# Medial and isospectral algebras

Yakov Krasnov  Department of Mathematics, Bar-Ilan University, Ramat-Gan,
52900, Israel<EMAIL_ADDRESS>and  Vladimir G. Tkachev
<EMAIL_ADDRESS>
###### Abstract.

The purpose of this paper is to give a systematic study of two new classes of
commutative nonassociative algebras, the so-called isospectral and medial
algebras. An isospectral algebra $\mathbb{A}$ is a generic commutative
nonassociative algebra whose idempotents have the same Peirce spectrum. A
medial algebra is algebra with identity $(xy)(zw)=(xz)(yw)$. We show that
these two classes are essentially coincide. We also prove that any medial
spectral algebra is isomorphic to a certain isotopic deformation of the
commutative associative quotient algebra $\mathbf{K}[z]/(z^{n}-1)$.

###### Key words and phrases:

Medial magma, Isospectral algebras, Quasigroups, Idempotents, Peirce
decomposition

##  1\. Introduction

By an algebra $\mathbb{A}$ we shall always mean a commutative nonassociative
finite dimensional algebra over a field $\mathbf{K}$ of
$\mathrm{char}(\mathbf{K})\neq 2,3$. If not explicitly stated otherwise, any
algebra will be commutative (but maybe non-associative) and the algebra
multiplication will normally be denoted by juxtaposition. An element
$c\in\mathbb{A}$ is called an idempotent if $c^{2}=c$. The set of nonzero
idempotents of $\mathbb{A}$ is denoted by $\operatorname{Idm}(\mathbb{A})$.

The spectrum $\operatorname{spec}(c)$ of an idempotent
$c\in\operatorname{Idm}(\mathbb{A})$ is the multiset of eigenvalues of the
multiplication operator $L_{c}:x\to cx$. The Peirce spectrum of an algebra is
the union of the Peirce spectra of its nonzero idempotents.

Inspired by recent progress in axial algebras [24], [46], [26], we introduce
and study here two new classes of algebras with remarkable properties. Recall
that algebras with a prescribed Peirce spectrum has been the object of
extensive investigation in the context of automorphisms of finite simple
groups. In [23], Griess proved the existence of the largest simple sporadic
group (Monster) by utilizing a nonassociative commutative algebra of dimension
$196883$, and showed that the automorphism group of this algebra is exactly
the Monster simple group. S. D. Smith studied in [51] nonassociative
commutative algebras for triple covers of 3-transposition groups and S. Norton
considered general transposition algebras with a natural permutation
representation [44]. Motivated by the work of Griess, K. Harada [27], [28]
considered the automorphism group of a vector space possessing a
nonassociative commutative algebra structure, which is closely related to a
natural permutation representation of a multiply transitive group. Some
further results were obtained by H.P. Allen [2], [3] and K. Narang [43] and
generalized by H. Suzuki in [54], [55]. Invariant commutative nonassociative
algebras for some sporadic simple groups were studied by A.J.E Ryba [47],
[48]. We also mention some related classes of algebras introduced very
recently in the work of D.J.F. Fox on nonassociative algebras of metric
curvature tensors [18], [19].

All the above algebras, including the Griess algebra, are generated by certain
idempotents having the same Peirce spectrum with distinguished fusion rules
(i.e. the multiplication rules between the corresponding Peirce eigenspaces).
These ideas are captured by Majorana and axial algebras in a recent study of
A. Ivanov, J. Hall, F. Rehren, S. Shpectorov in [29], [25], [24], [46], [30].
Note that many of axial algebras are metrized, i.e. carry an invariant
bilinear form $b(x,y)$, in other words, there holds

$b(xy,z)=b(x,yz)\quad\text{for all $x,y,z\in\mathbb{A}$.}$ (1)

Beside the group theoretic context, the algebras with prescribed spectral
properties also naturally appear in finite geometries, algebraic combinatorics
[12], [11], and differential geometry. We especially mention the class of
commutative algebras associated with cubic minimal cones which is relevant in
the context of isospectral algebras. More precisely, one has the following
definition [42, Chapter 6], [57], [60], [56]: a commutative metrized
nonassociative algebra $\mathbb{A}$ over $\mathbb{R}$ is called a Hsiang
algebra if it satisfies the trace free condition $\operatorname{tr}L_{x}=0$
and the splitting identity

$b(x^{2},x^{3})=k\,b(x,x)b(x^{2},x),\qquad\forall x\in\mathbb{A},$ (2)

where $k\in\mathbb{R}$ is some fixed nonzero constant and $b(x,y)$ is a
nondegenerate invariant bilinear form on $\mathbb{A}$. It follows immediately
from (2) that all idempotents in a Hsiang algebra have the same length
$b(e,e)=1/k$. A more delicate property is that (2) implies that in any Hsiang
algebra all idempotents have the same spectrum with eigenvalues in
$\\{-1,-\frac{1}{2},\frac{1}{2},1\\}$, where the multiplicity of each
eigenvalue is independent on a choice of an idempotent [42].

In the light of the above, the following definition is natural.

######  Definition 1.1.

A commutative nonassociative algebra is called isospectral if all nonzero
algebra idempotents have the same spectrum.

By the definition, the Peirce spectrum of an isospectral algebra coincides
with the spectrum of any idempotent of the algebra. In this paper we shall
primarily concern with generic isospectral algebras over algebraically closed
fields, i.e. those containing the maximal possible finite number of distinct
idempotents ($=2^{\dim\mathbb{A}}$); see section 2.1 for concise definitions.
Note, that the Hsiang algebras above are isospectral but they normally are not
generic and contain infinitely many idempotents if the dimension $\geq 4$. A
classification of all non-generic commutative nonassociative isospectral
algebras is more involved and will be considered elsewhere.

Coming back to the generic case, we have proved in [35] that any finite
dimensional isospectral generic algebra $\mathbb{A}$ over $\mathbb{C}$
necessarily has the cyclotomic Peirce spectrum, i.e. the finite set generated
by a primitive root of unity of degree $n=\dim\mathbb{A}$.

######  Remark 1.2.

The latter implies that except for the case of dimension $\dim\mathbb{A}=2$,
the spectrum of any isospectral generic algebra over $\mathbb{C}$ contains
complex numbers, therefore a isospectral generic algebra of dimension $\geq 3$
is not metrized! Indeed, a metrized algebra must have a real spectrum because
the multiplication operator $L(x)$ is self-adjoint for any element
$x\in\mathbb{A}$) [58], [59].

Below we consider two motivating examples of isospectral algebras, in
dimension 2 and 3 respectively.

######  Example 1.3.

Let $\mathbb{A}_{2}(\mathbf{K})$ be the vector space over a field $\mathbf{K}$
of $\mathrm{char}(\mathbf{K})\neq 2,3$ generated by two vectors $c_{1}$ and
$c_{2}$. Let us introduce a commutative algebra structure on
$\mathbb{A}_{2}(\mathbf{K})$ by requiring

$c_{1}c_{1}=c_{1},\quad c_{2}c_{2}=c_{2},\quad
c_{1}c_{2}=c_{2}c_{1}=-c_{1}-c_{2}.$ (3)

This immediately implies that $c_{1}$ and $c_{2}$, as well as
$c_{3}=-c_{1}-c_{2}$ are distinct idempotents in $\mathbb{A}_{2}(\mathbf{K})$.
Furthermore, an easy verification reveals that under the made assumptions,
there exists exactly $3=2^{2}-1$ nonzero idempotents in
$\mathbb{A}_{2}(\mathbf{K})$ and they are closed under multiplication:

$c_{i}c_{j}=c_{k},\quad\text{where $\\{i,j,k\\}=\\{1,2,3\\}$}.$ (4)

The latter identity also implies that the spectrum of each idempotent $c_{i}$
is $\\{1,-1\\}$, therefore $\mathbb{A}_{2}(\mathbf{K})$ is an isospectral
algebra (see also Example 3.21 below for the particular case
$\mathbf{K}=\mathbb{F}_{7}$). Next, we have $\operatorname{tr}L_{c_{i}}=1-1=0$
for any idempotent $c_{i}$, thus $\operatorname{tr}L_{x}=0$ holds true for any
$x\in\mathbb{A}_{2}$. Furthermore, one readily verifies that any element
$x\in\mathbb{A}_{2}$ satisfies the identity

$x^{3}=\beta_{2}(x)x,$ (5)

where $\beta_{2}(x):\mathbb{A}_{2}\to\mathbf{K}$ is a certain multiplicative
homomorphism:

$\beta_{2}(xy)=\beta_{2}(x)\beta_{2}(y),\qquad\forall x,y\in\mathbb{A}_{2}.$
(6)

Explicitly, $\beta_{2}$ is given by the positive definite quadratic form

$\beta_{2}(x)=x_{1}^{2}-x_{1}x_{2}+x_{2}^{2},\quad\text{where
$x=x_{1}c_{1}+x_{2}c_{2}$.}$

The identity (6) also shows that $\mathbb{A}_{2}$ is a commutative composition
nonunital algebra [52, § 1.2]. The bilinear form
$b(x,y)=\beta_{2}(x+y)-\beta_{2}(x)-\beta_{2}(y)$ is positive definite and
invariant, i.e. satisfies (1). In other words, $\mathbb{A}$ is a metrized
algebra, see [59].

Furthermore, by (5)

$b(x^{2},x^{3})=b(x^{2},\beta_{2}(x)x)=\beta_{2}(x)b(x^{2},x)=\frac{1}{2}b(x,x)b(x^{2},x),$

which implies that $\mathbb{A}_{2}$ is a Hsiang algebra. In fact,
$\mathbb{A}_{2}$ is the Hsiang algebra of a one-dimensional minimal cone
obtained by union of two perpendicular lines in $\mathbb{R}^{2}$ (Chap. 6,
[42]). It is not difficult to show that $\mathbb{A}_{2}$ is the only
isospectral generic algebra in dimension two (in any characteristic $\neq 2$).

######  Remark 1.4.

The algebra $\mathbb{A}_{2}(\mathbf{K})$ is isomorphic to the two-dimensional
Harada algebra defined [28]; its automorphism group is exactly the permutation
group of the idempotents:

$\operatorname{Aut}(\mathbb{A}_{2}(\mathbf{K}))=\mathrm{Perm}(\operatorname{Idm}(\mathbb{A}_{2}))\cong
S_{3},$

is isomorphic to the symmetric group of degree three. The general Harada
algebras appeared very recently in the context of simplicial algebras [20] and
cyclic elements in semisimple Lie algebras [13]. Note, however, that the
general Harada algebras in dimensions $n\geq 3$ are not isospectral.

######  Example 1.5.

The above example can be appropriately generalized for dimension 3. The
corresponding three dimensional isospectral algebra
$\mathbb{A}_{3}(\mathbb{C})$ over the field of complex numbers $\mathbb{C}$
has been announced in [35, Sec. 5.1]. More precisely, define
$\mathbb{A}_{3}(\mathbb{C})$ as the free commutative algebra over $\mathbb{C}$
spanned by three elements $c_{1},c_{2},c_{3}$ subject to $c_{i}c_{i}=c_{i}$,
$1\leq i\leq 3$, and

$\displaystyle c_{5}:=c_{1}c_{2}$ $\displaystyle=(\gamma-1)c_{1}-\gamma
c_{2}+\gamma c_{3},$ $\displaystyle c_{6}:=c_{2}c_{3}$ $\displaystyle=\gamma
c_{1}+(\gamma-1)c_{2}-\gamma c_{3},$ $\displaystyle c_{7}:=c_{3}c_{1}$
$\displaystyle=-\gamma c_{1}+\gamma c_{2}+(\gamma-1)c_{3},$

where $\gamma$ is the Kleinian integer unit [10], i.e. a root of

$2\gamma^{2}-\gamma+1=0.$

Define also

$c_{4}=-\gamma(c_{1}+c_{2}+c_{3}).$

Then one can show that $c_{i}$, $1\leq i\leq 7$ are the only idempotents of
$\mathbb{A}_{3}$, in particular, the algebra $\mathbb{A}_{3}$ is generic. A
straightforward examination reveals that all idempotents have the same
(cyclotomic) spectrum

$\sigma(c_{i})=\\{1,\begin{matrix}\frac{-1-\sqrt{-3}}{2}\end{matrix},\begin{matrix}\frac{-1+\sqrt{-3}}{2}\end{matrix}\\},\qquad
1\leq i\leq 7.$

Furthermore, as in Example 1.3, one can show that all nonzero idempotents in
$\mathbb{A}_{3}(\mathbb{C})$ are closed under the algebra multiplication, i.e.
$c_{i}c_{j}\in\operatorname{Idm}(\mathbb{A}_{3}(\mathbb{C}))$ for any $1\leq
i,j\leq 7$. More precisely, the multiplication rules between idempotents
$c_{i}$ follow the pattern in the left part of Table 1.

| 1 | 2 | 3 | 4 | 5 | 6 | 7
---|---|---|---|---|---|---|---
1 | 1 | 5 | 7 | 3 | 6 | 2 | 4
2 | 5 | 2 | 6 | 1 | 4 | 7 | 3
3 | 7 | 6 | 3 | 2 | 1 | 4 | 5
4 | 3 | 1 | 2 | 4 | 7 | 5 | 6
5 | 6 | 4 | 1 | 7 | 5 | 3 | 2
6 | 2 | 7 | 4 | 5 | 3 | 6 | 1
7 | 4 | 3 | 5 | 6 | 2 | 1 | 7

| 1 | 2 | 3 | 4 | 5 | 6 | 7
---|---|---|---|---|---|---|---
1 | 1 | 5 | 2 | 6 | 3 | 7 | 4
2 | 5 | 2 | 6 | 3 | 7 | 4 | 1
3 | 2 | 6 | 3 | 7 | 4 | 1 | 5
4 | 6 | 3 | 7 | 4 | 1 | 5 | 2
5 | 3 | 7 | 4 | 1 | 5 | 2 | 6
6 | 7 | 4 | 1 | 5 | 2 | 6 | 3
7 | 4 | 1 | 5 | 2 | 6 | 3 | 7

Table 1. The left multiplication table corresponds to $c_{i}$ for Example 1.5,
the right table illustrates the medial law (9) after permutation (8)

For example, $c_{2}c_{5}=c_{4}$ and $c_{3}c_{5}=c_{1}$. It is easy to see that
the latter multiplication table is a Latin square, i.e. each index occurs
precisely one time in each column and row. In fact, there is another
remarkable property of this Latin square: the authors were completely
surprised to notice that the table also satisfies the medial magma identity

$(xy)(zw)=(xz)(yw),\qquad\forall
x,y,z,w\in\operatorname{Idm}(\mathbb{A}_{3}).$ (7)

We point out, that, although it is easy to verify (7) for particular cases, it
is quite nontrivial to examine (7) for all quadruples $(x,y,z,w)$ (taking into
account the commutativity, one have to verify totally
$\frac{1}{2}(\frac{7\cdot 6}{2})\cdot((\frac{7\cdot 6}{2}-1)=210$ quadruples).
In fact, we will show in this paper that (7) holds for all isospectral
algebras including $\mathbb{A}_{3}$. But in this particular case, we can
explain how to check (7) right now. To this end, let us consider the
permutation of $c_{i}$ given accordingly to

$\left(\begin{array}[]{cccccccc}1&2&3&4&5&6&7\\\
1&5&2&3&6&4&7\end{array}\right)$ (8)

Then the resulting multiplication (the so-called isotopy of the Latin square)
corresponds to the right part of Table 1. This updated table is more
illuminating and informative. In particular, one can easily verify that the
entries in the right table satisfy the following nice rule:

$i\circledast j=\frac{1}{2}(i+j)\mod 7,$ (9)

i.e. the multiplication of idempotents (after permutation (8)) satisfies
$c_{i}c_{j}=c_{i\circledast j}$. This new multiplication $\circledast$ on
$\mathbb{Z}_{7}$ is commutative but nonassociative. But, since

$(i\circledast j)\circledast(k\circledast l)\equiv\frac{1}{4}(i+j+k+l)\mod 7,$

the new multiplication $\circledast$ obviously satisfies the medial magma
identity (7).

Since $\mathbb{A}_{3}(\mathbb{C})$ is generated by the idempotents, it follows
that any quadruple of elements of $\mathbb{A}_{3}(\mathbb{C})$ also satisfies
(7). This motivates the following

######  Definition 1.6.

A commutative algebra $\mathbb{A}$ satisfying

$(xy)(zw)=(xz)(yw),\qquad\forall x,y,z,w\in\mathbb{A}$ (10)

is called medial.

Finally, one can show that any element of $\mathbb{A}_{3}(\mathbb{C})$
satisfies the identity generalizing (5):

$x^{4}=\beta_{3}(x)x,\qquad\forall x\in\mathbb{A}_{3}(\mathbb{C}),$ (11)

where $\beta_{3}(x):\mathbb{A}_{3}(\mathbb{C})\to\mathbb{C}$ is a certain
multiplicative homomorphism.

The remarkable medial magma property for isospectral algebras
$\mathbb{A}_{2}(\mathbf{K})$ and $\mathbb{A}_{3}(\mathbb{C})$ has been one of
the starting points for the research presented in this paper. The medial magma
identity itself and its variations is very well-known in the context of
quasigroups, finite geometries and Steiner designs. The classical result of
Toyoda [61], Bruck [8] and Murdoch [41] characterizes a medial quasigroup is
an isotope of a certain abelian group. On the other hand, to the best of our
knowledge, the algebras satisfying the medial magma identity have not yet been
systematically studied; see, however, some related train algebras satisfying
polynomial identities studied in [36].

The paper is organized as follows. We recall the basic facts about the Peirce
decomposition, generic algebras and quasigroups in section 2.1. In section 2.2
we formulate our main results. In section 3 we discuss medial algebras and
their Peirce decomposition. In particular, we show that any medial algebra is
a direct product of its radical and an isospectral algebra. This establishes a
link between medial and isospectral algebras. Next, isospectral algebras are
discussed in section 5 and the subclass of medial isospectral algebras in
section 6. In section 7 we prove the existence of a medial isospectral algebra
in any dimension $n\geq 2$. The set of all idempotents in a isospectral medial
algebra has a natural structure of a (medial) qiuasigroup. The classical
result of Toyoda-Bruck-Murdoch characterizes a medial quasigroup is an isotope
of a certain abelian group. In fact, we establish in section 8 that the
idempotent quasigroup of isospectral medial alegrbas is much simpler: it is
just cyclic.

For the reader convenience, we list below some frequently used notation:

|  |
---|---|---
$\mathbb{A}$ | a commutative nonassociative algebra over a field $\mathbf{K}$
$L_{y}:x\to yx$ | the (left) multiplication operator by $y$
$\mathbb{A}_{c}(\lambda)$ | the $\lambda$-eigenspace of $L_{c}$
$\operatorname{Idm}(\mathbb{A})$ | the set of nonzero idempotents of $\mathbb{A}$
$\operatorname{spec}(c)$ | the spectrum of an idempotent $c$, i.e. the multiset of eigenvalues of $L_{c}$
$\sigma(c)$ | the Peirce spectrum of $c$, i.e. the distinct eigenvalues of $L_{c}$
$\epsilon_{n}:=e^{\frac{2\pi\sqrt{-1}}{n}}$ | the primitive root of unity order $n$.

##  2\. Main results

###  2.1. Preliminaries

We use the standard notation for nonassociative monomials generated by
$x\in\mathbb{A}$: $x^{2}=xx$, $x^{3}=x(xx)$. Note that in general
$x^{4}=xx^{3}$ and $x^{2}x^{2}$ maybe different.

We recall some basic terminology and concepts following to [31], [35], [60].
An element $c$ of an algebra $\mathbb{A}$ is called an idempotent (resp. a
$2$-nilpotent) if $c^{2}=c$ (resp. $c^{2}=0$). By
$\operatorname{Idm}(\mathbb{A})$ we denote the set of all nonzero idempotents
of $\mathbb{A}$. Sometimes it is convenient to denote by
$\operatorname{Idm}_{0}(\mathbb{A})$ the set of all idempotents (including
zero). Given an idempotent $c$, one can define the (left$=$right)
multiplication operator

$L_{c}:x\to cx.$

We abuse terminology by saying that a vector $x\neq 0$ is an eigenvector of an
idempotent $c$ with eigenvalue $\lambda$ if $cx=L_{c}x=\lambda x$. Any
eigenvalue $\lambda$ of $c$ is a root of the characteristic polynomial of
$L_{c}$ regarded as an endomorphism
$L_{c}\in\mathrm{End}_{\mathbf{K}}(\mathbb{A})$.

######  Definition 2.1.

The spectrum of an idempotent $c$, denoted by $\operatorname{spec}(c)$, is the
multi-set (a set with repeated elements) of all roots of the characteristic
polynomial of $L_{c}$. We also denote by $\sigma(c)$ the Peirce spectrum of
$c$, i.e. the set of all distinct eigenvalues of $L_{c}$. In other words, the
Peirce spectrum is the support of $\operatorname{spec}(c)$. Thus, two
idempotents may have distinct spectra while their Peirce spectra may be the
same. By

$\sigma(\mathbb{A})=\bigcup_{c\in\operatorname{Idm}(\mathbb{A})}\sigma(c)$

we denote the Peirce spectrum of an algebra $\mathbb{A}$, i.e. all possible
distinct eigenvalues of $L_{c}$, when $c$ runs over all idempotents in
$\operatorname{Idm}(\mathbb{A})$.

The classical example is the projection operator with (Peirce) spectrum $0$
and $1$, where the multiplicity of $1$ indicates the dimension of the
projection target space.

The spectrum of any idempotent is always nonempty because $1\in\sigma(c)$. An
idempotent $c$ is called primitive if the eigenvalue $1$ has multiplicity one.
An idempotent $c$ is called semi-simple if $\mathbb{A}$ splits into a direct
sum of $\lambda$-eigenspaces $\mathbb{A}_{c}(\lambda)$ of $L_{c}$, where
$\lambda\in\sigma(c)$:

$\mathbb{A}=\bigoplus_{\lambda\in\sigma(c)}\mathbb{A}_{c}(\lambda).$ (12)

This decomposition is also known as the Peirce decomposition of $\mathbb{A}$
relative to $c$. A fusion law is a map $\star:\sigma(c)\times\sigma(c)\to
2^{\sigma(c)}$ such that

$\mathbb{A}_{c}(\lambda)\mathbb{A}_{c}(\mu)\subset\bigoplus_{\nu\in\lambda\star\mu}\mathbb{A}_{c}(\nu).$
(13)

Recall that two algebras $\mathbb{A}$ and $\mathbb{B}$ over a field
$\mathbf{K}$ are isotopic if there exist three non-singular linear
transformations $f$, $g$ and $h$ from $\mathbb{A}$ to $\mathbb{B}$ such that

$f(x)g(y)=h(xy).$ (14)

The triple $(f,g,h)$ is called isotopism between the algebras $\mathbb{A}$ and
$\mathbb{B}$. If $f=g$, $\mathbb{A}$ and $\mathbb{B}$ are called strongly
isotopic, or $\mathbb{A}\simeq\mathbb{B}$. If $f=g=h$, the algebras
$\mathbb{A}$ and $\mathbb{B}$ are called isomorphic, or
$\mathbb{A}\cong\mathbb{B}$.

The isotopism of algebras is an important concept of nonassociative algebra
introduced by [1]; see also [16]. Partition of nonassociative algebras into
(strong) isotopism classes is very different from that the partition into
classes of non-isomorphic algebras, but it helps to recognize some very common
patterns which are usually not seen via isomorphisms. We refer the interested
reader to a nice survey of R. Falcón, and O. Falcón, and Núñez [17] about a
comprehensive discussion of the isotopy concept and its applications.

An important concept in the context of the present paper is generic
nonassociative algebras. This class has been recently introduced and
thoroughly studied in [35]. Below we recall some basic facts and refer the
interested reader to [35] for more details and the proof of Theorem 2.3 below.

The definition of a generic algebra is based on the Segre’s classical approach
[49] to idempotents in commutative algebras over the complex numbers (or, in
general, over an algebraically closed field $\mathbf{K}$) by purely algebraic
geometry methods. More precisely, Segre remarked that the set of idempotents
of an algebra can be characterized as the solution set of a system of
quadratic equations over $\mathbf{K}$. Since a generic (in the Zariski sense)
polynomial system has always Bézout’s number of solutions, this implies that a
generic algebra must have exactly $2^{\dim A}$ distinct idempotents. The
genericity here should be understood in the sense that the subset of
nonassociative algebra structures on a vector space $V$ is an open Zariski
subset in $V^{*}\otimes V^{*}\otimes V$. This suggests the following formal
definition.

Given an algebra $\mathbb{A}$ over a subfield $\mathbf{K}$ of complex numbers,
by $\mathbb{A}_{\mathbb{C}}$ we denote the complexification of $\mathbb{A}$
obtained in an natural way by extending the ground field to $\mathbb{C}$.

######  Definition 2.2.

An algebra $\mathbb{A}$ over $\mathbf{K}$ is called a generic nonassociative
algebra if its comlexification $\mathbb{A}_{\mathbb{C}}$ contains exactly
$2^{n}$ distinct idempotents, where $n=\dim\mathbb{A}$.

######  Theorem 2.3 (Theorem 3.3 in [35]).

If $\mathbb{A}$ is a commutative nonassociative generic algebra then
$\begin{matrix}\frac{1}{2}\end{matrix}\not\in\sigma(\mathbb{A})$. In the
converse direction: if
$\begin{matrix}\frac{1}{2}\end{matrix}\not\in\sigma(\mathbb{A})$ and
$\mathbb{A}$ does not contain $2$-nilpotents then $\mathbb{A}$ is generic.

The main result of [35] states that idempotents in a generic commutative
algebra must satisfy certain a priori constrains, syzygies. This phenomenon is
somewhat unexpected taking into account that an algebra in the context is
generic. Remarkably, the syzygies can be written explicitly. We combine the
results of Theorem 4.1 and Proposition 4.2 in [35] in the theorem below.

######  Theorem 2.4.

Let $\mathbb{A}$ be a generic commutative nonassociative algebra over $K$,
$\dim A=n$. Then

$\sum_{c\in\operatorname{Idm}(\mathbb{A})}\frac{\chi_{c}(t)}{\chi_{c}(\begin{matrix}\frac{1}{2}\end{matrix})}=2^{n}(1-t^{n}),\quad\forall
t\in\mathbb{R},$ (15)

where $\chi_{c}(t)=\det(L_{c}-\mathbf{1}t)$ is the characteristic polynomial
of $L_{c}$. Furthermore, let $H(x):\mathbf{K}^{n}\to\mathbf{K}^{s}$ be a
vector-valued polynomial map ($s\geq 1$) such that for each coordinate $\deg
H_{i}\leq n-1$, $1\leq i\leq n$. Then

$\displaystyle\sum_{c\in\operatorname{Idm}_{0}(\mathbb{A})}\frac{H(c)}{\chi_{c}(\frac{1}{2})}$
$\displaystyle=0,$ (16)

In particular,

$\displaystyle\sum_{c\in\operatorname{Idm}(\mathbb{A})}(\chi_{c}(\frac{1}{2}))^{-1}c$
$\displaystyle=0,$ (17)

For the reader convenience, we recall the key steps of the proof. Since
$\mathbb{A}$ is a commutative algebra over an infinite field, the
multiplication is uniquely determined by the square mapping
$\Psi(x)=x^{2}:\mathbb{A}\to\mathbb{A}$ by polarization

$xy=\frac{1}{2}(\Psi(x+y)-\Psi(x)-\Psi(y))=\frac{1}{2}D\Psi(x)\,y$

where $D\Psi(x)$ is the Jacobi matrix of $\Psi$ at $x$. In this notation,
idempotents $c\in\mathbb{A}$ are in one-to-one correspondence with zeroes of
the quadratic map $F(x)=\Psi(x)-x$. Note that

$D\Psi(x)=2L_{x}.$

Then the genericity condition on $\mathbb{A}$ is equivalent to saying that all
roots of $F(x)=0$ are nonsingular, i.e. $DF(x)$ has the maximal rank. Applying
the Euler-Jacobi formula [4, p. 106] (see also Theorem 4.3 in [5]), one
obtains

$0=\sum_{c:F(c)=0}\frac{H(c)}{\det[DF(c)]}=\sum_{c:\Psi(c)=c}\frac{H(c)}{\det[D\Psi(c)-1]}=\sum_{c\in\operatorname{Idm}(\mathbb{A})}\frac{H(c)}{2^{n}\det[L_{c}-\frac{1}{2}]}$

which proves (16). Applying (16) to $H(x)=$ the $i$th coordinate function (in
a certain fixed basis of $\mathbb{A}$) one arrives at (17).

Finally, let us recall some principal facts from quasigroup theory. Following
[6], [9], we define a quasigroup to be a set $Q$ with a binary operation
$\circ$ in which the equations $a\circ x=b$ and $y\circ a=b$ have unique
solutions $x$ and $y$ for any given $a$ and $b$ from $Q$. A Latin square is an
$N\times N$ array containing $N$ different entries, such that each entry
occurs exactly once in each row and once in each column. Thus, a set with a
binary operation is a quasigroup if and only if its operation table is a Latin
square. Two quasigroups $(Q,\circ)$ and $(S,\bullet)$ are called isotopic if
there exist bijections $f,g,h:Q\to S$ such that $f(x)\bullet g(y)=h(x\circ y)$
for any $x,y\in Q$.

The number of elements of a quasigroup is called its order and denoted by
$|Q|$. A quasigroup (Latin square) $(Q,\circ)$ is called commutative (resp.
idempotent) if $x\circ y=y\circ x$ (resp. the identity $x\circ x=x$ holds for
any $x\in Q$). Similarly, $Q$ and it is called medial [6, p. 33] if the
identity

$(x\circ y)\circ(z\circ w)=(x\circ z)\circ(y\circ w)$ (18)

holds. Note that the concept of a quasigroup with the medial identity was
originally introduced by Murdoch [41] by the name abelian quasigroups; see
also [8], [61]. We refer also to [53] for some equivalent descriptions of
medial quasigroups and their structure. Many examples of medial semigroups as
well as their discussion in the context of abstract means and mid-points can
be found in [21].

An important theorem of Bruck-Murdoch-Toyoda classifies medial quasigroups.

######  Theorem 2.5 (The Bruck-Murdoch-Toyoda Theorem [41], [8], [61]).

For any medial quasigroup $(Q,\circ)$, there exists an abelian group $(G,+)$,
a fixed element $g\in G$, and commuting automorphisms $\phi,\psi:G\to G$ such
that $x\circ y=g+\phi(x)+\psi(y)$.

This implies that every medial quasigroup is isotopic to an abelian group. For
the proof, see [6, Theorem 2.10] and also [50] for further developments.

###  2.2. Main results

Medial algebras constitutes a rather big class. In this paper, we completely
classify one particular case. More precisely, we shall prove the following
result.

######  Theorem 2.6.

For a fixed dimension $n\geq 2$, there exists exactly one isomorphy class of
medial generic isospectral algebras over an algebraically closed field
$\mathbf{K}$. Furthermore, any medial generic isospectral algebra $\mathbb{A}$
satisfies the identity

$x^{n+1}=\beta_{n}(x)x,\qquad x\in\mathbb{A},$ (19)

where $n=\dim\mathbb{A}$ and $\beta_{n}:\mathbb{A}\to\mathbf{K}$ is a
multiplicative homomorphism.

The key ingredient of the uniqueness part of the above theorem is the Peirce
decomposition of a medial isospectral algebra. Given an idempotent
$c\in\operatorname{Idm}(\mathbb{A})$ we prove that there exist
$w_{1}\in\mathbb{A}$ such that $\\{w_{1}^{i}\\}_{i=0}^{n-1}$ (where
$w_{1}^{0}=c$) and the algebra $\mathbb{A}$ decomposes into the direct sum of
one-dimensional eigespaces spanned on the principal powers $w_{1}^{i}$ (for
the definition, see (53) below):

$A=\bigoplus_{i=0}^{n-1}\langle w_{1}^{i}\rangle,\qquad
cw_{1}^{i}=\epsilon^{i}w_{1}^{i}.$

Then the multiplication between the eigenvectors is given by

$w_{1}^{i}w_{1}^{j}=\epsilon_{n}^{\delta_{ij,0}-(i-1)(j-1)}w_{1}^{i+j},$ (20)

where $\delta_{i,j}$ is the Kronecker delta. The latter relations determine
the multiplication structure on a medial isospectral algebra up to an
isomorphism.

The existence part follows from the following result which we prove in Section
7.

######  Theorem 2.7.

Let $n\geq 2$. Consider the quotient

$\mathscr{C}_{n}=\mathbf{K}[z]/(z^{n}-1)$

with a new multiplication $\circ$ on $\mathscr{C}_{n}$ defined by

$p(z)\circ q(z)=p(\epsilon_{n}z)q(\epsilon_{n}z)\mod z^{n}-1,$ (21)

where $\epsilon_{n}$ is a primitive root of unity of order $n$. Then the
algebra $\mathscr{C}_{n}$ is a medial isospectral algebra over $\mathbb{C}$ of
dimension $n$. Furthermore, the corresponding multiplicative homomorphism (19)
is given explicitly by

$\beta(p(z))=p(1)p(\epsilon_{n})p(\epsilon_{n}^{2})\cdots
p(\epsilon_{n}^{n-1}):\mathscr{C}_{n}\to\mathbf{K}.$ (22)

###  2.3. Idempotents as a quasigroup

In other words, any medial isospectral algebra is isomorphic to some
$\mathscr{C}_{n}$. It is interesting to study the idempotents of such
algebras. While the algebra structure on $\mathscr{C}_{n}$ and the
multiplicative homomorphism $\beta(z)$ are given very explicitly by (21) and
(22), the structure of idempotents of $\mathscr{C}_{n}$ is more sophisticated.
When $n\leq 6$, it is possible to describe the structure of
$\operatorname{Idm}(\mathscr{C}_{n})$ is a straightforward way. In order to
understand the relations between idempotents in a satisfactory way, one need
to look at $\operatorname{Idm}(\mathscr{C}_{n})$ as a quasigroup.

To this end, note that it follows from (7) and the fact that the Peirce
spectrum of an isospectral algebra does not contain $0$ that the product of
two nonzero idempotents in a medial algebra $\mathbb{A}$ is always a nonzero
idempotent and the relation $xy=z$ uniquely determines one idempotent if two
others are given. This implies that $\operatorname{Idm}(\mathbb{A})$ is a
commutative idempotent medial (CIM, for short) quasigroup with respect to the
algebra multiplication. Then the classical Bruck-Murdoch-Toyoda theorem
characterizes any medial quasigroup as an isotope of a certain abelian group
$G$. The structure of a CIM quasigroups and their endomorphisms has been
recently studied by A. Leibak and P. Puusemp [38], [37]. In [38] a refinement
of the Bruck-Murdoch-Toyoda theorem for CIM quasigroups has also been
established. A natural question arise: Given a medial isospectral algebra
$\mathbb{A}$, how to characterize the underlying abelian group $G$? We
establish that $G$ must be the cyclic group $\mathbb{Z}_{2^{n}-1}$,
$n=\dim\mathbb{A}$.

More precisely, we have

######  Theorem 2.8.

Let $\mathbb{A}$ be a medial isospectral algebra of dimension $n$. Then the
there exists a bijection
$\phi:\operatorname{Idm}(\mathbb{A})\to\mathbb{Z}_{2^{n}-1}$ such that

$\phi(xy)\equiv\frac{1}{2}(\phi(x)+\phi(y))\mod 2^{n}-1.$

##  3\. Medial algebras

###  3.1. General properties

Obviously, a zero algebra (i.e. an algebra with $\mathbb{A}\mathbb{A}=0$) is
medial. To avoid this trivial case, we shall always assume that all algebras
below are nonzero.

Let $\mathbb{A}$ be a medial algebra. Then the specialization $x=z$ and $y=w$
in (7) implies

$(xy)^{2}=x^{2}y^{2},\qquad\forall x,y\in\mathbb{A}.$ (23)

It turns out that the identity (23) is essentially equivalent to the medial
magma identity (7). More precisely, we have

######  Proposition 3.1.

If $\mathbf{K}$ is a field of characteristic not $2$ or $3$ then (23) and (7)
are equivalent.

###### Proof.

It suffices to show that (23) implies (7). To this end, note that the
linearization of (23) yields

$(xy)(xz)=x^{2}(yz),\qquad\forall x,y,z\in\mathbb{A},$ (24)

and a further linearization in $x$ yields

$(xy)(wz)+(wy)(xz)=2(xw)(yz),\qquad\forall x,y,z,w\in\mathbb{A}.$ (25)

Permuting the variables we obtain

$\displaystyle(xz)(wy)+(wz)(xy)$ $\displaystyle=2(xw)(yz)$
$\displaystyle(xz)(yw)+(yz)(xw)$ $\displaystyle=2(xy)(wz),$

and summing up the three latter identities yields

$2(xy)(wz)+3(wy)(xz)+(yz)(xw)=4(xw)(yz)+2(xy)(wz),$

hence $3(wy)(xz)=3(xw)(yz),$ which implies (7) (after a suitable renaming of
variables). ∎

Thus, when working with algebras over a field with characteristic $2$ or $3$
one need to be careful with the defining identities. In this paper, we work
with $\mathrm{char}(\mathbf{K})\neq 2,3$, therefore we shall not distinguish
the definitions.

The identity (23) is remarkable in several aspects. First note that it is
equivalent to that the square mapping $\Psi(x)=x^{2}$ is a multiplicative
homomorphism and (under various names) has attracted much attention, primarily
in the context of quasigroups. Furthermore, algebras with (23) appear, for
example, in the study of train algebras satisfying polynomial identities [36].
Also, we mention that (23) resembles the Norton inequality $\langle
x^{2};y^{2}\rangle\geq\langle xy;xy\rangle$ which appears as an axiom in
construction of the Griess-Conway-Norton algebra of the monster sporadic
simple group and also in general Majorana algebras, see [40], [29]. Note also
that the latter inequality is defined in the presence of an invariant bilinear
form $\langle;\rangle$.

We have the following elementary corollaries of the definitions.

######  Proposition 3.2.

A subalgebra of a medial algebra is medial. Furthermore, if some basis
$\\{e_{i}\\}_{i=1}^{n}$ of a commutative algebra $\mathbb{A}$ then
$\mathbb{A}$ is medial if and only if $\\{e_{i}\\}_{i=1}^{n}$ satisfies (7).

###### Proof.

The first claim is obvious. For the second claim it suffices to verify the
‘if’ condition. To this end, suppose that a basis $\\{e_{i}\\}_{i=1}^{n}$ of a
commutative algebra $\mathbb{A}$ satisfies (7). Then for any quadruple
$(x,y,z,w)$ we have

$\displaystyle(xy)(zw)$
$\displaystyle=\sum_{i,j,k,l=1}^{n}x_{i}y_{j}z_{k}w_{l}(e_{i}e_{j})(e_{k}e_{l})=\sum_{i,j,k,l=1}^{n}x_{i}y_{j}z_{k}w_{l}(e_{i}e_{k})(e_{j}e_{l})=(xz)(yw)$

where $x=\sum_{i=1}^{n}x_{i}e_{i}$ etc. Hence $\mathbb{A}$ satisfies (7), as
desired. ∎

######  Definition 3.3.

Let $\mathbb{A}$ be a medial algebra. Define

$\mathbb{A}^{\Box}:=\\{x\in\mathbb{A}:\quad\text{there exists $a\in\mathbb{A}$
with $x=a^{2}$}\\}.$

An immediate observation is

$\operatorname{Idm}(\mathbb{A})\subset\mathbb{A}^{\Box}.$ (26)

######  Proposition 3.4.

For any medial algebra $\mathbb{A}$, the set $\mathbb{A}^{\Box}$ is a
nontrivial multiplicative magma.

###### Proof.

Since $\mathbb{A}^{\Box}$ is nonzero algebra, there exists $a\in\mathbb{A}$
such that $a^{2}\neq 0$. Then $x=a^{2}\in\mathbb{A}^{\Box}$. Therefore
$\mathbb{A}^{\Box}$ is nonempty. If $x,y\in\mathbb{A}^{\Box}$ then $x=a^{2}$,
$y=b^{2}$, therefore $xy=(ab)^{2}\in\mathbb{A}^{\Box}$, i.e. $G$ is a
multiplicative magma. ∎

###  3.2. The quasigroup property

The set of idempotents in a medial algebra, if nonempty, has a very special
structure.

######  Proposition 3.5.

If $c_{1},c_{2}$ are idempotents in a medial algebra then so is $c_{1}c_{2}$
(maybe zero). In other words, the set of all idempotents in a medial algebra
is a multiplicative magma.

###### Proof.

Follows immediately from (23). ∎

######  Proposition 3.6.

Let $\mathbb{A}$ be a medial algebra. For any idempotent
$c\in\operatorname{Idm}(\mathbb{A})$:

  1. (a)

$L_{c}(xy)=L_{c}(x)L_{c}(y)$ for all $x,y\in\mathbb{A}$, i.e. the
multiplication operator $L_{c}:\mathbb{A}\to\mathbb{A}$ is an algebra
endomorphism;

  2. (b)

$\mathbb{A}_{c}(0)$ is an ideal of $\mathbb{A}$;

  3. (c)

$L_{c}(\mathbb{A})$ is a subalgebra of $\mathbb{A}$;

  4. (d)

$\mathbb{A}_{c}(1)$ is a subalgebra of $L_{c}(\mathbb{A})$ and
$\dim\mathbb{A}_{c}(1)\geq 1$;

  5. (e)

for any idempotents $c_{1},c_{2}\in\operatorname{Idm}(\mathbb{A})$ the
following transformation rule holds:

$L_{c_{2}}L_{c_{1}}=L_{c_{2}c_{1}}L_{c_{2}}$ (27)

###### Proof.

(a): the multiplication operator $L_{c}:\mathbb{A}\to\mathbb{A}$ is linear and
it follows from the medial magma identity that for any idempotent
$c\in\operatorname{Idm}(\mathbb{A})$

$\displaystyle L_{c}(xy)$
$\displaystyle=c(xy)=(cc)(xy)=(cx)(cy)=L_{c}x\,L_{c}y,$

hence $L_{c}$ is an algebra endomorphism. As the kernel of a homomorphism,
$\mathbb{A}_{c}(0)=\ker L_{c}$ is ideal and as the image of a homomorphism,
$\operatorname{im}L_{c}=L_{c}(\mathbb{A})$ is a subalgebra, which proves (b)
and (c) respectively. Further, $\mathbb{A}_{c}(1)=\\{x:L_{c}x=x\\}$ is the set
of fixed points of $L_{c}$, thus, is a subalgebra of $\mathbb{A}$. Since
$L_{c}$ stabilizes $\mathbb{A}_{c}(1)$, the latter is also a subalgebra of
$L_{c}(\mathbb{A})$. Also,

$\langle c\rangle\subset\mathbb{A}_{c}(1)\subset cA,$

hence $\dim\mathbb{A}_{c}(1)\geq 1$. This proves (d). Finally, (e) follows
from

$L_{c_{2}}L_{c_{1}}x=c_{2}(c_{1}x)=(c_{2}c_{2})(c_{1}x)=(c_{2}c_{1})(c_{2}x)=L_{c_{2}c_{1}}L_{c_{2}}x.$

∎

The set of idempotents with non-degenerated multiplication operator

$\operatorname{Idm}^{\times}(\mathbb{A}):=\\{c\in\operatorname{Idm}(A):\,\ker
L_{c}=0\\}\subset\operatorname{Idm}(\mathbb{A})$

plays a distinguished role.

######  Proposition 3.7.

$\operatorname{Idm}^{\times}(\mathbb{A})$ is a medial idempotent quasigroup.
Furthermore, all $c\in\operatorname{Idm}^{\times}(\mathbb{A})$ have the same
characteristic polynomial.

###### Proof.

If $c_{1},c_{2}\in\operatorname{Idm}^{\times}(\mathbb{A})$ then $c_{1}c_{2}$
is a nonzero idempotent, hence
$\operatorname{Idm}^{\times}(\mathbb{A})\operatorname{Idm}^{\times}(\mathbb{A})\subset\operatorname{Idm}(\mathbb{A})$.
Furthermore, since $L_{c_{1}}$ is invertible, by (e) in Proposition 3.5 we
obtain

$L_{c_{1}c_{2}}=L_{c_{1}}L_{c_{2}}L_{c_{1}}^{-1}$ (28)

is also invertible, hence in fact
$c_{1}c_{2}\in\operatorname{Idm}^{\times}(\mathbb{A})$. Therefore,
$\operatorname{Idm}^{\times}(\mathbb{A})\operatorname{Idm}^{\times}(\mathbb{A})\subset\operatorname{Idm}^{\times}(\mathbb{A})$
i.e. $\operatorname{Idm}^{\times}(\mathbb{A})$ is a multiplicative magma.
Furthermore, if $c_{2}c_{1}=c_{3}c_{1}$ for some
$c_{i}\in\operatorname{Idm}^{\times}(\mathbb{A})$ then
$c_{1}(c_{2}-c_{3})=L_{c_{1}}(c_{2}-c_{3})=0$, hence $c_{2}-c_{3}=0$. This
yields that $\operatorname{Idm}^{\times}(\mathbb{A})$ is a quasigroup. Since
$\operatorname{Idm}^{\times}(\mathbb{A})$ satisfies (7), it is also a medial
idempotent quasigroup. Next, it follows from (28) and $c_{1}c_{2}=c_{2}c_{1}$
that

$L_{c_{1}}L_{c_{2}}L_{c_{1}}^{-1}=L_{c_{2}}L_{c_{1}}L_{c_{2}}^{-1},$

i.e. $L_{c_{1}}$ and $L_{c_{2}}$ are conjugate linear endomorphisms, thus have
the same characteristic polynomials. The proposition follows. ∎

###  3.3. Medial and associative algebras

There are useful connections between medial and associative algebras. In one
direction, the relation is simple. Indeed, if $\mathbb{A}$ is associative and
commutative then $(xy)(zw)=xyzw=xzyw=(xz)(yw)$, hence

######  Proposition 3.8.

Any associative commutative algebra is medial.

In the converse direction, one has the following observation. If $A$ is a
unital medial algebra and $e$ is the algebra unit then

$x(yz)=(ex)(yz)=(ez)(xy)=z(xy)=(xy)z.$

This implies

######  Proposition 3.9.

If $\mathbb{A}$ is a unital medial algebra then $\mathbb{A}$ is associative.

######  Remark 3.10.

The above observations show that medial algebras are ‘nearly associative’. Of
course, a general medial algebra is not associative. The simplest example is
the two-dimensional algebra $\mathbb{A}_{2}$ from Example 1.3 above. Indeed,
the identity (4) implies that the basis $\\{c_{1},c_{2}\\}$ satisfies (7),
thus by Proposition 3.2 $\mathbb{A}_{2}$ is medial. Since

$(c_{1}c_{2})c_{3}=c_{3}c_{3}=c_{3}\neq c_{1}=c_{1}c_{1}=c_{1}(c_{2}c_{3}),$

$\mathbb{A}_{2}$ is nonassociative.

######  Remark 3.11.

In the above context, a natural question arises: Does a unitalization preserve
the medial property? The answer is ‘no’. Indeed, let $\mathbb{A}$ be any
nonassociative medial algebra. Then its unitalization
$\mathbb{A}^{\mathrm{uni}}$ contains $\mathbb{A}$ as a subalgebra, therefore
$\mathbb{A}^{\mathrm{uni}}$ is also nonassociative. But by Proposition(3.9),
any unital medial algebra is associative, hence $\mathbb{A}^{\mathrm{uni}}$ in
this case is not medial.

On the other hand, if a medial algebra has an idempotent with nondegenerate
multiplication (i.e. the set $\operatorname{Idm}^{\times}(\mathbb{A})$ is non-
empty) then such an algebra is a (strong) isotope of a commutative associative
unital algebra. More precisely, using Kaplansky’s Trick [45], [32] one obtains

######  Proposition 3.12.

Let $\mathbb{A}$ be a medial algebra, $c\in\operatorname{Idm}^{\times}(A)$.
Define the strong isotopy $(\mathbb{A},\circ)$ with the new multiplication

$x\circ y=L_{c}^{-1}(xy).$ (29)

Then $(\mathbb{A},\circ)$ is a unital commutative associative algebra.

###### Proof.

The algebra $(\mathbb{A},\circ)$ is commutative and $c$ is a unit in
$(\mathbb{A},\circ)$. Also, by (a), $L_{c}^{-1}$ is $\mathbb{A}$-isomorphism
and

$(x\circ y)\circ(z\circ
w)=L_{c}^{-1}(L_{c}^{-1}(xy)\,L_{c}^{-1}(zw))=L_{c}^{-2}((xy)(zw))$

which by the nongeneracy of $L_{c}$ proves that $(\mathbb{A},\circ)$ is medial
if and only if $\mathbb{A}$ is medial. By Proposition 3.9,
$(\mathbb{A},\circ)$ is associative. ∎

######  Remark 3.13.

Although relation (29) establishes the existence of an associative algebra
structure on the isotopy $(\mathbb{A},\circ)$, there is no canonical way to
reconstruct the original algebra structure from $(\mathbb{A},\circ)$. It would
be interesting to characterize all commutative algebras $\mathbb{A}$ whose
associative isotopes $(\mathbb{A},\circ)$ are isomorphic.

###  3.4. The determinant on a medial algebra

We have already seen in Examples 1.3 and 1.5 that medial algebras in dimension
2 and 3 carry a multiplicative homomorphism. As a corollary of Proposition
3.12 is the following result.

######  Theorem 3.14.

Let $\mathbb{A}$ be a medial algebra, $c\in\operatorname{Idm}^{\times}(A)$.
Then

$\phi(x)=\frac{1}{\det L_{c}}\det L_{x}$

is $\mathbb{A}$-multiplicative, i.e.

$\phi(xy)=\phi(x)\phi(y),\qquad\forall x,y\in\mathbb{A}.$

###### Proof.

Indeed, the algebra $(\mathbb{A},\circ)$ in Proposition 3.12 is associative.
Let $M_{x}:y\to x\circ y$ denote the (left) multiplication operator in
$(\mathbb{A},\circ)$. The multiplication operators in $(\mathbb{A},\circ)$ and
$\mathbb{A}$ are related by $M_{x}=L_{c}^{-1}L_{x}$. By the associativity we
have $M_{x\circ y}=M_{x}M_{y}$, hence by the multiplicative property of the
determinant

$\det M_{x\circ y}=\det(M_{x}M_{y})=\det M_{x}\det M_{y}.$ (30)

Next, by (a) we have the $\mathbb{A}$-homomorphism property:

$L_{c}^{-1}(xy)=L_{c}^{-1}(x)L_{c}^{-1}(y),\qquad x,y\in\mathbb{A}.$ (31)

With the latter relation in hand we obtain

$(L_{c}^{-1}z)x=(L_{c}^{-1}z)(L_{c}^{-1}L_{c}x)=L_{c}^{-1}(zL_{c}x)=L_{c}^{-1}L_{z}L_{c}x$

implying

$L_{L_{c}^{-1}z}=L_{c}^{-1}L_{z}L_{c}.$ (32)

Setting $z=xy$ and applying the determinant we obtain

$\det L_{L_{c}^{-1}(xy)}=\det L_{xy}.$ (33)

On the other hand, since $L_{c}^{-1}(xy)=x\circ y$ we get

$\det L_{x\circ y}=\det L_{xy},$ (34)

which yields

$\det M_{x\circ y}=\det L_{c}^{-1}L_{x\circ y}=\det L_{c}^{-1}\det L_{xy}=\det
L_{c}^{-1}L_{xy}=\det M_{xy}.$

Combining the latter with (30) we finally obtain

$\det M_{xy}=\det M_{x}\det M_{y}$

therefore $\phi(x)=\det M_{x}=\det L_{c}^{-1}\det L_{x}$ is multiplicative in
$\mathbb{A}$. ∎

###  3.5. Further properties of medial algebras

The following two propositions follow by straightforward examination of (7).

######  Proposition 3.15.

Let $\mathbb{A}$ and $\mathbb{B}$ be medial algebras. Then the direct product
$\mathbb{A}\times\mathbb{B}$ with

$\displaystyle(a,x)+(b,y)$ $\displaystyle=(a+b,x+y),$
$\displaystyle(a,x)\cdot(b,y)$ $\displaystyle=(ab,xy)$

is a medial algebra. The medial subalgebras $\mathbb{A}\times 0$ and
$0\times\mathbb{B}$ are ideals of $\mathbb{A}\times\mathbb{B}$.

######  Proposition 3.16.

Let $\mathbb{B}\subset\mathbb{A}$ be an ideal of a medial algebra. Then the
factor algebra $\mathbb{A}/\mathbb{B}$ is medial too.

There is another useful construction of medial algebras from a given one.

######  Proposition 3.17 (Twisted doubling).

Let $\mathbb{A}$ be a medial algebra and let $\zeta\in\\{-1,1\\}$. Define the
new algebra $(\mathbb{A}\times\mathbb{A})_{\zeta}$ with multiplication

$(x,y)\circ(z,w)=(xz+yw,\zeta(xw+yz)).$

Then $(\mathbb{A}\times\mathbb{A})_{\zeta}$ is medial and
$\mathbb{A}\cong\mathbb{A}\times 0$ is a medial subalgebra of
$(\mathbb{A}\times\mathbb{A})_{\zeta}$.

###### Proof.

Setting $u=(x,y)$ and $v=(z,w)$, we obtain $u\bullet u=(x^{2}+y^{2},2\zeta
xy)$, hence

$\displaystyle(u\bullet u)\bullet(v\bullet v)$
$\displaystyle=((x^{2}+y^{2})(z^{2}+w^{2})+4(xy)(zw),\,2(x^{2}+y^{2})zw+2(z^{2}+w^{2})xy)$
$\displaystyle(u\bullet v)\bullet(u\bullet v)$
$\displaystyle=((xz+yw)^{2}+(xw+yz)^{2},\,2(xz+yw)(xw+yz)).$

By (7), the right hand sides are equal, which implies (23), therefore
$(\mathbb{A}\times\mathbb{A})_{\zeta}$ is medial. Since
$(x,0)\circ(z,0)=(xz,0)$, $\mathbb{A}\times 0$ is a medial subalgebra of
$(\mathbb{A}\times\mathbb{A})_{\zeta}$. ∎

The twisted doubling is a particular case of the following general
construction. Suppose that $\zeta$ is a root of one (not necessarily
primitive) in $\mathbf{K}$ of degree $d$, i.e. $\zeta^{d}=1$. Let
$\mathbb{A}[z]$ be the (nonassociative commutative) polynomial ring over a
medial algebra $\mathbb{A}$. Obviously $\mathbb{A}[z]$ satisfies (7). Next,
define on $\mathbb{A}[z]$ a new multiplication by

$p(z)\bullet q(z)\equiv p(\zeta z)q(\zeta z)\mod z^{d}-1.$

Thus obtained algebra $\mathbb{A}^{\times d}_{\zeta}$ is a commutative
nonassociative algebra of dimension

$\dim\mathbb{A}^{\times d}_{\zeta}=d\dim\mathbb{A}.$

A less obvious property is the following

######  Proposition 3.18.

If $\mathbb{A}$ is medial so is $\mathbb{A}^{\times d}_{\zeta}$. The algebra
$\mathbb{A}^{\times d}_{1}$ is unital with unit $e=1$.

###### Proof.

Setting $w=\zeta^{2}z$ we obtain

$\displaystyle(p(z)\bullet q(z))\bullet(r(z)\bullet s(z))$
$\displaystyle\equiv(p(\zeta^{2}z)q(\zeta^{2}z))(r(\zeta^{2}z)s(\zeta^{2}z))\mod
z^{d}-1$ $\displaystyle\equiv(p(w)q(w))(r(w)s(w))\mod w^{d}-1$
$\displaystyle\equiv(p(w)r(w))(q(w)s(w))\mod w^{d}-1$
$\displaystyle\equiv(p(\zeta^{2}z)r(\zeta^{2}z))(q(\zeta^{2}z)s(\zeta^{2}z))\mod
z^{d}-1$ $\displaystyle=(p(z)\bullet r(z))\bullet(q(z)\bullet s(z)).$

as desired. Also, if $\zeta=1$ then $p(z)\bullet 1=p(\zeta z)=p(z)$, i.e. $1$
is the unit in $\mathbb{A}^{\times d}_{1}$. ∎

Comparing the above definitions reveals also that

$(\mathbb{A}\times\mathbb{A})_{\zeta}\cong\mathbb{A}^{\times 2}_{\zeta}.$

We shall see that the simplest case when $\mathbb{A}=\mathbf{K}$ plays an
essential role in the characterization of medial isospectral generic algebras
given below.

###  3.6. Some further examples of medial algebras

Medial algebras appear in several different contexts.

######  Example 3.19.

First note that there is a nice relation between the squared magma identity
(23) and genetic algebras. Recall that a commutative algebra is called baric
if there exists a nontrivial algebra homomorphism
$\omega:\mathbb{A}\to\mathbf{K}$, also called the weight function, which
carries a certain probabilistic information. A baric algebra is called a
Bernstein algebra of order zero if it satisfies the identity

$x^{2}=\omega(x)x.$ (35)

The latter equation models a population with random mating [22], [7], [15].
Note that a general commutative baric algebra $\mathbb{A}$ can be made up into
a Bernstein algebra of order zero by introducing the multiplication

$x\circ y=\frac{1}{2}(\omega(x)y+\omega(y)x).$

Indeed, since $\omega(x\circ y)=\omega(x)\omega(y)$, the algebra
$(\mathbb{A},\omega)$ is baric and $x\circ x=\omega(x)x$, as desired. Any
Bernstein order zero algebra satisfies

$(xy)^{2}=\omega(xy)xy=(\omega(x)x)(\omega(y)y)=x^{2}y^{2},$

which implies

######  Corollary 3.20.

Any Bernstein algebra $\mathbb{A}$ of order zero is medial.

On the other hand, the class of genetic algebras is too thin to cover all
medial algebras. The algebra $\mathbb{A}_{2}$ from Example 1.3 is medial but
not Bernstein. Indeed, if $\omega$ is a weight function on $\mathbb{A}_{2}$
then we obtain from (35) that $c_{i}^{2}=\omega(c_{i})c_{i}$, therefore
$\omega(c_{i})=1$ for $i=1,2,3$. Since $c_{i}$ span $\mathbb{A}_{2}$, we
deduce that $\omega\equiv 1$ on $\mathbb{A}_{2}$, i.e. $\omega$ is a trivial
homomorphism, a contradiction. In fact, as we know, $\mathbb{A}_{2}$ satisfies
a 3rd order identity (5).

######  Example 3.21.

Note that $\mathbf{K}$ is a one-dimensional medial (associative) algebra over
$\mathbf{K}$. Following Proposition 3.17, let us consider an algebra
$(\mathbf{K}\times\mathbf{K})_{\epsilon}$. When $\epsilon=1$, the element
$(1,0)$ is the algebra unit, hence by Proposition 3.9,
$(\mathbf{K}\times\mathbf{K})_{1}$ is associative. Let us consider
$\epsilon=-1$. Then the idempotent equation in
$(\mathbf{K}\times\mathbf{K})_{-1}$ for $u=(x,y)$ is equivalent to the system

$\left\\{\begin{array}[]{cc}x^{2}+y^{2}&=x\\\ 2xy+y^{2}&=0\end{array}\right.$

If $y=0$ then $c_{0}=(0,0)$ and $c_{1}=(1,0)$. If $y\neq 0$ then
$x=-\frac{1}{2}$ and $y^{2}=-\frac{3}{4}$. We have two cases:

  * (i)

If $t^{2}=-3$ is non-solvable in $\mathbf{K}$ then $c=(1,0)$ is the only
nonzero idempotent in $(\mathbf{K}\times\mathbf{K})_{-1}$.

  * (ii)

If $t^{2}=-3$ is solvable in $\mathbf{K}$ then one has exactly three nonzero
idempotents
$\\{(1,0),(-\frac{1}{2},-\frac{1}{2}t),(-\frac{1}{2},\frac{1}{2}t)\\}$, i.e.
$(\mathbf{K}\times\mathbf{K})_{-1}$ is generic.

Let us illustrate the above cases by the finite fileds
$\mathbf{K}=\mathbb{F}_{5}$ and $\mathbf{K}=\mathbb{F}_{7}$. In the first
case, $t^{2}\equiv-3\mod 5$ is non-solvable in $\mathbb{F}_{5}$, hence we have
(i). If $\mathbf{K}=\mathbb{F}_{7}$ then $t^{2}\equiv-3\equiv 4\mod 7$ is
solvable in $\mathbb{F}_{5}$, $t=\pm 2$. Since $-1/2\equiv 3\mod 7$, the
corresponding idempotents are $c_{2}=(3,1)$ and $c_{3}=(3,6)$. Furthermore,

$\displaystyle c_{1}c_{2}$ $\displaystyle=(1,0)\circ(3,1)=(3,6)=c_{3}$
$\displaystyle c_{2}c_{3}$ $\displaystyle=(3,1)\circ(3,6)=(1,0)=c_{1}$
$\displaystyle c_{3}c_{1}$ $\displaystyle=(3,6)\circ(1,0)=(3,1)=c_{2},$

and $c_{1}+c_{2}+c_{3}=0$ imply (3) and (4). This shows that

$\mathbb{A}_{2}(\mathbb{F}_{7})\cong(\mathbb{F}_{7}\times\mathbb{F}_{7})_{-1}.$

Note, however, that
$\mathbb{A}_{2}(\mathbb{F}_{5})\not\cong(\mathbb{F}_{5}\times\mathbb{F}_{5})_{-1}$
for the reasons indicated above.

In general, it is well known that $-3$ is a quadratic residue modulo $p$ if
and only if $p=3k+1$. Therefore, if $\mathbf{K}=\mathbb{F}_{p}$ then
$\mathbb{A}_{2}(\mathbb{F}_{p})\cong(\mathbb{F}_{p}\times\mathbb{F}_{p})_{-1}$
if and only if $p=3k+1$. Note, however, that if $p\neq 3k+1$ then
$\mathbb{A}_{2}(\mathbb{F}_{p})$ is a nontrivial isospectral algebra, but it
is not isomorphic to $(\mathbb{F}_{p}\times\mathbb{F}_{p})_{-1}$.

Finally, note that

$\mathbb{A}_{2}(\mathbb{C})\cong(\mathbb{C}\times\mathbb{C})_{-1}.$

##  4\. Idempotents in medial algebras

###  4.1. The zero Peirce eigenspace

The following example shows that there exist medial algebras with non-empty
set
$\operatorname{Idm}(\mathbb{A})\setminus\operatorname{Idm}^{\times}(\mathbb{A})$.

######  Example 4.1.

Let $\mathbb{A}$ be an arbitrary medial algebra over $\mathbf{K}$. Then the
direct product $\mathbb{A}\times\mathbf{K}$ is a medial algebra by Proposition
3.15. Clearly, $c_{0}=(0,1)$ is an idempotent in $\mathbb{A}\times\mathbf{K}$.
We also know by Proposition 3.15 that $\mathbb{A}\times 0$ and
$0\times\mathbf{K}$ are ideals. Then $\mathbb{A}\times 0=\ker L_{c_{0}}$. ∎

Let $\mathbb{A}$ be a medial algebra and let $f:\mathbb{A}\to\mathbb{A}$ be an
algebra endomorphism. Then $\ker f$ is a (medial algebra) ideal in
$\mathbb{A}$ and the factor $\mathbb{A}/\ker f$ obviously satisfies the medial
magma identity (7). We have the direct product of medial algebras

$\mathbb{A}\cong\ker f\times(\mathbb{A}/\ker f).$

Now, if $c$ is an idempotent in a medial algebra $A$ then by (a) in
Proposition 3.6, $L_{c}$ is an algebra homomorphism. Hence the zero Peirce
eigenspace

$\mathbb{A}_{c}(0)=\ker L_{c},$

is a (left=right) ideal of $\mathbb{A}$. Therefore,

$\mathbb{A}\cong\mathbb{A}_{c}(0)\oplus\mathbb{A}/\mathbb{A}_{c}(0),$

where $\mathbb{A}/\mathbb{A}_{c}(0)$ is medial (a trivial verification).
Moreover, the class $c^{\prime}=c+\mathbb{A}_{c}(0)$ is obviously an
idempotent in $\mathbb{A}/\mathbb{A}_{c}(0)$. We have

$\displaystyle\mathbb{A}_{c^{\prime}}(0)$
$\displaystyle=\\{x+\mathbb{A}_{c}(0):(c+\mathbb{A}_{c}(0))(x+\mathbb{A}_{c}(0))=\mathbb{A}_{c}(0)\\}$
$\displaystyle=\\{x+\mathbb{A}_{c}(0):cx\in\mathbb{A}_{c}(0)\\}$
$\displaystyle=\\{x+\mathbb{A}_{c}(0):L_{c}^{2}x=0\\}$
$\displaystyle\equiv\ker L_{c}^{2}\mod\mathbb{A}_{c}(0)$

######  Proposition 4.2.

The Peirce subspace $\mathbb{A}_{c}(0)$ is an ideal in $\mathbb{A}$ and the
factor $\mathbb{A}^{\prime}=\mathbb{A}/\mathbb{A}_{c}(0)$ is a medial algebra.
If additionally $c$ is semisimple then
$\mathbb{A}^{\prime}_{c^{\prime}}(0)=0$, where $c^{\prime}$ is the class of
$c$ in $\mathbb{A}^{\prime}$.

###### Proof.

If $x\in\mathbb{A}$ and $y\in\mathbb{A}_{c}(0)$ then applying (24) yields

$c(xy)=(cc)(xy)=(cx)(cy)=0,$

thus, $xy\in\mathbb{A}_{c}(0)$. This implies that
$A\,\mathbb{A}_{c}(0)\subset\mathbb{A}_{c}(0)$, i.e. $\mathbb{A}_{c}(0)$ is an
ideal of $\mathbb{A}$. Next, if $u_{i}=x_{i}+y_{i}$ with
$y_{i}\in\mathbb{A}_{c}(0)$, then

$u_{i}^{2}=x_{i}^{2}+z_{i},\qquad\text{where
}z_{i}=2x_{i}y_{i}+y_{i}^{2}\in\mathbb{A}_{c}(0),$

which readily yields

$u_{1}^{2}u_{2}^{2}=x_{1}^{2}x_{2}^{2}+z_{1}x_{2}^{2}+z_{2}x_{1}^{2}+z_{1}z_{2}=x_{1}^{2}x_{2}^{2}+z_{3},$

where $z_{3}\in\mathbb{A}_{c}(0)$. Similarly,

$(u_{1}u_{2})^{2}=(x_{1}x_{2}+x_{1}y_{2}+y_{1}x_{2}+y_{1}y_{2})^{2}=(x_{1}x_{2}+z_{4})^{2}=(x_{1}x_{2})^{2}+z_{5},$

where $z_{4},z_{5}\in\mathbb{A}_{c}(0)$. This shows that
$u_{1}^{2}u_{2}^{2}=(u_{1}u_{2})^{2}$ in
$\mathbb{A}^{\prime}=\mathbb{A}/\mathbb{A}_{c}(0)$, hence by Proposition 3.1
the factor algebra $\mathbb{A}^{\prime}$ is also medial.

Further, since $L_{c}$ annihilate $\mathbb{A}_{c}(0)$,

$(c+\mathbb{A}_{c}(0))^{2}=c^{2}+\mathbb{A}_{c}(0)^{2}=c+\mathbb{A}_{c}(0),$

therefore the class $c^{\prime}=c+\mathbb{A}_{c}(0)$ is an idempotent in
$\mathbb{A}^{\prime}$. Since $\mathbb{A}_{c}(0)$ is an ideal we have

$\displaystyle\mathbb{A}_{c^{\prime}}(0)$
$\displaystyle=\\{x+\mathbb{A}_{c}(0):(c+\mathbb{A}_{c}(0))(x+\mathbb{A}_{c}(0))=0^{\prime}=\mathbb{A}_{c}(0)\\}$
$\displaystyle=\\{x+\mathbb{A}_{c}(0):cx\in\mathbb{A}_{c}(0)\\}$
$\displaystyle=\\{x+\mathbb{A}_{c}(0):L_{c}^{2}x=0\\}$

Suppose additionally that $c$ is semisimple. Then we can decompose $x$ into a
direct sum $x=x_{0}+\sum_{\lambda\neq 0}x_{\lambda}$, where
$x_{\lambda}\in\mathbb{A}_{c}(\lambda)$. This yields

$0=L_{c}^{2}x=\sum_{\lambda\neq 0}\lambda^{2}x_{\lambda},$

hence $x_{\lambda}=0$ for all $\lambda\neq 0$, i.e.
$x=x_{0}\in\mathbb{A}_{c}(0)$. This implies $x^{\prime}=0^{\prime}$, hence the
proposition follows. ∎

######  Definition 4.3.

A medial algebra $\mathbb{A}$ is said to be reduced if
$\operatorname{Idm}^{\times}(\mathbb{A})=\operatorname{Idm}(\mathbb{A})$.

######  Remark 4.4.

Note that
$\dim\mathbb{A}/\mathbb{A}_{c}(0)=\dim\mathbb{A}-\dim\mathbb{A}_{c}(0)<\dim\mathbb{A}$
if $\mathbb{A}_{c}(0)$ is nontrivial. Hence, continuing in an obvious way
factorizing out of $\mathbb{A}$ nontrivial Peirce zero subspaces, one arrives
in finitely many steps to a reduced medial algebra $\tilde{A}$. We shall
consider reduced medial algebras in more detail below.

###  4.2. The Peirce structure of reduced medial algebras

Throughout this section we assume, unless stated otherwise, that $\mathbb{A}$
is a reduced medial algebra of dimension $\geq 2$. Since $\ker L_{c}=0$ for
any idempotent $c$, $L_{c}$ is an algebra automorphism. Given
$\lambda\in\mathbf{K}$ we denote

$\mathbb{A}_{c}(\lambda)=\ker(L_{c}-\lambda\mathbf{1})$

######  Proposition 4.5.

Let $c_{1},c_{2}\in\operatorname{Idm}(\mathbb{A})$. Then
$c_{1}c_{2}\in\operatorname{Idm}(\mathbb{A})$ and

$\displaystyle\mathbb{A}_{c_{1}}(\lambda_{1})\mathbb{A}_{c_{2}}(\lambda_{2})$
$\displaystyle\subset\mathbb{A}_{c_{1}c_{2}}(\lambda_{1}\lambda_{2}),$ (36)
$\displaystyle\mathbb{A}_{c_{1}}(\lambda_{1})\mathbb{A}_{c_{1}}(\lambda_{2})$
$\displaystyle\subset\mathbb{A}_{c_{1}}(\lambda_{1}\lambda_{2}).$ (37)

In particular,

$\sigma(c_{1})\cup\sigma(c_{2})\subset\sigma(c_{1}c_{2})$ (38)

and

$\dim\mathbb{A}_{c_{j}}(\lambda_{j})\leq\dim\mathbb{A}_{c_{i}c_{j}}(\lambda_{j}),\qquad\forall\lambda_{i}\in\sigma(c_{i}),i=1,2.$
(39)

###### Proof.

If some of $\mathbb{A}_{c_{1}}(\lambda_{1})$ and
$\mathbb{A}_{c_{2}}(\lambda_{2})$ is zero, (36), (37) are trivial. Suppose
that both $\mathbb{A}_{c_{1}}(\lambda_{1})$ and
$\mathbb{A}_{c_{2}}(\lambda_{2})$ are nonzero and let
$c_{i}x_{i}=\lambda_{i}x_{i}$, $x_{i}\neq 0$, $i=1,2$. Then $c_{3}=c_{1}c_{2}$
is an idempotent, and, since $\mathbb{A}$ is reduced, we have $c_{3}\neq 0$.
Using (7) obtain

$c_{3}(x_{1}x_{2})=(c_{1}c_{2})(x_{1}x_{2})=(c_{1}x_{1})(c_{2}x_{2})=\lambda_{1}\lambda_{2}x_{1}x_{2}.$

Thus $x_{1}x_{2}\in\mathbb{A}_{c_{3}}(\lambda_{1}\lambda_{2})$. This implies
(36). Setting $c_{2}=c_{1}$ we arrive at (37). Finally, since $x_{2}$ is a
$\lambda_{2}$-eigenvector of $c_{2}$,

$\lambda_{2}c_{1}x_{2}=c_{1}\lambda_{2}x_{2}=c_{1}(c_{2}x_{2})=(c_{1}c_{1})(c_{2}x_{2})=(c_{1}c_{2})(c_{1}x_{2})=c_{3}(c_{1}x_{2}).$

Since $\mathbb{A}$ is reduced, we have $c_{1}x_{2}\neq 0$. Therefore
$c_{1}x_{2}$ is a $\lambda_{2}$-eigenvector of $c_{3}$. This proves (38).
Furthermore, since $L_{c_{1}}$ is non-degenerated linear endomorphism, we also
deduce that

$0\neq
L_{c_{1}}(\mathbb{A}_{c_{2}}(\lambda_{2}))\subset\mathbb{A}_{c_{1}c_{2}}(\lambda_{2}).$
(40)

Since $\mathbb{A}$ is reduced, (40) implies (39). ∎

Note that the product in the left hand side maybe zero such that the inclusion
in (37) is trivial in that case.

######  Proposition 4.6.

If $c_{1}$ is a semi-simple idempotent then for any idempotent
$c_{2}\in\operatorname{Idm}(\mathbb{A})$, the idempotent $c_{1}c_{2}$ is also
semi-simple and

$\operatorname{spec}(c_{1})=\operatorname{spec}(c_{1}c_{2}).$

###### Proof.

By the assumption we have the following direct sum decomposition:

$A=\bigoplus_{\lambda_{i}\in\sigma(c_{1})}\mathbb{A}_{c_{1}}(\lambda_{i}),$

where all $\lambda_{i}$ are distinct. Since $\mathbb{A}$ is reduced,
$L_{c_{2}}$ is a nondegenerate endomorphism of $\mathbb{A}$, hence

$A=\bigoplus_{\lambda_{i}\in\sigma(c_{1})}L_{c_{2}}(\mathbb{A}_{c_{1}}(\lambda_{i})).$

Comparing this with (40) and (39) imply for dimensional reasons that

$L_{c_{2}}(\mathbb{A}_{c_{1}}(\lambda_{i}))=\mathbb{A}_{c_{1}c_{2}}(\lambda_{i}),$

hence

$A=\bigoplus_{\lambda_{i}\in\sigma(c_{1})}\mathbb{A}_{c_{1}c_{2}}(\lambda_{i}).$

yields that $c_{1}c_{2}$ is a semi-simple idempotent. Since
$\dim\mathbb{A}_{c_{1}c_{2}}(\lambda_{i})=\dim\mathbb{A}_{c_{1}}(\lambda_{i})$,
the spectra of $c_{1}c_{2}$ and $c_{1}$ are identical. ∎

######  Corollary 4.7.

If $c_{1}$ and $c_{2}$ are semi-simple idempotents then
$\operatorname{spec}(c_{1})=\operatorname{spec}(c_{2})$.

###### Proof.

It follows from Proposition 4.6 that the spectra of $c_{1}$, $c_{2}$ and
$c_{1}c_{2}$ are identical, that implies the desired conclusion. ∎

######  Corollary 4.8.

If all idempotents of a reduced medial algebra $\mathbb{A}$ are semi-simple
then $\mathbb{A}$ is isospectral.

######  Corollary 4.9.

If a reduced medial algebra $\mathbb{A}$ is generic and all its idempotents
are semi-simple then $\mathbb{A}$ is isospectral and its spectrum consists of
the primitive roots of unity of degree $\dim\mathbb{A}$:

$\operatorname{spec}(c)=\\{\epsilon_{n}^{k}:1\leq k\leq n=\dim
A\\},\qquad\forall c\in\operatorname{Idm}(\mathbb{A}).$

###### Proof.

Follows from the syzygy relation in Theorem 2.4. ∎

##  5\. Isospectral algebras

###  5.1. Basic properties

######  Definition 5.1.

A commutative algebra is called isospectral if all its nonzero idempotents
have the same spectrum.

We point out that we require in the definition that the spectra of each two
idempotents coincide as a multi-set, i.e. counting the multiplicities of all
eigenvalues.

In the generic case, one a priori knows that there exists exactly the maximal
number $2^{n}-1$ idempotents, where $n=\dim\mathbb{A}$, so that the condition
‘all idempotents’ has a clear sense. In the nongeneric case, however, there
may exist either nontrivial 2-nilpotent elements or infinitely many
idempotents, and thus, containing the exceptional value $\frac{1}{2}$ in the
spectrum, see [35], [60]. The constancy of the spectrum still has an exact
meaning, but the situation in the nongeneric case will be completely
different. For example, if there is only few distinct idempotents (but many
nilpotents) then the constancy spectrum condition is not so much restrictive.
On the other hand, there are nonisospectral generic algebras, for example,
Hsiang algebras (see [60] for the definition and basic properties of Hsiang
algebras). It would be interesting to characterize the Peirce spectrum of
nonisospectral generic algebras.

In this paper we shall primarily concern with the generic case. Then the most
natural choice of the ground field is the complex numbers $\mathbb{C}$ or an
algebraically closed field $\mathbf{K}$. Then by Corollary 5.2 in [35] the
spectrum of a generic algebra consists of the primitive roots of unity of
order $n=\dim\mathbb{A}$. More precisely, applying the syzygies constructed in
[35], one derives a more precise description.

######  Proposition 5.2.

Let $\mathbb{A}$ be a generic isospectral algebra over an algebraically closed
field $\mathbf{K}$. Let $n=\dim\mathbb{A}$. Then for any
$c\in\operatorname{Idm}(\mathbb{A})$

$\operatorname{spec}(c)=\\{\epsilon_{n}^{k}:1\leq k\leq n\\}$ (41)

and the following Peirce decomposition holds:

$A=\bigoplus_{k=0}^{n-1}\mathbb{A}_{c}(\epsilon_{n}^{k})=\bigoplus_{k=0}^{n-1}\langle
w_{k}\rangle,$

where $\\{w_{0},w_{1},\ldots,w_{n-1}\\}$ is an eigenbasis of $L_{c}$:
$cw_{k}=\epsilon_{n}^{k}w_{k}$, where the corresponding Peirce eigenspaces are
one-dimensional:

$\mathbb{A}_{c}(\epsilon_{n}^{k})=\langle w_{k}\rangle=\mathbf{K}w_{k}.$ (42)

In particular, the algebra $\mathbb{A}$ does not contain $2$-nilpotents and

$L_{c}^{n}=1,\qquad\forall c\in\operatorname{Idm}(\mathbb{A}).$ (43)

Furthermore, for any homogeneous polynomial map $H:\mathbb{A}\to\mathbb{A}$ of
degree $1\leq m\leq n-1$,

$\displaystyle\sum_{c\in\operatorname{Idm}(\mathbb{A})}H(c)$
$\displaystyle=0.$ (44)

In particular,

$\displaystyle\sum_{c\in\operatorname{Idm}(\mathbb{A})}c$ $\displaystyle=0.$
(45)

###### Proof.

By the assumptions, there exists exactly $2^{n}-1$ nonzero idempotents in
$\mathbb{A}$ and the characteristic polynomials $\chi_{c}(t)=P(t)$ for some
fixed polynomial $P$ of degree $n$. Applying Theorem 2.3, we obtain from (15)
that $(2^{n}-1)P(t)=2^{n}(1-t^{n})P(\frac{1}{2})$ which immediately implies
$P(t)=t^{n}-1$. Since the spectrum of each nonzero idempotent $c$ in
$\mathbb{A}$ is the multiset of all roots of $P(t)$, we conclude that $L_{c}$
has exactly $n$ distinct eigenvalues
$\\{1,\epsilon_{n},\ldots,\epsilon_{n}^{n-1}\\}$, where $\epsilon_{n}$ is an
$n$th primitive root of unity in $\mathbf{K}$. In particular, for dimension
reasons, we conclude that for each eigenvalue $\epsilon_{n}^{k}$ its
eigenspace is one-dimensional. Furthermore, (44) and (45) follows from (16)
and (17) respectively. ∎

Note that in Proposition 5.2 one can additionally assume that $w_{0}=c$. This
motivates

######  Definition 5.3.

Let $c\in\operatorname{Idm}(\mathbb{A})$. A basis $\\{w_{i}\\}_{0\leq i\leq
n-1}$ is called an eigenbasis associated with $c$ if $w_{0}=c$ and

$cw_{i}=\epsilon_{n}^{i}w_{i}\quad\text{for all $1\leq i\leq n-1$}.$

Let $y\in\mathbb{A}$ be a fixed element and $1\leq m\leq n-1$ be an integer.
Applying $H(x)=L_{x}^{m}y$ to (44) yields

$\sum_{c\in\operatorname{Idm}(\mathbb{A})}L_{c}^{m}y=0.$ (46)

######  Corollary 5.4.

Let $\mathbb{A}$ be an isospectral generic algebra. For any
$c\in\operatorname{Idm}(\mathbb{A})$ and integer $k$, there exists a linear
form $\theta_{c,k}(y):\mathbb{A}\to\mathbf{K}$ such that

$y+\epsilon_{n}^{-k}L_{c}y+\epsilon_{n}^{-2k}L_{c}^{2}y+\ldots+\epsilon_{n}^{-(n-1)k}L_{c}^{n-1}y=\theta_{c,k}(y)w_{k},$
(47)

and

$\theta_{c,k}(w_{j})=\left\\{\begin{array}[]{cl}n&\text{if $k=j$}\\\
0&\text{if $k\neq j$}.\end{array}\right.$

In particular, if $y\neq 0$ then $y,L_{c}y,L_{c}^{2}y,\ldots,L_{c}^{n-1}y$ is
a basis of $\mathbb{A}$ as a vector space and

$y+L_{c}y+L_{c}^{2}y+\ldots+L_{c}^{n-1}y=\theta_{c,0}(y)c.$ (48)

###### Proof.

Let $z$ denote the left hand side of (47). Applying $L_{c}$ to $z$ and
rearranging we obtain

$\displaystyle L_{c}z$
$\displaystyle=L_{c}y+\epsilon_{n}^{-k}L^{2}_{c}y+\epsilon_{n}^{-2k}L_{c}^{2}y+\ldots+\epsilon_{n}^{-(n-1)k}L_{c}^{n}y$
$\displaystyle=\epsilon_{n}^{k}L_{c}^{n}y+L_{c}y+\epsilon_{n}^{-k}L^{2}_{c}y+\ldots+\epsilon_{n}^{-(n-2)k}L_{c}^{n-1}y$
$\displaystyle=\epsilon_{n}^{k}(y+\epsilon_{n}^{-k}L^{2}_{c}y+\epsilon_{n}^{-2k}L_{c}^{2}y+\ldots+\epsilon_{n}^{-(n-1)k}L_{c}^{n-1}y)$
$\displaystyle=\epsilon_{n}^{k}z,$

hence $z\in\mathbb{A}_{c}(\epsilon_{n}^{k})$. By Proposition 5.2,
$\mathbb{A}_{c}(\epsilon_{n}^{k})=\langle w_{k}\rangle$, hence $z=\lambda
w_{k}$, for some complex number $\lambda$. Clearly, $\theta:y\to\lambda$ is a
linear form, and (47) follows. Substituting $y=w_{j}$ in (47) yields

$\displaystyle\theta_{c,k}(w_{j})w_{k}$
$\displaystyle=(1+\epsilon_{n}^{j-k}+\epsilon_{n}^{2(j-k)}+\ldots+\epsilon_{n}^{(n-1)(j-k)})w_{j}$
$\displaystyle=w_{j}\left\\{\begin{array}[]{cl}n&\text{if $k=j$}\\\
\frac{\epsilon_{n}^{n(j-k)}-1}{\epsilon_{n}^{j-k}-1}&\text{if $k\neq
j$}.\end{array}\right.$

Since $\epsilon_{n}^{n(j-k)}=1$, the sum vanishes whenever $j\neq k$, thus the
claim follows. Also, if we consider (47) as a system of $n$ equations (for
$0\leq k\leq n-1$) with respect to $y,L_{c}y,L_{c}^{2}y,\ldots,L_{c}^{n-1}y$
then its determinant is of Vandermond type and it is obviously different of
zero. Since the system $\\{w_{0},w_{1},\ldots,w_{n-1}\\}$ is a basis in $A$,
so is also $y,L_{c}y,L_{c}^{2}y,\ldots,L_{c}^{n-1}y$. ∎

Summing up the identities (48) for all $c\in\operatorname{Idm}(\mathbb{A})$,
we obtain using (46) that

######  Corollary 5.5.

For any $y\in\mathbb{A}$

$y=\frac{1}{(2^{n}-1)}\sum_{c\in\operatorname{Idm}(\mathbb{A})}\theta_{c,0}(y)c.$
(49)

In particular, $\operatorname{Idm}(\mathbb{A})$ spans $\mathbb{A}$, i.e. any
isospectral generic algebra is generated by idempotents (even on the level of
the vector space).

######  Remark 5.6.

Combining the latter with the property that any idempotent has the same
spectrum and fusion rules, one concludes that an isospectral algebra is axial
in the sense of definitions in [24], [46], [33], [39]. The axes in the present
context are all idempotents of $\mathbb{A}$.

###  5.2. The generic trace and the generic determinant

Let $c\in\operatorname{Idm}(\mathbb{A})$ and let $\\{w_{i}\\}_{0\leq i\leq
n-1}$ be an eigenbasis associated with $c$. Define for any
$x=\sum_{i=0}^{n-1}a_{i}w_{i}$ the generic trace

$T(x):=\sum_{i=0}^{n-1}a_{i}.$

We have

$L_{c}^{k}x=\sum_{i=0}^{n-1}a_{i}L_{c}^{k}w_{i}=\sum_{i=0}^{n-1}\epsilon_{n}^{ik}a_{i}w_{i},$

hence

$T(L_{c}^{k}x)=\sum_{i=0}^{n-1}\epsilon_{n}^{ik}a_{i}.$

Let us define

$\delta(x):=\prod_{k=0}^{n-1}T(L_{c}^{k}x)=\det X(x),$ (50)

where $X$ is the circulant matrix of $x$:

$X_{i,j}=x_{\bar{j}-\bar{i}+1},\qquad\bar{i},\bar{j}\in\mathbb{Z}_{n}.$

The product will be called the generic determinant of $x$ in $\mathbb{A}$.

###  5.3. The generic determinant

######  Theorem 5.7.

Let $\mathbb{A}$ be a isospectral generic algebra. Then it satisfies the
identity

$L_{x}^{n}y=\delta(x)y,$ (51)

where $\delta(x)$ is the generic determinant of $x$.

##  6\. Medial isospectral algebras

Throughout this section, $\mathbb{A}$ denotes a medial generic isospectral
algebra over an algebraically closed subfield $\mathbf{K}$ of $\mathbb{C}$.
Let $N=2^{n}-1$ and let $\\{c_{i}\\}_{1\leq i\leq N}$ denote the set the
distinct nonzero idempotents. The main result of this section is

######  Theorem 6.1.

If two medial generic isospectral algebras over an algebraically closed field
$\mathbf{K}$ have the same dimension they are isomorphic.

We begin with some elementary observations.

######  Proposition 6.2.

Let $\mathbb{A}$ be a medial generic isospectral algebra. Then

  1. (a)

Any idempotent $c\in\operatorname{Idm}(\mathbb{A})$ is semi-simple.

  2. (b)

For any two idempotents $c_{i},c_{j}\in\operatorname{Idm}(\mathbb{A})$ the
product $c_{i}c_{j}\in\operatorname{Idm}(\mathbb{A})$.

  3. (c)

For any $0\neq x\in\mathbb{A}$, $x^{2}\neq 0$.

  4. (d)

If $c\in\operatorname{Idm}(\mathbb{A})$ and
$\mathbb{A}_{c}(\epsilon_{n}^{i})=\langle w_{i}\rangle$ then

$w_{k}w_{j}\in\mathbb{A}_{c}(\epsilon_{n}^{k+j})=\langle
w_{j+k}\rangle,\qquad\forall k,j.$ (52)

###### Proof.

If $c\in\operatorname{Idm}(\mathbb{A})$ then $L_{c}^{n}=\mathbf{1}$ has only
simple roots. Since the field is algebraically closed, for any eigenvalue
$\epsilon_{n}^{k}$, $0\leq k\leq n-1$, there exists an eigenvector which
implies for dimension reasons that each eigenspace
$\mathbb{A}_{c}(\epsilon_{n}^{k})$ has dimension 1. This implies that
$\mathbb{A}$ decomposes into the direct sum $\bigoplus_{1\leq i\leq
n-1}\mathbb{A}_{c}(\epsilon_{n}^{k})$, hence (a) follows.

If $c_{i}c_{j}=0$ then $c_{j}$ is an eigenvector of $c_{i}$ with eigenvalue
$0$, a contradiction with $L_{c_{i}}^{n}=1$ follows. The fact that
$c_{i}c_{j}$ is an idempotent follows from Proposition 3.5. This proves (b)

Since $\mathbb{A}$ is generic, it does not contain 2-nilpotent elements (see
Remark 3.2 in [35]), hence (c) follows. Finally, by (36) and (42) we have

$w_{j}w_{k}\subset\mathbb{A}_{c}(\epsilon_{n}^{j})\mathbb{A}_{c}(\epsilon_{n}^{j})\subset\mathbb{A}_{c}(\epsilon_{n}^{j+k})=\langle
w_{j+k}\rangle.$

as desired. ∎

To proceed we need to show that the product in (52) never vanishes. First let
us recall some standard definitions.

A nonassociative monomial $z^{\alpha}$ is an element of the commutative
multiplicative groupoid generated by a single element $z$. The total degree
$\deg z^{\alpha}$ is the number of elements of $z$ used in the word
$z^{\alpha}$, counting multiplicities. For example, $\deg z^{2}z^{2}=4$. The
simplest nonassociative monomials are the principal powers, defined by
induction as

$z^{1}=z,\quad z^{n+1}=zz^{n},\quad n\geq 1.$ (53)

Then $\deg z^{m}=m$.

######  Proposition 6.3.

Let $\mathbb{A}$ be a medial generic isospectral algebra and let
$\mathbb{A}_{c}(\epsilon_{n})=\langle w_{1}\rangle$. Then

  1. (i)

$w_{k}=a_{k}w_{1}^{m}$, for some $0\neq a_{k}\in\mathbf{K}$;

  2. (ii)

$w_{k}w_{j}\neq 0$ for any $k,j$;

  3. (iii)

one can choose the Peirce eigenbasis such that $w_{m}=w_{1}^{m}$ for all
$0\leq m\leq n-1$, where $w_{1}^{0}=c$.

###### Proof.

Let $w_{k}$ and $w_{j}$ be eigenvectors of $c$ such that $w_{k}w_{j}=0$. Then
for any $x\in\mathbb{A}$ there holds $(xw_{k})w_{j}=w_{k}(xw_{j})=0$. Indeed,

$(w_{k}x)w_{j}=\frac{1}{\epsilon_{n}^{j}}(w_{k}x)(w_{j}c)=\frac{1}{\epsilon_{n}^{j}}(w_{k}w_{j})(xc)=0.$

We refer to this as the cancelation property.

We claim that any principal power of $w_{j}$ is nonzero. Indeed,
$w_{j}^{2}\neq 0$ by (c) in Proposition 6.2. Let $w_{j}^{m}=0$ for some
principal power $m$ and let $m$ be the smallest such number. Then $m\geq 3$.
Using the medial algebra relation,

$\displaystyle w_{1}^{m-1}w_{j}^{m-1}$
$\displaystyle=\frac{1}{\epsilon_{n}^{m-1}}(w_{j}^{m-1}c)(w_{j}w_{j}^{m-2})=\frac{1}{\epsilon_{n}^{m-1}}(w_{j}w_{j}^{m-1})(w_{j}^{m-2}c)$
$\displaystyle=\frac{1}{\epsilon_{n}^{m-1}}(w_{j}^{m})(w_{j}^{m-2}c)=0,$

a contradiction with $w_{j}^{m-1}\neq 0$ and (c) follows. This proves our
claim.

By (52), $w_{k}w_{j}\subset\mathbb{A}_{c}(\epsilon_{n}^{k+j})$. Moreover,
since the latter eigenspace is one-dimensional, $\langle
w_{k}w_{j}\rangle=\mathbb{A}_{c}(\epsilon_{n}^{k+j})$ whenever $w_{k}w_{j}\neq
0$. Since $w_{1}^{2}\neq 0$, $\langle
w_{1}^{2}\rangle=\mathbb{A}_{c}(\epsilon_{n}^{2})=\langle w_{2}\rangle$. Using
the fact that the principal powers of $w_{1}$ are nonzero we conclude by
indiction that

$\langle w_{1}^{m}\rangle=\mathbb{A}_{c}(\epsilon_{n}^{m})=\langle
w_{m}\rangle,\qquad\forall m.$

Hence (i) follows. Furthermore, scaling $w_{m}$, one may assume without loss
of generality that

$w_{m}=w_{1}^{m},\qquad\forall m,$ (54)

which proves (iii). Finally, suppose that (iii) holds and suppose by
contradiction that $w_{k}w_{m}=0$ for some $k,m$. Then $w_{1}^{k}w_{1}^{m}=0$
and $k\neq m$ by (c) in Proposition 6.2. Suppose that $k<m$. By the
cancelation property, $w_{1}^{k+1}w_{1}^{m}=(w_{1}w_{1}^{k})w_{1}^{m}=0$, etc,
implying in $m-k$ steps that $w_{1}^{m}w_{1}^{m}=0$, a contradiction with (c)
follows. This proves (ii). ∎

######  Remark 6.4.

Note that the subalgebra generated by any $w_{j}$ is not power-associative.
Indeed,

$w_{1}^{2}w_{1}^{2}=\frac{1}{\epsilon_{n}^{2}}(cw_{1}^{2})(w_{1}w_{1})=\frac{1}{\epsilon_{n}^{2}}(cw_{1})(w_{1}^{2}w_{1})=\frac{1}{\epsilon_{n}}w_{1}^{4},$

hence $w_{1}^{4}\neq w_{1}^{2}w_{1}^{2}$.

In fact, one has the following general observation.

######  Proposition 6.5.

The subalgebra generated by $w_{1}$ is weakly power-associative in the sense
that any nonassociative monomial is proportional to the corresponding
principal power:

$w_{1}^{\alpha}=bw_{1}^{\deg w_{1}^{\alpha}},\qquad b=\epsilon_{n}^{s},\quad
0\leq s\leq n-1.$ (55)

Furthermore, the product of two principal powers can be explicitly found by

$w_{1}^{k}w_{1}^{m}=\frac{1}{\epsilon_{n}^{(k-1)(m-1)}}w_{1}^{k+m},\qquad
k,m\geq 1.$ (56)

###### Proof.

The proof of (56) follows from the recurrence identity

$w_{1}^{k}w_{1}^{m}=\frac{1}{\epsilon_{n}^{m}}(w_{1}w_{1}^{k-1})(cw_{1}^{m})=\frac{1}{\epsilon_{n}^{m-1}}w_{1}(w_{1}^{k-1}w_{1}^{m})$

by induction. Since the general nonassociative monomial is a successive
product of principal power, one arrives at (55). ∎

###### Proof of Theorem 6.1.

Let $\mathbb{A}$ be a medial generic isospectral algebra. Given an idempotent
$c$, let $\\{w_{1}^{i}\\}_{0\leq i\leq n-1}$ be a Peirce eigenbasis chosen
accordingly (iii) in Proposition 6.3. The algebra structure of an algebra is
completely determined by product of its basis elements (56). This implies that
two medial generic isospectral algebras of equal dimensions are isomotphic. ∎

##  7\. The cyclotomic polynomial model

###  7.1. The definition of $\mathscr{C}_{n}$

Let $n\geq 2$. Consider the quotient

$\mathscr{C}_{n}=\mathbf{K}[z]/z^{n}-1$

with a (commutative but nonassociative) multiplication on $\mathscr{C}_{n}$
defined by

$p(z)\circ q(z)=p(\epsilon_{n}z)q(\epsilon_{n}z)\mod z^{n}-1,$

where $\epsilon_{n}$ is a primitive root of unity of order $n$. Denote the
corresponding algebra by $\mathscr{C}_{n}(z,\circ)$, or just $\mathscr{C}_{n}$
when there is no ambiguity. Such defined multiplication is obviously
commutative.

######  Remark 7.1.

The multiplication on $\mathscr{C}_{n}$ is an isotopy of the usual
(associative commutative) multiplication on
$\mathbb{A}_{n}:=\mathbf{K}[z]/z^{n}-1$. Indeed, define
$f:\mathbb{A}_{n}\to\mathscr{C}_{n}$ by $f(p(z))=p(\epsilon_{n}z)$, then $h$
is a nonsingular linear transformation and

$p(z)\circ q(z)=h(p(z)q(z)).$

Thus, $\mathscr{C}_{n}$ is an isotopy of $\mathbb{A}_{n}$ by a cyclic linear
substitution.

######  Example 7.2.

Let us demonstrate the multiplication by the table of multiplication of
monomials for $n=3$:

| $1$ | $z$ | $z^{2}$
---|---|---|---
$1$ | $1$ | $\epsilon_{3}z$ | $\epsilon_{3}^{2}z^{2}$
$z$ | $\epsilon_{3}z$ | $\epsilon_{3}^{2}z^{2}$ | $1$
$z^{2}$ | $\epsilon_{3}^{2}z^{2}$ | $1$ | $\epsilon_{3}z$

This in particular shows that

$1\circ(1\circ z)=1\circ\epsilon_{3}z=\epsilon_{3}^{2}z,$

while

$(1\circ 1)\circ z=1\circ z=\epsilon_{3}z.$

A similar property holds for any $n\geq 2$ which shows that the algebra
$\mathscr{C}_{n}$ is nonassociative.

The element $1$ is obviously an idempotent of $\mathscr{C}_{n}$ for any $n$
and furthermore

$L^{\circ}_{1}z^{k}=\epsilon_{n}^{k}z^{k},$

i.e. $z^{k}$ is an eigenvector of $L^{\circ}_{1}$ with eigenvalue
$\epsilon_{n}^{k}$. In other words, $\\{z^{k}\\}_{0\leq k\leq n-1}$ is a
Peirce eigenbasis of $1$. Each eigenvalue is simple.

The main result of this section is

######  Theorem 7.3.

The algebra $\mathscr{C}_{n}$ is medial. Furthermore,

$L^{\circ n}_{p(z)}=\Delta(p)\mathbf{1},$ (57)

where

$\Delta(p):=p(1)p(\epsilon_{n})p(\epsilon_{n}^{2})\cdots
p(\epsilon_{n}^{n-1}):\mathscr{C}_{n}\to\mathbf{K}$

is a multiplicative homomorphism.

###### Proof.

Indeed, $p(z)\circ q(z)=p(\epsilon_{n}z)q(\epsilon_{n}z)$ and $r(z)\circ
s(z)=s(\epsilon_{n}z)r(\epsilon_{n}z)$, therefore

$(p(z)\circ q(z))\circ(r(z)\circ
s(z))=p(\epsilon_{n}^{2}z)q(\epsilon_{n}^{2}z)r(\epsilon_{n}^{2}z)s(\epsilon_{n}^{2}z)\mod
z^{n}-1.$

The latter product is totally symmetric in $p,q,r,s$ which implies the first
claim. Next, note that

$L^{\circ 2}_{p(z)}q(z)=p(z)\circ(p(z)\circ q(z))\equiv
p(\epsilon_{n}z)p(\epsilon_{n}^{2}z)q(\epsilon_{n}^{2}z)\mod
z^{n}-1,\quad\text{etc}$

hence by induction

$\displaystyle L^{\circ n}_{p(z)}q(z)$ $\displaystyle\equiv
p(\epsilon_{n}z)p(\epsilon_{n}^{2}z)\cdots
p(\epsilon_{n}^{n}z)q(\epsilon_{n}^{n}z)\mod z^{n}-1$ $\displaystyle\equiv
p(\epsilon_{n}z)p(\epsilon_{n}^{2}z)\cdots p(z)q(z)\mod z^{n}-1$
$\displaystyle\equiv P(z)q(z)\mod z^{n}-1,$

where

$P(z)=p(z)p(\epsilon_{n}z)p(\epsilon_{n}^{2}z)\cdots p(\epsilon_{n}^{n-1}z).$

Since $P(\epsilon_{n}^{i}z)=P(z)$ for all $i$, there exists a polynomial $Q$
such that $P(z)=Q_{p}(z^{n})$. Therefore

$P(z)=Q_{p}(z^{n})\equiv Q_{p}(1)\equiv P(1)\mod z^{n}-1.$

This shows that

$L^{\circ n}_{p(z)}q(z)\equiv P(1)q(z)\mod z^{n}-1,$

which proves (57). Finally,

$\Delta(p(z)q(z))=\prod_{i=0}^{n-1}p(\epsilon_{n}^{i})q(\epsilon_{n}^{i})=\prod_{i=0}^{n-1}p(\epsilon_{n}^{i})\prod_{i=0}^{n-1}q(\epsilon_{n}^{i})=\Delta(p(z))\Delta(q(z)),$

hence $\Delta$ is a multiplicative homomorphism. ∎

It follows from the Sylvester formula for the resultant of two polynomials
that

$\Delta(p):=\prod_{z:z^{n}-1=0}p(z)=\mathcal{R}(p(z),z^{n}-1)=\det\mathrm{circ}(p),$
(58)

where $\mathrm{circ}(p)$ is the circulant matrix associated with $p$.

###  7.2. The idempotents in $\mathscr{C}_{n}$

We have seen above that the element $1$ is an idempotent of $\mathscr{C}_{n}$.
Below we consider the general idempotents of $\mathscr{C}_{n}$.

Let $p(z)\in\mathscr{C}_{n}$. Define the $\deg_{R}p(z)$ as the smallest degree
of a polynomial $q$ such that $p\equiv q\mod z^{n}-1$. It is easy to see that
there exists a unique polynomial of this kind.

Let $p(z)\in\mathscr{C}_{n}$ be an idempotent distinct from $1$. Then the
idempotency relation reads as follows:

$p(z)=p(z)\circ p(z)=p^{2}(\epsilon_{n}z)\mod z^{n}-1,$

which is equivalent to the existence of a polynomial $q(z)$ such that

$p^{2}(\epsilon_{n}z)-p(z)=Q(z)(z^{n}-1)$ (59)

Since $p(z)$ distinct from $1$, the polynomial $Q(z)$ is not identically zero.

######  Proposition 7.4.

If $p(z)\in\operatorname{Idm}(\mathscr{C}_{n})$ and $p(z)\not\equiv 1$ then
$\Delta(p)=1$. Moreover, the Peirce spectrum of $p(z)$ is exactly

$\sigma(p(z))=\\{\epsilon_{n}^{i}:0\leq i\leq n-1\\}.$ (60)

In particular, any idempotent $p(z)\in\mathscr{C}_{n}$ is semi-simple.

###### Proof.

Let $p(z)$ be a nonzero idempotent in $\mathscr{C}_{n}$. For any $0\leq k\leq
n-1$, $\epsilon_{n}^{k}$ is a root of the right hand side of (59), therefore

$p^{2}(\epsilon_{n}^{k+1})=p(\epsilon_{n}^{k}).$ (61)

Taking the product readily yields by virtue of the cyclicity of
$\epsilon_{n}^{k}$ that

$\Delta(p)=\prod_{k=0}^{n-1}p(\epsilon_{n}^{k})\in\\{0,1\\}.$

Let us show that $\Delta(p)=0$ is impossible. Indeed, if $\Delta(p)=0$ then
$p(\epsilon_{n}^{k})=0$ for some $k$. It follows from (59) that
$p^{2}(\epsilon^{k+1})=0$, thus $\epsilon^{k+1}$ also a root of $p$. This
implies that all $\epsilon^{i}$, $0\leq i\leq n-1$ are the roots of $p$, i.e.
$z^{n}-1$ divides $p(z)$, therefore implying that $p\equiv 0$ in
$\mathscr{C}_{n}$. Therefore, if $p(z)$ is an idempotent then $\Delta(p)=1$.
Next, by (57) $L^{\circ n}_{p(z)}=\mathbf{1},$ which implies that any
eigenvalue of $L^{\circ}_{p(z)}$ is a root of unity of degree $n$.

Let us show that the Peirce spectrum is given by (60). If
$p(z)\in\operatorname{Idm}(\mathscr{C}_{n})$ then by the above $\Delta(p)=1$,
thus (57) yields

$L^{\circ n}_{p(z)}=\mathbf{1}.$ (62)

This implies that if $\lambda\in\sigma(p(z))$ then $\lambda^{n}-1=0$. Suppose
that $\lambda=\epsilon_{n}^{k}$, $0\leq k\leq n-1$. Consider $q(z)=p(z)z^{k}$.
Then using (59)

$q(z)\circ p(z)\equiv
q(\epsilon_{n}z)p(\epsilon_{n}z)\equiv\epsilon_{n}^{k}z^{k}p(\epsilon_{n}z)p(\epsilon_{n}z)\equiv\epsilon_{n}^{k}z^{k}p(z)\equiv\epsilon_{n}^{k}q(z)\mod
z^{n}-1,$

which shows that $q(z)=z^{k}p(z)$ is a (nonzero in $\mathscr{C}_{n}$)
eigenvector with eigenvalue $\epsilon_{n}^{k}$. For dimensional reasons, the
equality (60) holds and each eigenvalue is simple. Finally, note that since
the eigenvectors $w_{k}=z^{k}p(z)$, $k=0,1,\ldots,n-1$, correspond to
different eigenvalues they form a basis in $\mathscr{C}_{n}$. Thus $p(z)$ is a
semi-simple idempotent. ∎

######  Proposition 7.5.

The algebra $\mathscr{C}_{n}$ is generic.

###### Proof.

First note that $\operatorname{Idm}(\mathscr{C}_{n})$ is nonempty because, for
example, $p_{0}(z)=1\in\operatorname{Idm}(\mathscr{C}_{n})$. We claim that
there no nontrivial 2-nilpotents in $\mathscr{C}_{n}$. Indeed, suppose that
$p(z)\circ p(z)=0$ in $\mathscr{C}_{n}$. The latter yields
$p^{2}(z\epsilon_{n})=(z^{n}-1)Q(z)$ for some polynomial $Q(z)$, or after
change of variables $\zeta=\epsilon_{n}z$ we have
$p^{2}(\zeta)=(\zeta^{n}-1)Q_{1}(\zeta)$. The latter implies that $p(\zeta)$
has $n$ distinct roots $\epsilon_{n}^{k}$, $0\leq k\leq n-1$, thus, by virtue
of $\deg p\leq n-1$, we conclude that $p(\zeta)\equiv 0$. This proves our
claim. Finally, by Proposition 7.4
$\begin{matrix}\frac{1}{2}\end{matrix}\not\in\sigma(p(z))$ for any idempotent
$p(z)$ of $\mathscr{C}_{n}$, and therefore
$\begin{matrix}\frac{1}{2}\end{matrix}\not\in\sigma(\mathscr{C}_{n})$.
Applying Theorem 2.3 we conclude that $\mathscr{C}_{n}$ is a generic algebra.
∎

######  Proposition 7.6.

The algebra $\mathscr{C}_{n}$ is isospectral.

###### Proof.

By Theorem 7.3 and Proposition 7.5, $\mathscr{C}_{n}$ is a medial generic
algebra and since $0\not\in\sigma(p(z))$ for any
$p(z)\in\operatorname{Idm}(\mathscr{C}_{n})$, we also conclude that
$\mathscr{C}_{n}$ is reduced. By Corollary 4.8, $\mathscr{C}_{n}$ is
isospectral. Alternatively, since $\mathscr{C}_{n}$ is generic it has the
maximal number of idempotents, and by Proposition 7.4 each idempotent is semi-
simple with the same spectrum. ∎

##  8\. The quasigroup structure of isospectral medial alegrbas

As we already seen above, idempotents of a medial algebra always form a
multiplicative magma. In fact, by the medial magma property one can deduce
that the set of idempotents is a quasigroup, see the definitions in section
2.1 above.

######  Proposition 8.1.

Let $\mathbb{A}$ be a medial generic isospectral algebra of dimension $n$.
Then $\operatorname{Idm}(\mathbb{A})$ is a idempotent medial commutative
quasigroup of order $2^{n}-1$.

###### Proof.

By (b) in Proposition 3.5, the product of two idempotents is a nonzero
idempotent. Let $c\in\operatorname{Idm}(\mathbb{A})$ be an arbitrary
idempotent. Then by the above,
$L_{c}:\operatorname{Idm}(\mathbb{A})\to\operatorname{Idm}(\mathbb{A})$ and
$L_{c}$ is a bijection because if $cc_{1}=cc_{2}$ for some $c_{1}\neq c_{2}$
then $c_{1}-c_{2}$ is a nonzero eigenvector of $L_{c}$ with eigenvalue $0$
which contradicts to (43). This proves that $\operatorname{Idm}(\mathbb{A})$
is a quasigroup. It is obviously idempotent medial and commutative. ∎

In fact we have a stronger conclusion

######  Proposition 8.2.

Let $\mathbb{A}$ be a medial generic isospectral algebra of dimension $n$ and
$c\in\operatorname{Idm}(\mathbb{A})$. Then
$Q:=(\operatorname{Idm}(\mathbb{A}),\boxplus)$ is an abelian group with the
identity $c$ with respect to the new binary operation $x\boxplus
y:=L_{c}^{-1}(xy)$.

###### Proof.

By Proposition 3.12, $x\circ y=L_{c}^{-1}(xy)$ is an isotopy of the algebra
$\mathbb{A}$ such that $(\mathbb{A},\circ)$ becomes a unital commutative
associative algebra with unit $c$. Note that by the definition $x\boxplus
y=x\circ y$ on the subset of nonzero idempotents of $\mathbb{A}$ and by
Proposition 8.1, $\boxplus$ preserves $\operatorname{Idm}(\mathbb{A})$, which
implies the desired conclusion. ∎

Proposition 8.2 shows the existence of a ‘hidden’ abelian group structure on
the set of idempotents of a medial algebra. This fact also follows from
Proposition 4.1 in the paper of A. Leibak and P Puusemp [38].

But, in fact, we have a much stronger conclusion.

######  Theorem 8.3.

The abelian group $Q:=(\operatorname{Idm}(\mathbb{A}),\boxplus)$ in
Proposition 8.2 is a cyclic group of order $2^{n}-1$.

###### Proof.

Given a fixed $c\in\operatorname{Idm}(\mathbb{A})$, we denote by
$T=L_{c}^{-1}:\mathbb{A}\to\mathbb{A}$ an algebra isomorphism. By Proposition
8.2, $Q:=(\operatorname{Idm}(\mathbb{A}),\boxplus)$ is an abelian group with
respect to $x\boxplus y:=T(xy)$ and with unity element $c$ of order $2^{n}-1$.
Since $T(xy)=T(x)T(y)$ on the algebra level and $T$ is invertible, we also
have

$xy=T^{-1}(x)\boxplus T^{-1}(y)=T^{-1}(x\boxplus y).$ (63)

In order to show that $Q$ is in fact cyclic, i.e.
$Q\cong\mathbb{Z}_{2^{n}-1}$, it suffices to show that there does not exist a
positive integer $1\leq m<2^{n}-1$ such that

$\underbrace{x\boxplus\ldots\boxplus x}_{\text{$m$ times}}=c\qquad\text{holds
for any element $x\in Q$}.$ (64)

We argue by contradiction and assume that there exist a positive integer
$1\leq m<2^{n}-1$ satisfying (64). Let us write
$m=2^{m_{1}}+\ldots+2^{m_{k-1}}+2^{m_{k}}$, $n>m_{k}>\ldots>m_{1}\geq 0$, the
binary decomposition of $m$. Using the doubling operator $T$ we obtain
$2\otimes x=x\boxplus x=T(xx)=T(x)$ for any $x\in Q$, hence $2^{s}\otimes
x=T^{s}(x)$ implying by virtue of (64)

$\underbrace{x\boxplus\ldots\boxplus x}_{\text{$m$
times}}=T^{m_{k}}(x)\boxplus\ldots\boxplus T^{m_{1}}(x)=c.$ (65)

Combining the terms into a nonassociative products we obtain using (63) and
then the fact that $T$ is an algebra isomorphism that

$T^{m_{2}}(x)\boxplus
T^{m_{1}}(x)=T(T^{m_{2}}(x)T^{m_{1}}(x))=T^{m_{2}+1}(x)T^{m_{1}+1}(x).$

Arguing similarly, we get

$T^{m_{3}}(x)\boxplus T^{m_{2}}(x)\boxplus
T^{m_{1}}(x)=T^{m_{3}+1}(x)\bigl{(}T^{m_{2}+2}(x)T^{m_{1}+2}(x)\bigr{)}\quad\text{etc}$

Define

$H(x):=T^{m_{k}+1}(x)\bigl{(}T^{m_{k-1}+2}(x)(\ldots(T^{m_{2}+k-1}(x)T^{m_{1}+k-1}(x)\bigr{)}.$

By the linearity of $T=L_{c}^{-1}$ on $\mathbb{A}$, $H(x)$ is a nonassociative
polynomial of degree $k\leq n-1$. Also $H(0)=0$ and by (65), $H(x)=c$ for all
$x\in\operatorname{Idm}(\mathbb{A})$. Hence by the syzygy relation (44) we
obtain

$(2^{n}-1)c=\sum_{x\in Q}H(x)=0,$

a contradiction finishes the proof.

∎

######  Corollary 8.4 (Theorem 2.8).

The quasigroup $\operatorname{Idm}(\mathbb{A})$ (with respect to the original
multiplication in $A$) is strongly isotopic to the quasigroup
$(\mathbb{Z}_{2^{n}-1},\bullet)$ with multiplication defined by

$u\bullet v\equiv\frac{1}{2}(u+v)\mod(2^{n}-1),\qquad\forall
u,v\in\mathbb{Z}_{2^{n}-1},$ (66)

###### Proof.

Given a fixed idempotent $c\in\operatorname{Idm}(\mathbb{A})$ let
$Q=(\operatorname{Idm}(\mathbb{A}),\boxplus)$ be the cyclic group of order
$2^{n}-1$ with unit $c$ and defined explicitly by $x\boxplus y=T(xy)$, where
$T=L_{c}^{-1}$. By Theorem 8.3 there exists an isomorphism $\phi$ of $Q$ to
the cyclic group $(\mathbb{Z}_{2^{n}-1},+)$ of integers with modular addition.
By the definition,

$\phi(T(xy))=\phi(x\boxplus y)=\phi(x)+\phi(y)\qquad\text{for any $x,y\in
Q$.}$ (67)

On the other hand, since $uu=u$ for any $u\in\operatorname{Idm}(\mathbb{A})$
then $T(u)=T(uu)=u\boxplus u$, hence $\phi(T(u))=2\phi(u)$, and combining this
with (67) we obtain

$2\phi(xy)=\phi(x)+\phi(y),$

i.e. using the notation in (66),

$\phi(xy)=\frac{1}{2}(\phi(x)+\phi(y))=\phi(x)\bullet\phi(y),$

which shows that
$\phi:\operatorname{Idm}(\mathbb{A})\to(\mathbb{Z}_{2^{n}-1},\bullet)$ is a
strong isotopy, hence implying the desired conclusion. ∎

##  9\. Discussion and final remarks

###  9.1. Medial algebra extensions

Let $(Q,\circ)$ be an idempotent medial commutative quasigroup, IMC-
quasigroup, for short, following [37]. Note that the order $|Q|$ of an IMC-
quasigroup is always an odd number, see, for example, Corollary 2.5 in [34].

######  Definition 9.1.

An algebra $\mathbb{A}$ over a field $\mathbf{K}$ is called an medial
extension of $Q$ if there exists a multiplicative monomorphism (injective
homomorphism) $f:Q\to\operatorname{Idm}(\mathbb{A})$ and $f(Q)$ generates
$\mathbb{A}$.

We see from Proposition 3.2 that any medial extension of a quasigroup is a
medial algebra.

It follows from the definition that $\dim
A\leq|Q|\leq\mathrm{card}(\operatorname{Idm}(\mathbb{A}))$. Furthermore, if
algebra $\mathbb{A}$ is generic then one also has

$\dim A\leq|Q|\leq\mathrm{card}(\operatorname{Idm}(\mathbb{A}))\leq 2^{\dim
A}-1.$

To distinguish the extremal cases in the latter inequality, we say that a
medial extension is minimal (resp. maximal) if $|Q|=2^{\dim A}-1$ (resp. if
$|Q|=\dim\mathbb{A}$).

A natural question in the present context arises: does an arbitrary IMC-
quasigroup obey a minimal medial extension?

The following proposition shows that any IMC-quasigroup obeys a maximal medial
extension.

######  Proposition 9.2.

Let $Q$ be an IMC-quasigroup and let $\mathbb{A}$ be the free module generated
by $Q$ as a basis. Define a commutative nonassociative algebra structure on
$\mathbb{A}$ by the natural extension of the quasigroup multiplication on $Q$.
Then $\mathbb{A}$ is a maximal medial extension of $Q$.

###### Proof.

Follows immediately from the definition and Proposition 3.2. ∎

###  9.2. Construction of a medial algebra from a medial quasigroup

Using Proposition 3.2, we are ready to produce some simple explicit of medial
algebras. As we already mentioned above, the order of any IMC-quasigroup is an
odd number. An example below shows that for any odd number there exists at
least one IMC-quasigroup.

Let $N$ be an odd positive integer and let
$\mathbb{Z}_{N}=\mathbb{Z}/N\mathbb{Z}$ denote the commutative ring of
integers modulo $N$. We denote by $\bar{x}$ the residue class containing
$x\in\mathbb{Z}$, or just by $x$ if this causes no ambiguity.

######  Proposition 9.3.

Then $\mathbb{Z}_{N}^{\circ}:=(\mathbb{Z}_{N},\circ)$ with multiplication
defined by

$\bar{x}\circ\bar{y}\equiv\frac{1}{2}(\bar{x}+\bar{y})\mod
N,\qquad\forall\bar{x},\bar{y}\in\mathbb{Z}_{N},$ (68)

is an IMC-quasigroup of order $N$.

###### Proof.

Since $2$ is invertible in $\mathbb{Z}_{N}$, the multiplication $\circ$ is
well defined and commutative. The medial magma identity follows immediately
from

$(\bar{x}\circ\bar{y})\circ(\bar{z}\circ\bar{w})=\frac{1}{4}(\bar{x}+\bar{y}+\bar{z}+\bar{w})$

whereas the left hand side is completely symmetric in all four variables.
Next, to verify the quasigroup property let
$\bar{x},\bar{y}\in\mathbb{Z}_{N}$. Then $\bar{w}=2\bar{y}-\bar{x}$ satisfies
$\bar{w}\circ\bar{x}=\bar{y}$. If $\bar{w}_{1}\circ\bar{x}=\bar{y}$ then
$\frac{1}{2}(\bar{w}_{1}-\bar{w})=0$, hence $\bar{w}=\bar{w}_{1}$. This shows
that $\circ$ satisfies the Latin square property, as desired. Finally,
$\bar{x}\circ\bar{x}\equiv\bar{x}$, thus all elements of
$\mathbb{Z}_{N}^{\circ}$ are idempotents. This shows that
$\mathbb{Z}_{N}^{\circ}$ is an IMC-quasigroup. Clearly,
$|\mathbb{Z}_{N}^{\circ}|=N$. ∎

######  Remark 9.4.

Comparing the above multiplication (68) with the Bruck-Murdoch-Toyoda theorem
(Theorem 2.5), one can identify the corresponding abelian group
$(\mathbb{Z}_{N},+)$ with $g=0$ and automorphisms $\phi=\psi$ defined as
multiplication by $\frac{1}{2}$. The latter choice is essentially the only
possible one as a straightforward verification by virtue of the Bruck-Murdoch-
Toyoda theorem shows; see also Proposition 4.1 in [38] for a general abelian
group. See some further discussion in [38], [37].

To proceed we need to characterize the $\circ$ multiplication in more detail.
To this end, recall that given two relative prime $a$ and $N$, the smallest
integer $n$ satisfying $a^{n}\equiv 1\mod N$ is called the multiplicative
order of $a$ modulo $N$, denoted by $\operatorname{ord}_{N}(a)$. It is well
known in the classical number theory that $\operatorname{ord}_{N}(a)$ divide
$\varphi(N)$, where $\varphi$ is Euler’s totient function. As a consequence of
Lagrange’s theorem, $\operatorname{ord}_{N}(a)$ always divides $\varphi(N)$.

Let ${x},{y}\in\mathbb{Z}_{N}^{\circ}$. Define the order
$\operatorname{ord}^{\circ}_{x}y$ of an element $y$ with respect to $x$ as the
minimal positive integer $p$ such that $L_{{x}}^{p}{y}={y}$.

######  Proposition 9.5.

The order is well defined and given by
$\operatorname{ord}^{\circ}_{x}y=\omega_{N}(x-y)$, where

$\omega_{N}(m):=\min\\{p\in\mathbb{N}:(2^{p}-1)m\equiv 0\mod N\\}.$ (69)

In particular, the order depends only on (the residue class of) $|x-y|$ and

$\operatorname{ord}^{\circ}_{x}y=\operatorname{ord}^{\circ}_{y}x.$ (70)

###### Proof.

First note that $\operatorname{ord}^{\circ}_{x}y$ is well defined. Indeed,
since $\mathbb{Z}_{N}^{\circ}$ is finite, then by the pigeonhole principle
there exists integers $q>k\geq 0$ such that $L_{x}^{q}y=L_{x}^{k}y$. By the
quasigroup cancelation property we have $L_{x}^{q-k}y=y$. Define $p$ as the
minimal positive integer such that $L_{x}^{p}y=y$ holds. Since

$L_{x}^{i}y\equiv\frac{1}{2^{i}}y+(\frac{1}{2^{i}}+\ldots+\frac{1}{2})x\equiv\frac{1}{2^{i}}y+(1-\frac{1}{2^{i}})x\mod
N,$

we have $\frac{2^{p}-1}{2^{p}}(x-y)\equiv 0\mod N$, or, since $N$ is odd,

$(2^{p}-1)(x-y)\equiv 0\mod N.$ (71)

This implies our claim by virtue of (69). Then (70) follows from (69). ∎

Since $2$ and $N$ are relatively prime, by Euler’s theorem
$2^{\varphi(N)}-1\equiv 0\mod N$. Thus, $p\leq\varphi(N)$. In fact, the above
argument also implies that
$\operatorname{ord}^{\circ}_{x}y\leq\operatorname{ord}_{N}(a)$, the
multiplicative order of $a$ modulo $N$.

######  Corollary 9.6.

If $N$ is an odd prime then
$\operatorname{ord}^{\circ}_{x}y=\operatorname{ord}_{N}(2)$ unless $x=y$ in
$\mathbb{Z}_{N}$.

###### Proof.

Indeed, for all distinct $x$ and $y$ in $\mathbb{Z}_{N}$, $x-y$ is coprime
with $N$, therefore (69) is equivalent to

$\operatorname{ord}^{\circ}_{x}y=\min\\{p\in\mathbb{N}:2^{p}\equiv 1\mod
N\\}=\operatorname{ord}_{N}(2),$

hence the conclusion follows. ∎

For composite odd $N$, the situation is more complicated. Since we are
primarily interested in the minimal medial extensions, the most relevant for
us case is when $N=2^{n}-1$, $n=2,3,\ldots$ We consider this case in the next
section.

###  9.3. The particular case $N=2^{n}-1$

We consider below the case $N=2^{n}-1$, where $n\geq 2$ is an integer. Then it
follows from (69) that

$\operatorname{ord}^{\circ}_{x}y\leq n.$

######  Proposition 9.7.

The set $P(n)$ of all possible orders $p=\operatorname{ord}^{\circ}_{x}y$,
where $x\neq y$, is exactly the set of all integers less or equal $n$ and
having a nontrivial common divisor with $n$.

###### Proof.

First note that if $k,m\geq 2$ then

$2^{km}-1=(2^{k}-1)(2^{k(m-1)}+\ldots+1)$

hence $2^{k}-1$ divides $2^{km}-1$.

Now let $s:=\gcd(p,n)>1$ and $p<n$. Then by the above, $r:=2^{s}-1$ is a
nontrivial common divisor of $2^{p}-1$ and $2^{n}-1$, therefore $p\in P(n)$.
In the converse direction, let $p\in P(n)$. Then it follows from (69) that
$2^{p}-1$ and $2^{n}-1$ have a nontrivial common divisor, say $r$. Then $r$ is
odd and divides $2^{n}-2^{p}$, thus also divides $2^{n-p}-1$. Arguing as in
the proof of the Euclidean algorithm, we conclude that $r$ also divides
$2^{s}-1$, where $s:=\gcd(p,n)$. Since $r$ is nontrivial, we have $s\geq 2$,
thus $p$ and $n$ also have a nontrivial divisor. The proposition follows. ∎

######  Example 9.8.

We illustrate the above by a composite number $N=15=2^{4}-1$, where
$\operatorname{ord}_{15}2=4$. Then $\frac{1}{2}\equiv 8\mod 15$, thus

$x\circ y\equiv 8(x+y)\mod 15.$

Let $x=1$ and $y=2$. Then the sequence $L_{x}^{k}y$ is given explicitly for
$k=0,1,2,3,\ldots$ by

$2,\,9,\,5,\,3,2\,\ldots\quad\Rightarrow\quad\operatorname{ord}^{\circ}_{1}2=4,$

in contrast to $x=1$ and $y=6$, where the corresponding $L_{x}^{k}y$ are

$6,\,11,6,\ldots\quad\Rightarrow\quad\operatorname{ord}^{\circ}_{1}6=2.$

By Proposition 9.7, $P(4)=\\{2,4\\}$. Indeed, for a general
$x,y\in\mathbb{Z}_{15}$ we have

$\operatorname{ord}^{\circ}_{x}y=\left\\{\begin{array}[]{cl}1&\text{if
$x-y=0$}\\\ 2&\text{if $(x-y,5)=5$}\\\ 4&\text{if
$(x-y,5)=1$}\end{array}\right.$

The action of $L_{1}$ on $\mathbb{Z}^{\circ}_{15}$ gives the following orbits:

$\begin{array}[]{ccccccc}1&&&\\\ 6&\to&11&&\\\ 2&\to&9&\to&5&\to&3\\\
4&\to&10&\to&13&\to&7\\\ 8&\to&12&\to&14&\to&15\\\ \end{array}$

###  9.4. An alternative model of $\mathbb{A}_{n}$

It would be very desirable to have some more explicit information about the
idempotents of $\mathscr{C}_{n}$ , see section 7. Unfortunately, the
idempotency relation (59) is not very illuminating and even in the lower-
dimensional case $n=3$ leads to a rather difficult algebraic manipulations. We
notice that the identities (61) uniquely determine the set of all idempotents
of $\mathscr{C}_{n}$. Indeed, let us consider the system

$\left\\{\begin{array}[]{ll}p^{2}(\epsilon_{n}^{1})&=p(1)\\\
p^{2}(\epsilon_{n}^{2})&=p(\epsilon_{n})\\\ \ldots&=\ldots\\\
p^{2}(\epsilon_{n}^{n})&=p(\epsilon_{n}^{n-1})\end{array}\right.$ (72)

Then the above system is a system of $n$ quadratic equations in variables
$(x_{0},x_{1},\ldots,x_{n-1})$. By Bézout’s theorem, (72) has either
infinitely many or less than $2^{n}$ complex solutions (counting the trivial
zero solution). It is not difficult to show that it has exactly $2^{n}$
distinct solutions using the fact that the Jacobian matrix is nonsingular at
any solution (cf. (9) in [35] which is essentially equivalent to that
$\frac{1}{2}\not\in\sigma(\mathscr{C}_{n})$). Due to the symmetry of (72), it
can be reduced to an algebraical equation of degree $2^{n}-1$. A more careful
analysis with help of Galois theory reveals that any idempotent $p$ is a
polynomial with coefficients in the cyclotomic extension
$\mathbb{Q}[\epsilon_{2^{n}-1},\epsilon_{n}]$

###  9.5. Iterations

Note also that since $\mathscr{C}_{n}$ is medial algebra it makes it easy to
construct idempotents if one knows some of them by combining the following
steps:

  1. (A)

If $p(z)$ and $q(z)$ are idempotents then
$p(z\epsilon_{n}^{k})q(z\epsilon_{n}^{k})$ is an idempotent too.

  2. (B)

Applying the above to $q=1$, $p(z\epsilon_{n}^{k})=L_{1}^{\circ k}p(z)$ is an
idempotent in $\mathscr{C}_{n}$.

Let $p(z)$ be an idempotent in $\mathscr{C}_{n}$ distinct from $1$ such that
let $\deg_{R}p=\deg p(z)$. By the medial property Proposition 3.5, $1\circ p$
is an idempotent too. This shows that the iterations

$p(z),\,p(\epsilon_{n}z),\,\ldots,\,p(\epsilon_{n}^{k}z),\,\ldots,\,p(\epsilon_{n}^{n-1}z)$

are also idempotents. Since $\deg p\leq n-1<n$, $p(\epsilon_{n}^{k}z)\neq
p(\epsilon_{n}^{m}z)$ as elements in $\mathscr{C}_{n}$ unless
$p(\epsilon_{n}^{k}z)=p(\epsilon_{n}^{m}z)$ as polynomials. Suppose that
$p(\epsilon^{k}z)=p(\epsilon^{m}p(\epsilon^{m}z)$ for some $0\leq k<m\leq
n-1$. Then $p(\epsilon^{k-m}z)=p(z)$. The latter is possible if and only if

$p(z)=p_{1}(z^{\nu}),\qquad\text{where $\nu=\frac{n}{\gcd(n,k-m)}$.}$

###  9.6. Appendix: Non-generic isospectral algebras

Finally we give an example of a three-dimensional non-generic isospectral
algebra which is non-Hsiang. This is a part of a more general construction, we
consider it in full generality elsewhere.

Let us define a three dimensional algebra $\mathbb{T}=\mathbf{K}^{3}$, where
$\mathbf{K}$ is any field of characteristic not 2 and 3, with multiplication

$x\circ
y=(x_{1}y_{1}-\frac{1}{2}x_{2}y_{3}-\frac{1}{2}x_{3}y_{2},\,x_{2}y_{2}-\frac{1}{2}x_{1}y_{3}-\frac{1}{2}x_{3}y_{1},\,x_{3}y_{3}-\frac{1}{2}x_{1}y_{2}-\frac{1}{2}x_{1}y_{2}),$
(73)

where $x=(x_{1},x_{2},x_{3})$ and $y=(y_{1},y_{2},y_{3})$ in a fixed basis of
$\mathbf{K}^{3}$.

######  Remark 9.9.

It is easy to see that any element of $\mathbb{T}$ satisfies the so-called
‘adjoint identity’

$(x\circ x)\circ(x\circ x)=N(x)x,$ (74)

with cubic form $N(x)$ given explicitly by

$N(x)=x_{1}^{3}+x_{2}^{3}+x_{3}^{3}-3x_{1}x_{2}x_{3}=(x_{1}+x_{2}+x_{3})(x_{1}^{2}-x_{1}x_{2}-x_{1}x_{3}+x_{2}^{2}-x_{2}x_{3}+x_{3}^{2}).$

In the general case, an algebra satisfying (74) is called in [14] an
admissible cubic algebra. We refer the interested reader to a recent paper of
Elduque and Okubo[14] for more details and classification. Any admissible
algebra carries an invariant symmetric bilinear form $b$, i.e. it is metrized.
In our case, the invariant bilinear form is just the Euclidean scalar product,
$\langle x;y\rangle=x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}$. One can examine this
property directly, alternatively we refer to the general formula (5) in [14].
We do not exploit the latter fact below but it explains why the Euclidean
scalar product is relevant for $\mathbb{T}$.

######  Proposition 9.10.

The algebra $\mathbb{T}$ is isospectral. If $\mathbf{K}$ is infinite, the
algebra $\mathbb{T}$ is non-generic. The set of nonzero idempotents of
$\mathbb{T}$ is the circle given by

$\operatorname{Idm}(\mathbb{T})=\\{(x_{1},x_{2},x_{3}):\,\sum_{i=1}^{3}x_{i}^{2}=\sum_{i=1}^{3}x_{i}=1\\}.$
(75)

Any idempotent $x\in\operatorname{Idm}(\mathbb{T})$ has spectrum
$\\{1,-\frac{1}{2},\frac{1}{2}\\}$.

###### Proof.

Let us first prove that the set of nonzero idempotents is exactly
$S:=\\{(x_{1},x_{2},x_{3}):\,\sum_{i=1}^{3}x_{i}^{2}=\sum_{i=1}^{3}x_{i}=1\\}$.
To see that $S\subset\operatorname{Idm}(\mathbb{T})$, we note that for any
point $x\in S$, there holds

$\displaystyle x_{1}^{2}-x_{2}x_{3}$
$\displaystyle=(1-x_{2}-x_{3})^{2}-x_{2}x_{3}=1+x_{2}x_{3}+x_{2}^{2}+x_{3}^{2}-2(x_{2}+x_{3})$
$\displaystyle=x_{2}x_{3}-x_{1}^{2}+2x_{1},$

implying $x_{1}^{2}-x_{2}x_{3}=x_{1}$, thus by symmetry implying $x\circ x=x$.
Conversely, if the latter idempotent identity holds true, then
$x_{i}^{2}-x_{j}x_{k}=x_{i}$ for any permutation $\\{i,j,k\\}=\\{1,2,3\\}$.
Summing up we obtain

$w-\frac{1}{2}(v^{2}-w)=v,\qquad w:=\sum_{i=1}^{3}x_{i}^{2},\quad
v:=\sum_{i=1}^{3}x_{i}.$ (76)

Next, subtracting the idempotents relations, one obtains

$(x_{i}-x_{j})(x_{1}+x_{2}+x_{3}-1)=0,\qquad\forall 1\leq i<j\leq 3.$

If $x_{1}+x_{2}+x_{3}=1$, we have $v=1$ and by (76) also $w=1$, hence $x\in
S$. If $x_{1}+x_{2}+x_{3}\neq 1$ then all $x_{i}$ are equal which immediately
implies $x=0$. This shows that $\operatorname{Idm}(\mathbb{T})\subset S$ and
finishes the proof of (75). In particular, the obtained algebra is non-
generic.

Next, set $e=(1,1,1)$. Then for any idempotent
$x\in\operatorname{Idm}(\mathbb{T})$ we have

$e\circ
x=(x_{1}-\frac{1}{2}x_{2}-\frac{1}{2}x_{3},x_{2}-\frac{1}{2}x_{1}-\frac{1}{2}x_{3},x_{3}-\frac{1}{2}x_{1}-\frac{1}{2}x_{2})=\frac{3}{2}x-\frac{1}{2}e,$

hence

$(e-x)\circ x=\frac{3}{2}x-\frac{1}{2}e-x=-\frac{1}{2}(e-x),$ (77)

i.e. $(e-x)$ is an eigenvector of $L^{\circ}_{x}$ with eigenvalue
$-\frac{1}{2}$. Similarly, define $u=e\times x$, where $\times$ is the
(formal) cross-product in $\mathbf{K}^{3}$, i.e.
$u=(x_{2}-x_{3},x_{3}-x_{1},x_{1}-x_{2}).$ Then

$x\circ u=\frac{x_{1}+x_{2}+x_{3}}{2}u=\frac{1}{2}u,$

hence $u$ is an eigenvector of $L^{\circ}_{x}$ with eigenvalue $\frac{1}{2}$.
Also, $x$ itself is an eigenvector of $L^{\circ}_{c}$ with eigenvalue $1$. In
summary, we have an eigenbasis of $L^{\circ}_{x}$ consisting of $x,e-x,e\times
x$. This implies that the spectrum of $L^{\circ}_{x}$ is
$\\{1,-\frac{1}{2},\frac{1}{2}\\}$. Thus, all idempotents of $\mathbb{T}$ have
the same spectrum, so the algebra is isospectral. Finally, since the trace
$\operatorname{tr}L^{\circ}_{x}=1\neq 0$ the algebra is not a Hsiang algebra.
∎

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