Representation of Group Isomorphisms: The Compact Case

Ferrer, Marita; Gary, Margarita; Hernández, Salvador

doi:https://doi.org/10.1155/2015/879414

Journal of Function Spaces

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Research Article | Open Access

Volume 2015 | Article ID 879414 | https://doi.org/10.1155/2015/879414

Representation of Group Isomorphisms: The Compact Case

Marita Ferrer,¹Margarita Gary,²and Salvador Hernández³

Academic Editor: Hugo Leiva

Received31 Oct 2014

Accepted12 Jan 2015

Published03 Feb 2015

Abstract

Let be a discrete group and let and be two subgroups of -valued continuous functions defined on two 0-dimensional compact spaces and . A group isomorphism defined between and is called separating when, for each pair of maps satisfying that , it holds that . We prove that under some mild conditions every biseparating isomorphism can be represented by means of a continuous function as a weighted composition operator. As a consequence we establish the equivalence of two subgroups of continuous functions if there is a biseparating isomorphism defined between them.

1. Introduction

Let be a discrete group and let and be topological spaces. If and are groups of -valued continuous maps, we say that and are equivalent when there is a homeomorphism and a continuous map such that the map defined as , , , is a group isomorphism of onto . Here is equipped with the pointwise convergence topology. We say in this case that is represented as a weighted composition operator. There are many results that are concerned with the representation of linear operators as weighted composition maps and the equivalence of specific groups of continuous functions in the literature, which is vast in this regard. We will only mention here the classic Banach-Stone theorem that, when is the field of real or complex numbers, establishes that if the Banach spaces of continuous functions and are isometric, then they are equivalent and the isometry can be represented as a weighted composition map (cf. [1–10]). Another important example appears in coding theory, where the well-known MacWilliams Equivalence Theorem asserts that when is a finite field and and are finite sets, two codes (linear subspaces) and of and , respectively, are equivalent when they are isometric for the Hamming metric (see [11–13]). This result has been generalized to convolutional codes in [14] and it also makes sense in other areas, for example, functional analysis and linear dynamical systems (cf. [14–18]). The main motivation of this research has been to extend MacWilliams Equivalence Theorem to more general settings and explore the possible application of these methods to the study of convolutional codes or linear dynamical systems. However, throughout this paper, we will only deal with -dimensional compact spaces and and a discrete group . We will look at the possible application of this abstract approach elsewhere. There are many precedents in the study of the representation of group homomorphisms for group-valued continuous functions. Among them, the following ones are relevant here (cf. [19–26]). Most basic facts and notions related to topological properties may be found in [27].

Throughout this paper all spaces are assumed to be Hausdorff 0-dimensional and compact. That is to say, we only deal with Hausdorff compact spaces that contain an open basis consisting of closed and open (clopen) subsets. If is a topological space and is a topological (discrete) group, we denote by the group of continuous functions from to . Let be the neutral element of . For the cozero of is the set and the zero of is the set . Since is discrete, and are both clopen subsets of .

Let be a subgroup of and set . Then denotes the minimum collection of subsets containing that is closed under finite unions and intersections (resp., and denotes the minimum collection of subsets containing that is closed under finite unions and intersections). It is said that separates points in if for every pair of distinct points there is a map such that and . It is said that strongly separates points in if, for every pair , there are maps such that , , and .

As group is finite, it could be thought that and should coincide. However, this is misleading as the following example shows. Obviously, by Morgan’s laws, it suffices to prove that .

Example 1. Let be the group with two elements and take equipped with the product topology. Clearly is a compact space homeomorphic to the Cantor set, which is -dimensional. Let be the th projection on and set as the subgroup of generated the collection . Take greater than and . It is easily verified that belongs to . However, take an arbitrary but fixed element . Denote by the element in such that if and and set . If , then there is such that . Therefore , which yields . On the other hand, if , take . We have that , which again yields . In either case, we obtain that . Thus , which completes the proof.

Denote by the evaluation map; that is, for every . It is said that is pointwise dense when is dense in for all . It is said that is controllable if for every and such that there exist a subset and a function such that , , and .

We now formulate our main results.

Theorem 2. Let and be -dimensional compact Hausdorff spaces and let be a discrete group. Suppose that and are controllable and pointwise dense subgroups of -valued continuous functions separating the points of and , respectively. If is a biseparating group isomorphism of onto , then there are continuous maps and satisfying the following properties.(1) is a homeomorphism of onto .(2)For each and every it holds (3) is a continuous isomorphism with respect to the pointwise convergence topology.(4) is a continuous isomorphism with respect to the compact open topology.

Corollary 3. Let and be -dimensional compact Hausdorff spaces and let be a discrete group. Suppose that and are controllable and pointwise dense subgroups of -valued continuous functions separating the points of and , respectively. If there is a biseparating group isomorphism of onto , then and are equivalent.

We notice that some of the requirements we have imposed on the previous results could be relaxed in general. However, this would take us to a wider setting in general. For instance, if we assume that does not separate points in , then there must be some point such that for all ; then we can replace by the largest subspace where separates points. This subspace is open but not necessarily closed in general. Thus, the study of subgroups that does not separate points leads us to consider locally compact spaces. We will discuss these spaces in a subsequent paper.

2. Basic Notions and Facts

The following lemma is easily verified using a standard compactness argument. Recall that we are assuming that all spaces are compact and -dimensional.

Lemma 4. Let be a family of clopen subsets of that is a subbase for the closed subsets of . Then for every disjoint nonempty closed subsets and of there are two disjoint subsets and in such that and .

Example 5. A specific example where Lemma 4 applies is given when separates points in , where .

Next proposition shows that the notions of separating and strongly separating points are equivalent for controllable subgroups.

Proposition 6. If is a controllable subgroup of that separates the points of , then strongly separates the points of .

Proof. Set and take two distinct elements in . Applying Lemma 4, since separates the points of , there are such that , , and . Take such that , . Since is controllable, we have and such that , , and . Therefore .
Applying the fact that is a subbase of closed subsets and using a compactness argument, we deduce that there is such that and . By the controllability of again, we have and such that , , and . Therefore , which yields . This completes the proof.

Definition 7. Let be a subgroup of and let be a group homomorphism. A subset of is said to be a support for if, given with , it holds that .

Some basic properties of support subsets are shown in the next proposition. Observe that since , we may assume without loss of generality that all support subsets are closed and therefore compact subsets of .

Proposition 8. Let be a nonnull group homomorphism. The following assertions hold.(1) is a support for .(2)If is a support for then .(3)If is a support for and then is a support for .(4)Let be a support for and such that . Then .If, in addition, is controllable and separates points in , then we have the following.(5)Let and be supports for ; then .

Proof. Assertions (1)–(4) are obvious.
(5) Let and be closed supports for . Suppose . Since separates points in , by Lemma 4 and Proposition 6, there are two disjoint subsets and in containing and , respectively. Take such that . Applying the controllability of , we obtain and such that , , and . This yields a contradiction as the evaluation of shows. Indeed, since for all , by item (4) it follows that . On the other hand, we have that for all , which imples . This contradiction completes the proof.

Definition 9. A map is said to be separating or disjointness preserving, if, for each pair of maps satisfying that , it holds that (equivalently, if implies for all ). In case is bijective, the map is said to be biseparating if both and are separating. Remark that this definition makes sense and extends naturally to maps .

Remark 10. Originally, separating maps for scalar-valued continuous functions were defined as those maps such that implies . If one interprets the null element as the identity of the group, then separating maps could be defined as those maps that satisfy for all which implies for all or that for all . Obviously this definition would be vacuous here since we are assuming that is a group homomorphism throughout this paper. Thus, this definition would take us to the more general question of representing group isomorphisms defined between groups of continuous functions without any further requirement. Unfortunately, this is not possible in general. Indeed, a remarkable result due to Milutin [28] establishes that if is an uncountable compact metric space, then is linearly isomorphic to . Therefore, it is essential to impose some extra algebraic or geometrical condition on the isomorphisms in order to be able to represent them by continuous maps defined on the compact spaces and . In this sense, the connection with separating isomorphisms stems from [8], where it was proved that every linear isometry is a separating isomorphism.

Next we will see that every nonnull separating group homomorphism , where is controllable, has the smallest possible compact support set. For that purpose, set There is a canonical partial order that can be defined on : , , if and only if . A standard argument shows that is an inductive set and, by Zorn’s lemma, contains a -minimal element. Furthermore, this minimal element is in fact a minimum because of the next proposition.

Proposition 11. If separates points in , then the minimum element consists of a singleton.

Proof. Let be a minimal element of , which is nonempty by Proposition 8. Suppose now that there are two different elements that are contained in . As is Hausdorff, we can select two disjoint open subsets in such that and . Since is minimal, the compact subset is not a support for . Hence, there are such that and , . Since is separating, it follows that is a nonempty compact subset of .
We claim that . Otherwise, pick up an element . If then and , which is a contradiction; but if then , which implies that and we get a contradiction again. Therefore . By Lemma 4, we can take two disjoint sets such that and . Applying the fact that is controllable to , and , we obtain a set and a map such that , , and . Then , , and . Since is separating the set , take . Then and ; that is, . As a consequence , which is a contradiction. Therefore we have proved that . This completes the proof.

3. Proof of Main Results

Along this section (resp., ) is a controllable subgroup of (resp., ) that separates points in (resp., ).

Let be a separating group homomorphism. The maps are separating group homomorphisms of into for all . Furthermore, since is controllable and separates points in , we can apply Proposition 11, in order to obtain that each partial ordered set has a minimum element, which is a singleton denoted by . Therefore, by sending to for every , we have defined the support map of into that is associated with .

Proposition 12. Let be a separating group homomorphism. Then the support map has the following properties.(1) is continuous.(2)If is open, , and , then .(3).(4)If is one-to-one, then is onto.Moreover, when is a bijection of onto , we have, in addition, the following.(5)If is biseparating, then is a homeomorphism of onto .

Proof. (1) Let be a net in converging to . By a standard compactness argument, we may assume without loss of generality that converges to . Reasoning by contradiction, suppose . Since is Hausdorff, we can take two disjoint open neighborhoods and of and , respectively. Using convergence, there is such that for all .
As every support subset for contains , for all , the subset may not be a support for . Therefore there exists such that and . Moreover, since is a continuous function, the net converges to and, because is discrete, there is such that for all . If we take and index such that , then the subset may not be a support for . Thus, there exists such that and . This means that and, since is a separating map, it follows that . But , which is disjoint from . This is a contradiction that completes the proof.
(2) Let be an open subset, , and . If we take , then is a nonempty compact subset that is not a support for . Then there is such that and . Therefore and . Since is separating, we have that . Therefore .
(3) Take . Then for some . Since is a support for , it follows that or, equivalently, we have .
(4) Suppose and take such that . Since is continuous and is compact, we have that is a proper compact subset of . Applying Lemma 4, there are two disjoint subsets such that and . Moreover, as separates points in , there exists such that . Again, by the controllability of , we may take a subset and a map such that , , and . As a consequence , , and for all . Then . Since is an injective group homomorphism, this yields , which is a contradiction.
(5) Since and are compact spaces, it will suffice to prove that is one-to-one. Suppose there are two elements in such that . Since separates the points of , there are such that , , and . Since , there is such that , for . Since is controllable, there are , such that , , and . Since , applying a compactness argument, there is such that and . Now, by the controllability of , there are and such that , , and . Hence, since , , and is biseparating, it follows that . On the other hand , , and by item (3) above, we have that , which is a contradiction.

We have just seen how a separating group homomorphism has associated a continuous map that assigns to each point the support subset of . Our next goal now is to obtain a complete representation of by means of the support map . Having this in mind, set which is a subgroup of for all , and denote by the set of all group homomorphisms on into . Consider now the set

We can think of the elements of as partial functions on . That is, functions whose domain is a (not necessarily proper) subset of . Since the group is discrete, we can equip with the product (or pointwise convergence) topology as follows.

Let be a basic neighborhood of a map . If now is a partial map, we can restrict this basic neighborhood to by letting be the set of all partial maps such that and , . It is easily verified that this procedure extends the pointwise convergence topology on (cf. [29]).

With this notation, we define by for each . We will see next that is well defined and continuous.

Proposition 13. With the terminology established above, the following assertions are true.(1) is a well-defined group homomorphism of into for all .(2) is continuous when is equipped with the pointwise convergence topology.

Proof. In order to prove that is well defined, take such that . By Proposition 8, we have . The verification that is a group homomorphism is easy and it is left to the reader.
Let be a net converging to in . If is an arbitrary element in , then for some . Since continuous by Proposition 12, and is a clopen neighborhood of . Since is discrete, there is such that for all . Thus for all . In like manner, as and , we have that is a clopen neighborhood of . As a consequence there is such that for all . Thus for all . This means that the net converges to in the pointwise convergence topology over .

Observe that since is discrete, the compact subsets in are all finite. Therefore, we have also proved that is also continuous if we consider the compact open topology on . We are in position now of establishing a main result in this paper.

Theorem 14. Let be a separating group homomorphism. Then there are continuous maps satisfying the following properties.(1)For each and every it holds (2) is continuous with respect to the pointwise convergence topology.(3) is continuous with respect to the compact open topology.(4)If is biseparating bijection of onto , then is a homeomorphism.

Proof. Item is consequence of the definition of and (2) follows from assertion (2) in Proposition 13. Thus, only needs to be verified.
Let be a net converging to in the compact open topology. If is a compact subset of , then is a compact subset in by the continuity of . Therefore is eventually the constant function on . Applying (1), it follows that is eventually on , which completes the proof.

Corollary 15. Let be a separating group homomorphism, where is pointwise dense. Then there are continuous maps and satisfying the following properties.(1)For each and every it holds (2) is continuous with respect to the pointwise convergence topology.(3) is continuous with respect to the compact open topology.(4)If is a biseparating bijection of onto , then is a homeomorphism.

We are now in position of establishing the results formulated at Introduction.

Proof of Theorem 2. After Theorem 14 and Corollary 15, we only need to verify that for all . Applying Corollary 15 to , we obtain maps such that for every and , we have
Thus, for every and , we have which means that the support subset of is both and . Since the minimum support is unique, this means that , which implies that is a right inverse of . Analogously, for every and , we have which means that the support subset of is both and . Again, this implies that is a left inverse of . Since is both a left and right inverse of , it follows that . Therefore
Applying the former equality to , it follows that for all , and from the latter, we also have that . This means that has left and right inverse and, therefore, it is an automorphism on , which completes the proof.

Proof of Corollary 3. It follows directly from Theorem 2.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors thank the referee for several valuable comments. The first and third listed authors acknowledge the partial support provided by the Generalitat Valenciana, Grant code: PROMETEO/2014/062, and by Universitat Jaume I, Grant P1.1B2012-05.

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Copyright © 2015 Marita Ferrer et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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