Left Regular Representation of Gyrogroups ^†

Abstract

In this article, we examine a subspace $L^{\mathrm{gyr}}\left(G\right)$ of the complex vector space, $L\left(G\right)=\{f:f \mathrm{is} \mathrm{a} \mathrm{function} \mathrm{from} \mathrm{G}\mathrm{to}\mathbb{C}\}$, where G is a nonassociative group-like structure called a gyrogroup. The space $L^{\mathrm{gyr}}\left(G\right)$ arises as a representation space for G associated with the left regular representation, consisting of complex-valued functions invariant under certain permutations of G. In the case when G is finite, we prove that ${\displaystyle \dim \left(L^{\mathrm{gyr}}\left(G\right)\right)={\displaystyle \frac{1}{\left|\gamma \left(G\right)\right|}}\sum _{\rho \in \gamma \left(G\right)}^{}\left|\mathrm{Fix}\left(\rho \right)\right|,}$ where $\gamma \left(G\right)$ is the subgroup of $\mathrm{Sym}\left(G\right)$ generated by a class of permutations of G and $\mathrm{Fix}\left(\rho \right)=\{a\in G:\rho \left(a\right)=a\}$.

1. Introduction

Let G be a nonempty set. Recall that the set of complex-valued functions with domain G, denoted by $L\left(G\right)$, forms a complex vector space under pointwise addition and scalar multiplication. In the case when G has a certain algebraic structure, the structure of $L\left(G\right)$ is enriched by that of G; and $L\left(G\right)$ becomes a representation space for G. For instance, if G is a finite group, then $L\left(G\right)$ possesses a Hermitian inner product defined by

$${\displaystyle \langle f,g\rangle ={\displaystyle \frac{1}{\left|G\right|}}\sum _{a\in G}^{}f\left(a\right)\overline{g\left(a\right)}}$$

(1)

for all $f,g\in L\left(G\right)$. Furthermore, G acts on $L\left(G\right)$ by $(a·f)\left(x\right)=f\left(a^{-1}x\right)$, $x\in G$, for all $a\in G,f\in L\left(G\right)$. This action induces a unitary representation of G on $L\left(G\right)$, known as the left regular representation of G. In this case, $L\left(G\right)$ decomposes into a direct sum of (irreducible) invariant subspaces; that is, $L\left(G\right)$ 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 $a\oplus x=b$ and $x\oplus a=b$ 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 $e\in G$ such that $e\oplus a=a$ for all $a\in G$.

(ii)

For each $a\in G$, there is an element $b\in G$ such that $b\oplus a=e$.

(iii)

For all a, $b\in G$, there is an automorphism $\mathrm{gyr}\left[a,b\right]\in Aut(G,\oplus )$ such that

$$a\oplus (b\oplus c)=(a\oplus b)\oplus \mathrm{gyr}\left[a,b\right]c$$

(2)

for all $c\in G$.

(iv)

For all a, $b\in G$, $\mathrm{gyr}\left[a\oplus b,b\right]=\mathrm{gyr}\left[a,b\right]$.

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 $\mathrm{gyr}\left[a,b\right]$ 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 $\ominus a$ denotes the (unique) two-sided inverse of an element a in a gyrogroup. Thus, $e=a\oplus (\ominus a)=(\ominus a)\oplus a$.

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 $·:G\times X\to X$, denoted by $(a,x)\mapsto a·x$, such that

(i)

$e·x=x$ for all $x\in X$ and

(ii)

$a·(b·x)=(a\oplus b)·x$ for all $a,b\in G,x\in X$.

A ‘permutation representation’ $\phi$ 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 $\mathrm{Sym}\left(X\right)$. This is equivalent to saying that $\phi :G\to \mathrm{Sym}\left(X\right)$ satisfies the condition

$$\phi (a\oplus b)=\phi \left(a\right)\circ \phi \left(b\right)$$

(3)

for all $a,b\in G$. 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 $\phi :a\mapsto \phi \left(a\right)$, where $\phi \left(a\right)\left(x\right)=a·x$ for $x\in X$, defines a permutation representation of G on X. Conversely, if $\phi :G\to \mathrm{Sym}\left(X\right)$ is a permutation representation, then the map · defined by $a·x=\phi \left(a\right)\left(x\right)$ for all $a\in G,x\in X$ 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 $v\mapsto a·v$, $v\in V$, is linear for all $a\in G$. 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 $(V,\phi )$ be a linear representation of a gyrogroup G. A subspace W of V is invariant if $\phi \left(a\right)\left(W\right)\subseteq W$ for all $a\in G$; that is, if $a·w\in W$ for all $a\in G,w\in W$, where · is the action induced by $\phi$. Furthermore, $\phi$ 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

$$L\left(G\right)=\{f:f \mathrm{is} \mathrm{a} \mathrm{function} \mathrm{from} \mathrm{G}\mathrm{to}\mathbb{C}\}.$$

(4)

Recall that $L\left(G\right)$ forms a complex vector space under pointwise addition and scalar multiplication. Furthermore, if G is finite, then $L\left(G\right)$ can be made into a complex inner product space by declaring a Hermitian inner product:

$${\displaystyle \langle f,g\rangle ={\displaystyle \frac{1}{\left|G\right|}}\sum _{a\in G}^{}f\left(a\right)\overline{g\left(a\right)}}$$

(5)

for all $f,g\in L\left(G\right)$. Here, $\overline{·}$ denotes the usual complex conjugation. We emphasize that Equation (5) makes sense since G is finite. In this case, $L\left(G\right)$ is finite-dimensional. In fact, the indicator functions:

$${\delta }_a\left(x\right)=\left\{\begin{array}{cc}1\hfill & \mathrm{if} x=a\hfill \\ 0\hfill & \mathrm{otherwise}\hfill \end{array}\right.$$

(6)

for all $a\in G$ form a basis for $L\left(G\right)$. Unfortunately, G does not, in general, act on the whole space $L\left(G\right)$ since the associative law fails to satisfy in G. Nevertheless, G acts linearly on a suitable subspace of $L\left(G\right)$, similar to the case of groups.

For each $a\in G$, let $L_a$ be the left gyrotranslation by a, defined by $L_a\left(x\right)=a\oplus x$ for all $x\in G$. By Theorem 18 of [6], $L_a$ is a permutation of G and $L_{\ominus a}=L_a^{-1}$ for all $a\in G$. Set

$$L^{\mathrm{gyr}}\left(G\right)=\{f\in L\left(G\right):f\circ (L_a\circ \mathrm{gyr}\left[x,y\right]\circ L_a^{-1})=f\mathrm{for}\mathrm{all}a,x,y\in G\}.$$

(7)

It is not difficult to check that $L^{\mathrm{gyr}}\left(G\right)$ forms a subspace of $L\left(G\right)$ and hence forms a complex vector space. The next lemma indicates that $L^{\mathrm{gyr}}\left(G\right)$ satisfies a condition for which G acts linearly on $L^{\mathrm{gyr}}\left(G\right)$.

Lemma 1.

Let $f:G\to \mathbb{C}$ be a function. Then $f\in L^{\mathrm{gyr}}\left(G\right)$ if and only if

$$f(a\oplus \mathrm{gyr}\left[x,y\right]z)=f(a\oplus z)$$

for all $a,x,y,z\in G$.

Proof. 

Suppose that $f\in L^{\mathrm{gyr}}\left(G\right)$ and let $a,x,y,z\in G$. Then

$$\begin{array}{cc}\hfill f(a\oplus z)& =(f\circ L_a\circ \mathrm{gyr}\left[x,y\right]\circ L_a^{-1})(a\oplus z)\hfill \\ \hfill & =f\left(L_a\left(\mathrm{gyr}\left[x,y\right]\left(L_a^{-1}(a\oplus z)\right)\right)\right)\hfill \\ \hfill & =f\left(L_a\left(\mathrm{gyr}\left[x,y\right]\left(L_{\ominus a}(a\oplus z)\right)\right)\right)\hfill \\ \hfill & =f\left(L_a\left(\mathrm{gyr}\left[x,y\right](\ominus a\oplus (a\oplus z\left)\right)\right)\right)\hfill \\ \hfill & =f\left(L_a\left(\mathrm{gyr}\left[x,y\right]z\right)\right)\hfill \\ \hfill & =f(a\oplus \mathrm{gyr}\left[x,y\right]z).\hfill \end{array}$$

To prove the converse, let $a,x,y,z\in G$. Since $L_a$ is surjective, there is an element $w\in G$ for which $L_a\left(w\right)=z$. Hence, $a\oplus w=z$ and $w=L_a^{-1}\left(z\right)$. By assumption,

$$\begin{array}{cc}\hfill f\left(z\right)& =f(a\oplus w)\hfill \\ \hfill & =f(a\oplus \mathrm{gyr}\left[x,y\right]w)\hfill \\ \hfill & =f(a\oplus \mathrm{gyr}\left[x,y\right]\left(L_a^{-1}\left(z\right)\right))\hfill \\ \hfill & =(f\circ L_a\circ \mathrm{gyr}\left[x,y\right]\circ L_a^{-1})\left(z\right).\hfill \end{array}$$

This proves that $f=f\circ (L_a\circ \mathrm{gyr}\left[x,y\right]\circ L_a^{-1})$. □

Theorem 1.

If G is a gyrogroup, then G acts linearly on $L^{\mathrm{gyr}}\left(G\right)$ by

$$a·f=f\circ L_{\ominus a}$$

(8)

for all $a\in G,f\in L^{\mathrm{gyr}}\left(G\right)$.

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 $L^{\mathrm{gyr}}\left(G\right)$. Its corresponding linear representation is called the left regular representation of G on $L^{\mathrm{gyr}}\left(G\right)$ over $\mathbb{C}$ [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 $L\left(G\right)$ unless G is a group. In fact, if G is a group, then $L^{\mathrm{gyr}}\left(G\right)=L\left(G\right)$ 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:

• $e·f=f$ for all $f\in L^{\mathrm{gyr}}\left(G\right)$;
• $a·(b·f)=(a\oplus b)·f$ for all $a,b\in G$, $f\in L^{\mathrm{gyr}}\left(G\right)$;
• for each $a\in G$, the map $\rho \left(a\right):f\mapsto a·f$ is an invertible linear transformation on $L^{\mathrm{gyr}}\left(G\right)$;
• the map $a\mapsto \rho \left(a\right)$ defines a gyrogroup homomorphism from G to $\mathrm{GL}\left(L^{\mathrm{gyr}}\left(G\right)\right)$, that is,

  $$\rho (a\oplus b)=\rho \left(a\right)\circ \rho \left(b\right)$$

  for all $a,b\in G$. Here, $\mathrm{GL}\left(L^{\mathrm{gyr}}\left(G\right)\right)$ stands for the general linear group of $L^{\mathrm{gyr}}\left(G\right)$.

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 $L^{\mathrm{gyr}}\left(G\right)$ is unitary; that is, $\rho \left(a\right)$ defines a unitary operator of $L^{\mathrm{gyr}}\left(G\right)$ for all $a\in G$:

$$\langle \rho \left(a\right)\left(f\right),\rho \left(a\right)\left(g\right)\rangle =\langle f,g\rangle$$

(9)

for all $f,g\in L^{\mathrm{gyr}}\left(G\right)$.

Corollary 1.

If ρ is the left regular representation of a finite gyrogroup G on $L^{\mathrm{gyr}}\left(G\right)$, then for each $a\in G$, there exist nonzero functions $v_1,v_2,\dots ,v_k$ in $L^{\mathrm{gyr}}\left(G\right)$ such that

$$\rho \left(a\right)=R_{v_1}\circ R_{v_2}\circ \cdots \circ R_{v_k},$$

where $R_v$ stands for the reflection on $L^{\mathrm{gyr}}\left(G\right)$ corresponding to v given by

$$R_v\left(w\right)=w-{\displaystyle \frac{2\langle w,v\rangle }{\langle v,v\rangle }}v$$

for all $w\in L^{\mathrm{gyr}}\left(G\right)$.

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 $L^{\mathrm{gyr}}\left(G\right)$ as well as the dimension of $L^{\mathrm{gyr}}\left(G\right)$ using the Cauchy–Frobenius lemma (also called the Burnside lemma) in group theory. Set

$$\kappa \left(G\right)=\{L_a\circ \mathrm{gyr}\left[x,y\right]\circ L_a^{-1}:a,x,y\in G\}$$

(10)

and

$$\gamma \left(G\right)=\{{\rho }_1\circ {\rho }_2\circ \cdots \circ {\rho }_n:{\rho }_i\in \kappa \left(G\right) \mathrm{for} \mathrm{all} i=1,2,\dots ,n\}.$$

(11)

It is not difficult to check that $\gamma \left(G\right)$ is the smallest subgroup of $\mathrm{Sym}\left(G\right)$ containing $\kappa \left(G\right)$. Furthermore, if G is finite, then $\gamma \left(G\right)$ is finite.

Theorem 3.

Let G be a gyrogroup and let $f\in L\left(G\right)$. Then the following are equivalent:

1.

$f\in L^{\mathrm{gyr}}\left(G\right)$;

2.

$f\left(\rho \right(z\left)\right)=f\left(z\right)$ for all $z\in G,\rho \in \kappa \left(G\right)$;

3.

$f\left(\rho \right(z\left)\right)=f\left(z\right)$ for all $z\in G,\rho \in \gamma \left(G\right)$.

Proof. 

The equivalence Equation (1) ⇔ Equation (2) follows immediately from Lemma 1. The implication Equation (3) ⇒ Equation (2) is clear since $\kappa \left(G\right)\subseteq \gamma \left(G\right)$. The implication Equation (2) ⇒ Equation (3) follows from Equation (11). □

Let G be a gyrogroup. Since $\gamma \left(G\right)$ is a group consisting of permutations of G, it follows that the group $\gamma \left(G\right)$ acts on the gyrogroup G by $\rho \star a=\rho \left(a\right)$ for all $\rho \in \gamma \left(G\right),a\in G$. This leads to the following equivalence relation:

$$x\sim y\mathrm{if} \mathrm{and} \mathrm{only} \mathrm{if}y=\rho \left(x\right) \mathrm{for} \mathrm{some} \rho \in \gamma \left(G\right)$$

(12)

for all $x,y\in G$.

Theorem 4.

Suppose that G is a gyrogroup and let $\mathcal{O}$ be the partition of G determined by the equivalence relation Equation (12). If $f\in L\left(G\right)$, then $f\in L^{\mathrm{gyr}}\left(G\right)$ if and only if f is constant on C for all $C\in \mathcal{O}$.

Proof. 

This is an application of Theorem 3.  □

Suppose that G is a gyrogroup and let $\mathcal{O}$ be the partition of G determined by the equivalence relation Equation (12). For each $C\in \mathcal{O}$, define the indicator function ${\delta }_C$ by

$${\delta }_C\left(x\right)=\left\{\begin{array}{cc}1\hfill & \mathrm{if} x\in C;\hfill \\ 0\hfill & \mathrm{if} x\in G\backslash C.\hfill \end{array}\right.$$

(13)

By Theorem 4, ${\delta }_C$ belongs to $L^{\mathrm{gyr}}\left(G\right)$ for all $C\in \mathcal{O}$. In fact, we obtain the following theorem.

Theorem 5.

Suppose that G is a gyrogroup and let $\mathcal{O}$ be the partition of G determined by the equivalence relation Equation (12). Then $\mathcal{B}=\{{\delta }_C:C\in \mathcal{O}\}$ is a subset of $L^{\mathrm{gyr}}\left(G\right)$ whose elements are linearly independent. If $\mathcal{O}$ is finite, then $\mathcal{B}$ forms a basis for $L^{\mathrm{gyr}}\left(G\right)$ so that $\dim \left(L^{\mathrm{gyr}}\left(G\right)\right)=\left|\mathcal{O}\right|$.

Let G be a gyrogroup. Recall that the right nucleus of G, denoted by $N_r\left(G\right)$, is defined as

$$N_r\left(G\right)=\{c\in G:\mathrm{gyr}\left[a,b\right]c=c \mathrm{for} \mathrm{all} a,b\in G\}.$$

(14)

By Theorem 3.1 of [9], $N_r\left(G\right)$ forms an L-subgyrogroup of G and a subgroup of G. In particular, $\ominus a\oplus z\in N_r\left(G\right)$ if and only if $a\oplus N_r\left(G\right)=z\oplus N_r\left(G\right)$ for all $a,z\in G$. Recall also that the set of fixed points of G corresponding to the group action of $\gamma \left(G\right)$ by ⋆ is defined as

$$\mathrm{Fix}(G,\star )=\{a\in G:\rho \star a=a \mathrm{for} \mathrm{all} \rho \in \gamma \left(G\right)\}.$$

(15)

It is clear that $\mathrm{Fix}(G,\star )=\{a\in G:\rho \left(a\right)=a \mathrm{for} \mathrm{all} \rho \in \gamma \left(G\right)\}$. Note that

$$\{\mathrm{gyr}\left[a,b\right]:a,b\in G\}\subseteq \gamma \left(G\right).$$

Hence, $\mathrm{Fix}(G,\star )\subseteq N_r\left(G\right)$.

Theorem 6.

If G is a finite gyrogroup, then

$${\displaystyle \dim \left(L^{\mathrm{gyr}}\left(G\right)\right)={\displaystyle \frac{1}{\left|\gamma \left(G\right)\right|}}\sum _{\rho \in \gamma \left(G\right)}^{}\left|\mathrm{Fix}\left(\rho \right)\right|,}$$

(16)

where $\gamma \left(G\right)$ is the permutation group defined by Equation (11) and

$$\mathrm{Fix}\left(\rho \right)=\{a\in G:\rho \left(a\right)=a\}$$

for all $\rho \in \gamma \left(G\right)$.

Proof. 

Let $\mathcal{O}=\{\mathrm{orb}x:x\in G\}$, where $\mathrm{orb}x=\{\rho \left(x\right):\rho \in \gamma \left(G\right)\}$. By Theorem 5, $\dim \left(L^{\mathrm{gyr}}\left(G\right)\right)=\left|\mathcal{O}\right|$. It follows from the Cauchy–Frobenius lemma that $\left|\mathcal{O}\right|$ equals ${\displaystyle {\displaystyle \frac{1}{\left|\gamma \left(G\right)\right|}}\sum _{\rho \in \gamma \left(G\right)}^{}\left|\mathrm{Fix}\left(\rho \right)\right|}$. □

Lemma 2.

Let G be a gyrogroup. If G is a group, then $\mathrm{Fix}(G,\star )=G$. If G is not a group, then $\mathrm{Fix}(G,\star )=\varnothing$.

Proof. 

Suppose that G is a group. Then $\mathrm{gyr}\left[a,b\right]={\mathrm{id}}_G$ for all $a,b\in G$. Hence, $\kappa \left(G\right)=\left\{{\mathrm{id}}_G\right\}$ and so $\gamma \left(G\right)=\left\{{\mathrm{id}}_G\right\}$. From Equation (15), it follows that $\mathrm{Fix}(G,\star )=G$.

Suppose that G is not a group. Hence, $N_r\left(G\right)$ must be a proper subset of G and so we can pick $a\in G\backslash N_r\left(G\right)$. Assume to the contrary that $\mathrm{Fix}(G,\star )\ne \varnothing$, say $z\in \mathrm{Fix}(G,\star )$. Note that $z\in N_r\left(G\right)$. This implies that $\ominus a\oplus z\notin N_r\left(G\right)$ since otherwise $a\oplus N_r\left(G\right)=z\oplus N_r\left(G\right)=e\oplus N_r\left(G\right)=N_r\left(G\right)$ and then $a\in N_r\left(G\right)$, a contradiction. Since $\ominus a\oplus z\notin N_r\left(G\right)$, there are elements $x,y\in G$ such that $\mathrm{gyr}\left[x,y\right](\ominus a\oplus z)\ne \ominus a\oplus z$. Hence, $(L_a\circ \mathrm{gyr}\left[x,y\right]\circ L_{\ominus a})\left(z\right)\ne z$. This is a contradiction since $L_a\circ \mathrm{gyr}\left[x,y\right]\circ L_{\ominus a}=L_a\circ \mathrm{gyr}\left[x,y\right]\circ L_a^{-1}\in \gamma \left(G\right)$ and the proof completes.  □

The following theorem relates the order of a gyrogroup G, the order of the group $\gamma \left(G\right)$, the dimension of $L^{\mathrm{gyr}}\left(G\right)$, and stabilizer subgroups.

Theorem 7.

Let G be a finite gyrogroup and let $n=\dim \left(L^{\mathrm{gyr}}\left(G\right)\right)$. If G is not a group, then

$${\displaystyle {\displaystyle \frac{\left|G\right|}{\left|\gamma \left(G\right)\right|}}=\sum _{i=1}^n{\displaystyle \frac{1}{|\mathrm{stab}{x}_i|}},}$$

(17)

where $x_1,x_2,\dots ,x_n$ are representatives of the distinct orbits in G induced by Equation (12) and

$$\mathrm{stab}{x}_i=\{\rho \in \gamma \left(G\right):\rho \left(x_i\right)=x_i\}$$

for all $i=1,2,\dots ,n$.

Proof. 

Recall that $\gamma \left(G\right)$ is a finite group acting on G by ⋆. According to Theorem 5, $\dim \left(L^{\mathrm{gyr}}\left(G\right)\right)=\left|\mathcal{O}\right|$, where $\mathcal{O}=\{\mathrm{orb}x:x\in G\}$. Hence, we may assume that

$$\mathcal{O}=\{\mathrm{orb}{x}_1,\mathrm{orb}{x}_2,\dots ,\mathrm{orb}{x}_n\},$$

where $x_i\nsim x_j$ for all $i,j$ with $i\ne j$. We claim that $\mathrm{orb}{x}_i$ is not a singleton for all $i=1,2,\dots ,n$. In fact, $\mathrm{orb}x=\left\{x\right\}$ if and only if $x\in \mathrm{Fix}(G,\star )$. By Lemma 2, $\mathrm{Fix}(G,\star )=\varnothing$. Hence, $\mathrm{orb}x\ne \left\{x\right\}$ for all $x\in G$. By the orbit decomposition theorem in group theory,

$${\displaystyle \left|G\right|=\left|\mathrm{Fix}(G,\star )\right|+\sum _{i=1}^n[\gamma \left(G\right):\mathrm{stab}{x}_i]=\sum _{i=1}^n{\displaystyle \frac{\left|\gamma \left(G\right)\right|}{|\mathrm{stab}{x}_i|}}.}$$

This proves Equation (17).  □

3.2. Orthogonal Decomposition

Next, we show that $L^{\mathrm{gyr}}\left(G\right)$ admits an orthogonal direct sum decomposition. The fixed subspace of $L^{\mathrm{gyr}}\left(G\right)$ is defined as

$$\mathrm{Fix}\left(L^{\mathrm{gyr}}\left(G\right)\right)=\{f\in L^{\mathrm{gyr}}\left(G\right):a·f=f \mathrm{for} \mathrm{all} a\in G\}.$$

(18)

For each $\alpha \in \mathbb{C}$, define a map $f_{\alpha }$ by

$$f_{\alpha }\left(x\right)=\alpha ,x\in G.$$

(19)

It is clear that $f_{\alpha }\in L^{\mathrm{gyr}}\left(G\right)$ for all $\alpha \in \mathbb{C}$. By Theorem 4.6 of [3], $\mathrm{Fix}\left(L^{\mathrm{gyr}}\left(G\right)\right)$ is an invariant subspace of $L^{\mathrm{gyr}}\left(G\right)$ under the action Equation (8) and

$$\mathrm{Fix}\left(L^{\mathrm{gyr}}\left(G\right)\right)=\{f_{\alpha }:\alpha \in \mathbb{C}\}.$$

(20)

Moreover, $\left\{f_1\right\}$ forms a basis for $\mathrm{Fix}\left(L^{\mathrm{gyr}}\left(G\right)\right)$. Therefore, $\mathrm{Fix}\left(L^{\mathrm{gyr}}\left(G\right)\right)$ is one-dimensional.

Theorem 8

(Theorem 4.2, [4]). If G is a finite gyrogroup, then

$$L^{\mathrm{gyr}}\left(G\right)=\langle f_1\rangle \oplus ker\sigma and\langle f_1\rangle \perp ker\sigma ,$$

(21)

where σ is the linear functional defined by ${\displaystyle \sigma \left(f\right)=\sum _{a\in G}^{}f\left(a\right)}$ for all $f\in L^{\mathrm{gyr}}\left(G\right)$. Furthermore, $ker\sigma$ is an invariant subspace of $L^{\mathrm{gyr}}\left(G\right)$.

Recall that an action of G on a vector space V is transitive if for all $u,v\in V$, there exists an element a in G such that $v=a·u$ (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 $L^{\mathrm{gyr}}\left(G\right)$ 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 $\mathrm{Fix}\left(L^{\mathrm{gyr}}\left(G\right)\right)$, then there is no element a in G for which $g=a·f$; otherwise, we would have $g=a·f=f$, a contradiction. Hence, the action is not transitive. If $f\in \mathrm{Fix}\left(L^{\mathrm{gyr}}\left(G\right)\right)$, then $\mathrm{stab}f=G$. This implies that the action is not fixed point free for $\mathrm{Fix}\left(L^{\mathrm{gyr}}\left(G\right)\right)\ne \varnothing$. □

An immediate application of Theorem 8 shows that

$$\dim (ker\sigma )=\dim \left(L^{\mathrm{gyr}}\left(G\right)\right)-1,$$

which implies that $ker\sigma$ is a (linear) hyperplane in $L^{\mathrm{gyr}}\left(G\right)$. 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

$${\displaystyle \sum _{a\in G}^{}tr\rho \left(a\right)=\left|G\right|.}$$

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Proof. 

Using Theorem 3.7 of [4], we obtain that ${\displaystyle \sum _{a\in G}^{}tr\rho \left(a\right)=\left|G\right|\dim \left(\mathrm{Fix}\left(L^{\mathrm{gyr}}\left(G\right)\right)\right)}$. By Equation (20), $\dim \left(\mathrm{Fix}\left(L^{\mathrm{gyr}}\left(G\right)\right)\right)=1$ and so ${\displaystyle \sum _{a\in G}^{}tr\rho \left(a\right)=\left|G\right|}$, as claimed.  □

Definition 1

(Gyrogroup characters). Let $(V,\phi )$ be a linear representation of a finite gyrogroup G of finite degree over $\mathbb{C}$. The character of φ is a function $\chi :G\to \mathbb{C}$ defined by

$$\chi \left(a\right)=tr\phi \left(a\right),a\in G.$$

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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 $L^{\mathrm{gyr}}\left(G\right)$.

Proof. 

Suppose that $\phi :G\to \mathrm{GL}\left(V\right)$ is a linear representation. Let $a,x,y,z\in G$. Using Proposition 32 (3) of [6], we obtain

$$\begin{array}{cc}\hfill \chi (a\oplus \mathrm{gyr}\left[x,y\right]z)& =tr\phi (a\oplus \mathrm{gyr}\left[x,y\right]z)\hfill \\ \hfill & =tr\left(\phi \right(a)\circ \mathrm{gyr}\left[\phi \left(x\right),\phi \left(y\right)\right]\phi \left(z\right))\hfill \\ \hfill & \stackrel{(\star )}{=}tr(\phi \left(a\right)\circ \phi \left(z\right))\hfill \\ \hfill & =tr\phi (a\oplus z)\hfill \\ \hfill & =\chi (a\oplus z),\hfill \end{array}$$

noting that (⋆) holds since any gyroautomorphism of $\mathrm{GL}\left(V\right)$ is the identity map on V. This proves $\chi \in L^{\mathrm{gyr}}\left(G\right)$.  □

The structure of $L^{\mathrm{gyr}}\left(G\right)$ does depend on the kernel of $\sigma$. In fact, knowing the structure of $ker\sigma$ amounts to knowing the structure of $L^{\mathrm{gyr}}\left(G\right)$ (cf. Theorem 8). Note that $ker\sigma$ is a proper invariant subspace of $L^{\mathrm{gyr}}\left(G\right)$. More precisely, if ${\chi }_{\rho }$ denotes the character of the left regular representation of G on $L^{\mathrm{gyr}}\left(G\right)$, then by Theorem 10,

$${\displaystyle \sigma \left({\chi }_{\rho }\right)=\sum _{a\in G}^{}{\chi }_{\rho }\left(a\right)=\sum _{a\in G}^{}tr\rho \left(a\right)=\left|G\right|\ne 0.}$$

Hence, ${\chi }_{\rho }$ is in $L^{\mathrm{gyr}}\left(G\right)$ but not in $ker\sigma$. 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 $ker\sigma$ is trivial.

Proof. 

If $\rho$ is irreducible, then the only invariant subspaces of $L^{\mathrm{gyr}}\left(G\right)$ are $\left\{0\right\}$ and $L^{\mathrm{gyr}}\left(G\right)$ itself. As mentioned previously, $ker\sigma$ is an invariant subspace of codimension 1 in $L^{\mathrm{gyr}}\left(G\right)$ and so $ker\sigma =\left\{0\right\}$. Conversely, if $ker\sigma =\left\{0\right\}$, then by Theorem 8, $L^{\mathrm{gyr}}\left(G\right)=\langle f_1\rangle$, which is a one-dimensional subspace. Thus, $\rho$ must be irreducible.  □