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\title{A Quasi-Affine Representation}
\author{Keith A. Kearnes}
\date{}

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\begin{document}
\maketitle

\abstract{We describe a quasi-affine representation 
for algebras having a binary central term.}

\section{Introduction}

An algebra \m a is said to be {\bf affine} if there
is a module \m m with the same universe
and the same polynomial operations as \m a.
\m a is said to be {\bf quasi-affine} 
if it is a subreduct (= a subalgebra of a reduct) 
of an affine algebra.
A homomorphism from \m a into a reduct of an
affine algebra will be called a 
{\bf quasi-affine representation} or just
a {\bf representation}. A representation 
is {\bf faithful} if it is 1-1. 

Not every algebra has a faithful 
quasi-affine representation, of course.
A non-trivial
semilattice is an example of an algebra
with no non-constant quasi-affine
representation.
To see this one can use a
direct calculation to show that it is impossible
for a semilattice operation
to be represented by a module polynomial.
But let us get to the root of the problem:
quasi-affine algebras satisfy the term condition
and semilattices do not.
(\m a is said to satisfy the {\bf term condition}
if whenever $s(x,\bar{y})$ is an $(n+1)$-ary term in the language
of \m a, $a, b\in A$
and $\bar{u}, \bar{v}\in A^n$, then the implication
$$s^{\sm a}(a,\bar{u}) = s^{\sm a}(a,\bar{v}) 
\rightarrow s^{\sm a}(b,\bar{u}) = s^{\sm a}(b,\bar{v})$$
holds. The groups which satisfy the term condition are precisely
the abelian groups so it is common to call any algebra
which satisfies the term condition an {\bf abelian algebra}
as is done in \cite{FM}, for example.)
Affine algebras satisfy the term condition and the
term condition is inherited by subreducts. 
Hence the implication
\begin{center}
{\rm quasi-affine $\Rightarrow$ abelian}
\end{center}
holds. To obtain a 
faithful quasi-affine representation of an algebra
\m a it is necessary for \m a to be abelian.
In general, this is not a sufficient condition.
It is shown in \cite{quack} that 
there is a list of implication schemes
generalizing and including the term condition 
which, when taken together, are 
necessary and sufficient conditions for 
an algebra to have a faithful quasi-affine representation.
It is further shown in \cite{quack} that 
no finite sublist of these schemes suffices (in general).
This paper is one among a few papers
that describe commonly occurring situations
where abelian algebras are quasi-affine. 
(Others are \cite{ch}, \cite{k1}, \cite{k2}
and Chapter 13 of \cite{HM}.)

If $f$ is an $m$-ary operation on $U$ and
$g$ is an $n$-ary operation on $U$, then
$f$ and $g$ are said to {\bf commute} if for all
$m\times n$ arrays $[u_i^j]$ of elements
of $U$:
$$\left[   \begin{array}{cccc}
         u_1^1 & u_1^2 & \cdots & u_1^n\\
         u_2^1 & u_2^2  &       & \\
        \vdots &        & \ddots & \\
         u_m^1 &       &  & u_m^n \end{array} \right],$$
it is the case that $f(\overline{g(\bar{u}_i)})$ =
$g(\overline{f(\bar{u}^j)})$ holds. If \m a is an algebra,
then we define a {\bf central} term operation for \m a 
to be a term 
operation which commutes with all the fundamental operations of 
\m a. It is a theorem of H. -P. Gumm that an algebra is affine
if and only if it has a central Mal'cev operation. A proof of this
can be found in \cite{FM}. In this paper we show that
central binary operations afford quasi-affine representations
of abelian algebras.
Our main result is the following.

\begin{th}
\label{MAIN1}
Let \m a be an abelian algebra with a
central term $t(x,y) = xy$. Then \m a has
a quasi-affine representation $\Phi$ 
with the property that $\Phi(a) = \Phi(b)$
if and only if for some $n$ it is the case that
$$(\cdots((aa)a)\cdots)a = (\cdots((ab)a)\cdots)a
\hspace{.2in}\mbox{\rm and} \hspace{.2in}
a(\cdots(a(aa))\cdots) = a(\cdots(a(ba))\cdots)$$
where there
are $n+1$ occurrences of $a$ on
the left side of each equation and $n$ occurrences of $a$
on the right.
\end{th}

\noindent
We list a few of the consequences
of this theorem.
First, we need some definitions.
The left cancellation law for a binary operation
$t(x,y) = xy$ is the law:
$$au = av \rightarrow u = v$$
and the right cancellation law is defined dually.
An operation satisfying the
left, right or both of the 
cancellation laws will be called 
{\bf left cancellative}, {\bf right cancellative} or just
{\bf cancellative}, respectively.

\begin{cor}
\label{COR1}
If \m a is an abelian algebra with a cancellative
binary central term, then \m a is quasi-affine.
\end{cor}

{\bf Proof:} Using right cancellation one may reduce
$(\cdots((aa)a)\cdots)a$ = $(\cdots((ab)a)\cdots)a$
to $aa = ab$. Using left cancellation one may reduce
$aa = ab$ to $a = b$. Hence the representation
of Theorem \ref{MAIN1} is faithful when
the central term is cancellative. $\Box$

\begin{cor}
\label{COR2}
If \m a is an algebra which has a homomorphism
$t:\m a^2\rightarrow \m a$ where
\begin{enumerate}
\item[$(i)$] $t$ commutes with itself on \m a and
\item[$(ii)$] {\rm ker} $t$ complements the coordinate
projection kernels in $\Cn a^2$,
\end{enumerate}
then \m a is quasi-affine.
\end{cor}

{\bf Proof:}
Expand \m a to $\m a'$ by adding $t$ as a new operation.
If we show that $\m a'$ is quasi-affine,
then \m a, as a reduct of $\m a'$, will
also be quasi-affine.

The hypotheses guarantee that $t$ is a 
cancellative central binary operation.
This result is a therefore a consequence of
Corollary \ref{COR1} if we can show that 
$\m a'$ is abelian.

To see that $\m a'$ is abelian
assume otherwise that 
$f(x,\bar{y})$ is an $(n+1)$-ary term, $a, b\in A$
and $\bar{u}, \bar{v}\in A^n$ and
$f^{\sm a}(a,\bar{u})$ = $f^{\sm a}(a,\bar{v})$
while $f^{\sm a}(b,\bar{u})\neq f^{\sm a}(b,\bar{v})$.
Set $\theta = \ker t$ and choose $r_0,\ldots,r_n\in A^2$ such that 
$r_0$ = $(a,a)$, $r_n$ = $(b,b)$,
$(r_{2i},r_{2i+1})\in\theta$ and 
$(r_{2i+1},r_{2i+2})\in\eta_1$. This is possible since
$\theta\vee\eta_1$ = 1 in $\Cn a^2$. Now in $\m a^2$ we have
$$f(r_0,\overline{(u_i,u_i)}) = f(r_0,\overline{(u_i,v_i)}), \hspace{.2in}
\mbox{\rm but}\hspace{.2in} 
f(r_n,\overline{(u_i,u_i)}) \neq f(r_n,\overline{(u_i,v_i)}).$$
Hence there is some $j$ such that 
$$f(r_j,\overline{(u_i,u_i)}) = f(r_j,\overline{(u_i,v_i)}) \hspace{.2in}
\mbox{\rm and}\hspace{.2in} 
f(r_{j+1},\overline{(u_i,u_i)}) \neq f(r_{j+1},\overline{(u_i,v_i)}).$$
If $j$ is even, then $(r_j,r_{j+1})\in\theta$, so 
in this case 
$$c = f(r_{j+1},\overline{(u_i,u_i)}) \;\theta\;
f(r_j,\overline{(u_i,u_i)}) = f(r_j,\overline{(u_i,v_i)})\;\theta\;
f(r_{j+1},\overline{(u_i,v_i)}) = d.$$
Also, we clearly have 
$$c = f(r_{j+1},\overline{(u_i,u_i)}) \;\eta_0\;
f(r_{j+1},\overline{(u_i,v_i)}) = d.$$
Hence $(c,d)\in\theta\wedge\eta_0$ = 0. This is impossible
since $c\neq d$. The case where $j$ is odd
is handled the same way. One argues that $(c,d)\in\eta_1$
and $(c,d)\in \eta_0$, so $(c,d)\in\eta_0\wedge\eta_1$ = 0;
a contradiction.
This finishes the proof. $\Box$

\bigskip

The hypotheses of Corollary \ref{COR2}
are not at all artificial. If \m a is a reduct
of a module \m m, then there will be homomorphisms
$t:\m a^2\rightarrow \m a$ satisfying 
hypotheses $(i)\;\&\;(ii)$ of Corollary \ref{COR2}.
For example, one can take $t(x,y)$ = $x+y$ or 
$t(x,y)$ = $x-y$ where + and $-$ are the addition and
subtraction of \m m (they need not be operations of \m a).

If \m a is an algebra 
defined with a single 
binary fundamental operation, then
we call \m a a {\bf binar}. 
An $m$-ary operation on a set $U$ is 
called {\bf idempotent} if 
$f(u,\ldots,u) = u$ holds for all $u\in U$.
Thus a binar is idempotent if it satisfies
the equation $xx = x$. 

\begin{cor}
\label{COR3}
If \m a is an idempotent abelian binar,
then \m a has a 
strongly abelian 
congruence $\theta$ such that
$\m a/\theta$ is quasi-affine.
\end{cor}

Definition \ref{str.abel} explains
what the phrase ``strongly abelian'' means.
In Corollary \ref{COR3} the statement that 
$\theta$ is strongly abelian implies that 
the subalgebra structure on each
$\theta$-class is that of a rectangular band,
but the strongly
abelian property says more than this.
We will not give the proof of Corollary
\ref{COR3} now, since we have deferred the 
explanation of what it means to be strongly abelian.
(For economy, we later combine the parts of
Corollary \ref{COR3} and Theorem
\ref{TH3} which refer to the strongly abelian congruence
$\theta$ into Theorem \ref{strong}.)

If \m a is an idempotent, abelian binar, 
it follows that
$$(\underline{y}y)(z\underline{\underline{z}}) = yz = 
(\underline{y}z)(y\underline{\underline{z}})$$
holds. Applying the term condition to
$\underline{y}$ and also to $\underline{\underline{z}}$
we deduce 
$$(xy)(zw) = (xz)(yw)$$
which is the statement that the term $xy$ commutes with itself.
(Change $\underline{y}$ to $x$ with one application
of the term condition and with a second application
change $\underline{\underline{z}}$ to $w$.)
Thus, any idempotent binar for which a faithful
quasi-affine 
representation exists satisfies $(xy)(zw) = (xz)(yw)$.
This shows that the 
fundamental operation of an idempotent abelian
binar is central.

\begin{th}
\label{TH3}
If \m a is a right cancellative binar which satisfies 
\begin{enumerate}
\item[$(i)$] $xx = x$ (Idempotent Law),
\item[$(ii)$] $(xy)(zu)$ = $(xz)(yu)$ (Entropic Law),
\end{enumerate}
then \m a has a strongly abelian congruence $\theta$
such that $\m a/\theta$ is quasi-affine and each 
$\theta$-class is the universe of a left-zero subsemigroup
of \m a.
\end{th}

The important point to notice in Theorem \ref{TH3}
is that it applies to non-abelian algebras.
(In the case that \m a is abelian we can drop the
assumption that the entropic law holds, since
this is a consequence of the idempotent law.
In this situation Theorem \ref{TH3} is  
a consequence of Theorem \ref{MAIN1}.)
The representation in Theorem \ref{TH3} is a slight
modification of representation in Theorem \ref{MAIN1}.
To see that the hypotheses of Theorem \ref{TH3}
can be satisfied by non-abelian algebras, we
describe a non-abelian, right cancellative 
binar satisfying
Theorem \ref{TH3} $(i) \& (ii)$
(the smallest example, in fact).
Let \m a be the
binar with operation table
$$
\begin{array}{|c||c|c|c|}
\hline
  &0&1&2\\
\hline
\hline
0&0&0&1\\
\hline
1&1&1&0\\
\hline
2&2&2&2\\
\hline
\end{array}
$$
\m a is a non-abelian binar ($\underline{2}0$ = 
$\underline{2}2\rightarrow {0}0$ = 
${0}2$ is false)
which is right cancellative
and satisfies the idempotent and
entropic laws. It has
no faithful quasi-affine representation,
since it is non-abelian, but it has 
only one non-zero congruence whose classes are left-zero
semigroups: the congruence $\cg(0,1)$. It follows from 
Theorem \ref{TH3} that $\theta$ = $\cg(0,1)$ is
strongly abelian and that $\m a/\theta$ is quasi-affine.

Since Theorem \ref{TH3} applies to non-abelian algebras,
it is clear that the cancellativity hypothesis is not 
extraneous. Any semilattice is an idempotent, entropic
binar, but only the 1-element semilattice satisfies the
conclusion of Theorem \ref{TH3}.

\medskip

One unfortunate feature of Theorem \ref{MAIN1}
is that the quasi-affine representation obtained
may not be faithful even when a faithful
representation exists.
In Corollary \ref{COR3}, for example, 
we do not always obtain a faithful representation;
and yet, we are not aware of a single
idempotent abelian algebra which is not
quasi-affine. (\cite{k1}
contains a proof that every {\it simple}
idempotent abelian algebra {\it is} quasi-affine.)
It is easily seen that a rectangular band is quasi-affine, 
but our representation theorem discovers
only the constant representation for any rectangular
band. It seems that this is a difficulty that cannot be 
avoided by any general affinization method. 
Rectangular bands (for example) are quasi-affine,
but they do not have 
enough structure to recover even a small part
of any representing affine algebra.
One way to improve the fidelity of the representation 
we describe in this paper is to choose a collection
of binary central terms, $\{t_i(x,y)\;|\;i\in I\}$,
and obtain a representation $\phi_i$ for each one.
Then the product of these representations leads 
to a single representation whose kernel is
$\bigcap_{i\in I} \theta_i$
where $\theta_i$ is the kernel of $\phi_i$.

\medskip

The theorems we prove belong to the
classical genre of embeddability theorems.
E.g., Ore domains can be embedded into division
rings, cancellative semigroups
satisfying a non-trivial equation
can be embedded into groups,
cancellative binars can be embedded
into quasigroups. What motivates
these results is the desire to
extend a given algebra to a ``nicer sort''
of algebra; one in which calculation
is easier. For example, problems in number theory
are ostensibly problems about the rational 
integers. But solutions of these problems often
require the fact that the rational integers
can be extended to number rings where certain
polynomial equations have solutions. Similarly,
some problems concerning semigroups or rings 
are solved by embedding the algebra
into a group or division ring respectively
and then using the inverse
operation in calculations.
(This is not possible for every semigroup
or ring, of course.)
What motivates us is the desire to 
express the term and polynomial operations 
of an abelian algebra 
as affine combinations of variables
to facilitate calculation.
For an excellent example of the usefulness
of this, the reader should peruse \cite{ch}.
In this fundamental paper, abelian algebras
in congruence modular varieties are shown to be affine 
in two steps. First, they are shown to be 
quasi-affine. Then, expressing Day terms
as linear combinations of variables, it is shown
that a central Mal'cev operation can be 
constructed by composition from these terms.

\section{The Representation}

Our task is to construct a quasi-affine
representation for \m a from the properties
of a fixed binary central term.
We shall do this as follows.
We shall show how to
\begin{enumerate}
\item[$(i)$] Construct an abelian group \m h.
\item[$(ii)$] Define a collection $P$ of endomorphisms of \m h.
\item[$(iii)$] Define a function $\phi:A\rightarrow H$.
\item[$(iv)$] For each fundamental operation $f$ of \m a, 
define an affine function 
$(\Sigma_{i=1}^n \hat{f}_i(x_i)) + h$
on \m h where each $\hat{f}_i\in P$ and $h\in H$.
\item[$(v)$] Show that 
$$\phi(f^{\sm a}(x_1,\ldots,x_n)) = (\Sigma_{i=1}^n \hat{f}_i(\phi(x_i))) + h$$
\end{enumerate}
Thus if we let $\widehat{\m h}$ denote the algebra
with universe $H$ and operations
$(\Sigma_{i=1}^n \hat{f}_i(\phi(x_i))) + h$, one such 
operation for each fundamental operation $f$ of \m a,
then $\widehat{\m h}$ is a reduct of the affine algebra
obtained from adjoining to \m h all constant operations
and all members of $P$ as
new unary operations. By $(v)$, the function $\phi$
is a homomorphism of \m a into $\widehat{\m h}$.
This will complete the representation. After this, all that
remains is to analyze the kernel of $\phi$
in order to establish Theorem \ref{MAIN1}.
Throughout this section we will assume that \m a is
an algebra which has a binary central term $t(x,y)$.
We will not assume that \m a is abelian without stating
this assumption.

Before getting into the proof we prove a
necessary lemma 
concerning binary term operations that commute
with themselves.

\begin{lm}
\label{calc}
Let $xy$ be a term operation of \m a
which commutes with itself.
The following equation is true.
$$\{[(xb)(da)][(ac)(aa)]\}\{[(cb)(cc)][(aa)(aa)]\} = \{[(xd)(bb)][(cc)(cc)]\}\{[(aa)(aa)][(aa)(aa)]\}.$$
If \m a is abelian and $f(x_1,\ldots,x_n)$ is an $n$-ary term operation
of \m a, then the following equation is true.
$$
\begin{array}{ccc}
\lefteqn{[(xf(a_1,\ldots,a_k,a_{k+1},u,\ldots,u))(f(u,\ldots,u)f(u,\ldots,u))] = }\\
\hspace{.2in}&\hspace{.2in}&
[(xf(a_1,\ldots,a_k,u,\ldots,u))(f(u,\ldots,u,a_{k+1},u,\ldots,u)f(u,\ldots,u))].
\end{array}
$$
\end{lm}

{\bf Proof:}
In the following derivation
we shall write $\underline{(p\underline{q})(\underline{r}s)}$
to indicate that we plan to use the entropic law
in the next step to infer from $(pq)(rs)$
that $(pr)(qs)$ holds.
If $E$ is the expression 
$$\{[(xb)(da)][(ac)(aa)]\}\{[(cb)(cc)][(aa)(aa)]\}$$
and $F$ is the expression 
$$\{[(xd)(bb)][(cc)(cc)]\}\{[(aa)(aa)][(aa)(aa)]\},$$
then we calculate that
$$
\begin{array}{rl}
E&=\{[(xb)(da)][(ac)(aa)]\}\{[(cb)(cc)][(aa)(aa)]\}\\
&=\{[\underline{(x\underline{b})(\underline{d}a)}][\underline{(a\underline{c})(\underline{a}a)}]\}\{[\underline{(c\underline{b})(\underline{c}c)}][(aa)(aa)]\}\\
&=\{[(xd)(ba)][(aa)(ca)]\}\{[(cc)(bc)][(aa)(aa)]\} \\
&=\{\underline{[(xd)\underline{(ba)}][\underline{(aa)}(ca)]}\}\{[(cc)(bc)][(aa)(aa)]\} \\
&=\{[(xd)(aa)][(ba)(ca)]\}\{[(cc)(bc)][(aa)(aa)]\} \\
&=\{[(xd)(aa)][\underline{(b\underline{a})(\underline{c}a)}]\}\{[(cc)(bc)][(aa)(aa)]\} \\
&=\{[(xd)(aa)][(bc)(aa)]\}\{[(cc)(bc)][(aa)(aa)]\} \\
&=\{\underline{[(xd)\underline{(aa)}][\underline{(bc)}(aa)]}\}\{[(cc)(bc)][(aa)(aa)]\} \\
&=\{[(xd)(bc)][(aa)(aa)]\}\{[(cc)(bc)][(aa)(aa)]\} \\
&=\underline{\{[(xd)(bc)]\underline{[(aa)(aa)]}\}\{\underline{[(cc)(bc)]}[(aa)(aa)]\}} \\
&=\{[(xd)(bc)][(cc)(bc)]\}\{[(aa)(aa)][(aa)(aa)]\} \\
&=\{\underline{[(xd)\underline{(bc)}][\underline{(cc)}(bc)]}\}\{[(aa)(aa)][(aa)(aa)]\} \\
&=\{[(xd)(cc)][(bc)(bc)]\}\{[(aa)(aa)][(aa)(aa)]\} \\
&=\{[(xd)(cc)][\underline{(b\underline{c})(\underline{b}c)}]\}\{[(aa)(aa)][(aa)(aa)]\} \\
&=\{[(xd)(cc)][(bb)(cc)]\}\{[(aa)(aa)][(aa)(aa)]\} \\
&=\{\underline{[(xd)\underline{(cc)}][\underline{(bb)}(cc)]}\}\{[(aa)(aa)][(aa)(aa)]\} \\
&=\{[(xd)(bb)][(cc)(cc)]\}\{[(aa)(aa)][(aa)(aa)]\}\\
&=F.
\end{array}
$$

For the second equation of the lemma, note that 
$$
\begin{array}{ccc}
\lefteqn{
[(xf({u},\ldots,{u},a_{k+1},u,\ldots,u))(f(u,\ldots,u,u,u,\ldots,u)f(u,\ldots,u))] 
= }\\
\hspace{.2in}&\hspace{.2in}&
[(xf({u},\ldots,u,u,u,\ldots,u))(f(u,\ldots,u,a_{k+1},u,\ldots,u)f(u,\ldots,u))]
\end{array}
$$
is a consequence of the fact that $xy$ commutes with itself.
Now we use the term condition to change $\underline{u}$ to $a_i$
in the $i^{\rm th}$ position of the left-most $f(-)$. We change
$$
\begin{array}{ccc}
\lefteqn{
[(xf(\underline{u},\ldots,\underline{u},a_{k+1},u,\ldots,u))
(f(u,\ldots,u,u,u,\ldots,u)f(u,\ldots,u))]
= }\\
\hspace{.2in}&\hspace{.2in}&
[(xf(\underline{u},\ldots,\underline{u},u,\ldots,u))(f(u,\ldots,u,a_{k+1},u,\ldots,u)f(u,\ldots,u))]
\end{array}$$
to 
$$
\begin{array}{ccc}
\lefteqn{
[(xf(a_1,\ldots,a_k,a_{k+1},u,\ldots,u))(f(u,\ldots,u)f(u,\ldots,u))]
= }\\
\hspace{.2in}&\hspace{.2in}&
[(xf(a_1,\ldots,a_{k},u,u,\ldots,u))(f(u,\ldots,u, a_{k+1}, u\ldots,u)f(u,\ldots,u))]
\end{array}
$$
which was the equality to be established.
This finishes the proof.
$\Box$

\bigskip

If \m a is an algebra with a binary
central term $t(x,y)$ = $xy$,
then we define a semigroup \m s
associated to \m a as follows.
For each $a\in A$ let 
$\rho_a: A\rightarrow A$ be the function
$x\mapsto xa$. We shall refer to $\rho_a$ as
``right translation by $a$''.
Let \m s be the semigroup of
self-maps of $A$ 
generated by $R$ = $\{\rho_a|a\in A\}$.
We compose right translations on the right,
so $\rho_a\rho_b$ denotes the polynomial 
$(xa)b$.

If \m t is a semigroup and $a, b\in T$, then we define
$a < b$ if $a$ is a proper left divisor of $b$.
This means that $a < b$ if and only if  
$b = ac$ for some $c\in T$. 
We call \m t a {\bf directed} semigroup
if for any $a, b\in T$ there is a $c\in T$ 
such that $a < c$ and $b < c$.
($a<a$ is allowed if $a = ac$ for some $c\in T$.)

\begin{lm}
\label{directed}
\m s is directed.
\end{lm}

{\bf Proof:}
First let's examine multiplication
in \m s. The fact that $xy$ commutes with 
all fundamental operation implies that it
commutes with itself. This may be written as
$$ (xu)(vw) = (xv)(uw)$$
or more suggestively as
$$\rho_u\rho_{vw} = \rho_v\rho_{uw}.$$
When $v = w$ this may be written as
$$\rho_u\rho_{vv} = \rho_v\rho_{uv}.$$
Hence for any $\rho_u, \rho_v\in R$
we have $\rho_u < \rho_u\rho_{vv}$ and
$\rho_v < \rho_u\rho_{vv}$.
Now suppose that 
$w$ = $\rho_{v_1}\rho_{v_2}\cdots\rho_{v_k}\in S$.
An easy induction on $k$, using 
equalities of the type 
$\rho_u\rho_{vv} = \rho_v\rho_{uv}$, establishes
that for any $\rho_u\in R$ we have
$$\rho_u\rho_{v_1v_1}\rho_{v_2v_2}\cdots\rho_{v_kv_k} = 
\rho_{v_1}\rho_{v_2}\cdots\rho_{v_k}\rho_c = w\rho_c$$
where $c$ = $(\cdots((uv_1)v_2)\cdots)v_k$.
Hence, for any $\rho_u\in R$ and $w\in S$
we have
$\rho_u < \rho_u\rho_{v_1v_1}\rho_{v_2v_2}\cdots\rho_{v_kv_k}$
and
$w < \rho_u\rho_{v_1v_1}\rho_{v_2v_2}\cdots\rho_{v_kv_k}$.

If we verify the following statement, then invoking induction
finishes the proof.

\medskip

\noindent
{\bf Claim.} 
For $w_1, w_2\in S$ and $\rho_v\in R$, if there
is a $y\in S$ such that $w_1 < y$ and $w_2 < y$, 
then there is a $z\in S$
such that $w_1 < z$ and $w_2\rho_v < z$. 

\medskip

(If we let 
$\ell(w)$ denote the length of 
a minimal representation of $w$ 
as a product of elements of $R$, then the
argument can be completed using the
above claim by induction on 
$\ell(w_1) + \ell(w_2)$.)
In the last paragraph
we proved the claim for $w_1\in R$. 
Now suppose that $w_1 < y$ and  $w_2 < y$. This means that
$w_1r = w_2s$ for some $r, s\in S$.
Also, by the first paragraph of the proof, 
$\rho_v < x$ and $s < x$ for some $x\in S$.
Hence there exist $p, q\in S$ such that 
$\rho_vp = sq$. Now we have
$$\begin{array}{rl}
w_1(rq) &= (w_1r)q \\
&= (w_2s)q\\
&= w_2(sq)\\
&= w_2(\rho_vp)\\
&= (w_2\rho_v)p
\end{array}
$$
which proves that $w_1 < w_1rq$
and $w_2\rho_v < w_1rq$.
Hence, the claim holds. The 
proof of the lemma is complete.
$\Box$

\bigskip

One usually says that a group \m g is a 
{\bf group of fractions} for a semigroup \m t
if \m g contains \m t as a subsemigroup
and \m g is generated as a group by $T$.
We shall say that a group \m g is a 
{\bf group of simple fractions} for \m t
if \m g contains \m t and every
$g\in G$ is expressible as $g$ = $ab^{-1}$
for some $a, b\in T$. For \m t to have a group of 
simple fractions
it is clearly necessary for \m t to have a group
of fractions,
but this condition is not sufficient.
The characterization of semigroups
which have a group of simple fractions is the
following classical result:

\begin{th}
\label{dub}
{\rm \cite{dubreil}}
A semigroup \m t has a group of simple fractions if and only if 
\m t is cancellative and directed. $\Box$
\end{th}

Incidentally, our definition of {\it directed} is not
left-right symmetric; neither is our
definition of a simple fraction $ab^{-1}$.
If we define directedness in terms of
right divisibility and write simple fractions
as $b^{-1}a$, then we get a theorem 
dual to Theorem \ref{dub} but not equivalent
to it.

The semigroup \m s is directed, but it may not 
cancellative. But any cancellative homomorphic
image of \m s will be both cancellative
and directed.

\begin{lm} 
\label{equiv}
The relation $\equiv$
on \m s defined by $a\equiv b$ if and only if  $aw = bw$
for some $w\in S$ is a congruence.
$\m s/\equiv$ is right
cancellative and directed.
If \m a is abelian or 
$t^{\sm a}(x,y)$ is idempotent, then
$\m s/\equiv$ is also left cancellative.
If, in addition, $t^{\sm a}(x,y)$ is 
right cancellative, then $\equiv$
is equality and 
\m s is cancellative and directed.
\end{lm}

{\bf Proof:}
The relation $\equiv$
is clearly reflexive and symmetric. To see that
it is transitive, assume that 
$a\equiv b$ and $b\equiv c$. 
Then $au = bu$ and $bv = cv$ for some $u, v\in S$.
Since \m s is directed, there is a $w\in S$ such that
$u < w$ and $v < w$. Say that $ur = w = vs$.
Then 
$$aw = aur = bur = bvs = cvs = cw,$$
so $a\equiv c$. Thus $\equiv$ is an equivalence
relation on \m s. Whenever $a, b, c\in S$ we have
$$
\begin{array}{rl}
a\equiv b &\leftrightarrow (\exists w) aw = bw\\
&\rightarrow (\exists w) caw = cbw\\
&\leftrightarrow ca\equiv cb.
\end{array}
$$
That is, $a\equiv b\rightarrow ca\equiv cb$.
This proves that $\equiv$ is a left congruence on \m s.
To see that it is a right congruence, assume that 
$a, b, c\in S$ and $a\equiv b$. Then there is some
$u\in S$ such that $au = bu$. Choose $w\in S$ such that
$c < w$ and $u < w$. Say that $cr = w = us$. Then 
$$(ac)r = aus = bus = (bc)r,$$
so $ac\equiv bc$. Thus $\equiv$ is a congruence
on \m s.

We can prove that $\m s/\equiv$ is right cancellative
by showing that 
$$ac\equiv bc \rightarrow a\equiv b.$$
Now if $ac\equiv bc$, then there is a $u\in S$
such that $acu = bcu$. But now with $w = cu$
we have $aw = bw$, so $a\equiv b$.
The proof of the lemma will be complete
if we show that $\m s/\equiv$ is left
cancellative. This part of the argument
depends on more than just the fact that \m s is directed.
We will argue separately the cases $(i)$ \m a is abelian and
$(ii)$ $t^{\sm a}(x,y)$ is idempotent.

We must show that for $a, b, c\in S$ we have
$$ca\equiv cb \rightarrow a\equiv b.$$
Assume that \m a is abelian.  
Since $ca\equiv cb$ we have that 
$caw = cbw$ for some $w\in S$.
Hence $(caw)(cbw) = (cbw)(caw)$.
Write each of the elements 
$a, b, c$ and $w$ as products of elements of $R$.
Say $a$ = $\rho_{a_1}\cdots\rho_{a_k}$, 
$b$ = $\rho_{b_1}\cdots\rho_{b_{\ell}}$,
$c$ = $\rho_{c_1}\cdots\rho_{c_{m}}$
and 
$w$ = $\rho_{w_1}\cdots\rho_{w_n}$
where $a_i, b_i, c_i, w_i\in A$. 
We have that 
$$caw = \rho_{c_1}\cdots\rho_{c_m}\rho_{a_1}\cdots\rho_{a_k}\rho_{w_1}\cdots\rho_{w_n}$$
and that $cbw$ can be similarly expressed. The polynomial 
$(caw)(cbw)$ is of the form
$$(\cdots((\cdots((\cdots((\cdots((\cdots((\cdots
(xc_1)c_2\cdots)a_1)a_2\cdots)w_1)w_2\cdots)c_1)c_2\cdots)b_1)b_2\cdots)w_1)w_2\cdots)w_n$$
and so is $(cbw)(caw)$. That is, 
$cawcbw$ and $cbwcaw$ are polynomials obtained from the
same term with different parameters.
We may apply the term condition to the
equality $\underline{c}awcbw = \underline{c}bwcaw$
to change the subexpression $(((xc_1)\cdots)c_m)$ which occurs
in the same position on both sides to the new expression $x$.
This yields that $awcbw = bwcaw$. Now we use the fact that 
$caw = cbw$ to obtain
$$awcbw = bw(caw) = bw(cbw).$$
But for $v = wcbw\in S$ this is just the statement $av = bv$.
Hence $a\equiv b$. We have shown that $ca\equiv cb\rightarrow a\equiv b$,
so $\m s/\equiv$ is left cancellative.
(We remark that what makes this argument work is that
$cawcbw$ and $cbwcaw$ are polynomials derived from the same
term with different parameters. Hence the term condition may be applied
in \m a to the equality $cawcbw = cbwcaw$.
This may not be the case with the equality $caw = cbw$ when 
$a$ and $b$ happen to be different-length products of elements of $R$.)

Now assume that $t^{\sm a}(x,y)$ = $xy$ is idempotent, 
but that \m a is not necessarily abelian.
We would like to show that $ca\equiv cb\rightarrow a\equiv b$.
It will be enough to show that
$$\rho_ua\equiv \rho_ub\rightarrow a\equiv b,$$
since if $c$ = $\rho_{c_1}\cdots\rho_{c_m}$, then we may use
the displayed rule to show for each $i$, $1\leq i\leq m$, that
$$\rho_{c_i}\cdots\rho_{c_m}a\equiv \rho_{c_i}\cdots\rho_{c_m}b\rightarrow
\rho_{c_{i+1}}\cdots\rho_{c_m}a\equiv \rho_{c_{i+1}}\cdots\rho_{c_m}b.$$
Taken together, these imply that $ca\equiv cb\rightarrow a\equiv b$.
Assuming that $\rho_ua\equiv \rho_ub$ we get that 
$\rho_uaw = \rho_ubw$ for some $w\in S$. Let
$f(x) = aw\in \poln 1 a$ and let $g(x) = bw\in \poln 1 a$.
Notice that $\rho_uaw$ is just
$f(xu)$ and that $\rho_ubw$ is just $g(xu)$.
We have assumed that $f(xu) = g(xu)$ holds.
Now $t^{\sm a}(x,y)$ is idempotent, so it commutes
with every constant operation. $t^{\sm a}(x,y)$ is
also central, so it commutes with every term operation. 
Together these facts show that $t^{\sm a}(x,y)$ 
commutes with every polynomial operation of \m a.
In particular, 
$$aw\rho_{f(u)} = f(x)f(u) = f(xu) = g(xu) = g(x)g(u) = bw\rho_{g(u)}.$$
$f(xu) = g(xu)$ for all values of $x$ and 
in particular for $x = u$, so
$f(u) = f(uu) = g(uu) = g(u)$. Thus
$\rho_{f(u)}$ = $\rho_{g(u)}$ and we conclude that
for $v$ = $w\rho_{f(u)}$ we have $av = bv$.
Hence $a\equiv b$ is a consequence of 
$\rho_ua\equiv \rho_ub$ as promised.

The final remark of the lemma is that if $t^{\sm a}(x,y)$
is right cancellative, then $\equiv$ is equality.
This follows from the fact that 
$$a\equiv b \leftrightarrow (\exists w) aw = bw\leftrightarrow a = b$$
where the second bi-implication is the statement of right cancellativity.
$\Box$

\bigskip

The congruence $\equiv$ has an especially simple description when 
\m a is abelian. We have $a\equiv b$ if and only if  there is some $w\in S$
such that $aw = bw$. Assume that 
$w = \rho_{w_1}\cdots\rho_{w_k}$ for some
$w_1, \ldots, w_k\in A$. If $a = f(x)\in\poln 1 a$ and
$b = g(x)\in\poln 1 a$, then  the
equality $aw = bw$ of polynomials
may be written 
$$(\cdots((f(x)\underline{w_1})\underline{w_2})\cdots)\underline{w_k}
= (\cdots((g(x)\underline{w_1})\underline{w_2})\cdots)\underline{w_k}.$$
Using the term condition in \m a 
to change each $\underline{w_i}$ to a fixed $v\in A$,
we obtain 
$$(\cdots((f(x)v)v)\cdots)v
= (\cdots((g(x)v)v)\cdots)v$$
or just $a\rho_v^n = b\rho_v^n$. Hence 
$a\equiv b$ is equivalent to $a\rho_v^n = b\rho_v^n$ for 
some $n$ and any $v\in A$ (independent of $n$) when \m a is abelian.

We get from Theorem \ref{dub} 
that $\m s/\equiv$ has a group of simple fractions which we shall
label \m g. 
For $w\in S$ we let $\tilde{w}$ denote
$w/\equiv$. 
Let \m h denote the subgroup
of \m g generated by $Q$ = $\{\tilde{\rho}_a\tilde{\rho}_b^{-1}|a, b\in A\}$.

\bigskip

\begin{lm}
\label{abelian}
\m h is abelian. 
\end{lm}

{\bf Proof:}
Our argument will be based
only on the fact that $xy$ commutes with itself.
Since \m h is the subgroup of \m g generated by
$Q$ = $\{{\tilde{\rho}}_a{\tilde{\rho}}_b^{-1}| a, b\in A\}$, 
we need only to prove that any element
of the form 
${\tilde{\rho}}_a{\tilde{\rho}}_b^{-1}$
commutes with any element of the form
${\tilde{\rho}}_c{\tilde{\rho}}_d^{-1}$. 
Our goal is to remove the `?' over the equality
in 
$$
1\stackrel{?}{=} 
[{\tilde{\rho}}_a{\tilde{\rho}}_b^{-1},{\tilde{\rho}}_c{\tilde{\rho}}_d^{-1}] \hspace{.2in}(:= 
{\tilde{\rho}}_b{\tilde{\rho}}_a^{-1}{\tilde{\rho}}_d{\tilde{\rho}}_c^{-1}{\tilde{\rho}}_a{\tilde{\rho}}_b^{-1}{\tilde{\rho}}_c{\tilde{\rho}}_d^{-1}).
$$
We shall simplify the expression which
is farthest right by a 
series of reversible maneuvers. This will lead
to an equivalent equality. Then we shall prove 
directly that the new equality holds.
Our tool will be the fact that
${\tilde{\rho}}_v^{-1}{\tilde{\rho}}_u$ = ${\tilde{\rho}}_{uv}{\tilde{\rho}}_{vv}^{-1}$,
a consequence of ${{\rho}}_u{{\rho}}_{vv}$ = ${{\rho}}_v{{\rho}}_{uv}$
which we established in the proof of Lemma \ref{directed}.
$$
\begin{array}{rl}
1&\stackrel{?}{=} 
{\tilde{\rho}}_b({\tilde{\rho}}_a^{-1}{\tilde{\rho}}_d)
({\tilde{\rho}}_c^{-1}{\tilde{\rho}}_a)({\tilde{\rho}}_b^{-1}{\tilde{\rho}}_c){\tilde{\rho}}_d^{-1}\\
&={\tilde{\rho}}_b({\tilde{\rho}}_{da}{\tilde{\rho}}_{aa}^{-1})
({\tilde{\rho}}_{ac}{\tilde{\rho}}_{cc}^{-1})({\tilde{\rho}}_{cb}{\tilde{\rho}}_{bb}^{-1}){\tilde{\rho}}_d^{-1}\\
&={\tilde{\rho}}_b{\tilde{\rho}}_{da}({\tilde{\rho}}_{aa}^{-1}{\tilde{\rho}}_{ac})
({\tilde{\rho}}_{cc}^{-1}{\tilde{\rho}}_{cb}){\tilde{\rho}}_{bb}^{-1}{\tilde{\rho}}_d^{-1}\\
&={\tilde{\rho}}_b{\tilde{\rho}}_{da}({\tilde{\rho}}_{(ac)(aa)}{\tilde{\rho}}_{(aa)(aa)}^{-1})
({\tilde{\rho}}_{(cb)(cc)}{\tilde{\rho}}_{(cc)(cc)}^{-1}){\tilde{\rho}}_{bb}^{-1}{\tilde{\rho}}_d^{-1}\\
&={\tilde{\rho}}_b{\tilde{\rho}}_{da}{\tilde{\rho}}_{(ac)(aa)}({\tilde{\rho}}_{(aa)(aa)}^{-1}{\tilde{\rho}}_{(cb)(cc)})
{\tilde{\rho}}_{(cc)(cc)}^{-1}{\tilde{\rho}}_{bb}^{-1}{\tilde{\rho}}_d^{-1}\\
&= {\tilde{\rho}}_b{\tilde{\rho}}_{da}{\tilde{\rho}}_{(ac)(aa)}({\tilde{\rho}}_{[(cb)(cc)][(aa)(aa)]}
{\tilde{\rho}}_{[(aa)(aa)][(aa)(aa)]}^{-1}){\tilde{\rho}}_{(cc)(cc)}^{-1}{\tilde{\rho}}_{bb}^{-1}{\tilde{\rho}}_d^{-1}\\
&= {\tilde{\rho}}_b{\tilde{\rho}}_{da}{\tilde{\rho}}_{(ac)(aa)}{\tilde{\rho}}_{[(cb)(cc)][(aa)(aa)]}
{\tilde{\rho}}_{[(aa)(aa)][(aa)(aa)]}^{-1}{\tilde{\rho}}_{(cc)(cc)}^{-1}{\tilde{\rho}}_{bb}^{-1}{\tilde{\rho}}_d^{-1}.
\end{array}
$$
The questionable equality is therefore equivalent to
$$
 {\tilde{\rho}}_b{\tilde{\rho}}_{da}{\tilde{\rho}}_{(ac)(aa)}{\tilde{\rho}}_{[(cb)(cc)][(aa)(aa)]}
\stackrel{?}{=} 
{\tilde{\rho}}_d{\tilde{\rho}}_{bb}{\tilde{\rho}}_{(cc)(cc)}{\tilde{\rho}}_{[(aa)(aa)][(aa)(aa)]}
$$
which may be written in terms of elements of \m s as
$$
 {{\rho}}_b{{\rho}}_{da}{{\rho}}_{(ac)(aa)}{{\rho}}_{[(cb)(cc)][(aa)(aa)]}
\stackrel{?}{\equiv} 
{{\rho}}_d{{\rho}}_{bb}{{\rho}}_{(cc)(cc)}{{\rho}}_{[(aa)(aa)][(aa)(aa)]}.
$$
But in fact it is true that in \m s we have
$$
 {{\rho}}_b{{\rho}}_{da}{{\rho}}_{(ac)(aa)}{{\rho}}_{[(cb)(cc)][(aa)(aa)]}
=
{{\rho}}_d{{\rho}}_{bb}{{\rho}}_{(cc)(cc)}{{\rho}}_{[(aa)(aa)][(aa)(aa)]}
$$
since this is a restatement of the equation
$$\{[(xb)(da)][(ac)(aa)]\}\{[(cb)(cc)][(aa)(aa)]\}
=
\{[(xd)(bb)][(cc)(cc)]\}\{[(aa)(aa)][(aa)(aa)]\}$$
which we proved in Lemma \ref{calc}. This proves that \m h is abelian.
$\Box$

\bigskip

\begin{lm}
\label{imp}
Assume that \m a is abelian.
If $r(x,\bar{y})\in {\rm Clo}_{n+1}\lb A;xy\rb$, 
$g(x)\in\poln 1 a$ and 
$\bar{a},\bar{b}\in A^n$, then the following implication
holds for any $x\in A$.
$$r^{\sm a}(x,a_1,\ldots,a_n) = r^{\sm a}(x,b_1,\ldots,b_n) 
\rightarrow
r^{\sm a}(x,g(a_1),\ldots,g(a_n)) = r^{\sm a}(x,g(b_1),\ldots,g(b_n)).$$
\end{lm}

{\bf Proof:}
Express $g(x)$ as $g(x) = s^{\sm a}(x,c_1, \ldots,c_k)$ for 
some $s\in {\rm Clo}_{k+1}\m a$ and $\bar{c}\in A^k$.
The fact that $xy$ is central implies that $r$ 
is central, so $r$ and $s$ commute on each of the matrices
$$
\left[  \begin{array}{cccc}
          x   & c_1 & \cdots & c_k\\
          a_1   & c_1 & \cdots & c_k\\
          a_2   & c_1 & \cdots & c_k\\
        \vdots &  \vdots & \ddots & \vdots  \\
          a_n   & c_1 & \cdots & c_k\end{array} \right],
\left[  \begin{array}{cccc}
          x   & c_1 & \cdots & c_k\\
          b_1   & c_1 & \cdots & c_k\\
          b_2   & c_1 & \cdots & c_k\\
        \vdots &  \vdots & \ddots & \vdots  \\
          b_n   & c_1 & \cdots & c_k\end{array} \right].
$$
The fact that $r$ and $s$ commute on the left matrix may be expressed as
$$r^{\sm a}(g(x),g(a_1),\ldots,g(a_n)) = s^{\sm a}(r^{\sm a}(x,\bar{a}), r^{\sm a}(c_1,\bar{c_1}),\ldots,r^{\sm a}(c_k,\bar{c_k})).$$
Here $\bar{c_i}$ is the $n$-tuple $(c_i,c_i,\ldots,c_i)$.
The fact that $r$ and $s$ commute on the right matrix 
from above may be expressed as
$$r^{\sm a}(g(x),g(b_1),\ldots,g(b_n)) = s^{\sm a}(r^{\sm a}(x,\bar{b}), r^{\sm a}(c_1,\bar{c_1}),\ldots,r^{\sm a}(c_k,\bar{c_k})).$$
If $r^{\sm a}(x,\bar{a}) = r^{\sm a}(x,\bar{b})$ for some specific value of $x$, then
the right-hand sides of the last two displayed equations are equal.
Hence for this value of $x$ we have
$$r^{\sm a}(\underline{g(x)},g(a_1),\ldots,g(a_n)) = r^{\sm a}(\underline{g(x)},g(b_1),\ldots,g(b_n)).$$
Using the term condition in \m a we may change $\underline{g(x)}$ to 
$x$ and obtain
$$r^{\sm a}(x,g(a_1),\ldots,g(a_n)) = r^{\sm a}(x,g(b_1),\ldots,g(b_n))$$
for this value of $x$. This proves the lemma.
$\Box$

\bigskip

\begin{lm}
\label{endo}
Assume that \m a is abelian.
If $g\in \poln 1 a$, then the function
$$\hat{g}:Q\rightarrow Q: {\tilde{\rho}}_a{\tilde{\rho}}_b^{-1}\mapsto 
{\tilde{\rho}}_{g(a)}{\tilde{\rho}}_{g(b)}^{-1}$$
is well-defined and extends uniquely to an
endomorphism of \m h.
\end{lm}

{\bf Proof:}
First we prove that $\hat{g}$ is well-defined.
For this we must show that if 
${\tilde{\rho}}_a{\tilde{\rho}}_b^{-1}$ = ${\tilde{\rho}}_c{\tilde{\rho}}_d^{-1}$, 
then ${\tilde{\rho}}_{g(a)}{\tilde{\rho}}_{g(b)}^{-1}$ = ${\tilde{\rho}}_{g(c)}{\tilde{\rho}}_{g(d)}^{-1}$.
For some $n$ and any $v\in A$ (independent of $n$) the following is true.
$$\begin{array}{rl}
{\tilde{\rho}}_a{\tilde{\rho}}_b^{-1} = {\tilde{\rho}}_c{\tilde{\rho}}_d^{-1}&\leftrightarrow
{\tilde{\rho}}_a{\tilde{\rho}}_b^{-1}{\tilde{\rho}}_d{\tilde{\rho}}_c^{-1} = 1\\
&\leftrightarrow {\tilde{\rho}}_a({\tilde{\rho}}_b^{-1}{\tilde{\rho}}_d){\tilde{\rho}}_c^{-1} = 1\\
&\leftrightarrow {\tilde{\rho}}_a({\tilde{\rho}}_{db}{\tilde{\rho}}_{bb}^{-1}){\tilde{\rho}}_c^{-1} = 1\\
&\leftrightarrow {\tilde{\rho}}_a{\tilde{\rho}}_{db} = {\tilde{\rho}}_c{\tilde{\rho}}_{bb}\\
&\leftrightarrow {\rho}_a{{\rho}}_{db} \equiv {{\rho}}_c{{\rho}}_{bb}\\
&\leftrightarrow {\rho}_a{{\rho}}_{db}\rho_v^n = {{\rho}}_c{{\rho}}_{bb}\rho_v^n.
\end{array}
$$
(We used ${\tilde{\rho}}_b^{-1}{\tilde{\rho}}_d$ = ${\tilde{\rho}}_{db}{\tilde{\rho}}_{bb}^{-1}$
which follows from the fact, established in Lemma
\ref{directed}, that ${\tilde{\rho}}_d{\tilde{\rho}}_{bb}$ = ${\tilde{\rho}}_b{\tilde{\rho}}_{db}$.)
The last equality may be written as
$(\cdots([(xa)(db)]v)\cdots)v$ 
= $(\cdots([(xc)(bb)]v)\cdots)v$.
The same argument with $g(u)$ substituted for $u$
for each occurrence of $u\in \{a, b, c, d, v\}$ yields that
${\tilde{\rho}}_{g(a)}{\tilde{\rho}}_{g(b)}^{-1}$ = ${\tilde{\rho}}_{g(c)}{\tilde{\rho}}_{g(d)}^{-1}$
is equivalent to  
$(\cdots([(xg(a))(g(d)g(b))]g(v))\cdots)g(v)$ = $(\cdots([(xg(c))(g(b)g(b))]g(v))\cdots)g(v)$.
Hence to show that $\hat{g}$ is well-defined, we must verify that
$$
\begin{array}{ccc}
\lefteqn{(\cdots([(xa)(db)]v)\cdots)v =(\cdots([(xc)(bb)]v)\cdots)v\rightarrow}\\
&&(\cdots([(xg(a))(g(d)g(b))]g(v))\cdots)g(v) = (\cdots([(xg(c))(g(b)g(b))]g(v))\cdots)g(v).
\end{array}$$
holds for all $x\in A$. This implication is a special
case of Lemma \ref{imp}.

Now we must show that $\hat{g}$ extends uniquely to 
an endomorphism of \m h.
Let \m f  be the free abelian group on the set 
$Q$. $\hat{g}$ extends uniquely to a homomorphism
$\nu:\m f\rightarrow \m h$. Let 
$\eta:\m f\rightarrow \m h$ be the 
canonical homomorphism; i.e., 
$\eta$ is the homomorphism
that extends the identity map from $Q$ to $Q$.
Now, if $\hat{g}:Q\rightarrow Q$ has an extension 
$G:\m h\rightarrow \m h$, then $G\circ\eta$ is
a map from \m f to \m h which agrees with 
$\nu$ on $Q$. It follows that $G\circ\eta = \nu$,
since $Q$ generates \m f. In particular,
${\rm ker}\;\eta\subseteq {\rm ker}\;\nu$.
But, conversely, if 
${\rm ker}\;\eta\subseteq {\rm ker}\;\nu$,
then the
first isomorphism theorem guarantees the
existence of a unique homomorphism 
$G:\eta(\m f)\rightarrow \m h$ such that
$G\circ\eta$ = $\nu$. 
The fact that $Q$ generates \m h implies that
$\eta(\m f)$ = \m h. The fact that 
$G\circ\eta$ = $\nu$ implies that
$G|_Q$ = $\hat{g}$. Hence
$G$ extends $\hat{g}$ in this case.
This shows that $\hat{g}$ has a unique
extension to \m h if and only if 
${\rm ker}\;\eta\subseteq {\rm ker}\;\nu$.
Let us establish this inclusion.

An element $q_1\cdots q_n\in F$ belongs
to ker $\eta$ if and only if  
$q_1\cdots q_n = 1$ in \m h. The same element
belongs to ker $\nu$ if and only if 
$\hat{g}(q_1)\cdots \hat{g}(q_n)$ = 1
in \m h. Hence we need to establish the implication
$$q_1\cdots q_n = 1\rightarrow \hat{g}(q_1)\cdots \hat{g}(q_n) = 1.$$
For $n = 1$ the argument is as follows.
$q_1$ = ${\tilde{\rho}}_a{\tilde{\rho}}_b^{-1}$ for some $a, b\in A$.
Hence $q_1 = 1$ is equivalent to  ${\tilde{\rho}}_a{\tilde{\rho}}_b^{-1} = 1$
or ${\tilde{\rho}}_a$ = ${\tilde{\rho}}_b$. This may be written as
$\rho_a\equiv\rho_b$. This is equivalent to the statement that 
for some $m$ and any $v\in A$, $\rho_a\rho_v^m = \rho_b\rho_v^m$ or just
$(\cdots([xa]v)\cdots)v = (\cdots([xb]v)\cdots)v$.
Similarly, $\hat{g}(q_1) = 1$
is equivalent to 
$(\cdots([xg(a)]g(v))\cdots)g(v) = (\cdots([xg(b)]g(v))\cdots)g(v)$. 
Thus we must show that
$$(\cdots([xa]v)\cdots)v = (\cdots([xb]v)\cdots)v\rightarrow
(\cdots([xg(a)]g(v))\cdots)g(v) = (\cdots([xg(b)]g(v))\cdots)g(v).$$
This implication follows from Lemma \ref{imp}.
The argument for $n = 2$ is identical
to the argument we used to show that $\hat{g}:Q\rightarrow Q$
is well-defined. Now we turn to higher values of $n$.

Our goal is to prove that
$${\tilde{\rho}}_{a_1}{\tilde{\rho}}_{b_1}^{-1}{\tilde{\rho}}_{a_2}{\tilde{\rho}}_{b_2}^{-1}\cdots {\tilde{\rho}}_{a_n}{\tilde{\rho}}_{b_n}^{-1} = 1
\rightarrow 
{\tilde{\rho}}_{g(a_1)}{\tilde{\rho}}_{g(b_1)}^{-1}{\tilde{\rho}}_{g(a_2)}{\tilde{\rho}}_{g(b_2)}^{-1}\cdots{\tilde{\rho}}_{g(a_n)}{\tilde{\rho}}_{g(b_n)}^{-1} = 1.$$
Our first step will be to 
rewrite 
${\tilde{\rho}}_{a_1}{\tilde{\rho}}_{b_1}^{-1}{\tilde{\rho}}_{a_2}{\tilde{\rho}}_{b_2}^{-1}\cdots {\tilde{\rho}}_{a_n}{\tilde{\rho}}_{b_n}^{-1}$ = 1
as an equality that is free of any occurrences of inversion.
The idea will be to move all inverted elements from left
to right using equalities of the form
$${\tilde{\rho}}_v^{-1}{\tilde{\rho}}_u = {\tilde{\rho}}_{uv}{\tilde{\rho}}_{vv}^{-1}.$$
Once we have all
inverted elements on the right, we can move them to the right
side of the ``='' using multiplication on the
right, thereby removing all occurrences of inversion.

Assume that we have an expression which is
a product of elements of $\tilde{R}$ = $\{{\tilde{\rho}}_a|a\in A\}$
and inverses of elements of $\tilde{R}$.
The {\bf active region} of the expression
is defined as follows.
If all of the
inverted elements appear to the right of all other elements,
then we shall say that the  {\bf active region is empty}.
Otherwise, the active region
will be the subproduct beginning at the left-most inverted
element and ending at the right-most non-inverted element.
For example, in the expression that concerns us now
we have
$${\tilde{\rho}}_{a_1}\underbrace{[{\tilde{\rho}}_{b_1}^{-1}{\tilde{\rho}}_{a_2}{\tilde{\rho}}_{b_2}^{-1}\cdots {\tilde{\rho}}_{a_n}]}_{\mbox{\rm the
active region}}{\tilde{\rho}}_{b_n}^{-1}.$$
(If $n = 1$ in this expression, then the active region is empty.)
Within the active region of this particular expression
we may pair up adjacent elements so that
we have a sequence of $n-1$ products of the form 
${\tilde{\rho}}_{b_i}^{-1}{\tilde{\rho}}_{a_{i+1}}$. We parenthesize to indicate
these pairs:
$${\tilde{\rho}}_{a_1}[({\tilde{\rho}}_{b_1}^{-1}{\tilde{\rho}}_{a_2})
({\tilde{\rho}}_{b_2}^{-1}{\tilde{\rho}}_{a_3})\cdots 
({\tilde{\rho}}_{b_{n-1}}^{-1}{\tilde{\rho}}_{a_n})]{\tilde{\rho}}_{b_n}^{-1}.$$
To each of these pairs we shall apply the rule 
${\tilde{\rho}}_v^{-1}{\tilde{\rho}}_u = {\tilde{\rho}}_{uv}{\tilde{\rho}}_{vv}^{-1}$.
Thinking of the inverted elements as marching
from left to right and the non-inverted elements as 
marching from right to left, we shall view a
parenthesized pair $({\tilde{\rho}}_{b_i}^{-1}{\tilde{\rho}}_{a_{i+1}})$
as pair of elements preparing to {\bf salute} each other
as they pass on their drill. The salute will 
involve $(i)$ switching positions and $(ii)$ altering
subscripts according to the rule
${\tilde{\rho}}_v^{-1}{\tilde{\rho}}_u = {\tilde{\rho}}_{uv}{\tilde{\rho}}_{vv}^{-1}$.
We mandate that all parenthesized pairs salute each other simultaneously.
Let us examine two rounds of the saluting process
for the case $n = 4$. From
$${\tilde{\rho}}_{a_1}\underbrace{[({\tilde{\rho}}_{b_1}^{-1}{\tilde{\rho}}_{a_2})({\tilde{\rho}}_{b_2}^{-1}{\tilde{\rho}}_{a_3})
({\tilde{\rho}}_{b_3}^{-1}{\tilde{\rho}}_{a_4})]}_{\mbox{\rm the
active region}}{\tilde{\rho}}_{b_4}^{-1}$$
we obtain
$${\tilde{\rho}}_{a_1}{\tilde{\rho}}_{a_2b_1}\underbrace{[({\tilde{\rho}}_{b_1b_1}^{-1}{\tilde{\rho}}_{a_3b_2})
({\tilde{\rho}}_{b_2b_2}^{-1}{\tilde{\rho}}_{a_4b_3})]}_{\mbox{\rm the
active region}}{\tilde{\rho}}_{b_3b_3}^{-1}{\tilde{\rho}}_{b_4}^{-1}$$
and afterwards
$${\tilde{\rho}}_{a_1}{\tilde{\rho}}_{a_2b_1}{\tilde{\rho}}_{(a_3b_2)(b_1b_1)}\underbrace{[({\tilde{\rho}}_{(b_1b_1)(b_1b_1)}^{-1}
{\tilde{\rho}}_{(a_4b_3)(b_2b_2)})]}_{\mbox{\rm the active region}}
{\tilde{\rho}}_{(b_2b_2)(b_2b_2)}^{-1}{\tilde{\rho}}_{b_3b_3}^{-1}{\tilde{\rho}}_{b_4}^{-1}.$$

After each round of salutes, we are left with a new product of elements,
a new active region (possibly empty) and new subscripts on some
elements. Let us say a few words about the subscripts that occur.
Let $Z$ = $\{a_1, \ldots,a_n,b_1,\ldots,b_n\}$. Define a sequence
of members of ${\rm Clo} \lb A;xy\rb$ as follows.
\begin{enumerate}
\item[$(i)$] $s_0(x_1) = x_1$, $s_1(x_1,x_2) = x_1x_2$, $s_2(x_1,x_2,x_3,x_4)$ = 
$(x_1x_2)(x_3x_4)$.
\item[$(ii)$] $s_{n+1}(X,Y)$ = $(s_n(X))(s_n(Y))$, where
$X$ = $(x_1,\ldots,x_{2^n})$ and 
$Y$ = $(x_{2^n+1},\ldots,x_{2^{n+1}})$.
\end{enumerate}
The following list includes the important points
concerning each round of salutes.
\begin{enumerate}
\item[$(i)$] The length of the entire product remains the same.
\item[$(ii)$] The length of the active region decreases by 2.
\item[$(iii)$] If two elements ${\tilde{\rho}}_u^{-1}$ and ${\tilde{\rho}}_v$
salute in the $i^{\rm th}$ round,
then their respective
subscripts are of the form $u$ = $s_{i-1}(\bar{p})$
and $v$ = $s_{i-1}(\bar{q})$ for some $\bar{p}, \bar{q}\in Z^{2^{i-1}}$
before the salute and their respective subscripts are 
$vv$ = $s_i(\bar{q},\bar{q})$ and $uv = s_i(\bar{p},\bar{q})$ afterwards.
\end{enumerate}
From $(iii)$ it is clear that the complexity of the subscript 
after $i$ rounds of saluting 
depends only on how long the element spends in the active
region. If an element spent $j$ of those
rounds in the active region, then its subscript
will be $s_j(\bar{u})$ for some $\bar{u}\in Z^{2^j}$.
Since any element salutes at most $n-1$ other elements
we see that after $n-1$ rounds of saluting, the active region is empty.
At this point, the complexity of the subscript of the element
that is $k^{\rm th}$ from the right equals the complexity 
of the element $k^{\rm th}$ from the left.
The final configuration looks like
$${\tilde{\rho}}_{c_1}{\tilde{\rho}}_{c_2c_3}{\tilde{\rho}}_{(c_4c_5)(c_6c_7)}\cdots{\tilde{\rho}}_{s_{n-1}(C)}{\tilde{\rho}}_{s_{n-1}(D)}^{-1}\cdots
{\tilde{\rho}}_{(d_7d_6)(d_5d_4)}^{-1}{\tilde{\rho}}_{d_3d_2}^{-1}{\tilde{\rho}}_{d_1}^{-1}$$
Where each $c_i, d_i\in Z$ and $C, D\in Z^{2^{n-1}}$. 

Now we return to our argument that
$${\tilde{\rho}}_{a_1}{\tilde{\rho}}_{b_1}^{-1}{\tilde{\rho}}_{a_2}{\tilde{\rho}}_{b_2}^{-1}\cdots {\tilde{\rho}}_{a_n}{\tilde{\rho}}_{b_n}^{-1} = 1
\rightarrow 
{\tilde{\rho}}_{g(a_1)}{\tilde{\rho}}_{g(b_1)}^{-1}{\tilde{\rho}}_{g(a_2)}{\tilde{\rho}}_{g(b_2)}^{-1}\cdots{\tilde{\rho}}_{g(a_n)}{\tilde{\rho}}_{g(b_n)}^{-1} = 1.$$
Assume that
${\tilde{\rho}}_{a_1}{\tilde{\rho}}_{b_1}^{-1}\cdots {\tilde{\rho}}_{a_n}{\tilde{\rho}}_{b_n}^{-1}$ = 1.
We may write this assumption as
$${\tilde{\rho}}_{c_1}{\tilde{\rho}}_{c_2c_3}{\tilde{\rho}}_{(c_4c_5)(c_6c_7)}\cdots{\tilde{\rho}}_{s_{n-1}(C)}{\tilde{\rho}}_{s_{n-1}(D)}^{-1}\cdots
{\tilde{\rho}}_{(d_7d_6)(d_5d_4)}^{-1}{\tilde{\rho}}_{d_3d_2}^{-1}{\tilde{\rho}}_{d_1}^{-1} = 1,$$
as we explained above, and by multiplying on the right to remove inversions
we get
$${\tilde{\rho}}_{c_1}{\tilde{\rho}}_{c_2c_3}{\tilde{\rho}}_{(c_4c_5)(c_6c_7)}\cdots{\tilde{\rho}}_{s_{n-1}(C)} = 
{\tilde{\rho}}_{d_1}{\tilde{\rho}}_{d_2d_3}{\tilde{\rho}}_{(d_4d_5)(d_6d_7)}\cdots{\tilde{\rho}}_{s_{n-1}(D)}.$$
To rewrite this as an equality between elements of \m s, we use the
same trick we have used earlier in this proof.
We $(i)$ replace all $\tilde{\rho}$s with $\rho$s
and equality  with $\equiv$ and then $(ii)$ multiply 
on the right with some $\rho_v^m$ for large enough $m$ so that equality is obtained.
We end up with  an equivalent equality
$${{\rho}}_{c_1}{{\rho}}_{c_2c_3}{{\rho}}_{(c_4c_5)(c_6c_7)}\cdots{{\rho}}_{s_{n-1}(C)}\rho_v^m =
{{\rho}}_{d_1}{{\rho}}_{d_2d_3}{{\rho}}_{(d_4d_5)(d_6d_7)}\cdots{{\rho}}_{s_{n-1}(D)}\rho_v^m.$$
This may be re-expressed as
$$
\begin{array}{ccc}
\lefteqn{\left(\cdots\left(\{\ldots[(xc_1)(c_2c_3)]
[(c_4c_5)(c_6c_7)]\ldots\}\{s_{n-1}(C)\}v\right)\cdots\right)v = }\\
&&\left(\cdots\left(\{\ldots[(xd_1)(d_2d_3)][(d_4d_5)(d_6d_7)]\ldots\}\{s_{n-1}(D)\}v\right)\cdots\right)v.
\end{array}$$
We need to prove from this assumption that
$${\tilde{\rho}}_{g(a_1)}{\tilde{\rho}}_{g(b_1)}^{-1}{\tilde{\rho}}_{g(a_2)}{\tilde{\rho}}_{g(b_2)}^{-1}\cdots
{\tilde{\rho}}_{g(a_n)}{\tilde{\rho}}_{g(b_n)}^{-1} = 1.$$
But the simplification procedure we used on the product
${\tilde{\rho}}_{a_1}{\tilde{\rho}}_{b_1}^{-1}\cdots {\tilde{\rho}}_{a_n}{\tilde{\rho}}_{b_n}^{-1}$
works exactly the same way on the product
${\tilde{\rho}}_{g(a_1)}{\tilde{\rho}}_{g(b_1)}^{-1}\cdots
{\tilde{\rho}}_{g(a_n)}{\tilde{\rho}}_{g(b_n)}^{-1}$
with $g(u)$ in place of $u$ for each $u\in Z\cup\{v\}$. Hence the implication
$${\tilde{\rho}}_{a_1}{\tilde{\rho}}_{b_1}^{-1}\cdots {\tilde{\rho}}_{a_n}{\tilde{\rho}}_{b_n}^{-1} = 1
\rightarrow 
{\tilde{\rho}}_{g(a_1)}{\tilde{\rho}}_{g(b_1)}^{-1}\cdots{\tilde{\rho}}_{g(a_n)}{\tilde{\rho}}_{g(b_n)}^{-1} = 1$$
is equivalent to the implication
$$
\begin{array}{ccc}
\lefteqn{
\left(\cdots\left(\{\ldots[(xc_1)(c_2c_3)][(c_4c_5)(c_6c_7)]\ldots\}\{s_{n-1}(C)\}v\right)\cdots\right)v = }\\
&&\left(\cdots\left(\{\ldots[(xd_1)(d_2d_3)][(d_4d_5)(d_6d_7)]\ldots\}\{s_{n-1}(D)\}v\right)\cdots\right)v
\rightarrow\\
&&\left(\cdots\left(\{\ldots[(xg(c_1))(g(c_2)g(c_3))][(g(c_4)g(c_5))(g(c_6)g(c_7))]\ldots\}
\{s_{n-1}(g(C))\}g(v)\right)\cdots\right)g(v) =\\
&&\left(\cdots\left(\{\ldots[(xg(d_1))(g(d_2)g(d_3))][(g(d_4)g(d_5))(g(d_6)g(d_7))]
\ldots\}\{s_{n-1}(g(D))\}g(v)\right)\cdots\right)g(v).
\end{array}
$$
(Here $g(C)$ and $g(D)$ mean the expressions obtained by applying
$g$ to each element from $Z$ that occurs in the sequences $C$ and $D$.)
This latter implication holds as it is a special case of Lemma \ref{imp}.
This finishes the proof.
$\Box$

\bigskip

Lemma \ref{endo} proves that $P$ = 
$\{\hat{g}|g\in \poln 1 a\}$
is a collection of endomorphisms of \m h.
This is the set $P$ mentioned at the
beginning of the section.

Now we describe the representing affine algebra
for the case where \m a is abelian.
Its universe will be $H$. Its fundamental operations
will be
\begin{enumerate}
\item[$(i)$] The group operations of \m h,
\item[$(ii)$] the constant operations on $H$ and
\item[$(iii)$] For each $g\in \poln 1 a$, the unary
endomorphism $\hat{g}$.
\end{enumerate}
We fix an element $u\in A$ and define 
a function $\phi:A\rightarrow H$ by the rule
$a\mapsto {\tilde{\rho}}_a{\tilde{\rho}}_u^{-1}$.
For each $n$-ary fundamental operation $f$ of \m a
we associate the following unary polynomials.
For each $i = 1, \ldots,n$, 
let $f_i(x)$ = $f^{\sm a}(u,u,\ldots,u,x,u,\ldots,u)$, $x$
in the $i^{\rm th}$ position only.
For each $n$-ary fundamental operation $f$ of \m a we 
associate the following $n$-ary operation on $H$:
$$\hat{f}_1(x_1) + \cdots + \hat{f}_n(x_n) + h$$
where for $U$ = $f^{\sm a}(u,u,\ldots,u)$
we define $h$ = ${\tilde{\rho}}_U{\tilde{\rho}}_u^{-1}\in H$.

\begin{lm} 
\label{homom}
For all $\bar{a}\in A^n$
$$\phi(f^{\sm a}(a_1,\ldots,a_n)) = (\Sigma_{i=1}^n \hat{f}_i(\phi(a_i))) + h.$$
\end{lm}

{\bf Proof:}
First, let us unravel the definitions involved.
Each $\phi(a_i)$ = ${\tilde{\rho}}_{a_i}{\tilde{\rho}}_u^{-1}$
and $\phi(f^{\sm a}(a_1,\ldots,a_n))$ = ${\tilde{\rho}}_{f(a_1,\ldots,a_n)}{\tilde{\rho}}_u^{-1}$.
Further, $\hat{f}_i({\tilde{\rho}}_{a_i}{\tilde{\rho}}_u^{-1})$ = 
${\tilde{\rho}}_{f_i(a_i)}{\tilde{\rho}}_{f_i(u)}^{-1}$. Hence, 
noting that $f_i(u) = f_j(u) = U$, we must prove that
$${\tilde{\rho}}_{f(a_1,\ldots,a_n)}{\tilde{\rho}}_u^{-1} = 
({\tilde{\rho}}_{f_1(a_1)}{\tilde{\rho}}_{U}^{-1})({\tilde{\rho}}_{f_2(a_2)}{\tilde{\rho}}_{U}^{-1})\cdots 
({\tilde{\rho}}_{f_n(a_n)}{\tilde{\rho}}_{U}^{-1})({\tilde{\rho}}_U{\tilde{\rho}}_u^{-1}).$$
Multiplying both sides on the right by ${\tilde{\rho}}_u{\tilde{\rho}}_U^{-1}$,
this simplifies to 
$${\tilde{\rho}}_{f(a_1,\ldots,a_n)}{\tilde{\rho}}_U^{-1} = 
({\tilde{\rho}}_{f_1(a_1)}{\tilde{\rho}}_{U}^{-1})({\tilde{\rho}}_{f_2(a_2)}{\tilde{\rho}}_{U}^{-1})\cdots 
({\tilde{\rho}}_{f_n(a_n)}{\tilde{\rho}}_{U}^{-1}).$$
We shall argue by induction in $i$ that 
$${\tilde{\rho}}_{f(a_1,\ldots,a_i,u,\ldots,u)}{\tilde{\rho}}_U^{-1} = 
({\tilde{\rho}}_{f_1(a_1)}{\tilde{\rho}}_{U}^{-1})({\tilde{\rho}}_{f_2(a_2)}{\tilde{\rho}}_{U}^{-1})\cdots 
({\tilde{\rho}}_{f_i(a_i)}{\tilde{\rho}}_{U}^{-1}).$$
The case $i = n$ is the statement of the lemma.

The case $i = 1$ is tautologous. Assume that the statement is true for
$i = k$. That is, assume that 
$${\tilde{\rho}}_{f(a_1,\ldots,a_k,u,\ldots,u)}{\tilde{\rho}}_U^{-1} = 
({\tilde{\rho}}_{f_1(a_1)}{\tilde{\rho}}_{U}^{-1})({\tilde{\rho}}_{f_2(a_2)}{\tilde{\rho}}_{U}^{-1})\cdots 
({\tilde{\rho}}_{f_k(a_k)}{\tilde{\rho}}_{U}^{-1})$$
holds for some $k\geq 1$. To show that the statement
then holds for $i = k+1$, we must prove that the first two
expressions below are equal.
$$
\begin{array}{rl}
{\tilde{\rho}}_{f(a_1,\ldots,a_{k+1},u,\ldots,u)}{\tilde{\rho}}_U^{-1} 
&\stackrel{?}{=} 
({\tilde{\rho}}_{f_1(a_1)}{\tilde{\rho}}_{U}^{-1})({\tilde{\rho}}_{f_2(a_2)}{\tilde{\rho}}_{U}^{-1})\cdots 
({\tilde{\rho}}_{f_{k+1}(a_{k+1})}{\tilde{\rho}}_{U}^{-1})\\
&= ({\tilde{\rho}}_{f(a_1,\ldots,a_k,u,\ldots,u)}{\tilde{\rho}}_U^{-1})({\tilde{\rho}}_{f_{k+1}(a_{k+1})}{\tilde{\rho}}_{U}^{-1}).
\end{array}
$$
The second two expressions are equal by hypothesis. It
 is enough for us to prove that the first expression equals the third:
$${\tilde{\rho}}_{f(a_1,\ldots,a_{k+1},u,\ldots,u)}{\tilde{\rho}}_U^{-1} \stackrel{?}{=} 
({\tilde{\rho}}_{f(a_1,\ldots,a_k,u,\ldots,u)}{\tilde{\rho}}_U^{-1})({\tilde{\rho}}_{f_{k+1}(a_{k+1})}{\tilde{\rho}}_{U}^{-1}).$$
(Again we adopt the convention of using 
$\stackrel{?}{=}$ to indicate an equality that is yet to be
established. As in the proof of Lemma \ref{abelian}, we shall
modify this equality to an equivalent one by a sequence
of reversible manuevers and then directly establish the final equality.)
Cancelling ${\tilde{\rho}}_U^{-1}$ on the right of each side of the
last displayed equality, we must prove
$${\tilde{\rho}}_{f(a_1,\ldots,a_{k+1},u,\ldots,u)} \stackrel{?}{=} 
{\tilde{\rho}}_{f(a_1,\ldots,a_k,u,\ldots,u)}{\tilde{\rho}}_U^{-1}{\tilde{\rho}}_{f_{k+1}(a_{k+1})}.$$
Note that ${\tilde{\rho}}_U^{-1}{\tilde{\rho}}_{f_{k+1}(a_{k+1})}$ = 
${\tilde{\rho}}_{f_{k+1}(a_{k+1})U}{\tilde{\rho}}_{UU}^{-1}$, so this may be rewritten as
$${\tilde{\rho}}_{f(a_1,\ldots,a_{k+1},u,\ldots,u)} \stackrel{?}{=} 
{\tilde{\rho}}_{f(a_1,\ldots,a_k,u,\ldots,u)}{\tilde{\rho}}_{f_{k+1}(a_{k+1})U}{\tilde{\rho}}_{UU}^{-1}$$
or just 
$${\tilde{\rho}}_{f(a_1,\ldots,a_{k+1},u,\ldots,u)}{\tilde{\rho}}_{UU} \stackrel{?}{=} 
{\tilde{\rho}}_{f(a_1,\ldots,a_k,u,\ldots,u)}{\tilde{\rho}}_{f_{k+1}(a_{k+1})U}.$$
In terms of elements of \m s, this is just
$${{\rho}}_{f(a_1,\ldots,a_{k+1},u,\ldots,u)}{{\rho}}_{f(u,\ldots,u)f(u,\ldots,u)}
 \stackrel{?}{\equiv} 
{{\rho}}_{f(a_1,\ldots,a_k,u,\ldots,u)}{{\rho}}_{f(u,\ldots,u,a_{k+1},u,\ldots,u)f(u,\ldots,u)},$$
since $U$ = $f(u,\ldots,u)$ and 
$f_{k+1}(a_{k+1})$ = $f(u,\ldots,u,a_{k+1},u,\ldots,u)$
with $a_{k+1}$ in the $(k+1)^{\rm rst}$ position. 
But is even true that in \m s we have
$${{\rho}}_{f(a_1,\ldots,a_{k+1},u,\ldots,u)}{{\rho}}_{f(u,\ldots,u)f(u,\ldots,u)} =
{{\rho}}_{f(a_1,\ldots,a_k,u,\ldots,u)}{{\rho}}_{f(u,\ldots,u,a_{k+1},u,\ldots,u)f(u,\ldots,u)},$$
since this is just the statement
$$
\begin{array}{ccc}
\lefteqn{[(xf(a_1,\ldots,a_{k+1},u,\ldots,u))(f(u,\ldots,u)f(u,\ldots,u))] = }\\
&&[(xf(a_1,\ldots,a_{k},u,u,\ldots,u))(f(u,\ldots,u, a_{k+1}, u\ldots,u)f(u,\ldots,u))]
\end{array}$$
which we proved in Lemma \ref{calc}.
This finishes the argument for the inductive step
and so completes the proof.
$\Box$

\bigskip

We define $\widehat{\m h}$ to be the algebra
whose universe is $H$ and whose 
fundamental operations are the collection of all
$(\Sigma_{i=1}^n \hat{f}_i(x_i)) + h$; one for each
fundamental operation $f$ of \m a.
Clearly $\widehat{\m h}$ is a reduct of the
unnamed affine algebra defined
in the paragraph preceding Lemma \ref{homom}.
The result of Lemma \ref{homom}
is that $\phi:\m a\rightarrow \widehat{\m h}$
is a homomorphism. From this we get our
main theorem.

\medskip

\noindent
{\bf THEOREM 1.1} {\sl 
Let \m a be an abelian algebra with a
central term $t(x,y) = xy$. 
\m a has a quasi-affine representation
$\Phi$ such that $\Phi(a) = \Phi(b)$
holds if and only if  for some $n$ it is the case that
$$(\cdots((aa)a)\cdots)a = (\cdots((ab)a)\cdots)a
\hspace{.2in}\mbox{\rm and} \hspace{.2in}
a(\cdots(a(aa))\cdots) = a(\cdots(a(ba))\cdots)$$
where there
are $n+1$ occurrences of $a$ on
the left side of each equation and $n$ occurrences of $a$
on the right.}

\medskip

{\bf Proof:} In this section we have described a representation
$\phi$ for any abelian algebra with a binary central term.
In our representation we have that for some $\ell$, any $v\in A$
and any $x\in A$ it is the case that
$$
\begin{array}{rl}
\phi(a) = \phi(b)&\leftrightarrow \tilde{\rho}_a\tilde{\rho}_u^{-1} = \tilde{\rho}_b\tilde{\rho}_u^{-1}\\
&\leftrightarrow \tilde{\rho}_a = \tilde{\rho}_b\\
&\leftrightarrow {\rho}_a \equiv {\rho}_b\\
&\leftrightarrow {\rho}_a\rho_v^{\ell} = {\rho}_b\rho_v^{\ell}\\
&\leftrightarrow (\cdots((xa)v)\cdots)v = (\cdots((xb)v)\cdots)v.
\end{array}
$$
In particular, using that \m a is abelian,
we set $x = v = a$ and obtain that for some $\ell$ 
$$\phi(a) = \phi(b) \leftrightarrow (\cdots((aa)a)\cdots)a = (\cdots((ab)a)\cdots)a$$
where there
are ${\ell}+1$ occurrences of $a$ on
the left side of each equation and $\ell$ occurrences of $a$
on the right. 

Similarly, we may construct a representation $\phi'$ based on the dual
term $t'(x,y) = t(y,x)$. We get that $\phi'(a) = \phi'(b)$ if and only if 
for some $m$ it is the case that
$a(\cdots(a(aa))\cdots)$ = $a(\cdots(a(ba))\cdots)$
where there
are $m+1$ occurrences of $a$ on
the left side of each equation and $m$ occurrences of $a$
on the right. If we choose $n = \max\{\ell, m\}$ and set 
$\Phi$ = $\phi\times \phi':\m a \rightarrow \widehat{\m h}\times\widehat{\m h}'$,
then $\Phi$ has the properties stated in the theorem.
$\Box$

\bigskip

Next we show that most of the arguments we have used
work equally well in some cases where \m a is non-abelian.

\begin{th}
\label{minor}
Let \m a = $\lb A;xy\rb$ be an idempotent, entropic binar. 
\m a has a quasi-affine representation 
$\Phi$ where $\Phi(a) = \Phi(b)$ if and only if 
there exist $c_1,\ldots, c_m, d_1,\ldots,d_n\in A$ such that 
$$(\forall x) (\cdots(xa)c_1\cdots)c_m = (\cdots(xb)c_1\cdots)c_m
\hspace{.2in}\mbox{\rm and}\hspace{.2in}
(\forall x) \;d_n(\cdots d_1(ax)\cdots) = d_n(\cdots d_1(bx)\cdots).$$
If $xy$ is right cancellative, then $\Phi(a) = \Phi(b)$ 
if and only if $(\forall x) \;xa = xb$. If $xy$ is also left cancellative, then
$\Phi$ is faithful.
\end{th}

{\bf Proof:}
As in the proof of Theorem \ref{MAIN1}, 
let $R$ = $\{\rho_a|a\in A\}$ and let
\m s be the semigroup of self maps of
\m a generated by $R$. The fundamental
operation of \m a is central and idempotent, 
by assumption,
so Lemmas \ref{directed}, \ref{equiv} and \ref{abelian} hold.
For a fixed $u\in A$ we have a function
$\phi:A\rightarrow H$ defined by 
$a\mapsto \tilde{\rho}_a\tilde{\rho}_u^{-1}$.
Our goal is to prove that $\phi$ extends to 
a representation. 

Consider the automorphism 
$$\alpha:\m g\rightarrow \m g: g\mapsto \tilde{\rho}_u^{-1}g\tilde{\rho}_u.$$
We claim that $\alpha(H)\subseteq H$.
To show this, it will suffice to prove that 
$\alpha(Q)\subseteq Q$ where $Q$ = $\{\tilde{\rho}_a\tilde{\rho}_b^{-1}|a, b\in A\}$
since $Q$ generates \m h.
We shall use the fact that 
for any $a, v\in A$ we have
$\rho_a\rho_{vv}$ = $\rho_v\rho_{av}$
which we established in Lemma \ref{directed}. Since
$xy$ is idempotent, $\rho_{vv}$ = $\rho_v$ and so
we have $\rho_a\rho_v$ = $\rho_v\rho_{av}$.
A fortiori, we have $\tilde{\rho}_a\tilde{\rho}_v$ = $\tilde{\rho}_v\tilde{\rho}_{av}$
which we choose to write as
$\tilde{\rho}_{av}$ = $\tilde{\rho}_v^{-1}\tilde{\rho}_a\tilde{\rho}_v$.
Using this equality in the case where $v = u$ we calculate that
$$
\begin{array}{rl}
\alpha(\tilde{\rho}_a\tilde{\rho}_b^{-1}) &= 
\tilde{\rho}_{u}^{-1}\tilde{\rho}_a\tilde{\rho}_b^{-1}\tilde{\rho}_{u}\\
&= (\tilde{\rho}_{u}^{-1}\tilde{\rho}_a\tilde{\rho}_{u})
(\tilde{\rho}_{u}^{-1}\tilde{\rho}_b^{-1}\tilde{\rho}_{u})\\
&= (\tilde{\rho}_{u}^{-1}\tilde{\rho}_a\tilde{\rho}_{u})
(\tilde{\rho}_{u}^{-1}\tilde{\rho}_b\tilde{\rho}_{u})^{-1}\\
&= \tilde{\rho}_{au}\tilde{\rho}_{bu}^{-1}\in Q.
\end{array}
$$
This shows that $\alpha(Q)\subseteq Q$ and so $\alpha(H)\subseteq H$.

\medskip

\noindent
{\bf Claim.} For all $a, b\in A$, $\phi(ab)$ = $\alpha(\phi(a))$ + $(1-\alpha)(\phi(b))$.

\medskip

We begin by computing in \m h, but after the first line 
we compute in \m g (so we change from additive
to multiplicative notation).
$$
\begin{array}{rl}
\alpha(\phi(a)) + (1-\alpha)(\phi(b))
&= \alpha(\phi(a)) - \alpha(\phi(b)) + \phi(b)\\
&= [\tilde{\rho}_u^{-1}(\tilde{\rho}_a\tilde{\rho}_u^{-1})\tilde{\rho}_u]
[\tilde{\rho}_u^{-1}(\tilde{\rho}_b\tilde{\rho}_u^{-1})\tilde{\rho}_u]^{-1}
[\tilde{\rho}_b\tilde{\rho}_u^{-1}]\\
&= [\tilde{\rho}_u^{-1}\tilde{\rho}_a\tilde{\rho}_u^{-1}\tilde{\rho}_u]
[\tilde{\rho}_u^{-1}\tilde{\rho}_u\tilde{\rho}_b^{-1}\tilde{\rho}_u]
[\tilde{\rho}_b\tilde{\rho}_u^{-1}]\\
&= [\tilde{\rho}_u^{-1}\tilde{\rho}_a\tilde{\rho}_b^{-1}\tilde{\rho}_u]
[\tilde{\rho}_b\tilde{\rho}_u^{-1}]
\end{array}
$$
We need to interrupt the argument in order to make the following observations:
If $c, d\in A$, $h\in H$, then 
$(i)$ $\tilde{\rho}_c\tilde{\rho}_d^{-1}\in H$ 
(since $\tilde{\rho}_c\tilde{\rho}_d^{-1}\in Q$) 
and $(ii)$ $\tilde{\rho}_c^{-1}h\tilde{\rho}_c$ = 
$\tilde{\rho}_d^{-1}h\tilde{\rho}_d$ (since this is equivalent to saying that
$\tilde{\rho}_c\tilde{\rho}_d^{-1}$ and $h$ commute. Both 
elements belong to \m h and \m h is commutative, so this is clear.)
From observation $(i)$ it follows that $\tilde{\rho}_a\tilde{\rho}_b^{-1}\in H$
in our previous derivation. Applying observation $(ii)$
to this, we get that 
$\tilde{\rho}_u^{-1}\tilde{\rho}_a\tilde{\rho}_b^{-1}\tilde{\rho}_u$ = 
$\tilde{\rho}_b^{-1}\tilde{\rho}_a\tilde{\rho}_b^{-1}\tilde{\rho}_b$. Now we resume
our derivation at the last line.
$$
\begin{array}{rl}
[\tilde{\rho}_u^{-1}\tilde{\rho}_a\tilde{\rho}_b^{-1}\tilde{\rho}_u]
[\tilde{\rho}_b\tilde{\rho}_u^{-1}] &=
[\tilde{\rho}_b^{-1}\tilde{\rho}_a\tilde{\rho}_b^{-1}\tilde{\rho}_b]
[\tilde{\rho}_b\tilde{\rho}_u^{-1}] \\
&=(\tilde{\rho}_b^{-1}\tilde{\rho}_a\tilde{\rho}_b)\tilde{\rho}_u^{-1} \\
&=\tilde{\rho}_{ab}\tilde{\rho}_u^{-1} \\
&=\phi(ab).
\end{array}
$$
This completes the proof of the claim

Let $\widehat{\m h}$ = $\lb H;\alpha(x)+(1-\alpha)(y)\rb$. The claim
proves that $\phi:\m a\rightarrow \widehat{\m h}$ is a homomorphism.
Clearly $\widehat{\m h}$ is a reduct of the affine algebra
obtained by adjoining the unary endomorphism $\alpha$
to the abelian group \m h. Hence $\phi$ is a quasi-affine
representation. For any $a, b\in A$ we have
$\phi(a) = \phi(b)$ if and only if 
$$\tilde{\rho}_a\tilde{\rho}_u^{-1} = \phi(a) = \phi(b) = \tilde{\rho}_b\tilde{\rho}_u^{-1}$$
which occurs exactly when $\tilde{\rho}_a$ = $\tilde{\rho}_b$.
This happens if and only if  $\rho_a\equiv \rho_b$ which, by Lemma \ref{equiv},
is equivalent to $\rho_aw = \rho_bw$ for some $w\in S$.
If we can express $w$ as $\rho_{c_1}\cdots\rho_{c_m}$
for $c_1,\ldots,c_m\in A$, then $\rho_aw = \rho_bw$  says that
the polynomials
$$(\cdots(xa)c_1\cdots)c_m\hspace{.2in}\mbox{\rm and}
\hspace{.2in} (\cdots(xb)c_1\cdots)c_m$$
are equal. Thus, $\phi(a) = \phi(b)$ if and only if  
$$(\forall x) (\cdots(xa)c_1\cdots)c_m = (\cdots(xb)c_1\cdots)c_m$$
holds.

We can finish the proof of the first claim of the 
theorem by using the same trick we used in the proof of Theorem \ref{MAIN1}.
We let $\phi':\m a\rightarrow \widehat{\m h}$ by a representation
based on $yx$ in place of $xy$ and set $\Phi$ = $\phi\times \phi'$.
For this $\Phi$ the first claim of the theorem holds.

For the second claim of the theorem, assume that 
$xy$ is right cancellative.
Then by Lemma \ref{equiv} we have that $\equiv$ is the equality
relation. Thus $\phi(a) = \phi(b)$ is equivalent to
$\rho_a = \rho_b$. This is just the statement that $(\forall x) xa = xb$.
If $\phi(a) = \phi(b)$, then for any $c\in A$ we have
$ca = cb$, so for all $x\in A$ we have $(ca)(cx) = (cb)(cx)$. With the entropic 
and idempotent laws we may write this as
$$(\forall x) c(ax) = c(bx)$$
which proves that $\phi'(a) = \phi'(b)$. It follows that
ker $\phi\subseteq {\rm ker}\;\phi'$ and so
$$\begin{array}{rl}
\Phi(a) = \Phi(b) &\leftrightarrow \phi(a) = \phi(b)\;\&\;\phi'(a) = \phi'(b)\\
&\leftrightarrow \phi(a) = \phi(b)\\
&\leftrightarrow (\forall x) xa = xb. 
\end{array}
$$

The third claim of the theorem easily follows from the second. $\Box$

\bigskip

The argument in Theorem \ref{minor} can be extended to the case
where \m a has several fundamental operations which need not be
binary. If it is the case that \m a is idempotent and entropic,
then our proof can be modified to construct a representation
in this more general setting.
For a given fundamental operation $f$ of \m a, one defines
the endomorphism which is the $i^{\rm th}$ 
coefficient of the corresponding
operation of $\widehat{\m h}$ in the way that we defined
$\alpha$ to be the first coefficient of the operation
corresponding to $xy$. In particular, it can be
shown that if \m a is an idempotent entropic algebra with
multiple cancellative binary operations, then \m a is quasi-affine.

Now we show that, in some circumstances,
quasi-affine representations have strongly abelian 
kernels.


\begin{df} 
\label{str.abel}
A congruence $\theta$ on \m a is said to be {\bf strongly
abelian} if for any $(n+1)$-ary term
$s(x,\bar{y})$
and all $a\;\theta\; b$, $u_i\;\theta\;v_i\;\theta\;w_i$
it is the case that 
$$s^{\sm a}(a,\bar{u}) = s^{\sm a}(b,\bar{v})\rightarrow 
s^{\sm a}(a,\bar{w}) = s^{\sm a}(b,\bar{w}).$$
\end{df}

\begin{th}
\label{strong}
If \m a is a right cancellative binar which satisfies 
\begin{enumerate}
\item[$(i)$] $xx = x$ (Idempotent Law),
\item[$(ii)$] $(xy)(zu)$ = $(xz)(yu)$ (Entropic Law),
\end{enumerate}
then \m a has a strongly abelian congruence $\theta$
such that $\m a/\theta$ is quasi-affine and each 
$\theta$-class is the universe of a left-zero subsemigroup
of \m a.

If \m a is an idempotent abelian binar,
then \m a has a 
strongly abelian 
congruence $\theta$ such that
$\m a/\theta$ is quasi-affine.
\end{th}

{\bf Proof:}
First assume that \m a is 
a right cancellative, idempotent,
entropic binar. Let $\theta$ =
ker $\Phi$ where $\Phi$ is the
representation defined
in Theorem \ref{minor}. As proven
in Theorem \ref{minor},
$(a,b)\in \theta$ if and only if  the polynomials
$\rho_a$ and $\rho_b$ are equal.
Now pick any $(n+1)$-ary term $s$ 
and $a\;\theta\; b$, $u_i\;\theta\;v_i\;\theta\;w_i$.
We shall argue by induction on the
complexity of $s$ that the implication 
in Definition \ref{str.abel} holds.
Clearly when $s$ is projection onto a variable,
the implication holds.
Assume that $s^{\sm a}(a,\bar{u}) = s^{\sm a}(b,\bar{v})$.
If $s(x,\bar{y})$ = $(s_1(x,\bar{y})s_2(x,\bar{y}))$,
then 
$$s_2(a,\bar{w})\;\theta\;s_2(a,\bar{u})\;\theta\;
s_2(b,\bar{v})\;\theta\;s_2(b,\bar{w}),$$
so 
$$\rho_{s_2(a,\bar{w})} = \rho_{s_2(a,\bar{u})} = 
\rho_{s_2(b,\bar{v})} = \rho_{s_2(b,\bar{w})}.$$
Hence we get 
$$
\begin{array}{rl}
(s_1^{\sm a}(a,\bar{u})s_2^{\sm a}(a,\bar{u})) &= s^{\sm a}(a,\bar{u}) \\
&= s^{\sm a}(b,\bar{v})\\
&= (s_1^{\sm a}(b,\bar{v})s_2^{\sm a}(b,\bar{v})) \\
&= [s_1^{\sm a}(b,\bar{v})]\rho_{s_2(b,\bar{v})} \\
&= [s_1^{\sm a}(b,\bar{v})]\rho_{s_2(a,\bar{u})} \\
&= (s_1^{\sm a}(b,\bar{v})s_2^{\sm a}(a,\bar{u})).
\end{array}
$$
Cancelling $s_2^{\sm a}(a,\bar{u}))$
from the the right in $s_1^{\sm a}(a,\bar{u})s_2^{\sm a}(a,\bar{u})$ = 
$s_1^{\sm a}(b,\bar{v})s_2^{\sm a}(a,\bar{u})$
we get $s_1^{\sm a}(a,\bar{u})$ = $s_1^{\sm a}(b,\bar{v})$.
Since $s_1$ has smaller complexity than $s$, we may use the
implication of Definition 
\ref{str.abel} to deduce that
$$s_1^{\sm a}(a,\bar{w}) = s_1^{\sm a}(b,\bar{w}).$$
Together with $\rho_{s_2(a,\bar{w})}$ = $\rho_{s_2(b,\bar{w})}$ we get 
$$
\begin{array}{rl}
s^{\sm a}(a,\bar{w}) &= (s_1^{\sm a}(a,\bar{w})s_2^{\sm a}(a,\bar{w}))\\
&= s_1^{\sm a}(a,\bar{w})\rho_{s_2(a,\bar{w})}\\
&= s_1^{\sm a}(b,\bar{w})\rho_{s_2(b,\bar{w})}\\
&= (s_1^{\sm a}(b,\bar{w})s_2^{\sm a}(b,\bar{w}))\\
&= s^{\sm a}(b,\bar{w}).
\end{array}
$$
This concludes the induction. Hence $\theta$
is strongly abelian. 

Let $U$ be a $\theta$-class. If $a, b\in U$, then 
the entropic law yields
$$(\underline{a}b)(\underline{a}\underline{a}) = (\underline{a}a)(\underline{b}\underline{a}).$$
Using the strong term condition to change all underlined
parameters to $a$, we get
$(ab)(aa)$ = $(aa)(aa)$.
We may cancel $(aa)$ on the right to obtain
$ab = aa$. Since \m a is idempotent, $ab = a$.
Thus $xy$ is a left-zero semigroup operation on $U$.
This establishes the first claim of the theorem.

Now assume that \m a is abelian.
As proved in the introduction,
any idempotent abelian binar
satisfies the entropic law.
Hence the fundamental operation
commutes with itself and with every
constant operation. It follows that 
for any $a\in A$, the polynomials
$\rho_a$ = $xa$  and $\lambda_a$ = $ax$
are endomorphisms of \m a.
By the term condition we get that 
$$\underline{a}x = \underline{a}y\leftrightarrow \underline{b}x = \underline{b}y.$$
That is, ker $\lambda_a$ = ker $\lambda_b$ for any $a, b\in A$.
Furthermore, 
$$\underline{a}(\underline{\underline{b}}x) = \underline{a}(\underline{\underline{b}}y) 
\leftrightarrow 
\underline{c}(\underline{\underline{d}}x) = \underline{c}(\underline{\underline{d}}y)$$
for any $a, b, c, d\in A$. Hence ker $\lambda_a\circ\lambda_b$ = 
ker $\lambda_c\circ\lambda_d$ for any $a, b, c, d\in A$.
In fact, any composition of $n$
functions of the form $\lambda_{a_i}$, $a_i\in A$, has the same kernel
as any other composition of $n$ such functions and that kernel is
just ker $\lambda_a^n$ for any fixed $a\in A$.
This statement is true with $\rho$ in place of $\lambda$ throughout.
Let us fix one choice of $a\in A$ for the rest of the proof.
Let $\beta$ be the congruence $\bigcup_{n<\omega} {\rm ker} \lambda_a^n$.
Let $\gamma$ be the congruence $\bigcup_{n<\omega} {\rm ker} \rho_a^n$.
We claim that 
\begin{enumerate}
\item[$(i)$] $\beta\wedge \gamma$ = $0_{\sm a}$.
\item[$(ii)$] \m b = $\m a/\beta$ is left cancellative and \m c = 
$\m a/\gamma$ is right cancellative.
\end{enumerate}

For $(i)$, choose $(b, c)\in \beta\wedge\gamma - 0_{\sm a}$.
This implies that $\lambda_a^m(b) = \lambda_a^m(c)$
for some $m$ and $\rho_a^n(b) = \rho_a^n(c)$ for some $n$.
We can replace $(b,c)$ with 
$(b',c') = (\lambda_a^k(b),\lambda_a^k(c))$ where
$k$ is chosen maximally so that 
$\lambda_a^k(b)\neq\lambda_a^k(c)$. Since $\lambda_a\in \poln 1 a$,
we still have $(b',c')\in \gamma$ but now $(b',c')\in {\rm ker} \lambda_a$.
With the same type of argument on the right, we can replace $(b',c')$
with some $(b'',c'')\in {\rm ker} \lambda_a\wedge {\rm ker} \rho_a - 0_{\sm a}$.
Changing notation, we assume that our original pair
$(b,c)$ was chosen from ${\rm ker} \lambda_a\wedge {\rm ker} \rho_a - 0_{\sm a}$.
But now we have a problem. We have 
$$b = \lambda_b(b) = \lambda_b(c) = \rho_c(b) = \rho_c(c) = c$$
since ker $\lambda_a$ = ker $\lambda_b$ and 
ker $\rho_a$ = ker $\rho_c$. This contradicts our 
assumption that $b\neq c$, so it must be that $\beta\wedge\gamma$ = $0_{\sm a}$.

To prove $(ii)$ we must show, for example, 
that $bx \;\beta\; by\rightarrow x\;\beta\;y$.
If $(bx,by)\in\beta$, then 
$\lambda_a^n\lambda_b(x) = \lambda_a^n\lambda_b(y)$ for some $n$.
Hence 
$$(x,y)\in {\rm ker }\; \lambda_a^n\lambda_b = \lambda_a^{n+1} \leq \beta.$$
The same argument with $\rho_a$
in place of $\lambda_a$ works for 
$\gamma$. 

Now $\m a/\gamma$ is a right cancellative, idempotent, entropic
binar. By the first part of the theorem, 
there is a congruence $\sigma\in \cn a$ such that
$\sigma/\gamma$ is a strongly abelian congruence on $\m a/\gamma$
and $(\m a/\gamma)/(\sigma/\gamma)$ is quasi-affine. Similarly,
there is a congruence $\varsigma\in \cn a$ such that
$\varsigma/\beta$ is a strongly abelian congruence and 
$(\m a/\beta)/(\varsigma/\beta)$ is quasi-affine. 
It follows that $\varsigma\times\sigma$ is a strongly
abelian congruence on $(\m a/\beta)\times(\m a/\gamma)$
and the corresponding  factor algebra is quasi-affine.
If we let $\theta$ = $\eta^{-1}(\varsigma\times\sigma)$ 
where $\eta:\m a\rightarrow (\m a/\beta)\times(\m a/\gamma)$
is the natural homomorphism, then $\theta$
is strongly abelian (since $(a)$ the strongly abelian implication 
is universally quantified and $(b)$
the fact that $\beta\wedge \gamma = 0$ implies that $\eta$
is 1-1). Furthermore, the induced homomorphism
$$\bar{\eta}:\m a/\theta\rightarrow (\m a/\varsigma)\times(\m a/\sigma)$$
is an embedding of $\m a/\theta$ into a product of quasi-affine
algebras. Hence $\m a/\theta$ is quasi-affine. This proves the
second statement of the theorem. $\Box$

\bigskip

Using tame congruence theory (see \cite{HM}) it can be
proved that any finite abelian algebra has a strongly 
solvable congruence $\theta$ such that $\m a/\theta$
is quasi-affine. ($\theta\in \cn a$
is {\bf strongly solvable}
if there is a chain of congruences 0 = $\theta_0 \leq \cdots \leq \theta_n$ = 
$\theta$ such that $\theta_{i+1}/\theta_i$ is strongly
abelian in $\m a/\theta_i$.)  I do not know
if every abelian algebra has a quasi-affine representation
with a strongly {\it abelian} kernel, but think it is likely
for finite algebras.

\section{Further Remarks}

In this section we discuss possible extensions
of Theorem \ref{MAIN1}. 
The reason for including
these remarks is that some of the 
ideas underlying our arguments 
are unlikely to be recognized after simply
reading through the calculations that lead up to 
the proof of Theorem \ref{MAIN1}. 

Assume that $\phi:\m a\rightarrow\m h$ is 
a quasi-affine representation where
\m h is a reduct of some affine algebra, \m m.
By replacing \m h and \m m if necessary, we may assume 
that \m a, \m h and \m m
have a close relationship, to be described shortly.
We shall assume that this relationship holds
throughout this paragraph and the next.
First, we may assume that
the clone of \m m is generated by $x - y + z$
and the operations of \m h.
This is possible since the clone so generated will be that of an affine 
algebra, $\m m'$, and \m h is also a reduct
of $\m m'$. We may further assume that
\m m is generated by the set 
$\phi(A)$. Finally, replacing \m m with a homomorphic image if
necessary, we may assume that every non-zero congruence on \m m
has a non-zero restriction to \m a.
With these assumptions in place we say that
\m a is {\bf minimally represented on \m m}.
When \m a is minimally represented on \m m, it is not
hard to see that 
any two term or polynomial operations of \m h which agree on 
$\phi(A)$
also agree on $H$ = $M$.
Such a representation of \m a on \m h where
\m h is a reduct of \m m
induces well-defined
clone homomorphisms
$$\epsilon: {\rm Clo}\;\m a\rightarrow {\rm Clo}\;\m h, \hspace{.2in}
\mu: {\rm Clo}\;\m h\rightarrow {\rm Clo}\;\m m$$
(and similar homomorphisms between polynomial clones).
Here $\epsilon$ is an epimorphism (in fact a surjection)
and $\mu$ is a monomorphism.
Clo \m m is generated by $\mu({\rm Clo}\;\m h)$
together with 
$x - y + z$, so $\mu({\rm Clo}\;\m h)$
might be considered to be
``large'' in Clo \m m, but $\mu$ 
does not have to be an epimorphism
of clones. (To eliminate \m h
from our discussion, note that 
$\mu$ is an epimorphism if and only if 
$\mu\epsilon$ is an epimorphism.)
For an example where $\mu$ is not an epimorphism, 
take \m a to be a 10-element set with no operations.
Then Clo \m m is generated by a ternary abelian
group operation $x - y + z$,
\m m has at least 10 distinct
elements and \m m is generated by those elements. 
Let $\sigma$ be any permutation of $M$
which does not commute with $x - y + z$
and let $C$ be the clone on $M$ generated
by the operations $x - y + z$ and
$\sigma^{-1}(\sigma(x) - \sigma(y) + \sigma(z))$.
There are two obvious maps from Clo \m m
to $C$: $f$ which sends $x - y + z$
to $x - y + z$ and $g$ which sends 
$x - y + z$ to $\sigma^{-1}(\sigma(x) - \sigma(y) + \sigma(z))$.
One has that $f\neq g$,
but $f\mu\epsilon = g\mu\epsilon$ = the unique clone homomorphism 
from Clo \m a to $C$. This shows that, no matter
how \m m is chosen, $\mu\epsilon$ will not
be an epimorphism
when \m a is a 10-element set with no operations.

However, for some quasi-affine representations
it {\it may} occur that $\mu\epsilon$ is an epimorphism.
The best known example where this occurs
is where \m a = \m n is the semigroup of natural numbers and 
\m m  = \m z is the group of integers. The classical
construction of \m z from \m n is perhaps the first 
``affinization'' argument. 
The construction of \m z from \m n  might
have been the blueprint
for some later affinization arguments, e.g., those found
in \cite{ch}, \cite{k1} and \cite{k2}.
In each of these papers where a
quasi-affine representation is obtained
for an algebra \m a 
it is possible to show that the clone
homomorphism from Clo \m a to 
the clone of the representing
algebra \m m is an epimorphic embedding of clones.
The main difference one finds in the present paper
is that we construct quasi-affine representations 
where the induced clone homomorphism does not need 
to be epimorphic, but as a consequence the
representations are not always faithful.

\medskip

Since every affine algebra is a reduct of a module
expanded by constants, the difficulty
in constructing a representation of \m a is
the problem of associating a module to \m a.
Let us look at a classical method
for constructing a module, and then discuss
a new view of essentially the same construction.

We begin with the following data: 
a pair $\lb \m g;\m h\rb$ where \m g is a 
group and \m h is an abelian subgroup of \m g.
The group \m h will be the underlying
abelian group of our module.
For each element $g\in G$ we have a
corresponding inner automorphism which,
restricted to \m h, gives
an endomorphism:
$$\alpha_g:\m h\rightarrow \m h: h\mapsto ghg^{-1}.$$
In this way, \m h may be viewed
as an \m r-module where \m r may be taken
to be the group ring of \m g with integer
coefficients (or the subring of 
{\rm End}(\m h) generated by the 
endomorphisms $\alpha_g$, $g\in G$).
So, as a general philosophy,
to come up with a module which we can
associate to an abelian algebra \m a, we might
begin by looking for some pair $\lb \m g;\m h\rb$
associated to \m a where \m g is a group
and \m h is an abelian subgroup of \m g.

Our other method for constructing a module
begins with the following data: $\lb \m g;\tau\rb$
where \m g is a group and $\tau$ is an abelian
tolerance on \m g. Group tolerances
are identical with group congruences and each 
congruence corresponds uniquely to a normal subgroup;
abelian tolerances/congruences correspond to abelian 
normal subgroups. Hence, the data required for this construction
is the same as the data for the previous construction.
The tolerance $\tau$ may be viewed as a 
subgroup of $\m g\times \m g$ which contains the
diagonal. If \m d is the diagonal
subgroup, then we will construct our module on
$\tau/\m d$: 
the set of left cosets of \m d in the group $\tau$.

Let \m h be the abelian normal subgroup
of \m g that corresponds to $\tau$.
Each left coset of \m d contains a unique element
of the form $\lb h,1\rb$ where $h\in H$.
E.g., the unique element in $\lb a,b\rb D$ of this
form is $\lb ab^{-1},1\rb$. Hence we may 
put abelian group operations on
$\tau/\m d$ by mimicking the operations
of \m h: to add $\lb a,b\rb D$ to $\lb c,d\rb D$ we first rewrite
these as $\lb h,1\rb D$ and $\lb k,1\rb D$
where $h = ab^{-1}$ and $k=cd^{-1}$.
Then we define
$$\lb a,b\rb D+\lb c,d\rb D = \lb h,1\rb D +\lb k,1\rb D = \lb hk,1\rb D = \lb ab^{-1}cd^{-1},1\rb D.$$
Negation is defined by 
$$-\lb a,b\rb D = -\lb h,1\rb D = \lb h^{-1},1\rb D = \lb ba^{-1},1\rb D = \lb b,a\rb D$$
and 0 is defined to be $\lb 1,1\rb D = D$.
The abelian group laws for $\tau/\m d$ follow from those
of \m h. 

Clearly, $\tau$ acts by multiplication (on the left)
on the  set of left cosets of \m d.
This action is affine with respect to the
described abelian group operations, so 
if we restrict the action 
to the subgroup $\m d<\tau$, then we get that
\m d acts on these cosets by left multiplication
in a way that preserves 0. Hence, \m d acts as
a group of endomorphisms of the 
abelian group structure on $\tau/\m d$.
This gives us a module structure on $\tau/\m d$.
This module is essentially the
same as the one defined earlier;
since the diagonal subgroup $\m d<\tau<\m g^2$ is isomorphic to 
\m g, the abelian group structure on
$\tau/\m d$ is isomorphic to \m h, and 
\m d acts on $\tau/\m d$ ``essentially'' by conjugation.
By this, we mean that if $\lb g,g\rb \in D$ acts on the
coset $\lb h,1\rb D$ by left multiplication, we get 
$$\lb g,g\rb (\lb h,1\rb D) = \lb gh,g\rb D = \lb ghg^{-1},1\rb D.$$
So in fact, the module we have described is 
essentially the same as the one described previously.

We introduced the second construction for the following reason.
It is nearly true that to each abelian algebra
we can associate a pair $\lb \m g;\tau\rb$ where
\m g is a group and $\tau$ is an abelian tolerance on
\m g. What is actually true is that it is always 
possible to associate
a pair $\lb \m s;\tau\rb$ where \m s is a
monoid and $\tau$ is an abelian tolerance on \m s.
(Here an {\bf abelian tolerance} is just 
what it sounds like:
$\tau$ is an abelian tolerance on the monoid \m s
if $s(x,\bar{y})$ is an $(n+1)$-ary monoid term,
$\lb f, g\rb \in\tau$
and $\lb u_i, v_i\rb\in \tau$, then the implication
$$s(f,\bar{u}) = s(f,\bar{v}) 
\rightarrow s(g,\bar{u}) = s(g,\bar{v})$$
holds.)
The monoid we associate to 
the abelian algebra \m a is just
$\m s = \poln 1 a$. The tolerance on this
monoid is just the {\bf twin relation}:
$\lb f(x),g(x)\rb \in\tau$ if there is a term
$t(x,\bar{y})$ such that 
$f(x) = t^{\sm a}(x,\bar{a})$ and 
$g(x)= t^{\sm a}(x,\bar{b})$ for some $\bar{a},\bar{b}\in A^n$.
The fact that $\tau$ is an abelian tolerance
follows easily from the fact that \m a is abelian.

Now, we have pointed out that to any abelian algebra
one may associate a pair $\lb \m s;\tau\rb$ 
where \m s is a monoid and $\tau$ is an abelian
tolerance on \m s. If \m s happened to be a group,
then we already know how to construct a module from
this data; but, of course, \m s is a group 
only when \m a is trivial. Nevertheless, this 
often doesn't matter. To see this, let's examine
what happens when \m a is an affine algebra.
For the rest of this paragraph,
\m a is affine, \m s equals $\poln 1 a$, and $\tau$ is the twin
relation. Let $T$ equal the set of unary
polynomials of \m a of the form
$x+a$, $a\in A$. I.e., $T$ is the set of
additive translations. Composition of translations 
is an abelian group operation on $T$, and 
under composition $T$ is
isomorphic to the underlying 
additive semigroup of \m a.
It turns out that $\tau$ is the submonoid of
$\m s^2 = (\poln 1 a)^2$ consisting of
pairs of unary polynomials of the
form $\lb \alpha(x) + m, \alpha(x) + n\rb$, $m, n\in A$. 
Each pair in $\tau$ may be thought of as representing
a translation, since there is exactly 
one $t\in T$ such that $t(\alpha(x)+m)$ = 
$\alpha(x)+n$: it is $t(x) = x + (n-m)$.
However, different pairs in $\tau$
may represent the same member of $T$.
For example,
$\lb \alpha(x) + m, \alpha(x) + n\rb$
and $\lb \beta(x) + r, \beta(x) + s\rb$
represent the same element of 
$T$ if $m - n = r - s$.
We factor the monoid $\tau$
by the relation $\sim$ that identifies
two pairs if they represent the same
translation. If $D$ is the diagonal
of $(\poln 1 a)^2$, then it turns out that 
$\sim$ is precisely the left congruence
on the monoid $\tau$ which is generated by $D\times D$.
Furthermore, $D$ is a $\sim$-class,
since a pair $\lb f,g\rb$ is in $D$ iff
$f = g$ iff  $\lb f,g\rb\in\tau$
represents the identity translation.
We can define abelian group operations on
$\tau/\sim$ in the expected way. Let $0 = D/\sim$.
Let $-\lb f,g\rb/\sim$ = 
$\lb g,f\rb/\sim$. 
If $f$ or $g$ is the identity function and 
$h$ or $k$ is the identity function, then define
$$\left(\lb f,g\rb/\sim\right) + \left(\lb h,k\rb/\sim\right) := \lb fh,gk\rb/\sim.$$
It turns out that +, $-$ and 0 are well-defined 
total abelian group operations
on $\tau/\sim$. Since $\sim$ is a left
congruence, the monoid $\tau$ acts 
by left multiplication on
$\tau/\sim$. The diagonal
submonoid $\m d<\tau$ acts as a monoid of abelian group
endomorphisms. We get that
$$\lb \tau/\sim; +, -, 0, d (d\in D)\rb$$
is a \m d-module. This module is polynomially equivalent 
to the underlying module of \m a.
The first main point to observe is that
{\em everything works just as before
even though \m s is not a group}.
The second main point is that {\em this procedure
always produces the correct
representing module if \m a is affine.}
This suggests how we might proceed when \m a
is not (known to be) affine. The problem is to decide
how to define $\tau$, $\sim$, +, -, 0 and \m d
in this generality.

We now relax our assumption that \m a is affine to
the assumption that \m a is abelian.
We still define $\m s = \poln 1 a$  and $\tau$ = 
the twin relation.
Next, it seems like a good idea to choose
$\sim$ to be the left congruence on 
$\tau$ generated by $D\times D$. In the affine case,
$\sim$ is the unique left congruence
on $\tau$ which has $D$ as a congruence
class, but there may be many such
left congruences when \m a is merely abelian.
If we choose $\sim$ to be any left congruence
other than the one generated by $D\times D$, our
choice for $\sim$ might be incompatible
with some or all of the possible
representations of \m a.
By choosing the least left congruence on $\tau$
which has $D$ as a class, we
do not exclude any representations.
This choice gives us a set $\tau/\sim$
on which to build a module structure.
There is no difficulty in defining
0, negation or a left \m d-action on this set
just as we did in the last paragraph.
But our definition for addition does not
always give a well-defined total operation. 
(In some cases it will. For example,
if \m a is a commutative cancellative
monoid, then $\lb \tau/\sim; +, -, 0, d (d\in D)\rb$
is a total algebra equal to the 
universal group of \m a.)
At best, we can only construct what might be called
a ``partial module.''

We have just described a failed 
attempt to construct a representing
module
for an arbitrary abelian algebra. However, it 
indicates a good outline for a successful representation
if one adopts certain additional hypotheses chosen 
so that the partial module obtained is total.
Rather than work with all of $\poln 1 a$, one might
think to restrict the above discussion
to a sub{\it group} \m g of $\poln 1 a$.
If one restricts the twin relation $\tau$ 
to \m g, one obtains an abelian tolerance on \m g.
As remarked earlier, group tolerances are congruences and 
abelian tolerances correspond to abelian normal 
subgroups. Hence, if one restricts to a subgroup
\m g of $\poln 1 a$, then there {\it will be} an abelian
normal subgroup $\m h\triangleleft \m g$ such that 
any family of mutually twin polynomials in \m g lie
in the same coset of \m h. This leads to an expansion of
\m h to a module and it suggests a mapping
of \m a into a coset of \m h: take a binary term $xy$
and map $a\mapsto \rho_a = xa$ (provided that each $\rho_a\in G$).
This is a function from $A$ into $G$ whose
image lies in a single coset of \m h, since any pair
$\lb \rho_a,\rho_b\rb$ is a pair of twins. The abelian group
operations on \m h may be translated to the coset
of \m h containing all $\{\rho_a\;|\;a\in A\}$
or the image of $\varphi$ may be translated to the
coset $H$ where we may directly use 
the abelian group operations of \m h.
We have the beginnings of a representation.
Hence, we are led to consider the
case of an abelian algebra which has a binary term
$xy$ where the right translations $\rho_a$
generate a subgroup of $\poln 1 a$. We may take \m g
to be the group so generated and the twin relation,
$\tau$, restricted to \m g
will be an abelian congruence. The function
$\varphi(a) = \rho_a$ is a mapping of $A$ into a single
congruence class in \m g. We can translate 
the image of $\varphi$ into $H$
by fixing $u\in A$ and defining 
$\phi(x)$ = $\varphi(x)\varphi(u)^{-1}$.
That is, $\phi:A\rightarrow H:a\mapsto \rho_a\rho_u^{-1}$
for a fixed $u\in A$. The reader will recognize that this
paragraph describes the essential
features of first half of the argument for Theorem \ref{MAIN1}.

But, assuming that the right translations $\{\rho_a\;|\;a\in A\}$
generate a subgroup of $\poln 1 a$ is too restrictive.
Here is why we say this. If $\poln 1 a$ has a subgroup
\m g which contains all $\rho_a = xa\in\poln 1 a$,
then for each $a\in A$ the function $x\mapsto xa$
has a polynomial inverse. It is hard to imagine
that this would ever happen naturally
except when the term $xy$ is invertible in its
first variable. Here we say that a polynomial
$f(x,\bar{y})$ is {\bf invertible in its
first variable} if there is a polynomial
$g(x,\bar{y})$ such that 
$$\m a\models f(g(x,\bar{y}),\bar{y}) = x = g(f(x,\bar{y}),\bar{y}).$$
Furthermore, if we hope that our representation
is faithful, then we will want distinct $a, b\in A$ 
to yield distinct operations $\rho_a = xa$ and $\rho_b = xb$.
Using the term condition, it is easy to see that this
is equivalent to cancellativity of $xy$ in its
second variable.
But now, with $xy$ invertible in its first variable
and cancellative in its second variable, we can only be talking about
an algebra which looks dangerously similar
to a division quasigroup.
In particular, any algebra
which has a polynomial which is invertible in
two distinct variables has division quasigroup polynomials,
as is easy to see.
What makes this situation ``dangerous'' is that 
any abelian algebra with division quasigroup polynomials
is affine; hence, constructing a quasi-affine representation
for such an algebra is certainly a step backwards.
Thus, it is our desire to avoid
trivial cases that makes us reluctant to restrict our
attention to subgoups of $\poln 1 a$. So, 
instead of dealing with a subgroup of $\poln 1 a$,
one might choose to deal with a submonoid embeddable
in a group or, as we have done in this paper, with a subsemigroup
of $\poln 1 a$ that has
a ``good'' homomorphism into a group. These generalizations
introduce a new interesting problem. 
When one begins with a subgroup $\m g\subseteq \poln 1 a$,
then the restriction of $\tau$ to \m g
is a congruence which has a congruence class
equal to an abelian subgroup $\m h\triangleleft \m g$. 
One may proceed to define a representation on an expansion
of \m h. But if one begins only with a  
subsemigroup $\m s\subseteq \poln 1 a$ 
which has an abelian tolerance $\tau$ and one
embeds \m s into a group \m g, will $\tau$ extend 
to an abelian congruence on \m g? The answer is `not
necessarily.' It is this difficulty which motivated
our decision to let the binary term $t(x,y) = xy$ 
be a {\it central} binary term
in Theorem \ref{MAIN1}.
The assumption that $xy$ is a central
term came in quite handy in the proofs of every lemma
that led up to the proof of Theorem \ref{MAIN1},
but primarily it was chosen make Lemma \ref{abelian}
work. If one begins with an abelian algebra \m a
and a binary term $xy$, one can construct the
semigroup \m s generated by the right translations.
The twin relation $\tau$ on \m s
is an abelian tolerance. If one wants to proceed
along the lines we have described 
{\it without} the assumption that $xy$ is central,
then we need a homomorphism of \m s into a group
which has an abelian congruence whose restriction
to \m s contains $\tau$. This leads us to the following
question.

\medskip

\noindent
{\bf Question.} If \m s is a semigroup which is embeddable into a group \m g
and $\tau$ is an abelian  tolerance on \m s, then is it possible to embed
\m s into a group $\m g'$
which has an abelian congruence whose restriction to 
\m s contains $\tau$?

\medskip

In general, one cannot always take \m g = $\m g'$. As an example,
let \m s be the free semigroup on the set $X$, $|X|>1$. The relation $\tau$ which relates
any two words of \m s which have the same length 
is an abelian congruence on \m s. Now \m s is embeddable into
the free group \m g generated by the set $X$ 
by a semigroup homomorphism which is the identity
on $X$. But \m g has no non-trivial abelian congruence,
hence \m g has no abelian congruence whose restriction
to \m s contains $\tau$. If \m s is embeddable into
a group $\m g'$ which has an abelian congruence 
$\alpha$ whose restriction
to \m s contains $\tau$, then $\m g'$ 
may be chosen to be a group of fractions for \m s.
Now, since $\tau\subseteq \alpha|_{\sm s}$, there is an induced 
homomorphism $h:\m s/\tau\rightarrow\m g'/\alpha$ where
$h(S/\tau)$ generates $\m g'/\alpha$. But $\m s/\tau$ is 1-generated,
so $\m g'/\alpha$ is cyclic. It follows that 
$\m g'$ must have
an abelian congruence $\alpha$ such that $\m g'/\alpha$
is cyclic. A positive answer to the question we listed would imply,
therefore, that any free semigroup is embeddable into a group $\m g'$
which has an abelian normal subgroup \m h such that $\m g'/\m h$
is cyclic. This statement, which turns out to be true, is not at 
all obvious. It hints that a positive answer to our
question will not be easily proven. 

\bigskip

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\end{document}
