Triply Transitive Natural Permutation Images of Idempotent Evolution Algebras

1. Introduction

An evolution algebra over a field $\mathbb{k}$ is a commutative, generally nonassociative algebra admitting a basis

$$\mathcal{B}=\{b_1,\dots ,b_n\}$$

such that $b_i{b}_j=0$ whenever $i\ne j$. Writing

$$b_i^2={\displaystyle \sum _{j=1}^n}{\mu }_{ij}{b}_j$$

defines the structure matrix $M_{\mathcal{B}}=\left({\mu }_{ij}\right)$. The algebra is called idempotent (also perfect or regular in parts of the literature) when $X^2=X$; in finite dimension this is equivalent to the invertibility of $M_{\mathcal{B}}$ ([7] Proposition 4.2). We write

$${\mu }_3\left(\mathbb{k}\right)=\{z\in {\mathbb{k}}^{\times }:z^3=1\}.$$

For an idempotent evolution algebra, natural bases are rigid up to a permutation and nonzero coordinate rescalings ([4] Theorem 2.4). Hence every automorphism has the form

$$\phi \left(b_i\right)={\lambda }_i{b}_{\sigma \left(i\right)},{\lambda }_i\in {\mathbb{k}}^{\times },\sigma \in S_n,$$

and one obtains a natural permutation representation

$${\rho }_X:Aut\left(X\right)⟶S_n,$$

well defined up to conjugacy in $S_n$. Compatibility with multiplication is equivalent to

$${\lambda }_j{\mu }_{ij}={\lambda }_i^2{\mu }_{\sigma \left(i\right)\sigma \left(j\right)}(1\le i,j\le n).$$

(1)

This identity is the source of the additional rigidity used below.

The automorphism groups of idempotent evolution algebras were studied earlier by Elduque–Labra and by Sriwongsa–Zou. In particular, Sriwongsa and Zou classified the n-dimensional idempotent evolution algebras whose full automorphism group is isomorphic to $S_n$ ([14] Theorem 3.2). Costoya, Mayorga, and Viruel subsequently studied the natural permutation representation ${\rho }_X$ [4]. Their Theorem A shows that, for $n\ge 3$, 2-transitivity of ${\rho }_X(Aut\left(X\right))$ already implies that ${\rho }_X$ is faithful, whereas, for $n\ge 5$, 4-transitivity forces

$${\rho }_X(Aut\left(X\right))=S_n.$$

What remains is the proper 3-transitive case. We classify it here.

The first step is to isolate the extra constraint coming from the evolution-algebra product. Once the permutation image is proper and 2-transitive, the results of [4] force zero diagonal and nonzero off-diagonal structure constants. They also imply uniqueness of the scalar lift of each realized permutation. If G is a 2-transitive subgroup of the image, this gives scalar functions ${\lambda }_i\left(g\right)$ satisfying

$${\lambda }_i\left(gh\right)={\lambda }_i\left(h\right){\lambda }_{h\left(i\right)}\left(g\right).$$

For any point $\alpha$, the assignment

$${\chi }_{\alpha }\left(h\right)={\lambda }_{\alpha }\left(h\right),h\in G_{\alpha },$$

defines a linear character of the point stabilizer. Stabilizer characters are classical in the theory of monomial covers; Goldberger and Dula develop this viewpoint systematically in cohomological language [10]. A related conjugation test appears in the theory of doubly transitive lines: for the projective ${PSL}_2\left(q\right)$ action, Iverson and Mixon compare a point-stabilizer character with its transform under inversion ([11] proof of Theorem 6.4). The additional constraint in the present setting is the exponent 2 imposed by the square in (1); both the stabilizer-character formalism and the use of inversion have precedents.

If $\beta =s\alpha$ and k fixes both $\alpha$ and $\beta$, then a direct comparison of the scalar cocycle with (1) yields

$${\chi }_{\alpha }\left(s^{-1}ks\right)={\chi }_{\alpha }{\left(k\right)}^2.$$

(2)

We call (2) the quadratic compatibility relation. This leads to a useful obstruction. If a 2-transitive subgroup has point stabilizer with no nontrivial ${\mathbb{k}}^{\times }$-valued character, a diagonal rescaling eliminates all scalar factors. Then 2-transitivity forces the structure matrix to be a scalar multiple of $J_n-I_n$, where $J_n$ and $I_n$ denote the $n\times n$ all-ones and identity matrices. Consequently the full symmetric group acts. Combining this obstruction with the classification of finite 3-transitive groups reduces the proper case to the projective family with socle ${PSL}_2\left(q\right)$.

In the projective case, the relation becomes decisive. On ${\mathbf{P}}^1\left({\mathbb{F}}_q\right)$ take $\alpha =\infty$, $\beta =0$, and $s:x\mapsto -1/x$. For a two-point stabilizer element $m_r:x\mapsto rx$ we have $s^{-1}{m}_{r}s=m_{r^{-1}}$. Hence (2) gives

$$\chi {\left(m_r\right)}^{-1}=\chi {\left(m_r\right)}^2,$$

and therefore $\chi {\left(m_r\right)}^3=1$. The character cannot be trivial, so a nontrivial cubic character is forced. In particular, $q\equiv 1(mod3)$ and the ground field must contain a primitive cube root of unity.

After these basis normalizations, the character determines the multiplication. Let $\theta :{\mathbb{F}}_q^{\times }\to {\mu }_3\left(\mathbb{k}\right)$ be the resulting nontrivial cubic character. For $x\in {\mathbb{F}}_q$ set $v_x={(x,1)}^T$ and $v_{\infty }={(1,0)}^T$, and define

$$c_{xx}=0,c_{xy}=\theta \left(\det (v_x,v_y)\right)(x\ne y).$$

(3)

After a diagonal change of natural basis and one global scalar normalization, every algebra in the proper 3-transitive case has structure matrix $C_{\theta }=\left(c_{xy}\right)$. Thus the automorphism equation forces the cubic-character form.

The underlying $C_3$-phase pattern is classical: it is the $d=1$ generalized Paley/projective-space pattern. Its standard complex realization is the phase-three generalized weighing matrix $GW(q+1,q;3)$, while $C_{\theta }$ is the corresponding $\mathbb{k}$-valued realization obtained from an identification $C_3\cong {\mu }_3\left(\mathbb{k}\right)$, up to the usual choices of projective representatives and diagonal normalization. This finite-geometric family already appears in Berman’s 1970s work; see [2] and the references there. Later treatments include Jungnickel–Tonchev [12] and the cohomological reconstruction of Goldberger–Dula ([10] Theorem 6.1 and Remark 6.2). In their complex 2-transitive census, Egan, Ó Catháin, and Švob list one classical equivalence class for each of the parameter sets $\mathrm{CGW}(5,4;3)$ and $\mathrm{CGW}(17,16;3)$ and record the corresponding matrix-theoretic monomial automorphism data ([6] Theorem 3.1 and Table 2). In the complex conference-matrix setting, Et-Taoui proves that the matrices attached to $\theta$ and its Frobenius twists are permutation equivalent [9]. These matrix-theoretic notions of automorphism and equivalence differ from the evolution-algebra condition (1). Accordingly, we do not claim novelty for the underlying phase pattern, its projective monomial symmetry, or the forward Frobenius equivalence. The additional rigidity supplied by (1) forces a cubic character under proper 3-transitivity; at the level of algebra isomorphism, it also forces two such characters to lie in the same Frobenius orbit.

A direct finite-field calculation gives

$$C_{\theta }{C}_{{\theta }^{-1}}=qI.$$

Hence $C_{\theta }$ is invertible over $\mathbb{k}$ exactly when $char\mathbb{k}\ne p$, where $q=p^m$. We then determine the full automorphism group. The group ${PGL}_2\left(q\right)$ always lifts to algebra automorphisms, and the only additional permutations are the Frobenius automorphisms preserving $\theta$. Thus

$$Aut\left(X_{\theta }\right)\cong {PGL}_2\left(q\right)⋊Gal{({\mathbb{F}}_q/{\mathbb{F}}_p)}_{\theta }.$$

Finally, using the monomial rigidity of natural bases ([7] Theorem 4.4) and the standard normalizer of ${PSL}_2\left(q\right)$ in the symmetric group ([5] Sec. 7.7, p. 245), we prove that, for every nontrivial cubic character $\eta :{\mathbb{F}}_q^{\times }\to {\mu }_3\left(\mathbb{k}\right)$,

$$X_{\theta }\cong X_{\eta }⟺\theta \mathrm{and}\eta \mathrm{are}\mathrm{in}\mathrm{the}\mathrm{same}\mathrm{Frobenius}\mathrm{orbit}.$$

The main classification is as follows.

Theorem 

(Main theorem). Let X be an n-dimensional idempotent evolution algebra over a field $\mathbb{k}$, with $n\ge 5$, and suppose that

$$G_X:={\rho }_X(Aut\left(X\right))\le S_n,G_X\ne S_n$$

is 3-transitive. Then there is a prime power $q=p^m$ such that

$$n=q+1,q\equiv 1(mod3),$$

the field $\mathbb{k}$ contains a nontrivial cube root of unity, and $char\mathbb{k}\ne p$. Moreover, X is isomorphic to a cubic-character algebra $X_{\theta }$ with structure matrix (3), for a nontrivial character

$$\theta :{\mathbb{F}}_q^{\times }⟶{\mu }_3\left(\mathbb{k}\right).$$

Conversely, every such $X_{\theta }$ is idempotent and has a proper 3-transitive natural permutation image. More precisely,

$$Aut\left(X_{\theta }\right)\cong {PGL}_2\left(q\right)⋊Gal{({\mathbb{F}}_q/{\mathbb{F}}_p)}_{\theta },$$

and, for every nontrivial cubic character $\eta :{\mathbb{F}}_q^{\times }\to {\mu }_3\left(\mathbb{k}\right)$,

$$X_{\theta }\cong X_{\eta }⟺\eta ={\theta }^{\left(p^r\right)}\mathrm{for}\mathrm{some}0\le r<m,$$

where ${\theta }^{\left(p^r\right)}\left(a\right)=\theta \left(a^{p^r}\right)$.

The case $q=4$ is particularly concrete. Since ${PGL}_2\left(4\right)={PSL}_2\left(4\right)\cong A_5$, the construction yields a unique five-dimensional isomorphism class with automorphism group $A_5$. Moreover, dimension 5 is minimal among all idempotent evolution algebras with automorphism group $A_5$.

The paper is organized as follows. sec:prelim records the precise evolution-algebra input. sec:characters develops stabilizer characters, quadratic compatibility, and the perfect-stabilizer obstruction. sec:reduction uses the classification of finite 3-transitive groups to reduce to projective type. sec:projective proves cubic forcing and reconstructs the structure matrix. sec:construction proves invertibility, constructs the projective automorphisms, and determines the full automorphism group. sec:isomorphism determines isomorphism classes and treats the minimal $A_5$ case.

2. Preliminaries on Idempotent Evolution Algebras

Throughout, $\mathbb{k}$ is a field and all algebras are finite-dimensional over $\mathbb{k}$.

Definition  

Anevolution algebrais a $\mathbb{k}$-algebra X admitting a basis $\mathcal{B}=\{b_1,\dots ,b_n\}$ such that

$$b_i{b}_j=0(i\ne j).$$

Such a basis is called anatural basis. Its structure matrix $M_{\mathcal{B}}=\left({\mu }_{ij}\right)$ is defined by

$$b_i^2={\displaystyle \sum _{j=1}^n}{\mu }_{ij}{b}_j.$$

The algebra isidempotentif $X^2=X$.

For finite-dimensional evolution algebras, idempotence is equivalent to invertibility of the structure matrix; see ([7] Proposition 4.2). We will also use the rigidity of natural bases.

Theorem 

(Elduque–Labra). Let X be an idempotent evolution algebra, and let $\mathcal{B}=\{b_1,\dots ,b_n\}$ and ${\mathcal{B}}^{\prime }=\{b_1^{\prime },\dots ,b_n^{\prime }\}$ be natural bases. Then there are a permutation $\sigma \in S_n$ and scalars $d_i\in {\mathbb{k}}^{\times }$ such that

$$b_i^{\prime }=d_i{b}_{\sigma \left(i\right)}.$$

This is ([7] Theorem 4.4), restated in ([4] Theorem 2.4). Consequently every automorphism is monomial in a natural basis.

Lemma 

(Automorphism equation). Let X be an evolution algebra with natural basis $\mathcal{B}=\{b_1,\dots ,b_n\}$ and structure matrix $\left({\mu }_{ij}\right)$. If

$$\phi \left(b_i\right)={\lambda }_i{b}_{\sigma \left(i\right)}$$

with ${\lambda }_i\in {\mathbb{k}}^{\times }$ and $\sigma \in S_n$, then φ is an automorphism if and only if

$${\lambda }_j{\mu }_{ij}={\lambda }_i^2{\mu }_{\sigma \left(i\right)\sigma \left(j\right)}(1\le i,j\le n).$$

(4)

Proof. 

The map $\phi$ is invertible and automatically preserves all products $b_i{b}_j=0$ with $i\ne j$. For each $i,j$, comparing the coefficient of $b_{\sigma \left(j\right)}$ in $\phi \left(b_i^2\right)$ with that in $\phi {\left(b_i\right)}^2$ gives exactly (4). Thus these identities are also sufficient. In the idempotent setting this is Eq. (3) of [4]. □

The permutation $\sigma$ in lem:automorphism-equation defines the natural representation

$${\rho }_X:Aut\left(X\right)⟶S_n.$$

Changing the natural basis conjugates this representation by an element of $S_n$.

For later use, we collect the consequences of Costoya–Mayorga–Viruel that recur throughout the paper.

Proposition 

(Costoya–Mayorga–Viruel). Let X be an n-dimensional idempotent evolution algebra with $n\ge 3$, and put $G_X={\rho }_X(Aut\left(X\right))$.

label=()

If $G_X$ is 2-transitive, then either all off-diagonal structure constants vanish or all of them are nonzero.

lbbel=()

If $G_X$ is transitive, then either all diagonal structure constants vanish or all of them are nonzero.

lcbel=()

If $G_X$ is 2-transitive and some diagonal structure constant is nonzero, then $G_X=S_n$.

ldbel=()

If $G_X$ is 2-transitive, then ${\rho }_X$ is faithful. In the case of nonzero off-diagonal support, the scalars ${\lambda }_i$ lifting a given realized permutation are uniquely determined.

lebel=()

If $n\ge 5$ and $G_X$ is 4-transitive, then $G_X=S_n$.

Proof. 

Items (i) and (ii) are Lemma 4.1 of [4]; item (iii) is Theorem 4.6 there. The faithfulness assertion in (iv) is Theorem A(i), while uniqueness of the scalar lift in the nonzero off-diagonal case follows from Lemma 4.3. Item (v) is Theorem A(ii). □

Corollary . 

Suppose $n\ge 3$, $G_X$ is 2-transitive, and $G_X\ne S_n$. Then, in every natural basis,

$${\mu }_{ii}=0\mathrm{for}\mathrm{all}i,{\mu }_{ij}\ne 0\mathrm{for}\mathrm{all}i\ne j.$$

Moreover, ${\rho }_X$ is faithful and every $\sigma \in G_X$ has a unique scalar lift.

Proof. 

If some diagonal coefficient were nonzero, prop:CMV(iii) would give $G_X=S_n$. Hence all diagonal coefficients vanish. If all off-diagonal coefficients also vanished, the structure matrix would be zero, contradicting idempotence. Thus prop:CMV(i) gives nonzero off-diagonal support. The final two assertions follow from prop:CMV(iv). □

3. Stabilizer Characters and Quadratic Compatibility

Let X satisfy the hypotheses of cor:proper-support, and let $G\le G_X$ be a 2-transitive subgroup. Because the scalar lift of each realized permutation is unique, for each $g\in G$ there is a uniquely determined automorphism

$$\widehat{g}\left(b_i\right)={\lambda }_i\left(g\right)b_{g\left(i\right)}.$$

Lemma 

(Scalar cocycle). For $g,h\in G$ and every i,

$${\lambda }_i\left(gh\right)={\lambda }_i\left(h\right){\lambda }_{h\left(i\right)}\left(g\right).$$

(5)

Fix α. Then

$${\chi }_{\alpha }:G_{\alpha }⟶{\mathbb{k}}^{\times },{\chi }_{\alpha }\left(h\right)={\lambda }_{\alpha }\left(h\right),$$

is a group homomorphism.

Proof. 

The composition $\widehat{g}\widehat{h}$ is a lift of $gh$. By uniqueness of lifts it equals $\widehat{gh}$, which gives (5). If $g,h\in G_{\alpha }$, then $h\left(\alpha \right)=\alpha$, so (5) becomes ${\chi }_{\alpha }\left(gh\right)={\chi }_{\alpha }\left(h\right){\chi }_{\alpha }\left(g\right)$. Since the target is abelian, this is the homomorphism law. □

This point-stabilizer character is part of the general monomial-cover formalism; compare ([10] Theorem 3.4). The next relation records the additional constraint imposed by the evolution-algebra square.

Proposition 

(Quadratic compatibility). Let $\alpha ,\beta$ be distinct points, choose $s\in G$ with $s\left(\alpha \right)=\beta$, and let $k\in G_{\alpha ,\beta }$. Then

$${\chi }_{\alpha }\left(s^{-1}ks\right)={\chi }_{\alpha }{\left(k\right)}^2.$$

(6)

Proof. 

Since k fixes $\beta =s\left(\alpha \right)$, the element $s^{-1}ks$ fixes $\alpha$. Using

$$ks=s\left(s^{-1}ks\right)$$

and applying (5) to both sides gives

$${\lambda }_{\alpha }\left(s\right){\lambda }_{\beta }\left(k\right)={\lambda }_{\alpha }\left(ks\right)={\lambda }_{\alpha }\left(s\left(s^{-1}ks\right)\right)={\chi }_{\alpha }\left(s^{-1}ks\right){\lambda }_{\alpha }\left(s\right).$$

Thus

$${\lambda }_{\beta }\left(k\right)={\chi }_{\alpha }\left(s^{-1}ks\right).$$

(7)

On the other hand, k fixes both $\alpha$ and $\beta$. Applying (4) to $(i,j)=(\alpha ,\beta )$ and using ${\mu }_{\alpha \beta }\ne 0$ yields

$${\lambda }_{\beta }\left(k\right)={\lambda }_{\alpha }{\left(k\right)}^2={\chi }_{\alpha }{\left(k\right)}^2.$$

Together with (7) this proves (6). □

Proposition 

(Stabilizer obstruction). Let $K\le G_X$ be 2-transitive, and suppose $G_X\ne S_n$. Then, for every point α, the induced stabilizer character

$${\chi }_{\alpha }:K_{\alpha }⟶{\mathbb{k}}^{\times }$$

from lem:cocycle is nontrivial. Consequently

$$Hom(K_{\alpha },{\mathbb{k}}^{\times })\ne 1.$$

In particular, a 2-transitive subgroup whose point stabilizer is perfect cannot be contained in a proper natural permutation image.

Proof. 

Suppose for contradiction that ${\chi }_{\alpha }$ is trivial.

For each point i, choose $t_i\in K$ with $t_i\left(\alpha \right)=i$ and set

$$d_i={\lambda }_{\alpha }\left(t_i\right).$$

The value $d_i$ is independent of the choice of $t_i$: if $t_i^{\prime }=t_{i}h$ with $h\in K_{\alpha }$, then (5) and ${\chi }_{\alpha }\left(h\right)=1$ give ${\lambda }_{\alpha }\left(t_i^{\prime }\right)={\lambda }_{\alpha }\left(t_i\right)$.

For $g\in K$, the elements $g{t}_i$ and $t_{g\left(i\right)}$ send $\alpha$ to the same point. Hence, again using the trivial stabilizer character,

$$d_{g\left(i\right)}={\lambda }_{\alpha }\left(g{t}_i\right)=d_i{\lambda }_i\left(g\right).$$

Replace the natural basis by $b_i^{\prime }=d_i{b}_i$. Then

$$\widehat{g}\left(b_i^{\prime }\right)=b_{g\left(i\right)}^{\prime }$$

for every $g\in K$, so all scalar factors are 1. Therefore the new structure constants satisfy

$${\mu }_{g\left(i\right)g\left(j\right)}^{\prime }={\mu }_{ij}^{\prime }.$$

Since K is 2-transitive and the diagonal is zero by cor:proper-support, there is a scalar $c\in {\mathbb{k}}^{\times }$ such that

$$M_{{\mathcal{B}}^{\prime }}=c(J_n-I_n).$$

Every permutation in $S_n$ preserves this matrix with scalar lift 1, so $G_X=S_n$, a contradiction. □

The zero-diagonal matrix $c(J_n-I_n)$ that appears in the proof is the symmetric-group family already present in the classification of Sriwongsa–Zou ([14] Theorem 3.2). Here it arises for a different purpose: as the obstruction produced by a trivial point-stabilizer character inside a 2-transitive subgroup.

4. Reduction to Projective Type

We now assume $n\ge 5$ and that

$$G_X\le S_n,G_X\ne S_n$$

is 3-transitive. By prop:CMV(v), $G_X$ is not 4-transitive.

We use the standard classification of finite 3-transitive permutation groups; see, for example, Cameron [3], Dixon–Mortimer ([5] Sec. 7.7), and the modern summary in Pantangi [13]. The affine cases are $AGL(d,2)$ and the exceptional group $2^4:A_7$; the almost simple cases have socle of alternating, projective, or Mathieu type. Here it is important to distinguish the permutation actions from the abstract group isomorphism types, because of the familiar exceptional isomorphisms $A_5\cong {PSL}_2\left(4\right)\cong {PSL}_2\left(5\right)$ and $A_6\cong {PSL}_2\left(9\right)$.

Theorem 

(Projective reduction). Under the preceding assumptions, there is a prime power $q\ge 4$ such that

$$n=q+1,$$

and, after conjugating the permutation action in $S_n$, the group $G_X$ is almost simple with socle ${PSL}_2\left(q\right)$ in the standard action on ${\mathbf{P}}^1\left({\mathbb{F}}_q\right)$.

Proof. 

We exclude the nonprojective cases.

In the alternating branch, the socle is $A_n$ in its natural degree-n action. If $n\ge 6$, then $A_n$ is at least 4-transitive, contrary to prop:CMV(v) and the assumption $G_X\ne S_n$. In degree 5, the natural action of $A_5$ is permutation-isomorphic to the projective-line action of ${PSL}_2\left(4\right)$, so it belongs to the projective branch. The exceptional abstract isomorphism $A_6\cong {PSL}_2\left(9\right)$ does not identify the natural degree-6 action with the degree-10 projective action; the former remains excluded here by 4-transitivity.

Suppose $G_X$ is affine. If $G_X=AGL(d,2)$, then $n=2^d$. Since $n\ge 5$, one has $d\ge 3$, and a point stabilizer is

$$GL(d,2)=PSL(d,2),$$

a nonabelian simple, hence perfect, group. This contradicts prop:stabilizer-obstruction. In the exceptional affine case $2^4:A_7$, the point stabilizer is $A_7$, again perfect, and the same contradiction applies.

The 4- and 5-transitive Mathieu actions are excluded by prop:CMV(v). The remaining 3-transitive Mathieu actions are the degree-12 action of $M_{11}$ and the degree-22 actions of $M_{22}$ and $Aut\left(M_{22}\right)$. The ATLAS records the point stabilizer in the degree-12 action of $M_{11}$ as ${PSL}_2\left(11\right)$, and the point stabilizer in the degree-22 action of $M_{22}$ as ${PSL}_3\left(4\right)$ ([1] ATLAS representation M11G1-p12B0 and subgroup program M22G1-max1W1); both are perfect. Thus prop:stabilizer-obstruction excludes $M_{11}$ and $M_{22}$. The group $Aut\left(M_{22}\right)$ contains the normal 3-transitive subgroup $M_{22}$, so it is excluded by applying the same obstruction to that subgroup.

The only remaining case in the classification is projective type with socle ${PSL}_2\left(q\right)$, acting on $q+1$ points. This socle formulation also covers the exceptional projective-type overgroups occurring in the standard classification; no assertion that the whole group lies in $\mathrm{P}\Gamma {\mathrm{L}}_2\left(q\right)$ is needed at this stage. □

5. Cubic Characters and Reconstruction

By thm:projective-reduction, after relabeling the natural basis we may identify the index set with

$$\Omega ={\mathbf{P}}^1\left({\mathbb{F}}_q\right)={\mathbb{F}}_q\cup \{\infty \}$$

and regard

$$L:={PSL}_2\left(q\right)\le G_X$$

as acting on $\Omega$ in the standard projective way. The group L is 2-transitive for every $q\ge 4$.

Let $B=L_{\infty }$. Write U for the translation subgroup. Thus

$$U=\{x\mapsto x+b:b\in {\mathbb{F}}_q\}.$$

The multiplier subgroup is

$$R_q=\left\{\begin{array}{cc}{\mathbb{F}}_q^{\times },\hfill & q\mathrm{even},\hfill \\ {\left({\mathbb{F}}_q^{\times }\right)}^2,\hfill & q\mathrm{odd},\hfill \end{array}\right.$$

and

$$B=U⋊R_q,$$

where $r\in R_q$ acts as $m_r:x\mapsto rx$. Let $\chi :B\to {\mathbb{k}}^{\times }$ be the stabilizer character from lem:cocycle.

Lemma  

One has $U\le [B,B]$. Consequently χ is trivial on U and is determined by its restriction to $R_q$.

Proof. 

Choose $r\in R_q$ with $r\ne 1$. Such an r exists for $q\ge 4$. If $u_c:x\mapsto x+c$, then

$$m_r{u}_c{m}_r^{-1}{u}_c^{-1}=u_{(r-1)c}.$$

Since multiplication by $r-1$ is a bijection of the additive group ${\mathbb{F}}_q$, every translation is a commutator. □

Proposition 

(Cubic forcing). The character χ is nontrivial and has image of order 3. Consequently

$$q\equiv 1(mod3),$$

and $\mathbb{k}$ contains a nontrivial cube root of unity.

Proof. 

By prop:stabilizer-obstruction, applied to the 2-transitive subgroup L, the character $\chi$ is nontrivial.

Set $\alpha =\infty$ and $\beta =0$. The projective transformation

$$s:x⟼-1/x$$

belongs to L and interchanges 0 and ∞. For $r\in R_q$,

$$s^{-1}{m}_{r}s=m_{r^{-1}}.$$

By quadratic compatibility,

$$\chi \left(m_{r^{-1}}\right)=\chi {\left(m_r\right)}^2.$$

Since $\chi$ is a character,

$$\chi \left(m_{r^{-1}}\right)=\chi {\left(m_r\right)}^{-1}.$$

Thus

$$\chi {\left(m_r\right)}^3=1$$

for every $r\in R_q$. The character is nontrivial, so its image is the full group of third roots of unity in $\mathbb{k}$.

If q is even, $|R_q|=q-1$, hence $3\mid (q-1)$. If q is odd, $|R_q|=(q-1)/2$, hence $3\mid (q-1)/2$. In either case $q\equiv 1(mod3)$. □

Lemma 

(Extension of the cubic character). There is a unique nontrivial character

$$\theta :{\mathbb{F}}_q^{\times }⟶{\mu }_3\left(\mathbb{k}\right)$$

whose restriction to $R_q$ equals ${\chi |}_{R_q}$.

Proof. 

If q is even, $R_q={\mathbb{F}}_q^{\times }$ and there is nothing to prove. Suppose q is odd. Then $R_q={\left({\mathbb{F}}_q^{\times }\right)}^2$ has index 2. Since ${\mathbb{F}}_q^{\times }$ is cyclic and the squaring map is an automorphism of the cyclic group ${\mu }_3\left(\mathbb{k}\right)$, restriction induces a bijection

$$Hom({\mathbb{F}}_q^{\times },{\mu }_3\left(\mathbb{k}\right))⟶Hom(R_q,{\mu }_3\left(\mathbb{k}\right)).$$

Thus the extension exists and is unique. □

Extend both $\theta$ and ${\theta }^{-1}$ to ${\mathbb{F}}_q$ by setting $\theta \left(0\right)={\theta }^{-1}\left(0\right)=0$. Since $\theta$ has odd order, $\theta (-1)=1$. For $x\in {\mathbb{F}}_q$ put

$$v_x=\left({\displaystyle \genfrac{}{}{0pt}{}{x}{1}}\right),v_{\infty }=\left({\displaystyle \genfrac{}{}{0pt}{}{1}{0}}\right),$$

and define the $(q+1)\times (q+1)$ matrix

$$C_{\theta }={\left(c_{xy}\right)}_{x,y\in \Omega },c_{xx}=0,c_{xy}=\theta \left(\det (v_x,v_y)\right)(x\ne y).$$

(8)

In particular,

$$c_{xy}=\theta (x-y)(x,y\in {\mathbb{F}}_q,x\ne y),c_{x\infty }=c_{\infty x}=1.$$

Let $E_{\theta }$ be the evolution algebra over $\mathbb{k}$ with natural basis

$${\mathcal{E}}_{\theta }=\{e_x:x\in \Omega \}$$

and structure matrix $C_{\theta }$. No idempotence assumption is made at this stage.

Lemma 

(Standard projective lift). Let $g\in L$ and choose $A\in {SL}_2\left(q\right)$ representing g. For every $x\in \Omega$, write

$$A{v}_x=a_x{v}_{g\left(x\right)},a_x\in {\mathbb{F}}_q^{\times },$$

and put

$${\nu }_x\left(g\right)=\theta {\left(a_x\right)}^{-1}.$$

Then

$${\nu }_y\left(g\right)c_{xy}={\nu }_x{\left(g\right)}^2{c}_{g\left(x\right)g\left(y\right)}$$

for all $x,y$. Hence

$$e_x⟼{\nu }_x\left(g\right)e_{g\left(x\right)}$$

defines an automorphism of $E_{\theta }$, whether or not $C_{\theta }$ is invertible. These automorphisms form a lift of the action of L, so the scalars ${\nu }_x\left(g\right)$ satisfy the cocycle identity (5). Moreover, the stabilizer character at ∞ is precisely χ.

Proof. 

The definition of ${\nu }_x\left(g\right)$ is independent of the chosen representative A: for odd q the two representatives in ${SL}_2\left(q\right)$ differ by $-I$, and $\theta (-1)=1$; for even q there is no ambiguity. Since $\det A=1$,

$$\det (v_{g\left(x\right)},v_{g\left(y\right)})=a_x^{-1}{a}_y^{-1}\det (v_x,v_y).$$

Therefore

$$c_{g\left(x\right)g\left(y\right)}=\theta {\left(a_x\right)}^{-1}\theta {\left(a_y\right)}^{-1}{c}_{xy}.$$

Since $\theta {\left(a_x\right)}^3=1$,

$${\nu }_x{\left(g\right)}^2{c}_{g\left(x\right)g\left(y\right)}=\theta {\left(a_x\right)}^{-2}\theta {\left(a_x\right)}^{-1}\theta {\left(a_y\right)}^{-1}{c}_{xy}=\theta {\left(a_y\right)}^{-1}{c}_{xy}={\nu }_y\left(g\right)c_{xy}.$$

The resulting monomial map is invertible, and the automorphism equation shows that it is an algebra automorphism; no nonsingularity assumption on $C_{\theta }$ is needed.

If $A_g,A_h$ represent $g,h\in L$, then $A_g{A}_h$ represents $gh$, up to the central factor $-I$ when q is odd. The independence just established gives

$${\nu }_x\left(gh\right)={\nu }_x\left(h\right){\nu }_{h\left(x\right)}\left(g\right),$$

so these automorphisms form a group lift of L.

Now suppose $g\in L_{\infty }$. We may represent it by

$$A=\left(\begin{array}{cc}a& b\\ 0& a^{-1}\end{array}\right).$$

Its affine multiplier is $r=a^2\in R_q$, while $a_{\infty }=a$. Hence

$${\nu }_{\infty }\left(g\right)=\theta {\left(a\right)}^{-1}=\theta {\left(a\right)}^2=\theta \left(a^2\right)=\theta \left(r\right)=\chi \left(g\right),$$

using the definition of $\theta$. □

Theorem 

(Reconstruction). Let X be as in thm:projective-reduction, and let θ be the cubic character obtained above. Then a diagonal change of natural basis, followed by a uniform rescaling, brings the structure matrix to $C_{\theta }$. Equivalently,

$$X\cong E_{\theta }.$$

Proof. 

Let ${\lambda }_x\left(g\right)$ be the scalar lift of $g\in L$ in X, and let ${\nu }_x\left(g\right)$ be the standard lift from lem:standard-L-lift. Define

$${\delta }_x\left(g\right)={\displaystyle \frac{{\lambda }_x\left(g\right)}{{\nu }_x\left(g\right)}}.$$

Both scalar systems satisfy the cocycle identity, so their quotient ${\delta }_x\left(g\right)$ also satisfies it. By lem:standard-L-lift, the two stabilizer characters at ∞ agree; hence

$${\delta }_{\infty }\left(h\right)=1(h\in L_{\infty }).$$

For each $x\in \Omega$, choose $t_x\in L$ with $t_x(\infty )=x$ and set

$$d_x={\delta }_{\infty }\left(t_x\right).$$

As in the proof of prop:stabilizer-obstruction, this is independent of the choice of $t_x$, and the cocycle identity gives

$$d_{g\left(x\right)}=d_x{\delta }_x\left(g\right).$$

Replace the natural basis by $b_x^{\prime }=d_x{b}_x$. In this basis, the scalar lift of every $g\in L$ is exactly ${\nu }_x\left(g\right)$.

Let $\left({\mu }_{xy}^{\prime }\right)$ be the resulting structure matrix. For $x\ne y$, divide the automorphism equation for $\left({\mu }_{xy}^{\prime }\right)$ by the corresponding equation for $C_{\theta }$, using the same scalar lift $\nu$. This gives

$${\displaystyle \frac{{\mu }_{g\left(x\right)g\left(y\right)}^{\prime }}{c_{g\left(x\right)g\left(y\right)}}}={\displaystyle \frac{{\mu }_{xy}^{\prime }}{c_{xy}}}.$$

The group L is 2-transitive on $\Omega$, so this quotient is constant, say $a\in {\mathbb{k}}^{\times }$, on all ordered pairs of distinct points. Both matrices have zero diagonal. Hence

$$M_{{\mathcal{B}}^{\prime }}=a{C}_{\theta }.$$

Finally, a uniform rescaling of the natural basis absorbs the scalar a: if $b_x^{\prime \prime }=a^{-1}{b}_x^{\prime }$, then the structure matrix becomes $C_{\theta }$. □

6. The Cubic-Character Algebras and Their Automorphism Groups

We now fix a prime power $q=p^m\equiv 1(mod3)$, a field $\mathbb{k}$ containing a nontrivial cube root of unity, and a nontrivial cubic character

$$\theta :{\mathbb{F}}_q^{\times }\to {\mu }_3\left(\mathbb{k}\right).$$

As above, extend both $\theta$ and ${\theta }^{-1}$ to ${\mathbb{F}}_q$ by setting $\theta \left(0\right)={\theta }^{-1}\left(0\right)=0$. Let $C_{\theta }$ be the matrix in (8), and let $E_{\theta }$ be the evolution algebra over $\mathbb{k}$ with natural basis ${\mathcal{E}}_{\theta }=\{e_x:x\in \Omega \}$ and structure matrix $C_{\theta }$.

Proposition 

(Character-sum inverse). The calculation below uses the finite-field character-sum identity that appears in Et-Taoui’s conference-matrix construction ([9] Theorem 1). In the present algebraic setting, this gives

$$C_{\theta }{C}_{{\theta }^{-1}}=q{I}_{q+1}.$$

(9)

Consequently, $C_{\theta }$ is invertible over $\mathbb{k}$ if and only if $char\mathbb{k}\ne p$.

Proof. 

Because $\theta$ is a nontrivial character of order 3, its three values occur equally often on ${\mathbb{F}}_q^{\times }$; hence

$$\sum _{t\in {\mathbb{F}}_q^{\times }}\theta \left(t\right)=0,\sum _{t\in {\mathbb{F}}_q^{\times }}{\theta }^{-1}\left(t\right)=0.$$

We compute the entries of the product. First let $x\in {\mathbb{F}}_q$. On the diagonal,

$${\left(C_{\theta }{C}_{{\theta }^{-1}}\right)}_{xx}=\sum _{z\in {\mathbb{F}}_q\setminus \left\{x\right\}}\theta (x-z){\theta }^{-1}(z-x)+1.$$

Because $\theta (-1)=1$, every summand in the finite sum is 1, giving $(q-1)+1=q$. The $\infty ,\infty$ entry is also q.

Now let $x,y\in {\mathbb{F}}_q$ with $x\ne y$. Put $d=x-y$ and write $z=y+du$. Then the finite part of the $(x,y)$ entry becomes

$$\sum _{u\in {\mathbb{F}}_q}\theta (1-u){\theta }^{-1}\left(u\right).$$

The terms at $u=0,1$ vanish. For the remaining terms,

$$\theta (1-u){\theta }^{-1}\left(u\right)=\theta (u^{-1}-1).$$

The map $u\mapsto u^{-1}-1$ bijects ${\mathbb{F}}_q\setminus \{0,1\}$ onto ${\mathbb{F}}_q\setminus \{0,-1\}$, so the sum equals

$$\sum _{t\in {\mathbb{F}}_q\setminus \{0,-1\}}\theta \left(t\right)=-\theta (-1)=-1.$$

The projective point $z=\infty$ contributes 1, so the off-diagonal entry is 0.

Finally, if $x\in {\mathbb{F}}_q$, then

$${\left(C_{\theta }{C}_{{\theta }^{-1}}\right)}_{x\infty }=\sum _{z\in {\mathbb{F}}_q\setminus \left\{x\right\}}\theta (x-z)=0,$$

and similarly for the transpose position. This proves (9).

If $char\mathbb{k}\ne p$, then $q\ne 0$ in $\mathbb{k}$ and (9) gives an inverse. If $char\mathbb{k}=p$, then the product in (9) is zero while $C_{{\theta }^{-1}}\ne 0$; hence $C_{\theta }$ cannot be invertible. □

Definition  

When $char\mathbb{k}\ne p$, write $X_{\theta }:=E_{\theta }$. By prop:inverse, this is an idempotent evolution algebra.

As noted in the Introduction, the abstract $C_3$-phase pattern of $C_{\theta }$ is the $d=1$ generalized Paley/projective-space pattern. Its standard complex realization is $GW(q+1,q;3)$, whereas $C_{\theta }$ is the corresponding $\mathbb{k}$-valued realization via an identification $C_3\cong {\mu }_3\left(\mathbb{k}\right)$; see [2,12] and ([10] Theorem 6.1 and Remark 6.2). The phase pattern itself is therefore classical; what the evolution-algebra square forces in the present setting is the cubic phase.

Proposition 

(Projective lift). The natural action of ${PGL}_2\left(q\right)$ on ${\mathbf{P}}^1\left({\mathbb{F}}_q\right)$ lifts to algebra automorphisms of $X_{\theta }$. In particular,

$${PGL}_2\left(q\right)\le {\rho }_{X_{\theta }}(Aut\left(X_{\theta }\right)).$$

Proof. 

Let $g\in {PGL}_2\left(q\right)$ and choose $A\in {GL}_2\left(q\right)$ representing it. Write

$$A{v}_x=a_x{v}_{g\left(x\right)}$$

with $a_x\in {\mathbb{F}}_q^{\times }$, and put

$$u_x=\theta \left(a_x\right),\delta =\theta \left(\det A\right).$$

Taking determinants gives

$$c_{g\left(x\right)g\left(y\right)}=\delta u_x^{-1}{u}_y^{-1}{c}_{xy}.$$

Define

$${\lambda }_x={\delta }^{-1}{u}_x^{-1}.$$

Since $u_x^3=1$,

$${\lambda }_x^2{c}_{g\left(x\right)g\left(y\right)}={\delta }^{-1}{u}_y^{-1}{c}_{xy}={\lambda }_y{c}_{xy}.$$

Thus (4) holds.

The lift is independent of the scalar representative of A. Indeed, replacing A by $cA$ multiplies $\delta$ by $\theta {\left(c\right)}^2$ and every $u_x$ by $\theta \left(c\right)$, so every ${\lambda }_x$ is multiplied by $\theta {\left(c\right)}^{-3}=1$. For the multiplier $m_r:x\mapsto rx$, the choice $A=diag(r,1)$ gives the scalar at $e_{\infty }$ as

$${\lambda }_{\infty }=\theta \left(r\right).$$

□

Let $F:x\mapsto x^p$ denote the Frobenius permutation of ${\mathbf{P}}^1\left({\mathbb{F}}_q\right)$, fixing ∞, and define

$$Gal{({\mathbb{F}}_q/{\mathbb{F}}_p)}_{\theta }=\{F^r:\theta \left(a^{p^r}\right)=\theta \left(a\right)\mathrm{for}\mathrm{all}a\in {\mathbb{F}}_q^{\times }\}.$$

Theorem 

(Full automorphism group). For every cubic-character algebra $X_{\theta }$,

$${\rho }_{X_{\theta }}(Aut\left(X_{\theta }\right))={PGL}_2\left(q\right)⋊Gal{({\mathbb{F}}_q/{\mathbb{F}}_p)}_{\theta }.$$

Since the natural representation is faithful,

$$Aut\left(X_{\theta }\right)\cong {PGL}_2\left(q\right)⋊Gal{({\mathbb{F}}_q/{\mathbb{F}}_p)}_{\theta }.$$

In particular,

$$Aut\left(X_{\theta }\right)\cong \left\{\begin{array}{cc}\mathrm{P}\Gamma {\mathrm{L}}_2\left(q\right),\hfill & p\equiv 1(mod3),\hfill \\ {PGL}_2\left(q\right)⋊\langle F^2\rangle ,\hfill & p\equiv 2(mod3).\hfill \end{array}\right.$$

In the second case m is even and $|\langle F^2\rangle |=m/2$.

Proof. 

Put

$$H={\rho }_{X_{\theta }}(Aut\left(X_{\theta }\right)).$$

By prop:PGL, H contains ${PGL}_2\left(q\right)$ and is therefore 3-transitive.

First, we show directly that $H\ne S_{q+1}$. Since every off-diagonal entry of $C_{\theta }$ is nonzero and H is 2-transitive, the scalar lift of each permutation in H is unique by prop:CMV(iv). Hence the scalar at $e_{\infty }$ defines a character

$${\psi }_{\infty }:H_{\infty }⟶{\mathbb{k}}^{\times }.$$

Restrict this character to $L_{\infty }$, where $L={PSL}_2\left(q\right)$. For $m_r\in L_{\infty ,0}$ with $r\in R_q$, the explicit lift in prop:PGL gives

$${\psi }_{\infty }\left(m_r\right)=\theta \left(r\right).$$

The restriction ${\theta |}_{R_q}$ is nontrivial: it is immediate when q is even, while for odd q the subgroup $R_q$ has index 2, which is coprime to the order 3 of $\theta$. Thus ${\psi }_{\infty }\left(m_r\right)$ has order 3 for some $r\in R_q$. If $H=S_{q+1}$, then $H_{\infty }\cong S_q$ with $q\ge 4$, and $S_q^{\mathrm{ab}}$ has exponent 2. Every character $S_q\to {\mathbb{k}}^{\times }$ therefore has image of exponent dividing 2, a contradiction.

Thus H is a proper 3-transitive group. Apply thm:projective-reduction to H. Its socle S is a copy of ${PSL}_2\left(q\right)$ in the projective-line action. Since the degree is $q+1$, the parameter is again q. Let

$$L={PSL}_2\left(q\right)\le {PGL}_2\left(q\right)\le H$$

be the standard subgroup already supplied by prop:PGL. Since H is almost simple with socle S, the quotient $H/S$ is a subgroup of the solvable outer automorphism group of S ([5] Sec. 7.7). The image of the perfect group L in $H/S$ is therefore trivial. Hence $L\le S$, and equality follows from $\left|L\right|=\left|S\right|$.

Now $S=L$ is normal in H. The standard normalizer formula for the projective-line action gives

$$N_{S_{q+1}}\left(L\right)=\mathrm{P}\Gamma {\mathrm{L}}_2\left(q\right)$$

([5] Sec. 7.7, p. 245). Therefore

$${PGL}_2\left(q\right)\le H\le \mathrm{P}\Gamma {\mathrm{L}}_2\left(q\right).$$

Since $\mathrm{P}\Gamma {\mathrm{L}}_2\left(q\right)={PGL}_2\left(q\right)⋊Gal({\mathbb{F}}_q/{\mathbb{F}}_p)$, every element of $\mathrm{P}\Gamma {\mathrm{L}}_2\left(q\right)$ can be written as $g{F}^r$ with $g\in {PGL}_2\left(q\right)$. Because ${PGL}_2\left(q\right)\le H$, one has $g{F}^r\in H$ if and only if $F^r\in H$.

It remains to decide which Frobenius powers lift. Suppose first that $F^r$ is the permutation part of an automorphism $\Phi$. Let $h\in L_{\infty }$. Uniqueness of lifts gives

$$\Phi \widehat{h}{\Phi }^{-1}=\widehat{F^{r}h{F}^{-r}}.$$

Because $F^r$ fixes ∞, the automorphism $\Phi$ preserves the line $\mathbb{k}{e}_{\infty }$ and acts on it by a scalar. Conjugation by $\Phi$ therefore leaves the stabilizer scalar unchanged, so

$${\psi }_{\infty }\left(F^{r}h{F}^{-r}\right)={\psi }_{\infty }\left(h\right).$$

For $h=m_a$ in the two-point stabilizer, the restriction computed above gives

$$\theta \left(a^{p^r}\right)=\theta \left(a\right)$$

on $R_q$. If q is odd, the restriction map for cubic characters is injective by lem:extension; hence the equality holds on all of ${\mathbb{F}}_q^{\times }$. Thus $F^r\in Gal{({\mathbb{F}}_q/{\mathbb{F}}_p)}_{\theta }$.

Conversely, if $F^r$ fixes $\theta$, then from (8)

$$c_{F^r\left(x\right),F^r\left(y\right)}=c_{xy}$$

for all $x,y$, so $F^r$ itself is a pure-permutation automorphism. This proves the group formula.

Finally, if $p\equiv 1(mod3)$, then every $p^r\equiv 1(mod3)$, so all Frobenius powers fix the two cubic characters. If $p\equiv 2(mod3)$, the condition $q=p^m\equiv 1(mod3)$ forces m even, and $p^r\equiv 1(mod3)$ exactly when r is even. Since H is 3-transitive, the faithfulness assertion follows from prop:CMV(iv). □

Combining thm:projective-reduction,thm:reconstruction,prop:inverse,thm:full-aut proves the existence and automorphism-group assertions of thm:main.

7. Isomorphism Classes and the Minimal ${\mathbf{A}}_{\mathbf{5}}$ Case

Isomorphisms between idempotent evolution algebras are monomial by ([7] Theorem 4.4); equivalently, the isomorphism classes can be described as orbits of the symmetric group and the diagonal torus on invertible structure matrices [15]. In the complex conference-matrix setting, Et-Taoui proved that Frobenius twists of multiplicative-character matrices are permutation equivalent [9]. That is a matrix-equivalence statement, whereas algebra isomorphisms here must satisfy the monomial change-of-natural-basis law imposed by the evolution product. The theorem below proves that isomorphic algebras in this family have defining characters in the same Frobenius orbit.

Theorem 

(Isomorphism classification). Let $q=p^m\equiv 1(mod3)$, let $\mathbb{k}$ be a field of characteristic different from p that contains a nontrivial cube root of unity, and let

$$\theta ,\eta :{\mathbb{F}}_q^{\times }⟶{\mu }_3\left(\mathbb{k}\right)$$

be nontrivial characters. Then

$$X_{\theta }\cong X_{\eta }⟺\eta ={\theta }^{\left(p^r\right)}\mathrm{for}\mathrm{some}0\le r<m,$$

where ${\theta }^{\left(p^r\right)}\left(a\right)=\theta \left(a^{p^r}\right)$. Consequently, the family has two $\mathbb{k}$-algebra isomorphism classes when $p\equiv 1(mod3)$, and one when $p\equiv 2(mod3)$.

Proof. 

Suppose $\Phi :X_{\theta }\to X_{\eta }$ is an isomorphism. By ([7] Theorem 4.4), the map $\Phi$ is monomial relative to natural bases. Its permutation part $\sigma \in S_{q+1}$ therefore conjugates the natural permutation image of $X_{\theta }$ to that of $X_{\eta }$. By thm:full-aut, both groups are almost simple with socle the standard subgroup

$$L={PSL}_2\left(q\right).$$

The socle is characteristic, hence

$$\sigma L{\sigma }^{-1}=L.$$

The standard normalizer formula

$$N_{S_{q+1}}\left(L\right)=\mathrm{P}\Gamma {\mathrm{L}}_2\left(q\right)$$

therefore gives $\sigma =g{F}^r$ for some $g\in {PGL}_2\left(q\right)$ and some Frobenius power $F^r$; see ([5] Sec. 7.7, p. 245). Composing $\Phi$ with an automorphism of $X_{\eta }$ having permutation part $g^{-1}$, we may assume $\sigma =F^r$.

Let ${\psi }_{\infty ,\theta }$ and ${\psi }_{\infty ,\eta }$ denote the point-stabilizer characters of $X_{\theta }$ and $X_{\eta }$, respectively, defined as in the proof of thm:full-aut. For $h\in L_{\infty }$, conjugating the unique lift of h by $\Phi$ shows that the stabilizer scalars satisfy

$${\psi }_{\infty ,\eta }\left(F^{r}h{F}^{-r}\right)={\psi }_{\infty ,\theta }\left(h\right).$$

On the multiplier subgroup, the computation in prop:PGL gives

$${\psi }_{\infty ,\theta }\left(m_a\right)=\theta \left(a\right),{\psi }_{\infty ,\eta }\left(m_a\right)=\eta \left(a\right).$$

Hence the preceding equality reads

$$\eta \left(a^{p^r}\right)=\theta \left(a\right)$$

for $a\in R_q$, and hence for all $a\in {\mathbb{F}}_q^{\times }$ by lem:extension. Equivalently,

$$\eta ={\theta }^{\left(p^s\right)}$$

for $s\equiv m-r(modm)$, so $\theta$ and $\eta$ lie in the same Frobenius orbit.

Conversely, suppose $\eta ={\theta }^{\left(p^s\right)}$. Choose r with $r+s\equiv 0(modm)$. Then $\eta \left(a^{p^r}\right)=\theta \left(a\right)$ for every $a\in {\mathbb{F}}_q^{\times }$, and the pure permutation $x\mapsto x^{p^r}$ satisfies

$$c_{F^r\left(x\right),F^r\left(y\right)}^{\eta }=c_{xy}^{\theta }.$$

Hence it defines an algebra isomorphism $X_{\theta }\to X_{\eta }$.

There are exactly two nontrivial cubic characters, $\theta$ and ${\theta }^{-1}$. If $p\equiv 1(mod3)$, Frobenius fixes both, giving two orbits. If $p\equiv 2(mod3)$, Frobenius interchanges them, giving one orbit. □

Together with thm:projective-reduction,thm:reconstruction,prop:inverse,thm:full-aut, this completes the proof of thm:main.

Corollary 

(Sharpness of the 4-transitivity threshold). There are infinitely many idempotent evolution algebras whose natural permutation image is proper and 3-transitive. Hence the 4-transitivity hypothesis in ([4] Theorem A(ii)) cannot, in general, be replaced by 3-transitivity.

Proof. 

Over any field that contains a primitive cube root of unity and has characteristic different from 2, take $q=4^r$ for $r\ge 1$. Then $q\equiv 1(mod3)$, so the algebras $X_{\theta }$ exist by prop:inverse; their natural permutation images contain the sharply 3-transitive group ${PGL}_2\left(q\right)$ and are proper by thm:full-aut. □

Corollary 

(The five-dimensional $A_5$ algebra). There exists a five-dimensional idempotent evolution algebra X over $\mathbb{k}$ with

$$Aut\left(X\right)\cong A_5$$

if and only if

$$char\mathbb{k}\ne 2\mathrm{and}\mathbb{k}\mathrm{contains}\mathrm{a}\mathrm{primitive}\mathrm{cube}\mathrm{root}\mathrm{of}\mathrm{unity}.$$

When these conditions hold, there is exactly one such five-dimensional algebra up to $\mathbb{k}$-algebra isomorphism. Moreover, dimension 5 is minimal among all idempotent evolution algebras with automorphism group $A_5$.

Proof. 

For existence, take $q=4$. Then

$${PGL}_2\left(4\right)={PSL}_2\left(4\right)\cong A_5.$$

The Frobenius automorphism of ${\mathbb{F}}_4/{\mathbb{F}}_2$ interchanges the two nontrivial cubic characters, so no nontrivial Frobenius power fixes a chosen $\theta$. Hence thm:full-aut gives

$$Aut\left(X_{\theta }\right)\cong A_5.$$

By thm:isomorphism, the two cubic characters lie in a single Frobenius orbit, so the five-dimensional algebra is unique up to isomorphism.

Conversely, suppose X is five-dimensional and $Aut\left(X\right)\cong A_5$. The kernel of ${\rho }_X$ is abelian; this follows, for example, from the diagonal-kernel exact sequence in ([8] Theorem 3.2) (see also [4]). Since $A_5$ is nonabelian simple, the kernel is trivial. Thus $A_5$ embeds into $S_5$ and its image is the standard alternating subgroup, which is 3-transitive. Applying thm:main gives $q=4$, $char\mathbb{k}\ne 2$, and the existence of a nontrivial cube root of unity in $\mathbb{k}$.

The same kernel argument works in every dimension: if $Aut\left(Y\right)\cong A_5$ for an idempotent evolution algebra Y of dimension d, then $A_5$ embeds in $S_d$. Since $|A_5|=60>|S_4|$, one must have $d\ge 5$. □

To make the five-dimensional case explicit, choose $\omega \in \mathbb{k}$ with ${\omega }^2+\omega +1=0$. With a suitable ordering of ${\mathbf{P}}^1\left({\mathbb{F}}_4\right)$, a representative of the unique isomorphism class in cor:A5 has the structure matrix

$$M_{A_5}=\left(\begin{array}{ccccc}0& 1& 1& {\omega }^2& \omega \\ 1& 0& 1& \omega & {\omega }^2\\ 1& 1& 0& 1& 1\\ {\omega }^2& \omega & 1& 0& 1\\ \omega & {\omega }^2& 1& 1& 0\end{array}\right).$$

(10)

A direct calculation gives

$$\det M_{A_5}=-32,$$

consistent with prop:inverse and the condition $char\mathbb{k}\ne 2$.

Remark  

The underlying $C_3$-phase pattern is classical. It is the phase-three generalized Paley/projective-space pattern. Its standard complex realization is $GW(q+1,q;3)$, while $C_{\theta }$ is the corresponding $\mathbb{k}$-valued realization via an identification $C_3\cong {\mu }_3\left(\mathbb{k}\right)$, up to the standard projective and diagonal normalizations. The finite-geometric family already appears in Berman’s 1970s work; see [2] and the references there. Later formulations are given by Jungnickel–Tonchev and Goldberger–Dula [10,12]; Goldberger–Dula also recover projective monomial lifts. In their complex 2-transitive census, Egan, Ó Catháin, and Švob list one classical equivalence class for each of the parameter sets $\mathrm{CGW}(5,4;3)$ and $\mathrm{CGW}(17,16;3)$, with underlying permutation groups $A_5$ and ${PSL}_2\left(16\right):C_2$, respectively ([6] Theorem 3.1 and Table 2). Frobenius twists were already known to give matrix equivalences in the complex conference setting [9]. Stabilizer characters and their conjugation constraints likewise have precedents in monomial-cover and doubly transitive line theories ([10] Theorem 3.4) ([11] proof of Theorem 6.4).

The cited matrix results concern matrix-theoretic monomial automorphisms or equivalences and do not involve the evolution-algebra automorphism relation (4), in which the source scalar appears quadratically. Accordingly, we do not claim novelty for the underlying phase pattern, its character-sum orthogonality, its projective monomial symmetry, the stabilizer-character framework, the corresponding small complex matrix classes, or the forward Frobenius equivalence. The additional ingredient in the present evolution-algebra setting is the exponent-2 compatibility forced by (4). It yields (6), forces a cubic character from proper 3-transitivity, reconstructs the evolution-algebra multiplication, determines the full evolution-algebra automorphism group, and shows that the defining characters of isomorphic algebras lie in the same Frobenius orbit.