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Let S denote the unit circle on the complex plane and be a continuous binary, associative and cancellative operation. From some already known results, it can be deduced that the semigroup is isomorphic to the group ; thus, it is a group, where · is the usual multiplication of complex numbers. However, an elementary construction of such isomorphism has not been published so far. We present an elementary construction of all such continuous isomorphisms F from into and obtain, in this way, the following description of operation ★: for . We also provide some applications of that result and underline some symmetry issues, which arise between the consequences of it and of the analogous outcome for the real interval and which concern functional equations. In particular, we show how to use the result in the descriptions of the continuous flows and minimal homeomorphisms on S.
Let and be groupoids (i.e., X is a nonempty set and are binary operations in X). If there exists a bijection such that
then we say that the pair induces ∘. It is easily seen that (1) is equivalent to the following property
which means that h is an isomorphism of into .
Recall yet that a groupoid is cancellative if operation ∘ is cancellative, i.e.,
for all with . A groupoid is a semigroup if operation ∘ is associative, i.e., for every .
Let S denote the unit circle of the complex plane and · be the usual multiplication of complex numbers. The main result of this paper is an elementary construction of homeomorphisms such that the pair induces a given binary, continuous, associative, and cancellative operation on S, where the topology in S is induced by the usual topology of the plane. As far as we know it is the only such proof.
This paper is partly of expository type because we also show several interesting applications of the main outcome and underline some symmetry issues concerning functional equations, which arise between those applications and some consequences of the following well known result of J. Aczél [1,2] (analogous result was obtained in Reference ; also see Reference  for a simpler proof).
Let I be a real nontrivial interval and be a binary operation that is continuous, associative, and cancellative. Then, there exist an infinite real interval J and a homeomorphism with
Note that property (4) means, in particular, that for every , i.e., for every . Therefore, J must be either the set of reals or of one of the following forms: , , , with some real .
Next, it is clear that every operation of form (4) must be cancellative. So, without the cancellativity assumption, the result is not true, and, for instance, the natural operation
which is associative and continuous, but not cancellative, cannot be represented in form (4). The same is true for for .
In the terms of functional equations, Theorem 1 is about continuous solutions (if we write ) of the following associativity equation:
In this case, the cancellativity of ∘ is equivalent to the injectivity of A with respect to either variable, which means the strict monotonicity with respect to either variable (because of the assumed continuity of A). Similar problem without the assumption of strict monotonicity (i.e., the cancellativity of the corresponding binary operation), was considered in Reference  (also see Reference ), and, under some additional assumptions, the following representation was obtained
with some continuous injection , where is a pseudoinverse of H (see Reference  for more details). Clearly, representation (4) is actually (7) with . For further, more general investigations of that subject, we refer to References [7,8,9].
Let us add that solutions to the associativity Equation (6) are important in statistical metric spaces and are called triangular norms or shortly t-norms (cf., e.g., References [6,10,11]); we also refer to Reference  for the notion of copulas.
A result analogous to Theorem 1, for the unit circle, has the subsequent form.
Let a binary operation be continuous, associative, and cancellative. Then, there exist exactly two homeomorphisms such that
In particular, for , where is the complex conjugate of x.
Theorem 2 can be derived (with some additional reasoning) from Reference  (Theorems 1.10 and 1.13) (continuous, compact and cancellative semigroup is a topological group) and Reference  (Theorem 2) (connected topological group, which contains a neighbourhood of the neutral element that is homeomorphic to an open real interval, is isomorphic either to the additive group of reals or to the factor group , where denotes the set of integers).
However, as we will see in the section with applications, it is useful to know how to construct the function (and thus also ). In the case of Theorem 1, the form of H can be deduced from the proofs in References [2,4]. We show in Section 5 that a symmetric reasoning, with a reasonably elementary and simple construction of the functions and , works for Theorem 2.
Finally, let us mention that the form of A in (7) reminds about a well-known problem of representing functions with several variables by functions in one variable, whether under some regularity assumptions or not (see References [5,15,16,17] for more details).
For further information on topological semigroups (also on historical background) we refer to References [13,18,19].
The paper is divided into 6 sections. The next section contains some observations concerning symmetry issue arising between Theorems 1 and 2. In Section 3, we present several applications of both theorems. Section 4, titled Auxiliary Results, contains a series of definitions, lemmas, remarks, and corollaries, leading to the final reasoning contained in Section 5, titled The Proof of Theorems 2, and being the proper proof of the theorem with a description of function . The sixth section (the last one) presents some concluding remarks.
2. Some Remarks on Symmetry Issues
Let us observe that the statements of Theorems 1 and 2 imply that both operations, ∘ and ★, must be commutative (symmetric). Moreover, in the next section we demonstrate that some applications of those theorems yield several somewhat symmetric results concerning functional equations. But, first, let us note some symmetry deficiencies between Theorems 1 and 2.
S is compact, while the interval J in Theorem 1 must be infinite and therefore not compact, which means that I cannot be compact. So, every continuous semigroup on a compact interval is not cancellative.
In Theorem 2 there exist exactly two functions, and , satisfying (8). The situation in the case of Theorem 1 (for a real interval) is somewhat different. Namely, let also be a continuous injection such that for (which means, in particular, that for ). Then
which means that the function is a homeomorphism satisfying
Consequently, with some real (see, e.g., Reference ), which means (with ) that
Note that, even if in Theorem 1, does not need to be a group, contrary to the situation on the circle. This happens in the case when .
Now, we show some simple applications of Theorems 1 and 2 in functional equations. They show some symmetries and some lack of symmetry between those two cases of the continuous cancellative semigroups on a real interval and on the unit circle. As they are clearly visible, we do not discuss them in detail.
Let us begin with the following auxiliary useful result in Reference  (p. 155).
Let . Then, a continuous function satisfies the functional equation
if and only if
in the case , there is with for ;
in the case , there are with for .
The next lemma seems to be well known, but, for the convenience of readers, we present a short proof of it.
Let be a continuous solution of the equation
Then, there is a real constant d such that for .
Define a function by for . Then, B is continuous and, by (14), for . Consequently, by Lemma 1, there is such that for . Since , the real part of a must be equal 0. So and consequently for . Hence, for . □
Now, we are in a position to present the mentioned before applications. As far as we know, they are new results, never published so far.
Let a binary operation be associative, continuous and cancellative. Let and be a homeomorphism with
Then, a continuous function fulfills the equation
if and only if there is such that,
in the case , for ;
in the case , for .
Let be a continuous function that fulfills Equation (16). Then, by (15),
whence the function , for , fulfills Equation (13). Hence, by Lemma 1, conditions (I) and (II) are valid. Clearly, for if and only if . Moreover, it is easy to check that for if and only if . This implies statements (a) and (b).
Now, suppose that has the form depicted by (a). Then, in view of (15),
The case of (b) is analogous. □
Let I be a nontrivial real interval and a binary operation be associative, continuous, and cancellative. Let and be a continuous injection satisfying (4). Then, the functional equation
has a non-constant continuous solution if and only if H is bijective.
Moreover, if H is bijective, then a continuous function fulfills Equation (19) if and only if there is such that,
in the case , for ;
in the case , for .
Let be a continuous solution of Equation (19). Then, the function , for , fulfills the equation
Moreover, h is continuous.
In the case there is such that for , which implies that either or H is surjective and for .
In the case there are such that for , where and denote the real and imaginary parts of a complex number s. Write . Then, it is easily seen that for , which implies that either or H is surjective and for .
Now, we see that, in the case where H is bijective, we obtain statements (i) and (ii). Moreover, it is easy to check functions depicted in statements (i) and (ii) are solutions to (19).
Next, suppose that Equation (19) has a non-constant continuous solution . Then, as we have already observed, in the case there is such that for . Since f is not constant and H is injective, this means that and consequently H is surjective. If , then we show the surjectivity of H in a similar way.
Finally, note that, if H is bijective, then functions depicted by statements (i) and (ii), with , are non-constant continuous solutions to (19). □
Let I be a nontrivial real interval and binary operations and be associative, continuous and cancellative. Let and be such that (4) and (15) hold. Then, the following two statements are valid.
A continuous function fulfills the functional equation
if and only if there is such that for .
Every continuous function fulfilling the functional equation
First we prove (A). So, let be a continuous function that fulfills functional Equation (21). Then, in view of (4) and (15),
which means that the function , for is continuous and satisfies the equation
Let be the solution of
such that for (we define by: for ). According to Proposition 1 there is such that for , whence for . Consequently,
The converse is easy to check.
Now, we prove (B). Let be a continuous solution to (22). Then, in view of (4) and (15),
whence the function , for , is continuous and satisfies the equation
Thus, the set is compact (because S is compact and h is continuous) and it is a subgroup of . The only such subgroup is the trivial one , which means that . Since is bijective and H is injective, f must be constant. □
Using Theorem 2 and some results from Reference  we also get the following proposition on the minimal homeomorphisms on S. Let us recall (see Reference  (Ch.1 §1)) that, if , then a set is called f-invariant provided . A homeomorphism is minimal if S does not contain any non-empty, proper, closed f-invariant subset (see References [23,24] for some related results).
A function is a minimal homeomorphism if and only if there exist and a continuous, associative, and cancellative binary operation such that
where and for ( denotes the set of positive integers).
Assume that is a minimal homeomorphism. Then, there exist an irrational real number c and a homeomorphism with for (see Reference  (Ch.3 §3 Th.1,3)). Define a binary operation by
and put . Then, for . Moreover, since is injective and c is irrational, we have
Now, suppose that is of form (29), where is a continuous, associative, and cancellative binary operation. In view of Theorem 2, there exists a homeomorphism such that (15) is valid. Consequently, T is a homeomorphism and for . Let be such that . Then,
Now, we show that c is irrational. So, for the proof by contradiction suppose that there are with . Then, . This is a contradiction to (30).
Thus, we have proved that c is irrational. Hence, it follows that the set is dense in S; thus, by (33), the set is dense in S for every , which means that T is a minimal homeomorphism. This ends the proof. □
Before our last proposition in this section, let us remind the notion of continuous flow on a topological space X. Namely (cf. Reference [13,25]), if for is a family of maps such that for all , is the identity map on X and the mapping , given by , is continuous, then we say that the family for is a continuous flow on X.
Observe that, if for is a continuous flow on a topological space X, then , , is a continuous solution of the translation functional equation
The last proposition in this section shows that the continuous, associative, and cancellative binary operations on S can be used in a description of the continuous flows on S (cf. References [26,27]) and therefore also in a description of the continuous solutions to the translation functional equation.
A family of continuous functions is a continuous flow such that either (the identity function on S) or for if and only if there exist a continuous, associative, and cancellative operation and a continuous solution of the functional equation
Let be a continuous flow such that either or for . Then, by  (Theorem 2), there exist and an orientation preserving homeomorphism with for , .
Let be the operation given by (31). Then, ★ is continuous and is a group. Next, the function , for , fulfills Equation (35) (because for ) and
Conversely, suppose that (36) holds. Then, it is easily seen that is a continuous flow. Further, by Theorem 2, is a group. So, if there is such that , then, by (36), , whence is the neutral element of . Thus, (36) implies that . This completes the proof. □
Finally, let us mention that somewhat related issues in particle physics can be found in Reference . Moreover, it seems that Theorem 2 can be applied in finding continuous solutions of some suitable functional equations analogously as Theorem 1 has been used in References [29,30,31] for real functions.
4. Auxiliary Results
In this section, we provide information and several observations necessary in the proof of Theorem 2 for the construction of function . They are presented in a series of lemmas, remarks, and corollaries. We start with some notations and definitions.
We define an order in S as follows:
for , where stands for the argument of the complex number u. For every , , we write
It is known (cf., e.g., Reference  (Chap. 2)) that, for every homeomorphism , there exists a unique homeomorphism satisfying for . If f is increasing (decreasing), then we say that homeomorphism T preserves (reverses) orientation. Note that every homeomorphism preserves or reverses orientation.
Now, we will prove several auxiliary lemmas, corollaries and remarks.
So, let be as in Theorem 2. Then, is a topological group (see Reference  (Theorems 1.10 and 1.13)). Denote by e the neutral element of and define by
It is easily seen that the mapping , for , is an isomorphism from onto . So, is a group with the neutral element equal 1. Next, γ is a homeomorphism and
whence is a topological group. For every , by , we always denote the inverse element to u in the group and is defined by for .
In what follows, and for and . Clearly, by the associativity of the operation, we have
Next, given , we define functions by the formulas:
Let . Since is a topological group, and are homeomorphisms without fixed points and therefore preserve orientation (see Reference  (Remark 3)).
There exists such that .
First, suppose that for every , that is for every . Let and . Then, by and Remark 5, we have , where . This means that .
On the other hand, , and, consequently, , a contradiction because . So, there exists such that . Putting , we have either or . □
If and , then .
It follows from Lemma 3 that there exists such that . Since and , J reverses orientation. Thus, the assertion follows from the fact that . □
Let , , and . Then, and .
Lemma 4 implies that . Thus, and . The lemma results now from Remark 5. □
From Lemma 5 (for ), we obtain the following
If and , then and .
Let , . Then:
if , , then .
Assertion (a) results from Corollary 1 (with ). For the proof of (b), fix . Then, by Lemma 4, we get . Hence, Lemma 6 gives
This ends the proof. □
Assume that and . Then, for every , , there exists exactly one , , such that . Moreover, .
Define a function , . Then, q is continuous and, on account of Lemma 6b, q is injective. Thus, and for every . Fix . By Lemma 4, . Consequently, in view of Lemma 6a, , which ends the proof. □
Let , . Then, for every , the set is non–empty and has the smallest element . Moreover, and for .
The proof is by induction with respect to n.
The case results from Lemma 7. Fix and assume that and there exists the smallest element of with . Then, according to Lemma 7, there is exactly one element such that . Note that . Thus, .
Suppose that there is with . Then, , whence . Moreover, by Lemma 6b, , a contradiction. Consequently, y is the smallest element of , , and . This completes the proof. □
Fix with (see Lemma 3). From Lemma 8, by induction, we get for , where . Hence, for , , we have
Note that, for every , there are unique , and such that and . Thus, we can define a function by
for , , , , where , , for and .
In what follows, we assume that is fixed and . Moreover, the function d is defined as above. Now, we prove some properties of d. We start with the following
Let D and be defined as in Remark 6. Then, for every .
In view of Remark 6, for every , there exist unique , , and with and . If , then we write and .
The proof is by induction with respect to . The case is trivial because, then,
So, fix and assume that the assertion holds for every with . Take with . Since, by (47), , it suffices to consider the following two cases.
Case I: Then,
with . Write
Clearly . So, in view of Remark 6, Lemma 8, and the induction hypothesis, we have
Case II: . Let and . Then, there exist , , and for , such that . Hence, by Remark 6 and the induction hypothesis, we obtain
Let D and be the same way as in Lemma 9. Then, for every , .
Clearly, . Next, for every , , there are unique , , , with .
The proof is by induction with respect to . The case results from Lemma 8. Fix and assume that the assertion holds for every , , with . Let , , and . From the induction hypothesis and Lemma 4, we obtain
because Lemma 8 yields . By Lemma 8, we also get Hence, in view of the induction hypothesis, Lemma 5 and Lemma 9, we have
This completes the proof. □
Let D and be the same way as in Lemma 9, and . Then, .
By Lemmas 10 and 4, we have . Thus, on account of Corollary 1 and Lemma 9, . This ends the proof. □
If M is defined by (57) with D and d the same way as in Lemma 9, then .
From Lemma 11, we get immediately . We prove that M is a finite number.
Suppose that for every . Then, on account of Corollary 1, for and consequently there exists .
If , then, in view of the continuity of the operation, we have
If , then, by Corollary 1, , a contradiction, too.
In this way we have proved that there is with . Thus, writing , we get for and . Observe that the set is connected, , and . Hence, and consequently, by Remark 5, , which means that . This ends the proof. □
Let D and be the same way as in Lemma 9 and M be given by (57). Then, the following two statements are valid.
For every , there is , , with .
The set is dense in S.
(a) Put (with respect to the order ≺) and suppose that . From Lemma 8 we get and, by Lemma 7 with , . Thus, there exists with . Hence, on account of Lemma 6b with , . By the definition of r and Lemma 8, there exists such that . Next, in view of Lemma 7, with , there exists with , whence, by Lemma 8, , which contradicts the definition of r.
(b) For the proof by contradiction, suppose that there exist such that and . First, consider the case where for every . Put (with respect to the order ≺) and . By a) and Lemma 11, there exists such that . Let and fix such that , . Put . Then, by Lemma 12, there exist and such that and , , which, in virtue of the definition of M and Lemma 4, means that ,
and . Thus, on account of Lemmas 5 and 9, we obtain
Since, according to the definition of M, for , , in this way, we have proved that for every with . This contradicts the definition of M.
Now, consider the case where there is such that . Put , (with respect to ≺), and . Let V be a neighbourhood (in S) of such that for every . Choose a neighbourhood of 1 such that . It results from a) and Lemma 11 that there exists with , , and . Thus, . Moreover, Lemma 4 and Corollary 1 give . Let be a neighbourhood of such that for every . By the definition of , there is with and , which means that . Clearly, for some . Since and, by Lemma 9, , we obtain a contradiction with the definition of and . □
Define a function by the formula
where D and are the same way as in Lemma 9 and M is given by (57).
As an immediate consequence of the definition and Lemma 13, we have the following.
Let be defined as above, where M is given by (57). Then, is a strictly increasing function (with respect to the order ≺ in S).
Let M and be as in Corollary 2. Then, is a continuous function.
Using Lemma 13a and Corollary 2, it is easily seen that is continuous at 0. Next, we show that is continuous at M. So, fix an increasing sequence in D such that . Then, according to the definitions of M and and Lemma 13b, for every , there is with , which means that . Hence, by Corollary 2, is continuous at M.
To complete the proof suppose that is discontinuous at a point . Then, by Corollary 2 and Lemma 13b, there exist with and , and a sequence in such that and
Hence, for . Since
we obtain a contradiction. This ends the proof. □
Let M and be as in Corollary 2 and . Then:
if , then ;
if , then .
(a) Let be sequences in such that for and , . By the continuity of ∘ and Lemmas 9 and 14, we have
(b) Let be increasing sequences in with , . Take a sequence in with and write for . Then, for and, by Lemmas 9 and 14, we may write
This ends the proof. □
5. Proof of Theorem 2
Now, we are in a position to present the final part of the proof of Theorem 2. In what follows, M and have the same meaning as in Corollary 2.
Define a function as follows:
Then, by Corollary 2, F is strictly increasing (with respect to the order ≺ in S) and, on account of Lemma 14, it is easily seen that F is continuous. Moreover, it results from Lemma 15 that, for every ,
Consequently, F is a homeomorphism fulfilling
Write , where is defined as in Remark 4. Then, is a homeomorphism and, by (68),
To complete the proof of Theorem 2 suppose that is also a homeomorphism such that for . Then, for every ,
whence Putting for we get and
Moreover, A is a homeomorphism. Hence, in view of Lemma 2, or . Since it is easily seen that (69) yields
this completes the proof.
Given a binary, continuous, associative, and cancellative operation , we presented an elementary construction of all continuous isomorphisms from the group onto the semigroup , where S is the unit circle on the complex plane and · is the usual multiplication of complex numbers. There are exactly two such isomorphisms , and they uniquely determine the form of operation ★ in the following way:
Moreover, for all .
Using this result, we have easily determined all continuous solutions of the functional equation
where is either the set of reals or the set of complex numbers .
We also provided some further applications of that result in functional equations and showed how to use it in the descriptions of the continuous flows and minimal homeomorphisms on S. In particular, we underlined some symmetry issues, which arise between the consequences of the result and of the analogous outcome for the real interval.
It would be interesting to investigate in the future to what extent the statements of Theorems 1 and 2 remain valid if the cancellativity is replaced by the one side cancellativity; for instance, by the left-cancellativity (a groupoid is left-cancellative if for all with ).
The next step might be to investigate how much the associativity assumption can be weakened; for instance, to what extent the assumption can be replaced by the square symmetry defined by the formula: . A natural example of square symmetric operation in , which is not associative, is given by: for , where are fixed and
Such new results would have interesting applications in functional equations in a similar way as Theorem 1 in References [29,30,31].
Conceptualization, M.B. and J.B.; methodology, M.B., J.B, E.-s.E.-h., and E.J.; software, M.B., J.B., E.-s.E.-h., and E.J.; validation, M.B., J.B., E.-s.E.-h., and E.J.; formal analysis, M.B., J.B., E.-s.E.-h., and E.J.; investigation, M.B., J.B., E.-s.E.-h., and E.J.; resources, M.B., J.B., E.-s.E.-h., and E.J.; data curation, M.B., J.B., E.-s.E.-h., and E.J.; writing—original draft preparation, M.B. and J.B; writing—review and editing, M.B., J.B., E.-s.E.-h., and E.J.; visualization, M.B., J.B., E.-s.E.-h., and E.J.; supervision, J.B.; project administration, E.-s.E.-h., and E.J.; funding acquisition, M.B., J.B., E.-s.E.-h., and E.J. All authors have read and agreed to the published version of the manuscript.
This research received no external funding.
Conflicts of Interest
The authors declare no conflict of interest.
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