\chapter{Rectangulation}\label{rectangulation1}
\index{rectangulation|(}

In this chapter we introduce 
a term condition for rectangulation and
prove a representation theorem for rectangular
tolerances. We explain how
the structure of rectangular 
tolerances improves in the presence 
of a join term. We give three different
Maltsev conditions that define the class of varieties that
omit rectangular tolerances.

\section{Rectangular Tolerances}

\begin{df}\label{rectangulation_defn}
Let $S$ and $T$ be tolerances
on an algebra $\m a$, let
$\sqsupseteq$ be a compatible quasi-order
on $\m a$ (i.e., $\sqsupseteq$
is a reflexive, transitive, binary
relation on $A$ that is a subuniverse
of $\m a^2$). 
Then {\bf $S$ rectangulates $T$ relative to $\sqsupseteq$} 
\index{rectangulation!defined}
if for all $u\in A$ and all 
\begin{equation}\label{S,T-matrix3}
\left[\begin{array}{cc}
p&q\\
r&s
\end{array}\right]\in M(S, T),
\end{equation}
it is the case that 
\begin{equation}\label{rectangulation_imp}
u\sqsupseteq q\quad\&\quad u\sqsupseteq r
\quad\mimplies \quad
u\sqsupseteq s.
\end{equation}
We indicate this by saying that 
{\bf $\C R(S,T;\sqsupseteq)$ holds.}
\index{$\C R(S,T;\sqsupseteq)$}
\end{df}

By interchanging the rows and columns
of matrices one sees that 
$\C {R}(S,T;\sqsupseteq)$ holds 
if and only if 
\begin{equation}\label{nine}
u\sqsupseteq q\quad\&\quad u\sqsupseteq r
\quad\miff \quad
u\sqsupseteq p \quad \&\quad u\sqsupseteq s
\end{equation}
for every $u\in A$ and every
$S,T$-matrix in (\ref{S,T-matrix3}).

\begin{thm}\label{basic_rectangulation}
Let $\m a$ be an algebra with tolerances 
$S, S', T, T'$, compatible quasi-orders $\sqsupseteq, \sqsupseteq_j$
and a congruence $\delta$.
The following are true.
\begin{enumerate}
\item[(1)] If $\C R(S,T;\sqsupseteq)$ holds 
and $S'\subseteq S$,
$T'\subseteq T$, then $\C R(S',T';\sqsupseteq)$ holds.

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

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

\item [(4)] 
If $\sqsupseteq$ is $\delta$-closed 
(cf.\ Section~\ref{the_term_condition}),
then $\C R(S, T; \sqsupseteq)$ holds in $\m a$ 
if and only if
$\C R(S/\delta, T/\delta; \sqsupseteq\!/\delta)$ 
holds in $\m a/\delta$.

\item[(5)] $\C R(S,T;\sqsupseteq)$ holds if and only if 
$\C {R}(T,S;\sqsupseteq)$ holds.

\item[(6)] $\C R(S,T;\delta)$ holds if and only if 
$\C {SR}(S,T;\delta)$ holds.
\end{enumerate}
\end{thm}

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

Item (5) follows from the fact that the
transpose of an $S,T$-matrix is a $T,S$-matrix,
and the definition of rectangulation is
invariant under transpose (as indicated by
(\ref{nine})).

Item (6) follows from the fact that 
the definition of 
$\C {SR}(S, T; \delta)$
is the restriction of the 
definition of $\C {R}(S, T; \sqsupseteq)$
to the case where $\sqsupseteq \wrel{=} \delta$ is a congruence.
\end{proof}

It follows from  Theorem~\ref{basic_rectangulation}~(2)
that for tolerances $S$ and $T$ on 
an algebra $\m a$ there
is a least compatible quasi-order $\sqsupseteq$
such that $\C R(S,T;\sqsupseteq)$ holds,
namely the intersection of all $\sqsupseteq_j$
such that 
$\C R(S,T;\sqsupseteq_j)$ holds.
If some quasi-order $\sqsupseteq_j$ 
is antisymmetric (equivalently, is a partial order),
then the intersection of all $\sqsupseteq_j$ will
also be antisymmetric. It is this situation
that interests us, so we make the following definition.

\begin{df}\label{rectangular_df}
A tolerance $S$ {\bf rectangulates} a tolerance
$T$ if $\C R(S,T;\sqsupseteq)$ holds for some
compatible partial order $\sqsupseteq$.
If $S=T$ we say that $T$ is {\bf rectangular}.
\index{tolerance!rectangular}
\index{rectangular}

If $\sqsupseteq$ is the least compatible quasi-order
such that $\C R(S,T;\sqsupseteq)$ holds, then 
$[S,T]_{\bf R}$ denotes the intersection of
$\sqsupseteq$ with its converse $\sqsubseteq$
( := $\sqsupseteq^{\cup}$).
\end{df}

If $S$ and $T$ are tolerances on $\m a$,
$\sqsupseteq$ is the least compatible quasi-order
such that $\C R(S,T;\sqsupseteq)$ holds, and
$\delta = [S,T]_{\bf R}$ ( = 
$\sqsupseteq \cap \sqsubseteq$), 
then $\delta$ is a congruence on $\m a$ 
%(since this is true for the intersection of
%any compatible quasi-order and its converse)
and $\sqsupseteq$ is $\delta$-closed.
Since $\C R(S,T;\sqsupseteq)$ holds, we get from
Theorem~\ref{basic_rectangulation}~(4) that 
$\C R(S/\delta,T/\delta;\sqsupseteq\!\!/\delta)$
holds.
Since $\sqsupseteq\!\!/\delta$ is a compatible partial
order on $\m a/\delta$, this means that 
$S/\delta$ rectangulates 
$T/\delta$ in $\m a/\delta$. 
Moreover, it follows from 
Theorem~\ref{basic_rectangulation}~(4) that 
$\delta = [S,T]_{\bf R}$ is the least congruence
on $\m a$ for which this statement is true.
We record this as:

\begin{thm}\label{least_congruence}
If $S$ and $T$ are tolerances on $\m a$, then
$[S,T]_{\bf R}$ is the least congruence
$\delta\in\Con(\m a)$
such that $S/\delta$ rectangulates $T/\delta$
in $\m a/\delta$.
\end{thm}

If $S$ and $T$ are disjoint tolerances, 
%($S\cap T  = 0$), 
then it follows from
Definition~\ref{rectangulation_defn} 
that $\C R(S,T;0)$ holds,
and therefore $[S,T]_{\bf R} = 0$. 
This describes a trivial
instance of rectangulation. 
If it happens that $S\cap T \neq 0$
yet still $[S,T]_{\bf R} = 0$, then the
intersection $S\cap T$ is a nontrivial 
rectangular tolerance
%that rectangulates itself 
(by Theorem~\ref{basic_rectangulation}~(1)). 
%Such objects have a good structure.
We spend the rest of this section describing
the structure of such objects.

\begin{df}
An operation $+$ on $A$ is a 
{\bf semilattice operation} 
on $A$ if it is 
\begin{enumerate}
\item[(i)] idempotent:
$\forall x\in A (x + x\approx x)$;
\item[(ii)] commutative:
$\forall x,y\in A
(x + y\approx y + x)$; and 
\item[(iii)] associative:
$\forall x,y,z\in A (x + (y + z) 
\approx (x + y) + z)$.
\end{enumerate}

If $T$ is a tolerance on an algebra $\m a$,
then a partial operation $x + y$ on $\m a$
is a {\bf compatible semilattice operation on $T$}
\index{compatible semilattice operation}
if 
\begin{enumerate}
\item[(i)] $+\colon T\to \m a$ is a homomorphism
(with $T$ considered as a subalgebra of $\m a^2$); and
\item[(ii)] $+$ is idempotent, commutative, and
associative on each $T$-block.
\end{enumerate}
\end{df}


\begin{thm}\label{un_rect_repn_thm}
Let $\m a$ be an algebra and let $\theta$ be a 
congruence on $\m a$. If $\theta$
supports a compatible semilattice
operation, then $\theta$ is rectangular.
\end{thm}

\begin{proof}
Let $+$ be a compatible
semilattice operation on $\theta$. 
Define a relation $\sqsupseteq$ by
\[
a\sqsupseteq b\wrel{\miff}(a,b)\in\theta\;\;\mbox{\rm and}\;\;a=a+b\,.
\]
This is a partial order on $A$ due to 
the facts that 
$\theta$ is reflexive and transitive and $+$ 
is a semilattice operation on $\theta$-blocks.
It is compatible because $\theta$ and $+$ are.

To show that $\m r(\theta,\theta;\sqsupseteq)$ holds, 
let 
\[
\left[\begin{array}{cc}
p&q\\
r&s\end{array}\right]=
\left[\begin{array}{cc}
f(\wec{a},\wec{u})&f(\wec{a},\wec{v})\\
f(\wec{b},\wec{u})&f(\wec{b},\wec{v})
\end{array}\right]
\]
be a $\theta,\theta$-matrix, and suppose that
$u\sqsupseteq q$ and $u\sqsupseteq r$ for
some $u\in A$. The definition of
$\sqsupseteq$ forces the elements 
$u, q$ and $r$ to 
belong to the same $\theta$-block,
and also forces $u = u+q = u+r$. 
This $\theta$-block is closed under $+$, so 
$(u,q+r)\in\theta$. Moreover
\[
u = u+u
= (u+q)+(u+r)
= u+(q+r)\,,
\]
proving that $u\sqsupseteq q+r$. But 
\[
\begin{array}{rl}
q+r&= f(\wec{a},\wec{v}) + f(\wec{b},\wec{u})\\
&= f(\wec{a}+\wec{b},\wec{v}+\wec{u})\\
&= f(\wec{a},\wec{u}) + f(\wec{b},\wec{v})\\
&= p+s \sqsupseteq s\,,
\end{array}
\]
so $u\sqsupseteq s$. This completes the proof.
\end{proof}


\begin{thm}\label{rect_repn_thm}
Let $\m a$ be an algebra and let $T$ be a tolerance
on $\m a$. The following conditions are equivalent
\begin{enumerate}
\item[(1)] $T$ is rectangular.
\item[(2)] There is an algebra $\widehat{\m a}\in {\vr V}(\m a)$
and a tolerance $\widehat{T}$ on $\widehat{\m a}$
such that 
\begin{enumerate}
\item[(i)] $\m a$ is a subalgebra of $\widehat{\m a}$;
\item[(ii)] $T\subseteq \widehat{T}|_A$;
\item[(iii)] $\widehat{T}$ is rectangular; and
\item[(iv)] $\widehat{T}$ supports a compatible semilattice 
operation.
\end{enumerate}
\end{enumerate}
\end{thm}

\begin{proof}
For (2)$\mimplies$(1), we leave it to the reader to 
check our definitions and verify
that whenever $\m a\leq \widehat{\m a}$, $\widehat{T}$
is a tolerance on $\widehat{\m a}$ for which
$T\subseteq \widehat{T}|_A$, and $\widehat{T}$
rectangulates itself with respect to the
partial order $\sqsupseteq$, then $T$ rectangulates
itself with respect to the restricted
partial order $\sqsupseteq|_A$. 
(In particular, (2)(iv) is not 
needed to prove that (2)$\mimplies$(1).)
The real content 
of the theorem is the claim that (1)$\mimplies$(2).

Assume that (1) holds, and that $\sqsupseteq$
is a compatible partial order on $\m a$ for which
${\bf R}(T,T;\sqsupseteq)$ holds. We will construct
$\widehat{\m a}$ from an algebra
$\m b$, defined as follows.
The universe of $\m b$ consists of the nonempty
order ideals
of the poset $\lb A; \sqsupseteq\rb$.
(A subset $I\subseteq A$ is an {\bf order ideal}
\index{ideal!order ---} if
$x\in I$ and $x\sqsupseteq y$ implies $y\in I$.)
If $f$ is a $k$-ary operation symbol
in the language of $\m a$, and $I_1, \ldots, I_k\in B$,
then $f^{\m b}(I_1,\ldots,I_k)$ is defined to be the order ideal
$I$ generated by the set of all elements of the form
$f^{\m a}(a_1,\ldots,a_k)$ with $a_i\in I_i$.
(Thus, $u\in I$ if and only if there exist 
$u_i\in I_i$ such that $f^{\m a}(u_1,\ldots,u_k)\sqsupseteq u$.)
If $a\in A$, write $\lb a\rb$ for the principal 
order ideal generated by $a$ in $\lb A; \sqsupseteq \rb$.

\begin{clm} \label{hom1}
The function $\m a\to \m b\colon x\mapsto \lb x \rb$
is a 1-1 homomorphism.
\end{clm}

The fact that $\lb\;\rb$ is a 1-1 function follows
from the fact that $\lb A;\sqsupseteq\rb$ is a partial order.
To show that a typical $k$-ary basic operation $f$ is
preserved by $\lb\;\rb$, choose $a_1,\ldots,a_k\in A$.
The ideal $f^{\m b}(\lb a_1\rb,\ldots,\lb a_k\rb)$
consists of all elements $u\in A$ for which there exist 
elements $u_i\in \lb a_i\rb$ 
such that $f^{\m a}(u_1,\ldots,u_k)\sqsupseteq u$.
But since $a_i\sqsupseteq u_i$ and $\sqsupseteq$
is compatible, this means that 
\[f^{\m a}(a_1,\ldots,a_k)\sqsupseteq 
f^{\m a}(u_1,\ldots,u_k)\sqsupseteq u.\]
Thus $u\in f^{\m b}(\lb a_1\rb,\lb a_2\rb,\ldots,\lb a_k\rb)$
if and only if 
$u\in \lb f^{\m a}(a_1,\ldots,a_k)\rb$, which proves
that $\lb\;\rb$ is a homomorphism.

\medskip

This claim shows that we may view $\m b$ as
an extension of $\m a$. In order to define a 
tolerance $S$ on $\m b$
whose restriction to $\m a$ contains $T$, we first define
\[S' = \{(I,J)\in B^2\;|\;(\forall u\in I)(\exists (v,w)\in 
(I\times J)\cap T)(v\sqsupseteq u)\}.\]
Now let $S$ be the intersection of $S'$
and its converse.

Let ${\vr L}$ be the language of $\m a$, and
let ${\vr L}(\cup)$ be the expansion
of the language to include the symbol $\cup$.
By interpreting $\cup$ in $\m b$
as union of ideals we get 
an ${\vr L}(\cup)$-algebra $\lb \m b;\cup\rb$.

\begin{clm} \label{tol}
$S$ is a tolerance on $\lb\m b;\cup\rb$. Moreover, if
$(a,b)\in T$, then $(\lb a\rb, \lb b\rb)\in S$.
\end{clm}

Since $S$ is the intersection of $S'$ 
and its converse, it will suffice to prove that
$S'$ is a compatible reflexive relation 
of $\lb \m b;\cup\rb$, and that if 
$(a,b)\in T$, then $(\lb a\rb, \lb b\rb)\in S'$.

Choose $I\in B$. For any $u\in I$ the pair 
$(u,u)\in (I\times I)\cap T$, so $(I,I)\in S'$.
This shows that $S'$ is reflexive.

Next, suppose that $(I_i,J_i)\in S', 1\leq i\leq k$,
and that $f$ is a $k$-ary $\vr L$-operation.
Let $I = f(I_1,\ldots,I_k)$ and 
$J = f(J_1,\ldots,J_k)$. To see that $(I,J)\in S'$
choose $u\in I$. There exist $u_i\in I_i$ such that 
$f(u_1,\ldots,u_k)\sqsupseteq u$, by the definition of
$\m b$, and there exist $(v_i,w_i)\in (I_i\times J_i)\cap T$
with $v_i\sqsupseteq u_i$, since $(I_i,J_i)\in S'$.
Since $\sqsupseteq$ is compatible, this means
that 
\[v := f(v_1,\ldots,v_k)\sqsupseteq f(u_1,\ldots,u_k)\sqsupseteq
u.\]
Since $T$ is also compatible, for $w:= f(w_1,\ldots,w_k)\in J$
we have
\[(v,w) = f((v_1,w_1),\ldots,(v_k,w_k))\in T.\]
Thus, for our $u\in I$ (which we chose arbitrarily)
there exist $(v,w)\in (I\times J)\cap T$ such that 
$v\sqsupseteq u$. Hence $(I,J)\in S'$, showing
that $S'$ is compatible with all $\vr L$-operations.

To show that $S'$ is compatible with $\cup$,
choose $(I',J'), (I'',J'')\in S'$ and 
let $(I,J) = (I'\cup I'', J'\cup J'')$.
Choose any $u\in I = I'\cup I''$. If $u\in I'$,  
then since $(I',J')\in S'$ there is a pair 
\[(v,w)\in ((I'\times J')\cap T) \subseteq 
((I\times J)\cap T)\]
such that $v\sqsupseteq u$. The same conclusion 
holds if $u\in I''$, so we have that 
$(I,J)\in S'$. Hence $S'$ is a compatible
reflexive relation of $\lb \m b;\cup\rb$.

To finish the proof of the
claim we must show that for $(a,b)\in T$
we have $(\lb a\rb,\lb b\rb)\in S'$.
To see that this is so, observe that for any 
$u\in \lb a\rb$ the pair $(v,w) = (a,b)$
satisfies the required properties:
$(v,w)\in (\lb a\rb \times \lb b\rb)\cap T$
and $v\sqsupseteq u$.

\medskip

We have constructed an algebra $\m b$,
a tolerance $S$, and we have a candidate
$\cup$ for the compatible
semilattice operation on $S$. 
For $\cup$ to be compatible on $S$ it 
must commute with the ${\vr L}$-operations on $S$.
This won't be true without some tinkering, so 
we factor by a congruence that collapses
all failures of compatibility. 
Specifically, let $\theta$ be the 
${\vr L}(\cup)$-congruence on $\m b$
generated by all pairs of the form
\[
\lb\, f(I_1,\ldots, I_k)\cup f(J_1,\ldots, J_k),\,\,
f(I_1\cup J_1,\ldots, I_k\cup J_k)\,\rb
\]
where $f\in {\vr L}(\cup)$ and each
$(I_i,J_i)\in S$.  Since $\theta$ is
defined to be a congruence of $\lb \m b;\cup\rb$, 
the quotient 
$\lb \m b;\cup\rb/\theta$ satisfies all equational
properties satisfied by $\lb \m b;\cup\rb$.
In particular, $\cup$ is a semilattice operation
of the quotient and all operations
are monotone with respect to
the semilattice order.

\begin{clm} \label{hom2}
If $\lb U,V\rb\in\theta$ and $u\in A$,
then $U \subseteq \lb u\rb\miff V \subseteq \lb u\rb$.
Hence if $\lb \lb u\rb, \lb v\rb\rb\in\theta$,
then $u = v$.
\end{clm}

To see that this is so, it suffices to check the claim only
in the case where $\lb U,V\rb$ is a polynomial image of a
generating pair for $\theta$. Here, by a polynomial, we mean
a unary polynomial of $\lb \m b;\cup\rb$, 
although in our argument we will only consider
special unary polynomials of $\lb\m b;\cup\rb$.

Let $\m c$ be any algebra. A
{\bf basic translation}\index{basic translation}
of $\m c$ is a unary
polynomial of the form 
$q(x,c_1,\ldots,c_n)$ where each $c_i\in C$, and 
$q$ is obtained from some basic operation
by permuting variables. It is known (and not
hard to see) that the congruence on $\m c$
generated by some set $G\subseteq C\times C$
can be described by first 
forming the closure of $G$ under all basic
translations, and then generating
an equivalence relation. Hence
$(x,y)\in \Cg C(G)$ if and only if there
is a sequence $x = x_0,\ldots , x_m = y$
where each $\{x_i, x_{i+1}\} = \{p(g),p(h)\}$
for some $(g,h)\in G$ and some unary polynomial
$p$ which is a composition of basic translations.

For us, the fact that for any $f\in {\vr L}$
we have 
\[
\begin{array}{rl}
f(x\cup W_1,W_2,\ldots, W_k) &=
f(x,W_2,\ldots, W_k) \cup f(W_1,W_2,\ldots, W_k)\\
&= f(x,W_2,\ldots, W_k) \cup Z
\end{array}
\]
when all $W_i\in \lb \m b;\cup\rb$
and $Z:= f(W_1,\ldots,W_k)$ implies
that if a composition of basic translations
of $\lb \m b;\cup\rb$ involves $\cup$, then 
it can be rewritten so that it is a composition
of basic translations not involving $\cup$,
followed by a single basic translation of the form
$p(x) = x\cup Z$. 
This observation shows that 
to prove the claim it suffices to show that 
if $(H,K)$ is a generating pair for $\theta$,
then 
\[p(H)\subseteq \lb u\rb\miff p(K)\subseteq\lb u\rb\]
when $p$ has the form
$p(x) = q(x,W_1,\ldots,W_k)\cup Z$ where
$q$ is a composition of basic translations
not involving $\cup$, and
$W_i, Z\in B$.

Since the generating pairs for $\theta$ are comparable
and the operations of $\lb \m b;\cup\rb$ are monotone,
to prove that 
\[p(H)\subseteq \lb u\rb\miff p(K)\subseteq\lb u\rb\]
for a generating pair $(H,K)$ it suffices to show
that 
\[
\begin{array}{rcl}
U:=&p(f(I_1,\ldots,I_n)\cup f(J_1,\ldots,J_n))\subseteq \lb u\rb
&\mimplies \\
V:=&p(f(I_1\cup J_1,\ldots,I_n\cup J_n))\subseteq \lb u\rb.&
\end{array}
\]
Assume that $U\subseteq \lb u\rb$
but that there is some $e\in V - \lb u\rb$.  
Then, since $p(x) = 
q(x,W_1,\ldots,W_k)\cup Z$
we get $Z\subseteq U\subseteq \lb u\rb$, so 
since 
\[e\in V-Z = 
q(f(I_1\cup J_1,\ldots,I_n\cup J_n),W_1,\ldots,W_k)-Z,\]
we get that 
$e\in q(f(I_1\cup J_1,\ldots,I_n\cup J_n),W_1,\ldots,W_k)$.
We can reorder the variables of $f$ so that
there are elements $s_i\in I_i\;(1\leq i\leq j)$,\; $t_i\in
J_i\;(j+ 1\leq i\leq n)$,\; $w_i\in W_i\;(1\leq i\leq k)$
such that
\[
e = q(f(s_1,\ldots,s_j,t_{j+1},\ldots,t_n),w_1,\ldots,w_k)
= q(f(\wec{s},\wec{t}),\wec{w})\,.
\]
For each $i$ between 1 and $j$
we use the facts that 
$s_i\in I_i$ and $(I_i,J_i)\in S$ to
choose $(s_i',t_i'')\in (I_i\times J_i)\cap T$
with $s_i'\sqsupseteq s_i$,
and similarly for $i$ between $j+1$ and $n$
we can choose $(t_i',s_i'')\in (J_i\times I_i)\cap T$
with $t_i'\sqsupseteq t_i$.
We use these elements to form tuples 
$\wec{s}' = (s_1',\ldots,s_j'), \wec{s}'' = 
(s_{j+1}'',\ldots,s_n'')$, $\wec{t}'$
and $\wec{t}''$.
Since $\wec{s}'$ and $\wec{t}''$ are $T$-related
coordinatewise, and the same is true for 
$\wec{s}''$ and $\wec{t}'$, we have 
\[
\left[\begin{array}{cc}
q(f(\wec{t}'',\wec{s}''),\wec{w})&q(f(\wec{t}'',\wec{t}'),\wec{w})\\
q(f(\wec{s}',\wec{s}''),\wec{w})&q(f(\wec{s}',\wec{t}'),\wec{w})
\end{array}\right] = 
\left[\begin{array}{cc}
p&q\\
r&s
\end{array}\right]\in M(T,T)
\]
with $q, r\in U\subseteq \lb u\rb$.
Since we have assumed that $T$ rectangulates 
itself with respect to $\sqsupseteq$, and $u\sqsupseteq q, r$,
we must have 
\[u\sqsupseteq s = 
q(f(\wec{s}',\wec{t}'),\wec{w})
\sqsupseteq 
q(f(\wec{s},\wec{t}),\wec{w}) = e,\]
which is contrary to our choice of $e$.
This proves the first sentence of the claim.
The second sentence follows from the first
by taking $U = \lb u\rb$, $V = \lb v\rb$, and using
the fact that $\sqsupseteq$ is a partial order.

\medskip

The relation $S/\theta = \{(I/\theta,J/\theta)\in 
(B/\theta)^2\;|\; (I,J)\in S\}$ is a tolerance
on $\m b/\theta$, hence a subalgebra
of $\m b/\theta\times \m b/\theta$.

\begin{clm} \label{comp_semi}
The function $\cup\colon S/\theta\to \m b/\theta$
is a homomorphism.
\end{clm}

We must show that if
$(I_i/\theta,J_i/\theta)\in S/\theta$
and $f\in {\vr L}$, then in $\m b/\theta$ we have
\begin{eqnarray*}
\lefteqn{f(I_1/\theta\cup J_1/\theta,\ldots, I_k/\theta\cup J_k/\theta)}\\
&&= f(I_1/\theta,\ldots,I_k/\theta)\cup
f(J_1/\theta,\ldots,J_k/\theta).
\end{eqnarray*}
Equivalently, we must show that if
$(I_i,J_i)\in S$ and $f\in {\vr L}$, then 
in $\m b$ we have
\[f(I_1\cup J_1,\ldots,I_k\cup J_k) \equiv_{\theta} 
f(I_1,\ldots,I_k)\cup
f(J_1,\ldots,J_k).\]
This is immediate from the definition of $\theta$.

\medskip

Claims \ref{hom1} and \ref{hom2}
prove that the composite
$x\mapsto \lb x\rb\mapsto \lb x\rb/\theta$ 
is an embedding:
$\m a\to \m b\to \m b/\theta$.
We identify $\m a$ with its image under this
homomorphism.

Let $\vr C$ be the set of all 
algebras $\m c\in {\vr V}(\m a)$ 
that are intermediate extensions:
$\m a\leq {\m c}\leq \m b/\theta$.
This set contains $\m a$, so it is nonempty,
and it is closed under unions of chains.
Therefore there is an algebra
$\widehat{\m a}\in {\vr C}$ which 
is maximal among intermediate extensions
that belong to ${\vr V}(\m a)$.
Let $\widehat{T} = S/\theta|_{\widehat{A}}$
be the restriction of $S/\theta$ to $\widehat{\m a}$.

\begin{clm}
$\widehat{T}$ is a tolerance on $\widehat{\m a}$,
and conditions 2(i)--(iv) hold
for $\widehat{T}$ and $\widehat{\m a}$.
\end{clm}

$\widehat{T}$ is the restriction of the
tolerance $S/\theta$ on $\m b/\theta$
to $\widehat{\m a}$, so it is a tolerance
on $\widehat{\m a}$. It is a consequence
of Claim~\ref{tol} that 
$T\subseteq \widehat{T}|_A$.
What we have left to show
is that $\widehat{T}$ is rectangular, and 
supports a compatible semilattice 
operation. We show the latter first.

Claim \ref{comp_semi} proves that 
$\cup\colon S/\theta\to \m b/\theta$
is a homomorphism. This homomorphism 
can be restricted to $\widehat{T}$.
The image $\cup(\widehat{T})$
of the restricted homomorphism
contains $\widehat{A}$, since 
$\widehat{T}$ is reflexive and
$a\cup a = a$ for all $a\in \widehat{A}$.
The image 
$\cup(\widehat{T})$
is a quotient of $\widehat{T}$,
and $\widehat{T}$ is a subalgebra of
$\widehat{\m a}\times \widehat{\m a}
\in {\vr V}(\m a)$,
so the image is in ${\vr V}(\m a)$.
The maximality of $\widehat{\m a}$
prevents it from being 
properly contained in $\cup(\widehat{T})$,
so $\cup(\widehat{T}) = \widehat{\m a}$.
This shows that  $\cup$ is a homomorphism 
from $\widehat{T}$ to $\widehat{\m a}$.

Let $D$ be a $\widehat{T}$-block.
If $a, b, c\in D$, then $(a,c),(b,c)\in \widehat{T}\subseteq
S/\theta$ so, since $S/\theta$ is 
a tolerance of $\lb \m b;\cup\rb/\theta$,
\[(a\cup b,c)=(a,c)\cup (b,c)\in S/\theta|_{\widehat{A}}=\widehat{T}.\]
This shows that $a\cup b$ is $\widehat{T}$-related to
any $c\in D$, so $a\cup b\in D$ by the fact that 
$D$ is a block. Since $\cup$
is a homomorphism from $\widehat{T}$ to
$\widehat{\m a}$ that 
is the restriction of a semilattice
operation on $\m b/\theta$, and since
each $\widehat{T}$-block is closed under
$\cup$, it follows that 
$\cup$ is a compatible semilattice operation
on $\widehat{T}$.

Now we prove that $\widehat{T}$ is rectangular. 
To check this we need a compatible
partial order on $\widehat{\m a}$.
We choose it to be the restriction
of the $\cup$-semilattice order on $\m b/\theta$,
which we denote by $\gg$ and define by
\[a\gg b \wrel{\miff} a = a\cup b.\]
We have already observed that the operations
of $\m b/\theta$ are monotone with respect
to the semilattice order, and this fact
is inherited by $\widehat{\m a}$, hence
$\gg$ is a compatible partial order.
To see that ${\bf R}(\widehat{T},\widehat{T};\gg)$ holds, 
choose $M \in M(\widehat{T},\widehat{T})$,
\[
M = 
\left[\begin{array}{cc}
p(\wec{a},\wec{u})&p(\wec{a},\wec{v})\\
p(\wec{b},\wec{u})&p(\wec{b},\wec{v})
\end{array}\right] = 
\left[\begin{array}{cc}
p&q\\
r&s
\end{array}\right],
\]
with $u\gg q, r$ for some $u\in \widehat{A}$.
All four elements of this matrix are $\widehat{T}$-related,
so 
\[u\gg u\cup u\gg q\cup r=
p(\wec{a},\wec{v})\cup
p(\wec{b},\wec{u}) = 
p(\wec{a}\cup\wec{b},\wec{v}\cup\wec{u})\gg
p(\wec{b},\wec{v}) = s.\]
Thus, $u\gg q, r$ implies $u\gg s$, as required.
This completes the proof of the claim,
and also of the theorem.
\end{proof}

It is worth recording that the proof produces
from $\m a$ and $T$ two algebras of interest:
$\widehat{\m a}$ and $\m b/\theta$.
The properties of the two algebras
are roughly the same with respect to
rectangulation, but $\widehat{\m a}$
is in the variety generated by $\m a$
while $\m b/\theta$ may not be.
On the other hand, the compatible semilattice operation
that we constructed is only a partial operation
on $\widehat{\m a}$ (it is
defined only on $\widehat{T}$-blocks),
while it is a total operation on $\m b/\theta$
and the semilattice order is 
a compatible partial order of $\m b/\theta$.


\section{Rectangular Tolerances and Join Terms}

We introduced join terms in Section~\ref{join_term},
but did not investigate their properties. It is
now important that we do so. In this section 
we explore the structure of 
rectangular tolerances on algebras 
in varieties with a join term.
For such tolerances we
prove that the representation given in
Theorem~\ref{rect_repn_thm} is simpler
in that one may take $\widehat{\m a} = \m a$
and $\widehat{T}=T$. We then describe
connections with the relations $\geq_T$
and $\sim_T$ that were
introduced in Definition~\ref{ge_defn}


\begin{thm}\label{join_prop}
Let $\vr V$ be a variety with a join term $+$, and
let $T$ be a tolerance on an algebra~$\m A\in\vr V$. 
\begin{enumerate}
\item[(1)] If $B$ is a $T$-block and
$w\geq_B u$ and $w\geq_B v$, then $w\geq_B u+v$.
Similarly, if 
$w\ge_T u$ and $w\ge_T v$, then $w\ge_T u+v$.
\item[(2)] If $B$ is a $T$-block, then 
$\geq_B$ is a join semilattice ordering
on $B$ modulo $\sim_B$, and $+$ is the join
with respect to that ordering.
\item[(3)] If $B$ is a $T$-block containing
all entries of 
$
\left[\begin{array}{cc}
a&b\\
c&d\end{array}\right]\in M(T,T)\,,
$
then $a+d\wrel{\sim_B} b+c$.
\item[(4)] If $(q,r;s)$ is a $(T,T)$-triple, then $q+r\ge_T s$.
\item[(5)] 
If $p$ is a polynomial
of $\m a$ and $\wec{u}\wrel{T}\wec{v}$,
then 
\[
p(\wec{u}+\wec{v}) 
\wrel{\sim_T} p(\wec{u}) + p(\wec{v})\,.
\]
\end{enumerate}
\end{thm}

\begin{proof}
From $w\wrel{\geq_B} u$ and $w\wrel{\geq_B} v$ it follows that 
$w = w+w\wrel{\geq_B} u+v$, since $+$ is
idempotent and $\geq_B$ is compatible with 
idempotent polynomials (Lemma~\ref{basic_prop}~(1)).
The statement about $\geq_T$ follows in the same way
from Lemma~\ref{basic_prop}~(2).

If $B$ is a $T$-block,
then $B$ is closed under $+$ since $+$ is idempotent.
From Corollary~\ref{omit_1_cor}, 
$u+v\wrel{\geq_B} u, v$, while 
item~(1) proves conversely that if
$w\wrel{\geq_B} u, v$, then $w\wrel{\geq_B} u+v$. 
Thus $\geq_B$ is indeed a semilattice order
modulo $\sim_B$ on $B$, and $+$ is the join 
with respect to this order.

For item~(3), Lemma~\ref{ge_elementary}~(3)
proves that 
$b+c\wrel{\ge_B} (b+d)+(a+c)$. But from 
Corollary~\ref{omit_1_cor}
we have $b+d\wrel{\ge_B} d$ and $a+c\wrel{\ge_B} a$, so
\[
b+c\wrel{\ge_B} (b+d)+(a+c)\wrel{\ge_B} d+a\,.
\]
Since $\geq_B$ is transitive on $B$,
we have $b+c\wrel{\geq_B} d+a$. If we interchange
the two rows of this $T,T$-matrix, we derive
conversely that $d+a\wrel{\geq_B} b+c$, hence
that $b+c\wrel{\sim_B} d+a$.
It follows from part (2) that 
$d+a\wrel{\sim_B} a+d$, so from the transitivity
of $\sim_B$ on $B$ we get 
$a+d\wrel{\sim_B} b+c$, as claimed.

For (4), if $(q,r;s)$ is a $(T,T)$-triple, then
its entries belong to a $T,T$-matrix. 
All four entries of any $T,T$-matrix
are pairwise $T$-related, hence a block $B$
exists that contains all four entries.
Applying item (3) to this block we obtain 
$q+r\wrel{\sim_B} p+s$ for some $p\in B$,
hence from item (1) it follows that 
$q+r\wrel{\geq_B} s$. This proves $q+r\wrel{\geq_T} s$.

We first prove item (5) in the case
where $p=f$ is a basic operation of $\m a$.
Choose $T$-blocks $B_i$ such that 
$u_i, v_i\in B_i$ for all $i$,
and choose a $T$-block $B$ such that
$f(B_1,\ldots,B_n)\subseteq B$.
According to part (2) of this theorem
we have 
\begin{equation}\label{sim_stuff}
u_i+v_i\wrel{\sim_{B_i}} v_i+u_i
\end{equation}
for all $i$. Now consider the $T,T$-matrix
\[
\left[\begin{array}{cc}
f(\wec{u}+\wec{u})&
f(\wec{u}+\wec{v})\\
f(\wec{v}+\wec{u})&
f(\wec{v}+\wec{v})
\end{array}\right]\,.
\]
All entries lie in $B$, so from part~$(3)$ we get that
\[
f(\wec{u}+\wec{u})+f(\wec{v}+\wec{v})\wrel{\sim_B}
f(\wec{u}+\wec{v})+f(\wec{v}+\wec{u})\,.
\]
The left hand side is $f(\wec{u})+f(\wec{v})$,
since $+$ is idempotent. The right side is 
$\sim_B$-related to $f(\wec{u}+\wec{v})$
since, by (\ref{sim_stuff}), Claim~\ref{compatibility_claim}
and the idempotence of $+$ we have
\[
f(\wec{u}+\wec{v})+f(\wec{v}+\wec{u})\wrel{\sim_B}
f(\wec{u}+\wec{v})+f(\wec{u}+\wec{v}) = f(\wec{u}+\wec{v})\,.
\]
Since $\sim_B$ is transitive on $B$ we get 
$f(\wec{u})+f(\wec{v}) \wrel{\sim_B} f(\wec{u} + \wec{v})$,
hence
$f(\wec{u})+f(\wec{v}) \wrel{\sim_T} f(\wec{u}+\wec{v})$.

Now we prove (5) for polynomials.
Consider the relation $\rho\subseteq A^3$
consisting of all triples $(u,v,w)$
for which there is a $T$-block $B$ 
such that $u, v, w\in B$ and $u+v\sim_B w$.
It follows from the fact that item (5)
is true when $f$ is a basic operation,
together with Claim~\ref{compatibility_claim}
and the transitivity of $\sim_{B'}$
on $B'$ for any block $B'$,
that $\rho$ is compatible with the basic operations
of $\m a$. It is compatible with all constant
operations on $A$ because $+$ is idempotent.
Thus $\rho$ is compatible with all polynomial
operations. Since $\rho$ contains all triples
of the form $(u,v,u+v)$, it therefore contains
$(p(\wec{u}),p(\wec{v}),p(\wec{u}+\wec{v}))$,
establishing that $p(\wec{u}+\wec{v})\sim_T
p(\wec{u})+p(\wec{v})$.
\end{proof}

We can prove more if $T$ is rectangular.

\begin{thm}\label{join_props_2}
Let $\vr V$ be a variety with a join term $+$, and
let $T$ be a tolerance on an algebra~$\m A\in\vr V$. 
If $\sqsupseteq$ is a compatible
partial order of $\m a$ for which 
$\C R(T,T;\sqsupseteq)$ holds, then 
the following are true.
\begin{enumerate}
\item[(1)] The relation 
$\sim_T$ is the equality relation on $\m a$.
\item[(2)] 
$\geq_T$ equals $\sqsupseteq \cap T$, and also equals
$\{(u,v)\in T\;|\;u=u+v\}$.
\item[(3)] If
$
\left[\begin{array}{cc}
a&b\\
c&d\end{array}\right]\in M(T,T)\,,
$
then $a+d = b+c$.
\item[(4)] 
If $f$ is a polynomial of $\m a$ and $\wec{u}\wrel{T}\wec{v}$,
then 
\[
f(\wec{u}+\wec{v}) = f(\wec{u}) + f(\wec{v})\,.
\]
\item[(5)] $+\colon T\to \m a$ is the unique compatible semilattice
operation supported by $T$.
\end{enumerate}
\end{thm}

\begin{proof}
If $(r,s)$ is a $(T,T;0)$-pair, then there is a matrix
\[
\left[\begin{array}{cc}
p&q\\
r&s
\end{array}\right]\in M(T,T)
\]
with $q=r$.
Since $r\sqsupseteq r$ and $r\sqsupseteq q\;( = r)$, 
we get $r\sqsupseteq s$ because 
$\C R(T,T;\sqsupseteq)$ holds.
This shows that the relation 
$\sqsupseteq$ contains all
$(T,T;0)$-pairs, hence contains 
$\geq_T$. Therefore, the reflexive
relation $\sim_T$ is contained
in the intersection of $\sqsupseteq$ and its converse,
which forces $\sim_T$ to be the equality relation.
This proves item (1), and together with 
Theorem~\ref{join_prop}~(3) and~(5) also proves 
items~(3) and~(4).

We will prove that $\geq_T$ is contained in
$\sqsupseteq\cap T$, which is contained in 
$\{(u,v)\in T\;|\;u=u+v\}$, which is contained
in $\geq_T$.
It follows from Definition~\ref{ge_defn}
that $\geq_T$ is contained in $T$,
and it follows from the argument of the previous paragraph
that $\geq_T$ is contained in $\sqsupseteq$,
so $\geq_T$ is contained in $\sqsupseteq\cap T$.
Next, choose any pair $(u,v)$ in $\sqsupseteq\cap T$.
Since $\sqsupseteq$ is reflexive and
compatible, and since $u\sqsupseteq v$, we get that 
$u=u+u\sqsupseteq u+v$.
But $u+v\geq_T u$, according to 
the definition of a join term, so $u+v\sqsupseteq u$
by the argument of the first paragraph.
This shows that both $(u,u+v)$ and $(u+v,u)$
belong to the partial order
$\sqsupseteq$, so $u=u+v$. Hence 
$\sqsupseteq\cap T$ is contained in
$\{(u,v)\in T\;|\;u=u+v\}$.
Finally, choose any $(u,v)\in T$ such that 
$u=u+v$. If $B$ is a $T$-block containing
$u$ and $v$, then $u=u+v\geq_B v$, since $\geq_B$ 
is the join order for $+$ on $B$
(Theorem~\ref{join_prop}~(2)). Therefore
$u\geq_T v$, proving that 
$\{(u,v)\in T\;|\;u=u+v\}$ is contained in $\geq_T$.
This completes the proof of item~(2).

For item~(5), it follows from
Theorem~\ref{join_prop}~(2) and
part~(1) of this theorem that $+$
is a semilattice operation
on the blocks of $T$. 
The property from item~(4) of
this corollary, in the cases
where $f$ is a basic operation
of $\m a$, is exactly the property that
$+\colon T\to \m a$ is a homomorphism.
Thus $+$ is a compatible semilattice operation
on $T$. If $s\colon T\to \m a$ is another compatible
semilattice operation, then using the
commutativity and idempotence of $s$ and $+$,
and the fact that $s\colon T\to \m a$
is a homomorphism, and that $+$ is a term,
we have for any $(u,v)\in T$ 
\[
\begin{array}{rl}
s(u,v) &= s(u,v) + s(v,u)\\
&= s(u+v,v+u) \\
&= s(u+v,u+v)\\
&= u+v.
\end{array}
\]
Thus $+$ is unique.
\end{proof}


\begin{cor}\label{str_ab_thm3}
Let $\vr V$ be a variety. The following are equivalent.
\begin{enumerate}
\item[(1)] $\vr V$ has no member with a nonzero
congruence that is simultaneously rectangular
and abelian.
\item[(2)] $\vr V$ has no member with a nonzero
tolerance that is simultaneously rectangular
and abelian.
\item[(3)] $\vr V$ satisfies a nontrivial 
idempotent Maltsev condition.
\end{enumerate}
\end{cor}

\begin{proof}
Since tolerances are congruences,
(2)$\mimplies$(1). 

A strongly abelian congruence is
both rectangular and abelian, so 
if (1) holds, then 
$\vr V$ omits strongly abelian 
abelian congruences. Hence
$\vr V$ satisfies a nontrivial idempotent
Maltsev condition, according to 
Theorem~\ref{str_ab_thm1}, showing
that (1)$\mimplies$(3). 

Finally, if $\vr V$ satisfies a nontrivial 
idempotent Maltsev condition, then it has a join term, $+$.
Suppose that $\m a\in {\vr V}$ has a tolerance $T$ that
is rectangular and abelian. 
By Theorem~\ref{join_props_2}~(5), 
$+$ is a semilattice
operation on $T$-blocks. Therefore, if 
$(a,b)\in T$ and $a\ge b$ in the semilattice
order, then 
\[
\left[
\begin{array}{cc}
a&a\\
a&b
\end{array}\right]=
\left[
\begin{array}{cc}
a+a&a+b\\
b+a&b+b
\end{array}\right]\in M(T,T)\,.
\]
But since $T$ is also abelian, 
the equality in the first row
of this matrix implies
equality in the second row. Hence
$T = 0$, and (3)$\mimplies$(2).
\end{proof}

One can use tame congruence theory, \cite{hobby-mckenzie},
to prove that a finite simple algebra has a 
nontrivial tolerance that is 
(a) simultaneously 
rectangular and abelian if and only 
if it has one that is (b) strongly rectangular
if and only 
if it has one that is (c) strongly abelian.
So perhaps it is not too surprising 
that the varieties that omit tolerances
of type (a) are the same as those
that omit type (b) or (c). However, (a), (b)
and (c) are truly different properties
of tolerances, even for finite algebras,
as we explain in the next example.

\begin{exmp}\label{A(M)_2}
This example is an extension of Example~\ref{A(M)}.
As in that example, 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 $X$.
Define $\m a({\vr M})$ as in Example~\ref{A(M)}.
Suppose that $\gg$ is a quasi-order defined
on the set $X$ such that whenever 
$M = \left[\begin{array}{cc} p&q\\r&s\end{array}\right]$
belongs to $\vr M$, then 
\[
u\gg q\quad\&\quad u\gg r
\quad\miff \quad
u\gg s\quad\&\quad u\gg p
\]
for any $u\in X$.
Let $\sqsupseteq$ be the least extension of 
$\gg$ to a quasi-order on $A = X\cup Y\cup \{0\}$
(namely, $\sqsupseteq$ is the union of $\gg$
with the equality relation on $A$).
Then $\sqsupseteq$ is a compatible quasi-order of
$\m a({\vr M})$, and 
it is clear that 
$\C R(R,S;\sqsupseteq)$ and $\C R(T,T;\sqsupseteq)$
both hold (since for $u\neq v$ we have
$u\sqsupseteq v$ if and only if $u\gg v$). 
Thus, we are again in the situation that
we may ignore algebraic considerations and think
set-theoretically when considering 
relationships between rectangulation 
and the term conditions introduced earlier.

To see, for example, that there is a finite algebra
with a tolerance that is rectangular and abelian
but not strongly rectangular (hence not strongly
abelian), it is entirely sufficient to
write down the single matrix 
$M = 
\left[\begin{array}{cc}
1&3\\
3&2
\end{array}\right]$.
Then, for ${\vr M}=\{M\}$, 
and $\m a({\vr M})$ and $T$ defined as in 
Example~\ref{A(M)}, the set of nontrivial 
$T,T$-matrices are just those that can be obtained
from $M$ by permuting its entries by some
combination of row or column interchanges
or transposes.
As a $T, T$-matrix, $M$ formally conflicts 
with $\C {SR}(T,T;0)$. But it 
does not formally conflict with
$\C {C}(T,T;0)$ or $\C R(T,T;\gg)$
where $\gg$ is the partial order on $X$
generated by the pairs $(3,1)$ and $(3,2)$.
Therefore, on the algebra 
$\m a({\vr M})$, the tolerance $T$ 
is rectangular and abelian but not 
strongly rectangular.
\end{exmp}

The construction of Example~\ref{A(M)_2}
can be used to show that 
it is possible for the transitive closure
of the $(T,T;0)$-pairs to be
a partial order even when $T$ is not rectangular.
For example, this occurs when 
\[
{\vr M} = \left\{
\left[\begin{array}{cc}
1&3\\
3&2
\end{array}\right], 
\left[\begin{array}{cc}
1&4\\
4&2
\end{array}\right]
\right\}\,.
\]
Generalizations of this
example show that the construction
of the rectangular commutator of
two tolerances can be arbitrarily complicated
in general. The situation is better
%for algebras 
in varieties
with join terms.

\begin{thm}\label{quotients}
Let $\vr V$ be a variety with a join term $+$, and
let $T$ be a tolerance on an algebra~$\m A\in\vr V$. 
Let $\rho$ be any congruence on $\m a$
that contains $\sim_T$, and let 
$\m c = \m a/\rho$ and $S = T/\rho$.
\begin{enumerate}
\item[(1)] If $a_1\ge_S a_2\ge_S\cdots \ge_S a_n\ge_S a_1$
in $\m c$, then $a_1 = a_2 = \cdots = a_n$.
\item[(2)]  If $\sqsupseteq$ is the transitive closure of $\geq_S$,
then $\sqsupseteq$ is a partial order of $C$ whose restriction 
to any $S$-block agrees with $\geq_S$.
\item[(3)] $\C R(S,S;\sqsupseteq)$ holds.
\item[(4)] The transitive closure of $\sim_T$ is $[T,T]_{\m r}$.
\end{enumerate}
\end{thm}

\begin{proof}
First we prove~$(1)$. We may assume that all $(a_i,a_{i+1})$
(with the understanding that $n+1=1$) are
$(S,S;0)$-pairs. Now choose $n\ge 2$ minimal among such cycles
for which the statement is false. We argue that
$a_1+a_2=a_2+a_1=a_1$. 
As $(a_1,a_2)$ is an $(S,S;0)$-pair,
there exists a $(T,T)$-matrix
\[
\left[\begin{array}{cc}
a&b\\
c&d\end{array}\right]
\]
such that $b/\rho=c/\rho=a_1$ and $d/\rho=a_2$. 
Choose a $T$-block $B$ containing these four elements.
By Theorem~\ref{join_prop}
we have that $b+c\sim_B a+d$, and
that $+$ is a semilattice operation on $B$
modulo~$\sim_B$, which implies that $(b+c)+d\sim_B
d+(b+c)\sim_B b+c$. The relation $\sim_B$ is contained
in~$\rho$, and therefore in $\m a/\rho$ we have
$(a_1+a_1)+a_2=a_2+(a_1+a_1)=a_1+a_1$. But $+$ is
idempotent, so $a_1+a_2=a_2+a_1=a_1$. A similar
argument shows that $a_i+a_{i+1}=a_{i+1}+a_{i}=a_i$ for all $i$. 

Consider
\[
a_1+a_2\ge_S a_2+a_3\ge_S a_2+a_4\ge_S\dots
\ge_S a_2+a_n\ge_S a_2+a_1\,.
%=a_1+a_2
\]
This is a cycle of length $n-1$. We show that every link in
it is an $(S,S;0)$-pair. Indeed, the sum of two
$(S,S)$-matrices is also an $(S,S)$-matrix, hence the sum of
two $(S,S)$-triples is an $(S,S)$-triple as well. Now
$(a_1,a_1;a_2)$ and $(a_2,a_2;a_3)$ are $(S,S)$-triples, and
so their sum $(a_1+a_2,a_1+a_2;a_2+a_3)$ is an
$(S,S)$-triple, proving that the first link is indeed an
$(S,S;0)$-pair. Similarly, $(a_2,a_2;a_2)$ and
$(a_i,a_i;a_{i+1})$ are $(S,S)$-triples, and their sum shows
that all the other links in the chain are
$(S,S;0)$-pairs,~too.

The minimality of~$n$ then forces this chain to be trivial,
that is, all links are equal. Thus $a_1+a_2=a_2+a_3$. But we
know that $a_1+a_2=a_1$ and $a_2+a_3=a_2$. Therefore
$a_1=a_2$. This contradicts the minimality of~$n$ in the
original chain. This argument works for every~$n>1$, and so
$(1)$ is proved.

To prove~$(2)$ suppose that $u\sqsupseteq v$ and $v\sqsupseteq u$. 
Then there is a $\ge_S$-chain from~$u$ to~$v$, and from~$v$
to~$u$. Thus the statement in~$(1)$ shows that
$u=v$. Therefore~$\ge$ is indeed a partial order. To prove
the other statement, suppose that
\[
b_1\ge_S b_2\ge_S b_3\ge_S \dots \ge_S b_n\,,
\]
where $(b_1,b_n)\in S$. Consider
\begin{equation}\label{cycle_of_bees}
b_1+b_1\ge_S b_2+b_1\ge_S b_3+b_1\ge_S 
\dots \ge_S b_n+b_1\,.
\end{equation}
Here $b_n+b_1\ge_S b_1=b_1+b_1$ by the properties of~$+$,
since $b_n$ and $b_1$ are~$S$-related. Hence 
(\ref{cycle_of_bees}) is a cycle,
which must be trivial by~$(1)$. Therefore $b_1=b_n+b_1\ge_S
b_n$, proving~$(2)$. 

Now we prove that $\C R(S,S;\sqsupseteq)$ holds. 
Suppose that
\[
\left[\begin{array}{cc}
a&b\\
c&d\end{array}\right]\in M(S,S)\,,
\]
and $u\in C$ satisfies that 
$u\sqsupseteq b$ and $u\sqsupseteq c$. 
If $B$ is an $S$-block that contains these four
elements, then Theorem~\ref{join_prop}~(3) guarantees that
$b+c\sim_B a+d$.
But the relation $\sim_B$ is the intersection of $\geq_B$
and its converse, the relation $\geq_B$ is contained in $\geq_S$,
and $\geq_S$ is antisymmetric by item~(2).
Thus $\sim_B$ is equality on $B$, yielding
$b+c=a+d$. As $+$ is idempotent and $\sqsupseteq$
is compatible, we get that $u=u+u\sqsupseteq b+c = a+d$. 
But $a+d\geq_S d$ by the definition of a join
term, so $u\sqsupseteq a+d\sqsupseteq d$, proving that
$\C R(S,S;\sqsupseteq)$ holds. 

For item (4), let $\rho$ be the transitive closure of
$\sim_T$ in $\m a$. Since $\sim_T$ is a tolerance,
its transitive closure $\rho$ is a congruence. 
Now suppose that $\gg$ is the least
compatible partial order on $\m a$
such that $\C R(T,T;\gg)$ holds.
If
\[
\left[\begin{array}{cc}
p&q\\
r&s\end{array}\right]\in M(T,T)\,,
\]
and $q=r$, then $r\gg r$ and $r\gg q (=r)$,
so $r\gg s$. Thus $\gg$ contains all $(T,T;0)$-pairs,
and therefore contains $\geq_T$.
According to Definition~\ref{rectangular_df}, 
$[T,T]_{\m r}$ is the
intersection of $\gg$ and its converse,
hence $[T,T]_{\m r}$ 
contains the intersection of $\geq_T$
and its converse, hence contains $\sim_T$.
Since $\rho$ is the least congruence containing
$\sim_T$, we get $\rho \leq [T,T]_{\m r}$.

On the other hand, according to 
Theorem~\ref{least_congruence},
$[T,T]_{\m r}$ is the least congruence
$\delta$ for which $T/\delta$ is
a rectangular tolerance of $\m a/\delta$.
By items (2) and (3) of this theorem,
$\rho$ is a congruence with this property.
Therefore $[T,T]_{\m r} = \rho$.
\end{proof}

The following consequence of Theorem~\ref{quotients}
is worth stating separately.

\begin{cor}\label{sim_is_equality}
Let $\vr V$ be a variety with a join term, and
let $T$ be a tolerance on some $\m a\in {\vr V}$.
$T$ is rectangular if and only if $\sim_T$
is the equality relation.
\end{cor}

\begin{proof}
If $T$ is rectangular, then 
$\sim_T$ is the equality relation
according to Theorem~\ref{join_props_2}~(1).
On the other hand, if $\sim_T$ is the equality relation,
then Theorem~\ref{quotients}~(4) guarantees
that $[T,T]_{\m r} = 0$.
\end{proof}

Corollary~\ref{sim_is_equality}
implies
the following characterization of 
rectangular tolerances in varieties
with join terms, which is a refinement
of Theorem~\ref{un_rect_repn_thm}.

\begin{cor}\label{rect_with_+}
Let $\vr V$ be a variety with a join term $+$, and
let $T$ be a tolerance on an algebra~$\m A\in\vr V$. 
$T$ is rectangular if and only if 
$+$ is a compatible semilattice operation
on $T$. 
\end{cor}

\begin{proof}
If $T$ is rectangular, then 
$+$ is a compatible semilattice operation
on $T$ by Theorem~\ref{join_props_2}~(5).
Conversely, suppose that $+$ is a compatible
semilattice operation on $T$. If 
$(r,s)$ is a $(T,T;0)$-pair, 
then there is a matrix
\[
\left[\begin{array}{cc}
p&q\\
r&s\end{array}\right]=
\left[\begin{array}{cc}
f(\wec{a},\wec{u})&f(\wec{a},\wec{v})\\
f(\wec{b},\wec{u})&f(\wec{b},\wec{v})
\end{array}\right]
\in M(T,T)
\]
with $q=r$. In this situation 
\[
q+r 
%=f(\wec{a},\wec{v}) + f(\wec{b},\wec{u})
= f(\wec{a}+\wec{b},\wec{v}+\wec{u}) 
%= f(\wec{a},\wec{u}) + f(\wec{b},\wec{v}) 
= p+s\,,
\]
and all elements $p, q, r, s, q+r$ and $p+s$
lie in a single $T$-block. Therefore
$r = q+r = p+s\ge s$ in the semilattice order 
on that block. This means that the equation
$r+s = r$ holds, which is a property that is
independent of the block that we are considering.
Thus if $(r,s)$ is a $(T,T;0)$-pair, then
$r\ge s$ in the semilattice order on any
block containing $r$ and $s$.
This conclusion implies that on any $T$-block
the relation $\ge_T$ is contained in the semilattice
order, and therefore
that $\sim_T$ is the equality relation.
Hence $T$ is rectangular by
Corollary~\ref{sim_is_equality}.
\end{proof}

The following corollary will be used in Section~\ref{ACD}.

\begin{cor}\label{ACD_cor}
Let $\vr V$ be a variety with a join term $+$, and
let $T$ be a rectangular
tolerance on an algebra~$\m A\in\vr V$. 
Assume that $a$ and $b$
are distinct $T$-related elements of $\m a$.
If $\rho$ is a congruence on $\m a$
that is maximal with respect to $(a,b)\notin\rho$,
then $T/\rho$ is a nontrivial
rectangular tolerance on the subdirectly
irreducible algebra $\m a/\rho$.
\end{cor}

\begin{proof}
If $T$ is rectangular, then $\sim_T$ is the equality
relation according to Theorem~\ref{join_props_2}~(1).
Therefore every congruence $\rho$
on $\m a$ contains 
%the transitive closure of 
$\sim_T$. From Theorem~\ref{quotients},
$T/\rho$ is a rectangular congruence
of $\m a/\rho$ for any $\rho$.
Choosing $\rho$ as described in
the statement of this corollary guarantees
that $T/\rho$ is nontrivial and
that $\m a/\rho$ is subdirectly irreducible.
\end{proof}


\section{Varieties Omitting Rectangular Tolerances}\label{omit_rect_tol}


The purposes of this section are to show that
a variety omits rectangular tolerances
if and only if it omits rectangular congruences,
and then to give three different Maltsev characterizations
of the class of varieties with these properties.


\begin{lm}\label{free_rect2}
Let $\vr V$ be a variety with a join term and 
let $\m F = \m f_{\vr V}(x,y)$ 
be the free $\vr V$-algebra generated by $\{x,y\}$.
Let $T = \Tg f(x,y)$, let $\theta = \Cg f(T)$, and
let $B = x/\theta$.
The relations $\geq_{\theta}$ 
and $\geq_T$ have the same restriction to $B$.
\end{lm}

\begin{proof}
It follows from
Lemma~\ref{basic_prop}~(4) that 
$\geq_T|_B = {\rm tr.cl.}(S|_B)$, where
$S$ is the set of $(T,T;0)$-pairs of $\m f$,
and that 
$\geq_{\theta}|_B = {\rm tr.cl.}(\overline{S}|_B)$, where
$\overline{S}$ 
is the (larger) set of $(\theta,\theta;0)$-pairs of $\m f$.
Thus, it is enough to show that if 
$(r,s)\in B^2$ is a $(\theta,\theta;0)$-pair,
then $(r,s)$ is in the transitive closure
of the set of $(T,T;0)$-pairs.
So choose a $\theta,\theta$-matrix of the form
\[
\left[\begin{array}{cc}
p&r\\
r&s\end{array}\right]
=
\left[\begin{array}{cc}
t(\wec{a},\wec{c})&
t(\wec{a},\wec{d})\\
t(\wec{b},\wec{c})&
t(\wec{b},\wec{d})
\end{array}\right]\,.
\]
Let $h\colon \m f\to\m f$ be the endomorphism 
determined by $x\mapsto x, y\mapsto x$.
The conditions $\wec{a}\equiv_\theta \wec{b}$ 
and 
$\wec{c}\equiv_\theta \wec{d}$ 
are equivalent
to $h(\wec{a})=h(\wec{b})$
and $h(\wec{c})=h(\wec{d})$.
As in the proof of Lemma~\ref{free_rect},
consider the
following~$4\times 4$ matrix:
\[
\left[\begin{array}{cccccc}
t(\wec{a},\wec{c})&
t(\wec{a},h(\wec{c}))&=&
t(\wec{a},h(\wec{d}))&
t(\wec{a},\wec{d})\\
t(h(\wec{a}),\wec{c})&
t(h(\wec{a}),h(\wec{c}))&=&
t(h(\wec{a}),h(\wec{d}))&
t(h(\wec{a}),\wec{d})\\
\Vert&\Vert&&\Vert&\Vert\\
t(h(\wec{b}),\wec{c})&
t(h(\wec{b}),h(\wec{c}))&=&
t(h(\wec{b}),h(\wec{d}))&
t(h(\wec{b}),\wec{d})\\
t(\wec{b},\wec{c})&
t(\wec{b},h(\wec{c}))&=&
t(\wec{b},h(\wec{d}))&
t(\wec{b},\wec{d})
\end{array}\right]\,.
\]
Recall from that earlier proof that
the four elements in the middle are
equal to~$x$ and that 
the four $2\times 2$ matrices in the
corners are $T,T$-matrices.
All elements of this $4\times 4$ matrix are contained
in~$B$, because all elements are $\theta$-related,
$r, s\in B$ by assumption,
and $B$ is a block of~$\theta$.

Proceeding as we did in the proof of Lemma~\ref{free_rect},
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~$k$, and apply~$r(x,y)$ to these two
matrices, the first matrix in the first argument, the
second matrix in the second argument. 
As before, we get 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 of the form:
\[
J = \left[\begin{array}{ccc}
p&i&r\\
j&r&\ell\\
r&m&s\\
\end{array}\right]\,.
\]
The four $2\times 2$ matrices in 
the corners of the $3\times 3$ matrix $J$
are $T,T$-matrices.
From the matrix in
the upper right corner we get that $r\ge_T \ell$. From the
matrix in the lower left corner we get that $r\ge_T m$. From
the matrix in the lower right corner we get, using
Theorem~\ref{join_prop}~(4), that $\ell + m\ge_T s$. But $+$ is
idempotent, $\ge_T$ is compatible, and $\ge_T$ is transitive
on~$B$, so we see that $r=r+r\ge_T \ell + m\ge_T s$.
\end{proof}


\begin{thm}\label{rect_cong}
Let $\vr V$ be a variety. The following are equivalent.
\begin{enumerate}
\item[(1)] 
No member of $\vr V$ has a nontrivial
rectangular congruence.
\item[(2)] 
No member of $\vr V$ has a nontrivial
rectangular tolerance.
\end{enumerate}
\end{thm}

\begin{proof}
Clearly (2)$\mimplies$(1),
since congruences are tolerances.
The same type of
argument as that used in the proof of
Lemma~\ref{4_eq_conditions}
shows that to prove the converse
it is enough to show that 
\[
(x,y)\in [\theta,\theta]_{\m r}
\wrel{\mimplies}
(x,y)\in [T,T]_{\m r}
\]
where $\theta = \Cg f(x,y)$ and $T = \Tg f(x,y)$
are the congruence and tolerance generated
by $(x,y)$ in the free algebra 
$\m f = \m f_{\vr V}(x,y)$.
So assume that $(x,y)\in [\theta,\theta]_{\m r}$.
This assumption, which is equivalent to
the assumption that $\vr V$ has no
member with a rectangular congruence, implies that
$\vr V$ has no algebra with a strongly
rectangular congruence.
Theorems~\ref{omit_1} and \ref{str_ab_thm2}
now guarantee that $\vr V$ has a join term.
This puts us into position to use
Theorem~\ref{quotients}.
By part~(4) of that theorem, the condition
that $(x,y)\in [\theta,\theta]_{\m r}$ 
means that  $(x,y)$ lies in the transitive
closure of $\sim_{\theta}$.
But according to Lemma~\ref{basic_prop}~(5),
the fact that $\theta$ is a congruence implies
that $\sim_{\theta}$ is transitive.
Therefore $x\sim_{\theta} y$. 
This implies that 
$x\ge_{\theta} y$ and 
$y\ge_{\theta} x$, so by 
Lemma~\ref{free_rect2} we get that 
$x\ge_T y$ and 
$y\ge_T x$. Since $B=x/\theta$ is the unique
$T$-block containing $x$ and $y$ we have
$x\geq_B y$ and $y\geq_B x$; i.e., $x\sim_B y$
and hence $x\sim_T y$.
Theorem~\ref{quotients}~(4) 
now guarantees that $(x,y)\in [T,T]_{\m r}$.
\end{proof}


The proof of Theorem~\ref{rect_cong} shows that
a variety $\vr V$ contains no member with a rectangular
tolerance or congruence if and only if
$x\sim_T~y$ in $\m f = \m f_{\vr V}(x,y)$ for
$T = \Tg f (x,y)$. This property is easily seen
to be expressible by an idempotent
Maltsev condition, which we record here.


\begin{thm}\label{rect_Malt_cond1}
Let $\vr V$ be a variety.
The following are equivalent.
\begin{enumerate}
\item[(1)] No member of $\vr V$ has a nontrivial
rectangular tolerance.
\item[(2)] $x\sim_T y$ in $\m f = \m f_{\vr V}(x,y)$ for
$T = \Tg f (x,y)$. 
\item[(3)] 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)] $x\wrel{\approx}f_1(y,y,x,x,x,y)$,
\item[(ii)] $f_m(y,x,x,x,y,y)\wrel{\approx} y$,  
\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}
\item[(4)] For any algebra $\m a\in {\vr V}$, tolerance
$S$ on $\m a$, and $(a,b)\in S$, 
$a\ge_S b$ holds.
\item[(5)] For any algebra $\m a\in {\vr V}$, tolerance
$S$ on $\m a$, and $(a,b)\in S$, 
$a\sim_S b$ holds.
\end{enumerate}
\end{thm}

\begin{proof}
The proof of Theorem~\ref{rect_cong}
shows that (1) and (2) are equivalent.

If (2) holds, then $x\ge_T y$ in $\m f$, 
so according 
to Lemma~\ref{ge_in_free2} there
exists a sequence
$f_1,\ldots,f_m$ of sixary
terms satisfying the identities in~(3).

Now suppose that (3) holds. Choose
$\m a\in {\vr V}$, a tolerance $S$ on $\m a$,
an $S$-block $B$ and $a, b\in B$.
Since $B$ is a block it is closed under
all the idempotent terms that appear in 
item (3). By substituting $a$ and $b$ 
for $x$ and $y$ in
the identities of (3) we obtain 
elements of $S$ witnessing that $a\geq_B b$.
Namely, the fact that $(a,b)\in S$
implies that for each $i$ the matrix
\[
\left[\begin{array}{cc}
p_i&q_i\\
r_i&s_i
\end{array}\right] = 
\left[\begin{array}{cc}
f_i(a,b,a,b,a,b)&
f_i(a,a,a,b,b,b)\\
f_i(b,b,a,a,a,b)&
f_i(b,a,a,a,b,b)
\end{array}\right]
\]
is in $M(S,S)$. Identities
of type (iii) imply that $q_i=r_i$,
so each $(r_i,s_i)$ is an $(S,S;0)$ pair.
The other identities imply that
the sequence of these
pairs is a connected
chain from $a$ to $b$.
Thus $a\geq_S b$, and item (4) is established.

The argument of the previous paragraph
showed that $a\geq_B b$ for some $S$-block $B$, and a similar
one shows that $b\geq_B a$ for the same block $B$. Hence
$a\sim_B b$, which implies that $a\sim_S b$.
This establishes item (5).

Item~$(2)$ is a special case of $(5)$.
\end{proof}

Our next task will be to connect the foregoing
with the Hobby--McKenzie term.
First we need a lemma.

\begin{lm}\label{2-elt-semi}
If $\m c$ is a nontrivial 
idempotent algebra with a compatible
semilattice term, then there is an algebra
$\m d$ that is a homomorphic image
of a subalgebra of $\m c$
and is term equivalent to the $2$-element
semilattice.
\end{lm}

\begin{proof}
Denote the semilattice operation by~$+$,
and call it join.
Since the semilattice order is equationally
definable by 
\[x\geq y\wrel{\miff} x=x+y\,,
\]
and $+$
is compatible, it follows that
the semilattice order
is a compatible relation.
As $C$ is not a singleton, there
exist two distinct elements $0<1$ of~$\m C$. 
If $c_1,\ldots,c_k$ belong to 
the interval $[0,1]$ (with respect
to the semilattice ordering), and 
$f$ is a term operation of $\m c$, then
\[
1 = f(1,\ldots,1)\ge f(c_1,\ldots,c_k)\ge f(0,\ldots,0) = 0
\]
by the idempotence of $f$ and the compatibility of $\ge$.
Thus $[0,1]$ is a nontrivial subalgebra of $\m c$.
Replacing $\m c$ by this subalgebra,
we may henceforth assume that $\m c$
has least and largest elements $0$ and $1$.

If $f$ is a $k$-ary term operation of $\m c$, then since 
$+$ is compatible the operation $f$ must
commute with $x_1+\cdots + x_k$ on the matrix
\[
\left[\begin{array}{cccc}
x_1 & 0 & \cdots & 0\\
0  & x_2 & \cdots & 0\\
\vdots & \vdots & & \vdots\\
0  & 0 & \cdots & x_k
\end{array}\right]\,.
\]
This says exactly that 
$f(x_1,\ldots,x_k) = f_1(x_1) + \cdots + f_k(x_k)$
where $f_i(x_i):=f(0,\dots,0,x_i,0\dots,0)$, with $x_i$ in
the $i$-th position. As $f$ is idempotent, we have that
\[
f_1(x)+\dots +f_n(x)=x
\]
for every $x$. In particular, $f_i(x)\le x$ for every
$x$. The functions $f_i$ are of course $+$-endomorphisms,
too, since $f$ is compatible with $+$ and $0+0=0$.

Choose a maximal 
semilattice ideal $I$ of $\m c$
that does not contain~$1$. Let $F=C-I$ be
the complementary filter. Let $\theta$
be the equivalence relation on $C$ whose
classes are $I$ and $F$.
We claim that $\theta$ is a congruence of~$\m C$, and
that $\m C/\theta$ is term equivalent to a semilattice.
To show this, represent an arbitrarily chosen 
term $f$ 
%be a term of~$\m C$, and represent it 
as a sum of unary polynomials as
described in the previous paragraph.
The condition $f_i(x)\le x$ implies that $f_i(I)\subseteq
I$. We prove that either $f_i(F)\subseteq F$, or
$f_i(F)\subseteq I$. Assume 
that $f_i(F)\not\subseteq F$, and choose
$u\in F$ so that $f_i(u)\in I$. By the
maximality of~$I$, there exists an element $a\in I$ such
that $a+u=1$. Since $f_i$ is a $+$-endomorphism
we have $f_i(1)=f_i(a)+f_i(u)\in I$. This shows that if
$f_i(u)\in I$, then $f_i(1)\in I$, hence $f_i(C)\subseteq I$.
Consequently each $f_i$ preserves $\theta$.
Since $+$ also preserves $\theta$, it follows that $f$ preserves it,
so $\theta$ is a congruence of~$\m C$.

The quotient $\m d = \m c/\theta$ is a $2$-element
idempotent algebra with a compatible semilattice term $+$.
Any such algebra is term equivalent
to the $2$-element semilattice, as we now show.
Suppose that the elements of $D$
are denoted $0$ and $1$, and that $0<1$
in the $+$-order.
As above, each term operation of $\m d$ may be
represented as $f(x_1,\ldots,x_k) = f_1(x_1)+\cdots + f_k(x_k)$
where each $f_i$ is a 
decreasing $+$-endomorphism.
But since $D = \{0,1\}$ this forces each $f_i$ to
be the identity function or the constant function
with value $0$. If $J \subseteq \{1,\ldots,k\}$
is the set of indices $j$ such that 
$f_j$ is the identity, then $J\neq\emptyset$,
since $f$ is idempotent, and $f(x_1,\ldots,x_k) = 
\sum_J x_j$. Hence $f$ equals a semilattice term operation.
\end{proof}

Now we are in position to give a second
Maltsev characterization of the class
of varieties that omit rectangular tolerances.

\begin{thm}\label{rect_Malt_cond2}
Let $\vr V$ be a variety.
The following are equivalent.
\begin{enumerate}
\item[(1)] No member of $\vr V$ has a nontrivial
rectangular tolerance.
\item[(2)] $\vr V$ satisfies an idempotent
(linear) Maltsev condition that fails in the 
variety of semilattices.
\item[(3)] $\vr V$ has a Hobby--McKenzie term.
\end{enumerate}
\end{thm}


\begin{proof}
The equivalence of (2) and (3) is Theorem~\ref{hobby-mckenzie_term}.

To show that (1)$\mimplies$(2) we argue
that the idempotent linear Maltsev
condition from Theorem~\ref{rect_Malt_cond1}~(3)
fails in the variety of semilattices.
We give two proofs of this.

First, the Maltsev condition 
from Theorem~\ref{rect_Malt_cond1}
holds only in varieties omitting
rectangular congruences.
A calculation shows that the largest congruence on the
2-element semilattice is rectangular. Therefore the Maltsev
conditon fails in the variety of semilattices.

For the second proof, observe that if
$f_1(x_1,\ldots,x_6)$ is a semilattice term,
then identity (i) of
Theorem~\ref{rect_Malt_cond1}~(3),
$x\approx f_1(y,y,x,x,x,y)$,
implies that $f_1$ does not involve the
variables $x_1, x_2$ or $x_6$.
(Otherwise, by putting the absorbing element 
of the 2-element semilattice 
in for these variables and the nonabsorbing
element for the other variables
one obtains the contradiction that
these two elements are equal.)
This, together with the first instance of
identity (iii), 
$f_{1}(x,x,x,y,y,y)\approx f_{1}(y,y,x,x,x,y)\;(\approx x)$,
shows that $f_1$ does not involve $x_4, x_5$
or $x_6$ either. Therefore, by idempotence,  
$f_1(x_1,\ldots, x_6) \approx x_3$.
This argument propagates down the chain
of $f_i$'s via identity (iv), showing that each 
term satisfies
$f_i(x_1,\ldots,x_6)\approx x_3$.
Identity (ii) then reduces to $x\approx y$,
which fails in the variety of semilattices.

Now we prove (2)$\mimplies$(1). 
Assume that $\vr V$ satisfies a
(not necessarily linear) idempotent
Maltsev condition, called $W$, that fails
in the variety of semilattices.
Let $T = \tg(x,y)$ be the tolerance generated by 
$(x,y)$ on the free algebra $\m F=\m F_{\vr V}(x,y)$,
and let $B$ be the $T$-block of idempotent
binary terms.
$B$ is closed under all idempotent terms of~$\vr V$,
so let $\m b$ be the algebra
with universe $B$ and basic operations all
idempotent terms of~$\vr V$. $\m b$ is an
idempotent algebra 
that satisfies the idempotent Maltsev 
condition $W$.
The relation~$\sim_T$ restricted to~$B$ is a
congruence~$\theta$ of~$\m B$.

Since the Maltsev condition 
$W$ is nontrivial, the
variety generated by $\m b$
has a join term $+$. 
By~Theorem~\ref{join_prop}~(2) and (5), 
$+$ is a compatible semilattice term
of the idempotent algebra~$\m C=\m B/\theta$.
If $\m c$ is nontrivial, then 
Lemma~\ref{2-elt-semi} guarantees
that some homomorphic image $\m d$ of 
some subalgebra $\m c$ is term
equivalent to the $2$-element
semilattice. The Maltsev condition $W$ 
is satisfied by $\m c$,
hence by $\m d$, hence by the
variety of semilattices.
But we have assumed the contrary,
so we conclude that $\m c$ is trivial.
This proves that $\m b$ is a single
$\sim_T$ class, and therefore that 
$x\sim_T y$ in $\m f_{\vr V}(x,y)$.
According to Theorem~\ref{rect_Malt_cond1},
this means that $\vr V$ contains
no algebra with a 
nontrivial rectangular tolerance.
\end{proof}

In the preceding proof the implication
(3)$\mimplies$(1) is verified indirectly.
Before proceeding, we would like
to describe a direct proof
of this implication.
Suppose that $f(x_1,\ldots,x_n)$
is a Hobby--McKenzie term for $\vr V$.
Recall from Chapter~\ref{centrality}
that for a subset $U\subseteq N := \{1,\ldots,n\}$
the expression $f_U(x,y)$ denotes the term
obtained from $f$ by substituting 
$x$ for $x_i$ if $i\in U$ and
$y$ for $x_i$ otherwise.
The statement that $f$ is a Hobby--McKenzie term
for $\vr V$ is precisely the statement
that $f$ is idempotent and for any 
nonempty subset $U\subseteq N$ there exist
subsets $V, W\subseteq N$ with $U\subseteq V$,
$U\not\subseteq W$ and ${\vr V}\models f_V(x,y)\approx f_W(x,y)$.

As usual, let $\m F=\m F_{\vr V}(x,y)$, 
$T = \Tg f(x,y)$, and $B$ be the $T$-block of 
idempotent binary terms.
Our aim is to prove that $x\sim_T y$ using
the equations of the Hobby--McKenzie term. 
Recall that the restriction of $\sim_T$ 
to $B$ is transitive.
Let $U\subseteq N$ be minimal under inclusion with respect
to the property that $f_U(x,x+y)\sim_T x$.  (Such a set~$U$
exists, since $f_N(x,x+y) = x \sim_T x$.) If $U$ is not empty,
then there exist $V, W\subseteq N$ such that $U\subseteq V$
but $U\not\subseteq W$, and $f_V(x,x+y) = f_W(x,x+y)$. Hence
by Theorem~\ref{join_prop}, 
\begin{equation}\label{HMpart1}
f_U(x,x+y)+f_W(x,x+y)\sim_T
f_{U\inter W}(x,x+y)\,,
\end{equation}
and
\begin{equation}\label{HMpart2}
f_U(x,x+y)+f_V(x,x+y)\sim_T
f_{U\inter V}(x,x+y)=
f_U(x,x+y)\,.
\end{equation}
Since $f_V(x,x+y)=f_W(x,x+y)$, the left
hand sides of lines (\ref{HMpart1})
and (\ref{HMpart2}) are the same. Therefore 
\[
f_{U\inter W}(x,x+y) \sim_T 
f_U(x,x+y) \sim_T x\,.
\]
This contradicts the minimality of~$U$ unless
$U = \emptyset$, which leads to the conclusion that 
$x\sim_T f_{\emptyset}(x,x+y) = x+y$. A symmetric
argument shows that $y\sim_T x+y$,
and therefore $x\sim_T y$. At this
point, Theorem~\ref{rect_Malt_cond1}
may be invoked to deduce that
no algebra in $\vr V$
has a rectangular tolerance.

\bigskip

We turn to our third
Maltsev characterization 
of the class of varieties that
omit rectangular tolerances.
This new Maltsev condition 
involves four variable terms,
and somewhat resembles
Alan Day's Maltsev condition
characterizing the class of
congruence modular varieties
(cf.\ \cite{day1}).
%This new condition will be used
%in Chapters~\ref{rectangulation2}
%and \ref{congruence_sdj}.

\begin{df} \label{D2}
Let $\m a$ be an algebra.
If $\Conb(\m a)$ has a sublattice
isomorphic to the lattice $\m d_2$
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(30,30)
\put(0,10){\circle*{1.2}}
\put(0,20){\circle*{1.2}}
\put(15,0){\circle*{1.2}}
\put(15,20){\circle*{1.2}}
\put(15,30){\circle*{1.2}}
\put(30,10){\circle*{1.2}}
\put(30,20){\circle*{1.2}}

\put(0,10){\line(3,-2){15}}
\put(0,10){\line(3,2){15}}
\put(0,10){\line(0,1){10}}
\put(15,30){\line(-3,-2){15}}
\put(15,30){\line(0,-1){10}}
\put(15,30){\line(3,-2){15}}
\put(30,10){\line(-3,-2){15}}
\put(30,10){\line(-3,2){15}}
\put(30,10){\line(0,1){10}}

\put(14,15){$\alpha$}
%\put(14,32){$\tau$}
\put(-4.5,9){$\mu$}
\put(-4.5,19){$\nu$}
\put(32,9){$\delta$}
\put(32,19){$\theta$}
\end{picture}
\end{center}
\caption{$\m d_2$}\label{fig20}
\end{figure}
with $(\theta\circ\mu)\cap (\nu\circ\delta) \not\subseteq \alpha$,
then we call this sublattice a {\bf special $\m d_2$}
\index{special $\m d_2$}
in $\Conb(\m a)$. We say that 
{\bf ${\vr V}$ omits special $\m d_2$'s} 
if for no $\m a\in {\vr V}$ 
does $\Conb(\m a)$ have a sublattice that is
a special $\m d_2$.
\end{df}

As an aid to recognizing copies of
$\m d_2$ in congruence lattices, we record
the following.

\begin{lm} \label{presentation}
A presentation of $\m d_2$ relative
to the variety of all lattices is
$\lb G\mid R\rb$ where 
$G = \{x, y, z\}$
and $R$ consists of the relations:
\begin{enumerate}
\item[(i)] $x\wedge  z\leq  y$, 
%\item[(ii)] $x\vee  z\geq  y$, 
\item[(ii)] $x\leq  z\vee( y\wedge x)$, 
\item[(iii)] $ z\leq x\vee( y\wedge  z)$, and 
\item[(iv)] $ y\leq (x\wedge  y)\vee( y\wedge  z)$.
\end{enumerate}
Moreover any lattice generated by $G$ and
satisfying the relations in $R$ 
and also satisfying $x\not\leq  y$ 
is isomorphic to $\m d_2$.
\end{lm}

\begin{proof} 
The reader can easily verify
that the copy of $\m d_2$ 
in Figure~\ref{fig21}
is generated by $x, y$ and $ z$, and
that the relations in the statement
of the lemma hold for these elements.
To show that it is a presentation
of $\m d_2$ we must prove from the
relations (and the laws of lattice theory)
that the seven elements $x$, $ y$, $ z$,
$p:= x\wedge y$, 
$q:=  z\wedge y$, 
$s:= x\wedge y\wedge z$, and 
$t:= x\vee y\vee z$ are
closed under meet and join.

\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(30,43)
\put(0,15){\circle*{1.2}}
\put(0,25){\circle*{1.2}}
\put(15,5){\circle*{1.2}}
\put(15,25){\circle*{1.2}}
\put(15,35){\circle*{1.2}}
\put(30,15){\circle*{1.2}}
\put(30,25){\circle*{1.2}}

\put(0,15){\line(3,-2){15}}
\put(0,15){\line(3,2){15}}
\put(0,15){\line(0,1){10}}
\put(15,35){\line(-3,-2){15}}
\put(15,35){\line(0,-1){10}}
\put(15,35){\line(3,-2){15}}
\put(30,15){\line(-3,-2){15}}
\put(30,15){\line(-3,2){15}}
\put(30,15){\line(0,1){10}}

\put(14,0){$s$}
\put(14,20){$y$}
\put(14,37){$t$}
\put(-4.5,14){$p$}
\put(-4.5,24){$x$}
\put(32,14){$q$}
\put(32,24){$z$}
\end{picture}
\end{center}
\caption{ \ }\label{fig21}
\end{figure}

It is clear that the seven 
elements are ordered as they should
be, and consequently $ s$ and $ t$
are the bottom and top elements of the sublattice
they generate. The only incomparable doubletons are
$\{x, y\}$, $\{x, z\}$, $\{x,  q\}$,
$\{y, z\}$, $\{p, z\}$ and $\{p, q\}$.
By symmetry, we do not need to consider the
meets or joins of 
$\{y, z\}$ or $\{p, z\}$.
For the remaining cases:
\begin{itemize}
\item
$\{x,  y\}$: $x\wedge y =  p$ by
the definition of $p$. $x\vee y$ is above
$ z$ by relation (iii). Since $x\vee y$ 
is formally below
$x\vee y\vee z =  t$, it must equal it.
\item
$\{x, z\}$: $x\wedge z$ is below $ y$
by relation (i). This implies 
$x\wedge z = x\wedge y\wedge z =  s$,
as desired. $x\vee z$ is above 
$(x\wedge y)\vee(y\wedge z)$, since 
corresponding joinands are larger, and
$(x\wedge y)\vee(y\wedge z)\geq y$ by (iv).
Therefore $x\vee z = x\vee y\vee z =  t$.
\item
$\{x,  q\}$: $x\wedge q = x\wedge( y\wedge z)
=  s$ by the definition of $ q$. 
$x\vee q = x\vee( y\wedge z)$ is above
$ z$ by (iii), therefore it is above $x\vee z =  t$
and hence equal to $ t$.
\item
$\{ p, q\}$: $ p\wedge q$ is below $x\wedge z$,
which has already been shown to be the bottom element, so 
$p\wedge q =  s$. By (iv) we have 
$y\leq (x\wedge y)\vee(y\vee z) = p\vee q\leq y$,
so $p\vee q =  y$.
\end{itemize}

This proves that $\lb G\mid R\rb$ presents $\m d_2$.
For the last claim of the lemma, 
any lattice generated
by $G$ satisfying the relations $R$ which
is not isomorphic to $\m d_2$ must 
be a proper quotient of $\m d_2$. 
Since $\m d_2$ is a subdirectly irreducible
lattice whose least congruence collapses
$p$ and $x$, any such lattice will satisfy
$x =  p\leq  y$. Therefore, if a lattice
is generated by $G$ and satisfies $R$ but
does not satisfy $x\leq  y$, then
the lattice is isomorphic to $\m d_2$.
\end{proof}


%The next 
%Maltsev characterization of the class of varieties
%that omit rectangular tolerances that is strong
%enough to use in our proof of Lemma~\ref{leadsto_cond}.

\begin{thm} \label{equiv}
The following are equivalent
for a variety ${\vr V}$.
\begin{enumerate}
\item[(1)] ${\vr V}$ satisfies an idempotent
Maltsev condition that fails in the variety
of semilattices.
\item[(2)] Whenever $\theta, \mu, \nu$ and
$\delta$ are congruences on some $\m a\in {\vr V}$,
then $(\theta\circ\mu)\cap (\nu\circ\delta) \subseteq 
[(\theta\vee\delta)\wedge(\nu\vee\mu)]
\vee \delta\vee\mu$.
\item[(3)] ${\vr V}$ has a sequence of
terms
$f_i(x,y,u,v), 0\leq i\leq 2m+1$, such that 
\begin{enumerate}
\item[(i)] ${\vr V}\models f_0(x,y,u,v)\approx x$;
\item[(ii)] ${\vr V}\models f_i(x,y,y,y)\approx
f_{i+1}(x,y,y,y)$ for even $i$;
\item[(iii)] ${\vr V}\models f_i(x,x,y,y)\approx
f_{i+1}(x,x,y,y)$ and 
${\vr V}\models f_i(x,y,x,y)\approx f_{i+1}(x,y,x,y)$ 
for odd $i$;
\item[(iv)] ${\vr V}\models f_{2m+1}(x,y,u,v)\approx v$.
\end{enumerate}
\item[(4)] ${\vr V}$ omits special $\m d_2$'s.
\end{enumerate}
\end{thm}

\begin{proof}
We first establish the equivalence of
(2), (3) and (4), and connect
them to item (1) afterward.

Assume that (2) holds.
On $\m f = \m f_{\vr V}(x,y,u,v)$ let
$\theta = \Cg f(x,u)$, 
$\mu = \Cg f(u,v)$, 
$\nu = \Cg f(x,y)$, and 
$\delta = \Cg f(y,v)$.
Since 
$(x,v)\in (\theta\circ\mu)\cap (\nu\circ\delta)$, 
and (2) holds:
\[
(\theta\circ\mu)\cap (\nu\circ\delta) \subseteq 
[(\theta\vee\delta)\wedge(\nu\vee\mu)]
\vee \delta\vee\mu\,,
\]
we have 
$(x,v)\in [(\theta\vee\delta)\wedge(\nu\vee\mu)]
\vee \delta\vee\mu$.
Therefore, there is a sequence $f_0, f_1,\ldots,
f_{2m+1}$ of elements of $\m f$ such that 
$f_0 = x, f_{2m+1} = v,
(f_i,f_{i+1})\in \delta\vee \mu$
for even $i$ while 
$(f_i,f_{i+1})\in (\theta\vee\delta)\wedge(\nu\vee\mu)$
for odd $i$. If one chooses 4--ary terms representing 
the $f_i$'s, it is straightforward to see
that the equations of (3) hold.


Assume that (3) holds and that
some $\m a\in {\vr V}$ 
has congruences $\mu, \nu, \alpha, \delta$ and $\theta$
that generate a sublattice of $\Conb(\m a)$
isomorphic to $\m d_2$ labeled as in Figure~\ref{fig20}.
Choose any $(a,d)\in (\theta\circ \mu)\cap (\nu\circ\delta)$.
There exist elements $b, c\in A$ such that 
$a\wrel{\equiv_{\nu}}b\wrel{\equiv_{\delta}}d$
and 
$a\wrel{\equiv_{\theta}}c\wrel{\equiv_{\mu}}d$.
The pairs
$(f_i(a,b,c,d), f_{i+1}(a,b,c,d))$
belong to $\delta\vee\mu = \alpha$
for even $i$ and 
$(\theta\vee\delta)\wedge(\nu\vee\mu)\leq \alpha$
for odd $i$.
Hence $a = f_0(a,b,c,d)$ is $\alpha$-related
to $f_{2m+1}(a,b,c,d) = d$.
Since $(a,d)\in (\theta\circ \mu)\cap (\nu\circ\delta)$
was chosen arbitrarily, this implies that
$(\theta\circ \mu)\cap (\nu\circ\delta)\subseteq\alpha$.
Therefore this copy of $\m d_2$ is 
not a special $\m d_2$. The argument
works for every $\m d_2$ that appears
in a congruence lattice of a member
of $\vr V$, so $\vr V$ omits special $\m d_2$'s.

Next we argue the contrapositive of (4)$\mimplies$(2),
so assume that some $\m a\in {\vr V}$ has
congruences
$\mu, \nu, \delta$ and $\theta$ such that 
%\begin{equation}\label{noninclusion}
$$
(\theta\circ\mu)\cap (\nu\circ\delta) 
\not\subseteq 
[(\theta\vee\delta)\wedge(\nu\vee\mu)]
\vee \delta\vee\mu\,.
$$
%\end{equation}
Choose $(a,d)\in 
(\theta\circ\mu)\cap (\nu\circ\delta)$
satisfying 
$(a,d)\notin 
[(\theta\vee\delta)\wedge(\nu\vee\mu)]
\vee \delta\vee\mu$, as well as elements
$b, c\in A$ such that 
$a\wrel{\equiv_{\nu}}
b\wrel{\equiv_{\delta}}d$ and 
$a\wrel{\equiv_{\theta}}
c\wrel{\equiv_{\mu}}d$.
Let $\theta^*:=\Cg a ((a,c),(b,d))$, 
$\nu^*:= \Cg a((a,b),(c,d))$, and
$\alpha^*:= \Cg a((b,d),(c,d))\vee (\theta^*\wedge \nu^*)$.
We will show that the congruences $x:=\theta^*$,
$y:=\alpha^*$, and $z:= \nu^*$ satisfy the
presentation for $\m d_2$ given in Lemma~\ref{presentation}.
%Let $\delta^* = \alpha^*\wedge \theta^*$ and $\mu^* = \alpha^*\wedge \nu^*$.

Item (i) of Lemma~\ref{presentation} requires that 
$\theta^*\wedge \nu^*\leq \alpha^*$, which is
an immediate consequence of the definition of
$\alpha^*$.

Item (ii) requires that 
$\theta^*\leq \nu^* \vee (\alpha^* \wedge \theta^*)$.
Since $\nu^*$ contains $(a,b)$ and $(c,d)$ while
$\alpha^*\wedge \theta^*$ contains $(b,d)$, the join
is the congruence generated by $X\times X$ where 
$X = \{a,b,c,d\}$. Hence the join contains the generators
$(a,c)$ and $(b,d)$ of $\theta^*$. Item (iii) is 
symmetric to item (ii).

For item (iv), the congruence 
$\theta^*\wedge \alpha^*$ contains 
$\{(b,d)\}\cup (\theta^*\wedge\nu^*)$,
since this is true of each of $\theta^*$ and $\alpha^*$.
Similarly $\nu^*\wedge \alpha^*$ contains 
$\{(c,d)\}\cup (\theta^*\wedge\nu^*)$. This shows that
%although $(\theta^*\wedge \alpha^*)\vee (\alpha^*\wedge \nu^*)$ 
%is contained in $\alpha^*$,
$(\theta^*\wedge \alpha^*)\vee (\alpha^*\wedge \nu^*)$ 
contains all the generators of $\alpha^*$.
Hence $\alpha^* \leq (\theta^*\wedge \alpha^*)\vee 
(\alpha^*\wedge \nu^*)$.

Finally we must 
show that $\theta^*\not\leq \alpha^*$.
Assume that this is not the case.
Since $(a,c)\in\theta^*$, this assumption forces 
$(a,c)\in\alpha^*$.
By definition, $\theta^*\leq (\theta\vee\delta)$, 
$\nu^*\leq (\nu\vee\mu)$, $\Cg a(b,d)\leq\delta$ and 
$\Cg a(c,d)\leq \mu$,
so 
$$
\alpha^* = (\theta^*\wedge \nu^*)\vee\Cg a(c,d)\vee\Cg a(b,d)
\leq 
[(\theta\vee\delta)\wedge(\nu\vee\mu)]
\vee \delta\vee\mu\,.
$$
Combining the facts that 
$(a,c)\in\alpha^*$ and $(c,d)\in \mu$, 
we obtain that the right
hand side of the previous displayed line
contains $(a,d)$. This is contrary
to the choice of $(a,d)$, so we are forced to conclude that
$\theta^*\not\leq \alpha^*$.
 This completes the proof that 
$\{\theta^*, \alpha^*, \nu^*\}$ generates 
a copy of $\m d_2$ in $\Conb(\m a)$.
Indeed, this is a {\em special} $\m d_2$ since 
$a\wrel{\equiv_{\nu^*}} b\wrel{\equiv_{\alpha^*\wedge\theta^*}} d$
and
$a\wrel{\equiv_{\theta^*}} c\wrel{\equiv_{\alpha^*\wedge\nu^*}} d$,
so
$$
(a,d)\in 
(\theta^*\circ (\alpha^*\wedge\nu^*))\cap 
(\nu^*\circ(\alpha^*\wedge \theta^*)) - \alpha^*.
$$
This proves that 
item (4) of the theorem fails to hold.

Now we turn our attention to item (1).
If $\m 2 = \lb \{0,1\}; +\rb$ is the 
$2$-element join semilattice, 
$\theta, \nu\in \Con(\m 2\times \m 2)$
are the projection kernels, and 
$\alpha\in \Con(\m 2\times \m 2)$
is the kernel of the homomorphism
$+\colon \m 2^2\to \m 2$,
then $\theta, \nu$ and $\alpha$ generate
a special $\m d_2$. Therefore the variety
of semilattices does not satisfy
the idempotent Maltsev condition
in (3). This shows that (3)$\mimplies$(1).
To complete the proof, we 
must prove that (1)$\mimplies$(3).

Both (1) and (3) are statements
about the idempotent Maltsev conditions 
satisfied by $\vr V$.
The idempotent reduct $\Id({\vr V})$
satisfies the same idempotent Maltsev
conditions as $\vr V$, so to
prove (1)$\mimplies$(3)
we may (and do) assume that $\vr V$ is idempotent.

Suppose that (3) fails.
Then $\vr V$ does not have terms satisfying
the identities listed in (3), so
the method that we used to construct those terms
in our proof that (2)$\mimplies$(3) must fail.
Recall from that part of the proof
that we defined $\m f = \m f_{\vr V}(x,y,u,v)$,
$\theta = \Cg f(x,u)$, 
$\mu = \Cg f(u,v)$, 
$\nu = \Cg f(x,y)$, and 
$\delta = \Cg f(y,v)$.
We argued that if 
\[(x,v)\in 
[(\theta\vee\delta)\wedge(\nu\vee\mu)]
\vee \delta\vee\mu,
\] 
then 
(3) holds. Therefore, in our current
situation, the congruence
$\alpha:=[(\theta\vee\delta)\wedge(\nu\vee\mu)]
\vee \delta\vee\mu$ does not contain the
pair $(x,v)$.
Our plan is to use this fact to construct
a nontrivial algebra in $\vr V$ 
that has a compatible semilattice operation.

Let $X = \{x,y,u,v\}$, so that $\m f = \m f_{\vr V}(X)$. 
Let $\m g = \m f_{\vr V}(x,v)$ be the subalgebra
of $\m f$ generated by $\{x,v\}$.
The functions 
$\tau_i\colon X\to X$, 
$i = 1,2$, defined by 
$\tau_1(x) = \tau_1(y) = x = \tau_2(x) = \tau_2(u)$
and 
$\tau_1(u) = \tau_1(v) = v = \tau_2(y) = \tau_2(v)$
are transversals for the equivalence
relations $E_1 = \Eq X((x,y),(u,v))$ and
$E_2 = \Eq X((x,u),(y,v))$ respectively.
These transversals induce homomorphisms
$h_i\colon \m f\to \m f$
whose kernels are $\theta\vee\delta$ and
$\nu\vee\mu$ respectively (according
to Lemma~\ref{T_theta}~(1)), and whose
images are both $\m g$.
%  Both
%$h_1$ and $h_2$ have image equal to
%$\Sg f(x,v) = \m f_{\vr V}(x,v)$, which
%we will denote by $\m g$.
Therefore the kernel of 
\[
h:=h_1\times h_2\colon\m f\to \m g\times \m g \colon 
t(x,y,u,v)\mapsto (t(x,x,v,v),t(x,v,x,v))
\]
is $[(\theta\vee\delta)\wedge(\nu\vee\mu)]$.
The idempotence of $\vr V$ implies that
this homomorphism is surjective. Namely, 
if $(p,q) = (p(x,v), q(x,v))\in G\times G$,
then $(p\circ q)(x,y,u,v) := p(q(x,y),q(u,v))\in F$
is a preimage of $(p,q)$ under $h$.
This shows that
$\m g\times \m g$ is generated by 
the four elements $h(x) = (x,x)$, 
$h(y) = (x,v)$, 
$h(u) = (v,x)$, and 
$h(v) = (v,v)$. 
%It also shows that 
%$h$ induces an isomorphism
%\[
%\bar{h}\colon \m f/[(\theta\vee\delta)\wedge(\nu\vee\mu)]
%\to \m g\times\m g\,.
%\]
Since the kernel of $h$
%this homomorphism 
is contained in
$\alpha=[(\theta\vee\delta)\wedge(\nu\vee\mu)]\vee \delta\vee\mu$, 
we may use $h$ to recast our earlier assumption
that $(x,v)\notin\alpha$
as a statement about $\m g\times\m g$.
That assumption is equivalent to the assumption
that $h(x,v) = ((x,x),(v,v))$ does not belong to
the congruence 
\[
\begin{array}{rl}
h(\alpha) &= h([(\theta\vee\delta)\wedge(\nu\vee\mu)]\vee \delta\vee\mu)\\
&= \Cg {\m G\times \m G}(h(\delta\vee\mu))\\
&= \Cg {\m G\times \m G}(h(y,v), h(u,v))\\
&= \Cg {\m G\times \m G}(((x,v),(v,v)), ((v,x),(v,v)))\,.
\end{array}
\]
Therefore, if we let $\beta = h(\alpha)$, then 
$\beta = \Cg {\m G\times \m G}(Y\times Y)$
where $Y = \{(x,v), (v,x), (v,v)\}$, and
our earlier assumption that $(x,v)\notin\alpha$
in $\m f$ 
is equivalent to the assumption that
$((x,x),(v,v))\notin \beta$ in 
$\m g\times \m g$.

For any $(p,q)\in G^2$ there is an $r\in G$
such that $(p,q)\wrel{\beta}(r,r)$.
Indeed, if $p=p(x,v)$ and $q = q(x,v)$,
then for $r = p(q(x,v),v)$ we have
\begin{equation}\label{complicated_stuff}
\begin{array}{rl}
(p,q)&=(p(x,v),q(x,v))\\
&=(p(q(x,x),q(v,v)),p(q(x,v),q(x,v)))\\
&= (p\circ q)((x,x),(x,v),(v,x),(v,v))\\
&\equiv_{\beta} (p\circ q)((x,x),(v,v),(v,v),(v,v))\\
&=(p(q(x,v),q(v,v)),p(q(x,v),q(v,v)))\\
&=(p(q(x,v),v),p(q(x,v),v))\\
&=(r,r)\,.
\end{array}
\end{equation}
Of course, the element $r$ need not
be unique, since it may happen
that $(p,q)\wrel{\beta}(r,r)\wrel{\beta}(s,s)$ 
for some $s\neq r$.
To account for this nonuniqueness,
let $\gamma$ be the congruence on $\m g$
defined by 
\[
r\wrel{\gamma} s\wrel{\miff}
(r,r)\wrel{\beta}(s,s).
\]
%\[
%(r,s)\in\gamma\wrel{\miff}
%((r,r),(s,s))\in\beta.
%\]
This is indeed a congruence,
as it is $\iota^{-1}(\beta)$ for
$\iota\colon \m g\to \m g^2\colon p\mapsto (p,p)$
equal to the diagonal embedding. 
Write $\overline{G}$ for $G/\gamma$
and $\overline{p}$ for $p/\gamma$ if $p\in G$.
In this notation, we have shown in (\ref{complicated_stuff})
that there is a well defined function
%\[
%\ast\colon G\times G\to \overline{G}
%\]
\[
\ast\colon G\times G\to\overline{G}\colon 
({p},{q})\mapsto \overline{r}\;\;
\mbox{\rm if $(p,q)\wrel{\beta}(r,r)$.}
\]
%defined by $p\ast q = \overline{r}$ if 
%$(p,q)\wrel{\beta} (r,r)$. 
Explicitly, the value of $p\ast q$
is $\overline{p(q(x,v),v)}$, as we have shown.

If $q\wrel{\gamma} q'$, then 
$p(q(x,v),v)\wrel{\gamma} p(q'(x,v),v),$
since $p$ is a term and $\gamma$ is a congruence.
Hence
\[
p\ast q = \overline{p(q(x,v),v)} =  \overline{p(q'(x,v),v)} = p\ast q',
\]
showing that 
left $\ast$-multiplication by any $p\in G$ is
compatible with $\gamma$.
Since the generators of $\beta$ are
invariant under switching coordinates, 
the property 
$(p,q)\wrel{\beta} (r,r)$ is equivalent
to 
$(q,p)\wrel{\beta} (r,r)$, so $p\ast q = q\ast p$.
This commutativity implies that
right $\ast$-multiplication is also
compatible with $\gamma$. Consequently $\ast$
induces a well defined binary operation
(also called $\ast$) on $\overline{G}$:
\[
\ast\colon \overline{G}\times \overline{G}\to\overline{G}\colon 
(\overline{p},\overline{q})\mapsto \overline{r}\;\;
\mbox{\rm if $(p,q)\wrel{\beta}(r,r)$.}
\]
We have already established the commutativity
of this operation. The idempotence of $\ast$
follows from the fact that 
$(p,p)\wrel{\beta}(p,p)$ for any $p\in G$.
Associativity is clear, too, since 
\[
(\overline{p}\ast \overline{q})\ast \overline{r} 
= \overline{p(q(r(x,v),v),v)} = 
\overline{p}\ast(\overline{q}\ast\overline{r})\,.
\]
Thus $\ast$ is a semilattice operation
on the set $\overline{G}$.
Moreover, $\ast$ is a compatible semilattice
operation of the algebra $\overline{\m g} := \m g/\gamma$.
For suppose that $f$ is an $n$-ary 
basic operation of $\overline{G}$, and
that $\overline{p}_i, \overline{q}_i\in \overline{G}$
are chosen arbitrarily. Choose $r_i$ such that 
$(p_i,q_i)\wrel{\beta}(r_i,r_i)$ for all $i$.
Then, since $\beta$ is a congruence, 
$(f(\wec{p}),f(\wec{q}))\wrel{\beta}
(f(\wec{r}),f(\wec{r}))$. This shows that 
\[
f(\overline{\wec{p}})\ast f(\overline{\wec{q}}) = 
f(\overline{\wec{r}}) = f(\overline{\wec{p}}\ast\overline{\wec{q}})\,.
\]
Finally (and critically), $\overline{\m g}$ is nontrivial.
Indeed, since $((x,x),(v,v))\notin\beta$, we get
that $(x,v)\notin\gamma$, so $\overline{x}$ and $\overline{v}$
are distinct elements of $\overline{\m g}$.

This proves
that $\overline{\m g}\in\vr V$ is a nontrivial algebra
with a compatible semilattice operation.
Theorem~\ref{un_rect_repn_thm} applies
to show that
the total relation on $\overline{\m g}$
is rectangular. From Theorem~\ref{rect_Malt_cond2}
we derive that any idempotent Maltsev 
condition satisfied by $\vr V$ is also
satisfied by the variety of semilattices.
Thus (1) fails, and the proof is complete.
\end{proof}

In Theorem~\ref{omit_d2_sub} we will learn 
that if $\vr V$ satisfies an idempotent Maltsev
condition that fails in the variety of semilattices,
then $\vr V$ omits all $\m d_2$'s,
not just the special ones.

%   *********************************
