> To the memory of Yakov Krasnov (1956-2023)
>
> my friend and colleague.

# Inner isotopes associated with automorphisms of commutative associative
algebras

Vladimir G. Tkachev  Department of Mathematics, Linköping University,
Linköping, 58183, Sweden<EMAIL_ADDRESS>
###### Abstract.

The principal observation of the present paper is that an inner isotopy (i.e.
a principal isotopy defined by an algebra endomorphism) is a very helpful
instrument in constructing and studying interesting classes of nonassociative
algebras. By using methods developed in the paper, we define a new class of
commutative nonassociative algebras obtained by inner isotopy from commutative
associative polynomial algebras. There is a natural bijection between
isomorphism classes of our algebras and integer partitions of the algebra
dimensions. Among the interesting features of the nonassociative algebras
constructed are that these algebras are generic, some of examples are axial
and metrized algebras. We completely describe both the set of algebra
idempotents and their spectra.

###### Key words and phrases:

Polynomial rings, Nonassocative algebras, Medial algebras, Idempotents,
Integer Partitions

smalltableaux

##  1\. Introduction

By an algebra $(\mathbb{A},\ast)$ we understand a nonassociative algebra over
a field $\mathbf{K}$ of $\mathrm{char}(\mathbf{K})\neq 2,3$ with
multiplication $\ast$. An element $c\in\mathbb{A}$ is called an idempotent if
$c\ast c=c$. The set of nonzero idempotents of $(\mathbb{A},\cdot)$ is denoted
by $\operatorname{Idm}(\mathbb{A},\ast)$. Given an element
$a\in(\mathbb{A},\ast)$, we denote by $L_{\ast}(a)x\to a\ast x$ the left
multiplication operator by $a$ and $\mathbb{A}_{\lambda}(a)$ the kernel of
$(\lambda\mathds{1}_{\mathbb{A}}-L_{\ast}(a))$, where
$\mathds{1}_{\mathbb{A}}$ is the identity operator.

A commutative algebra $(\mathbb{A},\ast)$ is called isospectral if the
spectrum of $L_{\ast}(c)$ is the same for any nonzero idempotent $c$. By using
the syzygy method, it was established in [16] that if an isospectral algebra
$(\mathbb{A},\ast)$ is generic (see Definition 1.1 below) then the common
spectrum of the algebra idempotents consists of simple eigenvalues satisfying
$\lambda^{\dim\mathbb{A}}=1$. A further analysis given in [17] reveals that
under some mild conditions, an isospectral generic algebra must be medial,
i.e. the algebra multiplication associates on pairs:

$(x\ast y)\ast(z\ast t)=(x\ast z)\ast(y\ast t),$

and, moreover, such an algebra is an inner isotopy of a certain commutative
associative algebra. Recall that given an algebra $(\mathbb{A},\ast)$, its
inner isotope $(\mathbb{A},\ast_{h})$ is the vector space $\mathbb{A}$ with
the new multiplication

$x\ast_{h}y=h(x\ast y)=h(x)\ast h(y),$

where $h\in\operatorname{Aut}(\mathbb{A},\ast)$ is an automorphism of
$(\mathbb{A},\ast)$. More precisely, the medial isospectral generic algebras
discussed in [17] are exactly the inner isotopes of the quotient polynomial
algebra $\mathbf{K}[z]/(z^{n}-1)$ under the automorphism
$\tau\in\operatorname{Aut}(\mathbf{K}^{n},\bullet)$ acting by substitution
$[p(z)]\to[p(\epsilon_{n}z)]$, where $\epsilon_{n}$ is a primitive root of
unity of order $n$. The corresponding inner isotope algebra
$(\mathbf{K}[z]/(z^{n}-1),\bullet_{\tau})$ has many distinguished properties
(see Section 10.2 below for $n=3$ and Section 7 for the general case).

In this regard, it is natural to find the full automorphism group
$\operatorname{Aut}(\mathbf{K}^{n},\bullet)$ and characterize the
corresponding (isomorphy classes of) inner isotopes of
$(\mathbf{K}^{n},\bullet)$. A similar approach is also relevant for an
arbitrary algebra $(\mathbb{A},\ast)$ and its inner isotopes
$(\mathbb{A},\ast_{h})$.

These questions were an original motivation for the present paper. As we shall
see, an inner isotopy is a very helpful instrument for constructing of
interesting classes of nonassociative algebras. In particular, this approach
is already fruitful for the simplest possible case when the initial algebra is
commutative and associative. In that case, any nontrivial inner isotopy
destroys the associativity but not so much: any inner isotope of a
(commutative) associative algebra is always a medial algebra, see Corollary
5.2.

A relevant in the present context is the mentioned above concept of generic
nonassociative algebras [16]. The definition comes back to Segre’s observation
[24] that idempotents in an $n$-dimensional commutative algebra
$(\mathbf{K}^{n},\ast)$ over an algebraically closed field $\mathbf{K}$ with a
basis $\\{e_{i}\\}_{1\leq i\leq n}$ can be interpreted as the set of common
zeros of $n$ quadratic polynomials

$\Phi_{k}(x):=(x\ast x-x)_{k}=\sum_{1\leq i,j\leq
n}a_{ijk}x_{i}x_{j}-x_{k}=0,\qquad 1\leq k\leq n,$

in $\mathbf{K}^{n}$, where $x=\sum_{1\leq i\leq n}x_{i}e_{i}$ and $a_{ijk}$
are the structural constant of $\ast$ in the basis $\\{e_{i}\\}$. By the
Bézout theorem the number of intersection points (i.e. the algebra
idempotents) properly counted in the projective space
$\mathbf{K}\mathbb{P}^{n}$ is either $2^{n}$ or infinite. An intersection
point $c$ is simple if the quadrics are in relative general position at $c$.
On the algebra level, the latter is equivalent to that the idempotent $c$ is
regular [32], i.e. the Jacobian of the quadratic endomorphism
$\Phi(x):\mathbf{K}^{n}\to\mathbf{K}^{n}$ is nonzero:
$\det(D\Phi(x))=\det(L_{\ast}(c)-\frac{1}{2}\mathds{1}_{\mathbb{A}})\neq 0$.
Then the Bézout estimate holds:

$\text{the number of {regular} idempotents of $\mathbb{A}$ }\leq
2^{\dim\mathbb{A}}.$ (1)

In general, if the ground field $\mathbf{K}$ is not algebraically closed then
the algebra $(\mathbb{A},\ast,\mathbf{K}^{\prime})$ over an algebraic
extension $\mathbf{K}^{\prime}$ of $\mathbf{K}$ has the same dimension, and
any idempotent regular in $(\mathbb{A},\ast,\mathbf{K})$ is a regular
idempotent in $(\mathbb{A},\ast,\mathbf{K}^{\prime})$. Applying (1) to
$(\mathbb{A},\ast,\mathbf{K}^{\prime})$, this implies that (1) also holds in
$(\mathbb{A},\ast,\mathbf{K})$. This motivates the following definition.

######  Definition 1.1.

A commutative nonassociative algebra $\mathbb{A}$ over an arbitrary field
$\mathbf{K}$ is called generic if it has exactly $2^{\dim\mathbb{A}}$ distinct
regular idempotents.

######  Remark 1.2.

An algebra on a vector space $V$ is uniquely identified with a point of the
set of all bilinear multiplication $V^{*}\otimes V^{*}\otimes V$ on $V$. In
this sense, the subset of generic algebras is a Zariski open subset of
$V^{*}\otimes V^{*}\otimes V$. The above definition has appeared in [16], [15]
and it should not be confused with some similar analogues, for example in
[27], [28].

In this paper, we show that under some natural assumptions any inner isotope
of a commutative associative algebra is generic and, moreover, we explicitly
characterize the set of idempotents and their spectral properties.

Another relevant concept here is the so-called axial algebras, i.e. the
algebras generated by a finite subset of idempotents which satisfy a common
fusion law. More precisely, a fusion law is a set
$\mathcal{F}\subset\mathbf{K}$ together with a symmetric binary map
$\theta:\mathcal{F}\times\mathcal{F}\to 2^{\mathcal{F}}$.

######  Definition 1.3 ([14]).

Given a fusion law $\mathcal{F}$, a commutative algebra $(\mathbb{A},\ast)$
over $\mathbf{K}$ together with a distinguished subset of elements $X$ (the
called axes) is an $\mathcal{F}$-axial algebra if $(\mathbb{A},\ast)$ is
generated by $X$, for each $c\in X$, $c$ is a semisimple idempotent, namely
$\mathbb{A}=\oplus_{\lambda\in\mathcal{F}}\mathbb{A}_{\lambda}(c)$ and for
$\lambda,\mu\in\mathcal{F}$:

$\mathbb{A}_{\lambda}(c)\ast\mathbb{A}_{\mu}(c)\subset\mathbb{A}_{\theta(\lambda,\mu)}(c).$

An axial algebra is called primitive if each idempotent in $X$ is primitive,
i.e. $\dim\mathbb{A}_{1}(a)=1$ for any $a\in X$.

The isospectral generic algebras considered in [16] and [17] are axial
algebras. The axial algebra concept is an important tool in understanding of
finite groups: it appears that many interesting groups (for example,
3-transposition groups including certain simple sporadic groups, in
particular, the monster group) arise as automorphism groups of cubic forms on
suitable modules [25], [19], [26], [22], [8] by virtue of a correspondence
between certain involutions generating a group and a distinguished family of
idempotents in an appropriate (non)associative commutative algebra. Such a
correspondence normally is very individual and drastically depends on a source
group/algebra, as well as its combinatorial or geometrical realizations [3],
[6]. Some recent developments in the axial algebra project can be found [21],
[10], [4], [7], [14], [9], [2] and the references therein.

The algebras discussed in the present paper fit perfectly this context and
provide us with new examples of axial algebras with known automorphism groups.
We only outline some partial results in this direction (see especially the
explicit examples in Section 10), while the general discussion will be
addressed elsewhere in the second part of this paper.

######  Remark 1.4.

We are very grateful to the referees for pointing us out the existence of the
isomorphism (17). In the original version of our paper [31], we used a
different exposition based on the polynomial model $\mathbf{K}[z]/P$. The
isomorphism (17) simplifies several proofs below and the structure of the
idempotent set becomes more transparent when written in this form. Still, we
believe that the polynomial model $\mathbf{K}[z]/P$ is also of interest,
especially for some particular choices of the polynomial $P$; see fore
example, the criterion given in Proposition 9.9 below. We refer an interested
reader to [31] for the original presentation and more details concerning the
polynomial model.

The paper is organized as follows. In Section 2, we recall some general
concepts and facts used in the paper. In section 3, we outline the general
properties of inner isotopes of an arbitrary algebra and in section 4 we
specify these results for commutative associative algebras. In particular, we
show that an inner isotope of a commutative associative algebra must be
medial. In section 5, we develop an appropriate category-theoretical context
for our considerations. The main result of this section states that the
categories of calibrated special commutative medial algebras is isomorphic to
the category of calibrated commutative associative algebras. In section 6 we
discuss the properties of idempotents in an arbitrary medial algebra. In
section 7 we study the automorphism group of a quotient polynomial algebra and
determine its inner isotopes. We characterize the set of idempotents and
theirs spectra in Section 8. The automorphism groups of the obtained algebras
are studied in Section 9. Finally, in section 10 we illustrate our results in
the case $n=3$.

##  2\. Preliminaries

By $\bar{n}$ we denote the set of indices $\\{1,2,\ldots,n\\}$. We recall some
standard definitions following [23], see also [1, p.149].
$(\mathbb{A},\bullet)$ denotes an algebra with a multiplication $\bullet$ on a
vector space $\mathbb{A}$. All algebras below are assumed to be commutative
but maybe nonassociative.

If $(\mathbb{B},\bullet)$ and $(\mathbb{C},\bullet)$ are ideals of an algebra
$(\mathbb{A},\bullet)$ such that $\mathbb{A}$ as a vector space is the direct
sum of $\mathbb{B}$ and $\mathbb{C}$ then $(\mathbb{A},\bullet)$ is called the
direct sum:

$(\mathbb{A},\bullet)=(\mathbb{B},\bullet)\oplus(\mathbb{C},\bullet).$

Note that in this case $\mathbb{B}\bullet\mathbb{C}=0$, so that
$(\mathbb{B},\bullet)$ and $(\mathbb{C},\bullet)$ are orthogonal.

Given any two arbitrary algebras $(\mathbb{B},\star)$ and $(\mathbb{C},*)$
over a field $\mathbf{K}$, one can construct an algebra $(\mathbb{A},\bullet)$
over $\mathbf{K}$ such that $(\mathbb{A},\bullet)$ is the direct sum
$(\mathbb{A},\bullet)=(\mathbb{B}^{\prime},\bullet)\oplus(\mathbb{C}^{\prime},\bullet)$
of ideals $(\mathbb{B}^{\prime},\bullet)$ and $(\mathbb{C}^{\prime},\bullet)$
which are isomorphic respectively to $(\mathbb{B},\star)$ and
$(\mathbb{C},*)$: $\mathbb{A}$ as a vector space is the Cartesian product of
$\mathbb{B}$ and $\mathbb{C}$ with the multiplicative structure $\bullet$
defined by the coordinate-wise multiplication
$(b_{1},c_{1})\bullet(b_{2},c_{2})=(b_{1}\star b_{2},c_{1}*c_{2})$ for
elements $(b_{1},c_{1}),(b_{2},c_{2})\in\mathbb{B}\times\mathbb{C}$. Then, for
example, $(\mathbb{B},\star)$ is isomorphic to the ideal
$(\mathbb{B}^{\prime},\bullet)=((\mathbb{B},0),\bullet)$ of
$(\mathbb{A},\bullet)$. The resulting algebra is called the direct product of
algebras $(\mathbb{B},\star)$ and $(\mathbb{C},*)$. In the above notation,

$(\mathbb{B},\star)\times(\mathbb{C},*)\cong(\mathbb{B}\times\mathbb{C},\bullet)\cong(\mathbb{B}^{\prime},\bullet)\oplus(\mathbb{C}^{\prime},\bullet)$

As in the case of vector spaces, the notion of direct sum extends to an
arbitrary set of summands. We shall have occasion to use only finite direct
sums.

In particular, rhe field $\mathbf{K}$ is a commutative associative algebra
over itself, denoted by $(\mathbf{K},\cdot)$. The direct summa of $n\geq 1$
copies of $\mathbf{K}$ with the coordinate-wise multiplication is denoted by
$(\mathbf{K}^{n},\bullet)$. The $\bullet$-idempotents

$e_{i}=(0,\ldots,1,\ldots,0),\qquad 1\leq i\leq n$ (2)

form the standard basis of $\mathbf{K}^{n}$. Let $x_{i}$ denote the
corresponding coordinate of $x\in\mathbf{K}^{n}$. Given any index
$j\in\\{1,2,\ldots,n\\}$, the subspaces

$\langle e_{i}\rangle:=\\{x\in\mathbf{K}^{n}:\,x_{i}=0\,\text{ for all $i\neq
j$}\\}$ (3)

are (pairwise orthogonal) ideals of $(\mathbf{K}^{n},\bullet)$ and

$(\mathbf{K}^{n},\bullet)=\bigoplus_{j=1}^{n}\langle e_{j}\rangle.$

Below we shall need the following observation.

######  Proposition 2.1.

Let $(\mathbb{A},\bullet)=\bigoplus_{j=1}^{r}(\mathbb{B}_{j},\bullet)$, where
$(\mathbb{B}_{j},\bullet)$ are ideals. Then $c$ is an idempotent in
$(\mathbb{A},\bullet)$ if and only if there are (uniquely determined) pairwise
orthogonal idempotents $c_{j}\in(\mathbb{B}_{j},\bullet)$ such that
$c=\sum_{j=1}^{r}c_{j}$. Moreover, in this case

$\det(\lambda\mathds{1}_{\mathbb{A}})-L_{\bullet}(c))=\prod_{j=1}^{r}\det(\lambda\mathds{1}_{\mathbb{B}_{j}}-L_{\bullet}(c_{j}))$
(4)

The algebra $(\mathbb{A},\bullet)$ is generic if and only each ideal
$(\mathbb{B}_{j},\bullet)$ is so.

###### Proof.

It follows for the assumptions that $\mathbb{B}_{i}\bullet\mathbb{B}_{j}=0$
for $i\neq j$. Let $x\in\mathbb{A}$ and let $x=\sum_{j=1}^{r}x_{j}$ be the
corresponding orthogonal decomposition, $x_{j}\in\mathbb{B}_{j}$. Then
$x\bullet x=\sum_{j=1}^{r}x_{j}\bullet x_{j}$, where $x_{j}\bullet
x_{j}\in\mathbb{B}_{j}$ and $x_{i}\bullet x_{j}=0$ whenever $i\neq j$. It
follows that $x$ is an idempotent on $(\mathbb{A},\bullet)$ if and only if
$x_{j}$ is an idempotent in $(\mathbb{B}_{j},\bullet)$ for all $1\leq j\leq
r$, thus implying the first part of the proposition.

Next, let $(\mathbb{A},\bullet)$ be generic. Then $(\mathbb{A},\bullet)$
contains exactly $2^{n}$ distinct regular idempotents, where
$n=\dim\mathbb{A}$. It follows from (4) that all idempotents in each ideal
$\mathbb{B}_{j}$ are regular. If $\dim\mathbb{B}_{j}=n_{j}$ then
$n=n_{1}+\ldots+n_{r}$. It follows from the above idempotent decomposition
that $m_{1}\cdot\ldots\cdot m_{r}=2^{n}$, where $m_{j}$ is the cardinality of
the set of idempotents in $\mathbb{B}_{j}$, hence $m_{j}=2^{k_{j}}$ for some
nonnegative integer $k_{j}$. On the other hand by the Bezout inequality (1) we
have $m_{j}=2^{k_{j}}\leq 2^{n_{j}}$, hence $2^{n}=m_{1}\cdot\ldots\cdot
m_{r}\leq 2^{n_{1}+\ldots+n_{r}}=2^{n}$ which implies that in fact
$m_{j}=2^{k_{j}}=2^{n_{j}}$, i.e. each $\mathbb{B}_{j}$ is generic. In the
converse direction, if each ideal $\mathbb{B}_{j}$ is generic then
$(\mathbb{A},\bullet)=\bigoplus_{j=1}^{r}(\mathbb{B}_{j},\bullet)$ has by
Proposition 2.1 at least $2^{n_{1}}\cdot\ldots 2^{n_{r}}=2^{n}$ distinct
regular idempotents, hence it is generic. ∎

We also fix some standard terminology and facts from permutation theory. Any
permutation $\sigma\in S_{n}$ can be written in cyclic form (or a disjoint
cycle decomposition)

$\sigma=\sigma_{1}\ldots\sigma_{r}\in S_{n}.$ (5)

For any $1\leq i,j\leq r$, the cycles $\sigma_{i}$ and $\sigma_{j}$ commute,
therefore the order in (5) in inessential. Then cycles $\sigma_{j}$ can be
naturally thought of as orbits of a faithful action of the cyclic group
$\langle\sigma\rangle$ generated by $\sigma$ on the et of indices
$\bar{n}:=\\{1,\ldots,n\\}$. To differ a cycle $\sigma_{j}$ as a group element
and as an orbit, we denote the latter by $[\sigma_{j}]\subset\bar{n}$. By
$|\sigma_{i}|$ we denote the length of the cycle $\sigma_{i}$, i.e. the
cardinality of the orbit $[\sigma_{i}]$. The type of a permutation is the
integer partition of $n$,

$|\sigma_{1}|+\ldots+|\sigma_{r}|=n,$ (6)

formed from the cycle; we write it by
$\operatorname{type}(\sigma)=(|\sigma_{1}|,\ldots,|\sigma_{r}|)$.

######  Definition 2.2.

Given $\sigma\in S_{n}$ let $\sigma=\sigma_{1}\ldots\sigma_{r}$ be its
disjoint cycle decomposition and $s_{i}=|\sigma_{i}|$. A field $\mathbf{K}$
will be said $\sigma$-admissible, if it is a splitting field for all
polynomials $z^{t}-1$, where $t\in
S:=\\{s_{1},\ldots,s_{r},2^{s_{1}}-1,\ldots,2^{s_{r}}-1\\}$. If $\mathbf{K}$
has a finite characteristic, to avoid complications, we additionally assume
that the characteristic is co-prime with all numbers in $S$,

##  3\. Inner isotopes

Two algebras $(\mathbb{A},\diamond)$ and $(\mathbb{B},\cdot)$ are called
isotopic, if there is an isotopism $(\alpha,\beta,\gamma)$, i.e. a triple of
nondegenerate linear maps $\mathbb{A}\to\mathbb{B}$ such that
$\alpha(x)\cdot\beta(y)=\gamma(x\diamond y)$ holds for any $x,y\in\mathbb{A}$.
If $\alpha=\beta$ the isotopy is called strong. If $\mathbb{A}=\mathbb{B}$ and
$\gamma=\mathds{1}_{\mathbb{A}}$ in the identity map, then an isotopy is
called a principal autotopy.

######  Definition 3.1.

Given an algebra $(\mathbb{A},\bullet)$ and an algebra endomorphism
$h\in\operatorname{End}(\mathbb{A},\bullet)$, we define a new algebra
$(\mathbb{A},\bullet_{h})$ on the vector space $\mathbb{A}$ by

$x\bullet_{h}y=h(x)\bullet h(y)=h(x\bullet y).$ (7)

The new algebra $(\mathbb{A},\bullet_{h})$ is called a weak inner isotope of
$(\mathbb{A},\bullet)$. If $h\in\operatorname{Aut}(\mathbb{A},\bullet)$,
$(\mathbb{A},\bullet_{h})$ is called an inner isotope.

The above definition is a particular case of a strong principal autotopy; more
precisely, the algebra $(\mathbb{A},\bullet_{h})$ is an principal isotopic of
$(\mathbb{A},\bullet)$ with $(h,h,\mathds{1}_{\mathbb{A}})$ where the map $h$
is an $\bullet$-algebra homomorphism.

######  Remark 3.2.

In the converse direction, if an endomorphism $h$ is invertible then
$(\mathbb{A},\bullet)$ is an inner isotope of $(\mathbb{A},\bullet_{h})$.
Indeed, by virtue of (7), $x\bullet y=h^{-1}(x\bullet_{h}y)$, therefore the
linear endomorphism $h^{-1}$ is also an $\bullet_{h}$-algebra endomorphism.

We distinguish the case when $h$ is an bijective, i.e. $h$ is an automorphism
of $(\mathbb{A},\bullet)$ and $(\mathbb{A},\bullet_{h})$ is an inner isotopy.
In this case, if $h,f\in\operatorname{Aut}(\mathbb{A},\bullet)$ then
$(\mathds{1}_{\mathbb{A}},\mathds{1}_{\mathbb{A}},f\circ h^{-1})$ is an
isotopy between the corresponding inner isotopes:

$x\bullet_{f}y=f(x\bullet y)=f\circ h^{-1}(x\bullet_{h}y),$ (8)

where $\circ$ here and in what follows denote the composition of two maps.

Comparing (8) with (7) a natural question arises: when the algebra
$(\mathbb{A},\bullet_{f})$ is an inner isotope of $(\mathbb{A},\bullet_{h})$?
Note that $f\circ h^{-1}\in\operatorname{Aut}(\mathbb{A},\bullet)$, but it is
not clear whether $f\circ
h^{-1}\in\operatorname{Aut}(\mathbb{A},\bullet_{h})$, see the diagram below

${(\mathbb{A},\bullet)}$${(\mathbb{A},\bullet_{h})}$${(\mathbb{A},\bullet_{f})}$$\scriptstyle{h}$$\scriptstyle{f}$$\scriptstyle{f\circ
h^{-1}}$

The next proposition reveals that this is true for commuting automorphisms
only.

######  Proposition 3.3.

Let $(\mathbb{A},\bullet)$ satisfy

$\mathbb{A}^{\bullet 2}:=\mathbb{A}\bullet\mathbb{A}=\mathbb{A}$ (9)

and let $h,f\in\operatorname{Aut}(\mathbb{A},\bullet)$. Then
$(\mathbb{A},\bullet_{f})$ is an inner isotope of $(\mathbb{A},\bullet_{h})$
if and only if $f$ and $h$ commute.

###### Proof.

First suppose that $(\mathbb{A},\bullet_{f})$ is an inner isotope of
$(\mathbb{A},\bullet_{h})$, i.e.

$x\bullet_{f}y=g(x\bullet_{h}y)=g(x)\bullet_{h}g(y),\qquad\forall
x,y\in\mathbb{A},$ (10)

holds for some $\bullet_{h}$-algebra endomorphism $g$. Comparing this with
(8), we obtain

$f(x\bullet y)=f\circ h^{-1}(x\bullet_{h}y)=g(x\bullet_{h}y)=(g\circ
h)(x\bullet y).$

Since the latter holds for any $x,y\in\mathbb{A}$ we conclude that $f=g\circ
h$ on $\mathbb{A}^{\bullet 2}$, and thus by (9) on $\mathbb{A}$. Therefore
$g=f\circ h^{-1}$. In particular, $g$ is an $\bullet$-automorphism too.
Therefore, using the second identity in (10) we get

$\displaystyle f(x\bullet y)$ $\displaystyle=g(x)\bullet_{h}g(y)=h(g(x)\bullet
g(y))$ $\displaystyle=(h\circ g(x))\bullet(h\circ g(y))$
$\displaystyle=(h\circ g)(x\bullet y)\qquad\forall x,y\in\mathbb{A},$

implying that $f=h\circ g=h\circ f\circ h^{-1}$ on $\mathbb{A}^{\bullet
2}=\mathbb{A}$, hence $f\circ h=h\circ f$ as desired.

Conversely, if $f\circ h=h\circ f$ then using (8)

$x\bullet_{f}y=g(x\bullet_{h}y),$ (11)

where $g=f\circ h^{-1}=h^{-1}\circ
f\in\operatorname{Aut}(\mathbb{A},\bullet)$. We have

$g(x\bullet_{h}y)=g\circ h(x\bullet y)=f\circ h^{-1}\circ h(x\bullet
y)=f(x\bullet y),$

and on the other hand,

$g(x)\bullet_{h}g(y)=h\circ(g(x)\bullet g(y))=(h\circ g)(x\bullet
y)=f(x\bullet y),$

implying $g(x\bullet_{h}y)=g(x)\bullet_{h}g(y)$, hence
$g\in\operatorname{Aut}(\mathbb{A},\bullet_{h})$, i.e. it follows from (11)
that $(\mathbb{A},\bullet_{f})$ is an inner isotope of
$(\mathbb{A},\bullet_{h})$. ∎

The next two propositions explain when two inner isotopes are isomorphic.

######  Proposition 3.4.

Let $f:(\mathbb{A},\bullet)\to(\mathbb{A}^{\prime},\bullet^{\prime})$ be an
algebra isomorphism and let $h\in\operatorname{Aut}(\mathbb{A},\bullet)$. Then
$h^{\prime}:=f\circ h\circ
f^{-1}\in\operatorname{Aut}(\mathbb{A}^{\prime},\bullet^{\prime})$ and
$f:(\mathbb{A},\bullet_{h})\to(\mathbb{A}^{\prime},\bullet^{\prime}_{h^{\prime}})$
is an algebra isomorphism.

###### Proof.

We have $f(x\bullet y)=f(x)\bullet^{\prime}f(y)$ and $h(x)\bullet
h(y)=h(x\bullet y)$ for any $x,y\in\mathbb{A}$, therefore

$\displaystyle f\circ h\circ f^{-1}(x^{\prime}\bullet^{\prime}y^{\prime})$
$\displaystyle=f\circ h(f^{-1}(x^{\prime})\bullet f^{-1}(y^{\prime}))$
$\displaystyle=f(h(f^{-1}(x^{\prime}))\bullet h(f^{-1}(y^{\prime})))$
$\displaystyle=f(h(f^{-1}(x^{\prime}))\bullet^{\prime}f(h(f^{-1}(y^{\prime})),$

readily implying that $h^{\prime}:=f\circ h\circ
f^{-1}\in\operatorname{Aut}(\mathbb{A}^{\prime},\bullet^{\prime})$.
Furthermore,

$f(x\bullet_{h}y)=f\circ h(x\bullet y)=f\circ h\circ
f^{-1}(f(x)\bullet^{\prime}f(y))=f(x)\bullet^{\prime}_{h^{\prime}}f(y)$

implying that
$f:(\mathbb{A},\bullet_{h})\to(\mathbb{A}^{\prime},\bullet^{\prime}_{h^{\prime}})$
is an algebra isomorphism.

∎

######  Proposition 3.5.

Let $h,f\in\operatorname{Aut}(\mathbb{A},\bullet)$. If $h$ and $f$ conjugate
in $\operatorname{Aut}(\mathbb{A},\bullet)$ then $(\mathbb{A},\bullet_{h})$ is
isomorphic to $(\mathbb{A},\bullet_{f})$. In the converse direction, if
$g:(\mathbb{A},\bullet_{h})\to(\mathbb{A},\bullet_{f})$ is an isomorphism and
$g\in\operatorname{Aut}(\mathbb{A},\bullet)$ then $h$ and $f$ conjugate in
$\operatorname{Aut}(\mathbb{A}^{\bullet 2},\bullet)$.

###### Proof.

If $f$ and $h$ conjugate in $\operatorname{Aut}(\mathbb{A},\bullet)$ then
$f=g\circ h\circ g^{-1}$ for some
$g\in\operatorname{Aut}(\mathbb{A},\bullet)$, hence

$g(x\bullet_{h}y)=g\circ h(x\bullet y)=f\circ g(x\bullet y)=f(g(x)\bullet
g(y))=g(x)\bullet_{f}g(y),$

hence $g:(\mathbb{A},\bullet_{h})\to(\mathbb{A},\bullet_{f})$ is an algebra
isomorphism. Conversely, by our assumptions we have

$g\circ h(x\bullet y)=g(x\bullet_{h}y)=g(x)\bullet_{f}g(y)=f(g(x)\bullet
g(y))=f\circ g(x\bullet y),$

hence $g\circ h=f\circ g$ on $\mathbb{A}^{\bullet 2}$. ∎

######  Corollary 3.6.

Let $(\mathbb{A},\bullet)$ satisfy (9) and
$h,f\in\operatorname{Aut}(\mathbb{A},\bullet)$. Then
$(\mathbb{A},\bullet_{h})$ is isomorphic to $(\mathbb{A},\bullet_{f})$ if and
only if $h$ and $f$ conjugate in $\operatorname{Aut}(\mathbb{A},\bullet)$.

##  4\. Inner isotopes of commutative associative algebras

So far, we have not specified any additional algebraic structure on
$(\mathbb{A},\bullet)$. Below we shall focus on the simplest case when the
original algebra $(\mathbb{A},\bullet)$ is commutative and associative.
Sometimes we shall also require that the algebra $(\mathbb{A},\bullet)$ is
unital. Note that any unital algebra satisfies automatically (9).

If $(\mathbb{A},\bullet)$ is a commutative associative algebra then any inner
isotope $(\mathbb{A},\bullet_{h})$ is obviously commutative but it maybe non-
associative, because

$x\bullet_{h}(y\bullet_{h}z)=h(x\bullet h(y\bullet z))=h(x)\bullet
h^{2}(y)\bullet h^{2}(z)$

and

$(x\bullet_{h}y)\bullet_{h}z=h^{2}(x)\bullet h^{2}(y)\bullet h(z)$

are not equal in general. On the other hand, such an inner isotope is nearly
associative, namely, it is medial. We recall the definition.

######  Definition 4.1.

An algebra $(\mathbb{A},\bullet)$ is called medial if

$(x\bullet y)\bullet(z\bullet w)=(x\bullet z)\bullet(y\bullet w),\qquad\forall
x,y,z,w\in\mathbb{A}.$ (12)

An important corollary of the definition is

$(x\bullet y)\bullet(x\bullet y)=(x\bullet x)\bullet(y\bullet y).$

which immediately implies

######  Proposition 4.2.

Product of two idempotents in a medial algebra is an idempotent again.

###### Proof.

If $c_{i}\bullet c_{i}=c_{i}$, $i=1,2$ then

$(c_{1}\bullet c_{2})\bullet(c_{1}\bullet c_{2})=(c_{1}\bullet
c_{1})\bullet(c_{2}\bullet c_{2})=c_{1}\bullet c_{2}.$

∎

######  Proposition 4.3.

If an algebra $(\mathbb{A},\bullet)$ is commutative and associative,
$h\in\operatorname{End}(\mathbb{A},\bullet)$, then the inner isotope
$(\mathbb{A},\bullet_{h})$ is a commutative medial algebra. Furthermore, if
$h\in\operatorname{Aut}(\mathbb{A},\bullet)$ and $(\mathbb{A},\bullet)$ is
additionally a unital algebra with unity $e$ then $e$ is an idempotent in
$(\mathbb{A},\bullet_{h})$ and $L_{\bullet_{h}}(e)$ is an invertible operator.

###### Proof.

We have

$\displaystyle(x\bullet_{h}y)\bullet_{h}(z\bullet_{h}w)$
$\displaystyle=h((x\bullet_{h}y)\bullet_{h}(z\bullet_{h}w))$
$\displaystyle=h(h(x\bullet y)\bullet h(z\bullet w))$
$\displaystyle=h^{2}(x\bullet y\bullet z\bullet w),$

where the right hand side is symmetric under any permutation of factors,
implying (12).

Next, assume that $(\mathbb{A},\bullet)$ is additionally a unital algebra with
unity $e$. Since an automorphism stabilizes the unity element, we have

$e\bullet_{h}e=h(e\bullet e)=h(e)=e$

therefore $e$ becomes an idempotent in $(\mathbb{A},\bullet_{h})$.
Furthermore, $e\bullet_{h}x=h(e\bullet x)=h(x)=0$ if and only if $x=0$, thus
$L_{\bullet_{h}}(e)$ is an invertible operator.

∎

######  Proposition 4.4.

Any associative commutative algebra is medial. A unital commutative medial
algebra is associative.

###### Proof.

Indeed, if $(\mathbb{A},\bullet)$ is an associative commutative algebra then
it is medial:

$(x\bullet y)\bullet(z\bullet w)=x\bullet y\bullet z\bullet w=x\bullet
z\bullet y\bullet w=(x\bullet z)\bullet(y\bullet w).$

On the other hand, if $(\mathbb{A},\bullet)$ is a unital commutative medial
algebra and $e$ is the algebra unity then

$x\bullet(y\bullet z)=(e\bullet x)\bullet(y\bullet z)=(e\bullet
z)\bullet(x\bullet y)=z\bullet(x\bullet y)=(x\bullet y)\bullet z,$

hence $(\mathbb{A},\bullet)$ is associative. ∎

We refer to [17] for general medial algebras and their spectral theory.

##  5\. Categories of calibrated medial algebras

We denote by $\operatorname{Idm}^{\times}(\mathbb{A},\ast)$ the subset of
nonzero idempotents in an algebra $(\mathbb{A},\ast)$ such that the left
multiplication operator $L_{\ast}(c)$ is invertible.

######  Definition 5.1.

A medial algebra $(\mathbb{A},\ast)$ is called special if the set
$\operatorname{Idm}^{\times}(\mathbb{A},\ast)$ is non-empty.

The terminology ‘special’ here correspond exactly to what is called ‘medial
class (iii) algebras’ in [17].

It follows from Proposition 4.3 above that if $(\mathbb{A},\bullet)$ unital
commutative associative algebra then its unity element $e$ of becomes an
idempotent in the inner isotope $(\mathbb{A},\bullet_{h})$,
$h\in\operatorname{Aut}(\mathbb{A},\bullet)$ and furthermore
$e\in\operatorname{Idm}^{\times}(\mathbb{A},\bullet_{h})$. This proves

######  Corollary 5.2.

Any inner isotope $(\mathbb{A},\bullet_{h})$ of a unital commutative
associative algebra $(\mathbb{A},\bullet)$ is a special medial algebra.

Idempotents in $\operatorname{Idm}^{\times}(\mathbb{A},\ast)$ are
distinguished in many aspects and a particular choice of such an idempotent
can be thought of as a calibration or pointing of an algebra. To put this
observation into an appropriate context, we define a concept of calibrations
for special medial algebras and unital commutative associative algebras. An
important ingredient in our constructions is Kaplansky’s trick [13], [20].

######  Definition 5.3.

A special (commutative) medial algebra $(\mathbb{A},\ast)$ with a
distinguished idempotent $c\in\operatorname{Idm}^{\times}(\mathbb{A},\ast)$ is
called calibrated and denoted by $(\mathbb{A},\ast,c)$. An algebra
homomorphism between two special medial algebras
$f:(\mathbb{A},\ast,a)\to(\mathbb{B},\circledast,b)$ is called a calibrated if
$f(a)=b$.

######  Definition 5.4.

A unital commutative associative algebra $(\mathbb{A},\diamond,e)$ with
algebra unity $e$ and a distinguished automorphism
$h\in\operatorname{Aut}(\mathbb{A},\diamond,e)$ is called calibrated and
denoted by $(\mathbb{A},\diamond,e,h)$. An algebra homomorphism between two
calibrated commutative associative algebras
$f:(\mathbb{A},\diamond,e,h)\to(\mathbb{A}^{\prime},\diamondsuit,e^{\prime},h^{\prime})$
is called a calibrated homomorphism if $h^{\prime}\circ f=f\circ h$ (note that
$f(e)=e^{\prime}$).

We denote two calibrated isomorphic algebras by
$\mathbb{A}\stackrel{{\scriptstyle\centerdot}}{{\cong}}\mathbb{A}^{\prime}$.
Of course,
$\mathbb{A}\stackrel{{\scriptstyle\centerdot}}{{\cong}}\mathbb{A}^{\prime}$
implies that $\mathbb{A}\cong\mathbb{A}^{\prime}$.

Denote by $\mathfrak{M}$ (respectively by $\mathfrak{A}$) the class of
calibrated special commutative medial algebras (respectively calibrated
commutative associative unital algebras). These classes are categories in an
obvious way, where the corresponding morphisms are calibrated homomorphisms.

######  Theorem 5.5.

The functor $\Phi:\mathfrak{A}\to\mathfrak{M}$ given by
$\Phi(\mathbb{A},\diamond,e,h)=(\mathbb{A},\ast,e)$, where $x\ast
y=h(x\diamond y)$ and $e\in\operatorname{Idm}^{\times}(\mathbb{A},\ast,e)$, is
a category isomorphism. The inverse functor $\Psi=\Phi^{-1}$ is given by
$\Psi(\mathbb{A},\ast,c)=(\mathbb{A},\diamond,c,h)$, where $x\diamond
y=L_{\ast}^{-1}(x\ast y)$, $c$ is a unity in $(\mathbb{A},\diamond)$ and
$h=L_{\ast}(c)\in\operatorname{Aut}(\mathbb{A},\diamond)$.

###### Proof.

The proof is divided into three steps.

Step 1. The map $\Phi$ is a functor $\mathfrak{A}\to\mathfrak{M}$.

Our first claim is that given $(\mathbb{A},\diamond,e,h)\in\mathfrak{A}$, a
new multiplication on $\mathbb{A}$ by

$x\ast y=h(x\diamond y)=h(x)\diamond h(y)$ (13)

defines a calibrated medial algebra $\mathbb{A},\ast,e)\in\mathfrak{M}$ of
class (iii) with an idempotent
$e\in\operatorname{Idm}^{\times}(\mathbb{A},\ast)$. Indeed, since
$h\in\operatorname{Aut}(\mathbb{A},\diamond)$ we have

$(x\ast y)\ast(z\ast w)=h((x\ast y)\diamond(z\ast w))=h(h(x\diamond y)\diamond
h(z\diamond w))=(h\circ h)(x\diamond y\diamond z\diamond w)$

where the right hand side is obviously symmetric for any permutation of
factors. Since $e$ is the unitity element and $h$ is an algebra automorphism
in $(\mathbb{A},\diamond)$ then $h(e)=e$, hence $e\ast e=h(e\diamond e)=e$,
i.e. $e$ is an idempotent in $(\mathbb{A},\ast)$. Also, $e\ast x=h(e\diamond
x)=h(x)$, hence $L_{\ast}(e)=h$ is a bijection, i.e.
$e\in\operatorname{Idm}^{\times}(\mathbb{A},h,\ast)$.

Next, we prove that $\Phi$ acts correspondingly on morphisms. To this end
consider a calibrated $\mathfrak{A}$-algebra homomorphism
$f:(\mathbb{A},\diamond,e,h)\to(\mathbb{A}^{\prime},\diamondsuit,e^{\prime},h^{\prime})$.
Then $f$ is an algebra homomorphism and $h^{\prime}\circ f=f\circ h$. Let
$\Phi(\mathbb{A},\diamond,e,h)=(\mathbb{A},\ast,e)$ and
$\Phi(\mathbb{A}^{\prime},\diamondsuit,e^{\prime},h^{\prime})=(\mathbb{A}^{\prime},\circledast,e^{\prime})$.
We consider $\Phi(f)=f$. Then $f(e)=e^{\prime}$ and

$f(x\ast y)=(f\circ h)(x\diamond y)=(h^{\prime}\circ f)(x\diamond
y)=h^{\prime}(f(x)\diamondsuit f(y))=f(x)\circledast f(y),$

hence $f:(\mathbb{A},\ast,e)\to(\mathbb{A}^{\prime},\circledast,e^{\prime})$
is a calibrated $\mathfrak{M}$-algebra homomorphism. The fact that $\Phi$
preserves identity morphisms and composition of morphisms trivially follows by
its definition.

Step 2. The map $\Psi$ is a functor $\mathfrak{M}\to\mathfrak{A}$.

Let $(\mathbb{A},\ast,c)\in\mathfrak{M}$. Define a new multiplication on
$\mathbb{A}$ by

$x\diamond y=L_{\ast}(c)^{-1}(x\ast y).$ (14)

The algebra $(\mathbb{A},\diamond)$ is commutative and $c\diamond
x=L_{\ast}(c)^{-1}(c\ast x)=L_{\ast}(c)^{-1}L_{\ast}(c)x=x$, hence $c$ is a
unit in $(\mathbb{A},\diamond)$. By 1 in Proposition 6.1, $L_{\ast}(c)^{-1}$
is an algebra isomorphism of $(\mathbb{A},\ast,c)$ hence

$(x\diamond y)\diamond(z\diamond
w)=L_{\ast}(c)^{-1}\bigl{(}L_{\ast}(c)^{-1}(x\ast y)\ast
L_{\ast}(c)^{-1}(z\ast w)\bigr{)}=L_{\ast}(c)^{-2}((x\ast y)\ast(z\ast w))$

which implies that $(\mathbb{A},\diamond)$ is medial. By Proposition 4.4,
$(\mathbb{A},\diamond)$ is associative. Since $L_{\ast}(c)$ is an algebra
isomorphism of $(\mathbb{A},\ast,c)$, it is a bijection. Furthermore,

$\displaystyle L_{\ast}(c)(x\diamond y)$
$\displaystyle=L_{\ast}(c)L_{\ast}(c)^{-1}(x\ast y)=x\ast y$ $\displaystyle
L_{\ast}(c)(x)\diamond L_{\ast}(c)(y)$
$\displaystyle=L_{\ast}(c)^{-1}\bigl{(}L_{\ast}(c)(x)\ast
L_{\ast}(c)(y)\bigr{)}=x\ast y,$

hence $L_{\ast}(c)(x\diamond y)=L_{\ast}(c)(x)\diamond L_{\ast}(c)(y)$, i.e.
$L_{\ast}(c)\in\operatorname{Aut}(\mathbb{A},\diamond,c)$. Then
$\Psi(\mathbb{A},\ast,c)=(\mathbb{A},\diamond,c,L_{\ast}(c))\in\mathfrak{A}$.

Next, we prove that $\Phi$ acts correspondingly on morphisms. Let
$f:(\mathbb{A},\ast,c)\to(\mathbb{A}^{\prime},\circledast,c^{\prime})$ be a
calibrated $\mathfrak{M}$-algebra homomorphism, i.e. $f(c)=c^{\prime}$. Let
$\Psi(\mathbb{A},\ast,c)=(\mathbb{A},\diamond,c,L_{\ast}(c))$ and
$\Psi((\mathbb{A}^{\prime},\circledast,c^{\prime})=(\mathbb{A}^{\prime},\diamondsuit,c^{\prime},L_{\circledast}(c^{\prime}))$.
Then $f:(\mathbb{A},\diamond)\to(\mathbb{A}^{\prime},\diamondsuit)$ is vector
space homomorphism and

$(L_{\circledast}(c^{\prime})\circ f)(x)=c^{\prime}\circledast
f(x)=f(c)\circledast f(x)=f(c\ast x)=(f\circ L_{\ast}(c))(x),$

hence $L_{\circledast}(c^{\prime})\circ f=f\circ L_{\ast}(c)$ and therefore
$f\circ L_{\ast}(c)^{-1}=L_{\circledast}(c^{\prime})^{-1}\circ f$, and it
follows that

$\displaystyle f(x\diamond y)$ $\displaystyle=\bigl{(}f\circ
L_{\ast}(c)^{-1}\bigr{)}(x\ast
y)=\bigl{(}L_{\circledast}(c^{\prime})^{-1}\circ f\bigr{)}(x\ast y)$
$\displaystyle=L_{\circledast}(c^{\prime})^{-1}(f(x)\circledast
f(y))=f(x)\diamondsuit f(y)$

i.e. $f$ is a calibrated $\mathfrak{A}$-algebra homomorphism. The fact that
$\Psi$ preserves identity morphisms and composition of morphisms readily
follows by its definition.

Step 3. $\Psi\circ\Phi=\mathrm{id}_{\mathfrak{A}}.$

We have $\Phi(\mathbb{A},\diamond,e,h)=(\mathbb{A},\ast,e)$ and
$\Psi(\mathbb{A},\ast,e)=(\mathbb{A},\diamondsuit,e,L_{\ast}(e))$, where

$\displaystyle x\ast y$ $\displaystyle=h(x\diamond y),$ $\displaystyle
x\diamondsuit y$ $\displaystyle=L_{\ast}(e)^{-1}(x\ast y).$

We have $e\ast z=h(e\diamond y)=h(e)\diamond h(z)=h(z)$, i.e. $L_{\ast}(e)=h$,
therefore $x\diamondsuit y=L_{\ast}(e)^{-1}(x\ast y)=h^{-1}(x\ast y)=x\diamond
y$. This implies that $\Psi(\mathbb{A},\ast,e)=(\mathbb{A},\diamond,e,h)$,
hence $\Psi\circ\Phi=\mathrm{id}_{\mathfrak{A}}$. The theorem follows. ∎

We have several important corollaries of the above result.

######  Corollary 5.6.

$\Psi(\mathbb{A})\stackrel{{\scriptstyle\centerdot}}{{\cong}}\Psi(\mathbb{A}^{\prime})$
if and only if
$\mathbb{A}\stackrel{{\scriptstyle\centerdot}}{{\cong}}\mathbb{A}^{\prime}$
for $\mathbb{A},\mathbb{A}^{\prime}\in\mathfrak{M}$ and
$\Phi(\mathbb{A})\stackrel{{\scriptstyle\centerdot}}{{\cong}}\Phi(\mathbb{A}^{\prime})$
if and only if
$\mathbb{A}\stackrel{{\scriptstyle\centerdot}}{{\cong}}\mathbb{A}^{\prime}$
for $\mathbb{A},\mathbb{A}^{\prime}\in\mathfrak{A}$.

The following propositions describe how the functors
$\Phi:\mathfrak{A}\to\mathfrak{M}$ and $\Psi:\mathfrak{M}\to\mathfrak{A}$
depend on a particular choice of calibrating.

######  Proposition 5.7.

If $(\mathbb{A},\ast)$ is a special commutative medial algebra and
$c_{1},c_{2}\in\operatorname{Idm}^{\times}(\mathbb{A},\ast)$ then
$(\mathbb{A},\ast,c_{1})\stackrel{{\scriptstyle\centerdot}}{{\cong}}(\mathbb{A},\ast,c_{2})$.
In particular, given a special commutative medial algebra, there exists a
unique calibrated isomorphy class of $\mathbb{A}$.

###### Proof.

By 1 in Proposition 6.1, $f=L_{\ast}(c_{1})^{-1}L_{\ast}(c_{2})$ is a
$\ast$-algebra automorphism of $(\mathbb{A},\ast)$. Furthermore,
$f(c_{1})=L_{\ast}(c_{1})^{-1}L_{\ast}(c_{2})(c_{1})=L_{\ast}(c_{1})^{-1}(c_{1}\ast
c_{2})=c_{2},$ hence $f$ a calibrated isomorphism of
$f:(\mathbb{A},\ast,c_{1})\to(\mathbb{A},\ast,c_{2})$. ∎

######  Proposition 5.8.

Let $(\mathbb{A},\diamond)$ be a unital commutative associative algebra,
$h_{1},h_{2}\in\operatorname{Aut}(\mathbb{A},\diamond)$. Then
$(\mathbb{A},\diamond,e,h_{1})\stackrel{{\scriptstyle\centerdot}}{{\cong}}(\mathbb{A},\diamond,e,h_{2})$
if and only if $h_{1}$ and $h_{2}$ are conjugate in
$\operatorname{Aut}(\mathbb{A},\diamond)$.

###### Proof.

By the definition,
$(\mathbb{A},\diamond,e,h_{1})\stackrel{{\scriptstyle\centerdot}}{{\cong}}(\mathbb{A},\diamond,e,h_{2})$
if and only if there exists an isomorphism
$f:(\mathbb{A},\diamond)\to(\mathbb{A},\diamond)$ such that $h_{2}\circ
f=f\circ h_{1}$, i.e. $f\in\operatorname{Aut}(\mathbb{A},\diamond)$ and
$h_{2}=f\circ h_{1}\circ f^{-1}$, which is equivalent to that $h_{1}$ and
$h_{2}$ are conjugate in $\operatorname{Aut}(\mathbb{A},\diamond)$. ∎

######  Corollary 5.9.

Given a unital commutative associative algebra $(\mathbb{A},\diamond)$, there
is a natural bijection between its calibrated isomorphy classes and the
conjugacy classes of its automorphism group:

$\mathbb{A}/_{\stackrel{{\scriptstyle\centerdot}}{{\cong}}}=\operatorname{Aut}(\mathbb{A})/_{\mathrm{conj}}$

##  6\. Idempotents in inner isotopes

We start with a general result which holds for any commutative medial algebra.

######  Proposition 6.1 ([17]).

Let $(\mathbb{A},\ast)$ be commutative medial algebra. If
$c_{1},c_{2}\in\operatorname{Idm}(\mathbb{A},\ast)$ then so is $c_{1}\ast
c_{2}$. In other words, the set of all idempotents
$\operatorname{Idm}(\mathbb{A},\ast)\cup\\{0\\}$ is a multiplicative magma.
Furthermore, for any idempotent $c\in\operatorname{Idm}(\mathbb{A},\ast)$:

  1. (1)

$L_{\ast}(c)$ is a $(\mathbb{A},\ast)$-algebra endomorphism;

  2. (2)

The $0$-Peirce subspace $\ker L_{\ast}(c)$ is an ideal of $(\mathbb{A},\ast)$
and the image $L_{\ast}(c)(\mathbb{A})$ is a subalgebra of $\mathbb{A}$;

  3. (3)

The $1$-Peirce subspace $\\{x\in\mathbb{A}:L_{\ast}(c)x=x\\}$ is a subalgebra
of the image $L_{\ast}(c)(\mathbb{A})$ and $\dim\mathbb{A}_{c}(1)\geq 1$;

  4. (4)

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

$L_{\ast}(c_{2})L_{\ast}(c_{1})=L_{\ast}(c_{1}\ast c_{2})L_{\ast}(c_{2})$ (15)

###### Proof.

The first claim is an immediate corollary of the medial magma identity (12).
Furthermore, the multiplication operator $L_{\ast}(c):\mathbb{A}\to\mathbb{A}$
is linear and it follows from (12) that for any idempotent
$c\in\operatorname{Idm}(\mathbb{A},\ast)$

$\displaystyle L_{\ast}(c)(x\ast y)$ $\displaystyle=c\ast(x\ast y)=(c\ast
c)\ast(x\ast y)=(c\ast x)\ast(c\ast y)$ $\displaystyle=L_{\ast}(c)x\,\ast
L_{\ast}(c)y,$

hence $L_{\ast}(c)$ is an algebra endomorphism. As the kernel of a
homomorphism, $\mathbb{A}_{c}(0)=\ker L_{\ast}(c)$ is an ideal and as the
image of a homomorphism, $L_{\ast}(c)(\mathbb{A})$ is a subalgebra. Further,
$\mathbb{A}_{c}(1)=\\{x:L_{\ast}(c)x=x\\}$ is the set of fixed points of the
algebra homomorphism $L_{\ast}(c)$, thus it is a subalgebra of $\mathbb{A}$.
Since $L_{\ast}(c)$ stabilizes $\mathbb{A}_{c}(1)$, the latter is also a
subalgebra of $L_{\ast}(c)(\mathbb{A})$. Also, the one-dimensional subspace
$\mathrm{span}(c)\subset L_{\ast}(c)(\mathbb{A}),$ hence
$\dim\mathbb{A}_{c}(1)\geq 1$. Finally, (15) follows from

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

∎

Now let $(\mathbb{A},\bullet)$ be a commutative algebra and
$(\mathbb{A},\bullet_{h})$ its inner isotope,
$h\in\operatorname{Aut}(\mathbb{A},\bullet)$. Then $c$ is an idempotent in
$(\mathbb{A},\bullet_{h})$ if $c=c\bullet_{h}c$, which implies

$\operatorname{Idm}(\mathbb{A},\bullet_{h})\cup\\{0\\}=\\{c\in\mathbb{A}:h(c\bullet
c)=c\\}.$ (16)

Combining Proposition 4.3 and Proposition 6.1 we obtain

######  Corollary 6.2.

Let $(\mathbb{A},\bullet)$ be commutative associative algebra. The set of all
idempotents $\operatorname{Idm}(\mathbb{A},\bullet_{h})\cup\\{0\\}$ is a
multiplicative magma.

The latter makes it natural to ask when the set of all nonzero idempotents
$\operatorname{Idm}(\mathbb{A},\bullet_{h})$ is a quasigroup. Note that this
property does not hold in general because the product of two idempotents may
be zero. But, under some natural assumptions, one has the desired property.

######  Corollary 6.3.

If $(\mathbb{A},\bullet)$ is a commutative associative division algebra and
$h\in\operatorname{Aut}(\mathbb{A},\bullet)$ then $(\mathbb{A},\bullet_{h})$
is also a division algebra and the set of nonzero idempotents
$\operatorname{Idm}(\mathbb{A},\bullet_{h})$ is a commutative idempotent
medial quasigroup.

###### Proof.

Indeed, given $x,y\in(\mathbb{A},\bullet_{h})$, $x\bullet_{h}y=0$ if and only
if $h(x\bullet y)=0$, where the latter by the bijectivity of $h$ is equivalent
to $x\bullet y=0$, therefore $(\mathbb{A},\bullet_{h})$ is also a division
algebra. This implies by Proposition 6.1 that for any
$c_{1},c_{2}\in\operatorname{Idm}(\mathbb{A},\bullet_{h})$ in fact
$c_{1}\bullet_{h}c_{2}\in\operatorname{Idm}(\mathbb{A},\bullet_{h})$. Suppose
that $c_{1}\bullet_{h}c_{2}=c_{1}\bullet_{h}c_{3}$ for some
$c_{1},c_{2},c_{3}\in\operatorname{Idm}(\mathbb{A},\bullet_{h})$. Then
$c_{1}\bullet_{h}(c_{2}-c_{3})=0$, therefore $c_{2}-c_{3}=0$, hence
$\operatorname{Idm}(\mathbb{A},\bullet_{h})$ is in fact a quasigroup, which is
obviously commutative idempotent and medial, the proposition follows. ∎

######  Proposition 6.4.

Let $(\mathbb{A},\bullet)$ be a commutative associative division algebra and
$h\in\operatorname{Aut}(\mathbb{A},\bullet)$. Then all idempotents
$c\in\operatorname{Idm}(\mathbb{A},\bullet_{h})$ have the same characteristic
polynomial.

###### Proof.

By Corollary 6.3, if
$c_{1},c_{2}\in\operatorname{Idm}(\mathbb{A},\bullet_{h})$ then
$c_{1}\bullet_{h}c_{2}\in\operatorname{Idm}(\mathbb{A},\bullet_{h})$ and
$L_{\bullet_{h}}(c_{1})$ is invertible, hence by Proposition 6.1 we obtain

$L_{\bullet_{h}}(c_{1}\bullet_{h}c_{2})=L_{\bullet_{h}}(c_{1})L_{\bullet_{h}}(c_{2})L_{\bullet_{h}}(c_{1})^{-1}.$

and similarly

$L_{\bullet_{h}}(c_{2}\bullet_{h}c_{1})=L_{\bullet_{h}}(c_{2})L_{\bullet_{h}}(c_{1})L_{\bullet_{h}}(c_{2})^{-1}.$

Since $c_{1}\bullet_{h}c_{2}=c_{2}\bullet_{h}c_{1}$ we obtain from the last
two relations that the characteristic polynomials of $L_{\bullet_{h}}(c_{2})$
and $L_{\bullet_{h}}(c_{1})$ are equal. The proposition follows. ∎

######  Proposition 6.5.

Let $(\mathbb{A},\bullet)$ be a unital commutative associative division
algebra with unity $e$ and let $h\in\operatorname{Aut}(\mathbb{A},\bullet)$ be
an automorphism of finite order $d$. Then $c^{\bullet(2^{d}-1)}=e$ for any
$c\in\operatorname{Idm}(\mathbb{A},\bullet_{h})$.

###### Proof.

By (16) $c=c\bullet_{h}c=h(c\bullet c)=h(c^{\bullet 2})$, therefore for all
$k=1,2,\ldots$

$\displaystyle c$ $\displaystyle=h(c)\bullet h(c)=h(h(c^{\bullet 2}))\bullet
h(h(c^{\bullet 2}))=h^{2}(c^{\bullet 2^{2}})=\ldots=h^{k}(c^{\bullet 2^{k}}).$

Since $h^{d}=\mathrm{id}$, we obtain $c=h^{d}(c^{\bullet 2^{d}})=c^{\bullet
2^{d}}$, i.e. $c\bullet(c^{\bullet(2^{d}-1)}-e)=0$, hence by the assumptions
$(\mathbb{A},\bullet)$ does not contain divisors of zero, we conclude that
$c^{\bullet(2^{d}-1)}=e$. ∎

##  7\. Inner isotopes of a quotient polynomial algebra

It worthy to point out that although the algebras $\mathbf{K}[z]/P$ considered
in the introduction originate from polynomials, their multiplicative structure
does not depend on a particular choice of a polynomial or set of its roots,
but only on the dimension $n$ (i.e. the number of distinct roots) and the
choice of a conjugacy class of permutation $\sigma\in S_{n}$. Furthermore,
under the splitting condition, the algebra $\mathbf{K}[z]/P$ turns out
isomorphic the direct product algebra of $n=\deg P$ copies of the ground field
$\mathbf{K}$, the theorem below claims.

######  Theorem 7.1.

Let a polynomial $P\in\mathbf{K}[z]$ split over $\mathbf{K}$ and have $n=\deg
P$ distinct roots. Then

$\mathbf{K}[z]/P\cong(\mathbf{K}^{n},\bullet).$ (17)

The automorphism group

$\operatorname{Aut}(\mathbf{K}[z]/P)\cong\operatorname{Aut}(\mathbf{K}^{n},\bullet)\cong
S_{n}$

is isomorphic to the symmetric group $S_{n}$. Furthermore, two inner isotopes
of $(\mathbf{K}^{n},\bullet_{\sigma})$ and $(\mathbf{K}^{n},\bullet_{\tau})$,
$\sigma,\tau\in S_{n}$ are isomorphic if and only if $\sigma$ and $\tau$
conjugate in $S_{n}$.

###### Proof of Theorem 7.1.

Under our assumptions, $P=(z-a_{1})\ldots(z-a_{n})$, where
$\\{a_{1},\ldots,a_{n}\\}$ are the distinct roots of $P$. Then the Chinese
Remainder Theorem [5, Propoisition 15] gives an explicit isomorphism

$\displaystyle\mathbf{K}[z]/P$
$\displaystyle\cong\mathbf{K}[z]/(z-a_{1})\cdot\ldots\cdot(z-a_{n})$
$\displaystyle\cong(\mathbf{K}[z]/(z-a_{1}))\times\ldots\times(\mathbf{K}[z]/(z-a_{n}))$
$\displaystyle\cong\underbrace{(\mathbf{K},\cdot)\times\ldots\times(\mathbf{K},\cdot)}_{n\text{
times}}$ $\displaystyle\cong(\mathbf{K}^{n},\bullet).$

implying (17), where $\bullet$ is the standard coordinate-wise multiplication
on $\mathbf{K}^{n}$. Let $e_{i}=(0,\ldots,1,\ldots,0)$, $1\leq i\leq n$ be the
standard basis of $\mathbf{K}^{n}$ (see section 2). Then it follows that
$e_{i}\bullet e_{i}=e_{i}$, hence $e_{i}$ are idempotents of
$(\mathbf{K}^{n},\bullet)$, and, moreover, the partial sums

$e_{I}:=\sum_{i\in I}e_{i},\qquad I\in 2^{\\{1,2,\ldots,n\\}}$ (18)

are the only nonzero idempotents in
$\operatorname{Idm}(\mathbf{K}^{n},\bullet)$. Furthermore, since $e_{i}\bullet
e_{j}=0$ for any $1\leq i<j\leq n$, $\\{e_{i}\\}_{1\leq i\leq n}$ are the only
primitive (i.e. indecomposable) idempotents in
$\operatorname{Idm}(\mathbf{K}^{n},\bullet)$. If
$\phi\in\operatorname{Aut}(\mathbf{K}^{n},\bullet)$ is an algebra automorphism
then $\phi(x\bullet x)=\phi(x)\bullet\phi(x)$, hence $\phi$ is a permutation
of the set of nonzero idempotents
$\operatorname{Idm}(\mathbf{K}^{n},\bullet)$. Since
$\phi(x+y)=\phi(x)+\phi(y)$, $\phi$ preserve primitive idempotents, thus
$\phi$ is a permutation of the set $\\{e_{i}\\}_{1\leq i\leq
n}\subset\operatorname{Idm}(\mathbf{K}^{n},\bullet)$. This implies that
$\operatorname{Aut}(\mathbf{K}^{n},\bullet)$ is a subgroup of $S_{n}$. On the
other hand, if $\sigma\in S_{n}$ is an arbitrary permutation, then the linear
map

$\psi_{\sigma}:(x_{1},\ldots,x_{n})\to(x_{\sigma(1)},\ldots,x_{\sigma(n)})$
(19)

is an isomorphism of $(\mathbf{K}^{n},\bullet)$ implying that in fact
$\operatorname{Aut}(\mathbf{K}^{n},\bullet)\cong S_{n}$. Finally, since
$(\mathbf{K}^{n},\bullet)$ trivially satisfies (9), we conclude by Corollary
3.6 that two inner isotopes of $(\mathbf{K}^{n},\bullet_{\sigma})$ and
$(\mathbf{K}^{n},\bullet_{\tau})$, $\sigma,\tau\in S_{n}$ are isomorphic if
and only if $\sigma$ and $\tau$ conjugate in $S_{n}$. ∎

######  Remark 7.2.

It follows from Theorem 7.1 that it suffices to consider some specific
polynomial $P(z)$ in each degree $n$. The circular polynomials $P(z)=z^{n}-1$
are distinguished in many respects. The corresponding polynomial quotient
algebra $(\mathbf{K}^{n},\bullet):=(\mathbf{K}[z]/(z^{n}-1),\bullet)$ and its
inner isotopes $(\mathbf{K}^{n},\bullet_{\sigma})$ have originally been
introduced and studied in [17] in the particular case when
$\sigma=(2\,3\,\ldots\,n\,1)$ (in cycle notation). The corresponding algebra
is isospectral and as a corollary of the syzygy relation [16], the spectrum of
each idempotent is the set of roots of $z^{n}-1$.

Below we consider the general case of a permutation $\sigma$ with several
cycles. It turns out that the resulting algebra decomposes as a direct sum of
the ideals corresponding to the decomposition of $\sigma$ into disjoint
permutations. By abuse of notation, we shall write

$\langle\sigma_{j}\rangle=\mathrm{span}(\\{e_{i}:i\in[\sigma_{j}]\\})=\bigoplus_{i\in[\sigma_{j}]}\langle
e_{i}\rangle,$ (20)

where $[\sigma_{j}]\subset\bar{n}=\\{1,2,\ldots,n\\}$ is the orbit of the
cycle $\sigma_{j}$. Each cycle $\sigma_{j}$ acts as a cyclic permutation of
order $|\sigma_{j}|$ on the orbit $[\sigma_{j}]$ such that the set of indices
$\bar{n}$ is a disjoint union of the orbits $[\sigma_{j}]$, $1\leq j\leq r$.

######  Proposition 7.3.

Let a permutation $\sigma\in S_{n}$ have a disjoint cycle decomposition
$\sigma=\sigma_{1}\ldots\sigma_{r}$. Then

$(\mathbf{K}^{n},\bullet_{\sigma})\cong\bigoplus_{j=1}^{r}(\langle\sigma_{j}\rangle,\bullet_{\sigma}),\qquad(\langle\sigma_{j}\rangle,\bullet_{\sigma})\cong(\mathbf{K}^{|\sigma_{j}|},\bullet_{\sigma_{j}}).$
(21)

###### Proof.

In the above notation, we have

$(\langle\sigma_{j}\rangle,\bullet)\cong(\mathbf{K}^{|\sigma_{j}|},\bullet).$
(22)

The corresponding isomorphism $\psi_{\sigma}$ of $(\mathbf{K}^{n},\bullet)$ in
(19) decomposes in the direct sum $\psi_{\sigma}=\bigoplus_{1\leq j\leq
r}\psi_{\sigma_{j}}$, where each
$\psi_{\sigma_{j}}\in\operatorname{End}(\langle\sigma_{j}\rangle)$ is
determined as the restriction of $\psi_{\sigma}$ to $\langle\sigma_{j}\rangle$
and moreover

$(\mathbf{K}^{n},\bullet)=\bigoplus_{j=1}^{r}(\langle\sigma_{j}\rangle,\bullet).$
(23)

It follows from the definitions that
$\langle\sigma_{i}\rangle\bullet\mathbb{A}\subset\langle\sigma_{i}\rangle$ and
$\langle\sigma_{i}\rangle\bullet\langle\sigma_{j}\rangle=0$ for $i\neq j$,
therefore (23) is a decomposition into the direct sum of ideals. Let
$\pi_{j}:\mathbf{K}^{n}\to\langle\sigma_{j}\rangle$ denote the canonical
projection (a linear homomorphism). Then
$\pi_{j}:(\mathbf{K}^{n},\bullet)\to(\langle\sigma_{j}\rangle,\bullet)$ is
also an algebra homomorphism and the following commutative diagram holds:

$\begin{CD}(\mathbf{K}^{n},\bullet)@>{\pi_{j}}>{}>(\langle\sigma_{j}\rangle,\bullet)\\\
@V{}V{\psi_{\sigma}}V@V{}V{\psi_{\sigma_{j}}}V\\\
(\mathbf{K}^{n},\bullet)@>{\pi_{j}}>{}>(\langle\sigma_{j}\rangle,\bullet)\end{CD}$
(24)

where the vertical arrows are algebra isomorphisms. Since
$\psi_{\sigma_{j}}\in\operatorname{Aut}(\langle\sigma_{j}\rangle,\bullet)$,
arguing as above we conclude that

$(\mathbf{K}^{n},\bullet_{\sigma})=\bigoplus_{j=1}^{r}(\langle\sigma_{j}\rangle,\bullet_{\sigma}),$
(25)

where
$(\langle\sigma_{j}\rangle,\bullet_{\sigma})\cong(\mathbf{K}^{|\sigma_{j}|},\bullet_{\sigma_{j}})$
are pairwise orthogonal ideals in $(\mathbf{K}^{n},\bullet_{\sigma})$,
implying (21). ∎

By virtue of Proposition 7.3 and Proposition 2.1, it suffices to study inner
isotopes $(\mathbf{K}^{n},\bullet_{\tau})$ for the case of a single cycle
$\tau$, we will consider this in the next section.

##  8\. The structure of idempotents

First we consider the case $(\mathbf{K}^{n},\bullet_{\tau})$ (which is
equivalent to the construction mentioned in Remark 7.2) in more detail, i.e.
we assume that $\tau$ contains a single cycle. More precisely, let $n\geq 2$
be an integer and $\tau=(2\,3\,\ldots\,n\,1)\in S_{n}$ be the right cyclic
shift (in cycle notation). We shall suppose that a field $\mathbf{K}$ is
$\tau$-admissible (see Definition 2.2 above), i.e. there are primitive roots
of unity of orders $n$ and $2^{n}-1$ in $\mathbf{K}$; denote them by
$\epsilon$ and $\zeta$, respectively.

######  Proposition 8.1.

The set of nonzero idempotents of $(\mathbf{K}^{n},\bullet_{\tau})$ can be
parameterized by

$\\{c_{k}=(\zeta^{2^{n-1}k},\zeta^{2^{n-2}k},\ldots,\zeta^{2k},\zeta^{k}):\quad
k\in\mathbb{Z}/(2^{n}-1)\mathbb{Z}\\},$ (26)

where all $c_{k}$ are pairwise distinct. The idempotents satisfy

$c_{i}\bullet_{\tau}c_{j}=c_{i\circledast j},$ (27)

where the binary operation $\circledast$ on $\mathbb{Z}/(2^{n}-1)\mathbb{Z}$
is defined by

$i\circledast j\equiv\frac{1}{2}(i+j)\equiv 2^{n-1}(i+j)\mod(2^{n}-1).$ (28)

###### Proof.

The multiplication of $c=(x_{1},\ldots,x_{n})$ in
$(\mathbf{K}^{n},\bullet_{\tau})$ is given by

$c\bullet_{\tau}c=(x_{\tau(1)}^{2},\ldots,x_{\tau(n)}^{2})=(x_{2}^{2},x_{3}^{2},\ldots,x_{n}^{2},x_{1}^{2}),$

thus $c$ is an idempotent if and only if

$x_{i+1}^{2}=x_{i}\quad\text{ for each $i\in\mathbb{Z}/n\mathbb{Z}$.}$ (29)

Iterating the latter relations $n$ times yields $x_{i}^{2^{n}}=x_{i}$ for any
$i$. Together with (29) this implies that either all of $x_{i}$ are zero (and
in that case $c=0$), or all $x_{i}$ are nonzero, and in the latter case they
satisfy

$x_{i}^{2^{n}-1}-1=0,\qquad\forall i\in\mathbb{Z}/n\mathbb{Z}.$ (30)

It follows from (29) that any nonzero idempotent $x$ can be written as

$x=(t^{2^{n-1}},t^{2^{n-2}},\ldots,t^{2^{1}},t),$ (31)

where $t$ is a primitive root of unity of order $2^{n}-1$, hence (31) readily
implies (26). Finally, we have

$c_{i}\bullet_{\tau}c_{j}=(\zeta^{2^{n-2}(i+j)},\zeta^{2^{n-3}(i+j)},\ldots,\zeta^{(i+j)},\zeta^{2^{n-1}(i+j)})$

implying (27).

∎

Identity (27) expresses the fact that the product of any two nonzero
idempotents in $\operatorname{Idm}(\mathbf{K}^{n},\bullet_{\tau})$ is a
nonzero idempotent again, in other words, the set
$\operatorname{Idm}(\mathbf{K}^{n},\bullet_{\tau})$ is a multiplicative magma.
Furthermore, (27) implies a magma isomorphism

$\operatorname{Idm}(\mathbf{K}^{n},\bullet_{\tau})\cong(\mathbb{Z}/(2^{n}-1)\mathbb{Z},\,\circledast)$

Moreover, we have

######  Proposition 8.2.

The multiplicative magma $(\mathbb{Z}/(2^{n}-1)\mathbb{Z},\,\circledast)$ is a
commutative idempotent medial quasigroup.

###### Proof.

The quasigroup property can be seen as follows: if
$s,t,r\in(\mathbb{Z}/(2^{n}-1)\mathbb{Z},\,\circledast)$ are such that
$s\circledast r=s\circledast t$ then $(t-r)/2\equiv 0\mod(2^{n}-1)$, hence
$t=r$ in $(\mathbb{Z}/(2^{n}-1)\mathbb{Z},\,\circledast)$. Also, given
arbitrary $s,t\in(\mathbb{Z}/(2^{n}-1)\mathbb{Z},\,\circledast)$, there exists
precisely one solution $r:=2t-s$ to the following equation:

$s\circledast r=r\circledast s=\frac{1}{2}(s+2t-s)=t.$

Next, $s\circledast s=s$ for all
$s\in(\mathbb{Z}/(2^{n}-1)\mathbb{Z},\,\circledast)$, thus the quasigroup is
idempotent. Finally, since

$(i\circledast j)\circledast(k\circledast
l)\equiv\frac{1}{2}(i+j+k+l)\mod(2^{n}-1)$

is totally symmetric in all variables,
$(\mathbb{Z}/(2^{n}-1)\mathbb{Z},\,\circledast)$ is a medial quasigroup. ∎

######  Proposition 8.3.

The algebra $(\mathbf{K}^{n},\bullet_{\tau})$ has exactly $2^{n}-1$ distinct
regular idempotents $c_{k}$, i.e. it is generic. Each idempotent
$c_{k}\in\operatorname{Idm}(\mathbf{K}^{n},\bullet_{\tau})$ has the spectrum
$\epsilon,\epsilon^{2},\ldots,\epsilon^{n}$, each eigenvalue has multiplicity
one. In other words, the characteristic polynomial of
$L_{\bullet_{\tau}}(c_{k})$ is given by

$\det(\lambda\mathds{1}-L_{\bullet_{\tau}}(c_{k}))=\lambda^{n}-1.$ (32)

In particular,

$(L_{\bullet_{\tau}}(c_{k}))^{n}=\mathds{1}.$ (33)

###### Proof.

Given $p\in\mathbb{Z}/n\mathbb{Z}$ and $k\in\mathbb{Z}/(2^{n}-1)\mathbb{Z}$,
we define

$\eta_{k,p}:=(z_{1},\epsilon^{p}z_{2},\,\epsilon^{2p}z_{3},\ldots,\epsilon^{(n-1)p}z_{n}),$
(34)

where $z_{i}$ will be specified later. By (26) to
$c_{k}=(\zeta^{2^{n-1}k},\zeta^{2^{n-2}k},\ldots,\zeta^{2^{1}k},\zeta^{k})$,
hence

$\displaystyle c_{k}\bullet_{\tau}\eta_{k,p}$
$\displaystyle=(\epsilon^{p}\zeta^{2^{n-2}k}z_{2},\,\epsilon^{2p}\zeta^{2^{n-3}k}z_{3},\,\ldots\,\epsilon^{(n-1)p}\zeta^{2^{0}k}z_{n},\,\epsilon^{np}\zeta^{2^{n-1}k}z_{1})$
$\displaystyle=\epsilon^{p}\cdot(\zeta^{2^{n-2}k}z_{2},\,\zeta^{2^{n-3}k}\epsilon^{p}z_{3},\,\ldots\,\zeta^{2^{0}k}\epsilon^{(n-2)p}z_{n},\,\zeta^{2^{n-1}k}\epsilon^{(n-1)p}z_{1}),$

therefore, setting

$z_{i}:=\zeta^{-(2^{n-2}+\ldots+2^{n-i})k}z_{1}\quad\text{ for
$i=2,3,\ldots,n$}$

we see that

$\zeta^{2^{n-i}k}z_{i}=\zeta^{2^{n-i}k}\cdot\zeta^{-(2^{n-2}+\ldots+2^{n-i})k}z_{1}=z_{i-1},\quad
2\leq i\leq n.$

Since $2^{n-2}+\ldots+2^{1}+2^{0}=2^{n-1}-1\equiv-2^{n-1}\mod(2^{n}-1)$, we
get

$z_{n}=\zeta^{-(2^{n-2}+\ldots+2^{1}+2^{0})k}z_{1}=\zeta^{2^{n-1}k}z_{1}$

implying together with the above that

$c_{k}\bullet_{\tau}\eta_{k,p}=\epsilon^{p}\cdot(z_{1},\epsilon^{p}z_{2},\,\epsilon^{2p}z_{3},\ldots,\epsilon^{(n-1)p}z_{n})=\epsilon^{p}\eta_{k,p}.$
(35)

In other words,

$\eta_{k,p}=(1,\epsilon^{p}\zeta^{-2^{n-2}k},\,\epsilon^{2p}\zeta^{-(2^{n-2}+2^{n-2})k},\ldots,\epsilon^{(n-1)p}\zeta^{-(2^{n-2}+\ldots+2^{1}+2^{0})k})$
(36)

is an eigenvector of $L_{\bullet_{\tau}}(c_{k})$ with eigenvalue
$\epsilon^{p}$, for any $p\in\mathbb{Z}/n\mathbb{Z}$. Since all
$\epsilon,\epsilon^{2},\ldots,\epsilon^{n}$ are pairwise distinct, for the
dimension reasons this implies that each eigenvalue $\epsilon^{p}$ is simple,
and moreover the eigen-decomposition of $L_{\bullet_{\tau}}(c_{k})$ holds:

$(\mathbf{K}^{n},\bullet_{\tau})=\bigoplus_{p=1}^{n}\mathrm{span}(\eta_{k,p}).$
(37)

This also implies the explicit form of the characteristic polynomial of
$L_{\bullet_{\tau}}(c_{k})$ is given by (32). This implies that
$\det(\frac{1}{2}\mathds{1}-L_{\bullet_{\tau}}(c_{k}))\neq 0$, therefore
$(\mathbf{K}^{n},\bullet_{\tau})$ is a generic algebra. ∎

######  Remark 8.4.

It follows from (32) that the generic algebra
$(\mathbf{K}^{n},\bullet_{\tau})$ is isospectral. On the other hand, there are
nongeneric commutative isospectral algebras containing infinitely many
idempotents. Then their Peirce spectrum (i.e. the total set of eigenvalues of
all idempotents) can have a completely different structure. This holds for the
Hsiang algebras that appear in the context of cubic minimal cones; we refer an
interested reader to [18], [30], [29] for more details.

We need the following auxiliary property

######  Lemma 8.5.

$\sum_{k=1}^{2^{n}-1}(L_{\bullet_{\tau}}(c_{k}))^{s}=0$ for any
$s\in\\{1,2,\ldots,n-1\\}$.

###### Proof.

Note that by the definition of isotopy, $x\bullet_{\tau}y=\tau(x\bullet y)$,
where $\tau(x_{1},\ldots,x_{n-1},x_{n})=(x_{2},\ldots,x_{n},x_{1})$ and
$\bullet$ is the (commutative associative) coordinate-wise multiplication on
$\mathbf{K}^{n}$. Write (26) as $c_{k}=(a_{1}^{k},\ldots,a_{n}^{k})$, where
$a_{i}=\zeta^{2^{n-i}}$ do not depend on $k$. Then iterating the definition of
$\bullet_{\tau}$ we obtain

$\displaystyle(L_{\bullet_{\tau}}(c_{k}))^{s}x$
$\displaystyle=\tau^{s}(c_{k})\bullet\tau^{s-1}(c_{k})\bullet\ldots\bullet\tau(c_{k})\bullet\tau^{s}(x)$
$\displaystyle=\omega_{k}\bullet\tau^{s}(x),$

where
$\omega_{k}=\tau^{s}(c_{k})\bullet\tau^{s-1}(c_{k})\bullet\ldots\bullet\tau(c_{k})$.
Since $\tau$, …, $\tau^{s}$ are cyclic coordinate shifts
$\omega_{k}=(\zeta^{km_{1}},\zeta^{km_{2}},\ldots,\zeta^{km_{n}})$, where
$m_{i}\in\mathbb{Z}_{2^{n}-1}$ do not depend on $k$, more precisely

$m_{j}=2^{n-j-1}+\ldots+2^{n-j-s}=2^{n-j-s}\cdot(2^{s}-1)\neq 0\quad\text{in
$\mathbb{Z}_{2^{n}-1}$}$

Therefore $\zeta^{km_{j}}\neq 1$ and we have

$\displaystyle\sum_{k=1}^{2^{n}-1}(L_{\bullet_{\tau}}(c_{k}))^{s}x$
$\displaystyle=\sum_{k=1}^{2^{n}-1}\omega_{k}\bullet\tau^{s}(x)$
$\displaystyle=\biggl{(}\sum_{k=1}^{2^{n}-1}(\zeta^{km_{1}},\zeta^{km_{2}},\ldots,\zeta^{km_{n}})\biggr{)}\bullet\tau^{s}(x)=0$

because
$\sum_{k=1}^{2^{n}-1}\zeta^{km_{j}}=\zeta^{m_{j}}(\zeta^{m_{j}(2^{n}-1)}-1)(\zeta^{m_{j}}-1)^{-1}=0$
for any $m_{j}$, the claim follows. ∎

######  Corollary 8.6.

$\mathrm{span}(\operatorname{Idm}(\mathbf{K}^{n},\bullet_{\tau}))=\mathbf{K}^{n}$.

###### Proof.

Let $x\in\mathbf{K}^{n}$ and
$z:=\sum_{s=0}^{n-1}(L_{\bullet_{\tau}}(c_{k}))^{s}x$. Then using (33) we find

$L_{\bullet_{\tau}}z=\sum_{s=1}^{n}(L_{\bullet_{\tau}}(c_{k}))^{s}x=\sum_{s=0}^{n-1}(L_{\bullet_{\tau}}(c_{k}))^{s}x=z,$

therefore $z$ is an eigenvector of $L_{\bullet_{\tau}}(c_{k})$ with eigenvalue
$1$, therefore by Proposition 8.3, $z\in\mathrm{span}(c_{k})$. This yields
$\sum_{s=0}^{n-1}(L_{\bullet_{\tau}}(c_{k}))^{s}x=\mu_{k}c_{k}$ for some
$\mu_{k}\in\mathbf{K}$. Summing up the obtained identities and applying Lemma
8.5, we get

$\displaystyle\sum_{k=1}^{2^{n}-1}\mu_{k}c_{k}$
$\displaystyle=\sum_{k=1}^{2^{n}-1}\sum_{s=0}^{n-1}(L_{\bullet_{\tau}}(c_{k}))^{s}x$
$\displaystyle=\sum_{s=0}^{n-1}\sum_{k=1}^{2^{n}-1}(L_{\bullet_{\tau}}(c_{k}))^{s}x$
$\displaystyle=\sum_{k=1}^{2^{n}-1}x+\sum_{s=1}^{n-1}\sum_{k=1}^{2^{n}-1}(L_{\bullet_{\tau}}(c_{k}))^{s}x$
$\displaystyle=(2^{n}-1)x,$

and since $2^{n}-1\neq 0$ in $\mathbf{K}$, we arrive at the desired
conclusion. ∎

######  Theorem 8.7.

$\mathbb{A}:=(\mathbf{K}^{n},\bullet_{\tau})$ is an axial algebra with the
cyclic fusion law

$\mathbb{A}_{\epsilon^{p}}(c_{k})\ast\mathbb{A}_{\epsilon^{q}}(c_{k})=\mathbb{A}_{\epsilon^{p+q}}(c_{k}),\quad\forall
p,q\in\mathbb{Z}_{n}.$ (38)

###### Proof.

By Corollary 8.6, $\mathbb{A}$ is spanned by the set of nonzero idempotents
$\operatorname{Idm}(\mathbb{A})$ and by Proposition 8.3 all nonzero
idempotents in $(\mathbf{K}^{n},\bullet_{\tau})$ have the same spectrum
$\epsilon,\epsilon^{2},\ldots,\epsilon^{n}$, each eigenvalue $\epsilon^{p}$ is
simple, and the eigen-decomposition (37) holds. Moreover, applying (36) we
obtain for the corresponding eigenvectors

$\displaystyle\eta_{k,p}\bullet_{\tau}\eta_{k,q}$
$\displaystyle=(\epsilon^{p+q}\zeta^{-2^{m-1}k},\epsilon^{2(p+q)}\zeta^{-(2^{m-1}+2^{m-2})k},\ldots)=\epsilon^{p+q}\zeta^{-2^{m-1}k}\eta_{k,p+q},$

which gives
$\mathrm{span}(\eta_{k,p})\bullet_{\tau}\mathrm{span}(\eta_{k,q})=\mathrm{span}(\eta_{k,p+q})$
and thereby implies the fusion law (38). ∎

Now we are ready to formulate our main result for the case of an arbitrary
$\sigma\in S_{n}$. Combining the above results with Proposition 7.3 and
Proposition 2.1 we arrive at the following general conclusion:

######  Theorem 8.8.

Let a permutation $\sigma\in S_{n}$ have the disjoint cycle decomposition
$\sigma=\sigma_{1}\ldots\sigma_{r}$ and a field $\mathbf{K}$ be
$\sigma$-admissible. Then the following properties hold:

  * (a)

There are exactly $2^{n}$ distinct regular idempotents in
$(\mathbf{K}^{n},\bullet_{\sigma})$ naturally stratified in $2^{r}$ classes
$I_{\alpha}$, enumerated by binary codes $\alpha\in\mathbb{F}_{2}^{r}$.

  * (b)

For each $\alpha\in\mathbb{F}_{2}^{r}$, all idempotents in $I_{\alpha}$ have
the same spectrum. More precisely,

$\det(\lambda\mathds{1}-L_{\bullet_{\tau}}(c))=\prod_{i=1}^{r}(\lambda^{|\sigma_{i}|}-\alpha(i)),\qquad\forall
c\in I_{\alpha}.$ (39)

  * (c)

The algebra $(\mathbf{K}^{n},\bullet_{\sigma})$ is generic.

##  9\. Automorphisms

In order to describe the automorphism group for
$(\mathbf{K}^{n},\bullet_{\sigma})$, we recall some definitions. For two
groups $G$, $H$, and an action $f:H\to\operatorname{Aut}(G)$, the
corresponding semidirect product $G\rtimes_{f}H$ is defined by the group
multiplication on $G\times H$ given by
$(g_{1},h_{1})(g_{2},h_{2})=(g_{1}f_{h_{1}}(g_{2}),h_{1}h_{2}).$ In
particular, if $H=\operatorname{Aut}(G)$ with $f=\mathrm{id}$ then one obtains
the classical notion of the holomorph of a group $G$ is the semi-direct
product $G\rtimes_{\mathrm{id}}\operatorname{Aut}(G)$ with the multiplication
given by

$(g_{1},\alpha_{1})\cdot(g_{2},\alpha_{2})=(g_{1}\alpha_{1}(g_{2}),\,\alpha_{1}\alpha_{2})$
(40)

The automorphism group of the additive cyclic group
$\mathbb{Z}_{N}:=(\mathbb{Z}_{N},+)$ is isomorphic to the multiplicative group
$(\mathbb{Z}_{N})^{\times}=(\mathbb{Z}_{N}^{\times},\cdot)$ of integers modulo
$N$ (the group of multiplicative units):

$\operatorname{Aut}(\mathbb{Z}_{N})\cong(\mathbb{Z}_{N})^{\times},$

where we write for short

$\mathbb{Z}_{k}:=(\mathbb{Z}/k\mathbb{Z},+),\qquad(\mathbb{Z}_{N})^{\times}=(\mathbb{Z}_{N}^{\times},\cdot).$

The group $(\mathbb{Z}_{N})^{\times}$ is not cyclic in general, but by the
fundamental theorem of finite abelian groups, it is isomorphic to a direct
product of cyclic groups of prime power orders. For our analysis the relevant
case is when $N=2^{n}-1$, $n\in\mathbb{Z}^{+}$. Then
$(\mathbb{Z}_{N})^{\times}$ is the direct product of the groups corresponding
to each of the (odd) prime power factors $N=p_{1}^{k_{1}}\ldots
p_{s}^{k_{s}}$:

$(\mathbb{Z}_{N})^{\times}=(\mathbb{Z}_{p_{1}^{k_{1}}})^{\times}\times\ldots\times(\mathbb{Z}_{p_{s}^{k_{s}}})^{\times}\cong
C_{p_{1}^{k_{1}}-p_{1}^{k_{1}-1}}\times\ldots
C_{p_{s}^{k_{s}}-p_{s}^{k_{s}-1}},$

where $C_{m}$ denote a cyclic group of order $m$. For example, for $n=6$,
$B=2^{6}-1=63=3^{2}\cdot 7$, hence

$(\mathbb{Z}_{63})^{\times}=C_{6}\times C_{6}.$

A relevant in the present context is the general affine group of
$\mathbb{Z}_{N}$ which is isomorphic to the holomorph of $\mathbb{Z}_{N}$:

$\begin{split}\mathrm{Aff}(\mathbb{Z}_{N})&\cong\mathbb{Z}_{N}\rtimes_{\mathrm{id}}(\mathbb{Z}_{N})^{\times}\\\
&\cong\\{\left(\begin{array}[]{cc}m&k\\\ 0&1\\\
\end{array}\right):\,m\in\mathbb{Z}_{N}^{\times},\,\,k\in\mathbb{Z}_{N}\\}\\\
&\cong\\{\psi_{m,k}(i)=mi+k:\,m\in\mathbb{Z}_{N}^{\times},\,\,k\in\mathbb{Z}_{N}\\},\end{split}$
(41)

where the last line is the group of compositions of affine functions
$\psi_{m,k}:\mathbb{Z}_{N}\to\mathbb{Z}_{N}$,

$\psi_{m,k}\circ\psi_{m^{\prime},k^{\prime}}=\psi_{mm^{\prime},mk^{\prime}+k}.$
(42)

The exponential map

$\delta(i):=2^{i},\quad\delta:(\mathbb{Z}_{n},+)\to(\mathbb{Z}_{2^{n}-1}^{\times},\cdot)\cong\operatorname{Aut}((\mathbb{Z}_{2^{n}-1},+)),$
(43)

is a well-defined injective group homomorphism, hence there holds the
following exact sequence of abelian groups:

$0\longmapsto(\mathbb{Z}_{n},+)\stackrel{{\scriptstyle\delta}}{{\longmapsto}}(\mathbb{Z}_{2^{n}-1}^{\times},\cdot)\stackrel{{\scriptstyle\mu}}{{\longmapsto}}(\mathbb{Z}_{2^{n}-1}^{\times},\cdot)/(\mathbb{Z}_{n},+)\longmapsto
0.$ (44)

Denote

$\Delta_{n}=\operatorname{im}\delta=\\{1,2,\ldots,2^{n-1}\\}\subset\mathbb{Z}_{2^{n}-1}.$

Note that $\Delta_{n}\cong(\mathbb{Z}_{n},+)\cong C_{n}$ is a multiplicative
subgroup of $(\mathbb{Z}_{2^{n}-1})^{\times}$. We shall also need the semi-
direct product

$(\mathbb{Z}_{2^{n}-1},+)\rtimes_{\delta}(\mathbb{Z}_{n},+)\cong\\{\psi_{2^{q},k}(i):\,q,k\in\mathbb{Z}_{n}\\}$
(45)

######  Remark 9.1.

Notice that any algebra automorphism stabilizes the algebra idempotents.
Therefore the automorphism group of an algebra $\mathbb{A}$ is a subgroup of
the group of symmetries of nonzero idempotents of $\mathbb{A}$. As above, it
suffices to consider the case when $\sigma=\tau\in S_{n}$ is a single cycle
element(the right shift permutation). In this case, Proposition 8.2 yields
that the set of nonzero idempotents is a multiplicative quasigroup. Below we
completely characterize its automorphism group.

######  Theorem 9.2.

Let $\tau=(2\,3\,\ldots\,n\,1)\in S_{m}$ be the right cyclic shift. Then the
idempotent quasigroup is

$\operatorname{Aut}(\operatorname{Idm}(\mathbf{K}^{n},\bullet_{\tau}))\cong\mathbb{Z}_{2^{n}-1}\rtimes_{\mathrm{id}}\mathbb{Z}_{2^{n}-1}^{\times}\cong\mathrm{Aff}(\mathbb{Z}_{2^{n}-1}).$
(46)

###### Proof.

Given a pair $m\in\mathbb{Z}_{2^{n}-1}^{\times}$ and
$k\in\mathbb{Z}_{2^{n}-1}$, we define a map $\psi_{m,k}(c_{i}):=c_{mi+k}$,
$i\in\mathbb{Z}_{2^{n}-1}$, of
$\operatorname{Idm}(\mathbf{K}^{n},\bullet_{\tau}))$ to itself. Then for any
pair $i,j\in\mathbb{Z}_{2^{n}-1}$ we obtain using (27)–(28)

$\displaystyle\psi_{m,k}(c_{i}\bullet_{\tau}c_{j})$
$\displaystyle=\psi_{m,k}(c_{{2^{n-1}}(i+j)})=c_{{2^{n-1}}m(i+j)+k}=c_{{2^{n-1}}m(i+j)+{2^{n}}k}$
$\displaystyle=c_{{2^{n-1}}(mi+k+mj+k)}=c_{mi+k}\bullet_{\tau}c_{mj+k}$
$\displaystyle=\psi_{m,k}(c_{i})\bullet_{\tau}\psi_{m,k}(c_{j})$

i.e.
$\psi_{m,k}\in\operatorname{Aut}(\operatorname{Idm}(\mathbf{K}^{n},\bullet_{\tau}))$.

In the converse direction, if
$g\in\operatorname{Aut}(\operatorname{Idm}(\mathbf{K}^{n},\bullet_{\tau}))$
then $g(c_{i})=c_{h(i)}$ for some bijection
$h:\mathbb{Z}_{2^{n}-1}\to\mathbb{Z}_{2^{n}-1}$, hence for any pair
$i,j\in\mathbb{Z}_{2^{n}-1}$

$c_{h({2^{n-1}}(i+j))}=g(c_{{2^{n-1}}(i+j)})=g(c_{i}\bullet_{\tau}c_{j})=g(c_{i})\bullet_{\tau}g(c_{j})=c_{{2^{n-1}}(h(i)+h(j))},$

implying $h({2^{n-1}}(i+j))={2^{n-1}}h(i)+{2^{n-1}}h(j).$ Multiplying this by
$2$ and setting $i=j+2$ yields in view of $2^{n}=1$ in $\mathbb{Z}_{2^{n}-1}$
that

$2h(2^{n}j+2^{n})=2h(j+1)=h(j+2)+h(j),$

hence

$h(j+2)-h(j+1)=h(j+1)-h(j)=\ldots=h(2)-h(1)=:m.$ (47)

Since $h$ is an injection, $m\neq 0$ in $\mathbb{Z}_{2^{n}-1}$, thus
$m\in\mathbb{Z}_{2^{n}-1}^{\times}$. Therefore (47) implies
$h(j)=m\cdot(j-1)+h(1)=mj+k$, where $k:=h(1)-m\in\mathbb{Z}_{2^{n}-1}$. This
yields $g=\psi_{m,k}$ and (41) establishes the desired isomorphisms in (46).

∎

Now we consider the automorphism group of the ambient algebra
$(\mathbf{K}^{n},\bullet_{\tau})$. Recall that by Remark 9.1,
$\operatorname{Aut}(\mathbf{K}^{n},\bullet_{\tau})$ is a subgroup of
$\operatorname{Aut}(\operatorname{Idm}(\mathbf{K}^{n},\bullet_{\tau}))$. A
part of $\operatorname{Aut}(\mathbf{K}^{n},\bullet_{\tau})$ can be identified
by the definitions. More precisely, let $k,q\in\mathbb{Z}_{n}$ and consider
the maps given by

$\displaystyle\alpha_{q}(x_{1},x_{2},\ldots,x_{n})$
$\displaystyle=(x_{1-q},x_{2-q},\ldots,x_{n-q})$
$\displaystyle\beta_{k}(x_{1},x_{2},\ldots,x_{n})$
$\displaystyle=(\zeta^{2^{n-1}k}x_{1},\zeta^{2^{n-2}k}x_{2},\ldots,\zeta^{k}x_{n})$

where $\alpha_{q}$ is the right cyclic shift of the coordinates (understood as
elements of $\mathbb{Z}/n\mathbb{Z}$), for example,
$\alpha_{1}(x)=(x_{n},x_{1},x_{2},\ldots,x_{n-1})$ etc.

######  Lemma 9.3.

In the notation of (41),

$\displaystyle\alpha_{q}$
$\displaystyle=\psi_{2^{q},0}\in\operatorname{Aut}(\mathbf{K}^{n},\bullet_{\tau})$
(48) $\displaystyle\beta_{k}$
$\displaystyle=\psi_{1,k}\in\operatorname{Aut}(\mathbf{K}^{n},\bullet_{\tau})$
(49)

Furthermore,
$\langle\alpha_{1},\beta_{1}\rangle\cong\mathbb{Z}_{2^{n}-1}\rtimes_{\delta}\mathbb{Z}_{n}$,
where $\delta$ is defined in (43).

###### Proof.

By their definitions, both $\alpha_{q}$ and $\beta_{k}$ are linear
isomorphisms of $\mathbf{K}^{n}$. An easy verification implies the relation
$\alpha_{q}(x\bullet_{\tau}y)=\alpha_{q}(x)\bullet_{\tau}\alpha(y)$.
Furthermore,

$\displaystyle\beta_{k}(x\bullet_{\tau}y)$
$\displaystyle=\beta((x_{2}y_{2},\ldots,x_{n}y_{n},x_{1}y_{1}))$
$\displaystyle=(\zeta^{2^{n-1}k}x_{2}y_{2},\ldots,\zeta^{2^{1}k}x_{n}y_{n},\,\zeta^{2^{0}k}x_{1}y_{1}),$

and on the other hand,

$\displaystyle\beta_{k}(x)\bullet_{\tau}\beta_{k}(y)$
$\displaystyle=(\zeta^{2^{n-2}k}x_{2}\cdot\zeta^{2^{n-2}k}y_{2},\ldots,\,\zeta^{2^{n-1}m}x_{1}\cdot\zeta^{2^{n-1}m}y_{1}).$

Comparing the obtained expressions yields
$\beta_{k}(x\bullet_{\tau}y)=\beta_{k}(x)\bullet_{\tau}\beta_{k}(y)$, hence
both $\alpha_{q}$ and $\beta_{k}$ are automorphisms of
$(\mathbf{K}^{n},\bullet_{\tau})$. Applying the definitions to (26) implies
(48) and (49). Furthermore, elements $\alpha_{1},\beta_{1}$ generate a
subgroup in $\mathrm{Aff}(\mathbb{Z}_{2^{n}-1})$ consisting of all elements of
the kind $\psi_{2^{q},k}$, $q,k\in\mathbb{Z}_{n}$, which by virtue of (45)
implies the last claim of the lemma. ∎

Recall that the cyclotomic polynomial $\Phi_{N}(z)$ is the unique irreducible
polynomial with integer coefficients that is a divisor of $z^{N}-1$ and is not
a divisor of $z^{k}-1$ for any $k<N$. Its roots are all $N$th primitive roots
of unity. It is well known that

$\prod_{d|N}\Phi_{d}(z)=z^{N}-1.$ (50)

######  Definition 9.4.

Let $n\geq 2$ be an integer and

$\Lambda_{n}(z):=z+z^{2}+z^{4}+\ldots+z^{2^{n-1}}.$

The number $n$ is said to be regular if the cyclotomic polynomial
$\Phi_{2^{n}-1}(z)$ does not divide $\Lambda_{n}(z^{m})-\Lambda_{n}(z)$ for
all $m\in\mathbb{Z}_{2^{n}-1}^{\times}\setminus\Delta_{n}$.

######  Theorem 9.5.

If $n\geq 2$ is a regular integer then
$\operatorname{Aut}(\mathbf{K}^{n},\bullet_{\tau})\cong\mathbb{Z}_{2^{n}-1}\rtimes_{\mathrm{\delta}}\mathbb{Z}_{n}.$

###### Proof.

An algebra automorphism stabilizes the set of nonzero idempotents, inducing an
automorphism on the idempotent quasigroup. By Corollary 8.6,
$\mathrm{span}(\operatorname{Idm}(\mathbf{K}^{n},\bullet_{\tau}))=\mathbf{K}^{n}$,
hence if some $f\in\operatorname{Aut}(\mathbf{K}^{n},\bullet_{\tau})$
stabilize each nonzero idempotent in
$\operatorname{Idm}(\mathbf{K}^{n},\bullet_{\tau})$ then $f=\mathds{1}$. This
implies that $\operatorname{Aut}(\mathbf{K}^{n},\bullet_{\tau})$ is a subgroup
of $\operatorname{Aut}(\operatorname{Idm}(\mathbf{K}^{n},\bullet_{\tau})),$ in
particular, any algebra automorphism has the form $\psi_{m,k}$, where
$m\in\mathbb{Z}_{2^{n}-1}^{\times}$, $k\in\mathbb{Z}_{2^{n}-1}$.

Therefore we need to identify only those
$\psi_{k,m}\in\mathrm{Aff}(\mathbb{Z}_{2^{n}-1})$ which can be extended to a
linear isomorphism of $\mathbf{K}^{n}$. To this end, we note that since
$m\in\mathbb{Z}_{2^{n}-1}^{\times}$, then $s:=m^{-1}k$ is well-defined and by
Lemma 9.3 $\psi_{1,s}\in\operatorname{Aut}(\mathbf{K}^{n})$, therefore using
(42) we conclude that
$\psi_{m,k}\circ\psi_{1,s}=\psi_{m,0}\in\operatorname{Aut}(\mathbf{K}^{n},\bullet_{\tau})$.
In other words, we can assume without loss of generality that $k=0$.

So let us assume that
$\psi_{m,0}\in\operatorname{Aut}(\mathbf{K}^{n},\bullet_{\tau})$. By Lemma
9.3, it suffices to show that $m\in\Delta_{n}=\\{1,2,\ldots,2^{n-1}\\}$. We
argue by contradiction and assume that
$m\in\mathbb{Z}_{2^{n}-1}^{\times}\setminus\Delta_{n}$. Then by virtue of (26)
we find $c_{2^{n}-1}=(1,1,\ldots,1)$ and also

$H_{l}:=c_{l}+c_{2l}+c_{4l}+\ldots+c_{2^{n-1}l}=\Lambda_{n}(\zeta^{l})c_{2^{n}-1},\qquad\forall
l\in\mathbb{Z}_{2^{n}-1}.$ (51)

Since $\psi_{m,0}(c_{i})=c_{mi}$ for any $i\in\mathbb{Z}_{2^{n}-1}$ (in
particular, $\psi_{m,0}(c_{2^{n}-1})=c_{2^{n}-1}$) we have
$\psi_{m,0}(H_{l})=H_{ml},$ implying by virtue of (51) that

$\Lambda_{n}(\zeta^{l})=\Lambda_{n}(\zeta^{lm}),\quad\forall
l\in\mathbb{Z}_{2^{n}-1}.$

Since the latter holds for any primitive root $\zeta$ of unity of order
$2^{n}-1$, we conclude that by the definition, the cyclotomic polynomial
$\Phi_{2^{n}-1}(z)$ divides $\Lambda_{n}(z^{lm})-\Lambda_{n}(z^{l})$ for any
$l\in\mathbb{Z}_{2^{n}-1}$, in particular, for $l=1$, which implies that $n$
is not a regular integer, a contradiction. ∎

######  Remark 9.6.

Conjecturally, all positive integer numbers are regular, an application of
Galois theory of cyclotomic polynomials would be helpful to establish this
conjecture, but we now are not able to prove this conjecture in the full
generality. There are however several particular cases when the verification
can be easily done, for example, for any Mersenne prime $2^{n}-1$, $n$ is a
regular number. In practice, a verification of that a given $n$ is regular can
be done using the resultant, as it shown in the example below.

######  Example 9.7.

Let $n=2$, then $\mathbb{Z}_{3}^{\times}\setminus\Delta_{2}=\emptyset$, hence
$n=2$ is a regular integer, implying that

$\operatorname{Aut}(\mathbf{K}^{2},\bullet_{\tau})\cong\mathbb{Z}_{3}\rtimes_{\mathrm{\delta}}\mathbb{Z}_{2}\cong
S_{3}.$

The latter is the well-known fact that $S_{3}$ is an internal semi-direct
product of the subgroups $\mathbb{Z}_{3}$ and $\mathbb{Z}_{2}$, where
$\mathbb{Z}_{3}$ is the subgroup generated by one of the two 3-cycles and
$\mathbb{Z}_{2}\cong C_{2}$ is the subgroup generated by any transposition.

######  Example 9.8.

Let $n=3$, then $\Lambda_{3}(z)=z+z^{2}+z^{4}$ and
$\mathbb{Z}_{7}^{\times}\setminus\Delta_{3}=\\{3,5,6\\}$. An easy verification
shows that the resultant

$R(\frac{\Lambda_{3}(z^{q})-\Lambda_{3}(z)}{z(z-1)},\,\Phi_{7}(z))=7^{2}\neq
0,\qquad\forall q\in\\{3,5,6\\},$

hence $\Lambda_{3}(z^{q})-\Lambda_{3}(z)$ does not have common divisors with
$\Phi_{7}(z)$ (note that $z=0,1$ cannot be common zeros). Therefore $n=3$ is a
regular integer. Similarly, for $\Lambda_{4}(z)=z+z^{2}+z^{4}+z^{8}$,
$\mathbb{Z}_{15}^{\times}\setminus\Delta_{3}=\\{7,11,13,14\\}$ and

$R(\frac{\Lambda_{4}(z^{q})-\Lambda_{4}(z)}{z(z-1)},\,\Phi_{15}(z))=3^{4}\cdot
5^{4}\neq 0,\qquad\forall q\in\Delta_{3},$

Finally, we point out the following useful observation. Let $n\geq 2$ be an
integer, and $\Phi_{n}(z)$ be the cyclotomic polynomial of degree $n$.
Consider the quotient polynomial algebra

$\mathbb{T}_{n}:=(\mathbf{K}[z]/\Phi_{2^{n}-1}(z),\bullet)\cong(\mathbf{K}^{\phi(2^{n}-1)},\bullet)$

where $\phi$ is Euler’s totient function. Then
$\operatorname{Aut}(\mathbb{T}_{n})\cong S_{\phi(2^{n}-1)}$, where any
automorphism is a substitution $P(z)\to P(t_{\alpha}(z))$, with $\alpha\in
S_{n}$ and $t_{\alpha}(z)$ is the Lagrange polynomial of degree
$\leq\phi(2^{n}-1)$ uniquely determined the relations
$t_{\alpha}(\zeta^{k})=\zeta^{\alpha(k)}$ for any
$k\in\mathbb{Z}_{2^{n}-1}^{\times}$, where $\zeta$ is a fixed primitive root
of unity of degree $2^{n}-1$. Note that
$h_{m}(z):=z^{m}\in\operatorname{Aut}(\mathbb{T}_{n})$, where
$m\in\mathbb{Z}_{2^{n}-1}^{\times}$ ($\mathbb{Z}_{2^{n}-1}^{\times}$ is an
abelian subgroup of $S_{\phi(2^{n}-1)}$).

Now, let us consider $\Lambda_{n}(z)$ as an element in $\mathbb{T}_{n}$. By
(50), $\Phi_{2^{n}-1}(z)|(z^{2^{n}-1}-1)$, hence

$\Lambda_{n}(z^{2})-\Lambda_{n}(z)=z^{2^{n}}-z\equiv 0\mod\Phi_{2^{n}-1}(z),$

therefore $\Lambda_{n}(z^{2})=\Lambda_{n}(z)$, in other words we conclude that

######  Proposition 9.9.

$\Lambda_{n}$ is a fixed point of the natural action of $\Delta_{n}$ by
substitutions. Moreover, the integer $n$ is regular if and only if the
stabilizer subgroup of $\Lambda_{n}$ in $\mathbb{Z}_{2^{n}-1}^{\times}$ is
exactly $\Delta_{n}$.

##  10\. Three examples for $n=3$

The main goal of this section is to illustrate our results for
$(\mathbf{K}^{n},\bullet_{\sigma})$ in the particular case $n=3$. We shall
assume that $\mathbf{K}$ is splitting field of polynomial $P(z)=z^{3}-1$ and
$\epsilon$ will denote a primitive root of unity of degree 3 (in section 10.2
we additionally assume that also a primitive root of unity of degree 7
exists). By Theorem 7.1, there are exactly three distinct (isomorphy classes
of) inner isotopes coded by the conjugacy classes of $S_{3}$, which are in a
one-to-one correspondence with integer partitions of $3$, i.e.

$\begin{split}3&=1+1+1\\\ &=2+1\\\ &=3.\end{split}$ (52)

Each of the three-dimensional isotope algebras will be considered below.

###  10.1. The case $"1+1+1"$: a unital commutative associative algebra

In this case we have the trivial three cycle partition $1+1+1$ which uniquely
determines the unity in $S_{3}$: $e=(1)(2)(3)$ (in the cyclic notation), thus

$(\mathbf{K}^{3},\bullet_{e})\cong(\mathbf{K}^{3},\bullet)$

i.e. the corresponding inner isotope is the associative (direct product)
algebra $(\mathbf{K}^{3},\bullet)$ itself. The multiplication structure is a
uniquely determined by the multiplication table in the standard basis
$\\{e_{1},e_{2},e_{3}\\}$ (2):

$e_{i}\bullet e_{j}=e_{i+j(\mathrm{mod}\,3)}.$

The automorphism group is given by Theorem 7.1

$\operatorname{Aut}(\mathbf{K}^{3},\bullet_{e})\cong S_{3}.$

The algebra $(\mathbf{K}^{3},\bullet_{e})$ is generic and any algebra
idempotent can be written as $c_{k}=(\alpha_{1},\alpha_{2},\alpha_{3})$, where
$(\alpha_{1},\alpha_{2},\alpha_{3})\in\mathbb{F}_{2}^{3}$ is the binary
decomposition of $k$, $0\leq k\leq 7$. For example, the binary codes
$1=001_{2}$, $2=010_{2}$ and $4=100_{2}$ correspond to the three standard
basis idempotents $e_{3},e_{2},e_{1}$. The idempotent
$c_{7}=e_{1}+e_{2}+e_{3}$ is the algebra unity. This yields the multiplication
table (Table 1 below) and the characteristic polynomials $\chi_{i}$ of
$L_{\bullet_{e}}(c_{i})$ are given by (cf. with (32))

$\begin{split}\chi_{0}&=\lambda^{3},\qquad\chi_{1}=\chi_{2}=\chi_{4}=(\lambda-1)\lambda^{2},\\\
\chi_{3}=\chi_{4}=\chi_{6}&=(\lambda-1)^{2}\lambda,\qquad\chi_{7}=(\lambda-1)^{3}.\end{split}$
(53)

$\bullet_{\tau}$ | 1 | 2 | 3 | 4 | 5 | 6 | 7
---|---|---|---|---|---|---|---
1 | 1 | 0 | 1 | 0 | 1 | 0 | 1
2 | 0 | 2 | 2 | 0 | 0 | 2 | 2
3 | 1 | 2 | 3 | 0 | 1 | 2 | 3
4 | 0 | 0 | 0 | 4 | 4 | 4 | 4
5 | 1 | 0 | 1 | 4 | 5 | 4 | 5
6 | 0 | 2 | 2 | 4 | 4 | 6 | 6
7 | 1 | 2 | 3 | 4 | 5 | 6 | 7
Table 1. The multiplication table of idempotents $i\sim c_{i}$ in
$(\mathbf{K}^{3},\bullet_{e})$ with $e=(1)(2)(3)\in S_{3}$

Note that $(\mathbf{K}^{3},\bullet_{e})$, as an associative algebra is axial
in the sense of Definition 1.3 above, it follows from the classical results
due to Benjamin Peirce. Indeed,
$(\mathbf{K}^{3},\bullet_{e})=\mathrm{span}(\\{e_{1},e_{2},e_{4}\\})$, these
idempotents are primitive and satisfye the same fusion law:

$\begin{array}[]{r||rr}\bullet_{e}&\mathbb{A}_{1}&\mathbb{A}_{0}\\\
\hline\cr\hline\cr\mathbb{A}_{1}&\mathbb{A}_{1}&\mathbb{A}_{0}\\\
\mathbb{A}_{0}&0&\mathbb{A}_{0}\\\ \end{array}$ (54)

However, in contrast to the single cycle case (see section 10.2 below), the
set of nonzero idempotents is not a multiplicative magma, see table 1.

###  10.2. The single cycle case $"3"$: a commutative isospectral medial
algebra

The one-cycle partition corresponds to the conjugacy class of
$\tau=(2\,3\,1)\in S_{3}$, i.e. shifts. Such algebras have been introduced and
studied first for $n=3$ in [16] in the context of isospectral algebras and
later for any $n\geq 2$ in [17] in the polynomial setting. It is natural to
assume that the ground field $\mathbf{K}$ additionally contains a primitive
root of unity of order $7=2^{3}-1$, denote it by $\zeta$.

The standard basis elements are no longer idempotents, for example
$e_{1}\bullet_{\tau}e_{1}=e_{2}$. By Proposition 8.1, the algebra
$(\mathbf{K}^{3},\bullet_{\tau})$ is generic and contains 7 distinct nonzero
idempotents, which can be explicitly written by

$\operatorname{Idm}(\mathbf{K}^{3},\bullet_{\tau})=\\{c_{k}=(\zeta^{4k},\zeta^{2k},\zeta^{k}):\quad
k\in\mathbb{Z}/7\mathbb{Z}\\}.$ (55)

The multiplication rule (27) between idempotents takes the form

$c_{i}\bullet_{\tau}c_{j}=c_{4(i+j)}=c_{i\circledast j},$ (56)

where $\circledast$ on $\mathbb{Z}/7\mathbb{Z}$ is defined by

$i\circledast j\equiv 4(i+j)\mod 7.$ (57)

The characteristic polynomials are given by

$\det(\lambda\mathds{1}-L_{\bullet_{\tau}}(c_{k}))=\lambda^{3}-1.$ (58)

Therefore $(\mathbf{K}^{3},\bullet_{\tau})$ is isospectral and furthermore it
is an axial algebra with the following fusion law:

$\begin{array}[]{c||ccc}\bullet_{\omega}&\mathbb{A}_{1}&\mathbb{A}_{\epsilon}&\mathbb{A}_{\epsilon^{2}}\\\
\hline\cr\hline\cr\mathbb{A}_{1}&\mathbb{A}_{1}&\mathbb{A}_{\epsilon}&\mathbb{A}_{\epsilon^{2}}\\\
\mathbb{A}_{\epsilon}&\mathbb{A}_{\epsilon}&\mathbb{A}_{\epsilon^{2}}&\mathbb{A}_{1}\\\
\mathbb{A}_{\epsilon^{2}}&\mathbb{A}_{\epsilon^{2}}&\mathbb{A}_{1}&\mathbb{A}_{\epsilon}\end{array}$
(59)

Combining Theorem 9.2 and Theorem 9.5 with Example 9.8, we obtain

######  Theorem 10.1.

The idempotent quasigroup and the algebra automorphism groups are
respectively:

$\displaystyle\operatorname{Aut}(\operatorname{Idm}(\mathbf{K}^{3},\bullet_{\tau}))$
$\displaystyle\cong\mathbb{Z}_{7}\rtimes_{\mathrm{id}}\mathbb{Z}_{7}^{\times}$
(60) $\displaystyle\operatorname{Aut}(\mathbf{K}^{3},\bullet_{\tau})$
$\displaystyle\cong\mathbb{Z}_{7}\rtimes_{\mathrm{\delta}}\mathbb{Z}_{3},$
(61)

where $\mathbb{Z}_{7}\rtimes_{\mathrm{\delta}}\mathbb{Z}_{3}$ is the smallest
non-abelian group of odd order.

Furthermore, $(\mathbf{K}^{3},\bullet_{\tau})$ has many other remarkable
properties (see [17] for a more detailed discussion), for example it satisfies
the algebra identity

$(x\bullet_{\tau}(x\bullet_{\tau}(x\bullet_{\tau}y)))=\Delta(x)y,\qquad\forall
x,y\in(\mathbf{K}^{3},\bullet_{\tau}),$

where $\Delta$ is a multiplicative homomorphism of degree $3$ given explicitly
by a circulant:

$\Delta(a_{0}e_{0}+a_{1}e_{1}+a_{2}e_{2})=\left|\begin{array}[]{ccc}a_{0}&a_{2}&a_{1}\\\
a_{1}&a_{0}&a_{2}\\\ a_{2}&a_{1}&a_{0}\\\
\end{array}\right|:(\mathbf{K}^{3},\bullet_{\tau})\to(\mathbf{K},\bullet).$

###  10.3. The case $[2+1]$

Finally, we consider the two-cycle partition $[2+1]$ corresponding to the
conjugacy class of transpositions. Without loss of generality we can assume
that $\omega=(2\,1)(3)\in S_{3}$. As in section 10.1, the algebra
$(\mathbf{K}^{3},\bullet_{\omega})$ is decomposable, more precisely:

$(\mathbf{K}^{3},\bullet_{\omega})\cong(\mathbf{K}^{2},\bullet_{\tau})\times(\mathbf{K},\bullet_{e}).$
(62)

The second factor is trivial and the first factor is the two-dimensional
Harada algebra [11], i.e. a uniquely determined up to isomorphism two-
dimensional commutative algebra generated by two distinct idempotents $c_{1}$
and $c_{2}$ subject to the condition $c_{1}\bullet c_{2}=-c_{1}-c_{2}$ (we
don’t need this characterization later and leave the details to an interested
reader).

Combining (62) with Example 9.7, we get

######  Proposition 10.2.

There holds $\operatorname{Aut}(\mathbf{K}^{3},\bullet_{\omega})\cong S_{3}.$

##  11\. Final remarks and questions

There are at least two natural questions that remained unanswered in this
article: 1) Determine the automorphism group of the obtained inner isotopes of
commutative associative algebras $(\mathbf{K}^{n},\bullet)$ for the general
dimensions, and 2) Which of the constructed above inner isotopes are axial
algebras. Note that in Section 10 we completely discussed these two questions
in the case $n=3$ (the case $n=2$ is trivial). A further analysis reveals that
by the same methods are applicable to $n=4$; also some partial results were
mentioned in [17].

Another interesting natural question is how to apply the present methods to
general nonassociative commutative algebras or at least to inner isotopes of
the 2nd order:

$(\mathbb{A},\star)\rightsquigarrow(\mathbb{A},\star_{h})\rightsquigarrow(\mathbb{A},{\star_{h}}_{g})\rightsquigarrow\ldots$

where $h\in\operatorname{Aut}(\mathbb{A},\star)$,
$g\in\operatorname{Aut}(\mathbb{A},\star_{h})$ etc?

Finally, we mention that some methods and ideas of the present paper (inner
isotopies) can be useful in the case $\mathbf{K}=\mathbb{C}$ in the study, for
example, of algebras of holomorphic functions in the unit disk like Bergman
and Bloch spaces of holomorphic functions defined on the open unit disc in the
complex plane [12].

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