\chapter{Residually Small Varieties}\label{RS_chap}

A variety is {\bf residually 
large}\index{residually large (small)} 
if it has a proper class of subdirectly
irreducible members up to isomorphism, otherwise it is {\bf
  residually small}.  Our primary goal in this chapter is to
prove that a residually small variety satisfies a congruence
identity if and only if it is congruence modular.  (David
Hobby and Ralph McKenzie proved essentially the same result
for locally finite varieties in Chapter~10
of~\cite{hobby-mckenzie}.)\index{Hobby, D.}\index{McKenzie, R.}  
Our result nearly completes the classification of
congruence varieties associated to residually small
varieties.  In Section~\ref{ACD}, we apply the result to
show that almost congruence distributive varieties cannot
have a Taylor term.

\section{Residual Smallness and Congruence Modularity}


\begin{lm}\label{RS_hypotheses_lm}
  Let\/ $\vr V$ be a variety that satisfies a nontrivial
  congruence identity. If\/ $\m a\in {\vr V}$ has
  congruences $\alpha$, $\beta$ and $\gamma$ that generate a
  sublattice isomorphic to\/ $\m n_5$ in\/~$\Con(\m a)$, as
  depicted in Figure~\ref{fig40},
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(20,30)
\put(0,15){\circle*{1.2}}
\put(10,0){\circle*{1.2}}
\put(10,30){\circle*{1.2}}
\put(20,10){\circle*{1.2}}
\put(20,20){\circle*{1.2}}

\put(10,0){\line(-2,3){10}}
\put(10,0){\line(1,1){10}}
\put(10,30){\line(-2,-3){10}}
\put(10,30){\line(1,-1){10}}
\put(20,20){\line(0,-1){10}}
%
\put(-4.5,13){$\beta$}
\put(22,9){$\delta$}
\put(22,19){$\alpha$}
\end{picture}
\end{center}
\caption{ \ }\label{fig40}
\end{figure}
then
\begin{enumerate}
\item[(1)] $\C C(\alpha,\beta;\delta)$ holds, and
\item[(2)] $\C C(\beta, \alpha;\delta)$ fails in the
  following strong way: There exists a
  $(\beta,\beta;0)$-pair $(a,b)$ (cf.\
  Definition~\ref{triples&pairs}), a polynomial\/
  $f(x,\wec{y})$ and tuples $\wec{c}\wrel{\alpha}\wec{d}$
  such that
  \[
  f(b,\wec{c}) \equiv_{\delta} f(b,\wec{d})\,,
  \]
  but
  \[
  f(a,\wec{c}) \not\equiv_{\delta} f(a,\wec{d})\,.
  \]
\end{enumerate}
\end{lm}

\begin{proof}
  Item (1) follows from Theorem~\ref{basic_centrality}~(8),
  since
  \[
  \beta\wedge
  \big(\alpha\vee(\beta\wedge\delta)\big)\leq\delta\,.
  \]

  For (2), assume instead that
  \[
  f(b,\wec{c}) \equiv_{\delta} f(b,\wec{d})
  \wrel{\mimplies}
  f(a,\wec{c}) \equiv_{\delta} f(a,\wec{d})
  \]
  holds for every $(\beta,\beta;0)$-pair $(a,b)$. Then by
  transitivity this implication holds whenever $a\ge_{\beta}
  b$.  But since $\vr V$ has no algebra with a rectangular
  tolerance, Theorem~\ref{rect_Malt_cond1}~(1)$\miff$(4)
  guarantees that the set of pairs $(a,b)$ satisfying
  $a\ge_{\beta} b$ contains $\beta$.  Thus, if this
  implication holds for every $f$ and every choice of tuples
  $\wec{c}\wrel{\alpha}\wec{d}$ whenever $(a,b)\in\beta$,
  then $\C C(\beta,\alpha;\delta)$ holds.  But $\C
  C(\beta,\alpha;\delta)$ does not hold, as we proved in
  Theorem~\ref{taylor_forbidden}~(2).  Therefore $f$,
  $\wec{c}$ and $\wec{d}$ exist.
\end{proof}

As it happens, the properties described in conditions (1)
and (2) of Lemma~\ref{RS_hypotheses_lm} are all that are
needed to establish the existence of a proper class of
subdirectly irreducibles in~$\vr V$.  (We no longer need the
assumption that $\vr V$ satisfies a nontrivial congruence
identity.)

\begin{thm}\label{kkvm-germ}
  Let\/ $\vr V$ be a variety.  Assume that\/ $\m a\in {\vr
    V}$ has congruences $\delta < \alpha$ and a tolerance
  $T$ such that
  \begin{enumerate}
  \item[(1)] $\C C(\alpha,T;\delta)$ holds, and
  \item[(2)] For some $(T,T;0)$-pair $(a,b)$ there is a
    polynomial\/ $f(x,\wec{y})$ and tuples
    $\wec{c}\wrel{\alpha}\wec{d}$ such that
    \[
    v := f(b,\wec{c}) \equiv_{\delta} f(b,\wec{d}) =: v'\,,
    \]
    but
    \[
    u := f(a,\wec{c}) \not\equiv_{\delta} f(a,\wec{d}) =: u'\,.
    \]
  \end{enumerate}
  Then $\vr V$ is residually large.
\end{thm}

\begin{proof}
  Both (1) and (2) continue to hold if we factor by
  $\delta$, so there is no loss in generality in assuming
  that $\delta = 0$.  (Under this assumption~$v=v'$.)  We
  now explain how to construct a subdirectly irreducible
  algebra in~$\vr V$ of cardinality $\geq \kappa$ for any
  infinite cardinal $\kappa$.

  Let $\m b$ be the subalgebra of $\m a^{\kappa}$ consisting
  of those $\kappa$-tuples $\wec{a}$ where $(a_i, a_j)\in T$
  for all coordinates $i < j < \kappa$.  For any $a\in A$
  let $\widehat{a}$ denote the constant $\kappa$-tuple
  satisfying $(\widehat{a})_i = a$ for all $i$.  Clearly
  $\widehat{a}\in B$ for all $a\in A$.  Let $N=\{u,v\}$ and
  let $N'=\{u',v'\}$.  Since $(a,b)\in T$ it follows that
  $(u,v) = \big(f(a,\wec{c}),f(b,\wec{c})\big)\in T$ and
  $(u',v')\in T$, so $N^{\kappa}\cup (N')^{\kappa}\subseteq
  B$.  For an element~$\wec{z}$ of~$N^\kappa$ denote
  by~$\wec{z}'$ the element of~$(N')^\kappa$ obtained
  from~$\wec{z}$ by priming the entry in each
  coordinate. Furthermore, let
  \[
  G = \big\{(\wec{z},\wec{z}')\in B^2\mid
  \wec{z}\in N^{\kappa}\setminus\{\widehat{u}\}\big\}\,,
  \]
  and let $\psi = \Cg b(G)$.

\begin{clm}
  The pair $(\hat{u},\hat{u}')$ is not contained in~$\psi$.
\end{clm}

As $u\ne u'$, it is sufficient to prove that
$\{\widehat{u}\}$ is a singleton $\psi$-class. For this we
must show that if $(\wec{z},\wec{z}')\in G$ and $p$ is a
unary polynomial of $\m B$, then
\[
p(\wec{z}) = \widehat{u} \wrel{\miff} p(\wec{z}') = \widehat{u}\,.
\]
Since $\wec{z}\in N^{\kappa}\setminus \{\widehat{u}\}$, at
least one coordinate of $\wec{z}$ is $v$.  If $z_i = v$,
then $z'_i = v' = v$, so $\wec{z}$ and $\wec{z}'$ agree in
the $i$-th coordinate. This implies that $p(\wec{z})$ and
$p(\wec{z}')$ agree in the $i$-th coordinate.  We now argue
that these tuples also agree in the $j$-th coordinate for
every other $j<\kappa$.  There is nothing to prove for those
coordinates where $z_j = v = v' = z'_j$, by the above
argument, so we consider only those $j$ where $z_j = u$ and
$z'_j = u'$.

The polynomial $p$ of~$\m B$ comes from a
term~$g(x,\wec{y})$, where appropriate parameters from $\m
b$ are substituted in place of~$\wec{y}$ in every
coordinate. Denote by~$\wec{r}$ and~$\wec{s}$ the parameters
occurring at the~$i$-th and the $j$-th coordinate,
respectively. Then $\wec{r}\wrel{T}\wec{s}$ holds by the
definition of~$\m B$. We have to prove that
$g(u,\wec{s})=g(u',\wec{s})$, and we have either
\begin{equation}\label{first}
g(u,\wec{s})=u=g(v,\wec{r})
\end{equation}
(in case $p(\wec{z})=\widehat{u}$), or 
\begin{equation}\label{second}
g(u',\wec{s})=u=g(v',\wec{r})
\end{equation}
(in case $p(\wec{z}')=\widehat{v}$). However $C(\alpha,T;0)$
holds and $\wec{c}\wrel{\alpha}\wec{d}$, so
\[
\big(g(u,\wec{s}) = \big)\quad
g\big(f(a,\underline{\wec{c}}),\wec{s}\big)=
g\big(f(b,\underline{\wec{c}}),\wec{r}\big)\quad
\big( = g(v,\wec{r})\big)
\]
is equivalent to
\[
\big(g(u',\wec{s}) = \big)\quad
g\big(f(a,\underline{\wec{d}}),\wec{s}\big) =
g\big(f(b,\underline{\wec{d}}),\wec{r}\big)\quad
\big( = g(v',\wec{r})\big)\,.
\]
Since at least one of (\ref{first}) or (\ref{second}) holds,
we get that $g(u,\wec{s})=g(v,\wec{r})$ and
$g(u',\wec{s})=g(v',\wec{r})$ both hold.  Moreover, $v=v'$
implies that $g(v,\wec{r})=g(v',\wec{r})$, so in fact
\[
g(u,\wec{s}) = g(v,\wec{r})=g(v',\wec{r})=g(u',\wec{s})\,.
\]
Thus $g(u,\wec{s})=g(u',\wec{s})$ indeed holds, completing
the argument that $p(\wec{z}) = \widehat{u}\miff p(\wec{z}')
= \widehat{u}$ holds. This proves the claim.

\medskip

Since $(\widehat{u}, \widehat{u}')\notin\psi$ we may extend
$\psi$ to a congruence $\psi_0\geq \psi$ that is maximal
with respect to $(\widehat{u}, \widehat{u}')\notin\psi_0$.
Then~$\m B/\psi_0$ is a subdirectly irreducible algebra in
$\vr V$.  We claim that $|B/\psi_0|\geq\kappa$.

In order to prove that $\m b/\psi_0$ is this large, we need
only to exhibit $\kappa$~elements of $\m b$ that are
pairwise incongruent modulo $\psi_0$.  We introduce the
following notation for certain elements of $A^{\kappa}$: for
$i<j<\kappa$, and for arbitrary elements $a,b,c\in A$ let
$[a,b,c]^{i,j}$ denote the tuple whose $k$-th coordinate is
$a$ for $0\le k<i$, is $b$ for $i\le k<j$, and is $c$ for
$j\le k<\kappa$.  (Note that $[a,b,c]^{i,j}\in B$ if $a$,
$b$ and $c$ are pairwise $T$-related.)  Similarly, if
$\wec{a}, \wec{b}, \wec{c}\in A^n$, then let $[\wec{a},
\wec{b}, \wec{c}]^{i,j}$ denote the member of $B^n$ whose
$\ell$-th coordinate is $[a_\ell,b_\ell,c_\ell]^{i,j}$ for
$1\le \ell\le n$.  We shall omit the superscripts ${i,j}$ if
we think it is clear from the context what they are.

As $(a,b)$ is a $(T,T;0)$-pair, there exists a $T,T$-matrix
\[
\left[
\begin{array}{cc}
f(\wec{p},\wec{r})&f(\wec{p},\wec{s})\\
f(\wec{q},\wec{r})&f(\wec{q},\wec{s})
\end{array}
\right] = 
\left[
\begin{array}{cc}
* & a\\
a & b
\end{array}
\right]\,.
\]

\begin{clm}\label{tuples_are_different}
  For each~$0< i < j<\kappa$ the tuples
  $[\wec{q},\wec{p},\wec{p}]$ and
  $[\wec{q},\wec{q},\wec{p}]$ are not $\psi_0$-congruent in
  every coordinate.
\end{clm}

Suppose that they are. Then in the algebra~$\m B$ we have that
\[
f([\wec{q},\wec{p},\wec{p}],[\wec{r},\wec{s},\wec{s}])
\wrel{\psi_0}
f([\wec{q},\wec{q},\wec{p}],[\wec{r},\wec{s},\wec{s}])\,,
\]
that is,
\[
[a,a,a]\wrel{\psi_0}[a,b,a]\,.
\]
Therefore
\[
[u,u,u]=f([a,a,a],[\wec{c},\wec{c},\wec{c}])
\wrel{\psi_0}
f([a,b,a],[\wec{c},\wec{c},\wec{c}])=[u,v,u]\,,
\]
and
\[
[u',u',u']=f([a,a,a],[\wec{d},\wec{d},\wec{d}])
\wrel{\psi_0}
f([a,b,a],[\wec{d},\wec{d},\wec{d}])=[u',v',u']\,.
\]
But then
\[
\widehat{u} = [u,u,u]
\wrel{\psi_0} [u,v,u]
\wrel{\psi} [u',v',u']
\wrel{\psi_0} [u',u',u'] = \widehat{u}'\,.
\]
By transitivity and the fact that $\psi_0\ge \psi$, this
yields $(\widehat{u},\widehat{u}')\in \psi_0$, contradicting
the choice of~$\psi_0$. So the claim is proved.

\medskip

To finish the proof, let~$m$ be the length of the
tuples~$\wec{p}$ and~$\wec{q}$, and define a mapping
$h\colon \kappa\to (B/\psi_0)^m$ in the following way. Let
$i<k<\kappa$, and
\[
h(i)=[\wec{q},\wec{p},\wec{p}]^{i,k}/\psi_0^m\,.
\]
The value of $h(i)$ does not depend on~$k$ because the last
two blocks of coordinates of the tuple
$[\wec{q},\wec{p},\wec{p}]$ are both $\wec{p}$.  We claim
that $h$ is injective. Indeed, suppose that
$i<j<\kappa$. Then
\[
h(i)=[\wec{q},\wec{p},\wec{p}]^{i,j}/\psi_0^m
\]
and
\[
h(j)=[\wec{q},\wec{q},\wec{p}]^{i,j}/\psi_0^m\,,
\]
and these tuples are different by
Claim~\ref{tuples_are_different}.  Therefore $h$ is indeed
injective, so $\kappa\le|B/\psi_0|^m = |B/\psi_0|$, proving
the theorem.
\end{proof}

We obtain the main theorem of this chapter as a combination
of Lemma~\ref{RS_hypotheses_lm} and Theorem~\ref{kkvm-germ}.

\begin{thm}\label{RS_congid}
  A residually small variety satisfies a congruence identity
  if and only if it is congruence modular.\qed
\end{thm}

This theorem generalizes Theorem 10.4 of
\cite{hobby-mckenzie} from locally finite varieties to
arbitrary varieties.

\begin{rem}
  We describe a three-step plan for classifying all
  congruence varieties associated to residually small
  varieties.
  \begin{enumerate}
  \item[(1)] Prove that if $\vr V$ is residually small and
    satisfies a nontrivial congruence identity, then $\vr V$
    is congruence modular.
  \item[(2)] Prove that if $\vr V$ is residually small and
    congruence modular, then $\vr V$ is congruence
    distributive or $\CON(\vr V)=\CON(\vr M)$ for some
    variety $\vr M$ of modules.
  \item[(3)] Classify the congruence varieties of varieties
    of modules.
  \end{enumerate}
  Theorem~\ref{RS_congid} accomplishes Step~(1).  Step~(3)
  was accomplished by 
  G\'{a}bor Cz\'{e}dli\index{Cz\'{e}dli, G.}  
  and George Hutchinson\index{Hutchinson, G.}  in
  \cite{hutchinson-czedli}.  Partial results on Step~(2)
  were obtained by Alan Day\index{Day, A.}  and Emil~W.~Kiss
  in \cite{day-kiss}.  Namely, Day and Kiss accomplished
  Step~(2) under the further assumption that $\vr V$~is
  either locally finite or locally solvable.  Their ideas
  extend a little further than these two cases, but fall
  short of fully accomplishing Step~(2).  Therefore, we pose
  the problem of accomplishing Step~(2) in full generality.

\begin{prb}\label{day-kiss_prb}
  Show that if $\vr V$ is residually small and congruence
  modular, then $\vr V$ is congruence distributive or
  $\CON(\vr V)=\CON(\vr M)$ for some variety $\vr M$ of
  modules.
\end{prb}

If this is proved, then we will know that the congruence
varieties $\CON(\vr V)$ of residually small varieties $\vr
V$ are precisely
\begin{itemize}
\item the variety of all lattices --- e.g. when $\vr V$ is
  the (residually small) variety of sets,
\item $\CON(\vr M)$ where $\vr M$ is a variety of modules,
\item the variety of distributive lattices --- e.g. when $\vr V$
is the (residually small) variety of distributive lattices, and
\item the variety of trivial lattices (e.g. when $\vr V$ is
  a trivial variety).
\end{itemize}
\end{rem}

\section{Almost Congruence Distributive Varieties}\label{ACD}

A variety is {\bf affine 
complete}\index{affine complete variety} 
if every congruence preserving function on every
member is a polynomial.  In \cite{kaarli-pixley}, Kalle
Kaarli\index{Kaarli, K.}  and Alden Pixley\index{Pixley, A.}
announced that any affine complete variety is residually
finite and has the property that every finitely generated
subvariety is congruence distributive.  At the time it was
not known whether the entire variety had to be congruence
distributive. Pixley then introduced the following concept.

\begin{df}\label{acd_df}
  A variety $\vr V$ is called {\bf almost congruence
    distributive},\index{almost congruence distributive (ACD)} 
  or {\bf ACD}, if it is residually finite, every
  finitely generated subvariety is congruence distributive,
  but $\vr V$ itself is not congruence distributive.
\end{df}

Pixley asked if there exists an ACD variety.  A negative
answer would, of course, show that every affine complete
variety is congruence distributive.

Kaarli and Ralph McKenzie\index{McKenzie, R.}  proved in
\cite{kaarli-mckenzie} that every affine complete variety is
congruence distributive, but it remains open whether there
is an ACD variety.  This is still an interesting question,
since we know very little about how local properties of a
variety, like the structure of the subdirectly irreducible
algebras, imply global properties, such as the satisfaction
of a Maltsev condition.

In this section we prove that an ACD variety satisfies no
nontrivial idempotent Maltsev condition.  This implies that
if an ACD variety exists, then some member has a nontrivial
strongly abelian congruence, but for some reason such
congruences do not show up in any subdirectly irreducible
algebra or even in any subvariety generated by a finite set
of subdirectly irreducible algebras.

\begin{thm}\label{acd_thm}
  Let\/ $\vr V$ be a variety satisfying a nontrivial
  idempotent Maltsev condition.
  \begin{enumerate}
  \item[(1)] If every subdirectly irreducible algebra in
    $\vr V$ generates a subvariety satisfying a nontrivial
    congruence identity, then $\vr V$ satisfies a nontrivial
    congruence identity.
  \item[(2)] If every subdirectly irreducible algebra in
    $\vr V$ generates a congruence join semidistributive
    subvariety, then $\vr V$ is congruence join
    semidistributive.
\end{enumerate}
Now assume also that\/ $\vr V$ is residually small.
\begin{enumerate}
\item[$(1)^{\prime}$] If every subdirectly irreducible
  algebra in~$\vr V$ generates a subvariety satisfying a
  nontrivial congruence identity, then $\vr V$ is congruence
  modular.
\item[$(2)^{\prime}$] If every subdirectly irreducible
  algebra in~$\vr V$ generates a congruence join
  semidistributive subvariety, then $\vr V$ is congruence
  distributive.
\end{enumerate}
\end{thm}

\begin{proof}
  If each subdirectly irreducible algebra in~$\vr V$
  generates a subvariety satisfying a nontrivial congruence
  identity, then by Theorem~\ref{main1} each subdirectly
  irreducible algebra in~$\vr V$ has a Hobby--McKenzie
  term. By Theorem~\ref{rect_Malt_cond2} the rectangular
  tolerances on subdirectly irreducible members of $\vr V$
  are trivial.  Corollary~\ref{ACD_cor} guarantees that no
  algebra in~$\vr V$ has a rectangular tolerance. Reusing
  Theorems~\ref{rect_Malt_cond2} and \ref{main1}, we get
  that $\vr V$ satisfies a nontrivial congruence
  identity. This proves (1).

  For (2), if every subdirectly irreducible in~$\vr V$
  generates a congruence join semidistributive subvariety,
  then no subdirectly irreducible algebra has a nontrivial
  rectangular or abelian tolerance. By part (1), this
  implies that $\vr V$ satisfies a nontrivial congruence
  identity, hence $\vr V$ has a weak difference term
  according to Corollary~\ref{justify_weak_cor}.  This shows
  that the hypotheses of Corollary~\ref{ACD_cor2} are met,
  so from the fact that no subdirectly irreducible algebra
  has an abelian tolerance we may derive that no algebra in
  $\vr V$ has an abelian tolerance.  By
  Theorem~\ref{sdj-thm}~(5)$\mimplies$(1) it follows that
  $\vr V$ is congruence join semidistributive.

  Items $(1)^{\prime}$ and $(2)^{\prime}$ follow from items
  (1) and (2) using Theorem~\ref{RS_congid}.
\end{proof}


\begin{cor}
  An ACD variety satisfies no nontrivial idempotent Maltsev
  condition.
\end{cor}

\begin{proof}
  This follows from Theorem~\ref{acd_thm}~$(2)^{\prime}$ and
  the definition of an ACD variety.
\end{proof}

Theorem~\ref{acd_thm} suggests the following problem.

\begin{prb}\label{SI_mod}
  Suppose that $\vr V$ satisfies a nontrivial idempotent
  Maltsev condition, and every subdirectly irreducible
  member of $\vr V$ generates a congruence modular
  subvariety.  Must $\vr V$ be congruence modular?  (We are
  interested in the case where $\vr V$ is residually large.)
\end{prb}

The answer to Problem~\ref{SI_mod} is affirmative if $\vr V$
is locally finite.  Indeed, it follows from tame congruence
theory that the following conditions are equivalent for a
locally finite variety:
\begin{enumerate}
\item[(1)] $\vr V$ is congruence modular.
\item[(2)] The minimal sets of the finite algebras are of
  type $\atyp, \btyp$ or $\ltyp$ and have empty tails.
\item[(3)] The minimal sets of the finite subdirectly
  irreducible algebras are of type $\atyp, \btyp$ or $\ltyp$
  and have empty tails.
\item[(4)] Each finite subdirectly irreducible algebra
  generates a congruence modular subvariety.
\end{enumerate}

\noindent
However, the modular law is special in this regard, because
of its link with the empty tails condition. We expect that
for identities weaker than modularity the problem analogous
to Problem~\ref{SI_mod} will have a negative answer.

Another problem is suggested by the proof of
Theorem~\ref{acd_thm}.

\begin{prb}\label{9.12}
  Suppose that $\vr V$ has a Taylor term.  If some algebra
  in~$\vr V$ has a nontrivial abelian tolerance, must some
  subdirectly irreducible algebra in~$\vr V$ have a
  nontrivial abelian tolerance?
\end{prb}

