\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 ``rectangulation''
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 rectangulation concept
that we will call in this monograph ``strong
rectangulation''.  In \cite{k-kiss} we developed a
commutator theory for strong rectangulation, too. In this
chapter we explore these two theories further.  In
Chapter~\ref{rectangulation1} we will introduce a new
definition for McKenzie's original rectangulation 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 ``rectangulation'' in \cite{k-kiss} for what will
henceforth be called ``strong rectangulation''.

\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 $r\equiv_{\delta} s$
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 ---} 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.}  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.

\begin{exmp}\label{A(M)} Let $X$ be a set 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 $Y = \bigcup_{M\in {\vr M}} Y_M$. 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 $\m a({\vr M}) \!=\! \lb A; \{f_M\mid M\in {\vr M}\}\rb$,
$R = {\Tg{a({\vr M})}\big(\{(a_M,b_M)\mid M\in {\vr M}\}\big)}$,
$S = \Tg{a({\vr M})}\big(\{(u_M,v_M)\mid M\in {\vr
  M}\}\big)$, and $T = \Tg{a({\vr M})}(R\cup S)$.

\begin{clm}\label{A(M)_clm}
  The following hold.
  \begin{enumerate}
  \item[(1)] The set $M(R,S)$ consists only of the trivial
    $R,S$-matrices, namely those of the form
    \[
    \left[\begin{array}{cc} p&p\\ q&q
      \end{array}\right]\quad\quad\mbox{\rm and}\quad\quad
    \left[\begin{array}{cc} r&s\\ r&s
      \end{array}\right]
    \]
    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: 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 $M =
\left[\begin{array}{cc} p&q\\ r&s
\end{array}\right]= \left[\begin{array}{cc} 0&1\\ 1&0
\end{array}\right]$ 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 $\m a$, 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 $T\cap \big(((S\circ T)\cap\delta)\circ
    S\big) \subseteq \delta$, then $\C {SR}(S,T;\delta)$
    holds. If $T\cap\big(S\circ (T\cap\delta)\circ
    S\big) \subseteq\delta$ 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 $r\wrel{(S\circ
    T)\cap\delta}q$.  Hence $r\wrel{(S\circ
    T)\cap\delta} q\wrel{S} s$ and $r\wrel{T} s$ imply
  $(r,s)\in T\cap \big(((S\circ T)\cap\delta)\circ
  S\big)\subseteq \delta$.  Since $M$ was chosen
  arbitrarily, $\C {SR}(S,T;\delta)$ holds.  If now $T\cap
  \big(S\circ (T\cap\delta)\circ S\big) \subseteq\delta$
  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, $\m f = \m f_{\vr V}(x,y)$ the
  free $\vr V$-algebra over $X = \{x,y\}$, and $\theta = \Cg
  f(x,y)$. If\/ ${\bf Q}={\bf S}$ or\/ ${\bf SR}$, then 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 $\lambda =
  h^{-1}(\beta)$.  As \hbox{$\big(h(x),h(y)\big) =
    (a,b)\in\beta\setminus 0$} 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.  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 $E$ be an equivalence relation on $X$, and $\tau\colon
  X\to X$ 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 $\tau\colon X\to X$.
  \item[(2)] $\big(a,{h}(a)\big)\in T$ for any $a\in F$.
  \item[(3)] $T\circ T = \theta$.
  \item[(4)] If\/ $B := \{ u(\wec{x})\in F\mid \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 $\big(x_i,\tau(x_i)\big)$
  for every $i$, it also contains
  $t\big((x_1,\tau(x_1)),\ldots,(x_n,\tau(x_n))\big) =
  \big(t(\wec{x}),{h}(t(\wec{x}))\big) =
  \big(a,{h}(a)\big)$.

  For $(1)$, the fact that ${h}$ is induced by an idempotent
  function 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 $\big(a,h(a)\big)\in T\subseteq \theta$ 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 (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
  $\big(x,u(x_1,\ldots,x_n)\big)\in\theta = \ker({h})$ if
  and only~if
  \[
  u(x,x,\ldots,x)={h}\big(u(x_1,\ldots,x_n)\big)
  = {h}(x) = x
  \]
  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
  \[
  u\big(v(x_{11},\ldots,x_{1n}),
  \ldots,v(x_{n1},\ldots,x_{nn})\big)\,,
  \]
  and the fact that $(x_i,x_j)\in E\subseteq T$ for all $i$
  and $j$, we derive that
  \[
  \begin{array}{r@{}l} u(x_1,\ldots,x_n) &{} =
    u\big(v(x_1,\ldots,x_1),\ldots,v(x_n,\ldots,x_n)\big)\\
    &{} \equiv_T
    u\big(v(x_1,\ldots,x_n),\ldots,v(x_1,\ldots,x_n)\big)\\
    &{} = v(x_1,\ldots,x_n)\,.
  \end{array}
  \]
  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}

\begin{lm}\label{free_rect}
  Let\/ $\vr V$ be a variety and\/ $\m F = \m f_{\vr
    V}(x,y)$ the free $\vr V$-algebra generated by
  $\{x,y\}$.  Let $\theta = \Cg f(x,y)$ and $T$ the
  tolerance of\/~$\m F$ generated by $(x,y)$.  Let $B =
  x/\theta$.
  \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:
  \[
  \left[\begin{array}{cccccc}
      t(\wec{a},\wec{c})&
      t\big(\wec{a},h(\wec{c})\big)&=&
      t\big(\wec{a},h(\wec{d})\big)&
      t(\wec{a},\wec{d})\\
      \noalign{\vskip 1pt}
      t\big(h(\wec{a}),\wec{c}\big)&
      t\big(h(\wec{a}),h(\wec{c})\big)&=&
      t\big(h(\wec{a}),h(\wec{d})\big)&
      t\big(h(\wec{a}),\wec{d}\big)\\
      \Vert&\Vert&&\Vert&\Vert\\
      t\big(h(\wec{b}),\wec{c}\big)&
      t\big(h(\wec{b}),h(\wec{c})\big)&=&
      t\big(h(\wec{b}),h(\wec{d})\big)&
      t\big(h(\wec{b}),\wec{d}\big)\\
      \noalign{\vskip 1pt}
      t(\wec{b},\wec{c})&
      t\big(\wec{b},h(\wec{c})\big)&=&
      t\big(\wec{b},h(\wec{d})\big)&
      t(\wec{b},\wec{d})
    \end{array}\right]\,.
  \]
  We claim that the four elements in the middle are equal
  to~$x$. Indeed, the assumption $p, q, r, s\in B$ implies
  $h(p)=h(q)=h(r)=h(s)=x$. Therefore, for example,
  \[
  t\big(h(\wec{b}),h(\wec{d})\big)=
  h\big(t(\wec{b},\wec{d})\big)=h(s)=x\,.
  \]
  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.  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 $\C {SR}(T,T;\rho)$ 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
  rectangulation 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 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 every~$\tau_n$.
  Since $\tau_n$ is a symmetric, transitive relation and is
  closed under the application of unary polynomials, it is a
  congruence.

\begin{clm}
  $\tau_n\leq \tau_{n+1}$ for all $n$.
\end{clm}

The argument is by induction. It is clear that $\tau_0\leq
\tau_1$, since $\tau_0$ is the equality relation. Suppose
that $\tau_k\leq \tau_{k+1}$ holds for some $k$.  Suppose
that $(r,s)$ is a generator of $\tau_{k+1}$ because, say,
there is some
\[
\left[\begin{array}{cc} p&q\\ r&s\end{array}\right]
\in M(S,T)\,,
\]
such that $q\equiv_{\tau_k} r$ (reason (1) preceding the
statement of the lemma). Then since $\tau_k\leq \tau_{k+1}$
we have $q\equiv_{\tau_{k+1}} r$, which shows that the same
matrix witnesses that $(r,s)$ is a generator
for~$\tau_{k+2}$.  Since $\tau_{k+2}$ contains the
generators of $\tau_{k+1}$, we have $\tau_{k+1}\leq
\tau_{k+2}$, proving the claim.

\medskip

Since $\tau_n\leq \tau_{n+1}$ for every $n$, the union $\tau
= \bigcup_{n<\omega}\tau_n$ 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
$(r,s)\in\delta$. Thus $\delta$~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 can 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 $B = x/\theta$.  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, $\m F = \m f_{\vr V}(x,y)$ 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
  \[
  r\big((x,y),(y,x),(x,x),(y,y)\big) = 
  \big(r(x,y,x,y), r(y,x,x,y)\big)
  \]
  for some $4$-ary term~$r$. Therefore a typical $T,
  T$-matrix may be written as
  \[
  \left[\begin{array}{cc}
      g\big(\wec{r}(x,y,x,y),\wec{s}(x,y,x,y)\big)&
      g\big(\wec{r}(x,y,x,y),\wec{s}(y,x,x,y)\big)\\
      g\big(\wec{r}(y,x,x,y),\wec{s}(x,y,x,y)\big)&
      g\big(\wec{r}(y,x,x,y),\wec{s}(y,x,x,y)\big)
    \end{array}\right]\,.
  \]
  If we define $f(x_1,\ldots,x_6) :=
  g\big(\wec{r}(x_1,x_4,x_3,x_6),\wec{s}(x_5,x_2,x_3,x_6)\big)$,
  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}
    \big(f_i(w,w,w,z,z,z),f_i(z,z,w,w,w,z)\big) &=
    (q_i,r_i)\in\tau_n,
    \quad\mbox{\rm or}\label{side_conditions1}\\
    \big(f_i(w,z,w,z,w,z),f_i(w,w,w,z,z,z)\big) &=
    (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)] $p_i(x,y)\approx f_i(x,y,x,y,x,y)$,
    $q_i(x,y)\approx f_i(x,x,x,y,y,y)$,
  \item[] $r_i(x,y)\approx f_i(y,y,x,x,x,y)$, 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
  \item[] $q_i(x,y)\approx f_i(x,x,x,y,y,y)$ and
  \item[] $r_i(x,y)\approx f_i(y,y,x,x,x,y)$.
  \item[(v)] The identities in $\Sigma(p_i,q_i)$ for each
    $i$ whose side condition is (\ref{side_conditions2}),
    along with
  \item[] $p_i(x,y)\approx f_i(x,y,x,y,x,y)$ and
  \item[] $q_i(x,y)\approx f_i(x,x,x,y,y,y)$.
  \end{enumerate}
  The identities in (i) and (ii) express that the pairs
  $\big(r_i^{\circ}(w,z),s_i^{\circ}(w,z)\big)$ form a
  directed chain connecting $u$ to $v$, the identities in
  (iii) express that
  $\big(r_i^{\circ}(w,z),s_i^{\circ}(w,z)\big)$ 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
  $\big(r_i(w,z),s_i(w,z)\big)$ 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
  $\big(u(a,b),v(a,b)\big)$ 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 $(w,z)\in [T,T]_{\bf
    S} = \bigcup_{n<\omega} \tau_n$, that is,
  $(w,z)\in\tau_{n+1}-\tau_n$ 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 $\big(w(c,d),z(c,d)\big)=(c,d)$ 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
  every $1\leq i\leq n$, 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
  \[
  \begin{array}{r@{}l}
    a &{} = f(a,a,a,\ldots,a) \\
    &{} = f(b,a,a,\ldots,a) \\
    &{} = f(b,b,a,\ldots,a) \\
    &{} \hphantom{f(aaaaaa)}\vdots\\
    &{} = f(b,b,b,\ldots,b) = b,
  \end{array}
  \]
  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
that 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 $b\equiv_{\delta} c$.
\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 exist $(T,T;0)$-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 quasi-order 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
    quasi-order 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 quasi-order 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, $p(B,\ldots,B)=B$.  As $S$ 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
  \[
  f(a_1,\ldots,a_n)\geq_B f(b_1,\ldots,b_n)\,.
  \]
  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 our
assumption $a_i\geq_{B_i} b_i$, so there exist
$(T,T;0)$-pairs $(u_i^1,u_i^2),\ldots,(u_i^{m-1},u_i^m)$
such that $a_i=u_i^1$, $b_i=u_i^m$, 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 $m$ 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}
\begin{array}{r@{}lllcr@{}l} 
a_1= {}&u_1^1 & u_1^2 & u_1^3 & \cdots & u_1^m&{} =b_1\\
a_2= {}&u_2^1 & u_2^2 & u_2^3 & \cdots & u_2^m&{} =b_2\\
&&&\;\vdots &&&\\
a_n= {}&u_n^1 & u_n^2 & u_n^3 & \cdots & u_n^m&{} =b_n\\
\noalign{\vskip 1pt}
\hline
\noalign{\vskip 2pt}
f(\wec{a})= {}&f(\wec{u}^1) &
f(\wec{u}^2) & f(\wec{u}^3) & \cdots & f(\wec{u}^m) &{} =
f(\wec{b})
\end{array}
\end{equation}
Each pair $\big(f(\wec{u}^j),f(\wec{u}^{j+1})\big)$ is a
$(T,T;0)$-pair, since the set $S$ of all $(T,T;0)$-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
if and only if all $(T,T;0)$-pairs are trivial.

To prove item (4) observe that 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) \;\;=\;\; (\geq_B) \;\;\subseteq\;\;
(\geq_T)|_B \;\;\subseteq\;\; \big({\rm tr.cl.}(S)\big)|_B.
\end{equation}
But since $B$ is also assumed to be a block of $\Cg a(T)$ in
item (4), we get that ${\rm tr.cl.}(S|_B) = \big({\rm
  tr.cl.}(S)\big)|_B$, so all 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~$\m a$.  But each element of $\m a$ is
contained in a unique block of the congruence $T$, so
$\sim_T$ 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
    $\big(p(c_1,\dots,c_n),p(d_1,\dots,d_n)\big)$ 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
    \[
    f(b,c)\geq_B f\big(f(b,d),f(a,c)\big)\,.
    \]
  \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
    \[
    q\big(\wec{a},q(\wec{b},c)\big)
    \geq_B q\big(\wec{a},q(\wec{a},c)\big)\,.
    \]
  \item[$(5)$] Let $F$ be an $n$-ary idempotent polynomial
    of\/~$\m A$. Let\/ $B$ be a block of\/ $T$ containing
    elements $u_{ij}$ (where $1\le i,j\le n$) 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
    \[
    F\big(F(\wec{u}^1),F(\wec{u}^2),
    \dots,F(\wec{u}^n)\big)\geq_B w\,.
    \]
  \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. We get a $T,
  T$-matrix whose off-diagonal entries are
  $p(b_1,\dots,b_n)$ and $p(c_1,\dots,c_n)$.  If these
  entries are equal, as is assumed in (2), then
  $\big(p(c_1,\dots,c_n), p(d_1,\dots,d_n)\big)$ is a
  $(T,T;0)$-pair by definition.

  To prove~$(3)$, we will apply item~$(2)$ to the polynomial
  \[
  p(x_1, x_2, x_3, x_4) = 
  f\big(f(x_1, x_2), f(x_3, x_4)\big)\,.
  \]
  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
  \[
  p(b, b, c, c) = f\big(f(b,b), f(c,c)\big) 
  = f\big(f(b,c), f(b,c)\big) = p(b, c, b, c)
  \]
  we have that $\big(p(b, c, b, c),p(b, d, a, c)\big)$ is a
  $(T,T;0)$-pair. But $B$ is closed under $f$, since $f$ is
  idempotent and $B$ is a tolerance block, and
  $\big(f(b,c),f(f(b, d), f(a, c))\big)=
  \big(p(b, c, b, c),p(b, d, a, c)\big)$, so
  \[
  f(b,c) \geq_B f\big(f(b, d), f(a, c)\big)\,,
  \]
  as claimed.

  For item~$(4)$, let $t=q\big(\wec{a},q(\wec{b},c)\big)$
  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~$B$.  Apply item $(2)$ using $p(\wec{x},
  \wec{y}, z) = q\big(\wec{x}, q(\wec{y}, z)\big)$ and the
  $2n-1$ $(T, T)$-triples $(a_1, t; a_1)$, \ldots,
  $(a_{n-1}, t; a_{n-1})$, $(b_1, a_1; a_1), \ldots,
  (b_{n-1}, a_{n-1}; a_{n-1})$, $(c, q(\wec{b},c); c)$
  (these are $(T, T)$-triples, by item $(1)$).  Item~$(2)$
  guarantees that since
  \[
  \begin{array}{r@{}l}
    p(\wec{a},\wec{b},c) 
    &{} = q\big(\wec{a}, q(\wec{b},c)\big)\\
    &{} = t \\
    &{} = q(\wec{t},t)\\
    &{} = q\big(\wec{t}, q(\wec{a}, q(\wec{b},c))\big)\\
    &{} = p\big(\wec{t},\wec{a}, q(\wec{b},c)\big)
  \end{array}
  \]
  the pair $\big(q(\wec{a}, q(\wec{b},c)),q(\wec{a},
  q(\wec{a},c))\big) = \big(p(\wec{t},\wec{a},
  q(\wec{b},c)),p(\wec{a}, \wec{a}, c)\big)$ is a
  $(T,T;0)$-pair. Since these elements belong to $B$ we have
  \[
  q\big(\wec{a}, q(\wec{b},c)\big)\geq_B
  q\big(\wec{a}, q(\wec{a},c)\big)\,.
  \]
  This proves $(4)$.

  Finally we prove~$(5)$. We show that
  \[
  F\big(F(\wec{u}^1),\dots,F(\wec{u}^i),w,w,\dots,w\big)
  \geq_B w
  \]
  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 considered will be
  in~$B$, because $F$ is idempotent and $B$ is a $T$-block.

  As an inductive assumption suppose that
  \[
  F\big(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),w,w,\dots,w\big)
  \geq_B w
  \]
  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
  \[
  \begin{array}{r@{}l} 
    \wec{a} &{} = 
    \big(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),w,\dots,w\big)\,, \\
    \wec{b} &{} = 
    \big(u_{i1},\dots, u_{i(i-1)},u_{i(i+1)},\dots,u_{in}\big)\,,
  \end{array}
  \]
  and choose $c = u_{ii}$.  With these choices $q(\wec{b},c)
  = F(\wec{u}^i)$.  The conclusion of $(4)$ informs us that
  \[
  q\big(\wec{a}, q(\wec{b}, c)\big) =
  F\big(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),
  F(\wec{u}^i),w,\dots,w\big)
  \]
  is $\geq_B q\big(\wec{a}, q(\wec{a}, c)\big)$, which
  equals
  \[
  F\big(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),
  F(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}), \underline{u_{ii}},
  w,\dots,w) ,w,\dots,w\big)\,.
  \]
  Using the assumption $u_{ii}\geq_B w$, we see that this
  element is $\geq_B$-related to the element
  \[
  F\big(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),
  F(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}), \underline{w},
  w,\dots,w) ,w,\dots,w\big)\,.
  \]
  Using the inductive assumption twice we find that this
  element,
  \[
  F\big(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),
  \underline{F(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}), w,
    w,\dots,w)} ,w,\dots,w\big)\,,
  \]
  is $\geq_B$-related to
  \[
  F\big(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),
  \underline{w},w,\dots,w\big) \geq_B w\,.
  \]
  By the transitivity of $\geq_B$ on $B$ we get that
  \[
  F\big(F(\wec{u}^1),\dots,F(\wec{u}^{i-1}),F(\wec{u}^i),
  w,\dots,w\big)\geq_B w\,,
  \]
  which completes the proof 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:
  \begin{enumerate}
  \item[(i)] $u(x,y)\approx f_1(y,y,x,x,x,y)$,
  \item[(ii)] $f_m(y,x,x,x,y,y)\approx v(x,y)$,
  \item[(iii)] $f_{i}(x,x,x,y,y,y) \approx
    f_{i}(y,y,x,x,x,y)$, $1\le i\le m$,
  \item[(iv)] $f_i(y,x,x,x,y,y)\approx
    f_{i+1}(y,y,x,x,x,y)$, $1\le i \le m-1$.
  \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
  $\big(f(y,y,x,x,x,y),f(y,x,x,x,y,y)\big)$ for which $\vr
  V$ satisfies the identity $f(x,x,x,y,y,y)\approx
  f(y,y,x,x,x,y)$.  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)] $s(x,y)\approx f_1(y,y,x,x,x,y)$,
    \item[(ii)] $f_m(y,x,x,x,y,y)\approx x$,
    \item[(iii)] $f_{i}(x,x,x,y,y,y) \approx
      f_{i}(y,y,x,x,x,y)$, $1\le i\le m$,
    \item[(iv)] $f_i(y,x,x,x,y,y)\approx
      f_{i+1}(y,y,x,x,x,y)$, $1\le i \le m-1$,
    \item[(v)] $s(x,y)\approx g_1(y,y,x,x,x,y)$,
    \item[(vi)] $g_n(y,x,x,x,y,y)\approx y$,
    \item[(vii)] $g_{i}(x,x,x,y,y,y) \approx
      g_{i}(y,y,x,x,x,y)$, $1\le i\le n$,
    \item[(viii)] $g_i(y,x,x,x,y,y)\approx
      g_{i+1}(y,y,x,x,x,y)$, $1\le i \le n-1$\,.
    \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
  \[
  f(x_{i1},\dots,x_{in}) \approx f(y_{i1},\dots,y_{in})\,,
  \]
  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
  \[
  s(x,y)= f\big(f(x_{11},\dots,x_{1n}),\dots,
  f(x_{n1},\dots,x_{nn})\big)\,.
  \]
  The term $s$ is idempotent, since $f$ is.  Applying
  Lemma~\ref{ge_elementary}~(5) to $F = f$, a block~$B$ 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 imply that the identity
  \[
  s(x,y) \approx f\big(f(y_{11},\dots,y_{1n}),\dots,
  f(y_{n1},\dots,y_{nn})\big)
  \]
  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}
  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
  above theorem is true when $\delta = 0$, then we get that
  $\C C(\alpha, \alpha; 0)$ holds for the congruence $\alpha
  := \Cg {\m a/\delta} (T/\delta)$. 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), and write $\Delta$ for the
  congruence on $\m t$ generated by the set
  \[
  \big\{\lb(a,a), (b,b)\rb\mid (a,b)\in T\big\}\,.
  \]
  Define tolerances on $\m t$ by $R_1 := \{ \lb (a,b), (a,
  c)\rb \in T^2\mid (b,c)\in T\}$ and $R_2 := \{ \lb
  (a,c), (b, c)\rb \in T^2\mid (a, b)\in T\}$.

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

It is sufficient to prove this claim 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 that) $(x_i,y_i)=(x,y)$.  Partition
$N = \{1,\ldots,n\}$ into blocks $\{B_1, \ldots, B_5\}$,
some of which might be empty, as follows:
\begin{enumerate}
\item[(1)] $B_1$ consists of those $j\in N$ for which
  $(x_j,y_j)= (x,x)$,
\item[(2)] $B_2$ consists of those $j\neq i$ for which
  $(x_j,y_j)= (x,y)$,
\item[(3)] $B_3=\{i\}$,
\item[(4)] $B_4$ consists of those $j\in N$ for which
  $(x_j,y_j)= (y,x)$, and
\item[(5)] $B_5$ consists of those $j$ for which $(x_j,y_j)=
  (y,y)$.
\end{enumerate}
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
$f_i(z_1,z_2,z_3,z_4,z_5)$.  With this notation, 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~$f$.  Define unary polynomials of $\m t$~by
\[
p_i\big((x,y)\big) = 
f_i^{\m t}\big((a,a), (a,b), (x,y), (b,a), (b,b)\big)\,.
\]
The $i$-th Taylor identity ensures that $p_i\big((a,b)\big)$
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
$p_i\big((a,b)\big)\wrel{\Delta} p_i\big((a,c)\big)$, we get
that $p_i\big((a,c)\big)$ lies on the diagonal of $\m t$.
Since $p_i\big((a,b)\big)$ and $p_i\big((a,c)\big)$ have the
same first coordinate, and both lie on the diagonal, we get
$p_i\big((a,b)\big) = p_i\big((a,c)\big)$. 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 right hand
side in the $i$-th position. We show 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 $\C
C\big(\Cg t(R_1\cup R_2), \Delta; 0\big)$ holds. Suppose
that $(a, b)\in T$. 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 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 $i\colon
\m a\to \m t$, where $a\mapsto (a,a)$.  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 $\m d$. But $i\colon \m a\to \m d$ is
an isomorphism, so 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, proving Theorem~\ref{abel_tol=abel_cong}.
\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.
  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 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\big(\Cg a(S), \Cg a(T); \delta\big)
  \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
  \[
  \C C(\alpha,\beta; \delta)\mimplies \C
  C(\alpha\vee\delta, \beta\vee\delta; \delta)
  \]
  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 a nontrivial abelian
congruence interval.
\end{rem}

