\chapter{A Theory of Solvability}\label{solvability}

%As is well known, 
A normal subgroup $N$ of a group $G$ is {\bf solvable}
if there is a finite chain of normal subgroups
of $G$,
$$
\{1\} = N_0 \subseteq N_1 \subseteq \cdots \subseteq N_k = N,
$$
such that each factor $N_{i+1}/N_i$ is abelian,
or equivalently such that
$[N_{i+1}, N_{i+1}] \subseteq N_i$ for all $i$.
An interval $I[M,N]$ in the normal subgroup
lattice of $G$ is a {\bf solvable interval}
if $N/M$ is a solvable normal subgroup of $G/M$.
Normal subgroups $M$ and $N$ of $G$ 
are {\bf solvably related}, written
$M\wrel{\solv} N$, if they belong 
to a solvable interval.
%Equivalently, $M$ is solvably related to $N$ if
%$I[M\cap N,M\vee N]$ is a solvable interval.
%This defines a binary relation 
%%on normal subgroups is a congruence 
%on the normal subgroup lattice of $G$. 
We will call the relation $\wrel{\solv}$ 
``the solvability relation'', and refer
to the study of this solvability relation as the
``theory of solvability'' for groups.
Among the important features of the
theory of solvability for groups are:
\begin{itemize}
\item The solvability relation is a congruence on the
normal subgroup lattice of any group.
\smallskip

\item The solvability relation is preserved
by homomorphisms in the sense that 
if $M, N$ and $P$ are normal subgroups of $G$
with $P\leq M\cap N$, then 
$M$ is solvably related to $N$ if and only if 
$M/P$ is solvably related to $N/P$ in $G/P$.
\smallskip

\item If $N$ is a minimal normal subgroup
of $G$, then $N$ is solvable if and only if 
it is abelian.
\end{itemize}

The ring-theoretic 
concept that corresponds directly to that of
a solvable normal subgroup is that of a nilpotent
ideal. In the early days of ring theory
%(about 100 years ago),
the concept of solvability was developed
in analogy with the group theoretical concept
via the Wedderburn radical ( = the largest
nilpotent ideal, if it exists).
Other radicals soon made their appearance,
which permitted new approaches to solvability
not bound to {\em finite} chains of congruences
with abelian factors. Moreover, in the case
of some radicals, the resulting 
``solvability theory'' was no longer 
%linked closely 
bound to the abelian property.
For example, if $\m r$ is a 
subdirectly irreducible ring that is not simple,
then the Brown-McCoy radical ( = the intersection
of maximal two-sided ideals)
contains the monolith 
{\em whether or not the monolith is abelian}.
Thus, it can happen that a minimal nonabelian
congruence is contained in the Brown-McCoy radical,
hence is ``solvable'' in the Brown-McCoy sense.

A third approach to solvability, 
valid for arbitrary finite algebras,
is developed by 
D.~Hobby and R.~McKenzie in Chapter~7 of 
\cite{hobby-mckenzie}.\index{Hobby, D.}\index{McKenzie, R.}
This approach does not use chains of congruences with
abelian factors nor does it use radicals
to define the solvability concept.
Rather, one starts by defining
a pair $(0,1)\in A^2$ from an algebra
$\m a$ to be a {\bf $2$-snag} if there is a
polynomial 
$p(x,y)\in \Pol_2(\m a)$ such that 
$p(0,0) = p(0,1) = p(1,0) = 0$ and $p(1,1) = 1$.
Hobby and McKenzie observe that 
if $\m a$ is a finite algebra
and $\alpha \prec \beta$
are congruences on $\m a$, then $\C c(\beta,\beta;\alpha)$
holds if and only if $\beta - \alpha$
contains no $2$-snag. 
From this it is clear that a chain of congruences:
$$
\alpha = \alpha_0\prec \alpha_1\prec\cdots\prec\alpha_k = \beta
$$
has abelian factors $\alpha_{i+1}/\alpha_i$
if and only if $\beta - \alpha$ contains
no $2$-snag. Starting from this
observation, it can be argued that the relation
$\alpha\wrel{\solv}\beta$, defined by
the condition that $\alpha$ and $\beta$ contain
the same $2$-snags, is a complete congruence
on $\Conb(\m a)$ for any locally finite algebra.

Our aim in this chapter is to 
synthesize ideas from the solvability
theories for groups, rings, and 
finite algebras in order to 
develop a solvability
theory for a wide class of varieties.
The theory resembles the theory
for groups in that it is defined
in terms of chains of congruences with abelian factors.
However, as we will show in Section~\ref{radical_approach}, 
the theory may viewed as an extension of the theory of 
solvability for rings that is based on the
prime radical ( = the intersection of 
all prime congruences), since for rings 
our theory defines an ideal to be solvable
precisely when it is contained in the prime radical.
Finally, anyone familiar with the theory developed 
by Hobby and McKenzie in \cite{hobby-mckenzie}
will recognize that our 
``solvability obstructions''
(Definition~\ref{XXXdefn}) play the
role of their $2$-snags, while 
Theorems~\ref{HMa}, \ref{3permute}
and \ref{weak_omit} of this
chapter are analogues of
Theorem~7.9~(3), Corollary~7.13
and Lemma~6.5  of \cite{hobby-mckenzie}.

\section{Varieties With A Weak Difference Term}
\index{weak difference term|(}

There is an
% serious 
obstacle to the development of a 
very general solvability theory
that we must deal with first.
Any absolutely free algebra 
in any given signature is abelian in the
sense of Chapter~\ref{centrality}. 
If there were a solvability
theory that applied to all such algebras,
where the solvability relation is a congruence
preserved by homomorphisms, then 
such a theory must identify every 
congruence on every algebra as solvable.
Such a theory would, of course,  be trivial.
We cannot avoid this type of problem even if we 
restrict our attention to varieties satisfying
a nontrivial idempotent Maltsev condition,
since there exist algebras (such as
$\m a = \left\lb [0,1]; \frac{x+y}{2}\right\rb$, the unit
interval of the real line under the averaging
operation) that are abelian in the sense
of Chapter~\ref{centrality}, 
have a Taylor term ($f(x,y) = \frac{x+y}{2}$),
but which have simple nonabelian
homomorphic images (in this case, $\m a/\theta$
where $\theta = [0,1)\times [0,1)\cup \{(1,1)\}$).
This phenomenon, that solvable algebras
have nonsolvable homomorphic images,
prevents the development
of a powerful solvability theory.
%so we will make assumptions
%to avoid it. 

As it happens, there is a weakest
idempotent Maltsev condition that guarantees
that homomorphic images of solvable algebras
are solvable, and this Maltsev condition
is a strong enough hypothesis for the theorems
that we have in mind. We start this chapter by discussing
the idempotent Maltsev condition on which 
the solvability 
theory is based, and then we discuss the theory.

\bigskip

\begin{df} \label{weak_difference}
Let ${\vr V}$ be a variety.
A ternary term $d(x,y,z)$ is a 
{\bf weak difference term} 
\index{weak difference term!defined}
for ${\vr V}$
if whenever $\m a\in {\vr V}$, 
$\theta\in \Con(\m a)$ and $(a,b)\in\theta$
it is the case that
\[d(a,a,b) \equiv b \equiv d(b,a,a) \pmod{[\theta,\theta]}.\]
\end{df}

The following theorem is included
to show that our development
of a solvability theory only for varieties
with a weak difference term is reasonable.

\begin{thm} \label{justify_weak}
The class of varieties with a weak difference term
is definable by a nontrivial 
idempotent Maltsev condition.
This idempotent Maltsev condition
is the weakest conjunction of
idempotent Maltsev conditions 
that implies that homomorphic images of solvable 
algebras are solvable.
It is also the weakest conjunction of 
idempotent Maltsev conditions that implies that abelian
algebras are affine.
\end{thm}

\begin{proof}
The first and third statements
are proved in Theorem~4.8 of
\cite{k-szendrei2}. We prove the second
only. 

If $\vr V$ has a weak difference term,
then it can be shown
(by using  Lemma~\ref{lift_delta}
and following the same argument
that works for groups)
that homomorphic images of solvable
algebras are solvable.

Conversely assume that $\vr V$ satisfies
some conjunction of 
idempotent Maltsev conditions that implies
that homomorphic images of solvable algebras
are solvable. At least one of these
Maltsev conditions
must be nontrivial, since there exist
solvable algebras with nonsolvable
homomorphic images 
(e.g. $\left\lb [0,1]; \frac{x+y}{2}\right\rb$).
The idempotent reduct $\Id({\vr V})$
of $\vr V$ satisfies the same idempotent
Maltsev conditions as $\vr V$,
so homomorphic images of solvable algebras
in $\Id({\vr V})$ are again solvable.
Let $\vr A$ be the subvariety of 
$\Id({\vr V})$ that is generated
by the abelian algebras in 
$\Id({\vr V})$. Since the class of abelian 
algebras in any language is axiomatizable
by the universal Horn sentences of the term condition,
this class is closed under the formation of
subalgebras and products. From our assumption
that homomorphic images of solvable algebras are
solvable we get that $\vr A$ consists of solvable
algebras.
In particular, $\vr A$ contains no subvariety
term equivalent to the variety of semilattices.
Since $\vr A$ also satisfies the nontrivial 
idempotent Maltsev conditions satisfied by $\vr V$,
it follows from Lemma~2.5 of \cite{idem}
that $\vr A$ must satisfy an idempotent Maltsev 
condition that fails in the variety of
semilattices. From Theorem~4.10 of \cite{k-szendrei2},
it follows that the variety $\vr A$
is affine, and in particular congruence permutable.
If $d(x,y,z)$ is a Maltsev operation for 
$\vr A$, then it will be a weak difference term
for $\vr V$, as is explained in the proof 
of Theorem~4.8 of \cite{k-szendrei2}.
\end{proof}

This theorem has a corollary which will prove useful
in what follows.

\begin{cor} \label{justify_weak_cor}
If a variety has a Hobby--McKenzie term,
then it has a weak difference term.
\end{cor}

\begin{proof}
Theorem~4.10 of \cite{k-szendrei2}
proves that if $\vr V$ satisfies an
idempotent Maltsev condition that fails
in the variety of semilattices, 
then abelian algebras in $\vr V$ are
affine. 
Theorem~\ref{rect_Malt_cond2} proves
that the hypothesis of this statement
is equivalent to the existence of a Hobby--McKenzie
term, while Theorem~\ref{justify_weak}
proves that the conclusion of the
statement implies the existence of a weak difference term.
\end{proof}

The concept of a weak difference term entered mathematics
during the development of commutator theory.
Recall from the introduction that
A.~I.~Maltsev 
\index{Maltsev, A.~I.}
showed in \cite{maltsev}
that a variety $\vr V$ has permuting
congruences if and only if it has a term 
$d(x,y,z)$ such that 
${\vr V}\models d(x,x,y) \approx y \approx d(y,x,x)$.
In \cite{herrmann}, C.~Herr\-mann 
\index{Herrmann, C.}
proved
that abelian algebras in a congruence modular
variety $\vr V$ are affine by constructing a 
term $d(x,y,z)$ from the Day terms
for congruence modularity and showing that 
$d(x,y,z)$ is a Maltsev term for any 
abelian algebra in $\vr V$.\footnote{This fact, 
together with the fact that the 
class of congruence modular varieties
is definable by an idempotent Maltsev condition,
already implies that $d(x,y,z)$ is a weak difference term
for $\vr V$, as the proof of 
Theorem~\ref{justify_weak} shows.}
Then H.~P.~Gumm 
\index{Gumm, H.~P.}
observed in \cite{gumm}
that the term constructed by Herrmann 
has the stronger properties that 
the identity $d(x,x,y)\approx y$ holds in
$\vr V$, and $d(b,a,a) \equiv b\pmod{[\theta,\theta]}$
for any $\m a\in {\vr V}$,
$\theta\in \Con(\m a)$, 
and $(a,b)\in\theta$.
Such a term is now called a 
%{\bf Gumm difference term},
%or simply a 
{\bf difference term}.
Difference terms play a central role
in most developments of modular commutator theory
and in some extensions of the theory to
nonmodular varieties (see \cite{diff,lipparini3,lipparini2}).
Weak difference terms were
first identified in \cite{hobby-mckenzie}.
They occur far more frequently
than ordinary difference terms, and 
are equally useful in most circumstances.


\begin{exmps}
As noted in Theorem~\ref{justify_weak}, any variety 
with a weak difference term satisfies a nontrivial idempotent
Maltsev condition, and therefore has a Taylor term.
The converse is not true in general,
as we explained in the first paragraph
of this section, but the converse does
hold for certain classes of varieties.
We describe some of these classes now,
thereby providing 
a number of examples of varieties with
weak difference terms.

Theorem~9.6 of \cite{hobby-mckenzie}
proves the converse for locally finite varieties:
that is, any locally
finite variety that has a Taylor term
has a weak difference term.

Any idempotent abelian (or solvable)
variety that has a Taylor term
has a weak difference term.
(The proof of this is contained
in the proof of Theorem~\ref{justify_weak}.)

A variety of semigroups
that has a Taylor term has a weak difference term.
For if $f(x_1,\ldots,x_n) = x_{i_0}\cdots x_{i_k}$
is a Taylor term for $\vr V$, 
then necessarily $k\geq 1$. The idempotence of $f$
implies that ${\vr V}\models x^{k+1}\approx x$.
Any semigroup $\m s$ satisfying this equation
has the property that $(s^k)^2 = s^{k-1}s^{k+1} 
=s^{k-1}s=s^k$ for
any $s\in S$, so $k$--th powers of
elements of $\m s$ are idempotent.
If $e, f\in S$ are idempotent, and $ef\neq fe$,
then it can be shown that 
the subsemigroup of $\m s$ 
generated by $\{e,f\}$
has a homomorphism onto a nontrivial left zero
or right zero semigroup.
But a nontrivial left zero semigroup
fails the $i_0$--th Taylor identity
for $f(x_1,\ldots,x_n) = x_{i_0}\cdots x_{i_k}$, and 
a nontrivial right zero semigroup
fails the $i_k$--th Taylor identity.
It follows that $\vr V$ has no nontrivial
left or right zero semigroups, and hence that
no $\m s\in {\vr V}$ has a pair of
noncommuting idempotents.
This fact may be expressed by 
saying that the
identity $x^ky^k\approx y^kx^k$ holds
in $\vr V$. Now we argue that 
the term
$d(x,y,z) = xy^{2k-1}z$ is a weak difference term
for the variety of
semigroups axiomatized by $x^{k+1}\approx x$
and $x^ky^k\approx y^kx^k$.
To justify this, it suffices to show
that if $\m a$ is a semigroup in $\vr V$
and $a,b\in A$ are elements for which
$\theta:=\Cg a(a,b)$ is abelian,
then $d(a,a,b) = b = d(b,a,a)$.
To show that 
$b = d(a,a,b) = a^{2k}b = a^kb$ and $b = ba^k$
it is enough to prove that $a^k = b^k$.
Since $e = a^k$ and $f = b^k$ are idempotent
and $(e,f) = (a^k,b^k)\in \theta$, it suffices
to prove that idempotents related by an
abelian congruence are equal.
The $\theta,\theta$-term condition applied
to $ee\underline{f} = ef = ef\underline{f}$
allows us to replace the underlined $f$ with $e$
and obtain $eee = efe$. Since idempotents
commute, this is equivalent to $e = ef$.
The same argument shows that $f = ef$,
so $e=f$.

It is easily seen that any 
congruence $n$-permutable variety satisfies a nontrivial 
idempotent Maltsev condition, hence
has a Taylor term. Less obvious is the
theorem of P.~Lipparini 
\index{Lipparini, P.}
in \cite{lipparini1}
that any congruence $n$-permutable
variety has a weak difference term. 
\end{exmps}

\section{$\XXX$-Solvability}

We now begin the development of
a solvability theory for varieties 
with a weak difference term.
Any succesor ordinal, considered as a lattice
under its usual ordering, is a meet continuous
lattice.
By a {\bf continuous sequence of congruences}
\index{continuous sequence of congruences}
on an algebra $\m a$ we mean a 
meet continuous lattice homomorphism
$\sigma: {\kappa+1}\rightarrow \Conb(\m a)$,
We write such a sequence as
$(\sigma_{\lambda})_{\lambda<\kappa+1}$
where $\sigma_{\lambda}:=\sigma(\lambda)$.
It is not hard to verify that 
$(\sigma_{\lambda})_{\lambda<\kappa+1}$
is a continuous sequence of congruences
if and only if $\sigma_{\mu}\leq \sigma_{\lambda}$
for $\mu<\lambda$ and 
$\sigma_{\lambda} = \bigcup_{\mu<\lambda} \sigma_{\mu}$
for all limit ordinals $\lambda$.


\begin{df} \label{XXXdefn}
Let $\m a$ be an algebra with congruences 
$\alpha, \beta,
\theta, \theta_{\lambda}, \delta$ and $\delta'$.

If $\delta\leq \theta$ and $\C c(\theta,\theta;\delta)$ holds
then we say that {\bf $\theta$ is abelian over $\delta$},
\index{abelian!over}
and write $\delta\wrel{\lhd}\theta$. 

If $\kappa$ is an ordinal, and
$(\theta_{\lambda})_{\lambda<\kappa+1}$
is a continuous sequence of congruences such that 
%\begin{enumerate}
%\item[(1)] 
$\theta_{\lambda}\wrel{\lhd} \theta_{\lambda+1}$
for all $\lambda < \kappa$,
%and
%\item[(2)] $\theta_{\lambda} = \bigcup_{\mu<\lambda}
%\theta_{\mu}$ for all limit ordinals $\lambda < \kappa+1$,
%\end{enumerate}
then we say that {\bf $\theta_{\kappa}$ is $\kappa$-step
solvable over $\theta_0$}.
\index{solvable!$\kappa$-step} 

We say that {\bf $\theta$ is $\XXX$-solvable over
$\delta$}
\index{$\XXX$--solvable!over} 
\index{solvable!$\XXX$-} 
if
$\theta$ is $\kappa$-step solvable over $\delta$
for some $\kappa$. We may denote this by 
$\delta\wrel{\lhdlhd}\theta$.
\index{$\lhdlhd$}
If $0\wrel{\lhdlhd}\theta$, then 
we say that {\bf $\theta$ is $\XXX$--solvable}.
\index{$\XXX$--solvable!congruence}
If $0\wrel{\lhdlhd} 1$, then 
say that {\bf $\m a$ is $\XXX$--solvable}.
\index{$\XXX$--solvable!algebra}

If $\alpha\wedge\beta\wrel{\lhdlhd}\alpha\vee\beta$, 
then we say that 
{\bf $\alpha$ is solvably related to $\beta$}
\index{solvably related to}
and denote this by $\alpha\wrel{\solv}\beta$.
\index{$\solv$}

If $\delta < \theta$, and $\delta\wrel{\lhd} \delta'\leq \theta$
implies $\delta = \delta'$, then we call
the congruence interval $I[\delta,\theta]$ 
a {\bf solvability obstruction} 
\index{solvability obstruction} 
and we denote
this by $\delta\blacktriangleleft \theta$.
\index{$\blacktriangleleft$}
\end{df}

Note that for comparable congruences
$\delta\leq\theta$  the relations
$\delta\wrel{\lhdlhd}\theta$ 
and $\delta\wrel{\solv}\theta$ are the same.

\begin{exmp}
%To give substance to the previous definition,
%we now exhibit a variety of algebras 
%whose members are all $\XXX$-solvable,
%but which have 
%arbitrarily large (infinite) solvability degree.
%
If $\vr V$ is a variety whose members have the property
that their principal congruences are abelian, then
every member of $\vr V$ is $\XXX$--solvable.
For if $\m a\in {\vr V}$,
$(a_{\lambda})_{\lambda<\kappa}$ is an enumeration
of $\m a$, and $\theta_{\lambda}$ is the
congruence generated by 
$\{(a_{0},a_{\mu})\;|\;\mu<\lambda\}$, then 
it is not hard to see that 
\begin{enumerate}
\item[(i)] $\theta_0=0$ 
and $\theta_{\kappa} = 1$,
\item[(ii)] 
$\theta_{\lambda}\wrel{\lhd} \theta_{\lambda+1}$
for all $\lambda < \kappa$, and
\item[(iii)] 
$\theta_{\lambda} = \bigcup_{\mu<\lambda}
\theta_{\mu}$
for all limit ordinals.
\end{enumerate}
Items (i) and (iii) follow immediately from
the definition of $\theta_{\lambda}$.
For item (ii), note that $\theta_{\lambda+1} = 
\Cg a(a_0,a_{\lambda})\vee\theta_{\lambda}$.
Thus $\theta_{\lambda+1}/\theta_{\lambda}$
is a principal (hence abelian) congruence
on $\m a/\theta_{\lambda}$. This implies
that $\theta_{\lambda}\wrel{\lhd}\theta_{\lambda+1}$
for all $\lambda<\kappa$.

An example of a variety in which 
principal congruences are abelian is
the variety $\vr V$ of nonunital rings satisfying
$x^2 \approx 0$. This identity implies that 
$0\approx (x+y)^2 \approx 
x^2 + xy + yx + y^2\approx xy+yx$,
so the
%which is equivalent to the 
anticommutative law $xy \approx -yx$
holds in $\vr V$.
This law implies that, for any
$\m r\in {\vr V}$, 
%In any such ring, $\m r$, 
all principal ideals have the
form $(a) = R\cdot a$. Hence 
$(a)^2 = RaRa = 
\sum \{ra\cdot sa = -rsa^2\;|\;r, s\in R\} = (0).$
Since congruences correspond to ideals,
and the commutator operation corresponds
to ideal multiplication, this is indeed a variety
in which principal congruences are abelian.
Hence $\vr V$ consists entirely of $\XXX$--solvable algebras.

However, there is no cardinal bound on
solvability degree in $\vr V$.
In fact, if $\kappa$ is any
infinite cardinal, 
$\vr V$ contains rings of solvability 
degree exactly $\kappa$.
One such ring, $\m r_{\kappa}$, may be presented 
by generators and relations relative to 
$\vr V$ as follows.
%the subvariety ${\vr W}\subseteq {\vr V}$ 
%of {commutative} nonunital rings satisfying 
%both $x \approx -x$ and $x^2 \approx 0$ as follows.
Let $S$ be an infinite set of cardinality $\kappa$.
Take as generators for $\m r_{\kappa}$
all symbols $x_U$ where 
$U$ is a nonempty subset of $S$.
Impose the relations
\begin{itemize}
\item $2x_U = 0$ for all nonempty $U\subseteq S$, 
\item $x_U\cdot x_V = 0$
if $U\cap V \neq \emptyset$, and 
\item $x_U\cdot x_V = x_{U\cup V}$ if $U\cap V = \emptyset$.
\end{itemize}
These relations force each element of $\m r_{\kappa}$ 
to be expressible as a finite
sum of generators $x_{U_1}+\cdots+x_{U_n}$
where $U_i\neq U_j$ if $i\neq j$.
We leave it as an exercise to prove
that the ring $\m r_{\kappa}$
is $\kappa$-step solvable, but not 
$\lambda$-step solvable for any ordinal
$\lambda<\kappa$.
\end{exmp}


\begin{lm} \label{tol=cong}
Let $\vr V$ be a variety with a weak difference term. 
If $\m a\in {\vr V}$
has a tolerance $T$ and a congruence
$\delta$ such that $\C C(T,T;\delta)$ holds,
then $T':=\delta\circ T\circ \delta$
is the smallest congruence on $\m a$ containing 
$T\cup \delta$. Furthermore, 
$\C C(T',T';\delta)$ holds.
\end{lm}

\begin{proof} 
We know from the basic properties
of the centralizer relation 
(enumerated in Theorem~\ref{basic_centrality})
that if
$\C C(T,T;\delta)$ holds and 
$T':=\delta\circ T\circ \delta$,
then $\C C(T',T';\delta)$ holds.
Since $(T\cup \delta)\subseteq T' = 
\delta\circ T\circ \delta\subseteq \Cg a(T\cup\delta)$, 
the only thing left to show is that $T'$ is a congruence.
For this we only need to establish
that $T'\circ T'\subseteq T'$. 

Let $\theta = \Cg a(T')$. Since $\C C(T',T';\delta)$ holds,
it follows from Theorems~\ref{abel_tol=abel_cong}
and \ref{justify_weak} that $\C C(\theta,\theta;\delta)$ holds.
If $(a,c)\in T'\circ T'$, then there is 
an element $b$ such that $(a,b), (b,c)\in T'\subseteq \theta$.
Since 
$a\equiv_{[\theta,\theta]} d(a,b,b)\equiv_{T'} d(b,b,c)
\equiv_{[\theta,\theta]} c$ and $[\theta,\theta]\leq\delta$,
we get that $(a,c)\in \delta\circ T'\circ \delta = T'$.
Since $(a,c)\in T'\circ T'$ was chosen
arbitrarily, $T'\circ T'\subseteq T'$.
\end{proof}


\begin{lm} \label{lift_delta}
Let $\vr V$ be a variety with a weak difference term. 
If $\m a\in {\vr V}$
has congruences $\theta$
and $\delta$ such that 
$[\theta,\theta]\leq \delta$,
then $\C C(\theta,\theta;\delta)$ holds.
\end{lm}

\begin{proof} 
Assume that $[\theta,\theta]\leq\delta$. If 
$\C C(\theta,\theta;\delta)$ does not hold,
then there are a polynomial 
$p(\wec{x},\wec{y})$ and tuples
$\wec{a} \wrel{\theta}\wec{b}$ and 
$\wec{u} \wrel{\theta}\wec{v}$ such that 
\[
p(\wec{a},\wec{u})\wrel{\equiv_{\delta}}
p(\wec{a},\wec{v})
\]
and 
\begin{equation}\label{line_zero}
p(\wec{b},\wec{u})\wrel{\not\equiv_{\delta}}
p(\wec{b},\wec{v}).
\end{equation}
Let 
$p'(\wec{x},\wec{y}) = d(p(\wec{x},\wec{y}),p(\wec{x},\wec{u}),p(\wec{b},\wec{u}))$.
Using that 
%$[\theta,\theta]\leq \delta$ and that
$d$ is an idempotent term that is 
Maltsev on $\theta$--classes modulo 
$[\theta,\theta]$, we get that
\[
p'(\wec{a},\underline{\wec{u}}) = 
d(p(\wec{a},\wec{u}),p(\wec{a},\wec{u}),p(\wec{b},\wec{u}))
\wrel{\equiv_{[\theta,\theta]}}
p(\wec{b},\wec{u})=
%d(p(b,\wec{u}),p(b,\wec{u}),p(b,\wec{u}))=
p'(\wec{b},\underline{\wec{u}}).
\]
Since $\C C(\theta,\theta;[\theta,\theta])$ holds, 
we can 
change the underlined $\wec{u}$ to $\wec{v}$
without changing the fact that the left hand side is 
$[\theta,\theta]$--related to the right hand side:
\begin{equation}\label{line_one}
p'(\wec{a},\underline{\wec{v}}) 
\wrel{\equiv_{[\theta,\theta]}}
p'(\wec{b},\underline{\wec{v}}).
\end{equation}
Moreover we have that 
\begin{equation}\label{line_two}
\begin{array}{rl}
p'(\wec{a},\wec{v}) &= 
d(p(\wec{a},\wec{v}),p(\wec{a},\wec{u}),p(\wec{b},\wec{u}))\\
&\equiv_{\delta}
d(p(\wec{a},\wec{u}),p(\wec{a},\wec{u}),p(\wec{b},\wec{u}))
\wrel{\equiv_{[\theta,\theta]}}
p(\wec{b},\wec{u}).
\end{array}
\end{equation}
Here 
\[
d(p(\wec{a},\wec{v}),p(\wec{a},\wec{u}),p(\wec{b},\wec{u}))
\wrel{\equiv_{\delta}}
d(p(\wec{a},\wec{u}),p(\wec{a},\wec{u}),p(\wec{b},\wec{u}))
\]
is a consequence of the fact that 
$p(\wec{a},\wec{v})\wrel{\equiv_{\delta}}
p(\wec{a},\wec{u})$, and
\[
d(p(\wec{a},\wec{u}),p(\wec{a},\wec{u}),p(\wec{b},\wec{u}))
\wrel{\equiv_{[\theta,\theta]}}
p(\wec{b},\wec{u})
\]
is a consequence of 
the facts that $\wec{a}\wrel{\theta}\wec{b}$ and
that $d$ is Maltsev on $\theta$--classes
modulo $[\theta,\theta]$.
We also have that 
\begin{equation}\label{line_three}
p'(\wec{b},\wec{v}) = 
d(p(\wec{b},\wec{v}),p(\wec{b},\wec{u}),p(\wec{b},\wec{u}))
\wrel{\equiv_{[\theta,\theta]}}
p(\wec{b},\wec{v}).
\end{equation}
since $\wec{u}\wrel{\theta}\wec{v}$
and $d$ is Maltsev on $\theta$--classes
modulo $[\theta,\theta]$.
Combining lines (\ref{line_one}), (\ref{line_two}), 
and (\ref{line_three}) yields
\[
p(\wec{b},\wec{u})
\wrel{\equiv_{\delta}}
p'(\wec{a},\wec{v})
\wrel{\equiv_{[\theta,\theta]}}
p'(\wec{b},\wec{v})
\wrel{\equiv_{[\theta,\theta]}}
p(\wec{b},\wec{v}),
\]
or simply 
$p(\wec{b},\wec{u})\wrel{\equiv_{\delta}}p(\wec{b},\wec{v})$
(since $[\theta,\theta]\leq\delta$).
This contradicts line (\ref{line_zero}), 
%our earlier assumption that
%$
%p(\wec{b},\wec{u})\wrel{\not\equiv_{\delta}}
%p(\wec{b},\wec{v}),
%$
thereby completing the proof.
\end{proof}

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

\begin{cor}\label{ACD_cor2}
Let $\vr V$ be a variety with a 
weak difference term.
Assume that $T$ is an abelian tolerance
on some $\m a\in {\vr V}$, and that $a$ and $b$
are distinct $T$-related elements.
If $\rho$ is a congruence on $\m a$
that is maximal with respect to $(a,b)\notin\rho$,
then $T/\rho$ is a nontrivial
abelian congruence on the subdirectly
irreducible algebra $\m a/\rho$.
\end{cor}

\begin{proof}
It follows from Lemma~\ref{tol=cong}
(using $\delta = 0$) that $T$ is an
abelian congruence on $\m a$.
Then, using Lemma~\ref{lift_delta}
in the case where $\theta = T$ and $\delta = \rho$,
we obtain that $\C C(T,T;\rho)$ holds,
so $T/\rho$ is a nontrivial abelian tolerance
on the subdirectly irreducible algebra
$\m a/\rho$. Using Lemma~\ref{tol=cong},
we find that $T/\rho$ is a congruence.
\end{proof}


\begin{lm} \label{meet+join}
Let $\vr V$ be a variety with a weak difference term.
Assume that $\m a\in {\vr V}$ 
has congruences $\alpha, \beta$,
and $\gamma$. 
\begin{enumerate}
\item[(1)] 
If $\alpha\wrel{\lhd} \beta$, then
$(\alpha\wedge\gamma)\wrel{\lhd} (\beta\wedge\gamma)$ and
$(\alpha\vee\gamma)\wrel{\lhd} (\beta\vee\gamma)$.
\item[(2)] If $\alpha\wrel{\lhdlhd} \beta$, then
$(\alpha\wedge\gamma)\wrel{\lhdlhd} (\beta\wedge\gamma)$ and
$(\alpha\vee\gamma)\wrel{\lhdlhd} (\beta\vee\gamma)$.
\end{enumerate}
\end{lm}

\begin{proof}
To prove the part of item (1) that concerns the
compatibility of $\lhd$ with meet, 
recall that 
$\alpha\wrel{\lhd} \beta$ means
that $\C C(\beta,\beta;\alpha)$ holds.
By item~(1) of
Theorem~\ref{basic_centrality}
we have that 
$\C C(\beta\wedge\gamma,\beta\wedge\gamma;\alpha)$ 
holds.
By item~(8) of
Theorem~\ref{basic_centrality}
we have that 
$\C C(\beta\wedge\gamma,\beta\wedge\gamma;\gamma)$ 
holds. Item~(6) of Theorem~\ref{basic_centrality}
now yields that 
$\C C(\beta\wedge\gamma,\beta\wedge\gamma;\alpha\wedge\gamma)$
holds, which means that 
$(\alpha\wedge\gamma)\wrel{\lhd} (\beta\wedge\gamma)$.

To prove the compatibility of $\lhd$ with join,
assume that $\C C(\beta,\beta;\alpha)$ holds.
Then $[\beta,\beta]\leq\alpha\leq\alpha\vee\gamma$,
so by Lemma~\ref{lift_delta}
we get that $\C C(\beta,\beta;\alpha\vee\gamma)$ holds.
Using Lemma~\ref{tol=cong}
with $T = \beta$ and $\delta = \alpha\vee\gamma$
we deduce that 
$(\alpha\vee\gamma)\circ\beta\circ(\alpha\vee\gamma)
= \beta\vee(\alpha\vee\gamma) = \beta\vee\gamma$
and that 
$\C C(\beta\vee\gamma,\beta\vee\gamma;\alpha\vee\gamma)$
holds. This shows that 
$(\alpha\vee\gamma)\wrel{\lhd}(\beta\vee\gamma)$,
so item (1) holds.

For (2), 
assume that $(\theta_{\lambda})_{\lambda<\kappa+1}$
is a continuous 
sequence of congruences for which 
$\alpha = \theta_0, \beta = \theta_{\kappa}$,
$\theta_{\lambda}\wrel{\lhd}\theta_{\lambda+1}$
for all $\lambda$.
%, and 
%$\theta_{\lambda} = \bigcup_{\mu<\lambda}\theta_{\mu}$
%for successor ordinals.
The sequences
$(\theta_{\lambda}\wedge \gamma)_{\lambda<\kappa+1}$
and 
$(\theta_{\lambda}\vee \gamma)_{\lambda<\kappa+1}$
are also continuous, since meeting or joining
with the fixed element $\gamma$ preserves
order and unions of chains.
Therefore it follows from item (1) of this lemma
that $(\theta_{\lambda}\wedge \gamma)_{\lambda<\kappa+1}$
and 
$(\theta_{\lambda}\vee \gamma)_{\lambda<\kappa+1}$
are sequences witnessing 
$(\alpha\wedge\gamma)\wrel{\lhdlhd}(\beta\wedge\gamma)$ 
and 
$(\alpha\vee\gamma)\wrel{\lhdlhd}(\beta\vee\gamma)$, respectively.
%The fact that these new sequences begin and end
%at the right places is trivial, 
%the fact that successor steps are in $\lhd$-relation
%follows from the first part of the theorem,
%and the fact that 
\end{proof}


\begin{thm}\label{solv_is_join_comp}
Let $\vr V$ be a variety with a weak difference term.
If $\m a\in {\vr V}$, then 
$\solv$ is a congruence on $\Conb(\m a)$
that is compatible with the complete join operation.
\end{thm}

\begin{proof}
Recall that $\alpha\wrel{\solv}\beta$ holds
if and only if 
$(\alpha\wedge\beta)\wrel{\lhdlhd}(\alpha\vee\beta)$.
It is easy to see that this relation is
reflexive and symmetric.
To show that it is transitive,
assume that $\alpha\wrel{\solv}\beta\wrel{\solv}\gamma$.
Then 
$(\alpha\wedge\beta)\wrel{\lhdlhd}(\alpha\vee\beta)$
and 
$(\beta\wedge\gamma)\wrel{\lhdlhd}(\beta\vee\gamma)$,
so by the second statement of 
Lemma~\ref{meet+join} we get that 
$(\alpha\wedge\beta\wedge\gamma) = 
(\alpha\wedge\beta)\wedge(\beta\wedge\gamma) 
\wrel{\lhdlhd}
(\alpha\wedge\beta)\wedge(\beta\vee\gamma) =
(\alpha\wedge\beta)$ and 
$(\alpha\vee\beta) = 
(\alpha\vee\beta)\vee(\beta\wedge\gamma)
\wrel{\lhdlhd}
(\alpha\vee\beta)\vee(\beta\vee\gamma) =
(\alpha\vee\beta\vee\gamma)$. Thus we
have 
$$
(\alpha\wedge\beta\wedge\gamma)
\wrel{\lhdlhd}
(\alpha\wedge\beta)
\wrel{\lhdlhd}
(\alpha\vee\beta)
\wrel{\lhdlhd}
(\alpha\vee\beta\vee\gamma).
$$
The well--ordered chains of congruences
that witness these three instances of 
$\XXX$--solvability may be concatenated
to produce a single such chain from 
$\alpha\wedge\beta\wedge\gamma$
to $\alpha\vee\beta\vee\gamma$,
thus proving that 
$(\alpha\wedge\beta\wedge\gamma)\wrel{\lhdlhd}
(\alpha\vee\beta\vee\gamma)$.
Since 
$(\alpha\wedge\beta\wedge\gamma)\leq
\alpha\wedge\gamma\leq\alpha\vee\gamma\leq
(\alpha\vee\beta\vee\gamma)$,
we can deduce that 
$(\alpha\wedge\gamma)\wrel{\lhdlhd}
(\alpha\vee\gamma)$ if we show first that 
$\delta\leq \delta'\leq\theta'\leq\theta$
and $\delta\wrel{\lhdlhd}\theta$ imply
$\delta'\wrel{\lhdlhd}\theta'$.
We show this by using 
Lemma~\ref{meet+join} again:
joining $\delta\wrel{\lhdlhd}\theta$
with $\delta'$
we obtain
$$
\delta' = 
(\delta' \vee \delta)
\wrel{\lhdlhd}
(\delta' \vee \theta)
= \theta.$$
%$$
%(\alpha\wedge\gamma)=
%(\alpha\wedge\gamma)
%\vee
%(\alpha\wedge\beta\wedge\gamma)
%\wrel{\lhdlhd}
%(\alpha\wedge\gamma)
%\vee
%(\alpha\vee\beta\vee\gamma)=
%(\alpha\vee\beta\vee\gamma).$$
Meeting with 
$\theta'$
%$\alpha\vee\gamma$, 
we find that
$$\delta' = 
(\delta' \wedge \theta')
\wrel{\lhdlhd}
(\theta\wedge\theta')=\theta'.$$
%$$(\alpha\wedge\gamma)=
%(\alpha\wedge\gamma)
%\wedge
%(\alpha\vee\gamma)
%\wrel{\lhdlhd}
%(\alpha\vee\beta\vee\gamma)
%\wedge
%(\alpha\vee\gamma)=
%(\alpha\vee\gamma).$$
This reasoning completes the
argument that $\solv$
is transitive, hence an equivalence relation,
and also shows that 
subintervals of $\XXX$--solvable intervals
are again $\XXX$--solvable.
Hence $\solv$--classes are convex 
sublattices of $\Conb(\m a)$.
This implies that $\alpha\wrel{\solv}\beta$
if and only if $\alpha$ and $\beta$
belong to some $\XXX$--solvable interval.
This reformulation of the definition of
the $\solv$--relation makes it 
easy to see that it is a congruence:
$\alpha\wrel{\solv}\beta$ if and only if
$\delta:= \alpha\wedge\beta\wrel{\lhdlhd}
\alpha\vee\beta =:\theta$. 
But $\delta\wrel{\lhdlhd}\theta$
implies 
$\delta\wedge\gamma\wrel{\lhdlhd}\theta\wedge\gamma$ and
$\delta\vee\gamma\wrel{\lhdlhd}\theta\vee\gamma$ 
by Lemma~\ref{meet+join}.
Since 
$\alpha\wedge\gamma, \beta\wedge\gamma\in 
I[\delta\wedge\gamma,\theta\wedge\gamma]$
and 
$\alpha\vee\gamma, \beta\vee\gamma\in 
I[\delta\vee\gamma,\theta\vee\gamma]$,
we conclude that 
$\alpha\wrel{\solv}\beta$ implies that 
$(\alpha\wedge\gamma)\wrel{\solv}(\beta\wedge\gamma)$ and
$(\alpha\vee\gamma)\wrel{\solv}(\beta\vee\gamma)$.

It remains to show that $\solv$ is compatible with 
the complete join operation of $\Conb(\m a)$.
Suppose that $\alpha_i\wrel{\solv}\beta_i$ for all
$i\in I$. Let $\delta_i = \alpha_i\wedge\beta_i$
and $\theta_i = \alpha_i\vee\beta_i$. 
Then $\delta_i\wrel{\lhdlhd}\theta_i$
for all $i$. If $\Delta:=\bigvee_{i\in I}\delta_i$
and $\Theta:=\bigvee_{i\in I}\theta_i$,
then both $\bigvee_{i\in I}\alpha_i$
and $\bigvee_{i\in I}\beta_i$ belong
to the interval 
$I[\Delta,\Theta]$.
We will be done if we show that 
$\Theta$ is $\XXX$--solvable over
$\Delta$.
%$\Delta\wrel{\lhdlhd}\Theta$.

Consider the collection $\mathcal P$ of all sequences
of the form
$\sigma = (\sigma_{\lambda})_{\lambda<\kappa+1}$ 
where $\sigma_{\lambda}\in\Con(\m a)$ and
%consisting of congruences on $\m a$ such that 
\begin{enumerate}
\item[(0)] $\sigma_0 = \Delta$,
%\bigvee_{i\in I}\delta_i$,
$\sigma_{\kappa}\in I[\Delta,\Theta]$,
%I[\bigvee_{i\in I}\delta_i,\bigvee_{i\in I}\theta_i]$,
\item[(1)] $\sigma_{\lambda}\wrel{\lhd} \sigma_{\lambda+1}$
for all $\lambda < \kappa$, and
\item[(2)] $\sigma_{\lambda} = \bigcup_{\mu<\lambda}
\sigma_{\mu}$
for limit ordinals $\lambda$.
\end{enumerate}
The elements of $\mathcal P$ are the continuous
sequences of congruences that witness the 
$\XXX$--solvability over $\Delta$ 
%$\bigvee_{i\in I}\delta_i$ 
of some elements 
$\sigma_{\kappa}\in I[\Delta,\Theta]$.
%I[\bigvee_{i\in I}\delta_i,\bigvee_{i\in I}\theta_i]$.
$\mathcal P$ is nonempty, since it contains 
the one-term sequence $(\Delta)$.
$\mathcal P$ 
can be partially ordered by the rule:
$\sigma\leq \sigma'$ if $\sigma'$ is
an extension of $\sigma$.
That is, $\sigma\leq \sigma'$
if the domain of $\sigma$
is contained in the domain of $\sigma'$
and the functions agree on their common domain.
This is an inductive ordering of $\mathcal P$,
as we now explain.
Assume that $(\sigma^{\mu})_{\mu<\nu}$ is 
an increasing sequence of
sequences in $\mathcal P$.
Define 
$\Sigma'$ to be the union of
the functions $\sigma^{\mu}$.
$\Sigma'$ is a function
from the ordinal 
$\rho:= \bigcup_{\mu<\nu} {\rm dom}(\sigma^{\mu})$
into $\Conb(\m a)$ that preserves
all existing joins.
$\Sigma'$ may not be in $\mathcal P$
because elements of $\mathcal P$
are functions from successor ordinals
into $\Conb(\m a)$ and $\rho$
may be a limit ordinal.
But if it is,
then there is a unique meet continuous extension
of $\Sigma'$ to $\rho+1$:
it is the sequence
$\Sigma = (\Sigma_{\lambda})_{\lambda<\rho+1}$
where $(\Sigma_{\lambda})_{\lambda<\rho} = \Sigma'$
and $\Sigma_{\rho}$ is defined to be
$\bigcup_{\lambda<\rho} \Sigma_{\lambda}$.
Since $\Sigma$ is a meet continuous homomorphism
from a successor ordinal into $\Conb(\m a)$,
$\Sigma$ extends each $\sigma^{\mu}$,
and $\Sigma_{\lambda} = \sigma^{\mu}_{\lambda}\lhd
\sigma^{\mu}_{\lambda+1} = \Sigma_{\lambda+1}$
for some $\mu$ for each $\lambda<\rho$,
$\Sigma$ is an upper bound in $\mathcal P$
for all the sequences $\sigma^{\mu}$.

Since $\mathcal P$ is nonempty and inductively ordered, 
Zorn's lemma applies to show that
$\mathcal P$ 
has a maximal element $(\psi_{\lambda})_{\lambda<\kappa+1}$.
The final congruence in this sequence,
$\psi_{\kappa}$, 
is $\XXX$--solvable over $\Delta$
and is a maximal 
congruence in the interval
$I[\Delta,\Theta]$ among those 
that are
$\XXX$--solvable over $\Delta$.
For each $i\in I$, 
$\delta_i\wrel{\lhdlhd}\theta_i$
and 
$\delta_i\leq \Delta\leq\psi_{\kappa}$,
so
by Lemma~\ref{meet+join} we have
$$
\psi_{\kappa} = (\psi_{\kappa}\vee\delta_i)
\wrel{\lhdlhd} (\psi_{\kappa}\vee\theta_i)\leq \Theta.
$$
The maximality condition on $\psi_{\kappa}$ implies
that $\psi_{\kappa} = \psi_{\kappa}\vee\theta_i$,
or $\psi_{\kappa}\geq \theta_i$. Since
this is true for all $i\in I$,
$\psi_{\kappa}\geq \bigvee_{i\in I} \theta_i = \Theta$.
This proves that $\Delta\wrel{\lhdlhd}\psi_{\kappa}=\Theta$,
as desired.
\end{proof}

\begin{df}
Assume that $\vr V$ is a variety with a 
weak difference term,
and $\m a\in {\vr V}$. If $\alpha\in\Con(\m a)$,
then 
$$
\rad(\alpha) := \bigvee \{\beta\in\Con(\m a)\;|\;\alpha\wrel{\solv}\beta\}
$$
is the {\bf $\XXX$--solvable radical of $\alpha$}.\index{rad@$\rad(\alpha)$}
The {\bf $\XXX$--solvable radical of $\m a$}\index{$\XXX$--solvable!radical}\index{solvable radical!$\XXX$--}
is
$\rad(0)$. A {\bf radical congruence}\index{radical congruence} 
is a congruence
$\alpha$ such that $\rad(\alpha) = \alpha$.
\end{df}

\begin{thm}\label{rad_stuff}
Let $\vr V$ be a variety with a weak difference term.
Assume that $\m a\in{\vr V}$ and that 
$\alpha,\beta\in\Con(\m a)$. Then
\begin{enumerate}
\item[(1)] $\rad(\alpha)$ is the largest congruence
$\solv$--related to $\alpha$.
\item[(2)] $\rad(\rad(\alpha)) = \rad(\alpha)$.
\item[(3)] $\rad(\alpha\wedge\beta) =
\rad(\alpha)\wedge\rad(\beta)$.
\item[(4)] $\alpha\wrel{\obst}\beta$
if and only if $\alpha <\beta$
and $\alpha = \beta\wedge\rad(\alpha)$.
\item[(5)] If ${\rm Rad}(\m a)$ 
\index{Rad@${\rm Rad}(\m a)$}
denotes
the meet subsemilattice of $\Conb(\m a)$
consisting of the radical congruences, then 
the restriction to ${\rm Rad}(\m a)$ 
of the natural homomorphism
$
\nu:~\Conb(\m a)\rightarrow~\Conb(\m a)/\solv:
\alpha\mapsto \alpha/\solv
$
to ${\rm Rad}(\m a)$ is an isomorphism
onto $\Conb(\m a)/\solv$.
\item[(6)] The natural homomorphism $\nu$
from item (5) is meet continuous, and 
its image $\Conb(\m a)/\solv$ 
meet semidistributive.
\end{enumerate}
\end{thm}

\begin{proof}
For item (1), 
the fact that $\solv$ is compatible with the 
complete join operation implies that 
the join of all 
congruence $\solv$--related to $\alpha$
is $\solv$--related to $\alpha$,
and it is necessarily the largest
congruence $\solv$--related to $\alpha$.

For (2), 
$\alpha\wrel{\solv}\rad(\alpha)$
and $\rad(\alpha)\wrel{\solv}\rad(\rad(\alpha))$
hold by (1), so 
$\alpha\wrel{\solv}\rad(\rad(\alpha))$.
Since $\rad(\alpha)$ is the largest congruence
$\solv$--related to $\alpha$, 
it follows that $\rad(\rad(\alpha))\leq \rad(\alpha)$.
The reverse inclusion follows from the
facts that $\rad(\alpha)\wrel{\solv}\rad(\alpha)$
and $\rad(\rad(\alpha))$ is the largest 
congruence $\solv$--related to $\rad(\alpha)$.

For (3),
$\alpha\wrel{\solv}\rad(\alpha)$ and 
$\beta\wrel{\solv}\rad(\beta)$, so
$\alpha\wedge\beta\wrel{\solv}\rad(\alpha)\wedge\rad(\beta)$
since $\solv$ is a congruence.
Since $\rad(\alpha\wedge\beta)$ is the 
largest congruence $\solv$--related to $\alpha\wedge\beta$
we get that 
$\rad(\alpha)\wedge\rad(\beta)\leq 
\rad(\alpha\wedge\beta)$.
On the other hand, 
$\alpha\wedge\beta\wrel{\solv}\rad(\alpha\wedge\beta)$,
so 
$$
\alpha = \alpha\vee(\alpha\wedge\beta)
\wrel{\solv}\alpha\vee\rad(\alpha\wedge\beta)\leq
\rad(\alpha)
$$
from the definition of $\rad(\alpha)$.
Thus, $\rad(\alpha\wedge\beta)\leq \rad(\alpha)$,
and similarly 
$\rad(\alpha\wedge\beta)\leq \rad(\beta)$, so 
$\rad(\alpha\wedge\beta)\leq \rad(\alpha)\wedge\rad(\beta)$.
This completes the proof of (3).

For (4), assume that $\alpha \wrel{\obst} \beta$.
Then $\alpha<\beta$ according to the definition of
$\obst$. The congruence
$\alpha' := \beta\wedge\rad(\alpha)$
satisfies $\alpha\leq \alpha'\leq \rad(\alpha)$
and is in $I[\alpha,\beta]$, 
so it is a congruence in 
$I[\alpha,\beta]$ that is $\XXX$--solvable
over $\alpha$. If this fact is witnessed by
a chain of congruences 
$(\theta_{\lambda})_{\lambda<\kappa+1}$
starting at $\theta_0 = \alpha$
and ending at $\theta_{\kappa} = \alpha'$,
and satisfying the conditions of
Definition~\ref{XXXdefn}, 
then using the fact that 
$\alpha\wrel{\obst}\beta$ and induction
it follows that $\theta_{\lambda}=\alpha$
for all $\lambda$, so $\alpha = \alpha'=
\beta\wedge\rad(\alpha)$.
Conversely, if 
$\alpha\wrel{\obst}\beta$ fails to hold
and $\alpha < \beta$,
then there is a congruence $\alpha'\in I[\alpha,\beta]$
such that $\alpha\lhd\alpha'$ and 
$\alpha<\alpha'$. Since $\alpha\wrel{\solv}\alpha'$
we get that $\alpha < \alpha'\leq 
\beta\wedge\rad(\alpha)$, so 
$\alpha = \beta\wedge\rad(\alpha)$ fails to hold.


For (5), 
it is clear that 
$\nu:{\rm Rad}(\m a)\rightarrow~\Conb(\m a)/\solv$
is an order-preserving function between 
meet semilattices. To prove that it is
an isomorphism it suffices to show that 
it is bijective, or equivalently
that the elements of ${\rm Rad}(\m a)$
form a transversal for the congruence $\solv$.
The fact 
that $\alpha\wrel{\solv}\rad(\alpha)$
(proved in (1)) together with the fact 
that $\rad(\alpha)\in {\rm Rad}(\m a)$
(an equivalent form of (2))
shows that every $\solv$--class contains
a radical congruence.
But the fact proved in (1), that
every radical element is the largest element
of its $\solv$--class, implies that each $\solv$--class
has at most one radical congruence. 
This shows 
that the elements of ${\rm Rad}(\m a)$
form a transversal for $\solv$.


For (6), the fact that $\solv$ is compatible with 
the complete join operation 
implies that it is a meet continuous
congruence, hence the quotient
modulo this congruence
is meet continuous and the natural homomorphism
$\nu:\Conb(\m a)\rightarrow \Conb(\m a)/\solv$
is a meet continuous homomorphism.
To prove that $\Conb(\m a)/\solv$ is meet semidistributive,
suppose that $(\alpha/\solv)\wedge(\beta/\solv)
= (\alpha/\solv)\wedge(\gamma/\solv) = (\delta/\solv)$.
From item (5)
we get that
$\rad(\alpha)\wedge\rad(\beta)
= \rad(\alpha)\wedge\rad(\gamma) = \rad(\delta)$
in ${\rm Rad}(\m a)$.
Since ${\rm Rad}(\m a)$ is a meet 
subsemilattice of $\Conb(\m a)$, these equalities
hold in $\Conb(\m a)$. 
Theorem~\ref{basic_centrality}~(8)
implies that 
$\C C(\rad(\beta),\rad(\alpha);\rad(\delta))$ and
$\C C(\rad(\gamma),\rad(\alpha);\rad(\delta))$ hold.
Theorem~\ref{basic_centrality}~(5)
%(join semidistributivity
%of the centralizer relation in its first variable)
now yields that
$\C C(\rad(\beta)\vee\rad(\gamma),\rad(\alpha);\rad(\delta))$ holds.
If 
$$\theta := \rad(\alpha)\wedge(\rad(\beta)\vee\rad(\gamma)),$$
%(\rad(\beta)\vee\rad(\gamma))\wedge\rad(\alpha)$,
then $\theta\geq \rad(\delta)$ and
$\C C(\theta, \theta;\rad(\delta))$ holds
according to Theorem~\ref{basic_centrality}~(1).
%(monotonicity of the centralizer).
This shows that $\rad(\delta)\lhd\theta$.
But $\rad(\delta)\wrel{\obst} 1$ or else
$\rad(\delta)=1$, by items (2) and (4)
of this theorem, and in either case we get
$$
\rad(\delta) = \theta = 
\rad(\alpha)\wedge(\rad(\beta)\vee\rad(\gamma)).
$$
Using the isomorphism from (5) once again,
we conclude that 
$
(\alpha/\solv)\wedge((\beta/\solv)\vee(\gamma/\solv)) 
= (\delta/\solv)$.
This proves that $\Conb(\m a)/\solv$ is meet semidistributive.
\end{proof}


\begin{thm} \label{solv=omit_obst}
Let $\vr V$ be a variety with a weak difference term.
If $\m a\in {\vr V}$ has congruences $\delta\leq\theta$,
then $\theta$ is $\XXX$--solvable over $\delta$
if and only if the interval $I[\delta,\theta]$
contains no solvability obstruction.
\end{thm}

\begin{proof}
Assume that $\theta$ is $\XXX$--solvable over 
$\delta$, and that $I[\alpha,\beta]$
is an arbitrarily chosen 
but nontrivial subinterval
of $I[\delta,\theta]$.
%$\delta\leq\alpha < \beta\leq \theta$.
Then $\delta = \delta\wedge\alpha\wrel{\solv}
\theta\wedge\alpha = \alpha$, so 
$\rad(\alpha) = \rad(\delta)\geq\theta\geq\beta$.
This shows that 
$\alpha < \beta = \beta\wedge\rad(\alpha)$.
By item (4) of Theorem~\ref{rad_stuff},
$I[\alpha,\beta]$ is not a solvability obstruction.

Conversely, assume that 
$I[\delta,\theta]$ contains no
solvability obstruction. 
Let $\alpha = \theta\wedge\rad(\delta)$.
Then since $\delta\leq\alpha\leq\rad(\delta)$
we get that $\delta\wrel{\solv}\alpha$,
and hence that $\rad(\alpha) = \rad(\delta)$.
This means that $\alpha = \theta\wedge\rad(\alpha)$.
According to item (4) of Theorem~\ref{rad_stuff},
either $\alpha\wrel{\obst}\theta$ or else $\alpha = \theta$.
The first possibility is ruled out by
the assumption that $I[\delta,\theta]$ contains
no solvability obstruction, so
it must be that  $\alpha = \theta$. Since 
$\delta\wrel{\solv}\alpha = \theta$, the proof
is complete.
\end{proof}

\begin{lm} \label{weak_permute} 
Let $\vr V$ be a variety with a weak difference term.
Assume that $\m a\in {\vr V}$ has congruences 
$\gamma, \gamma', \delta$ and $\delta'$ such that 
$\gamma'\lhd \gamma$ and $\delta'\lhd \delta$. Then
$\gamma\circ\delta\subseteq \gamma'\circ\delta\circ\gamma\circ\delta'$.
\end{lm}

\begin{proof}
Choose any $(a,c)\in \gamma\circ\delta$. There is a 
$b\in A$ such that $a\equiv_{\gamma} b\equiv_{\delta} c$, 
so 
\[a\equiv_{\gamma'} d(a,b,b)
\equiv_{\delta} d(a,b,c)
\equiv_{\gamma} d(b,b,c)
\equiv_{\delta'} c\]
shows that $(a,c)\in \gamma'\circ\delta\circ\gamma\circ\delta'$.
\end{proof}


\begin{thm} \label{sim_cong} 
Let $\vr V$ be a variety with a weak difference term.
If $\m a\in {\vr V}$, then 
$\solv$--classes of $\Conb(\m a)$
consist of permuting congruences.
In particular, these classes are modular
sublattices.
\end{thm}

\begin{proof} 
We assume that the theorem is false and argue
to a contradiction.
If some 
$\solv$--class of $\Conb(\m a)$ contains
a pair of nonpermuting
congruences, then there is a subinterval
$I[\sigma,\tau]$ of that $\solv$--class that
contains the two congruences.
(For example, we could take
$\sigma$ to be the meet 
and $\tau$ to be the join
of the two nonpermuting
congruences.)
Let $A$ denote the ordered
set of all congruences 
$\theta\in I[\sigma,\tau]$ such that 
every congruence in $I[\sigma,\theta]$
permutes with every congruence in $I[\sigma,\tau]$:
\[A = \{\theta\in I[\sigma,\tau]\;|\;\forall\theta'\in I[\sigma,\theta],
\forall\psi\in I[\sigma,\tau]\;
(\theta'\circ \psi = \psi\circ\theta')\}.\]
$A$ is nonempty (since $\sigma\in A$)
and it is closed under updirected unions,
therefore $A$ has maximal elements.
Let $\alpha$ be a maximal element of $A$.
It must be that $\alpha < \tau$ 
since otherwise all
congruences in $I[\sigma,\tau]$
permute, contrary to assumption.
Since $I[\sigma,\tau]$ is $\XXX$--solvable
and $\sigma\leq \alpha < \tau$,
there is a congruence $\widehat{\alpha}$ such that 
$\alpha < \widehat{\alpha} \leq \tau$ and
$\alpha\lhd \widehat{\alpha}$.
%since $I[\alpha,\tau]$ is
%not a solvability obstruction.
The set 
\[B = \{\theta\in I[\sigma,\tau]\;|\;\forall\theta'\in I[\sigma,\theta],
\forall\psi\in I[\sigma,\widehat{\alpha}]\;
(\theta'\circ \psi = \psi\circ\theta')\}\]
of all $\theta\in I[\sigma,\tau]$ such that 
every congruence in $I[\sigma,\theta]$
permutes with every congruence in $I[\sigma,\widehat{\alpha}]$ 
is also nonempty and closed under updirected unions, so
$B$ has a maximal element $\beta$.
We cannot have $\beta = \tau$ since 
this would force $\widehat{\alpha}\in A$,
contradicting the the maximality of $\alpha$.
Since $\beta < \tau$,
there is a congruence $\widehat{\beta}$
such that $\beta < \widehat{\beta}\leq \tau$
and $\beta\lhd \widehat{\beta}$.

By the maximality of $\beta$ and the
fact that $\beta < \widehat{\beta}$, there is some 
$\delta\leq \widehat{\beta}$ which 
fails to permute with some $\gamma\leq \widehat{\alpha}$.
Every congruence below $\beta$ permutes with
every congruence below $\widehat{\alpha}$, so
$\delta\not\leq \beta$. Every congruence
below $\alpha$  permutes with
every congruence in $I[\sigma,\tau]$, so 
$\gamma\not\leq \alpha$.
Therefore $\gamma' = \alpha\wedge\gamma$
is strictly less than $\gamma$
and $\delta' = \beta\wedge\delta$
is strictly less than $\delta$.
The four congruences $\gamma, \gamma', \delta$ and $\delta'$
have the following properties:
\begin{enumerate}
\item[(i)] $\gamma' < \gamma\leq \widehat{\alpha}$, $\gamma'\leq \alpha$;
\item[(ii)] $\delta' < \delta\leq \widehat{\beta}$, $\delta'\leq \beta$;
\item[(iii)] $\gamma'\lhd \gamma$, $\delta'\lhd \delta$; and 
\item[(iv)] any two congruences in $\{\gamma,\gamma',\delta,\delta'\}$
permute, except $\gamma\circ\delta\not\subseteq \delta\circ\gamma$.
\end{enumerate}
(The first, second and fourth items
follow from the way we chose
$\alpha, \widehat{\alpha}, \beta, \widehat{\beta},
\gamma, \gamma', \delta$, and $\delta'$.
The third item follows from Lemma~\ref{meet+join}
since, for example, $\alpha\lhd \widehat{\alpha}$
so $\gamma' = \alpha\wedge \gamma\lhd 
\widehat{\alpha}\wedge \gamma = \gamma$.)
%But these items contradict one another:
By item (iii) and Lemma~\ref{weak_permute} we have
$\gamma\circ\delta\subseteq
\gamma'\circ\delta\circ\gamma\circ\delta'$.
But by items (i), (ii) and (iv) we have
\[\gamma'\circ\delta\circ\gamma\circ\delta' =
\delta\circ\gamma'\circ\gamma\circ\delta' =
\delta\circ\gamma\circ\delta' =
\delta\circ\delta'\circ\gamma = 
\delta\circ\gamma.\]
Thus $\gamma\circ\delta\subseteq \delta\circ\gamma$,
in contradiction with (iv). This is a contradiction 
to the assumption that $I[\sigma,\tau]$
contains a pair of nonpermuting congruences.

For the final statement of the theorem, recall that
any lattice of permuting equivalence relations is modular.
\end{proof}

This result may be used to establish 
a new characterization of
varieties with a weak difference term. 
In the following theorem, a lattice
$\m l$ has an $\sdm/\Mod$ 
{\bf factorization}
\index{factorization}
in a category $\vr C$ of lattices if
there is a $\vr C$-morphism of $\m l$
onto a meet semidistributive lattice whose kernel
classes are modular.
%(For a similar characterization for locally finite varieties,
%see Theorems~7.9 and 7.12 of \cite{hobby-mckenzie}.)

\begin{thm} \label{HMa}
The following are equivalent
for a variety $\vr V$.
\begin{enumerate}
\item[(1)] $\vr V$ has a weak difference term.
\item[(2)] For every $\m a\in {\vr V}$, $\Conb(\m a)$
has an $\sdm/\Mod$ factorization 
in the category of meet continuous lattices.
\item[(3)] For every $\m a\in {\vr V}$, $\Conb(\m a)$
has an $\sdm/\Mod$ factorization 
in the category of ordinary lattices.
\item[(4)] For every $\m a\in {\vr V}$,
each $\sdm$-failure in $\Conb(\m a)$
is modular.
\item[(5)] If $\m l$ belongs to the meet continuous
congruence variety of $\vr V$, then each 
$\sdm$-failure in $\m l$ is modular.
\end{enumerate}
\end{thm}

\begin{proof}
If $\vr V$ has a weak difference term, then
the natural homomorphism 
$\nu\colon \Conb(\m a)\to \Conb(\m a)/\solv$
is a homomorphism of
meet continuous lattices whose image 
is meet semidistributive according
to Theorem~\ref{rad_stuff}~(6).
The kernel classes are modular
according to Theorems~\ref{sim_cong}. This proves
that (1)$\mimplies$(2).

The implications (2)$\mimplies$(3)$\mimplies$(4)
are straightforward; the first since
any factorization in the category of
meet continuous lattices is a factorization in the
category of lattices, and the
second because $\sdm$-failures
must be collapsed by any homomorphism
into a meet semidistributive lattice.

Assume that $\sdm$-failures
of $\Conb(\m a)$ are
modular for any $\m a\in {\vr V}$. 
As explained in the third
paragraph of Remark~\ref{sdm_rem},
the class of meet continuous lattices
whose $\sdm$-failures are modular
is a definable by a meet continuous identity.
If (4) holds, then the congruence
lattices of algebras in $\vr V$ satisfy this identity,
so the meet continuous congruence variety of $\vr V$
also satisfies it, hence (5) holds.

According to Theorem~\ref{pixley-wille_extension}
the class of varieties $\vr U$
that satisfy the the meet continuous congruence
identity expressing the fact that 
$\sdm$-failures are modular 
is definable by idempotent Maltsev conditions.
To show that (5)$\mimplies$(1)
we must show that these Maltsev conditions are strong enough
to force the existence of a weak difference term.
By Theorem~\ref{justify_weak}, it is enough
to show that if $\vr V$ is a variety satisfying (5)
then the abelian algebras in $\vr V$ are affine.
So let $\vr V$ be any variety satisfying
condition (5), and let $\vr A$ be the
subvariety of $\vr V$ 
that is generated by the abelian algebras
of $\vr V$. $\vr A$ must also satisfy
item (5), and in particular (since
the variety of sets
does not satisfy (5)) this 
implies that $\vr A$ has a Taylor term.
Let $\m f = \m f_{\vr A}(4)$ be the
$4$-generated free algebra of this variety.
$\m f$ is abelian 
since $\vr A$ is generated by abelian algebras,
the abelian property is preserved by subalgebras and products,
and free algebras may be constructed as subalgebras of products
of generating algebras.
Since $\m f$ is abelian, 
there is a congruence $\Delta$ on $\m f^2$
that has the diagonal $D = \{(f,f)\;|\;f\in F\}$
as a class. 
It follows from Lemma~4.4 of \cite{k-szendrei2}
(in the case $\gamma=\delta=1$ of that result)
that since $\vr A$ has a Taylor term the congruence
$\Delta$ is a lattice complement
of the coordinate projection kernels
$\eta_1, \eta_2\in \Con(\m f^2)$.
Since $\Delta\wedge \eta_1 = \Delta\wedge \eta_2 = 0$,
the interval $I[0,\Delta\vee(\eta_1\vee\eta_2)] = I[0,\Delta]$
is an $\sdm$-failure. By item (5), this interval
is modular. Since the diagonal 
$D$ is a $\Delta$-class, it follows from 
Lemma~2.4 of \cite{hobby-mckenzie} that
restriction is a lattice homomorphism from
the interval $I[0,\Delta]$ to $\Conb(\m f^2|_D)$, 
where $\m f^2|_D$ is the algebra 
induced on $D$ by $\m f^2$,
so $\Conb(\m f^2|_D)$ is modular. 
The operations of $\m f^2|_D$
are the polynomial operations of $\m f^2$
under which $D$ is closed, 
each of which agrees with some polynomial operation
of $\m f$ acting coordinatewise on $D$.
Thus, $\m f^2|_D$ is isomorphic as a nonindexed
algebra to the algebra $\m f|_F$ that $\m f$
induces on itself. But the congruence lattices
of $\m f$ and $\m f|_F$ are the same,
so $\m f = \m f_{\vr A}(4)$ 
has a modular congruence lattice.
By Day's Theorem, \cite{day1}, this implies that 
$\vr A$ is congruence modular. Since $\vr A$ is
generated by abelian algebras, modular commutator
theory implies that all algebras in $\vr A$ are affine.
\end{proof}

\begin{exmp}
There exists a purely lattice-theoretic proof
that (2)$\mimplies$(3)$\mimplies$(4)
in Theorem~\ref{HMa}, but not for the reverse
implications. Those implications are false
even for congruence lattices of 
individual algebras. 

For example, if $\m l$ is a lattice of height $\leq 3$,
then any $\sdm$-failure has height $\leq 2$,
hence is modular. Thus, any finite,
nonmodular, simple lattice of height $\leq 3$
has the properties that
it is algebraic (hence isomorphic to the congruence
lattice of some algebra), has modular $\sdm$-failures,
but has no $\sdm/\Mod$ factorization.
There are many graph lattices satisfying
the required conditions (cf.\ Section~\ref{4.3.5}).

There also exist algebraic lattices with $\sdm/\Mod$
factorizations in the category of ordinary lattices
and no such factorization in the category of 
meet continuous lattices, 
such as the one in Figure~\ref{fig22}.

\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(45,45)
\put(0,5){\circle*{1.2}}
\put(5,0){\circle*{1.2}}
\put(5,5){\circle*{1.2}}
\put(5,10){\circle*{1.2}}
\put(10,5){\circle*{1.2}}
\put(5,0){\line(-1,1){5}}
\put(5,0){\line(0,1){5}}
\put(5,0){\line(1,1){5}}
\put(5,10){\line(-1,-1){5}}
\put(5,10){\line(0,-1){5}}
\put(5,10){\line(1,-1){5}}

\put(0,15){\circle*{1.2}}
%\put(5,10){\circle*{1.2}}
\put(5,15){\circle*{1.2}}
\put(5,20){\circle*{1.2}}
\put(10,15){\circle*{1.2}}
\put(5,10){\line(-1,1){5}}
\put(5,10){\line(0,1){5}}
\put(5,10){\line(1,1){5}}
\put(5,20){\line(-1,-1){5}}
\put(5,20){\line(0,-1){5}}
\put(5,20){\line(1,-1){5}}

\put(0,25){\circle*{1.2}}
%\put(5,20){\circle*{1.2}}
\put(5,25){\circle*{1.2}}
\put(5,30){\circle*{1.2}}
\put(10,25){\circle*{1.2}}
\put(5,20){\line(-1,1){5}}
\put(5,20){\line(0,1){5}}
\put(5,20){\line(1,1){5}}
\put(5,30){\line(-1,-1){5}}
\put(5,30){\line(0,-1){5}}
\put(5,30){\line(1,-1){5}}


\put(5,33){\circle*{.6}}
\put(5,35){\circle*{.6}}
\put(5,37){\circle*{.6}}
\put(5,40){\circle*{1.2}}

\put(45,20){\circle*{1.2}}
\put(45,20){\line(-2,1){40}}
\put(45,20){\line(-2,-1){40}}
\end{picture}
\end{center}
\caption{No $\sdm/\Mod$ factorization in ${\vr L}_{MC}$}\label{fig22}
\end{figure}
\end{exmp}

We next show that the 
relation $\solv$ behaves well with respect
to the class operators $\Ho, \Su$ and $\Pd_{fin}$.
%We now return to the development of 
%$\XXX$-solvability theory.

\begin{thm}\label{hom+sub+prod}
Let $\vr V$ be a variety with a weak difference term.
\begin{enumerate}
\item[(1)] 
If $\m a\in {\vr V}$ and $\alpha, \beta\geq \gamma$
are congruences on $\m a$, then 
$\alpha\wrel{\solv}\beta$ in $\Conb(\m a)$
if and only if $\alpha/\gamma\wrel{\solv}\beta/\gamma$
in $\Conb(\m a/\gamma)$.

\item[(2)] 
If $\m b$ is a subalgebra of $\m a\in {\vr V}$, then
$\alpha\wrel{\solv}\beta$ in $\Conb(\m a)$
implies that 
$\alpha|_B\wrel{\solv}\beta|_B$ in $\Conb(\m b)$.

\item[(3)] 
If $\m a\in{\vr V}$ has congruences $\alpha$ and $\beta$
and $\m b\in{\vr V}$ has congruence $\gamma$ and $\delta$,
then $\alpha\times\gamma\wrel{\solv}\beta\times\delta$
in $\Conb(\m a\times \m b)$
if and only if $\alpha\solv \beta$ in $\Conb(\m a)$
and $\gamma\solv\delta$ in $\Conb(\m b)$.
\end{enumerate}
In particular, the class of $\XXX$--solvable
algebras in $\vr V$ is closed under
the formation of homomorphic images, subalgebras
and finite products.
\end{thm}

\begin{proof}
It follows from Theorem~\ref{basic_centrality}~(10)
that if $\gamma\leq\delta\leq\theta$,
then $\delta\lhd\theta$ if and only if 
$\delta/\gamma\lhd \theta/\gamma$.
It is also true that a sequence 
$(\theta_{\lambda})_{\lambda<\kappa}$
of congruences on $\m a$, each containing $\gamma$,
is continuous if and only if 
$(\theta_{\lambda}/\gamma)_{\lambda<\kappa}$
is. The property
$\alpha\wrel{\solv}\beta$ in $\Conb(\m a)$
is the property that there exists a continuous
sequence of congruences witnessing $\XXX$-solvability
from $\alpha\wedge\beta$ to $\alpha\vee\beta$, which
is therefore equivalent to the existence of a continuous
sequence of congruences witnessing $\XXX$-solvability
from $(\alpha\wedge\beta)/\gamma = 
(\alpha/\gamma)\wedge(\beta/\gamma)$ to 
$(\alpha\vee\beta)/\gamma = (\alpha/\gamma)\vee(\beta/\gamma)$.
Hence $\alpha\solv \beta$ is equivalent to $\alpha/\gamma\solv \beta/\gamma$
when $\gamma\leq \alpha\wedge\beta$. This proves (1).

It follows from Theorem~\ref{basic_centrality}~(9)
that restriction to $\m b$ preserves the
relation $\lhd$ (i.e., $\delta\lhd\theta$
in $\Conb(\m a)$ implies
$\delta|_{\m b}\lhd \theta|_{\m b}$).
Since restriction to $\m b$ also preserves
unions of chains, it must preserve $\lhdlhd$.
Hence $\XXX$-solvable intervals in $\Conb(\m a)$
restrict to $\XXX$-solvable intervals in $\Conb(\m b)$.
Since $\alpha\solv \beta$ if and only if $\alpha$
and $\beta$ lie in a single $\XXX$-solvable
interval, $\solv$ is also preserved
under restriction to $\m b$. This proves (2).

Applying item (1) to $\m a\times \m b$
(with the coordinate projection kernels 
$\eta_1$ and $\eta_2$ playing
the role of $\gamma$ in (1)) we obtain
that $\alpha\solv \beta$ in $\Conb(\m a)$
if and only 
if $\alpha_1\solv \beta_1$ in $\Conb(\m a\times \m b)$
and similarly
$\gamma\solv \delta$ in $\Conb(\m b)$ if and only 
if $\gamma_2\solv \delta_2$
in $\Conb(\m a\times \m b)$.
Since $\XXX$-solvability is a congruence, 
the conjunction of $\alpha\solv \beta$ and 
$\gamma\solv \delta$ implies that 
$$
\alpha\times\gamma := \alpha_1\wedge \gamma_2
\wrel{\solv}\beta_1\wedge \delta_2 =: \beta\times\delta.
$$
Conversely, if 
$\alpha\times\gamma \wrel{\solv}\beta\times\delta$,
then again because $\XXX$-solvability is a congruence
we get 
$$
\alpha_1 = \eta_1\vee(\alpha\times\gamma) 
\wrel{\solv}\eta_1\vee(\beta\times\delta) = \beta_1,
$$
and similarly $\gamma_2\wrel{\solv}\delta_2$.
As proved above, these instances of $\solv$ 
hold if and only if $\alpha\wrel{\solv}\beta$ 
in $\Conb(\m a)$ and
$\gamma\wrel{\solv}\delta$ in $\Conb(\m b)$.
\end{proof}

By a {\bf partial Maltsev operation}
\index{Maltsev!partial --- operation}
\index{partial Maltsev operation}
on a set $A$ we mean a function 
$m\colon D\to A$, for some domain $D\subseteq A^3$
containing all triples $(x,y,y)$ and $(y,y,x)$,
$x, y\in A$, which satisfies 
$m(x,y,y) = x = m(y,y,x)$ for all $x, y\in A$.

\begin{thm}\label{tol=cong2_prep}
Let $\vr V$ be a variety with a weak difference term.
Assume that $\m a\in {\vr V}$, 
$\theta\in\Con(\m a)$ is $\XXX$-solvable, and
$S\subseteq A^n[\theta]$ is a subset.
If $S$ is compatible with the operations
of $\m a$, then $S$ is also
compatible with 
some partial Maltsev operation on $A$
\end{thm}

\begin{proof}
Recall from the opening paragraph of Chapter~\ref{elementary}
that $A^n[\theta]$ is defined to be
$
\{\wec{x}\in A^n\;|\;x_i\equiv_{\theta} x_j
\;\mbox{\rm for all $i, j$}\}.
$
Assume that $S\subseteq A^n[\theta]$ 
is compatible with the operations
of $\m a$. We will show that $S$ is 
compatible with the partial Maltsev operation on $A$
with smallest domain, which is
$$
D = \{(a,b,b)\;|\;a, b\in A\}\cup \{(a,a,b)\;|\;a, b\in A\}.
$$
We must show that if $\wec{a}, \wec{b}, \wec{c}\in S$ and
the tuple
$$
m(\wec{a}, \wec{b}, \wec{c}) = (m(a_1,b_1,c_1),m(a_2,b_2,c_2),
\ldots,m(a_n,b_n,c_n)) = :\wec{d}
$$
is defined in all coordinates, then $\wec{d}\in S$.
To be defined in all coordinates, it must be the case
that $a_i = b_i$ or $b_i = c_i$ (or both) for all $i$.
Assume that this is so.
Let $I \subseteq N:=\{1,\ldots,n\}$ be the
set of coordinates where $a_i = b_i$.
If $I = \emptyset$, then $\wec{b}=\wec{c}$, so 
$\wec{d} = \wec{a}\in S$.
If $I = N$, then $\wec{a}=\wec{b}$, so
$\wec{d} = \wec{c}\in S$. Thus the only
nontrivial cases are those where
$\emptyset\neq I\neq N$.
Restrict attention to those, and 
let $J = N - I$.

Let $\m s$ be the subalgebra of
$\m a^n$ that is supported by $S=A^n[\theta]$.
Elements $\wec{u}, \wec{v}\in S$ are related
by $\theta_i$ ( = $\pi_i^{-1}(\theta)$) if and only if
$u_i\equiv_\theta v_i$. Since $S\subseteq A^n[\theta]$
this implies that 
$u_j\equiv_{\theta} u_i\equiv_{\theta} v_i\equiv_{\theta} v_j$
for any $j$, hence that $\theta_i\subseteq \theta_j$
for any $i$ and $j$.
Let $\widehat{\theta}$ denote the congruence
$\theta_1 = \theta_2 = \cdots = \theta_n$ of $\m s$.
It follows from
Theorem~\ref{hom+sub+prod}~(1) and (2) that 
$\theta_i\solv \eta_i$ for each $i$, so 
$\widehat{\theta} = \bigwedge \theta_i\solv \bigwedge \eta_i = 0$,
proving that $\widehat{\theta}$ is $\XXX$-solvable.

If $i\in I$, then 
$\eta_I\leq \eta_i\leq \theta_i = \widehat{\theta}$,
and a similar argument proves that $\eta_J\leq \widehat{\theta}$.
Since $\widehat{\theta}$ is $\XXX$-solvable, 
$\eta_I\solv \widehat{\theta}\solv\eta_J$.
This and Theorem~\ref{sim_cong} prove that 
$\eta_I$ and $\eta_J$ are permuting congruences
of $\m s$. According to the definition
of $I$ and $J$, we have
$
\wec{a}\wrel{\eta_I}
\wec{b}\wrel{\eta_J}
\wec{c},
$
so there must exists a tuple $\wec{x}\in S$
such that 
$
\wec{a}\wrel{\eta_J}
\wec{x}\wrel{\eta_I}
\wec{c}.
$
The tuple $\wec{x}$ satisfies the condition 
that $x_i=c_i$ if $i\in I$ and 
$x_j = a_j$ if $j\in N-I=J$. That is, 
$x_i = c_i$ if $m(a_i,b_i,c_i) = m(a_i,a_i,c_i) = c_i$,
and $x_j = a_j$ if $m(a_j,b_j,c_j) = m(a_j,c_j,c_j) = a_j$.
This proves that $\wec{x} = \wec{d}$, hence 
$\wec{d} = \wec{x} \in S$ as desired.
\end{proof}

It is a general fact that if $\m a$ is an algebra,
$m$ is an operation on $A$, and for every cardinal $\kappa$
any relation $S\subseteq A^{\kappa}$ compatible with
the operations of $\m a$ is also
compatible with $m$, then $m$ is a term
operation of $\m a$. If instead one only
has that any {\it finitary} relation $S$
on $A$ that is compatible with the operations
of $\m a$ is also compatible with $m$,
then $m$ is only guaranteed to be
a {\bf local term operation}
\index{local term operation}
of $\m a$. This means that for any finite subset
$F\subseteq A$ there is a term operation
$t_F$ that agrees with $m$ on $F$.
Theorem~\ref{tol=cong2_prep} suggests
the possibility that if $\vr V$ has a weak difference term
and $\m a\in {\vr V}$ is $\XXX$-solvable,
then $\m A$ has a local Maltsev term operation.
Admittedly Theorem~\ref{tol=cong2_prep} speaks
only about partial Maltsev operations, 
but that is exactly what one would expect 
of a result like Theorem~\ref{tol=cong2_prep}
if the local Maltsev term operation were not unique.
These observations lead to speculations
recorded in the following problem.

\begin{prb}
Assume that $\vr V$ has a weak difference term,
and that $\m a\in {\vr V}$ is $\XXX$-solvable.
Does $\m a$ have a local Maltsev term operation?
Does $\m a$ have a Maltsev term operation?
If $\vr V$ is generated by $\XXX$--solvable algebras,
then must $\vr V$ be congruence permutable?
\end{prb}

Although affirmative answers to these 
questions seem
unlikely, it is true that for each finite $n$
there is term that is Maltsev on 
classes of $n$-step solvable congruences
of algebras in $\vr V$.
It would be interesting if this
property extended to congruences with higher
degrees of solvability.

\begin{cor}\label{tol=cong2}
Let $\vr V$ be a variety with a weak difference term.
If $\m a\in {\vr V}$, 
$\rho\wrel{\solv}0$ in $\Conb(\m a)$ and 
$S\subseteq \rho \;( = A^2[\rho])$
is a reflexive
compatible binary relation on $\m a$,
then $S$ is a congruence.
% on $\m a$.
\end{cor}

\begin{proof}
This is proved in the same way that one proves
that a reflexive compatible binary 
relation on a Maltsev algebra is a congruence,
since that argument works equally well with 
a partial Maltsev operation.

For the record, the details are as follows.
If $S\subseteq \rho$
is reflexive and compatible, 
then $S$ is compatible with a partial
Maltsev operation according to 
Theorem~\ref{tol=cong2_prep}.
Thus, if $(a,b)\in S$, then since
$S$ is reflexive 
$(a,a), (b,b)\in S$ also, so 
$(b,a) = m((a,a),(a,b),(b,b))\in S$.
This proves that $S$ is symmetric.
If $(a,b), (b,c)\in S$, then
$(a,c) = m((a,b),(b,b),(b,c))\in S$.
This proves that $S$ is transitive,
hence a congruence.
\end{proof}

\begin{thm}\label{3permute}
Let $\vr V$ be a variety with a weak difference term.
\begin{enumerate}
\item[(1)] 
If $\m a\in {\vr V}$ has congruences
$\alpha'\wrel{\lhdlhd}\alpha$ and 
$\beta'\wrel{\lhdlhd}\beta$, then 
$$\alpha\vee\beta = (\alpha'\vee\beta')\circ\alpha\circ
\beta\circ(\alpha'\vee\beta').$$
\item[(2)] 
If $\m a\in {\vr V}$ has congruences
$\alpha$ and $\beta$ 
with $\beta\wrel{\solv}\alpha\wedge\beta$,
then 
$$\alpha\vee\beta = \alpha\circ\beta\circ\alpha.$$
\end{enumerate}
\end{thm}

\begin{proof}
For item (1),
the relation $R = 
(\alpha'\vee\beta')\circ\alpha\circ
\beta\circ(\alpha'\vee\beta')$
is an $(\alpha'\vee\beta')$--closed 
compatible reflexive relation on $\m a$ whose
transitive closure is $\alpha\vee\beta$.
To prove this part of the 
theorem it suffices to 
show that $R$ is a congruence.

Since $\alpha'\wrel{\lhdlhd}\alpha$ and 
$\beta'\wrel{\lhdlhd}\beta$, 
it follows from Lemma~\ref{meet+join} 
and the transitivity of $\lhdlhd$ that
$(\alpha'\vee\beta')\wrel{\lhdlhd}
(\alpha\vee\beta)$.
In the quotient algebra $\m a/(\alpha'\vee\beta')$ we have that 
$$\rho:=(\alpha\vee\beta)/(\alpha'\vee\beta')
\wrel{\solv}(\alpha'\vee\beta')/(\alpha'\vee\beta') = 0$$
by Theorem~\ref{hom+sub+prod}. The relation 
$S := R/(\alpha'\vee\beta')$
is a compatible reflexive relation
of $\m a/(\alpha'\vee\beta')$, and $S\subseteq \rho$,
hence $S$ is a congruence on $\m a/(\alpha'\vee\beta')$
according to Corollary~\ref{tol=cong2}.
The fact that $R$ is $(\alpha'\vee\beta')$--closed
means that for the natural homomorphism
$\nu: \m a\rightarrow\m a/(\alpha'\vee\beta'): 
a\mapsto a/(\alpha'\vee\beta')$
we have $R = \nu^{-1}(\nu(R))$. But
$\nu(R) = R/(\alpha'\vee\beta') = S$ is a congruence, 
and the inverse image of a congruence
is a congruence. Hence $R$ is a congruence.

Item (2) is the special case of item (1)
in which $\alpha = \alpha'\geq \beta'=\alpha\wedge\beta$.
%=\alpha\wedge\beta$.
\end{proof}

%Theorem~\ref{3permute}~(1) 
%is a strengthening of Theorem~\ref{sim_cong},
%since the statement of 
Theorem~\ref{sim_cong} follows from the case of
Theorem~\ref{3permute}~(1) 
where $\alpha'=\beta'=\alpha\wedge\beta$.

\begin{thm} \label{pre-N5}
Let $\vr V$ be a variety with a weak difference term.
If $\m a\in {\vr V}$ has congruences
$\alpha$ and $\beta$ with $\beta\wrel{\solv}\alpha\wedge\beta$
and $\alpha\leq\gamma\leq\alpha\vee\beta$, then
$\gamma = \alpha\vee(\beta\wedge\gamma)$.
Hence the map $J: I[\alpha\wedge\beta,\beta]
\rightarrow I[\alpha,\alpha\vee\beta]:
x\mapsto x\vee \alpha$ is surjective.
\end{thm}

\begin{proof}
The hypotheses of Theorem~\ref{3permute}~(2) are met,
so $\alpha\vee\beta = \alpha\circ\beta\circ\alpha$.
Arbitrarily choose $(a,b)\in\gamma$.
Since $\gamma\leq 
\alpha\vee\beta = \alpha\circ\beta\circ\alpha$,
there are elements $u, v\in A$ such that 
\begin{equation}\label{a_u_v_b}
a\wrel{\equiv_{\alpha}}u
\wrel{\equiv_{\beta}}v
\wrel{\equiv_{\alpha}}b.
\end{equation}
This means that 
$u\wrel{\equiv_{\alpha}}a
\wrel{\equiv_{\gamma}}b
\wrel{\equiv_{\alpha}}v$, so 
$(u,v)\in\beta\wedge(\alpha\vee\gamma)=
\beta\wedge\gamma$.
This allows us to strengthen the
$\beta$--relation in 
%the previous displayed
line (\ref{a_u_v_b})
to a $(\beta\wedge\gamma)$--relation,
and in the strengthened form that line indicates
that $(a,b)\in\alpha\vee(\beta\wedge\gamma)$.
Since $(a,b)$ was chosen arbitrarily,
$\gamma\subseteq \alpha\vee(\beta\wedge\gamma)$.
The reverse inclusion follows from the fact
that $\alpha\leq\gamma$.

For the final statement of the theorem,
note that for any 
$\gamma\in I[\alpha,\alpha\vee\beta]$
the element $\beta\wedge\gamma$ is in
$I[\alpha\wedge\beta,\beta]$ and
we have proved above that  
$\gamma = J(\beta\wedge\gamma)$.
\end{proof}


\begin{thm} \label{weak_omit}
Let $\vr V$ be a variety with a weak difference term.
Assume that $\m a\in {\vr V}$.
\begin{enumerate}
\item[(1)]
If $\m n_5$, 
with the labeling indicated in Figure~\ref{fig23}, 
appears as a sublattice of $\Conb(\m a)$,
then $\beta\wrel{\not\!\solv}\beta\wedge\alpha$.

\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(20,35)
\put(0,20){\circle*{1.2}}
\put(10,5){\circle*{1.2}}
\put(10,35){\circle*{1.2}}
\put(20,15){\circle*{1.2}}
\put(20,25){\circle*{1.2}}

\put(10,5){\line(-2,3){10}}
\put(10,5){\line(1,1){10}}
\put(10,35){\line(-2,-3){10}}
\put(10,35){\line(1,-1){10}}
\put(20,25){\line(0,-1){10}}

\put(-4.5,18){$\beta$}
\put(22,14){$\alpha$}
\put(22,24){$\gamma$}
\put(5.5,0){$\beta\wedge \alpha$}
\end{picture}
\end{center}
\caption{$\beta\wrel{\not\!\!\solv}\beta\wedge\alpha$}\label{fig23}
\end{figure}

\item[(2)] If $\m d_2$
appears as a sublattice of $\Conb(\m a)$, then 
no two distinct congruences in the sublattice are
$\XXX$-solvably related.
\end{enumerate}
\end{thm}

\begin{proof}
In item (1), we have
$\alpha\leq\gamma\leq\alpha\vee\beta$.
Therefore, if we had
$\beta\wrel{\solv}\beta\wedge\alpha$,
then all of the 
the hypotheses of Theorem~\ref{pre-N5} 
would be met.
This would force $\gamma = \alpha\vee(\beta\wedge\gamma)$.
But $\alpha\vee(\beta\wedge\gamma) = \alpha\neq\gamma$,
so we cannot have
$\beta\wrel{\solv}\beta\wedge\alpha$.

For item (2), the relation $\solv$ restricts to a congruence
on $\m d_2$, which is a subdirectly irreducible lattice.
If the copy of $\m d_2$ under consideration is labeled
as in Figure~\ref{fig24}, then the critical intervals
are $I[\mu,\nu]$, 
$I[\alpha,\tau]$, and 
$I[\delta,\theta]$. Therefore, if any two distinct
elements of this sublattice are $\solv$--related then
$\mu\wrel{\solv}\nu$, $\alpha\wrel{\solv}\tau$, and
$\delta\wrel{\solv}\theta$.
\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{ \ }\label{fig24}
\end{figure}
Assume that this is so and choose $\alpha'$
so that $\alpha < \alpha'\leq\tau$ and 
$\alpha\lhd\alpha'$. By Theorem~\ref{pre-N5}
the maps 
$I[\mu,\nu]\to I[\alpha,\tau]\colon x\mapsto x\vee\alpha$
and
$I[\delta,\theta]\to I[\alpha,\tau]\colon x\mapsto x\vee\alpha$
are surjective, so there exist $\mu'$ and 
$\delta'$ such that $\mu < \mu'\leq\nu$, 
$\delta < \delta' < \theta$ and
$\mu'\vee\alpha = \alpha' = \delta'\vee\alpha$.
Since $I[\mu,\mu']$ and $I[\delta,\delta']$
are perspective with $I[\alpha,\alpha']$
we get that $\mu\lhd \mu'$ and $\delta\lhd\delta'$
from Theorem~\ref{perspective_abelian_taylor}
(or Lemma~\ref{meet+join}).

\begin{clm}
$\mu', \alpha$, and $\delta'$ and generate a sublattice 
isomorphic to $\m d_2$.
\end{clm}

Our argument uses Lemma~\ref{presentation}
with $x = \mu'$, $y = \alpha$ and $z = \delta'$,
as well as the fact that $\nu, \alpha$ and $\theta$
generate a copy of $\m d_2$.

For item (i) of Lemma~\ref{presentation}, 
\[
x\wedge y = \mu'\wedge\delta'\leq \nu\wedge\theta\leq \alpha = y\,.
\]

%For item (ii), 
%\[
%x\vee y = \mu'\vee\delta'\geq \mu\vee\delta=\alpha = y\,.
%\]

For item (ii), 
\[
z\vee(y\wedge x) = \delta'\vee(\alpha\wedge \mu')
= \delta'\vee \mu = \alpha'  = \alpha\vee \mu'\ge \mu' = x\,.
\]
Here we have used the facts that 
$I[\mu,\mu']$ and $I[\delta,\delta']$
are perspective up to $I[\alpha,\alpha']$, and
that $\delta'\vee\mu = \delta'\vee(\delta\vee\mu) = 
\delta'\vee\alpha = \alpha'$.

Item (iii) is similar to (ii). For (iv), 
\[
(x\wedge y)\vee(y\wedge z) = 
(\mu'\wedge \alpha)\vee(\alpha\wedge\delta') = 
\mu\vee\delta = \alpha\,.
\]

Since $x = \mu'\not\leq \alpha = y$, the sublattice
generated by $\{\mu', \alpha, \delta'\}$ is 
isomorphic to $\m d_2$. This prove the claim.

\medskip

From the assumption that $\Conb(\m a)$ has a sublattice
isomorphic to $\m d_2$ with $\XXX$-solvable
critical interval we have shown that 
$\Conb(\m a)$ has a sublattice
isomorphic to $\m d_2$ with abelian critical interval.
This contradicts Theorem~\ref{taylor_forbidden}~(2).
\end{proof}

The rest of the results in this section provide some
information about failures of join semidistributivity
in the congruence lattice of an algebra with a 
weak difference term.

\begin{lm} \label{sdj_failure_cent_lm}
Assume that $\vr V$ has a weak difference term.
If $\m a\in {\vr V}$ has congruences $\alpha, \beta$
and $\gamma$ satisfying 
\begin{enumerate}
\item[(1)] $\alpha\vee\beta\wrel{\solv}\alpha
\wrel{\solv}\alpha\vee\gamma$, and
\item[(2)] $\beta\wedge\gamma\leq \alpha$, 
\end{enumerate}
then 
$(\beta\vee(\alpha\wedge\gamma))\wedge
(\gamma\vee(\alpha\wedge\beta)) = (\alpha\wedge\beta)\vee
(\alpha\wedge\gamma)$.
\end{lm}

\begin{proof}
Since $\beta\wedge\gamma$ lies below all congruences
of interest, we may factor by it
and assume that $\beta\wedge\gamma = 0$.

In the next claim we label some of the congruences
in the sublattice of $\Conb(\m a)$ generated
by $\{\alpha, \beta, \gamma\}$, and prove
some of the more obvious relationships between them.

\begin{clm} \label{get_stuff_right}
Let $\delta = (\alpha\wedge\beta)\vee(\alpha\wedge\gamma)$, 
$\widehat{\beta} = \beta\vee\delta$,
$\widehat{\gamma} = \gamma\vee\delta$, 
$\theta = \widehat{\beta}\wedge\widehat{\gamma}$,
$\beta' = \beta\wedge\theta$, and $\gamma' = \gamma\wedge\theta$.
Then
\begin{enumerate}
\item[(i)] $\delta\leq \alpha$, 
\item[(ii)] $\beta\wedge\delta = \beta\wedge\alpha$ and
$\gamma\wedge\delta = \gamma\wedge\alpha$, 
\item[(iii)] $\beta\wrel{\solv} \beta\wedge\delta$ and 
$\gamma\wrel{\solv} \gamma\wedge\delta$, 
\item[(iv)] $\delta\leq\theta\leq\widehat{\beta}$ and 
$\delta\leq\theta\leq\widehat{\gamma}$,
\item[(v)] $\widehat{\beta}\wrel{\solv}\delta\wrel{\solv}\widehat{\gamma}$, 
\item[(vi)] $\widehat{\beta}\vee\alpha = \beta\vee\alpha$ and 
$\widehat{\gamma}\vee\alpha = \gamma\vee\alpha$, and
\item[(vii)] $\beta'\vee\delta = \theta = \gamma'\vee\delta$.
\end{enumerate}
\end{clm}

Some of these congruences and 
relationships are indicated in
Figure~\ref{fig25}.

\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}  %unit length increased from .8mm
\begin{picture}(30,30)
\put(0,15){\circle*{1.2}}
\put(10,5){\circle*{1.2}}
\put(10,25){\circle*{1.2}}
\put(15,0){\circle*{1.2}}
\put(15,10){\circle*{1.2}}
\put(15,20){\circle*{1.2}}
%\put(15,30){\circle*{1.2}}
\put(20,5){\circle*{1.2}}
\put(20,25){\circle*{1.2}}
\put(30,15){\circle*{1.2}}

\put(15,0){\line(-1,1){15}}
\put(15,0){\line(1,1){15}}
\put(15,10){\line(-1,-1){5}}
\put(15,10){\line(1,-1){5}}
\put(15,10){\line(0,1){10}}
\put(15,20){\line(-1,1){5}}
\put(15,20){\line(1,1){5}}
\put(0,15){\line(1,1){10}}
\put(30,15){\line(-1,1){10}}

\put(-5,14){$\beta$}
\put(32,14){$\gamma$}
\put(-4,3){$\alpha\wedge\beta$}
\put(23,3){$\alpha\wedge\gamma$}
\put(-21,25){$\widehat{\beta} = \beta\vee(\alpha\wedge\gamma)$}
\put(23,25){$\gamma \vee (\alpha\wedge\beta) = \widehat{\gamma}$}
\put(10.5,18.5){$\theta$}
\put(10.5,8.5){$\delta$}
\end{picture}
\end{center}
\caption{ \ }\label{fig25}
\end{figure}

We omit the proofs of the parts of the claim 
that concern $\gamma$ if those parts are symmetric to the
arguments we give for $\beta$.

For item (i) of Claim~\ref{get_stuff_right},
note that $\delta = (\alpha\wedge\beta)\vee (\alpha\wedge\gamma)$
is a join of two elements below $\alpha$, so $\delta\leq \alpha$.

For item (ii), since $\delta\leq \alpha$ we get that
$\beta\wedge\delta \leq \beta\wedge\alpha$. On the other hand,
both $\beta$ and $\delta$ are above $\beta\wedge\alpha$,
so $\beta\wedge\delta\geq \beta\wedge\alpha$.

For item (iii), start with 
$\alpha\wrel{\solv} \alpha\vee\beta$, 
which holds by assumption (1) of the theorem.
Using the fact that $\solv$ is a congruence, meet
both sides of 
$\alpha\wrel{\solv} \alpha\vee\beta$
with 
$\beta$ to obtain $\beta\wedge\alpha\wrel{\solv} \beta$.
Applying the result from (ii) of this claim yields
$\beta\wedge\delta\wrel{\solv} \beta$.

For item (iv), the fact that 
$\theta = \widehat{\beta}\wedge\widehat{\gamma} = 
(\beta\vee\delta)\wedge(\gamma\vee\delta)$ is a meet
of elements above $\delta$ is enough to
prove that $\delta\leq \theta$, while 
$\theta\leq\widehat{\beta}$ is immediate from
the definition $\theta = \widehat{\beta}\wedge\widehat{\gamma}$.

For item (v), join the relation 
$\beta\wedge\delta\wrel{\solv} \beta$ from (iii)
with $\delta$ to obtain 
$\delta\wrel{\solv}\beta\vee\delta = \widehat{\beta}$.

For item (vi), 
$\widehat{\beta}\vee\alpha = (\beta\vee\delta)\vee\alpha = 
\beta\vee (\delta\vee\alpha) = \beta\vee\alpha$,
where the last equality follows from item (i).

For item (vii), use the fact established in (iii)
that $\beta\wrel{\solv} \beta\wedge\delta$,
the fact established in (iv) that 
$\delta\leq\theta\leq\widehat{\beta} =\beta\vee\delta$, and 
Theorem~\ref{pre-N5} to deduce that 
$\theta = \delta\vee(\beta\wedge\theta)$.
Since $\beta' = \beta\wedge\theta$, this
yields
$\theta = \beta'\vee\delta$.
This completes the proof of Claim~\ref{get_stuff_right}.

By examining the definitions of $\delta$
and $\theta$, the reader will verify that the statement
of the next claim is precisely 
what is asserted in the statement of this lemma.

\begin{clm} \label{get_meet_right}
$\delta = \theta$.
\end{clm}

We assume instead that $\delta \neq \theta$ and show
that the sublattice of $\Conb(\m a)$ generated
by $\{\beta', \gamma', \delta\}$
is isomorphic to $\m d_2$.
In order to show this, we verify that 
the congruences
$x = \beta', y = \delta$, and $z = \gamma'$
satisfy the relations from the 
presentation of $\m d_2$ given in 
Lemma~\ref{presentation}. Throughout the argument
we will make free use of Claim~\ref{get_stuff_right}.

To establish relation (i) of Lemma~\ref{presentation}, we have
$$
x\wedge z = \beta'\wedge\gamma' 
%(\beta\wedge\theta)\wedge(\gamma\wedge\theta)
\leq \beta\wedge\gamma = 0\leq \delta = y.
$$
%For relation (ii), 
%$$
%\begin{array}{rl}
%x\vee z
%&= \beta'\vee\gamma'\\
%&= (\beta\wedge\theta)\vee(\gamma\wedge\theta)\\
%&= ((\beta\wedge\theta)\vee(\beta\wedge\delta))
%\vee((\gamma\wedge\theta)\vee(\gamma\wedge\delta))\\
%&= ((\beta\wedge\theta)\vee(\gamma\wedge\theta))\vee
%((\beta\wedge\delta)\vee(\gamma\wedge\delta))\\
%&= (\beta'\vee \gamma')\vee
%((\beta\wedge\alpha)\vee(\gamma\wedge\alpha))\\
%&= \beta'\vee(\gamma'\vee\delta)\\
%&= \beta'\vee\theta= \theta\geq \delta = y.
%\end{array}
%$$
For relation (ii), 
$$
\begin{array}{rl}
z\vee (y\wedge x) 
&= \gamma'\vee(\delta\wedge\beta')\\
&= (\gamma\wedge\theta)\vee(\delta\wedge(\beta\wedge\theta))\\
&= (\gamma\wedge\theta)\vee(\beta\wedge\delta)\\
&= (\gamma\wedge\theta)\vee(\gamma\wedge\delta)\vee
(\beta\wedge\delta)\\
&= \gamma'\vee((\gamma\wedge\alpha)\vee
(\beta\wedge\alpha))\\
&= \gamma'\vee \delta\\
&= \theta\geq \beta\wedge\theta = \beta' = x.
\end{array}
$$
The proof that $x\vee (y\wedge z) \geq z$ is similar
to the one just given.
For relation (iv), 
$$
\begin{array}{rl}
(x\wedge y)\vee(y\wedge z)&= 
(\beta'\wedge\delta)\vee (\delta\wedge\gamma')\\
&=((\beta\wedge\theta)\wedge\delta)\vee
(\delta\wedge(\gamma\wedge\theta))\\
&=(\beta\wedge\delta)\vee(\gamma\wedge\delta)\\
&=(\beta\wedge\alpha)\vee(\gamma\wedge\alpha)\\
&=\delta = y.
\end{array}
$$
Finally, $x = \beta'\not\leq \delta = y$ since
$\beta'\vee\delta = \theta$
by Claim~\ref{get_stuff_right}~(vii),
and we are assuming that $\delta\neq\theta$.

Thus, our  
assumption that $\delta \neq \theta$ 
and Lemma~\ref{presentation} together
%( = $\beta'\vee\delta$) 
imply that the sublattice generated by 
$\{\beta', \gamma', \delta\}$ is 
isomorphic to $\m d_2$. 
It follows from Claim~\ref{get_stuff_right}~(iv) and (v)
that $I[\delta, \theta]$
is an $\XXX$--solvable interval in this sublattice.
This is forbidden by Theorem~\ref{weak_omit},
hence Claim \ref{get_meet_right} and the lemma are proved.
\end{proof}


\begin{thm} \label{sdj_failure_cent}
Assume that $\vr V$ has a weak difference term.
If $\m a\in {\vr V}$ has congruences $\alpha, \beta$
and $\gamma$ satisfying 
\begin{enumerate}
\item[(1)] $\alpha\vee\beta\wrel{\solv}\alpha
\wrel{\solv}\alpha\vee\gamma$, and
\item[(2)] $\beta\wedge\gamma\leq \alpha$, 
\end{enumerate}
then $\C C(\alpha\vee\beta, \alpha\vee\gamma;\alpha)$ holds.
\end{thm}

\begin{proof}
As in the proof of Lemma~\ref{sdj_failure_cent_lm},
we may assume that $\beta\wedge\gamma = 0$.
As in that proof we let 
$\delta = (\alpha\wedge\beta)\vee(\alpha\wedge\gamma)$, 
$\widehat{\beta} = \beta\vee\delta$, and
$\widehat{\gamma} = \gamma\vee\delta$. 

\begin{clm} \label{get-meet-again}
$\alpha\wedge \widehat{\beta} = \alpha\wedge\widehat{\gamma}= 
\widehat{\beta}\wedge \widehat{\gamma}$.
\end{clm}

We proved in 
%Claim~\ref{get_meet_right} of 
Lemma ~\ref{sdj_failure_cent_lm} 
that $\delta = \widehat{\beta}\wedge\widehat{\gamma}$,
so it suffices (by symmetry) to prove only
that $\alpha\wedge\widehat{\beta} = \delta$.
Both $\alpha$ and $\widehat{\beta}$ are above
$\delta$, according
to Claim~\ref{get_stuff_right}~(i) and (iv)
of Lemma ~\ref{sdj_failure_cent_lm},
so we certainly have 
$\alpha\wedge \widehat{\beta}\geq \delta$.
This forces 
$\alpha\wedge \widehat{\beta}\in I[\delta,\widehat{\beta}]
= I[\delta,\beta\vee\delta]$.
The perspective interval 
$I[\beta\wedge\delta, \beta] = I[\beta\wedge\alpha,\beta]$
is $\XXX$--solvable by assumption, so Theorem~\ref{pre-N5} 
guarantees that
$$\alpha\wedge\widehat{\beta} = 
\delta\vee(\beta\wedge(\alpha\wedge\widehat{\beta})).
$$
But we can simplify the expression on the right 
as follows: 
$\delta\vee(\beta\wedge(\alpha\wedge\widehat{\beta})) = 
\delta\vee(\alpha\wedge(\beta\wedge\widehat{\beta})) = 
\delta\vee(\alpha\wedge\beta) = \delta$.
This proves the claim.

\begin{clm} \label{cent1}
$\C C(\widehat{\beta}, \widehat{\gamma};\alpha)$ holds.
\end{clm}

It follows from Claim~\ref{get-meet-again} that
$\widehat{\gamma}\wedge(\widehat{\beta}\vee(\widehat{\gamma}\wedge\alpha))
\leq \alpha$, so Claim~\ref{cent1} isa direct
application of Theorem~\ref{basic_centrality}~(8).

\begin{clm}
$\C C(\alpha\vee\beta, \alpha\vee\gamma; \alpha)$ holds.
\end{clm}

It follows from Theorem~\ref{basic_centrality}~(7)
or (8) that $\C C(\alpha, \widehat{\gamma}; \alpha)$ holds.
From this, Claim~\ref{cent1}, 
%$\C C(\widehat{\beta}, \widehat{\gamma};\alpha)$ 
and
Theorem~\ref{basic_centrality}~(5) we get that
$\C C(\alpha\vee\widehat{\beta}, \widehat{\gamma};\alpha)$ holds.
We established that $\alpha\vee\widehat{\beta} = 
\alpha\vee\beta$ in Claim~\ref{get_stuff_right}~(vi)
of Lemma ~\ref{sdj_failure_cent_lm},
so $\C C(\alpha\vee\beta, \widehat{\gamma};\alpha)$ holds.
Using Theorem~\ref{basic_centrality}~(3) we finally get that 
$\C C(\alpha\vee\beta, \alpha\circ\widehat{\gamma}\circ\alpha;\alpha)$ holds.
This claim, and this theorem, will be established if we show
that $\alpha\circ\widehat{\gamma}\circ\alpha 
= \alpha\vee\widehat{\gamma}$ ( = $\alpha\vee\gamma$).
For this, meet both sides of the relation 
$\alpha\vee\gamma\wrel{\solv}\alpha$,
which is part of assumption (1) of the theorem,
with the congruence $\widehat{\gamma}$. 
The result is that 
$\widehat{\gamma}\wrel{\solv} \alpha\wedge \widehat{\gamma}$.
From Theorem~\ref{3permute}~(2) we conclude that 
$\alpha\circ\widehat{\gamma}\circ\alpha 
= \alpha\vee\widehat{\gamma}$, as desired.
\end{proof}

There is a special case of the previous theorem 
that is worth recording.

\begin{cor} \label{sdj_failure=abel}
Assume that $\vr V$ has a weak difference term,
that $\m a\in {\vr V}$, and that 
$I$ is an $\sdj$-failure in $\Conb(\m a)$.
If $I$ is $\XXX$--solvable, then 
$I$ is abelian.
\end{cor}

\begin{proof}
An $\sdj$-failure is a nontrivial interval 
$I=I[\alpha,\alpha\vee\beta]$ determined by 
congruences $\alpha, \beta$ and $\gamma$ 
satisfying $\alpha\vee\beta=\alpha\vee\gamma$
and $\beta\wedge\gamma\leq \alpha$.
The assumption that $I$ is $\XXX$--solvable
means that 
$\alpha \wrel{\solv} \alpha\vee\beta 
\;\; ( = \alpha\vee\gamma)$.
This establishes hypothesis (1) of Theorem~\ref{sdj_failure_cent}.
Hypothesis (2) holds simply because $I$ is an $\sdj$-failure.
Hence we have the conclusion of Theorem~\ref{sdj_failure_cent},
that $\C C(\alpha\vee\beta, \alpha\vee\gamma; \alpha)$ 
(equivalently $\C C(\alpha\vee\beta, \alpha\vee\beta; \alpha)$) holds.
This is precisely what it means for $I$ to be abelian.
\end{proof}


\section{An Alternative Development}\label{radical_approach}

In the preceding section we developed the theory
of solvability in analogy with the theory of solvability
for groups, using chains of congruences
with abelian factors. In this short section we show
that it would have been possible to developed the theory
of solvability in analogy with the theory of solvability
for rings. Namely, we show that the basic notions of the 
preceding section, $\rad$, $\solv$ and $\lhdlhd$, can
be defined in terms of prime congruences.

\begin{df} \label{YYYdefn}
Let $\m a$ be an algebra.
% and let 
%$\theta$ and $\theta_i$ be congruences on $\m a$.
A congruence $\theta$ on $\m a$
is {\bf semiprime} 
\index{semiprime}
if $\theta \blacktriangleleft 1$;
it is {\bf prime}\index{prime} 
if it is semiprime and meet irreducible 
in $\Conb(\m a)$.

An {\bf $m$-sequence}\index{m-sequence@$m$-sequence} 
is a sequence
$\sigma = (\theta_{\lambda})_{\lambda<\omega}$ of principal 
congruences where 
$[\theta_{\lambda},\theta_{\lambda}]\geq \theta_{\lambda+1}$
holds for all $\lambda$. 
The $m$-sequence $\sigma$
{\bf avoids} a congruence $\theta$
if $\theta_{\lambda}\not\leq\theta$
for all $\lambda$, otherwise it 
{\bf intersects} $\theta$.
\end{df}

Our definitions for ``prime'' and ``semiprime''
agree with those from ring theory.
The concept of an $m$-sequence generalizes the concept
of a multiplicatively closed subset of a commutative
ring, or an $m$-system in a noncommutative ring.
All of these notions
are definable in terms of the centralizer relation.

Observe that if $(\theta_{\lambda})_{\lambda<\omega}$
is an $m$-sequence, then 
%$\theta_0\geq \theta_1 \geq\cdots$, 
it is descending, since
$\theta_{\lambda}\geq [\theta_{\lambda},\theta_{\lambda}]
\geq \theta_{{\lambda}+1}$
for all ${\lambda}$. 
%Observe also that 
%any subsequence of an $m$-sequence
%is an $m$-sequence. 

\begin{thm} \label{ring1} 
Let $\vr V$ be a variety with a weak difference term.
Assume that $\m a\in {\vr V}$, and $\theta\in \Con(\m a)$.
The following conditions concerning $\theta$ are equivalent.
\begin{enumerate}
\item[(1)] $\theta$ is semiprime.
\item[(2)] $\theta$ is an intersection of prime congruences.
\item[(3)] For every principal congruence
$\theta_0\not\leq\theta$ there is an $m$-sequence
$\sigma = (\theta_0,\theta_1,\ldots)$ starting
at $\theta_0$ that avoids $\theta$.
\item[(4)] $\theta$ is a radical congruence.
\end{enumerate}
\end{thm}

\begin{proof}
It follows immediately from the definitions
that (1) and (4) are equivalent, so we prove only that
(1)$\Longrightarrow$(3)$\Longrightarrow$(2)$\Longrightarrow$(1).

Assume that $\theta$ is semiprime 
and that $\theta_0\not\leq\theta$.
If $[\theta_0,\theta_0]\leq\theta$, then 
since $[\theta_0,\theta_0]\lhd\theta_0$
we get from Lemma~\ref{meet+join} that
$$
\theta = \theta\vee[\theta_0,\theta_0]\lhd
\theta\vee\theta_0.
$$
But since $\theta < \theta\vee\theta_0$,
this contradicts the assumption that
$\theta\blacktriangleleft 1$ (i.e., that
$\theta$ is semiprime). Thus it must be that 
$[\theta_0,\theta_0]\not\leq \theta$, and this
means that there is a principal congruence
$\theta_1\leq [\theta_0,\theta_0]$ such that 
$\theta_1\not\leq \theta$.
The same argument applied to $\theta_1$ 
produces a principal congruence $\theta_2$
such that $\theta_2\leq [\theta_1,\theta_1]$
and $\theta_2\not\leq \theta$. Repeating
the argument indefinitely produces
an $m$-sequence $(\theta_0,\theta_1,\ldots)$
starting at $\theta_0$ that avoids $\theta$.

Now suppose that Condition (3) holds
for $\theta$. Let $\theta'$ be the intersection
of the prime congruences containing $\theta$.
If $\theta < \theta'$, then there is a principal
congruence $\theta_0\leq \theta'$ such that 
$\theta_0\not\leq \theta$. By 
Condition (3), there is an $m$-sequence
$\sigma = (\theta_0,\theta_1,\ldots)$
that avoids $\theta$. Extend $\theta$ to a congruence
$\pi$ that is maximal for the property that $\sigma$
avoids $\pi$. (The fact that such a 
congruence exists uses Zorn's Lemma, and
relies on the fact that each $\theta_i$ in $\sigma$
is compact.) We claim that $\pi$ is a prime
congruence. To see that it is meet irreducible,
assume instead 
that $\pi = \alpha\wedge\beta$ where $\pi < \alpha,\beta$.
Then by the maximality of $\pi$ it must be that 
$\sigma$ intersects both $\alpha$ and $\beta$,
so $\theta_i\leq \alpha$ and $\theta_j\leq \beta$
for some $i$ and $j$. But then for $k = \max(i,j)$
we have $\theta_k\leq \alpha\wedge\beta = \pi$,
since $\sigma$ is a descending chain of congruences,
and this contradicts the fact that $\sigma$
avoids $\pi$. To show that $\pi$ is semiprime,
assume instead that there is a congruence $\delta$
such that $\pi \lhd \delta$ and $\pi < \delta$.
Then the maximality condition on $\pi$ forces
$\sigma$ to intersect $\delta$, so for some $i$
we have $\theta_i\leq\delta$. But now the monotonicity
of the commutator operation implies that 
$$
\theta_{i+1}\leq [\theta_i,\theta_i]\leq [\delta,\delta]\leq \pi.
$$
This contradicts the fact that $\sigma$ avoids $\pi$.
This concludes the proof that $\pi$ is prime.
Since $\pi$ extends $\theta$, and $\theta'$
is the intersection of prime congruences containing $\theta$,
it follows that $\theta'\leq \pi$. But now we have
the contradiction that $\theta_0\leq\theta'\leq \pi$
while $\pi$ avoids $\sigma = (\theta_{\lambda})_{\lambda<\omega}$.
This final contradiction proves that $\theta'=\theta$,
and hence that Condition (2) holds.

Finally, assume that Condition (2) holds.
Then $\theta = \bigcap_{i\in I} \pi_i$
where each $\pi_i$ is a prime congruence.
If $\theta\lhd \delta$, then 
(since $\theta\leq\pi_i$) we get
$\pi_i\lhd \pi_i\vee\delta$ for each $i$.
Since each $\pi_i$ is semiprime
we deduce that $\pi_i = \pi_i\vee\delta$,
or that $\delta\leq\pi_i$ for all $i$.
Hence $\delta\leq \bigcap_{i\in I} \pi_i = \theta$.
This shows that $\theta\lhd \delta$
implies that $\theta = \delta$, so $\theta\blacktriangleleft 1$.
Hence Condition (1) holds.
\end{proof}

\begin{cor}
Let $\vr V$ be a variety with a weak difference term.
Assume that $\m a\in {\vr V}$ has congruences
$\alpha$ and $\beta$.
\begin{enumerate}
\item[(1)] $\rad(\alpha)$ equals the intersection
of the prime congruences containing $\alpha$.
\item[(2)] $\alpha\solv \beta$ if and only if
$\alpha$ and $\beta$ are contained in the same
prime congruences.
\item[(3)] $\alpha\lhdlhd\beta$ if and only if
$\alpha\leq\beta$ and 
every $m$-sequence that intersects $\beta$
also intersects $\alpha$.
\end{enumerate}
\end{cor}

\begin{proof}
Item (1) follows from the fact, proved
in Theorem~\ref{ring1}, that the 
radical congruences 
%are exactly the semiprime congruences
are exactly the intersections
of prime congruences. Item (2)
follows from (1) and the fact that 
$\alpha\solv \beta$ if and only if 
$\rad(\alpha) = \rad(\beta)$.

For item (3), assume that $\alpha\lhdlhd\beta$
and that the $m$-sequence $\sigma$
%$\sigma =(\theta_{\lambda})_{\lambda<\omega}$
intersects $\beta$. If $\sigma$ avoids $\alpha$, then
$\alpha$ can be extended to a prime $\pi$
that $\sigma$ avoids, as in the proof of 
Theorem~\ref{ring1}~(3)$\mimplies$(2).
The prime $\pi$ cannot contain $\beta$
since $\sigma$ avoids $\pi$ but
intersects $\beta$. Therefore
$\pi$ contains $\alpha$ and not $\beta$.
Since $\alpha\lhdlhd\beta$, this contradicts
Item (2) of this corollary.
For the other direction, assume that 
$\alpha\not\!\!\!\lhdlhd\beta$.
Then, since $\alpha\leq \beta$,
Item~(2) guarantees that there is a prime $\pi$ containing
$\alpha$ and not containing $\beta$. If $\theta_0$
is a principal congruence satisfying $\theta_0\leq \beta$
and $\theta_0\not\leq\pi$, then Theorem~\ref{ring1} guarantees
the existence of an $m$-sequence 
starting at $\theta_0$ and avoiding $\pi$
since the prime $\pi$ is semiprime.
This $m$-sequence
also avoids $\alpha$ since $\alpha\leq\pi$,
but it does not avoid $\beta$ since $\theta_0\leq\beta$.
Thus (3) holds.
\end{proof}



