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\begin{document}

\title{An Easy Test For Congruence Modularity}

\thanks{This material is based upon work supported by
the Hungarian National Foundation for Scientific Research (OTKA)
grant no.\ K77409.
}

\author{Topaz Dent}
\address[Topaz Dent]{Department of Mathematics\\
University of Colorado\\
Boulder, CO 80309-0395\\
USA}
\email{Topaz.Dent@Colorado.EDU}

\author{Keith A. Kearnes}
\address[Keith Kearnes]{Department of Mathematics\\
University of Colorado\\
Boulder, CO 80309-0395\\
USA}
\email{Keith.Kearnes@Colorado.EDU}

\author{\'Agnes Szendrei}
\address[\'Agnes Szendrei]{Department of Mathematics\\
University of Colorado\\
Boulder, CO 80309-0395\\
USA}
\email{Agnes.Szendrei@Colorado.EDU}
\subjclass{}
\keywords{}

\begin{abstract}
Some notes.
\end{abstract}

\maketitle

\section{Notes}
Theorems XXX (3.2) and YYY (4.2) prove that a variety satisfies
a particular congruence condition iff it realizes some idempotent $\Sigma$
satisfying some specified properties. The difficulty in applying
the result is that it is not clear how to find the appropriate $\Sigma$.
In general, these theorems remain \emph{sufficent conditions} only.
In this section, we will establish that these theorems are 
necessary and sufficient conditions
\emph{if $\Sigma$ is linear}.

A term is \emph{linear} if it is a variable or a function
symbol applied to variables. An identity is \emph{linear}
if the terms on both sides are linear. 

Let $\Sigma$ be a consistent
set of linear identities and let $Y$ be a set of variables.
Theorem~3.2.2 of \cite{kearnes-kiss-szendrei}
constructs a 
model $\m v$ of $\Sigma$, depending on $Y$. 
The universe of the model is
$V:=\{[y]\;|\;y\in Y\}\cup \{[0]\}$. Let $F$
be a function symbol in the language.
If $([a_1],\ldots,[a_m])\in V^n$, then 
let 
$a_i' = a_i$ if $a_i\in Y$ and 
$a_i'=z$ if $a_i=0$, where $z$ is a variable not in $Y$.
Define
$
F^{\m v}([a_1],\ldots,[a_m]) = [t] 
$
if there exists $t\in Y$ such that
$
\Sigma\vdash F(a_1',\ldots,a_m') \approx t,
$
and define $F^{\m v}([a_1],\ldots,[a_m]) = [0]$
if there is no such $t$.
$\m v$ is the algebra with universe $V$ 
equipped with all operations of the form
$F^{\m v}$.

\begin{thm}
Let $\Sigma$ be a set of idempotent linear identities.
Every variety realizing $\Sigma$ is congruence modular iff
$\Sigma'$ is inconsistent.
\end{thm}

\begin{proof}
If $\Sigma'$ is inconsistent, then it follows from Theorem XXX
that every variety realizing $\Sigma$ is congruence modular.
Therefore, to prove this theorem, it is enough to derive a contradiction
from the assumptions that 
$\Sigma'$ is consistent and 
every variety realizing $\Sigma$ is congruence modular.

We will make use of the model $\m V$ of $\Sigma$ 
where $Y = \{x,y\}$. Hence the universe of $\m v$ is
$V = \{[x], [y], [0]\}$.

For simplicity we will write pairs $([a],[b])\in V\times V$ as $ab$.
Consider the congruence $\theta = \cg(0x,0y)$ 
on $\m v\times \m v$. If $\Sigma$ entails congruence modularity,
then $(xx,xy)\in\theta$.
There must be a basic translation relating
$xx$ to some other element of the form $xz$ where $z\neq x$.
If we write this translation to indicate all
elements that 
may appear as parameters (whether or not they actually appear), 
and we group together equal parameters, then we derive that 
there must be a basic operation $F$ such that 
\[
F^{\m v}(00,0x,0y,x0,xx,xy,y0,yx,yy,0x)=xx 
\]
or 
\[
F^{\m v}(00,0x,0y,x0,xx,xy,y0,yx,yy,0y)=xx,
\]
while 
\[
F^{\m v}(00,0x,0y,x0,xx,xy,y0,yx,yy,0x)\neq 
F^{\m v}(00,0x,0y,x0,xx,xy,y0,yx,yy,0y).
\]
From the definition of $\m v$, we obtain that 
\[
\Sigma\vdash 
F(z,z,z,x,x,x,y,y,y,z)=x
\]
in the first coordinate, and either 
\[
\Sigma\vdash 
F(z,x,y,z,x,y,z,x,y,x)=x 
\]
or
\[
\Sigma\vdash 
F(z,x,y,z,x,y,z,x,y,y)=x 
\]
in the second coordinate.
In either case, $\Sigma'$ proves that $F$ is independent
of all variables but the fifth, hence is projection
onto its fifth variable. On the other hand,
\[
F^{\m v}(00,0x,0y,x0,xx,xy,y0,yx,yy,0x)\neq 
F^{\m v}(00,0x,0y,x0,xx,xy,y0,yx,yy,0y)
\]
prevents this. 
Moreover, it prevents this in an equational way:
if, say, 
\[F^{\m v}(00,0x,0y,x0,xx,xy,y0,yx,yy,0x)=xx\]
and
\[F^{\m v}(00,0x,0y,x0,xx,xy,y0,yx,yy,0y)=x0, \]
then 
$\Sigma\vdash F(z,x,y,z,x,y,z,x,y,x)=x$ and
$\Sigma\vdash F(z,x,y,z,x,y,z,x,y,y)=z$.
Hence $\Sigma'$ proves that $F$ is 
independent of all variables. There are a number of other
case like this, but all assert that
$F$ is indepenedent of all variables
or depends only on the last. Either contradicts the conclusion that $F$
is projection onto its fifth variable.
\end{proof}


\begin{thm}
Let $\Sigma$ be a set of idempotent linear identities.
Every variety realizing $\Sigma$ satisfies a nontrivial 
congruence identity iff
$\Sigma^{(n)}$ is inconsistent for some $n$.
\end{thm}

\begin{proof}
If instead that $\Sigma^{(n)}$ is consistent
for all $n$, then the union $\Omega = \bigcup_{n<\omega} \Sigma^{(n)}$
is also consistent. Moreover, $\Omega$ 
is a set of idempotent linear identities satisfying $\Omega'=\Omega$.
It suffices to show that if such a set $\Omega$ is consistent,
then it is realized by the 2-element semilattice.

Let $\m v$ be the model of $\Omega$ corresponding to the
choice $Y = \{y\}$, a singleton. The universe of $\m v$ is 
$V=\{y,z]$ and 
$F^{\m v}([a_1],\ldots,[a_m]) = [y]$
exactly when 
$
\Omega\vdash F(a_1',\ldots,a_m') \approx y,
$
where each $a_i'\in \{y,z\}$.
Order the universe by $[z] < [y]$.

Since $\Omega'=\Omega$, we get 
$
\Omega\vdash F(a_1',\ldots,a_m') \approx y,
$
iff $a_i'=y$ for each place $i$ on which $F$
depends. Thus $F^{\m v}(x_1,\ldots,x_n)$ agrees with 
$\bigwedge x_i$, where the meet is over the variables 
on which $F$ depends. This gives a realization 
of $\Omega$ in the 2-element semilattice.
\end{proof}

\section{The case when $\Sigma$ is linear}

George McNulty proves in \cite{mcnulty} that the problem
of determining if a finite set of identities, $\Sigma$,
axiomatizes a congruence modular variety is undecidable.
He also shows that the problem of determining if the variety
axiomatized by $\Sigma$ satisfies a nontrivial congruence
identity is undecidable. In this section we will show that
our derivative test yields a decision procedure in the case
where $\Sigma$ is a finite set of idempotent linear identities.

We call a term $t$ \emph{linear} if it has at most one function symbol
and an identity $s\approx t$ \emph{linear} if both $s$ and $t$
are linear. 
Throughout this section $\Sigma$ will be a 
set of idempotent linear identities.
Examples of linear identities are those expressing that
a function symbol is weakly independent of its $i$-th place,
or fully independent of a place, and the identity $x\approx y$.

If $\Sigma$ is a consistent set of idempotent linear identities,
then a construction in \cite{kearnes-kiss-szendrei}
shows that $\Sigma$ has models of all nonzero cardinalities.
The proofs in this section will use only the 2-element
model, but we describe the general construction.
Let $Y=\{y_i\;|\;i\in I\}$ be a set (of `variables')
and let $z$ be an element (= `variable') not in $Y$.
Let $[Y] = \{[y]\;|\;y\in Y\}$.
The universe of the model is $V:= [Y]\cup \{[z]\}$.
Thus $V$ consists of `copies' of elements from the
set $Y\cup \{z\}$ of variables.
For each $n$-ary function symbol $F$ appearing in $\Sigma$
define 
\[
F^{\m v}([x_1],\ldots,[x_n]) = 
\begin{cases}
[y] & \textrm{if $\Sigma\vdash F(x_{1},\ldots,x_{n}) \approx y$
for $x_i,y\in Y\cup \{z\}$,}\\
[z] & \textrm{else}.
\end{cases}
\]
$\m v$ is the algebra whose universe is $V$ 
and whose basic operations are all 
operations $F^{\m v}$. That $\m v$ is a model of $\Sigma$
is proved in \cite{kearnes-kiss-szendrei}.

\begin{thm}
The following are equivalent for 
a set $\Sigma$ of idempotent linear identities.
\begin{enumerate}
\item
$\Sigma'$ is inconsistent.
\item
Any variety that realizes $\Sigma$ is congruence modular.
\item
The variety axiomatized by $\Sigma$ is congruence modular.
\end{enumerate}
\end{thm}

\begin{proof}
The implication (1)$\Rightarrow$(2) is from Theorem~\ref{main1}.
The implication (2)$\Rightarrow$(3) is trivial, since
the variety axiomatized by $\Sigma$ realizes $\Sigma$.
The important part of the proof is 
(3)$\Rightarrow$(1), which we prove in contrapositive form.

Assuming that $\Sigma'$ is consistent, we must show
that the variety axiomatized by $\Sigma$ is not
congruence modular. We will do this by showing that if $\m v$
is the 2-element model of $\Sigma'$ described before
the theorem statement, then $\m v\times \m v$
has a nonmodular congruence lattice.

The universe of $\m v$ is $\{[y], [z]\}$,
and for any $n$-ary function symbol $F$ and
any $n$-tuple $([x_1],\ldots,[x_n])\in V^n$ we have
$F^{\m v}([x_1],\ldots,[x_n]) = [y]$ if and only if
$\Sigma\vdash F(x_{1},\ldots,x_{n}) \approx y$.

For simplicity we will write a pair $([a],[b])\in V\times V$ as $ab$.
Let $\eta_1$ and $\eta_2$ be the kernels of the 
first and second coordinate projections respectively.
These are the equivalence relations on $V\times V$
whose associated partitions are 
$\{\{yy,yz\},\{zy,zz\}\}$ and 
$\{\{yy,zy\},\{yz,zz\}\}$ respectively.
Let $\theta$ denote the equivalence relation on $V\times V$
whose associated partition is $\{\{yy\}, \{yz\}, \{zy,zz\}\}$.
We claim that $\theta$ is a congruence of $\m v\times \m v$.
Since $\theta < \eta_1$, and both of these equivalence
relations are complementary to $\eta_2$, establishing this
claim will show that $\m v\times \m v$
is not congruence modular, thereby completing the proof.

If $\theta$ is not a congruence, then there must be a function
symbol $F$, a pair $(ab,a'b')\in\theta$
and elements $c_id_i$ from $V\times V$ such that 
\[
F^{\m v\times \m v}(\underline{ab},c_2d_2\ldots,c_nd_n)\not\equiv
F^{\m v\times \m v}(\underline{a'b'},c_2d_2\ldots,c_nd_n)\pmod{\theta},
\]
or the same condition
with the underlined place appearing elsewhere.
Since $\theta \subseteq \eta_1$, and $\eta_1$ is a congruence,
the left and right hand sides must be related by $\eta_1$
but not by $\theta$. Moreover, since
the two sides are different, $ab$ and $a'b'$ must be
unequal $\theta$-related elements.
These observations force either 
\begin{enumerate}
\item
$F^{\m v\times \m v}(zy,c_2d_2\ldots,c_nd_n)=yy$ and 
$F^{\m v\times \m v}(zz,c_2d_2\ldots,c_nd_n)=yz$, or 
\item
$F^{\m v\times \m v}(zy,c_2d_2\ldots,c_nd_n)=yz$ and 
$F^{\m v\times \m v}(zz,c_2d_2\ldots,c_nd_n)=yy$.
\end{enumerate}
The argument is both cases is the same, so we will assume that we are
in the first case. The condition 
$F^{\m v\times \m v}(zy,c_2d_2\ldots,c_nd_n)=yy$ implies that 
\[
\Sigma'\vdash F(z,c_2,\ldots,c_n)\approx y
\quad
\textrm{and}
\quad
\Sigma'\vdash F(y,d_2,\ldots,d_n)\approx y.
\]
\end{proof}

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\end{document}
