compactification). Recall that a completely regular Hausdorff topological space is called pseudocompact if every continuous function f : X → R is bounded.
In this paper, we prove that if X and Y are two G-Tychonoff spaces such that the product X × Y is pseudocompact, then β G (X × Y ) = β G X × β G X (see Theorem 2.2). This is a G-equivariant version of the well-known result of Glicksberg [16], which for G a locally compact group was proved earlier by de Vries in [10]. Note that even in the case of a locally compact acting group G, our proof is shorter than that of [10,Theorem 4.1]. It follows from Proposition 2.7 that the equality β G (X × Y ) = β G X × β G X does not imply, in general, the pseudocompactness of X × Y even if X and Y both are infinite (cf. [16,Theorem 1]). Theorem 2.10 says that if a pseudocompact group G acts continuously on a pseudocompact space X, then β G X = βX.
Let us introduce some terminology we will use in the paper. Throughout the paper, all topological spaces are assumed to be Tychonoff (i.e., completely regular and Hausdorff). The letter "G" will always denote a Hausdorff (and hence, completely regular) topological group unless otherwise stated.
For the basic ideas and facts of the theory of G-spaces or topological transformation groups, we refer the reader to [5,7,11]. However, we recall below some more special notions and facts we need in the paper. and f ∈ K. If K is equicontinuous at each point z 0 ∈ Z, then we will say that it is an equicontinuous set.
If additionally Z is a G-space for a group G, then one can define the following (in general not continuous) action of G on C(Z,R): (1.1) If G is locally compact, then this action is continuous, otherwise it may be discontinuous (see, e.g., [7, Chapter I, Section 2.1]). However, the following result is true.
Lemma 1.1. Let Z be a G-space and K an invariant equicontinuous subset of C(Z,R). Then the closure K is also an invariant set and the restriction of the action (1.1) to G × K is continuous.
Proof. For every g ∈ G, define the map g * : where gψ is defined as in (1.1). First we show that g * is a continuous map. Indeed, let C be a compact set in Z, U an open set in R, and M(C,U) = {ψ ∈ C(Z,R) | ψ(C) ⊂ U}. Since all the sets of the form M(C,U) constitute a subbase of the compactopen topology of C(Z,R) and g −1 * (M(C,U)) = M(g −1 C,U), we infer that g * is continuous. Now choose ϕ ∈ K and h ∈ G arbitrary. One needs to show that hϕ ∈ K. Let V be a neighborhood of gϕ. Since the above-defined map h * is continuous, the set h −1 as required. Thus, the proof that the closure K is an invariant subset is complete.
Next we observe that the closure of an equicontinuous set is again equicontinuous [17, Chapter 7, Theorem 14]; so K is an equicontinuous invariant subset of C(Z,R). Now the continuity of the restriction of the action (1.1) to G × K follows easily from the continuity of the evaluation map ω : K × Z → R defined by ω(ψ,z) = ψ(z), ψ ∈ K, z ∈ Z (see, e.g., [17,Chapter 7,Theorem 15]). We refer the reader to [2, Lemma 2] for more details.
We will need this lemma in the proof of Theorem 2.2.
In what follows, we will need also the following two characterizations of the maximal G-compactification β G X established in [8] (see also [4]).
Proposition 1.2. Let G be a group and X a G-Tychonoff space. Then the following hold.
(1) Each G-map f : X → B to a compact G-space has a unique G-extension F : β G X → B.
(2) Let bX be a G-compactification of X such that every G-map f : X → B to a compact G-space has a G-extension F : bX → B. Then bX is equivalent to β G X. Proposition 1.3. Let G be a group and X a G-Tychonoff space. Then the following hold.
(1) Each bounded G-uniform function f : X → R possesses a unique continuous exten- admits a continuous extension F : bX → R, then bX is equivalent to β G X.
4 On the maximal G-compactification

Main results
Proof. Define the map f : The continuity of f follows from the fact that the compact-open topology is proper (see [14,Theorem 3.4

.1]).
It is easy to see that the G-uniformness of f is just equivalent to the equicontinuity of the image f (X) in C(G,R). Since the restriction f | A is G-uniform, we infer that the set f (A) is equicontinuous. But closure of an equicontinuous set is again equicontinuous [17,Chapter 7,Theorem 14] Proof. According to Proposition 1.3, it suffices to prove that every bounded G-uniform The idea is first to extend f to a bounded G-uniform function ϕ : β G X × Y → R, and then to extend in a similar way ϕ to obtain the desired extension F. In the nonequivariant case, this is due to Todd [21].
Define the map f : Continuity of f follows from the fact that the compact-open topology is proper (see [13, Theorem 3.1]).
Proof of the claim. Let ε > 0 and (g 0 , y 0 ) ∈ G × Y . We have to show that there exist neighborhoods U of g 0 and V of y 0 such that Since f is a G-uniform function, one can choose a neighborhood U of the unity in G such that (2.4) Natella Antonyan 5 It follows from (2.3) that for all x ∈ X and g ∈ Ug 0 , we have It is known that the formula defines a continuous function ϕ : Y → R (see [15,Lemma 1.3]).
Since ϕ(g 0 y 0 ) = 0, we conclude that there is a neighborhood V of g 0 y 0 in Y such that By continuity of the action on Y , there exist neighborhoods O and W of g 0 and y 0 , respectively, such that OW ⊂ V and O ⊂ Ug 0 . Consequently, if g ∈ O and y ∈ W, then g y ∈ V and g y 0 ∈ V . Hence, (2.7) yields for all Now, (2.4), (2.5), and (2.8) imply for all g ∈ Ug 0 and y ∈ W that as required. Thus, f (X) is indeed an equicontinuous set, and the proof of the claim is complete.
Now we continue with the proof of Theorem 2.2. Consider G × Y as a G-space endowed with the action h * (g, y) = (gh −1 ,hy). Then the induced action (1.1) becomes the following action: (2.10) We claim that f is algebraically equivariant, that is, which means that h f (x) = f (hx). Consequently, f (X) is an invariant subset of C(G × Y ,R). By Lemma 1.1 and the above claim, the closure T = f (X) also is an invariant subset of C(G × Y ,R), and the restriction of the action (2.10) to G × T is continuous.
Further, since f (X) is a bounded subset of C(G × Y ,R), it follows from the Arzela-Ascoli theorem [13, Theorem 6.4] that T is compact.
Thus, T is a compact G-space. Next, since f : X → T is a G-map, by Proposition 1.2, f extends to a G-map F : β G X → T ⊂ C(G × Y ,R).

On the maximal G-compactification
Define the map φ : β G X × Y → R by the formula φ(z, y) = F (z)(e, y), where (z, y) ∈ β G X × Y and e is the unity of G. Clearly, φ is bounded.
Since the evaluation map ω : T × (G × Y ) → R defined by ω(ψ,t) = ψ(t), ψ ∈ T, t ∈ G × Y , is continuous (see, e.g., [17, Chapter 7, Theorem 15]), we infer that φ is also continuous. If Since the product of a pseudocompact space and a compact space is pseudocompact (see, e.g., [14,Corollary 3.10.27]), β G X × Y is a pseudocompact G-space. Consequently, by the same way, one can prove that the bounded G-uniform function φ : β G X × Y → R extends to a continuous function F : β G X × β G Y → R, which is the desired extension of f . This completes the proof.

Remark 2.4.
For G a locally compact group, Theorem 2.2 was proved earlier by de Vries in [10] in a different way. If G, as a topological space, is a k-space (i.e., a quotient image of a locally compact space) and X is a pseudocompact G-space, then β G X = βX (see [10,Lemma 5.5]). Hence, Theorem 2.2 follows in this case directly from the classical result of Glicksberg [16] (this is just [10,Corollary 5.7]).
In the following lemma, we just list two known important cases when the product of two pseudocompact spaces is pseudocompact.

two spaces is pseudocompact, if at least one of the following conditions is fulfilled:
(1) X is a pseudocompact k-space and Y is a pseudocompact space; (2) X is a pseudocompact topological group and Y is a pseudocompact space.  Let G be any group, H a closed subgroup of G such that G/H is compact, and let X be a Tychonoff space endowed with the trivial action of G.
Proof. Evidently, G/H × βX is a G-compactification of G/H × X. Hence, according to Proposition 1.3, it suffices to prove that every bounded G-uniform function f : G/H × X → R has a continuous extension F : Define a function f : Then f is continuous, and it follows from the G-uniformness of f that the image f (X) is an equicontinuous set in C (G/H,R). Besides, the set f (X)(t 0 ) = { f (x)(t 0 ) | x ∈ X} is bounded for all t 0 ∈ G/H. Consequently, by the Arzela-Ascoli theorem [13,Theorem 6.4], f (X) has a compact closure f (X) in C (G/H,R). Hence, f has a continuous extension Natella Antonyan 7 F : βX → f (X) ⊂ C (G/H,R). Define F : G/H × βX → R by F(t,z) = f (z)(t). The compactness of G/H insures that F is continuous (see, e.g., [14,Theorem 3.4.3]). It remains only to observe that F extends f .
Recall that a G-space X is called free if for every x ∈ X, the equality gx = x implies that g = e, the unity of G.
Below, we will need the following well-known result.
Lemma 2.8. Let G be a compact group and X a free G-space. Then Proof. The desired G-homeomorphism f : (G × X)/G → X is defined as follows: where G(g,x) stands for the G-orbit of the pair (g,x). It is easy to verify that f is continuous and bijective. The closedness of f follows from that of the map G × X → X, (g,x) → g −1 x (see [5,Chapter I,Theorem 1.2]).
If the action of G on X is not trivial, then Proposition 2.7 is no longer true. Namely, we have the following proposition.
Proposition 2.9. Let G be an infinite, compact, metrizable group and X a finite-dimensional, paracompact, noncompact, free G-space.
Proof. Suppose the contrary, that β G (G × X) = G × β G X. Passing to the orbit spaces, we have (2.13) Using the formula (β G Z)/G = β(Z/G) (see [4,Corollary 4.10]), we get (2.14) Hence, It is known that a finite-dimensional, paracompact, free G-space has a free G-compactification and in this case β G X is also a free G-space (see [3,Proposition 3.7]). Consequently, by virtue of Lemma 2.8, one has that (G × X)/G = X and (G × β G X)/G = β G X. In sum, we get βX = β G X, which implies that each bounded continuous function f : X → R is G-uniform. However, this is not true.
Indeed, since X is paracompact and noncompact, it is not countably compact [ 8 On the maximal G-compactification by continuity of the G-action at x n ∈ X, there exists an element g n ∈ O n such that g n is different from the unity of G and g n x n ∈ U n , n = 1,2,.... Since X is a free G-space, we see that g n x n = x n , n ≥ 1.
Now, let f n : X → [0,1] be a continuous function such that f n (x n ) = 1, f n (g n x n ) = 0 and f n ( ..} is disjoint and locally finite, f is a well-defined, continuous, bounded function X → R. Hence, it should be also G-uniform, which yields a neighborhood Q of the unity in G such that | f (gx) − f (x)| < 1/2 for all x ∈ X and g ∈ Q. We choose n ≥ 1 so large that O n ⊂ Q. This implies that g n ∈ Q, and hence 1 = | f (g n x n ) − f (x n )| < 1/2, a contradiction.
In general, if the acting group G is not discrete, an action G × X → X cannot be extended (continuously) to an action G × βX → βX; the natural rotation-action of the circle group on the plane R 2 provides a counterexample (see [19,Section 1.5]). However, the following result holds true.
Theorem 2.10. Let G be a pseudocompact group and X a pseudocompact G-space. Then X is G-Tychonoff and β G X = βX.
Further, the fact that α satisfies the two algebraic conditions of action implies easily that the map ϕ : βG × βX → βX satisfies these conditions as well. Thus, ϕ is an action, and hence βX is a βG-space. In particular, βX is a G-space. Consequently, βX is a Gcompactification of X, and hence X is a G-Tychonoff space. It is also clear that βX is the maximal G-compactification of X, that is, β G X = βX, as required.
Remark 2.11. It is worth to mention that there exists a pseudocompact group whose underlying topological space is not a k-space (see, e.g., [12,20]).