\chapter{Strong Term Conditions}\label{strong_rectangulation1}

In this chapter we introduce 
the term conditions that define strongly
abelian and strongly rectangular congruences.
Our purpose is to show that a variety
satisfies a nontrivial idempotent
Maltsev condition if and only if it has
no member with a nonzero strongly abelian
congruence, or equivalently has no
member with a nonzero strongly rectangular
congruence.

\section{Varieties Omitting Strongly Abelian Congruences}\label{str_ab}

Ralph McKenzie introduced the notion
of ``strong abelianness'' in \cite{mckenzie3},
and, with David Hobby, \index{Hobby, D.}
showed in \cite{hobby-mckenzie}
that it is a natural and useful concept in the
study of finite algebras.
McKenzie also discovered a quite different
abelianness concept, called
``rectangularity'' in \cite{kiss-prohle},
which made sense for finite subdirectly
irreducible algebras with nonabelian
monolith, and which has applications
to the study of residually small varieties.

In \cite{k-kiss}, 
we required a commutator
theory associated to the strong abelianness concept.
In the process of developing one, we discovered
a new natural and useful type of abelianness 
resembling McKenzie's rectangularity concept that we 
will call in this monograph ``strong rectangularity''. 
In \cite{k-kiss} we developed a commutator theory
for strong rectangularity, too. In this 
chapter we explore these two theories further.
In Chapter~\ref{rectangulation1}
we will introduce a new definition for McKenzie's original 
rectangularity concept that makes sense for
any algebra, and initiate the development
of a commutator theory for it.

It is important to draw attention to the fact
that we used the term ``rectangularity'' 
in \cite{k-kiss} for what 
will henceforth be called ``strong rectangularity''.

\begin{df}
Let $S$ and $T$ be tolerances
on an algebra $\m a$, and let 
$\delta$ be a congruence on $\m a$.
If $q\equiv_{\delta} r$ implies that 
$r\equiv_{\delta} s$
% \equiv_{\delta} r \equiv_{\delta} s$
whenever
\begin{equation}\label{S,T-matrix2}
\left[\begin{array}{cc}
p&q\\
r&s
\end{array}\right]\in M(S, T),
\end{equation}
then we say that 
{\bf $\C {SR}(S,T;\delta)$ holds},
\index{$\C {SR}(S,T;\delta)$}
or 
{\bf $S$ strongly rectangulates $T$ modulo $\delta$}.
\index{rectangulation!strong ---}
We say that 
{\bf $\C {S}(S,T;\delta)$ holds},
\index{$\C {S}(S,T;\delta)$}
or {\bf $S$ strongly centralizes $T$ modulo $\delta$}
\index{centralizes!strongly ---}
%\index{centralizer relation!strong}
if $\C {SR}(S,T;\delta)$ and $\C C(S,T;\delta)$ 
both hold.

A tolerance $T$ is {\bf strongly rectangular}
\index{rectangular!strongly ---}\index{strongly rectangular}
if $\C {SR}(T,T;0)$ holds, and is
{\bf strongly abelian}\index{abelian!strongly ---}
\index{strongly abelian}
if $\C {S}(T,T;0)$ holds.
\end{df}

By interchanging the rows and columns
of matrices one sees that 
$\C {SR}(S,T;\delta)$ holds 
if and only if 
\[
q\equiv_{\delta} r\wrel{\miff}
p\equiv_{\delta} q \equiv_{\delta} r \equiv_{\delta} s
\wrel{\miff} p\equiv_{\delta} s.
\]
for every $S,T$-matrix in (\ref{S,T-matrix2}).
It follows from this 
that $\C {SR}(S,T;\delta)$ is equivalent
to $\C {SR}(T,S;\delta)$.

Now, and later, we must consider the 
issue of whether nonobvious relationships 
hold among the term conditions that we introduce.
For example, is every abelian tolerance
strongly abelian?\footnote{No.}
%The answer in this instance is affirmative,
%as any abelian group witnesses, however 
There is an all-purpose example for 
answering this type of question,
which may be used to demonstrate
the independence of any combination of the
term conditions that we define in this monograph.

%Where should this example go?
\begin{exmp}\label{A(M)}
Let $X$ be a set 
%not containing $0$
and let ${\vr M}\subseteq X^{2\times 2}$
be a set of $2\times 2$ matrices with entries in $X$.
For $M\in {\vr M}$, let $Y_M = \{a_M, b_M, u_M, v_M\}$.
Let $Y = \cup_{M\in {\vr M}} Y_M$. Let 
$A$ be the disjoint union $X\cup Y\cup \{0\}$.
For each $M = \left[\begin{array}{rl}
p&q\\r&s\end{array}\right]
\in {\vr M}$ define a binary
operation $f_M$ on $A$ by 
\[
f_M(x,y):=\left\{\begin{array}{ll}
p&\mbox{\rm if $(x,y)=(a_M,u_M)$;}\\
q&\mbox{\rm if $(x,y)=(a_M,v_M)$;}\\
r&\mbox{\rm if $(x,y)=(b_M,u_M)$;}\\
s&\mbox{\rm if $(x,y)=(b_M,v_M)$; and}\\
0&\mbox{\rm otherwise.}
\end{array}
\right.
\]
Let $\m a({\vr M}) = \lb A; \{f_M\wrel{\mid}M\in M\}\rb$,
$R = \Tg a(\{(a_M,b_M)\wrel{\mid}M\in M\})$, 
$S = \Tg a(\{(u_M,v_M)\wrel{\mid}M\in M\})$,  and
$T = \Tg a(R\cup S)$.

\begin{clm} \label{A(M)_clm}
The following hold.

\begin{enumerate}
\item[(1)] $M(R,S)$ consists only of the trivial
$R,S$-matrices, namely those of the form
%\begin{equation}
\[
\left[\begin{array}{cc}
p&p\\
q&q
\end{array}\right]\quad\quad\mbox{\rm and}\quad\quad
\left[\begin{array}{cc}
r&s\\
r&s
\end{array}\right]
\]
%\end{equation}
with $(p,q)\in R$ and $(r,s)\in S$, 
together with the closure of $\vr M$ under
interchanging rows and columns.

\item[(2)] $M(T, T)$ consists only of the trivial
$T,T$-matrices
together with the closure of $\vr M$ under
interchanging rows and columns
and under transpose. 
\end{enumerate}
\end{clm}

We leave the proof of this claim as an exercise,
guided by the following comments. 
\begin{enumerate}
\item[(i)] The trivial matrices come from 
using a projection
operation as the term in the 
matrix. 
\item[(ii)] Any nontrivial composition
of $f_M$'s is zero, so only $f_M(x,y)$
and $f_M(y,x)$ can be used to create 
a nontrivial matrix. However $f_M(y,x)$ does not
produce any nontrivial $R,S$-matrices.
\end{enumerate}
We also leave the verification of the following 
as an exercise: the claim remains true
if you replace $R, S$ and $T$ by the
congruences they generate.
And, as a final exercise:
%verify that 
any failure of any term condition
introduced in this monograph must
involve a nontrivial $R,S$ or 
$T,T$-matrix.

What this means is that when considering
relationships between term conditions
we may ignore any algebraic considerations and
think completely set-theoretically.
If it is possible to write down matrices
whose patterns of entries 
formally conflict with the satisfaction 
of one term condition but do not 
formally conflict with the satisfaction
of another, then it is possible to build
an algebra that has matrices with these properties
(subject to the restrictions
of (1) and (2) of Claim~\ref{A(M)_clm}).

For example, to show that there exists an
algebra that has an abelian tolerance
that is not strongly abelian, it is enough to observe
that any $T,T$-matrix of the form
%\[
%\left[\begin{array}{cc}
%p&q\\
%r&s
%\end{array}\right]=
$M = 
\left[\begin{array}{cc}
p&q\\
r&s
\end{array}\right]=
\left[\begin{array}{cc}
0&1\\
1&0
\end{array}\right]$
%\]
formally conflicts 
with the satisfaction of $\C {SR}(T,T;0)$ 
(since $q=r\neq s$), but
does not formally conflict with
$\C {C}(T,T;0)$ (even after interchanging the rows
or columns, or taking the transpose).
This suggests that we take $X = \{0,1\}$ and 
${\vr M} = \{M\}$. By Claim~\ref{A(M)_clm},
the nontrivial $T,T$-matrices 
of $\m a({\vr M})$ are 
\[
\left[\begin{array}{cc}
0&1\\
1&0
\end{array}\right]\quad\quad\mbox{\rm and}\quad\quad
\left[\begin{array}{cc}
1&0\\
0&1
\end{array}\right]\,.
\]
This implies that the tolerance
$T$ on $\m a({\vr M})$ is abelian but
not strongly rectangular
(hence not strongly abelian).
\end{exmp}

In Lemma~2.2 of \cite{k-kiss} we state and prove
a result for 
$\C {SR}({\hphantom{S}},{\hphantom{T}};{\hphantom{T}})$
similar to Theorem~\ref{basic_centrality}
for 
$\C {C}({\hphantom{S}},{\hphantom{T}};{\hphantom{T}})$.
These two results can be combined to produce a 
similar result for 
$\C {S}({\hphantom{S}},{\hphantom{T}};{\hphantom{T}})$.
Here we describe only the results about
the strong centrality relation and the strong
rectangulation relation 
that will be used in this chapter.

\begin{thm}\label{basic_strong_centrality}
Let $\m a$ be an algebra with tolerances 
$S, S', T, T'$
and congruences $\delta,\delta',\delta_j$.
Let ${\bf Q} = {\bf S}$ or ${\bf SR}$.
The following are true.
\begin{enumerate}
\item[(1)] If $\C Q(S,T;\delta)$ holds 
and $S'\subseteq S$,
$T'\subseteq T$, then $\C Q(S',T';\delta)$ holds.

\item[(2)] If $\C Q(S,T;\delta_j)$ holds 
for all $j\in J$, then
$\C Q(S,T;\bigwedge_{j\in J}\delta_j)$ holds.

\item[(3)] 
Let $\m b$ be a subalgebra of $\m a$.
If $\C Q(S, T; \delta)$ holds in $\m a$, 
%and $\m b$ is a subalgebra of $\m a$, 
then
$\C Q(S|_{\m b}, T|_{\m b}; \delta|_{\m b})$ holds in $\m b$.

\item [(4)] 
If $\delta'\leq \delta$, then the relation
$\C Q(S, T; \delta)$ holds in $\m a$ if and only if
$\C Q(S/\delta', T/\delta'; \delta/\delta')$ 
holds in $\m a/\delta'$.

\item[(5)] If $T\cap (((S\circ T)\cap\delta)\circ S)\subseteq \delta$,
then $\C {SR}(S,T;\delta)$ holds. If 
$T\cap(S\circ (T\cap\delta)\circ S) \subseteq\delta$
also holds, then $\C S(S,T;\delta)$ also holds.
\end{enumerate}
\end{thm}

\begin{proof}
Items (1)--(4) are proved in exactly the same
way as items (1), (6), (9) and (10) of
Theorem~\ref{basic_centrality}.

For item (5), choose any $S,T$-matrix
$M = \left[\begin{array}{cc} p & q\\ r& s\end{array}\right]$. 
Assume that $r\equiv_{\delta} q$. Since
the elements in the same row of $M$ are $T$-related
and the elements in the same column are $S$-related,
we have 
$r\wrel{S} p\wrel{T} q$ and 
$r\wrel{\delta} q$, so  $r\wrel{(S\circ T)\cap\delta}q$.
Since $r\wrel{(S\circ T)\cap\delta} q\wrel{S} s$ and 
$r\wrel{T} s$ we have 
$(r,s)\in T\cap (((S\circ T)\cap\delta)\circ S)\subseteq \delta$.
Since $M$ was chosen arbitrarily, 
$\C {SR}(S,T;\delta)$ holds. 
If now $T\cap(S\circ (T\cap\delta)\circ S) \subseteq\delta$
also holds, then $\C C(S,T;\delta)$ also holds by 
Theorem~\ref{basic_centrality}~(7).
Hence $\C S(S,T;\delta)$ also holds.
\end{proof}

Theorem~\ref{basic_strong_centrality}~(2)
implies that for {\bf Q} = {\bf S}
or {\bf SR} there is a least $\delta$
such that $\C Q(S,T;\delta)$ holds,
and we denote the least such $\delta$ 
by $[S,T]_{\bf Q}$.

\begin{lm}\label{4_eq_conditions}
Let $\vr V$ be a variety, let $\m f = \m f_{\vr V}(x,y)$
be the free $\vr V$-algebra over $X = \{x,y\}$, and
let $\theta = \Cg f(x,y)$. Let ${\bf Q}={\bf S}$
or ${\bf SR}$. The following are
equivalent.
\begin{enumerate}
\item[(1)] There exists an algebra $\m a\in \vr V$ 
that has a nonzero ${\bf Q}$-abelian congruence 
(i.e., a congruence $\alpha$ such that 
$[\alpha,\alpha]_{\bf Q}=0$).
\item[(2)] $(x,y)\notin [\theta,\theta]_{\bf Q}$.
\end{enumerate}
Now let $T$ be the tolerance on
$\m f$ generated by the pair $(x,y)$. The
following are equivalent.
\begin{enumerate}
\item[(3)] There exists an algebra $\m a\in \vr V$ 
that has a nonzero ${\bf Q}$-abelian tolerance 
(i.e., a tolerance $S$ such that 
$[S,S]_{\bf Q}=0$).
\item[(4)] $(x,y)\notin [T,T]_{\bf Q}$.
\end{enumerate}
\end{lm}

\begin{proof}
Both halves of the lemma are proved in
the same way, so we prove the first half only.

Assume that $\alpha$ is a nonzero ${\bf Q}$-abelian
congruence on $\m a$. Choose $(a,b)\in\alpha$ with
$a\neq b$. Then from
Theorem~\ref{basic_strong_centrality}~(3) and (1) 
we get that if $\m b$ is the subalgebra
of $\m a$ generated by $\{a,b\}$ and 
$\beta = \Cg b(a,b)$, then $\beta$
is ${\bf Q}$-abelian. Let $h\colon \m f\to\m b$
be the homomorphism sending $x$ to $a$ and $y$ to $b$.
This homomorphism is surjective since $\m b$
is generated by $a$ and $b$. 
Let $\kappa = \ker(h)$ and let $\lambda = h^{-1}(\beta)$. 
Since $(h(x),h(y)) = (a,b)\in\beta\setminus 0$ we have
$(x,y)\in\lambda\setminus \kappa$, which means that
$\theta\leq\lambda$ and $\theta\not\leq\kappa$.
From Theorem~\ref{basic_strong_centrality}~(4)
we know that $\C Q(\lambda,\lambda; \kappa)$
holds in $\m f$ if and only if 
$\C Q(\lambda/\kappa,\lambda/\kappa; 0)$
holds in $\m f/\kappa$. But it does,
since $\lambda/\kappa$ corresponds to $\beta$
under the natural isomorphism of $\m f/\kappa$
with $\m b$.
Therefore, by 
Theorem~\ref{basic_strong_centrality}~(1),
we derive that $\C Q(\theta,\theta; \kappa)$ holds.
We have $\C Q(\theta,\theta; \theta)$ by 
Theorem~\ref{basic_strong_centrality}~(5),
and therefore 
$\C Q(\theta,\theta; \kappa\wedge\theta)$ by 
Theorem~\ref{basic_strong_centrality}~(2).
This proves that 
$[\theta,\theta]_{\bf Q} \leq \kappa\wedge\theta < \theta$,
so $(x,y)\notin [\theta,\theta]_{\bf Q}$.

On the other hand, if $(x,y)\notin [\theta,\theta]_{\bf Q}$,
then Theorem~\ref{basic_strong_centrality}~(4)
guarantees that $\theta/[\theta,\theta]_{\bf Q}$ is a 
${\bf Q}$-abelian
congruence on $\m f/[\theta,\theta]_{\bf Q}$ that relates
the distinct elements $x/[\theta,\theta]_{\bf Q}$ and 
$y/[\theta,\theta]_{\bf Q}$. This shows that 
$\vr V$ contains an algebra with a nonzero
${\bf Q}$-abelian congruence.
\end{proof}

Our goal in this section is to prove that 
the four conditions of Lemma~\ref{4_eq_conditions}
are equivalent to each other, and that
when ${\bf Q} = {\bf S}$
the conditions 
hold for a variety if and only if
the variety has no Taylor term (Definition~\ref{taylor_term}). 
This is also true when ${\bf Q} = {\bf SR}$,
and that will be proved in the next section.
%This will be proved when ${\bf Q} = {\bf SR}$
%in the next section.)
To prove the equivalence of (1) and (2) with
(3) and (4) 
we start by analyzing the relationship between 
the congruence $\theta$ and the tolerance $T$
from Lemma~\ref{4_eq_conditions}.
This is done in a little more generality
than is immediately necessary, 
in that the next lemma considers
free algebras on more than two variables,
but this will allow us 
to avoid the need to generalize the
result later.


\begin{lm} \label{T_theta}
Let $\vr V$ be any variety and 
let $\m F = \m f_{\vr V}(X)$ 
be the free $\vr V$-algebra generated by $X$.
%Let $h\colon X\to X$ be a function
%satisfying $h\circ h = h$, and let
%$E$ be the equivalence relation
%on $X$ that is the kernel of $h$.
Let $E$ be an equivalence relation on $X$,
and let $\tau\colon X\to X$ be a transversal for $E$.
That is, $\tau$ is a function from $X$ to $X$
satisfying 
\begin{enumerate}
\item[(i)] $x\wrel{E} \tau(x)$, and 
\item[(ii)] $x\wrel{E}y\mimplies \tau(x)=\tau(y)$.
\end{enumerate}
Let $\theta = \Cg f(E)$ 
and $T = \Tg f(E)$.
The following hold.
\begin{enumerate}
\item[(1)] $\theta$ is the kernel of the endomorphism 
${h}\colon \m f\to \m f$ induced by 
$\tau\colon X\to X$.
% is $\theta$.
\item[(2)] $(a,{h}(a))\in T$ for any $a\in F$.
\item[(3)] $T\circ T = \theta$.
\item[(4)] If 
$B := \{ u(\wec{x})\in F\wrel{|}\mbox{\rm 
$u$ is an idempotent term}\}$, 
$E = X\times X$, 
and $\tau(X) = \{x\}$, 
then $B = x/\theta$ and $B$ is the unique
$T$-block containing $x$.
\end{enumerate}
\end{lm}

\begin{proof}
Observe that the fact that $\tau$ is a transversal 
for $E$ implies that $\tau\circ \tau = \tau$ and
$E = \ker(\tau)$.

We prove (2) first.
If $a\in F$, then $a = t(x_1,\ldots,x_n)$ for some
term $t$. Since $T$ is
a compatible relation containing  
$(x_i,\tau(x_i))$
% = (x_i,\widehat{h}(x_i))$ 
for every $i$, it also contains 
$t((x_1,\tau(x_1)),\ldots,(x_n,\tau(x_n))) 
%= (t(\wec{x}),t(h(\wec{x}))) 
= (t(\wec{x}),{h}(t(\wec{x}))) = (a,{h}(a))$.

For $(1)$, the fact that ${h}$ is induced by 
an idempotent function 
%$\{x,y\}\to \{x,y\}$
on generators
implies that ${h}$ is an idempotent endomorphism,
therefore ${h}$ is a transversal for 
$\ker({h})$. Since 
$\ker(\tau)\subseteq \ker({h})$ we have
$\theta\subseteq \ker({h})$.
It follows that every $\theta$-block 
contains at most one element of $\im(h)$.
On the other hand, every $\theta$-block
contains at least one element of $\im(h)$,
since if $a$ is in a $\theta$-block $B$, then 
$h(a)$ is certainly in $\im(h)$,
while $(a,h(a))\in T\subseteq \theta$
by (2). Thus $h(a)\in \im(h)\cap B$.
This shows that ${h}$ is also a transversal for 
$\theta$. Since comparable equivalence relations
with a common transversal are equal, $\theta = \ker({h})$.

For (3), $T\circ T \subseteq \Cg f(T) = \theta$,
so it is enough to show that $T\circ T\supseteq \theta$.
If $(a,b)\in\theta$, then ${h}(a) = {h}(b)$.
% by (1).
By (2), we have 
\[a\wrel{T} {h}(a) = {h}(b)\wrel{T} b\,,\]
proving that $(a,b)\in T\circ T$, indeed.

For (4), it is clear that 
$(x,u(x_1,\ldots,x_n))\in\theta = \ker({h})$
if and only if 
\[u(x,x,\ldots,x)={h}(u(x_1,\ldots,x_n)) = {h}(x) = x\] 
in $\m f$,
which holds if and only if $u$ is idempotent.
Thus $B = x/\theta$, as claimed.
If $u(x_1,\ldots,x_n)$ and 
$v(x_1,\ldots,x_n)$ are arbitrarily 
chosen idempotent terms, then 
using the term
\[
u(v(x_{11},\ldots,x_{1n}),\ldots,v(x_{n1},\ldots,x_{nn}))\,,
\]
and the fact that $(x_i,x_j)\in E\subseteq T$ for all $i$ and $j$,
we derive that
\[
\begin{array}{rl}
u(x_1,\ldots,x_n) &= 
u(v(x_1,\ldots,x_1),\ldots,v(x_n,\ldots,x_n))\\
&\equiv_T
u(v(x_1,\ldots,x_n),\ldots,v(x_1,\ldots,x_n))\\
&= v(x_1,\ldots,x_n)\,.
\end{array}
\]
Thus $B$ is contained in a $T$-block. That it is
the unique $T$-block containing $x$ follows from
the fact that $B = x/\theta$ is a $\theta$-block
(so no element in $F-B$ is $T$-related
to any element of $B$).
\end{proof}

%Next we 

\begin{lm}\label{free_rect}
Let $\vr V$ be a variety and 
let $\m F = \m f_{\vr V}(x,y)$ 
be the free $\vr V$-algebra generated by $\{x,y\}$.
Let $\theta = \Cg f(x,y)$ 
and $T$ be the tolerance of\/~$\m F$ 
generated by $(x,y)$.
Let $B = x/\theta$.
%\{ u(x,y)\in F\wrel{|}\mbox{\rm $u$ is an idempotent binary term}\}$.
\begin{enumerate}
\item[$(1)$] Let $\rho$ be a congruence on\/~$\m F$
such that $\C {SR}(T,T;\rho)$ holds. If 
\[
\left[\begin{array}{cc}
p&q  \\
r&s\end{array}\right]\in M(\theta,\theta)
\]
and $p, q, r, s\in B$, then $q\equiv_\rho r\implies
r\equiv_\rho s$.
\item[$(2)$] 
Let $\rho$ be a congruence on\/~$\m F$
such that $\C {S}(T,T;\rho)$ holds. If 
\[
\left[\begin{array}{cc}
p&q\\
r&s\end{array}\right]\in M(\theta,\theta)
\]
and $p,q,r,s\in B$, then $p\equiv_\rho q\implies
r\equiv_\rho s$.
\end{enumerate}
\end{lm}

\begin{proof}
Choose any $\theta,\theta$-matrix
\[
\left[\begin{array}{cc}
p&q\\
r&s\end{array}\right]
=
\left[\begin{array}{cc}
t(\wec{a},\wec{c})&
t(\wec{a},\wec{d})\\
t(\wec{b},\wec{c})&
t(\wec{b},\wec{d})
\end{array}\right]\,.
\]
Let $h\colon \m f\to\m f$ be the endomorphism 
determined by $x\mapsto x, y\mapsto x$.
The assumption $\wec{a}\equiv_\theta \wec{b}$ is equivalent
to $h(\wec{a})=h(\wec{b})$, and similarly 
$\wec{c}\equiv_\theta \wec{d}$ is equivalent
to $h(\wec{c})=h(\wec{d})$ (where by
$h(\wec{a})$, etc., we mean that we apply the
function~$h$ componentwise). 
Consider the
following~$4\times 4$ matrix:
\[
\left[\begin{array}{cccccc}
t(\wec{a},\wec{c})&
t(\wec{a},h(\wec{c}))&=&
t(\wec{a},h(\wec{d}))&
t(\wec{a},\wec{d})\\
t(h(\wec{a}),\wec{c})&
t(h(\wec{a}),h(\wec{c}))&=&
t(h(\wec{a}),h(\wec{d}))&
t(h(\wec{a}),\wec{d})\\
\Vert&\Vert&&\Vert&\Vert\\
t(h(\wec{b}),\wec{c})&
t(h(\wec{b}),h(\wec{c}))&=&
t(h(\wec{b}),h(\wec{d}))&
t(h(\wec{b}),\wec{d})\\
t(\wec{b},\wec{c})&
t(\wec{b},h(\wec{c}))&=&
t(\wec{b},h(\wec{d}))&
t(\wec{b},\wec{d})
\end{array}\right]\,.
\]
We claim that the four elements in the middle are
equal to~$x$. Indeed, the assumption that $p, q, r, s\in B$
implies that $h(p)=h(q)=h(r)=h(s)=x$. Therefore, for example,
\[t(h(\wec{b}),h(\wec{d}))=
h(t(\wec{b},\wec{d}))=h(s)=x\,.
\]
Note also that the four $2\times 2$ matrices in the
corners are $T,T$-matrices, since $\wec{u}$
and $h(\wec{u})$ are $T$-related
componentwise by Lemma~\ref{T_theta}~(2). 
All elements of this $4\times 4$ matrix are contained
in~$B$, because all elements are $\theta$-related,
the corners are in $B$ by assumption,
and $B$ is a block of~$\theta$.

Next, let $k\colon \m f\to \m f$ be the endomorphism
of $\m f$ that sends $x$ to $y$ and $y$ to $y$.
Consider the analogous $4\times 4$ matrix with respect
to~$k$. All the statements above remain true
with $k$ in place of $h$, with the
exception that in this case the four elements in the middle
are equal to~$y$, not~$x$. 
Now apply~$q(x,y)$ to these two
matrices, the first matrix in the first argument, the
second matrix in the second argument. This produces a
$4\times 4$ matrix which, after deleting one of 
the doubled middle columns and one of the doubled middle rows,
is a $3\times 3$ matrix with certain known entries:
\[
J = \left[\begin{array}{ccc}
p&i&q\\
j&q&\ell\\
r&m&s\\
\end{array}\right]\,.
\]
Indeed, the middle element is
$q(x,y)=q$, the element in the upper left corner is
$q(p,p)=p$, since $q$ is idempotent, and a similar argument
proves that in the other corners we get $q$, $r$ and~$s$. 
It will not matter in this argument 
what $i$, $j$, $\ell$, $m$ are.
% except that they belong to~$B$. 
The four $2\times 2$ matrices in 
the corners of the $3\times 3$ matrix $J$
are still $T,T$-matrices.

To prove~$(1)$, we have assumed that $q\equiv_\rho r$. 
We have also assumed that $\C {SR}(T,T;\rho)$ holds,
so from the $T,T$-matrix 
$\left[\begin{array}{cc} i&q\\q&\ell\end{array}\right]$
in the
upper right corner of $J$ we derive that $q\equiv_\rho \ell$. 
From the matrix 
$\left[\begin{array}{cc} j&q\\r&m\end{array}\right]$
in the lower left corner of $J$ 
(and the assumption that $q\equiv_\rho r$)
we get that $r\equiv_\rho m$. 
Thus $\ell\equiv_\rho q \equiv_\rho r \equiv_\rho m$, 
so since the cross diagonal entries of the $T,T$-matrix 
$\left[\begin{array}{cc} q&\ell\\m&s\end{array}\right]$
in the lower right corner 
are $\rho$-related 
we deduce that $m\equiv_\rho s$. Thus
$r\equiv_\rho m \equiv_\rho s$, and $(1)$ is proved.

To prove~$(2)$, we have assumed that $p\equiv_\rho q$. 
We have also assumed that 
$\C {S}(T,T;\rho)$ holds, and therefore 
$\C {RS}(T,T;\rho)$ and 
$\C {C}(T,T;\rho)$ both hold.
From the $T,T$-matrix 
$\left[\begin{array}{cc} i&q\\q&\ell\end{array}\right]$
in the upper right corner of $J$ we
get by strong rectangularity that $q\equiv_\rho \ell$. From the
$T,T$-matrix 
$\left[\begin{array}{cc} i&p\\q&j\end{array}\right]$,
which is obtained from the $T,T$-matrix
in the upper left corner of $J$ by 
switching columns, we get 
(from $p\equiv_{\rho} q$) that $q\equiv_\rho j$.
Now that we have 
$j\equiv_\rho q$
and 
$q\equiv_\rho \ell$ , we apply $\C C(T,T;\rho)$
to the matrices 
\[\left[\begin{array}{cc} j&q\\r&m\end{array}\right]
\quad\quad\mbox{\rm and}\quad\quad
\left[\begin{array}{cc} q&\ell\\m&s\end{array}\right]
\]
from the bottom two corners of $J$ to obtain 
$r\equiv_\rho m$
and 
$m\equiv_\rho s$. Therefore 
$r\equiv_\rho m \equiv_\rho s$, and (2) is proved.
\end{proof}

Our next task is to describe 
how to construct the ${\bf Q}$-commutator 
$[S,T]_{\bf Q}$ of two tolerances
where ${\bf Q} = {\bf S}$ or ${\bf SR}$.
Fix the choice of ${\bf Q} \in \{{\bf S}, {\bf SR}\}$.
Let $\m A$ be an algebra, and let $S$ and
$T$ be tolerances of~$\m A$. 
Inductively define binary relations $\tau_n$ 
on $\m a$ as follows.
Let $\tau_0$ be the equality relation.
If $\tau_n$ is defined, let $\tau_{n+1}$ be the symmetric
and transitive closure of the 
%union of $\tau_n$ and the 
set of all pairs that can be obtained
in one of the following two ways.
For each matrix
\[
\left[\begin{array}{cc}
p&q\\
r&s\end{array}\right]\in M(S,T)\,,
\]
\begin{enumerate}
\item[$(1)$] if ${\bf Q} = {\bf S}$ or ${\bf SR}$ and
$q\equiv_{\tau_n} r$, then put the pair
$(r, s)$ into $\tau_{n+1}$;
\item[$(2)$] if ${\bf Q} = {\bf S}$ and 
$p\equiv_{\tau_n} q$, then also put the pair
$(r, s)$ into $\tau_{n+1}$.
\end{enumerate}
Let $\tau = \bigcup_{n<\omega} \tau_n$.

\begin{lm}\label{strong_commutator}
$[S,T]_{\bf Q} = \tau$.
\end{lm}

\begin{proof}
The set of~$S,T$-matrices is closed under the
componentwise application of 
unary polynomials of $\m a$, so inductively
the same property holds for each of the $\tau_n$'s.
Since $\tau_n$ is a symmetric, transitive
relation and is 
closed under the application of unary polynomials,
it is a congruence. Since 
%The $S,T$-matrices of the form 
\[
\left[\begin{array}{cc}
p&q\\
p&q\end{array}\right]\in M(S,T)\,,
\]
it follows that, if $(p,q)\in\tau_n$, then
$(p,q)\in\tau_{n+1}$, by item (1) of the
definition of $\tau_{n+1}$ above. 
Therefore 
$\tau_n\leq \tau_{n+1}$ for every $n$, 
which implies that the union 
$\tau = \bigcup_{n<\omega}\tau_n$ 
of this ascending chain of congruences 
on $\m a$ is also a congruence on $\m a$.

Let $\delta = [S,T]_{\bf Q}$.  We argue by
induction that $\tau_n\leq \delta$
for all $n$.  This is trivially true if
$n=0$, so suppose that $\tau_n\leq \delta$
for some fixed $n$ and that $(r,s)$ 
is one of the generators of $\tau_{n+1}$. 
If $(r,s)\in\tau_{n+1}$ 
because of reason (1) preceding the statement of this
lemma, then there is a matrix
\[
\left[\begin{array}{cc}
p&q\\
r&s\end{array}\right]\in M(S,T)\,,
\]
such that 
$q\equiv_{\tau_n} r$.  We have $q\equiv_{\delta} r$
(since $\tau_n\leq\delta$) and also
$\C {Q}(S,T;\delta)$ (hence $\C {SR}(S,T;\delta)$)
since $\delta = [S,T]_{\bf Q}$. 
Therefore we have $(r,s)\in\delta$.
If $(r,s)\in\tau_{n+1}$
because of reason (2), then we must have
${\bf Q} = {\bf S}$ and so 
$\C C(S,T;\delta)$ holds. Now 
a similar argument shows 
that $(r,s)\in\delta$. This shows that $\delta$
contains all the generators of
$\tau_{n+1}$, so $\tau_{n+1}\leq \delta$.
By induction $\tau_n\leq \delta$ for all $n$,
so $\tau\leq\delta = [S,T]_{\bf Q}$.

To finish the proof we must verify that 
$[S,T]_{\bf Q}\leq \tau$, and for
this it suffices to show that $\C Q(S,T;\tau)$ holds.
Choose any $S,T$-matrix
\[
\left[\begin{array}{cc}
p&q\\
r&s\end{array}\right]\,,
\]
and assume that 
$q\equiv_{\tau} r$. Then $q\equiv_{\tau_n} r$
for some $n$, so from item (1) preceding
the lemma we have $(r,s)\in\tau_{n+1}\leq \tau$.
Thus $\C {SR}(S,T;\tau)$ holds, and 
$[S,T]_{\bf Q} = \tau$ when ${\bf Q} = {\bf SR}$.
If ${\bf Q} = {\bf S}$, then 
a similar argument using (2) in place of (1)
shows that $\C {C}(S,T;\tau)$ holds, so
$\C {S}(S,T;\tau)$ holds (and so
$[S,T]_{\bf Q} = \tau$ when ${\bf Q} = {\bf S}$).
\end{proof}

Now we are in position to prove that the four
conditions of Lemma~\ref{4_eq_conditions}
are equivalent.

\begin{lm}\label{strrect_cong}
Let $\vr V$ be any variety, and let 
${\bf Q} = {\bf S}$ or ${\bf SR}$. 
The following are equivalent.
\begin{enumerate}
\item[(1)] Some algebra in~$\vr V$ has a
nonzero ${\bf Q}$-abelian congruence.
\item[(2)] Some algebra
in~$\vr V$ has a nonzero ${\bf Q}$-abelian tolerance.
\end{enumerate}
\end{lm}

\begin{proof}
Denote by~$\m F$ the free $\vr V$-algebra generated by
$\{x,y\}$, by~$T$ the tolerance of~$\m F$ generated
by~$(x,y)$, and by~$\theta$ the congruence of~$\m F$
generated by $(x,y)$. Let $B = x/\theta$.
% denote the $T$-block
%represented by the idempotent terms.
According to Lemma~\ref{4_eq_conditions},
we must establish that 
$(x,y)\notin [\theta,\theta]_{\bf Q}$ if and only
if $(x,y)\notin [T,T]_{\bf Q}$. 
Since $x, y\in B$, this will follow if we show that 
the restrictions of the relevant congruences 
to $B$ are equal, i.e., if we show that, 
\[[\theta,\theta]_{\bf Q}|_B = 
[T,T]_{\bf Q}|_B.\]
We proceed to do this.

Since $T\subseteq \theta$ we have 
$[T,T]_{\bf Q}|_B\subseteq [\theta,\theta]_{\bf Q}|_B$,
so we only need to verify the reverse inclusion.
Let $\rho = [T,T]_{\bf Q}$. 
Build $\tau=[\theta,\theta]_{\bf Q}$ according 
to the recipe described preceding 
Lemma~\ref{strong_commutator}.
We argue by induction that 
$\tau_n|_B\subseteq \rho$.
This is clear for $n=0$. Suppose that $n\geq 0$
and that $\tau_n|_B\subseteq \rho$.
Choose any $(u,v)\in\tau_{n+1}|_B$. 
By the definition of~$\tau_{n+1}$,
there exists a chain $u=u_1,u_2,\dots,u_k=v$ such that for
every $1\le i< k$ the pair $(u_i, u_{i+1})$ or the pair
$(u_{i+1}, u_i)$ is produced by rule $(1)$ or~$(2)$ of the
definition of~$\tau_{n+1}$. That is, there exists a
$\theta,\theta$-matrix
\[
\left[\begin{array}{cc}
p&q\\
r&s\end{array}\right]
\]
such that $\{u_i, u_{i+1}\}=\{r,s\}$, and either $(q,r)$ or
$(p, q)$ is contained in~$\tau_n$. We want to show that
$u_i\equiv_\rho u_{i+1}$.

As $u\in B$ and $\tau\le\theta$, every~$u_i$ belongs
to~$B$. Hence the entries of the $\theta,\theta$-matrix
above are in~$B$, too. In the first case, when
$(q,r)\in\tau_n$ we therefore have that $(q,r)\in\rho$ by
the induction assumption. Then $(1)$ of
Lemma~\ref{free_rect} shows that $(r,s)\in\rho$ as
desired. In the second case, when $(p,q)\in\tau_n$
(which only arises when ${\bf Q} = {\bf S}$) we use
$(2)$ of Lemma~\ref{free_rect} to reach the same conclusion.
Thus the induction step, and hence the proof of the
lemma is complete.
\end{proof}

The next task we set for ourselves is 
to show that a variety has no member
with a strongly abelian tolerance if and only if the
variety has a Taylor term. To prove this we need
a description of the $T,T$-matrices where 
$T$ is the tolerance on $\m f_{\vr V}(x,y)$
generated by $(x,y)$.

\begin{lm}\label{ge_in_free}
Let $\vr V$ be a variety, $\m F = \m f_{\vr V}(x,y)$ 
be the free $\vr V$-algebra
generated by $\{x,y\}$, 
and let $T$ be the tolerance of\/~$\m F$
generated by~$(x,y)$. 
The $T,T$-matrices in~$\m F$ are exactly the
matrices of the form
\[
\left[\begin{array}{cc}
f(x,y,x,y,x,y)&
f(x,x,x,y,y,y)\\
f(y,y,x,x,x,y)&
f(y,x,x,x,y,y)
\end{array}\right]\,,
\]
where~$f$ is a sixary term.
\end{lm}

\begin{proof} 
A typical $T, T$-matrix has the form 
\[
\left[\begin{array}{cc}
g(\wec{a},\wec{c})&
g(\wec{a},\wec{d})\\
g(\wec{b},\wec{c})&
g(\wec{b},\wec{d})
\end{array}\right]
\]
where $(a_i,b_i)\in T$, $(c_j,d_j)\in T$, 
and (we may assume) $g$ is a term.
Since $T$ is the subuniverse of $F^2$
generated by $(x,y)$, $(y, x)$, $(x, x)$
and $(y, y)$, 
a typical pair in~$T$ has the form
\[
r((x,y),(y,x),(x,x),(y,y)) = (r(x,y,x,y), r(y,x,x,y))
\]
for some $4$-ary term~$r$. Therefore a
typical $T, T$-matrix may be written as
\[
\left[\begin{array}{cc}
g(\wec{r}(x,y,x,y),\wec{s}(x,y,x,y))&
g(\wec{r}(x,y,x,y),\wec{s}(y,x,x,y))\\
g(\wec{r}(y,x,x,y),\wec{s}(x,y,x,y))&
g(\wec{r}(y,x,x,y),\wec{s}(y,x,x,y))
\end{array}\right]\,.
\]
If we define
$f(x_1,\ldots,x_6) := g(\wec{r}(x_1,x_4,x_3,x_6),\wec{s}(x_5,x_2,x_3,x_6))$,
then this matrix is the one described in the lemma.
\end{proof}


\begin{lm}\label{strabel_tol}
If no algebra in the variety $\vr V$ has
a nonzero strongly abelian tolerance, then $\vr V$
satisfies a nontrivial idempotent Maltsev condition.
\end{lm}

\begin{proof}
Let~$\m F$ be the free $\vr V$-algebra generated by
$\{w,z\}$, let $T$ be the tolerance of~$\m F$ generated
by~$(w,z)$, and let $B$ be the $T$-block $w/\Cg f(w,z)$.
For each element $u\in F$ choose and fix
a binary term $u(x,y)$ such that $u(w,z)=u$.
Build $\tau =\bigcup_{n<\omega} \tau_n = [T,T]_{\bf S}$ 
using the method described before Lemma~\ref{strong_commutator}. 
For each pair $(u,v)\in \tau$ 
we shall define a finite set of equations $\Sigma(u,v)$.
If $(u,v)\in\tau_0$, let $\Sigma(u,v)=\{u(x,y)\approx v(x,y)\}$.
To define $\Sigma(u,v)$ when $(u,v)\in \tau_{n+1}-\tau_n$,
choose and fix a finite sequence 
$u = u_1, u_2, \ldots,u_{k+1} = v$ of elements of $F$ 
such that for each $i$ either $(u_i,u_{i+1})$ or 
$(u_{i+1},u_i)$ is a generating pair for 
$\tau_{n+1}$. Generating pairs for $\tau_{n+1}$
come from $T,T$-matrices, and we know from 
Lemma~\ref{ge_in_free} that the $T,T$-matrices
in~$\m F$ are exactly the matrices of the form
\[
\left[\begin{array}{cc}
p & q\\
r & s
\end{array}\right] = 
\left[\begin{array}{cc}
f(w,z,w,z,w,z)&
f(w,w,w,z,z,z)\\
f(z,z,w,w,w,z)&
f(z,w,w,w,z,z)
\end{array}\right]\,.
\]
Therefore for each $i$ between $1$ and $k$
there is a sixary term $f_i$ such that 
\begin{equation}\label{u's_and_f's}
\{u_i,u_{i+1}\} = \{r_i,s_i\} = \{f_i(z,z,w,w,w,z), f_i(z,w,w,w,z,z)\}
\end{equation}
where one of the following ``side conditions''
holds: 
\begin{align}
(f_i(w,w,w,z,z,z),f_i(z,z,w,w,w,z)) 
&= (q_i,r_i)\in\tau_n, \quad\mbox{\rm or}\label{side_conditions1}\\
(f_i(w,z,w,z,w,z),f_i(w,w,w,z,z,z)) &= (p_i,q_i)
\in\tau_n.\label{side_conditions2}
\end{align}
Let $(r_i^{\circ},s_i^{\circ}) := (r_i,s_i)$ if $(r_i,s_i)=(u_i,u_{i+1})$
and let 
$(r_i^{\circ},s_i^{\circ}) := (s_i,r_i)$ if $(r_i,s_i)=(u_{i+1},u_{i})$.
In either case, $r_i^{\circ}=u_i$ and $s_i^{\circ}=u_{i+1}$
in $\m f$.

Define $\Sigma(u,v)$ to be the set of the 
following identities.
\begin{enumerate}
\item[(i)] $u(x,y)\approx r_1^{\circ}(x,y)$ and 
$s_k^{\circ}(x,y)\approx v(x,y)$.
\item[(ii)]
$s_i^{\circ}(x,y)\approx r_{i+1}^{\circ}(x,y)$ for each $1\leq i\leq k$.
\item[(iii)] 
$p_i(x,y)\approx f_i(x,y,x,y,x,y)$, $q_i(x,y)\approx f_i(x,x,x,y,y,y)$, 
\hfill\linebreak
$r_i(x,y)\approx f_i(y,y,x,x,x,y)$, and $s_i(x,y)\approx f_i(y,x,x,x,y,y)$.
\item[(iv)]
The identities in $\Sigma(q_i,r_i)$ for each $i$ whose side 
condition is (\ref{side_conditions1}), along with 
$q_i(x,y)\approx f_i(x,x,x,y,y,y)$
and $r_i(x,y)\approx f_i(y,y,x,x,x,y)$.
\item[(v)] 
The identities in
$\Sigma(p_i,q_i)$ for each $i$ whose side condition 
is (\ref{side_conditions2}), along with 
$p_i(x,y)\approx f_i(x,y,x,y,x,y)$ and 
$q_i(x,y)\approx f_i(x,x,x,y,y,y)$.
\end{enumerate}
The identities in (i) and (ii) 
express that the pairs $(r_i^{\circ}(w,z),s_i^{\circ}(w,z))$
form a directed chain connecting $u$ to $v$,
the identities in (iii) express that the pair
$(r_i^{\circ}(w,z),s_i^{\circ}(w,z))$ is the bottom
row of a $T,T$-matrix, while the identities
of type (iv) and (v) express that the relevant
$T,T$-matrices satisfy side conditions sufficient
to guarantee that
$(r_i(w,z),s_i(w,z))$ is a generator of $\tau_{n+1}$.
These identities hold in $\vr V$,
since they become equalities if we substitute 
the free generators $w$ and $z$ of $\m f$ for 
the variables $x$ and $y$. 

Induction on $n$ may be used to prove
that $\Sigma(u,v)$ is a finite set of identities
for every $(u,v)\in \tau_{n+1}-\tau_n$, and that if
$u\in B$ then all terms appearing in
the identities in $\Sigma(u,v)$ are idempotent.
Moreover, induction on $n$ also shows
that if $(u,v)\in\tau$, then 
for any pair $(a,b)$ from any
tolerance $S$ on an algebra $\m a\in {\vr V}$
the pair $(u(a,b),v(a,b))$ belongs to 
the strong commutator $[S,S]_{\bf S}$.

We have assumed that no algebra in $\vr V$
has a nontrivial strongly abelian tolerance,
so by Lemma~\ref{4_eq_conditions} we have
$(w,z)\in [T,T]_{\bf S} = \bigcup_{n<\omega} \tau_n$,
or $(w,z)\in\tau_{n+1}-\tau_n$ for some $n$. Therefore
$\vr V$ satisfies the finite
set of idempotent identities in $\Sigma(w,z)$.
These identities constitute an idempotent
Maltsev condition which, when satisfied by
a variety $\vr W$, forces the following property:
for any pair $(c,d)$ from any
tolerance $R$ on an algebra $\m b\in {\vr W}$
the pair $(w(c,d),z(c,d))=(c,d)$ belongs to
the strong commutator $[R,R]_{\bf S}$.
Using the monotonicity of the strong commutator
(Theorem~\ref{basic_strong_centrality}~(1)),
this is equivalent to the property
that $(c,d)\in [\Tg b(c,d),\Tg b(c,d)]_{\bf S}$.
Using Lemma~\ref{4_eq_conditions} again we derive that
if this Maltsev condition
holds in $\vr W$, then no algebra in
$\vr W$ has a nontrivial strongly abelian tolerance.
Since every tolerance on every member of the variety
of sets is strongly abelian, this 
idempotent Maltsev condition fails in the variety
of sets, hence is nontrivial.
\end{proof}

\begin{thm}\label{str_ab_thm1}
Let $\vr V$ be a variety. The following are equivalent.
\begin{enumerate}
\item[(1)] $\vr V$ has no member with a nonzero
strongly abelian congruence.
\item[(2)] $\vr V$ has no member with a nonzero
strongly abelian tolerance.
\item[(3)] $\vr V$ satisfies a nontrivial 
idempotent Maltsev condition.
\end{enumerate}
\end{thm}

\begin{proof}
We proved 
in Lemma~\ref{strrect_cong}
that items (1) and (2) are equivalent,
and in Lemma~\ref{strabel_tol} that 
(2)$\mimplies$(3).
It remains to prove that (3)$\mimplies$(1).

Assume that (3) holds, and let $f(x_1,\ldots,x_n)$
be a Taylor term for $\vr V$. 
Arbitrarily choose and fix an algebra $\m a\in {\vr V}$,
a strongly abelian congruence $\alpha\in \Con(\m a)$,
and a pair $(a,b)\in \alpha$. 
Let $f_i(x,\wec{y})$ be an
$n$-ary term obtained from $f$ by reordering
the variables so that $x_i$ is first.
(That is, $f(x_1,\ldots,x_n) = f_i(x_i,\wec{y})$
where $\wec{y}$ is an ordering of 
$\{x_1,\ldots,x_n\}\setminus\{x_i\}$.)
The $i$-th Taylor identity implies that 
$f_i(a,\wec{u}) = f_i(b,\wec{v})$ for some
$(n-1)$-tuples $\wec{u}$ and $\wec{v}$ 
consisting of $a$'s and $b$'s. Therefore 
\[
\left[\begin{array}{cc}
p & q\\
r & s
\end{array}\right] = 
\left[\begin{array}{cc}
f_i(a,\wec{v}) & f_i(b,\wec{v}) \\
f_i(a,\wec{u}) & f_i(b,\wec{u}) 
\end{array}\right]\in M(\alpha,\alpha)
\]
is an $\alpha,\alpha$-matrix with $q = r$.
Since $\alpha$ is strongly abelian
it is strongly rectangular, so 
$f_i(a,\wec{u}) = r = s = f_i(b,\wec{u})$.
Now choose $\wec{w}\in \{a,b\}^{n-1}$ arbitrarily.
The matrix 
\[
\left[\begin{array}{cc}
P & Q\\
R & S
\end{array}\right] = 
\left[\begin{array}{cc}
f_i(a,\wec{u}) & f_i(b,\wec{u}) \\
f_i(a,\wec{w}) & f_i(b,\wec{w}) 
\end{array}\right]\in M(\alpha,\alpha)
\]
is an $\alpha,\alpha$-matrix with 
$P=Q$. Since $\alpha$ is strongly abelian
it is abelian, so 
$f_i(a,\wec{w}) = R = S = f_i(b,\wec{w})$.

What this says about the original Taylor term
is that, for any $1\leq i\leq n$, 
if $\wec{c}, \wec{d}\in \{a,b\}^n$ are
tuples that agree in every coordinate except
possibly the $i$-th, then $f(\wec{c}) = f(\wec{d})$.
This statement is true for all $i$, so
\[
\begin{array}{rl}
a &= f(a,a,a,\ldots,a) \\
&= f(b,a,a,\ldots,a) \\
&= f(b,b,a,\ldots,a) \\
& \hphantom{f(aaaaaa)}\vdots\\
&= f(b,b,b,\ldots,b) = b,
\end{array}
\]
forcing $a=b$. Since $(a,b)$ was chosen arbitrarily,
we conclude that no
$\m a\in {\vr V}$ has a nontrivial strongly
abelian congruence.
\end{proof}


\section{Join Terms}\label{join_term}
\index{join term|(}

In the previous section we showed that a variety
omits strongly abelian congruences or tolerances
if and only if it satisfies a nontrivial idempotent 
Maltsev condition (Theorem~\ref{str_ab_thm1}). 
In this section we will prove the the varieties
with these properties are exactly the varieties
that omit strongly rectangular 
congruences or tolerances.
The obvious approach to this result 
is to note that if a variety omits
strongly rectangular congruences or tolerances,
then it must omit
strongly abelian congruences or tolerances,
and therefore must satisfy a nontrivial
idempotent Maltsev condition. Thus,
the only thing to show is that it is possible to modify the 
proof of Theorem~\ref{str_ab_thm1}~(3)$\mimplies$(1)
so that it only uses the assumption of strong rectangulation.

It must be possible to obtain the result in
this way, but in this section we choose another path.
We instead develop machinery for dealing with
stongly rectangular congruences and tolerances
that makes the proof of the desired result a triviality.
Although the path we choose does not lead
to the result more quickly or more simply 
than the path sketched in the previous paragraph,
we feel that developing the machinery 
for strong rectangulation 
is more important than the target result.

\begin{df}\label{triples&pairs}
Let $\m a$ be an algebra, let 
$T$ be a tolerance on $\m a$ and
let $\delta$ be a congruence on $\m a$.
A {\bf $(T,T)$-triple}\index{$(T,T)$-triple} is a triple 
$(b,c;d)$ such that there is a matrix
\[
\left[\begin{array}{cc}
a&b\\
c&d\end{array}\right]\in M(T,T)\,.
\]
A {\bf $(T, T; \delta)$-pair}\index{$(T, T; \delta)$-pair} 
is a pair $(c,d)\in A^2$ such that there is a
$(T,T)$-triple $(b,c;d)$ with
$b\equiv_{\delta} c$.
%A {\bf $(T, T)$-pair} is a 
%$(T,T;0)$-pair.
\end{df}

Observe that the set of all $(T,T)$-triples
is a subuniverse of $\m a^3$, and the
set of all $(T,T;\delta)$-pairs is a subuniverse
of $\m a^2$.

\begin{df}\label{ge_defn}
Let $T$ be a tolerance on an algebra~$\m A$.  
For any block $B$ of $T$ write
$a\geq_B b$\index{$\ge_B$} (or $b\leq_B a$)
to mean that there are
$(T,T;0)$-pairs $(u_1,u_2), \ldots, (u_{n-1}, u_n)$
such that $a=u_1$, $b=u_n$, and $u_i\in B$ 
for all $i$. 
Write $a\sim_B b$\index{$\sim_B$}
to mean $a\geq_B b$ and $a\leq_B b$.

Write $a\ge_T b$\index{$\geq_T$} 
(or $b\le_T a$) if there is some $T$-block $B$
for which $a\geq_B b$. Similarly, write 
$a\sim_T b$\index{$\sim_T$}  
if $\exists B(a\sim_B b)$.
\end{df}

There is a potential ambiguity in the notation $\geq_B$, 
because a subset $B$ may be a block of
more than one tolerance, but this issue will not
arise in this monograph.

\begin{lm}\label{basic_prop}
Let $T$ be a tolerance on an algebra~$\m A$.
The following hold. 
\begin{enumerate}
\item[$(1)$] 
If $B$ is a block of $T$,
then $\geq_B$ is a
quasiorder on~$B$, and $\sim_B$
is the induced equivalence relation on~$B$. 
Both relations are compatible with the idempotent
polynomials of $\m a$.
\item[$(2)$] 
The relation $\geq_T$ is a 
reflexive compatible relation of~$\m A$, 
and $\sim_T$ is a tolerance on $\m a$.
\item[$(3)$] 
If $\C {SR}(T,T;0)$ holds, then 
$\ge_T$ is the equality relation on $\m a$.
\item[$(4)$] If $B$ is a block of $T$ and is also
a block of $\Cg a(T)$, then $\geq_B$ equals the restriction to $B$ 
of $\geq_T$ and also equals the restriction to $B$ 
of the transitive closure of
the set of $(T,T;0)$-pairs.
\item[$(5)$] If $T$ is a congruence, then $\ge_T$ is a
quasiorder of~$\m A$, which is the transitive closure of
the set of $(T,T;0)$-pairs. The relation $\sim_T$ is 
the equivalence relation induced by $\geq_T$,
which is a congruence on~$\m A$ contained in~$T$.
\end{enumerate}
\end{lm}

\begin{proof}
For this proof, let $S$ denote the set of $(T,T;0)$-pairs
of $\m a$. 

For item (1), the matrix $\left[\begin{array}{cc}
u&u\\
u&u\end{array}\right]$
belongs to $M(T,T)$ for any $u\in A$, so 
the trivial pair $(u,u)$ is a
$(T,T;0)$-pair for any $u\in A$. This is enough
to show that $\geq_B$ is reflexive on $B$.
It is clear from Definition~\ref{ge_defn}
that $\geq_B$ equals ${\rm tr.cl.}(S|_B)$, hence 
$\geq_B$ is transitive as well. This proves that
$\geq_B$ is a quasiorder on $B$; it follows from
Definition~\ref{ge_defn} that 
$\sim_B$ is the induced equivalence relation.
If $p(x_1,\ldots,x_n)$ is an idempotent polynomial
of $\m a$, then $p(B,\ldots,B)$ is a subset of $A$
containing $B$ and consisting of pairwise $T$-related
elements. Since $B$ is a block, $p(B,\ldots,B)=B$.
Since $S$ is compatible with all polynomials
of $\m a$, the fact that $p(B,\ldots,B)=B$
implies that $S|_B$ is compatible with $p$.
Therefore $\geq_B = {\rm tr.cl.}(S|_B)$ 
is also compatible with $p$, as is 
the intersection $\sim_B$
of this relation with its converse.

Now we prove item (2). Since $\geq_T$ 
equals the union
over all blocks $B$ of the relations $\geq_B$,
it follows that $\geq_T$ is reflexive. 
Similarly, $\sim_T$ is reflexive
and symmetric. It is only necessary to prove that 
$\geq_T$ and $\sim_T$ are compatible relations.
We will prove the following slightly
stronger fact.

\begin{clm}\label{compatibility_claim}
Let $f(x_1,\ldots,x_n)$ be an $n$-ary 
basic operation of $\m a$. Assume that 
$B_1, \ldots, B_n, B$ are $T$-blocks
such that $f(B_1,\ldots,B_n)\subseteq B$.
If $a_i\geq_{B_i} b_i$ holds for all $i$,
then 
$$
f(a_1,\ldots,a_n)\wrel{\geq_B} f(b_1,\ldots,b_n).
$$
The same statement is true with $\sim$ in place of $\geq$.
\end{clm}

The arguments are essentially
the same for $\geq$ and $\sim$, so we prove
the claim only for $\geq$.
By assumption $a_i\geq_{B_i} b_i$, so 
there are $(T,T;0)$-pairs 
$(u_i^1,u_i^2),\ldots,(u_i^{n-1},u_i^n)$
such that $a_i=u_i^1$, $b_i=u_i^n$,
and all $u_i^j$ lie in $B_i$.
There is no a priori reason to expect that
the number of such pairs is the same for all $i$,
but this can be arranged to be so by 
adding trivial pairs of the form $(u,u)$
to the end of any sequence of pairs that is
shorter than the longest sequence.
For each $i$, the pairs $(u_i^j,u_i^{j+1})$
define a chain of $n$ elements in $B_i$ 
connecting $a_i$ to $b_i$. 
We write these chains as rows of
an array, and apply $f$ to the columns of the array.
\begin{equation}\label{T,T;0-array}
%\left[
\begin{array}{rlllcrl}
a_1&=u_1^1 & u_1^2 & u_1^3 & \cdots & u_1^n&=b_1\\
a_2&=u_2^1 & u_2^2 & u_2^3 & \cdots & u_2^n&=b_2\\
&&&\vdots &&&\\
a_m&=u_m^1 & u_m^2 & u_m^3 & \cdots & u_m^n&=b_m\\
\hline
f(\wec{a})&=f(\wec{u}^1) & f(\wec{u}^2) & f(\wec{u}^3) & 
\cdots & f(\wec{u}^n) &= f(\wec{b}) 
\end{array}
%\right]
\end{equation}
Each pair $(f(\wec{u}^j),f(\wec{u}^{j+1}))$
is a $(T,T;0)$-pair, 
since the set $S$ of $(T,T;0)$-pairs is a subalgebra
of $\m a^2$. 
The block $B$ contains $f(\wec{u}^j)$ for all $j$, so the
bottom row of (\ref{T,T;0-array}) is a sequence
that witnesses the fact that $f(\wec{a})\geq_B f(\wec{b})$.

\medskip

The compatibility of $\geq_T$ follows
immediately from this claim. For example, if 
$a_i\geq_T b_i$ for $i=1,\ldots,n$, then there
exist $T$-blocks $B_i$ such that $a_i\geq_{B_i} b_i$
for all $i$. If $f$ is an $n$-ary basic operation of $\m a$,
then the set $f(B_1,\ldots,B_n)$ consists
of $T$-related elements, hence can be extended to
a block $B$. The claim now applies to prove that 
$f(\wec{a})\geq_B f(\wec{b})$, hence that 
$f(\wec{a})\geq_T f(\wec{b})$.

Item (3) follows from the fact that $\C {SR}(T,T;0)$
holds iff all $(T,T;0)$-pairs are trivial.

For item (4), for any $T$-block $B$
the transitive closure of $S|_B$ equals $\geq_B$, which
is contained in $\geq_T$, which is contained
in the transitive closure of $S$. 
If we restrict these relations to $B$ we get 
\begin{equation}\label{relations_are_equal}
{\rm tr.cl.}(S|_B) \;\wrel{=}\; (\geq_B) \;\wrel{\subseteq}\;
(\geq_T)|_B \;\wrel{\subseteq}\;
({\rm tr.cl.}(S))|_B.
\end{equation}
But since $B$ is also assumed to be a block of $\Cg a(T)$
in item (4), we get that 
${\rm tr.cl.}(S|_B) = ({\rm tr.cl.}(S))|_B$, 
so all of the relations in (\ref{relations_are_equal}) 
are equal.

For item (5), the fact that $T$ is an equivalence
relation containing $\geq_T$, together with the fact
proved in item (4) that $\geq_T$ is transitive
on any block of $T=\Cg a(T)$, implies that $\geq_T$
is transitive on $\m a$. Since $T$ contains 
the set $S$ of all $(T,T;0)$-pairs
and the relation $\geq_T$ agrees with 
${\rm tr.cl.}(S)$ on each block of $T$, it follows
that $\geq_T$ agrees with 
${\rm tr.cl.}(S)$ on $\m a$.
Since each element of $\m a$ is contained in a unique 
block of the congruence $T$, it follows that $\sim_T$ agrees with 
$\sim_B$ on any block of $T$. Thus $a\sim_B b$ holds
if and only if $a\geq_T b$ and $a\leq_T b$, which implies
that $\sim_T$ is the equivalence relation induced
by $\geq_T$. It is a congruence by item (2), and is contained
in $T$ because all $(T,T;0)$-pairs are contained in $T$.
\end{proof}

\begin{lm}\label{ge_elementary}
Let $T$ be a tolerance on an algebra~$\m A$. 
The following hold.
\begin{enumerate}
\item[$(1)$] If $(b,c)\in T$, then $(b,c;b)$ and $(b,c;c)$
are $(T,T)$-triples.
\item[$(2)$] Let $p$ be a $k$-ary polynomial, and suppose that
$(b_i,c_i;d_i)$ are $(T,T)$-triples for $1\le i\le k$. 
If $p(b_1,\dots,b_n)=p(c_1,\dots,c_n)$, then 
$(p(c_1,\dots,c_n),p(d_1,\dots,d_n))$ is a
$(T,T;0)$-pair.
\item[$(3)$] Let $f$ be a binary idempotent 
polynomial of~$\m A$, and assume
\[
\left[\begin{array}{cc}
a&b\\
c&d\end{array}\right]\in M(T,T)\,.
\]
If $B$ is a block of $T$ containing $a, b, c$ and $d$,
then 
\[f(b,c)\geq_B f(f(b,d),f(a,c))\,.\]
\item[$(4)$] 
Let $q$ be an $n$-ary idempotent 
polynomial of~$\m A$. If $B$ is a block
of $T$ containing elements 
$a_1, \ldots, a_{n-1}, b_1, \ldots, b_{n-1}$ 
and $c$, then 
\[
q(\wec{a},q(\wec{b},c))\geq_B
q(\wec{a},q(\wec{a},c))\,.
\]
\item[$(5)$] Let $F$ be an $n$-ary idempotent 
polynomial of~$\m A$. Let $B$ be a block of $T$
containing elements 
$u_{ij}$ ($1\le i,j\le n$) and $w$, and 
let $\wec{u}^i=(u_{i1},\dots, u_{in})$.
If $u_{ii}\geq_B w$ for $1\le i\le n$, then 
\[
F(F(\wec{u}^1),F(\wec{u}^2),
\dots,F(\wec{u}^n))\geq_B w\,.
\]
\end{enumerate}
\end{lm}

\begin{proof} 
Item $(1)$ follows from the fact that
\[
\left[\begin{array}{cc}
c&b\\
c&b\end{array}\right]
\qquad\text{and}\qquad
\left[\begin{array}{cc}
b&b\\
c&c\end{array}\right]
\]
are $(T,T)$-matrices if $(b,c)\in T$. 

To prove~$(2)$ suppose that 
$(b_i,c_i;d_i)$ are $(T,T)$-triples 
for $1\le i\le k$. They come from
some $(T,T)$-matrices
\[
\left[\begin{array}{cc}
a_i&b_i\\
c_i&d_i\end{array}\right]\,.
\]
Apply $p$ to these matrices componentwise. The 
resulting matrix is a $T, T$-matrix whose off-diagonal
entries are 
$p(b_1,\dots,b_n)$ and $p(c_1,\dots,c_n)$.
If these are equal, as is assumed in (2), then 
$(p(c_1,\dots,c_n),p(d_1,\dots,d_n))$ is a
$(T,T;0)$-pair by definition.

To prove~$(3)$, 
we will apply item~$(2)$ to the polynomial
\[
p(x_1, x_2, x_3, x_4) = f(f(x_1, x_2), f(x_3, x_4)).
\]
The $(T, T)$-triples that we will use are
$(b,b;b)$, $(b,c;d)$, $(c, b; a)$ and 
$(c, c; c)$. The first and last of these triples
are $(T, T)$-triples by item~$(1)$, the
second is a $(T, T)$-triple since 
we have assumed that
\[
\left[\begin{array}{cc}
a&b\\
c&d\end{array}\right]\in M(T,T), 
\]
and the third is a $(T, T)$-triple
since we can interchange the rows and columns
of this matrix to obtain that 
\[
\left[\begin{array}{cc}
d&c\\
b&a\end{array}\right]\in M(T,T).
\]
Item $(2)$ guarantees that since
\[
p(b, b, c, c) = f(f(b,b), f(c,c)) = 
%f(b,c) = 
f(f(b,c), f(b,c)) = p(b, c, b, c)
\]
we have that 
$(p(b, c, b, c),p(b, d, a, c))$ is
a $(T,T;0)$-pair. But $B$ is closed
under $f$, since $f$ is idempotent and $B$
is a tolerance block, and
$(f(b,c),f(f(b, d), f(a, c)))=(p(b, c, b, c),p(b, d, a, c))$,
so 
$$f(b,c) \geq_B f(f(b, d), f(a, c)),$$ 
as claimed.

For item~$(4)$, let $t=q(\wec{a},q(\wec{b},c))$
and write $\wec{t}$ for the $(n-1)$-tuple
$(t, t, \ldots, t)$.
Since $q$ is idempotent and all $a_i, b_j$ and $c$
belong to $B$, we get that both 
$t$ and $q(\wec{b},c)$ belong to $B$.
Now apply item $(2)$ using 
$p(\wec{x}, \wec{y}, z) = q(\wec{x}, q(\wec{y}, z))$ 
and the $2n-1$
$(T, T)$-triples 
$(a_1, t; a_1)$, \ldots, $(a_{n-1}, t; a_{n-1})$, 
$(b_1, a_1; a_1)$, \ldots, $(b_{n-1}, a_{n-1}; a_{n-1})$, 
$(c, q(\wec{b},c); c)$ 
(which are $(T, T)$-triples, by item $(1)$).
Item~$(2)$ guarantees that since
\[
\begin{array}{rl}
p(\wec{a},\wec{b},c) &= q(\wec{a}, q(\wec{b},c))\\
&= t \\
&= q(\wec{t},t)\\
&= q(\wec{t}, q(\wec{a}, q(\wec{b},c)))\\
&= p(\wec{t},\wec{a}, q(\wec{b},c)))
\end{array}
\]
the pair $(q(\wec{a}, q(\wec{b},c)),q(\wec{a}, q(\wec{a},c))) = 
(p(\wec{t},\wec{a}, q(\wec{b},c))),p(\wec{a}, \wec{a}, c))$
is a $(T,T;0)$-pair. Since these elements belong to $B$
we have 
$$q(\wec{a}, q(\wec{b},c))\geq_B q(\wec{a}, q(\wec{a},c)).$$
This proves $(4)$.

Finally we prove~$(5)$. We show that
\[
F(F(\wec{u}^1),\dots,F(\wec{u}^i),w,w,\dots,w)\geq_B w
\]
by induction on~$i$. This statement follows from
the idempotence of $F$ when $i=0$, 
and for $i=n$ yields the desired statement. 
All elements that we shall consider will be in~$B$, 
because $F$ is idempotent and $B$ is a $T$-block.
%In particular, which means that we are allowed
%to use $(4)$ and the transitivity of~$\ge_T$ as well. 
%%By permuting the
%%variables of~$F$ it remains idempotent, and therefore we are allowed to
%%use~$(4)$ not just in the last variable of~$F$, but in any variable.

As an inductive assumption suppose that
\[
F(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),w,w,\dots,w)\geq_B w
\]
holds.
Now apply~$(4)$ in the $i$-th variable. 
By this we mean cyclically
permute the last $n-i+1$ variables of $F$
to obtain an $n$-ary idempotent polynomial $q$ whose
last variable corresponds to the $i$-th variable of $F$:
\[
q(x_1,\ldots,x_n) := F(x_1,\ldots,x_{i-1},x_n,x_{i}, \ldots, x_{n-1}),
\]
then apply $(4)$ to $q$. 
Choose $(n-1)$-tuples
\[
\begin{array}{rl}
\wec{a} &= (F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),w,\dots,w), \\
\wec{b} &= (u_{i1},\dots, u_{i(i-1)},u_{i(i+1)},\dots,u_{in})\,,
\end{array}
\]
and choose $c = u_{ii}$.
With these choices $q(\wec{b},c) = F(\wec{u}^i)$.
The conclusion of $(4)$
%that 
%\[
%q(\wec{a}, q(\wec{b}, c))\ge_T q(\wec{a}, q(\wec{a}, c))
%\]
%holds, 
informs us that 
\[
q(\wec{a}, q(\wec{b}, c)) = 
F(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),F(\wec{u}^i),w,\dots,w)
\]
is $\geq_B q(\wec{a}, q(\wec{a}, c))$, which equals
\[
F(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),
F(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),
\underline{u_{ii}},
w,\dots,w)
,w,\dots,w).
\]
Using the assumption that
$u_{ii}\geq_B w$, we find that this element is $\geq_B$-related
to the element
\[
F(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),
F(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),
\underline{w}, w,\dots,w)
,w,\dots,w).
\]
Using the inductive assumption 
%(\ref{inductive_assumption})
twice we find that this element, 
\[
F(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),
\underline{F(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),
w, w,\dots,w)}
,w,\dots,w),
\]
is $\geq_B$-related to
\[
F(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),\underline{w},w,\dots,w)\geq_B w.
\]
By the transitivity of $\geq_B$ on $B$ we get that
\[
F(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),F(\wec{u}^i),w,\dots,w)\geq_B w,
\]
which completes the proof of 
%the inductive step, hence of 
$(5)$.
\end{proof}

\begin{df}\label{join_term_df}
A {\bf join term}\index{join term!defined}
for a variety ~$\vr V$ is
an idempotent binary term $s(x,y)$ such that 
for every~$\m A\in\vr V$, 
tolerance $T$ of~$\m A$, and pair $(u,v)\in T$
it is the case that $s(u,v)\ge_T u$ and
$s(u,v)\ge_T v$.
\end{df}

In order to investigate this concept we need to
understand the relation $\geq_T$ in
the generic setting.

\begin{lm}\label{ge_in_free2}
Let $\vr V$ be a variety, let 
$\m F = \m f_{\vr V}(x,y)$ 
be the free $\vr V$-algebra generated by $\{x,y\}$, 
let $T$ be the tolerance of\/~$\m F$
generated by~$(x,y)$, let $\theta = \Cg f(x,y)$,
and let $B = x/\theta$.
If $u, v\in B$, 
then $u\ge_T v$ if and only if there exists an $m\ge 1$, and
sixary terms $f_1,\dots,f_m$ such that $\vr V$ satisfies the
following identities:
% for $1\le i\le m-1$:
\begin{enumerate}
\item[(i)] $u(x,y)\wrel{\approx}f_1(y,y,x,x,x,y)$,
\item[(ii)] $f_m(y,x,x,x,y,y)\wrel{\approx} v(x,y)$,  
\item[(iii)] $f_{i}(x,x,x,y,y,y) \wrel{\approx} f_{i}(y,y,x,x,x,y)$,
$1\le i\le m$
\item[(iv)] $f_i(y,x,x,x,y,y)\wrel{\approx} f_{i+1}(y,y,x,x,x,y)$,
$1\le i \le m-1$.
\end{enumerate}
\end{lm}

\begin{proof} 
By Theorem~\ref{T_theta}~(4), 
$B$ is a block of the congruence $\theta$.
By Lemma~\ref{basic_prop}~$(4)$,
the restriction of
$\ge_T$ to $B$ equals the restriction of
the transitive closure 
of the set of $(T, T; 0)$-pairs to $B$.
It follows from Lemma~\ref{ge_in_free}
that the 
$(T,T;0)$-pairs are exactly the pairs of the form
$(f(y,y,x,x,x,y),f(y,x,x,x,y,y))$ for which $\vr V$ 
satisfies the identity
$f(x,x,x,y,y,y)\wrel{\approx} f(y,y,x,x,x,y)$.
Therefore, the identities enumerated in the
statement of this lemma 
express exactly that $(u,v)$ is in the transitive
closure of the set of $(T, T; 0)$-pairs.
\end{proof}

\begin{thm}\label{omit_1}
The following are equivalent for a variety~$\vr V$.
\begin{enumerate}
\item[$(1)$] $\vr V$ satisfies a
nontrivial idempotent Maltsev condition.
\item[$(2)$] $\vr V$ has a join term.
\item[$(3)$] $\vr V$ satisfies the following 
Maltsev condition: there exist $m\ge 1$, $n\ge 1$, 
idempotent sixary
terms $f_1,\dots,f_m$, $g_1,\dots,g_n$, and a binary
term~$s$ such that the following identities hold in~$\vr V$:
\begin{enumerate}
\item[(i)] $s(x,y)\wrel{\approx}f_1(y,y,x,x,x,y)$,
\item[(ii)] $f_m(y,x,x,x,y,y)\wrel{\approx} x$,  
\item[(iii)] $f_{i}(x,x,x,y,y,y) \wrel{\approx} f_{i}(y,y,x,x,x,y)$,
$1\le i\le m$
\item[(iv)] $f_i(y,x,x,x,y,y)\wrel{\approx} f_{i+1}(x,x,x,y,y,y)$,
$1\le i \le m-1$.
\item[(v)] $s(x,y)\wrel{\approx}g_1(y,y,x,x,x,y)$,
\item[(vi)] $g_m(y,x,x,x,y,y)\wrel{\approx} y$,  
\item[(vii)] $g_{i}(x,x,x,y,y,y) \wrel{\approx} g_{i}(y,y,x,x,x,y)$,
$1\le i\le n$
\item[(viii)] $g_i(y,x,x,x,y,y)\wrel{\approx} g_{i+1}(x,x,x,y,y,y)$,
$1\le i \le n-1$.
\end{enumerate}
\end{enumerate}
\end{thm}

\begin{proof}
To prove $(1)\mimplies (2)$ assume that a variety~$\vr V$
has a Taylor term $f$. The Taylor identities 
may be written in the form 
\[
f(x_{i1},\dots,x_{in}) \wrel{\approx}
f(y_{i1},\dots,y_{in})\,,
\]
where $1\leq i \leq n$, 
$x_{ij}, y_{ij} \in \{x,y\}$, 
and $x_{ii} = x$ while $y_{ii} = y$.
Consider
the binary term
\[
s(x,y)=
f(f(x_{11},\dots,x_{1n}),\dots,
f(x_{n1},\dots,x_{nn}))\,.
\]
The term $s$ is idempotent, since $f$ is.
Applying Lemma~\ref{ge_elementary}~$(5)$ 
to $F = f$, $B$ a block of some
tolerance $T$ on some $\m a\in {\vr V}$,
and elements $u, v\in B$, shows that 
$s(u,v)\geq_B u$. 
On the other hand, the Taylor identities
show that the identity
\[
s(x,y) \wrel{\approx}
f(f(y_{11},\dots,y_{1n}),\dots,
f(y_{n1},\dots,y_{nn}))
\]
holds in~$\vr V$, and so Lemma~\ref{ge_elementary}~$(5)$ shows
that $s(u,v)\geq_B v$ holds as well. Thus, $s$ is a join term
in~$\vr V$, and $(2)$ is proved.

Now suppose that a variety~$\vr V$ has a join term~$s$.
Then we have $s(x,y)\ge_T x$ and $s(x,y)\ge_T y$ in the free
algebra~$\m F = \m f_{\vr V}(x,y)$ 
for the tolerance~$T$ generated by $(x,y)$.
It follows from Lemma~\ref{ge_in_free2} 
that $\vr V$ satisfies the idempotent
Maltsev condition defined in~$(3)$. This proves that
$(2)\mimplies (3)$. This Maltsev condition 
expresses the fact that $s$ is a join term in~$\vr V$. The
variety of sets does not have a join term, since the
relation~$\ge_T$ is the equality relation in
sets. Therefore this Maltsev condition is nontrivial,
and $(3)\mimplies(1)$.
\end{proof}

According to Definitions~\ref{ge_defn}
and \ref{join_term_df}, $s(x,y)$ is a join
term for $\vr V$ if it is idempotent
and for every
$\m a\in {\vr V}$, tolerance $T$ on $\m a$,
and pair $(u,v)\in T$, there exist blocks
$B_1$ and $B_2$ such that $s(u,v)\geq_{B_1} u$
and $s(u,v)\geq_{B_2} v$. It is a consequence
of the previous theorem
that the blocks $B_1$ and $B_2$ may be chosen to be equal,
as we now explain.

\begin{cor}\label{omit_1_cor}
Let $\vr V$ be a variety with a join term $s(x,y)$.
If $\m a\in {\vr V}$, $T$ is a tolerance on $\m a$,
and $B$ is any $T$-block containing $u$ and $v$, then
$s(u,v)\geq_B u$ and $s(u,v)\geq_B v$.
\end{cor}

\begin{proof}
Since $B$ is a $T$-block, it is closed under all 
idempotent term operations of $\m a$, in particular
it is closed under $s$ and all the $f_i$ and $g_i$
from Theorem~\ref{omit_1}~(3). If we substitute 
$u$ for $x$ and $v$ for $y$ in these term operations
we obtain a sequence of elements of $B$ which,
according to the equations of that theorem,
witness that $s(u,v)\geq_B u$ and $s(u,v)\geq_B v$.
\end{proof}

\begin{thm}\label{str_ab_thm2}
Let $\vr V$ be a variety. The following are equivalent.
\begin{enumerate}
\item[(1)] $\vr V$ has no member with a nonzero
strongly rectangular congruence.
\item[(2)] $\vr V$ has no member with a nonzero
strongly rectangular tolerance.
\item[(3)] $\vr V$ satisfies a nontrivial 
idempotent Maltsev condition.
\end{enumerate}
\end{thm}

\begin{proof}
%We proved 
%in Lemma~\ref{strrect_cong}
%that items (i) and (ii) are equivalent.
%Since (i) implies that $\vr V$
%omits strongly abelian congruences,
%it follows from Theorem~\ref{str_ab_thm2}
%that (i)$\mimplies$(iii). 
%It remains to prove that 
As noted at the beginning of this section, 
the only part of this theorem that remains to be
proved is (3)$\mimplies$(1).

Assume that (3) holds, and let $s$
be a join term for $\vr V$. 
Choose an algebra $\m a\in {\vr V}$,
a strongly rectangular congruence $\alpha$
on $\m a$, and a pair $(a,b)\in \alpha$.
Since $s$ is a join term
we have $s(a,b)\ge_{\alpha} a$ and 
$s(a,b)\geq_{\alpha} b$. By Lemma~\ref{basic_prop}~(3)
the relation $\geq_{\alpha}$ is the equality relation
when $\alpha$ is strongly rectangular, so $a=b$.
Hence $\alpha = 0$ and (1) holds.
\end{proof}
\index{join term|)}

\section{Abelian Tolerances and Congruences}

In the earlier part of this chapter we saw
that a variety has a Taylor term if and only
if its members have no strongly abelian tolerances
or congruences. Taylor terms do not conflict
with the existence of nontrivial
(ordinary) abelian tolerances and congruences,
but do improve their behavior.
In this section we will see that if a variety
has a Taylor term, then abelian tolerances
generate abelian congruences, 
and also that a congruence interval
perspective with an abelian congruence
interval is itself abelian.

\begin{thm} \label{abel_tol=abel_cong}
Assume that $\vr V$ satisfies a nontrivial idempotent 
Maltsev condition.
If $\m a\in \vr V$ has a tolerance $T$
and a congruence $\delta$ such that 
$\C C(T, T; \delta)$ holds, then
$\C C(\Cg a(T), \Cg a(T); \delta)$ holds.
\end{thm}

\begin{proof}
It suffices to prove this theorem in the
case when $\delta = 0$, for if 
$\C C(T, T; \delta)$ holds, then
$\C C(T/\delta, T/\delta; 0)$ holds
in $\m a/\delta$, according to 
Theorem ~\ref{basic_centrality}~(10).
If the theorem is true when $\delta = 0$,
then we get that
$\C C(\alpha, \alpha; 0)$ holds for
$\alpha := \Cg {\m a/\delta} (T/\delta)$. But then
$\alpha = \Cg {\m a/\delta}(\Cg a(T)/\delta)$, too,
so again using Theorem ~\ref{basic_centrality}~(10)
we get that 
%\[
$
\C C(\Cg a(T), \Cg a(T); \delta)
$
holds, which is the statement of the theorem.
We therefore assume that $T$ is an abelian tolerance,
and make it our task to prove that
$T$ generates an abelian congruence.

Write $\m t$ for the subalgebra of $\m a^2$
whose universe is 
$A^2[T]$, which is just the set $T$ 
(cf.\ the first paragraph
of Chapter 2),
%$T$, 
and write $\Delta$
for the congruence on $\m t$ generated by 
the set 
\[
\{\lb(a,a), (b,b)\rb\wrel{|}(a,b)\in T\}.
\]
Define tolerances on $\m t$ by 
$R_1 := \{ \lb (a,b), (a, c)\rb \in T^2\wrel{|} (b,c)\in T\}$
and let 
$R_2 := \{ \lb (a,c), (b, c)\rb \in T^2\wrel{|} (a, b)\in T\}$.

\begin{clm}\label{disjoint_Delta}
$\Delta\cap R_i$ is the equality relation
for both $i= 1$ and $2$.
\end{clm}

It suffices to prove this for $i=1$ only,
so choose a pair $\lb (a,b), (a,c)\rb\in \Delta\cap R_1$.
We must prove that $b = c$.
Let $f(x_1,\ldots, x_n)$ be a Taylor term
for $\vr V$.
The $i$-th Taylor identity
has the form 
\[
f(x_1,\ldots, x_n) \approx f(y_1,\ldots, y_n)
\]
where $(x_j,y_j)\in \{(x,x), (x,y), (y,x), (y,y)\}$
for all $j$, and (we may assume) $(x_i,y_i)=(x,y)$.
Partition $N$ into blocks $\{B_1, \ldots, B_5\}$,
some of which might be empty, as follows:
$B_1$ consists of those $j\in N$ for which
$(x_j,y_j)= (x,x)$, $B_2$ consists of
those $j\neq i$ for which 
$(x_j,y_j)= (x,y)$, $B_3=\{i\}$,
$B_4$ consists of
those $j\in N$ for which
$(x_j,y_j)= (y,x)$, and $B_5$
consists of those $j$ for which
$(x_j,y_j)= (y,y)$.
Now substitute the variable $z_1$
for the variable $x_j$ of $f(x_1,\ldots,x_n)$
if $j\in B_1$, $z_2$ if $j\in B_2$, and so on.
This produces a $5$-ary 
term $f_i(z_1,z_2,z_3,z_4,z_5)$.
The $i$-th Taylor identity is now
$f_i(x,x,x,y,y)=f_i(x,y,y,x,y)$, where the third argument
of~$f_i$ corresponds to the $i$-th argument of~$f$.
%The construction of $f_i$ from $f$
%ensures that if we substitute an element $v$ into the $i$-th
%argument of~$f$, and $u$ to all other arguments, then the
%result is $f_i(u,u,v,u,u)$.
Define unary polynomials of $\m t$ by
\[
p_i((x,y)) = f_i^{\m t}((a,a), (a,b), (x,y), (b,a), (b,b)).
\]
The $i$-th Taylor identity ensures that 
$p_i((a,b))$ lies on the diagonal of $\m t$.
The property that $T$ is abelian is equivalent
to the property that the diagonal 
of $\m t$ is a union of $\Delta$-classes, so,
since $p_i((a,b))\wrel{\Delta} p_i((a,c))$,
we get that 
$p_i((a,c))$ lies on the diagonal of $\m t$.
Since $p_i((a,b))$ and $p_i((a,c))$ have the same
first coordinate, and both lie on the diagonal,
we get $p_i((a,b)) = p_i((a,c))$. Equating second
coordinates yields 
\[
f_i(\underline{a}, b, b, \underline{a}, b) = 
f_i(\underline{a}, b, c, \underline{a}, b).
\]
Since $(a,b), (b,c)\in T$ and $T$ is abelian, we 
still have equality after changing
the underlined $a$'s to $b$'s:
\[
f_i(\underline{b}, b, b, \underline{b}, b) = 
f_i(\underline{b}, b, c, \underline{b}, b).
\]
For the original Taylor term this means that
\[
b = f(b, b, \ldots, b) = f(b, \ldots, b, c, b, \ldots, b) 
\]
where only one $c$ occurs, and it occurs
on the righthand side in the $i$-th
position. We argue by induction that, for all $i$,
$b = f(c, c, \ldots, c, b, \ldots, b)$,
where $c$ occurs in the first $i$ positions and
$b$ occurs elsewhere. For $i=1$ this follows from
the last displayed line. For later values
of $i$ we apply the $T, T$-term condition
to change $b$ to $c$ in the following 
underlined positions:
\[
f(\underline{b}, \underline{b}, \ldots, \underline{b}, b, b, \ldots, b) =
f(\underline{b}, \underline{b}, \ldots, \underline{b}, c, b, \ldots, b).
\]
We obtain 
\[
f(\underline{c}, \underline{c}, \ldots, \underline{c}, b, b, \ldots, b) =
f(\underline{c}, \underline{c}, \ldots, \underline{c}, c, b, \ldots, b),
\]
which shows that the desired result with
$c$ in the first $i-1$  positions implies
the result for $c$ in the first $i$ positions.

Continuing to the end, we finally have
$b = f(c, c, \ldots, c) = c$ by the idempotence
of $f$. This proves the claim.

\medskip

\begin{clm}\label{abelian_Delta}
$\Delta$ is abelian.
\end{clm}

From Claim~\ref{disjoint_Delta} and 
Theorem ~\ref{basic_centrality}~(8) we get that
$\C C(R_i, \Delta; 0)$ holds for 
$i = 1$ and $2$. This and 
Theorem~\ref{basic_centrality}~(2) and (5) imply that
$\C C(\Cg t(R_1\cup R_2), \Delta; 0)$ holds. 
If $(a, b)\in T$, then
$(a,a)\wrel{R_1}(a,b)\wrel{R_2}(b,b)$,
so $\lb (a,a), (b,b)\rb\in \Cg t(R_1\cup R_2)$.
This shows that each generator of $\Delta$ belongs to 
$\Cg t(R_1\cup R_2)$, so 
$\Cg t(R_1\cup R_2)\geq \Delta$. 
From the monotonicity of the centralizer in
its first two variables
%Theorem ~\ref{basic_centrality}~(1)
we get that $\C C(\Delta, \Delta; 0)$ holds
in $\m t$. This proves the claim.

\medskip

Let $\m d$ be the image of 
the diagonal
embedding $i\colon \m a\to \m t\colon a\mapsto (a,a)$.
Since $\Delta$ is an abelian congruence of $\m t$,
it follows from Theorem ~\ref{basic_centrality}~(9)
that that $\Delta|_{\m d}$ is an abelian
congruence of $\m d$. Since $i\colon \m a\to \m d$
is an isomorphism, the congruence
$i^{-1}(\Delta|_{\m d})$ is an abelian
congruence of $\m a$. 
Since $i(T)$ is the generating
set for $\Delta$, it follows that
$T\subseteq i^{-1}(\Delta|_{\m d})$, 
and therefore  $T$ is contained in the abelian 
congruence $i^{-1}(\Delta|_{\m d})$. 
Hence $\Cg a(T)$ is abelian.
\end{proof}

\begin{thm}\label{perspective_abelian_taylor}
Assume that $\vr V$ satisfies a nontrivial idempotent 
Maltsev condition and that $\m a\in \vr V$. 
If $\alpha, \beta\in \Con(\m a)$, then 
\[
\C C(\alpha,\alpha;\alpha\wedge\beta)
\wrel{\Longleftrightarrow}
\C C(\alpha\vee\beta,\alpha\vee\beta;\beta).
\]
\end{thm}

\begin{proof}
If 
$\C C(\alpha\vee\beta,\alpha\vee\beta;\beta)$ 
holds, then 
$\C C(\alpha,\alpha;\beta)$ holds, by 
monotonicity.
%Theorem ~\ref{basic_centrality}~(1).
From this and $\C C(\alpha,\alpha;\alpha)$
we get that 
$\C C(\alpha,\alpha;\alpha\wedge \beta)$ holds, by 
Theorem ~\ref{basic_centrality}~(4).

Now assume that 
$\C C(\alpha,\alpha;\alpha\wedge \beta)$ holds.
Using Theorem ~\ref{basic_centrality}~(4) again
we derive that $\C C(\alpha,\alpha;\beta)$ holds.
By Theorem~\ref{basic_centrality}~(8), 
$\C C(\beta,\alpha;\beta)$ holds, so we get that
$\C C(\alpha\vee\beta,\alpha;\beta)$
and then 
$\C C(\alpha\vee\beta,\beta\circ \alpha\circ \beta;\beta)$
hold by parts (5)
and (3)
of Theorem~\ref{basic_centrality}.
This and 
%Theorem~\ref{basic_centrality}~(1)
the monotonicity of the centralizer
yield  that $\C C(T, T;\beta)$ holds for 
$T = \beta\circ\alpha\circ\beta$.
Since $\Cg a(T) = \alpha\vee\beta$ we get 
$\C C(\alpha\vee\beta,\alpha\vee \beta;\beta)$ holds
from Theorem~\ref{abel_tol=abel_cong}.
\end{proof}


\begin{rem}
It is not possible to strengthen the implication in
Theorem~\ref{abel_tol=abel_cong} to
\begin{equation}\label{C(S,T;d)_imp}
\C C(S, T; \delta)\mimplies
\C C(\Cg a(S), \Cg a(T); \delta)
\end{equation}
assuming only that $\vr V$ has a
Taylor term. If implication~(\ref{C(S,T;d)_imp})
holds for $\vr V$, then arguing as
in the proof of Theorem~\ref{perspective_abelian_taylor}
we would have 
%\begin{equation}
%\label{C(a+d,b+d;d)_imp}
\[
\C C(\alpha,\beta; \delta)\mimplies
\C C(\alpha\vee\delta, \beta\vee\delta; \delta)
\]
%\end{equation}
for any $\alpha, \beta, \delta\in \Con(\m a)$, 
$\m a\in \vr V$.
In particular, since $\C C(\alpha,\beta;\delta)$
holds whenever $\alpha, \beta$ and $\delta$ generate
a sublattice isomorphic to $\m n_5$, labeled as in 
Figure~\ref{fig4}, 
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(20,30)
\put(0,15){\circle*{1.2}}
\put(10,0){\circle*{1.2}}
\put(10,30){\circle*{1.2}}
\put(20,10){\circle*{1.2}}
\put(20,20){\circle*{1.2}}

\put(10,0){\line(-2,3){10}}
\put(10,0){\line(1,1){10}}
\put(10,30){\line(-2,-3){10}}
\put(10,30){\line(1,-1){10}}
\put(20,20){\line(0,-1){10}}

\put(-4.5,13){$\beta$}
\put(22,9){$\delta$}
\put(22,19){$\alpha$}
\end{picture}
\end{center}
\caption{ \ }\label{fig4}
\end{figure}
we would also 
have $\C C(\alpha\vee\delta, \beta\vee\delta; \delta)$,
and therefore $\C C(\alpha, \alpha; \delta)$
by the monotonicity of the commutator.
Altogether this shows that if $\vr V$
has a Taylor term and satisfies
implication~(\ref{C(S,T;d)_imp}),
then the critical quotient of any
$\m n_5$ in a congruence lattice
of a member of $\vr V$ is abelian.
Thus, for example, the variety of semilattices
has the Taylor term $f(x,y) = x\wedge y$,
but cannot satisfy implication~(\ref{C(S,T;d)_imp})
since some semilattices have sublattices
isomorphic to $\m n_5$
in their congruence lattices but no semilattice
has an abelian congruence interval.
%We would get the same result if we replace
%$\alpha$ and $\delta$ by 
%$\alpha'$ and $\delta'$ where
%$\delta\leq \delta ' < \alpha' \leq \alpha$,
%since $\{\alpha', \beta, \delta'\}$ still generate
%a copy of $\m n_5$. This shows that 
%implication~(\ref{C(a+d,b+d;d)_imp})
%forces every subinterval of the critical quotient 
%of a pentagon in $\Conb(\m a)$ to be abelian. 
%If $\vr V$ has a Taylor term, then it will follow from 
%Theorem~\ref{taylor_forbidden}~(2) that no
%sublattice of the interval $I[\delta,\alpha]$
%is isomorphic to $\m n_5$. 
%%This proves that
%%the critical interval of any pentagon
%%$\Conb(\m a)$ has no sublattice isomorphic
%%to $\m n_6$. ********
%Altogether this shows that if $\vr V$ has a Taylor
%term and satisfies implication~(\ref{C(S,T;d)_imp}),
%then the critical interval of any pentagon
%%$\m n_6$ does not appear as a sublattice
%in $\Conb(\m a)$ is modular 
%for any $\m a\in \vr V$.
%It will follow from 
%Theorem~\ref{main1} (3)$\mimplies$(2)
%that the variety $\vr V$ must satisfy an
%idempotent Maltsev condition much stronger than
%than the one asserting that $\vr V$ has a Taylor term.
\end{rem}



%\begin{thm}
%Assume that $\vr V$ satisfies a nontrivial idempotent 
%Maltsev condition. If $\vr V$ contains an algebra
%with a nonzero abelian congruence, then
%it contains a subdirectly irreducible algebra
%with abelian monolith.
%\end{thm}
%
%\begin{proof}
%
%\end{proof}












