1. Introduction
Let
G be a nonempty set. Recall that the set of complex-valued functions with domain
G, denoted by
, forms a complex vector space under pointwise addition and scalar multiplication. In the case when
G has a certain algebraic structure, the structure of
is enriched by that of
G; and
becomes a representation space for
G. For instance, if
G is a finite group, then
possesses a Hermitian inner product defined by
for all
. Furthermore,
G acts on
by
,
, for all
. This action induces a unitary representation of
G on
, known as the left regular representation of
G. In this case,
decomposes into a direct sum of (irreducible) invariant subspaces; that is,
is completely reducible. The left regular representation arises naturally in studying group representations.
It is well known that the associative property is reflected in the notion of group actions via the action by left multiplication. Hence, it seems difficult to extend the notion of group actions to nonassociative structures. However, this can be done (in certain circumstances) by employing the equivalence of group actions and permutation representations, see, for instance, [
1,
2,
3,
4]. This motivates us to study the left regular representation in the setting of gyrogroups in more detail. It turns out that the left regular representation provides a prime example of a unitary representation of a finite gyroroup. Unlike the situation of groups, however, the left regular representation of a (finite) gyrogroup is so complicated that its appearance is not quite clear. It should be emphasized that the study of loop and quasigroup representations (in a different approach) can be found, for instance, in [
5].
2. Preliminaries
Gyrogroups are nonassociative algebraic structures that share many properties with groups [
1,
2,
4,
6,
7]. One of the main aspects of gyrogroup structures is that the Cayley table of a finite gyrogroup represents a ’Latin square’—an object studied in combinatorics and in experimental design. This follows from the fact that the equations
and
in the variable
x always have solutions in gyrogroups; see, for instance, Theorem 2.22 of [
7]. Important combinatorial results such as the orbit-stabilizer theorem and the Cauchy–Frobenius lemma remain valid in the case of gyrogroups [
1], among other things.
The formal definition of a gyrogroup resembles that of a group. By a gyrogroup we mean a nonempty set G endowed with a binary operation ⊕ on G that satisfies the following properties.
- (i)
There is an element such that for all .
- (ii)
For each , there is an element such that .
- (iii)
For all
a,
, there is an automorphism
such that
for all
.
- (iv)
For all a, , .
We remark that Properties (i)–(iv) have their right-counterparts. The algebraic identity Equation (
2) is called the ’left gyroassociative law’, an analogue of the left associative law in abstract algebra. The automorphism
in Property (iii) is called the ‘gyroautomorphism’ generated by
a and
b. For basic knowledge of gyrogroups, we refer the reader to [
6,
7]. Throughout the remainder of the article,
e denotes the (unique) two-sided identity of a gyrogroup, and
denotes the (unique) two-sided inverse of an element
a in a gyrogroup. Thus,
.
In [
1], we introduce a generalization of group actions. By an action of a gyrogroup
G on a nonempty set
X we mean a map
, denoted by
, such that
- (i)
for all and
- (ii)
for all .
A ‘permutation representation’
of a gyrogroup
G on a nonempty set
X is defined as a gyrogroup homomorphism from
G to the symmetric group on
X, denoted by
. This is equivalent to saying that
satisfies the condition
for all
. By Theorems 3.2 and 3.3 of [
1], the notions of gyrogroup actions and permutation representations are equivalent. In fact, if · is an action of
G on
X, then the map
, where
for
, defines a permutation representation of
G on
X. Conversely, if
is a permutation representation, then the map · defined by
for all
is a gyrogroup action. For more details, we refer the reader to [
1].
In [
2], the authors extend the study of gyrogroup actions and representations to a linear version. A gyrogroup action of
G on a vector space
V is linear if in addition every element of
G induces an (invertible) linear transformation on
V; that is, if the map
,
, is linear for all
. A linear representation of
G on
V is defined as a gyrogroup homomorphism from
G to the general linear group of
V. Of course, the notions of linear actions and representations of a gyrogroup are equivalent (cf. Theorems 3.4 and 3.5 of [
2]). A linear representation of
G on
V is of finite degree if
V is finite-dimensional. Let
be a linear representation of a gyrogroup
G. A subspace
W of
V is invariant if
for all
; that is, if
for all
, where · is the action induced by
. Furthermore,
is irreducible if the invariant subspaces of
V are only the trivial subspace and the whole space. For more details, we refer the reader to [
2].
3. The Left Regular Representation
Let
G be a (finite or infinite) gyrogroup and define
Recall that
forms a complex vector space under pointwise addition and scalar multiplication. Furthermore, if
G is finite, then
can be made into a complex inner product space by declaring a Hermitian inner product:
for all
. Here,
denotes the usual complex conjugation. We emphasize that Equation (
5) makes sense since
G is finite. In this case,
is finite-dimensional. In fact, the indicator functions:
for all
form a basis for
. Unfortunately,
G does not, in general, act on the whole space
since the associative law fails to satisfy in
G. Nevertheless,
G acts linearly on a suitable subspace of
, similar to the case of groups.
For each
, let
be the left gyrotranslation by
a, defined by
for all
. By Theorem 18 of [
6],
is a permutation of
G and
for all
. Set
It is not difficult to check that
forms a subspace of
and hence forms a complex vector space. The next lemma indicates that
satisfies a condition for which
G acts linearly on
.
Lemma 1. Let be a function. Then if and only iffor all . Proof. Suppose that
and let
. Then
To prove the converse, let
. Since
is surjective, there is an element
for which
. Hence,
and
. By assumption,
This proves that
. □
Theorem 1. If G is a gyrogroup, then G acts linearly on byfor all . Proof. The theorem is an application of Theorem 4.3 of [
3] and Lemma 1. □
The action described by Equation (
8) is called the left regular action of
G on
. Its corresponding linear representation is called the left regular representation of
G on
over
[
3,
4]. It should be emphasized that the term “regular” does not use in the usual sense of regularity in group actions. We remark that
G does not act on
unless
G is a group. In fact, if
G is a group, then
and Equation (
8) induces the well-known left regular representation in group theory. This justifies the use of the term “left regular representation” for the representation afforded by Equation (
8); see Item 4 below. As a consequence of Theorem 1, we obtain the following facts:
for all ;
for all , ;
for each , the map is an invertible linear transformation on ;
the map
defines a gyrogroup homomorphism from
G to
, that is,
for all
. Here,
stands for the general linear group of
.
The reason why we prefer complex-valued functions as a representation space for a finite gyrogroup is indicated in the following theorem.
Theorem 2 (Theorem 4.1, [
4]).
If G is a finite gyrogroup, then the left regular representation ρ of G on is unitary; that is, defines a unitary operator of for all :for all . Corollary 1. If ρ is the left regular representation of a finite gyrogroup G on , then for each , there exist nonzero functions in such thatwhere stands for the reflection on corresponding to v given byfor all . Proof. It is a standard result in linear algebra that any unitary operator on a finite-dimensional inner product space can be decomposed as a product of reflections; see, for instance, Theorem 10.17 of [
8]. □
3.1. Bases and Dimensions
In this section, we compute a basis of
as well as the dimension of
using the Cauchy–Frobenius lemma (also called the Burnside lemma) in group theory. Set
and
It is not difficult to check that
is the smallest subgroup of
containing
. Furthermore, if
G is finite, then
is finite.
Theorem 3. Let G be a gyrogroup and let . Then the following are equivalent:
- 1.
;
- 2.
for all ;
- 3.
for all .
Proof. The equivalence Equation (
1) ⇔ Equation (
2) follows immediately from Lemma 1. The implication Equation (
3) ⇒ Equation (
2) is clear since
. The implication Equation (
2) ⇒ Equation (
3) follows from Equation (
11). □
Let
G be a gyrogroup. Since
is a group consisting of permutations of
G, it follows that the group
acts on the gyrogroup
G by
for all
. This leads to the following equivalence relation:
for all
.
Theorem 4. Suppose that G is a gyrogroup and let be the partition of G determined by the equivalence relation Equation (
12)
. If , then if and only if f is constant on C for all . Proof. This is an application of Theorem 3. □
Suppose that
G is a gyrogroup and let
be the partition of
G determined by the equivalence relation Equation (
12). For each
, define the indicator function
by
By Theorem 4,
belongs to
for all
. In fact, we obtain the following theorem.
Theorem 5. Suppose that G is a gyrogroup and let be the partition of G determined by the equivalence relation Equation (
12)
. Then is a subset of whose elements are linearly independent. If is finite, then forms a basis for so that . Let
G be a gyrogroup. Recall that the right nucleus of
G, denoted by
, is defined as
By Theorem 3.1 of [
9],
forms an L-subgyrogroup of
G and a subgroup of
G. In particular,
if and only if
for all
. Recall also that the set of fixed points of
G corresponding to the group action of
by ⋆ is defined as
It is clear that
. Note that
Hence,
.
Theorem 6. If G is a finite gyrogroup, thenwhere is the permutation group defined by Equation (
11)
andfor all . Proof. Let , where . By Theorem 5, . It follows from the Cauchy–Frobenius lemma that equals . □
Lemma 2. Let G be a gyrogroup. If G is a group, then . If G is not a group, then .
Proof. Suppose that
G is a group. Then
for all
. Hence,
and so
. From Equation (
15), it follows that
.
Suppose that G is not a group. Hence, must be a proper subset of G and so we can pick . Assume to the contrary that , say . Note that . This implies that since otherwise and then , a contradiction. Since , there are elements such that . Hence, . This is a contradiction since and the proof completes. □
The following theorem relates the order of a gyrogroup G, the order of the group , the dimension of , and stabilizer subgroups.
Theorem 7. Let G be a finite gyrogroup and let . If G is not a group, thenwhere are representatives of the distinct orbits in G induced by Equation (
12)
andfor all . Proof. Recall that
is a finite group acting on
G by ⋆. According to Theorem 5,
, where
. Hence, we may assume that
where
for all
with
. We claim that
is not a singleton for all
. In fact,
if and only if
. By Lemma 2,
. Hence,
for all
. By the orbit decomposition theorem in group theory,
This proves Equation (
17). □
3.2. Orthogonal Decomposition
Next, we show that
admits an orthogonal direct sum decomposition. The fixed subspace of
is defined as
For each
, define a map
by
It is clear that
for all
. By Theorem 4.6 of [
3],
is an invariant subspace of
under the action Equation (
8) and
Moreover,
forms a basis for
. Therefore,
is one-dimensional.
Theorem 8 (Theorem 4.2, [
4]).
If G is a finite gyrogroup, thenwhere σ is the linear functional defined by for all . Furthermore, is an invariant subspace of . Recall that an action of
G on a vector space
V is transitive if for all
, there exists an element
a in
G such that
(cf. [
1] (Definition 3.18)). It is fixed point free if the stabilizer of any point in
V is trivial (cf. [
1] (Definition 3.19)). In view of Equation (
20), the left regular representation of
G on
is neither transitive nor fixed point free.
Theorem 9. The left regular representation of a nontrivial finite gyrogroup is neither transitive nor fixed point free.
Proof. If f and g are distinct elements of , then there is no element a in G for which ; otherwise, we would have , a contradiction. Hence, the action is not transitive. If , then . This implies that the action is not fixed point free for . □
An immediate application of Theorem 8 shows that
which implies that
is a (linear) hyperplane in
. The following theorem motivates the notion of gyrogroup characters.
Theorem 10. Let ρ be the left regular representation of a finite gyrogroup G afforded by Equation (
8)
and let tr denote the trace of a linear operator. Then Proof. Using Theorem 3.7 of [
4], we obtain that
. By Equation (
20),
and so
, as claimed. □
Definition 1 (Gyrogroup characters).
Let be a linear representation of a finite gyrogroup G of finite degree over . The character of φ is a function defined by The gyrogroup characters are a numerical invariant of linear representations of a gyrogroup of finite degree. In fact, one can show that equivalent representations of a gyrogroup have the same character. The following theorem emphasizes the importance of gyrogroup characters.
Theorem 11. If χ is the character of a linear representation of a finite gyrogroup G of finite degree on V, then χ belongs to .
Proof. Suppose that
is a linear representation. Let
. Using Proposition 32 (3) of [
6], we obtain
noting that (⋆) holds since any gyroautomorphism of
is the identity map on
V. This proves
. □
The structure of
does depend on the kernel of
. In fact, knowing the structure of
amounts to knowing the structure of
(cf. Theorem 8). Note that
is a proper invariant subspace of
. More precisely, if
denotes the character of the left regular representation of
G on
, then by Theorem 10,
Hence,
is in
but not in
. We emphasize that the character of the left regular representation of a finite gyrogroup is not quite as simple as in the case of groups. In summary, we obtain the following theorem.
Theorem 12. The left regular representation of a finite gyrogroup is irreducible if and only if is trivial.
Proof. If is irreducible, then the only invariant subspaces of are and itself. As mentioned previously, is an invariant subspace of codimension 1 in and so . Conversely, if , then by Theorem 8, , which is a one-dimensional subspace. Thus, must be irreducible. □