Ralph Freese http://math.hawaii.edu/ ralph

Dedicated to the memory of George F. McNulty

# Finitely Based Congruence Varieties

Paolo Lipparini  Department of Mathematics
University of Tor Vergata
Viale della Ricerca Scientivica
I-00133 Rome
ITALY<EMAIL_ADDRESS>
(Nov 13, 2023)

###### Abstract.

We show that for a large class of varieties of algebras, the equational theory
of the congruence lattices of the members is not finitely based.

###### Key words and phrases:

congruence lattice, congruence variety, finite (equational) basis, projective
lattices, higher Arguesian identities

###### 1991 Mathematics Subject Classification:

06B15, 06C05, 08B99

The Version of Record of this article is published in Algebra Universalis, and
is available online at https://doi.org/10.1007/s00012-023-00840-6. Also see
the first author’s website given at the end of this manuscript.

##  1\. Introduction

Let $\mathscr{V}$ be a variety of algebras and let

$\mathbf{Con}(\mathscr{V})=\\{\mathbf{Con}(\mathbf{A}):\mathbf{A}\in\mathscr{V}\\}.$
(1.1)

The variety of lattices, $\textit{\emphslb{V}}\,\mathbf{Con}(\mathscr{V})$,
generated by the congruence lattices of the members of $\mathscr{V}$, is
called the _congruence variety_ associated with $\mathscr{V}$. Congruence
varieties originated with Nation in his thesis [33]; he showed, among other
things, that the lattice variety generated by $\mathbf{N}_{5}$ (the 5 element
nonmodular lattice) is not a congruence variety; see [15, Theorem 6.99].

A lattice is _meet semidistributive_ if it satisfies the (universally
quantified) implication

$x\wedge y=x\wedge z\,\rightarrow\,x\wedge y=x\wedge(y\vee z)$

It is _join semidistributive_ if it satisfies the dual condition and it is
_semidistributive_ if it satisfies both.

In [13] Freese and Jónsson proved that every modular congruence variety
actually satisfies the arguesian identity. Since the arguesian identity is
properly stronger than the modular law (as witnessed by the lattice of
subspaces of any nonarguesian projective plane), this implies, for example,
that the variety of all modular lattices is not a congruence variety. That the
variety of all arguesian lattices is not a congruence variety was shown in
[11].

In [24, Problem 9.12] B. Jónsson asked if any nontrivial congruence variety
could be finitely based other than the variety of distributive lattices and
the variety of all lattices. For congruence modular varieties this question
was completely answered by the first author with the following theorem.

######  Theorem 1.1 ([10, Theorem 4]).

There is no nontrivial finitely based modular congruence variety other than
the variety of distributive lattices.

In this paper we come close to answering Jónsson’s problem by extending
Theorem 1.1 with the following theorem:

######  Theorem 1.2.

Let $\mathscr{V}$ be a variety of algebras such that
$\mathbf{Con}\;\mathbf{(}\mathscr{V})$ satisfies a nontrivial lattice
identity. Then, if $\mathbf{Con}\;\mathbf{(}\mathscr{V})$ has a finite basis
for its equations, $\mathbf{Con}(\mathscr{V})$ is semidistributive.

We now outline how we prove Theorem 1.2. For each field $\mathbf{F}$ with at
least 3 elements Haiman [18] has constructed a sequence of modular lattices
$\mathbf{H}_{n}(\mathbf{F})$, $n\geq 3$. When $n\geq 4$ these lattices are
Aguresian but cannot be represented as lattices of permutable equivalence
relations. In proving Theorem 1.1 we showed that for every $n\geq 3$ and every
field $\mathbf{F}$ with $|\mathbf{F}|>2$,

  1. (1)

$\mathbf{H}_{n}(\mathbf{F})$ lies in no modular congruence variety.

  2. (2)

For any modular, nondistributive congruence variety $\mathscr{K}$ there is a
field $\mathbf{F}$ such that a nonprincipal ultraproduct of the
$\mathbf{H}_{n}(\mathbf{F})$’s is in $\mathscr{K}$.

To prove Theorem 1.2 we strengthen these statements as follows. By a proper
congruence variety we mean one that is not the variety of all lattices.

######  Theorem 1.3.

Let $\mathbf{H}_{n}(\mathbf{F})$ be Haiman’s lattices for $\mathbf{F}$,
$|\mathbf{F}|>2$, and $n\geq 3$. Then

  1. (1′)

$\mathbf{H}_{n}(\mathbf{F})$ lies in no congruence variety, except the variety
of all lattices.

  2. (2′)

For any congruence variety $\mathscr{K}$ that is not join semidistributive,
there is a field $\mathbf{F}$ such that every nonprincipal ultraproduct of the
$\mathbf{H}_{n}(\mathbf{F})$’s is in $\mathscr{K}$.

That these statements imply Theorem 1.2 is standard; see [15, Theorem 8.52].

The ideas proving (2′) also yield interesting results about embedding lattices
into members of $\mathbf{Con}(\mathscr{V})$ for $\mathscr{V}$ a variety with a
weak difference term which is not congruence meet semidistributive. Such
varieties admit a large class of modular lattices, as is shown in §6.

##  2\. Preliminaries

Before we begin the proof of (1′) and (2′) we prove some basic facts and
introduce some notation. First we will (usually) use $\mathscr{V}$ to denote a
variety of algebras. $\mathscr{K}$ denotes a variety of lattices, usually a
congruence variety. This convention is essentially the opposite of the one
used in [10], but is more standard now-a-days. We let H, $\SS$, P, and
$\textit{\emphslb{V}}=\textit{\emphslb{H}}\SS\textit{\emphslb{P}}$ be the
usual class operators as defined in §4.10 of [32]. As mentioned in (1.1)
above, for $\mathscr{V}$ a class of algebras, we let
$\mathbf{Con}(\mathscr{V})$ simply be the class of all congruence lattices of
members of $\mathscr{V}$ and use the class operators for the congruence
variety:
$\textit{\emphslb{V}}\,\mathbf{Con}(\mathscr{V})=\textit{\emphslb{H}}\SS\textit{\emphslb{P}}\,\mathbf{Con}(\mathscr{V})$,
and also for the congruence prevariety
$\SS\textit{\emphslb{P}}\,\mathbf{Con}(\mathscr{V})$. Note by the next lemma
that the congruence variety is
$\textit{\emphslb{H}}\SS\,\mathbf{Con}(\mathscr{V})$ and the congruence
prevariety is $\SS\,\mathbf{Con}(\mathscr{V})$.

In most of our results we assume that $\mathscr{V}$ is a variety of algebras
but in almost all cases it is enough to assume it is closed under $\SS$ and P.

######  Lemma 2.1.

For every variety $\mathscr{V}$ of algebras, the following hold.

  1. (i)

$\textit{\emphslb{P}}\,\mathbf{Con}\;\mathbf{(}\mathscr{V})\subseteq\SS\,\mathbf{Con}\;\mathbf{(}\mathscr{V})$.

  2. (ii)

$\textit{\emphslb{V}}\,\,\mathbf{Con}\;\mathbf{(}\mathscr{V})=\textit{\emphslb{H}}\,\SS\,\mathbf{Con}\;\mathbf{(}\mathscr{V})$.

  3. (iii)

If $\mathscr{U}$ is the idempotent reduct of $\mathscr{V}$, then
$\textit{\emphslb{V}}\,\,\mathbf{Con}\;\mathbf{(}\mathscr{U})=\textit{\emphslb{V}}\,\,\mathbf{Con}\;\mathbf{(}\mathscr{V})$.

###### Proof.

(i) is the first statement in Proposition 4.5 of [14] (notice that the proof
does not require congruence modularity).

(ii) is immediate from (i).

(iii) follows from the well known Pixley-Wille algorithm [34, 36], which shows
the property that some variety of algebras satisfies a given congruence
identity is equivalent to a weak Mal’cev condition witnessed by idempotent
terms: see for example [15, Theorem 6.111]. ∎

### The centrality relation and the commutator

The _centrality relation_ on an algebra $\mathbf{A}$ is denoted
$\operatorname{C}(\alpha,\beta;\delta)$ for congruences on $\mathbf{A}$; see
[16, Definition 11.3 and Lemma 11.4] for its definition and basic properties.
We say $\alpha$ _centralizes $\beta$ modulo $\delta$_ whenever the relation
$\operatorname{C}(\alpha,\beta;\delta)$ holds.

The (term condition) _commutator_ , $[\alpha,\beta]$, is defined as the least
$\delta$ such that $\operatorname{C}(\alpha,\beta;\delta)$. The centrality
relation and the commutator have become a standard tool in universal algebra.
$\mathbf{M}_{3}$ denotes the five element modular, nondistributve lattice.

######  Lemma 2.2.

Suppose $\delta\leq\alpha,\beta,\gamma\leq\mu$ are congruences on an algebra
$\mathbf{A}$ that form a copy of $\mathbf{M}_{3}$. Then
$[\alpha,\alpha]\leq\delta$, $[\beta,\beta]\leq\delta$, and
$[\gamma,\gamma]\leq\delta$.

###### Proof.

By [16, Lemma 11.4(vi)]), $C(\alpha,\gamma;\alpha\wedge\gamma)$ holds. Since
$\delta=\alpha\wedge\gamma$, we have $C(\alpha,\gamma;\delta)$. Similarly
$C(\beta,\gamma;\delta)$. By [16, Lemma 11.4(iv) and (i)])
$\operatorname{C}(\alpha\vee\beta,\gamma;\delta)=\operatorname{C}(\mu,\gamma;\delta)$,
and so $\operatorname{C}(\gamma,\gamma;\delta)$. Using the definition of the
commutator, this yields $[\gamma,\gamma]\leq\delta$. The other two
inequalities follow from symmetry. ∎

### Weak difference terms

Let $\mathbf{A}$ be an algebra. A term $d(x,y,z)$ in the signature of
$\mathbf{A}$ is a _weak difference term_ for $\mathbf{A}$ if, for all $a$,
$b\in A$ and all $\theta\in\mathbf{Con}\;\mathbf{(}\mathbf{A})$ with $\langle
a,b\rangle\in\theta$,

$a\mathrel{[\theta,\theta]}d(a,b,b)\quad\text{ and }\quad
d(a,a,b)\mathrel{[\theta,\theta]}b.$ (2.1)

Most of the properties of the modular commutator hold in varieties with a weak
difference term. The next lemma is an illustration. It is special case of more
general permutability results given in [30]; see for example Theorem 3.5(i) of
that paper.

######  Lemma 2.3.

Let $\mathbf{A}$ be an algebra with a weak difference term $d$, and let
$\alpha$ and $\beta\in\mathbf{Con}\;\mathbf{(}\mathbf{A})$. Then.

  1. (1)

if $[\alpha,\alpha]\leq\beta$ and $[\beta,\beta]\leq\alpha$, then $\alpha$ and
$\beta$ permute;

  2. (2)

if $\alpha$ and $\beta$ are atoms of a sublattice of
$\mathbf{Con}\;\mathbf{(}\mathbf{A})$ isomorphic to $\mathbf{M}_{3}$, then
$\alpha$ and $\beta$ permute.

###### Proof.

For (1) suppose $a\mathrel{\alpha}b\mathrel{\beta}c$. Then

$a\mathrel{[\alpha,\alpha]}d(a,b,b)\mathrel{\beta}d(a,b,c)$

and since $[\alpha,\alpha]\leq\beta$, we have $a\mathrel{\beta}d(a,b,c)$. A
symmetric argument shows $c\mathrel{\alpha}d(a,b,c)$, and so
$a\mathrel{\beta}d(a,b,c)\mathrel{\alpha}c$. From this it follows that
$\alpha$ and $\beta$ permute. (2) follows from (1) and Lemma 2.2 ∎

### Mal’tsev conditions and Mal’tsev classes

A collection of varieties defined by a Mal’tsev condition is called a
_Mal’tsev class_. Examples include the classes of congruence modular,
congruence distributive, and congruence semidistributive varieties. Other
examples include varieties with a Taylor term and varieties with a Hobby-
McKenzie term. We are particularly concerned with varieties having a weak
difference term. By Kearnes and Szendrei [29] this is a Mal’tsev class; see
[16, Theorem 11.59]. Having a weak difference term is a relatively weak
property: all of the other conditions mentioned, except having a Taylor term,
imply the existence of a weak difference term. We are interested in varieties
whose congruence variety is not the variety of all lattices; that is,
varieties that satisfy a nontrivial congruence identity. These varieties form
a Mal’tsev class; in fact it is the class of varieties with a Hobby-McKenzie
term [27, Theorem A.2(11)].

An interval $I[\beta,\alpha]$ is _abelian_ if
$\operatorname{C}(\alpha,\alpha;\beta)$. This implies
$[\alpha,\alpha]\leq\beta$ but the converse is false. However, if
$\mathscr{V}$ has a weak difference term, $I[\beta,\alpha]$ is abelian if and
only if $[\alpha,\alpha]\leq\beta$. Consequently, subintervals of abelian
intervals are abelian in such varieties, [30, Proposition 4.2]; see also [16,
Theorem 11.29].

The _solvable series for $\alpha$_ is defined by $[\alpha]^{0}=\alpha$,
$[\alpha]^{n+1}=[[\alpha]^{n},[\alpha]^{n}]$. We say $\alpha$ is _solvable_ if
$[\alpha]^{n}=0$ for some $n$. For $\beta\leq\alpha$ we say the interval
between $\beta$ and $\alpha$ is _solvable_ in case there exists some $n$ such
that $[\alpha]^{n}\leq\beta$.

The next theorem records some facts we need from [30] on the behavior of the
commutator in varieties with a weak difference term. The first is from [30,
Theorem 5.1] and the second is from [30, p. 197]; see also [16, Theorem
11.87].

######  Theorem 2.4.

Suppose that the algebra $\mathbf{A}$ has a weak difference term. Then the
following hold in $\mathbf{Con}\;\mathbf{(}\mathbf{A})$.

  1. (i)

(Abelian and solvable intervals are preserved under transpositions) If the
interval $I[\beta,\alpha]$ is abelian (solvable), then the intervals
$I[\beta\wedge\gamma,\alpha\wedge\gamma]$ and
$I[\beta\vee\delta,\alpha\vee\delta]$ are abelian (solvable).

  2. (ii)

(Solvable intervals are intervals of permuting equivalence relations) If the
interval $I[\beta,\alpha]$ is solvable and $\gamma,\delta\in I[\beta,\alpha]$,
then $\gamma\circ\delta=\delta\circ\gamma$. Hence $I[\beta,\alpha]$ is
modular.

We list in the next theorem the results we need from [29, 27]. For $\alpha$,
$\beta$ and $\gamma$ elements of a lattice, define $\beta^{0}=\beta$,
$\gamma^{0}=\gamma$,

$\beta^{m+1}=\beta\wedge(\alpha\vee\gamma^{m})\qquad\gamma^{m+1}=\gamma\wedge(\alpha\vee\beta^{m})$
(2.2)

######  Theorem 2.5.

Suppose that $\mathscr{V}$ is a variety and that the congruence variety
$\mathscr{K}$ associated with $\mathscr{V}$ is not the variety of all
lattices. Then the following hold.

  1. (i)

(Kearnes and Szendrei [29, Corollary 4.12]) $\mathscr{V}$ has a weak
difference term.

  2. (ii)

(Kearnes and Kiss [27, Theorem 8.3]) There exists a positive integer $m$ such
that the congruence identity $\beta^{m}=\beta^{m+1}$ holds in $\mathscr{K}$.

  3. (iii)

(Kearnes and Kiss [27, Theorem 8.5]) Whenever $\mathbf{A}\in\mathscr{V}$ and
$\alpha$, $\beta$, $\gamma$ are congruences of $\mathbf{A}$ such that
$\alpha\vee\beta=\alpha\vee\gamma$ then the interval between
$\alpha\vee(\beta\wedge\gamma)$ and $\alpha\vee\beta$ is abelian.

##  3\. Haiman’s lattices and higher order Arguesian identities

In [18] M. Haiman studied a chain of stronger and stronger higher order
Arguesian identities ($\text{D}_{n}$), given below. He showed that
$\mathbf{H}_{n}(\mathbf{F})$ witnesses that the congruence identities
($\text{D}_{n}$) are properly increasing in strength. In Theorem 3.2 we
present an identity, ($\text{D}_{n}^{*}$), equivalent to ($\text{D}_{n}$) but
which more closely resembles Jónsson’s Arguesian identity and is easier to
work with. The equivalence of these identities is of independent interest.

######  Lemma 3.1.

Let $x_{0}$, $x_{0}^{\prime}$, $A$, $B$ be elements of a modular lattice such
that $x_{0}\wedge x_{0}^{\prime}\leq B$ and $x_{0}\vee x_{0}^{\prime}\geq A$.
Then $A\leq x_{0}^{\prime}\vee(x_{0}\wedge B)$ if and only if
$x_{0}\wedge(x_{0}^{\prime}\vee A)\leq B$.

###### Proof.

$A\leq x_{0}^{\prime}\vee(x_{0}\wedge B)$ is equivalent to $x_{0}^{\prime}\vee
A\leq x_{0}^{\prime}\vee(x_{0}\wedge B)$, which implies
$x_{0}\wedge(x_{0}^{\prime}\vee A)\leq
x_{0}\wedge(x_{0}^{\prime}\vee(x_{0}\wedge B))=(x_{0}\wedge
x_{0}^{\prime})\vee(x_{0}\wedge B)\leq B$ by modularity. The converse is the
dual argument (with $B$ in place of $A$ and $x_{0}$ in place of
$x_{0}^{\prime}$). ∎

The next theorem generalizes a result first obtained by A. Day and D.
Pickering [7] in the particular case $n=3$. By Corollary 1 on page 104 of
[15], ($\text{D}_{3}^{*}$) is equivalent to the Agruesian identity.

######  Theorem 3.2.

Let $y_{i}=(x_{i}\vee x_{i+1})\wedge(x_{i}^{\prime}\vee x_{i+1}^{\prime})$
where the indices are computed modulo $n$ so $y_{n-1}=(x_{n-1}\vee
x_{0})\wedge(x_{n-1}^{\prime}\vee x_{0}^{\prime})$. Then the following two
equations are equivalent.

$x_{0}\wedge(x^{\prime}_{0}\vee\bigwedge_{i=1}^{n-1}(x_{i}\vee
x_{i}^{\prime}))\leq x_{1}\vee[(x_{0}^{\prime}\vee
x_{1}^{\prime})\wedge\bigvee_{i=1}^{n-1}y_{i}]$ ($\text{D}_{n}$)

and

$\bigwedge_{i=0}^{n-1}(x_{i}\vee x_{i}^{\prime})\leq
x_{0}^{\prime}\vee(x_{0}\wedge(x_{1}\vee[(x_{0}^{\prime}\vee
x_{1}^{\prime})\wedge\bigvee_{i=1}^{n-1}y_{i}]))$ ($\text{D}^{*}_{n}$)

###### Proof.

Both of these equations imply modularity: for ($\text{D}_{n}^{*}$) we make the
substitutions $x_{0}\mapsto x$, $x_{i}\mapsto y\vee z$ for all $i>0$, and
$x^{\prime}_{i}\mapsto y\wedge z$ for all $i$. A similar substitution works
for ($\text{D}_{n}$).

We claim $x_{0}\wedge(x^{\prime}_{0}\vee\bigwedge_{i=1}^{n-1}(x_{i}\vee
x_{i}^{\prime}))=x_{0}\wedge(x^{\prime}_{0}\vee\bigwedge_{i=0}^{n-1}(x_{i}\vee
x_{i}^{\prime}))$. Indeed, modularity gives

$\displaystyle
x_{0}\wedge\big{(}x^{\prime}_{0}\vee\bigwedge_{i=0}^{n-1}(x_{i}\vee
x_{i}^{\prime})\big{)}$
$\displaystyle=x_{0}\wedge\big{(}x^{\prime}_{0}\vee[(x_{0}\vee
x_{0}^{\prime})\wedge\bigwedge_{i=1}^{n-1}(x_{i}\vee x_{i}^{\prime})]\big{)}$
$\displaystyle=x_{0}\wedge(x_{0}\vee
x^{\prime}_{0})\wedge[x_{0}^{\prime}\vee\bigwedge_{i=1}^{n-1}(x_{i}\vee
x_{i}^{\prime})]$
$\displaystyle=x_{0}\wedge\big{(}x^{\prime}_{0}\vee\bigwedge_{i=1}^{n-1}(x_{i}\vee
x_{i}^{\prime})\big{)}$

Now apply Lemma 3.1 with

$A=\bigwedge_{i=0}^{n-1}(x_{i}\vee x_{i}^{\prime})\quad\text{and}\quad
B=x_{1}\vee[(x_{0}^{\prime}\vee
x_{1}^{\prime})\wedge\bigvee_{i=1}^{n-1}y_{i}].$

Notice that $x_{0}\wedge x_{0}^{\prime}\leq B$, since $x_{0}\wedge
x_{0}^{\prime}\leq y_{n-1}=(x_{n-1}\vee x_{0})\wedge(x^{\prime}_{n-1}\vee
x^{\prime}_{0})$, and clearly $x_{0}\vee x_{0}^{\prime}\geq A$. Thus the
hypotheses of Lemma 3.1 hold. Using the claim we see that, for given $x_{i}$,
$x_{i}^{\prime}$, the conclusion of that lemma is exactly the statement that
($\text{D}_{n}$) holds if and only if ($\text{D}_{n}^{*}$) holds, as desired.
∎

Actually, the proof of Theorem 3.2 gives something more:

######  Corollary 3.3.

If $\mathbf{L}$ is a modular lattice and $x_{i}$,
$x_{i}^{\prime}\in\mathbf{L}$, then $x_{i}$, $x_{i}^{\prime}$ satisfy
($\text{D}_{n}$) if and only if $x_{i}$, $x_{i}^{\prime}$ satisfy
($\text{D}_{n}^{*}$).

Haiman showed that ($\text{D}_{n}$) holds in any lattice of permuting
equivalence relations, but we require the following strengthening of that
fact.

######  Lemma 3.4.

If $\mathbf{L}$ is a sublattice of the lattice of equivalence relations on a
set $A$, and if $\alpha_{i}$, $\alpha_{i}^{\prime}$ are elements of
$\mathbf{L}$, and for each $i$, $\alpha_{i}$ and $\alpha_{i}^{\prime}$
permute, then the instance of ($\text{D}_{n}^{*}$) in which $\alpha_{i}$ and
$\alpha_{i}^{\prime}$ are substituted for $x_{i}$, $x_{i}^{\prime}$,
$i=0,\ldots,n-1$, holds in $\mathbf{L}$.

###### Proof.

$a$$b$$c_{0}$$c_{n-1}$$c_{i}$$c_{1}$$c_{i+1}$$\alpha_{i}$$\alpha_{n-1}$$\alpha_{0}$$\alpha_{1}$$\alpha_{i+1}$$\gamma_{i}$$\alpha^{\prime}_{0}$$\alpha^{\prime}_{i}$$\alpha^{\prime}_{n-1}$$\alpha^{\prime}_{1}$$\alpha^{\prime}_{i+1}$
Figure 1.

Suppose that $\langle
a,b\rangle\in\bigwedge_{i=0}^{n-1}(\alpha_{i}\vee\alpha_{i}^{\prime})$, that
is, $\langle a,b\rangle\in\alpha_{i}\vee\alpha_{i}^{\prime}$, for every $i$.
Since $\alpha_{i}$ and $\alpha_{i}^{\prime}$ permute for every $i$, this
implies there exists $c_{i}$ such that $\langle a,c_{i}\rangle\in\alpha_{i}$
and $\langle c_{i},b\rangle\in\alpha_{i}^{\prime}$. Now let
$\gamma_{i}=(\alpha_{i}\vee\alpha_{i+1})\wedge(\alpha_{i}^{\prime}\vee\alpha_{i+1}^{\prime})$,
the element of $L$ corresponding to $y_{i}$. It follows that $\langle
c_{i},c_{i+1}\rangle\in\gamma_{i}$. The indices are computed modulo $n$ so
when $i=n-1$ we get $\langle
c_{n-1},c_{0}\rangle\in\gamma_{n-1}=(\alpha_{n-1}\vee\alpha_{0})\wedge(\alpha_{n-1}^{\prime}\vee\alpha_{0}^{\prime})$.
These relations are indicated in Figure 1. Hence
$c_{1}\mathrel{\gamma_{1}}c_{2}\mathrel{\gamma_{2}}c_{3}\cdots
c_{n-1}\mathrel{\gamma_{n-1}}c_{0}$, and so $\langle
c_{0},c_{1}\rangle\in\gamma_{1}\vee\cdots\vee\gamma_{n-1}$. Thus $\langle
c_{0},c_{1}\rangle\in(\alpha_{0}^{\prime}\vee\alpha_{1}^{\prime})\wedge\bigvee_{i=1}^{n-1}\gamma_{i}$
and so $\langle
a,c_{0}\rangle\in\alpha_{0}\wedge\big{(}\alpha_{1}\vee[(\alpha_{0}^{\prime}\vee\alpha_{1}^{\prime})\wedge\bigvee_{i=1}^{n-1}\gamma_{i}]\big{)}$.
Since $\langle b,c_{0}\rangle\in\alpha_{0}^{\prime}$, we get $\langle
b,a\rangle\in\alpha_{0}^{\prime}\vee\big{(}\alpha_{0}\wedge\big{(}\alpha_{1}\vee[(\alpha_{0}^{\prime}\vee\alpha_{1}^{\prime})\wedge\bigvee_{i=1}^{n-1}\gamma_{i}]\big{)}\big{)}$,
proving the lemma. ∎

From Lemmas 2.3 and 3.4 we immediately get the following corollary.

######  Corollary 3.5.

Let $\mathbf{A}$ be an algebra with a weak difference term. Suppose
$\alpha_{i}$ and $\alpha^{\prime}_{i}\in\mathbf{Con}\;\mathbf{(}\mathbf{A})$,
$i<n$, and also assume there is a congruence $\theta_{i}$ such that
$\alpha_{i}$, $\alpha^{\prime}_{i}$ and $\theta_{i}$ are the atoms of a
sublattice of $\mathbf{Con}\;\mathbf{(}\mathbf{A})$ isomorphic to
$\mathbf{M}_{3}$. Then ($\text{D}_{n}^{*}$) holds when $\alpha_{i}$ is
substituted for $x_{i}$ and $\alpha^{\prime}_{i}$ for $x^{\prime}_{i}$.

If $\mathbf{F}$ is a skew field we let $\mathscr{M}_{\mathbf{F}}$ be the
variety of left vector spaces over $\mathbf{F}$ and
$\mathscr{M}^{\text{fd}}_{\mathbf{F}}$ be the class of all finite dimensional
left vector spaces over $\mathbf{F}$.

######  Lemma 3.6.

Let $\mathbf{F}$ be a field with $|F|>2$ and let $\mathbf{H}_{n}(\mathbf{F})$
be Haiman’s lattice, $n\geq 3$.

  1. (1)

Every proper sublattice of $\mathbf{H}_{n}(\mathbf{F})$ can be embedded into
the lattice of subspaces of a $2n$-dimensional vector space over $\mathbf{F}$.

  2. (2)

Every sublattice of $\mathbf{H}_{n}(\mathbf{F})$ generated by less than $n$
elements is proper.

  3. (3)

There exist elements $x_{i}$ and
$x_{i}^{\prime}\in\mathbf{H}_{n}(\mathbf{F})$, $i<n$, that generate
$\mathbf{H}_{n}(\mathbf{F})$ and witness the failure of ($\text{D}_{n}^{*}$)
in $\mathbf{H}_{n}(\mathbf{F})$. Moreover, there are elements
$x_{i}^{\prime\prime}\in\mathbf{H}_{n}(\mathbf{F})$ such that $x_{i}$,
$x_{i}^{\prime}$ and $x_{i}^{\prime\prime}$ are the atoms of a sublattice
isomorphic to $\mathbf{M}_{3}$.

  4. (4)

Let $\mathbf{P}$ be the prime subfield of $\mathbf{F}$. Then any nonprincipal
ultraproduct of $\\{\mathbf{H}_{n}(\mathbf{F}):n=3,4,\ldots\\}$ lies in
$\SS\,\mathbf{Con}\;\mathbf{(}\mathscr{M}_{\mathbf{P}})$.

###### Proof.

(1) and (2) are Theorems 2 and 3 of [18], respectively. Haiman’s paper defines
$x_{i}$ and $x_{i}^{\prime}$ in $\mathbf{H}_{n}(\mathbf{F})$ to be atoms of an
interval, $I[p_{i},r_{i}]$, isomorphic to the lattice of subspaces of a three
dimensional vector space over $\mathbf{F}$. In such an interval the join of
any two distinct atoms contains a third atom and (3) follows. Haiman deals
with the identity ($\text{D}_{n}$), instead, but we can equivalently use
($\text{D}_{n}^{*}$) in view of Corollary 3.3 and since Haiman’s lattices
$\mathbf{H}_{n}(\mathbf{F})$ are modular.

$\SS\,\mathbf{Con}\;\mathbf{(}\mathscr{M}_{\mathbf{P}})$ is a quasivariety,
that is, it is defined by quasi-equations: by Lemma 2.1 it is closed under
$\SS$ and P. In addition this class is closed under ultraproducts; see [21]
and [31]. See the proof of Theorem 2.1(2) of [28] for a very approachable
proof. Consequently by Theorem 8.105 of [15],
$\SS\,\mathbf{Con}\;\mathbf{(}\mathscr{M}_{\mathbf{P}})$ is a quasivariety and
so is defined by a set $\Phi$ of quasi-equations. (We point out that it is not
always the case that $\SS\,\mathbf{Con}\;\mathbf{(}\mathscr{V})$ is a
quasivariety for a variety $\mathscr{V}$: Kearnes and Nation [28] show, for
example, that if $\mathscr{V}$ has a Taylor term but does not have a Hobby-
McKenzie term, then $\SS\,\mathbf{Con}\;\mathbf{(}\mathscr{V})$ is not a
quasivariety. So for example the class of lattices embeddable into the
congruence lattice of a semilattice is not elementary.) Let $\phi\in\Phi$. By
(1) and (2) of this lemma and the fact that the lattice of subspaces of a
vector space over $\mathbf{F}$ is embedded into the lattice of subspaces of a
vector space over its prime subfield, we see that if $\phi$ has at most $n$
variables, then it holds in $\mathbf{H}_{m}(\mathbf{F})$ for all $m>n$. So
$\phi$ holds in all but finitely many $\mathbf{H}_{m}(\mathbf{F})$’s. Thus
$\phi$ holds in any nonprincipal ultraproduct of the
$\mathbf{H}_{m}(\mathbf{F})$’s, proving (4). ∎

##  4\. Projectivity of $\mathbf{M}_{3}$ in congruence varieties

In order to use the results from the previous sections to prove that no
congruence variety, other than the variety of all lattices, contains any
$\mathbf{H}_{n}(\mathbf{F})$ (that is to prove Theorem 1.3($1^{\prime}$)), we
will show that $\mathbf{M}_{3}$ is projective for every proper congruence
variety. To make this precise we need a definition. If $\mathbf{L}$ is any
lattice, we say that _$\mathbf{M}_{3}$ is projective for $\mathbf{L}$_ if and
only if whenever $\varphi:\mathbf{L}\twoheadrightarrow\mathbf{M}_{3}$ is an
epimorphism, then $\mathbf{L}$ contains a sublattice isomorphic to
$\mathbf{M}_{3}$ which maps to $\mathbf{M}_{3}$ under $\varphi$. We say that
$\mathbf{M}_{3}$ is projective for a class $\mathscr{X}$ of lattices if
$\mathbf{M}_{3}$ is projective for every member of $\mathscr{X}$. If
$\mathbf{L}$ does not have $\mathbf{M}_{3}$ as homomorphic image, then
$\mathbf{M}_{3}$ is projective for $\mathbf{L}$ vacuously. $\mathbf{M}_{3}$ is
projective for the class of modular lattices as follows from Dedekind’s
description of the free modular lattice on three generators, [8]. On the other
hand, $\mathbf{M}_{3}$ is not projective for the class of all lattices. This
section will show that $\mathbf{M}_{3}$ is projective for every congruence
variety except for the variety of all lattices.

######  Theorem 4.1.

Let $\mathscr{K}$ be the congruence variety of a variety $\mathscr{V}$. Then
$\mathbf{M}_{3}$ is projective for $\mathscr{K}$ if and only if $\mathscr{K}$
is not the variety of all lattices.

###### Proof.

First assume $\mathscr{K}$ is the variety of all lattices. Then it contains
the free lattice on three generations which has a homomorphism onto
$\mathbf{M}_{3}$. But $\mathbf{M}_{3}$ is not a sublattice of the the free
lattice, see [12], so it is not projective for $\mathscr{K}$.

Now assume that $\mathscr{K}$ is not the variety of all lattices. Let
$\mathbf{L}\in\mathscr{K}$ and suppose
$\varphi:\mathbf{L}\twoheadrightarrow\mathbf{M}_{3}$ is an epimorphism. By
Lemma 2.1(ii) there is a lattice
$\mathbf{L}^{\prime}\in\SS\,\mathbf{Con}(\mathbf{A})$ and an epimorphism
$\psi:\mathbf{L}^{\prime}\twoheadrightarrow\mathbf{L}$, for some
$\mathbf{A}\in\mathscr{V}$. Summarizing:

$\mathbf{L}^{\prime}\stackrel{{\scriptstyle\psi}}{{\twoheadrightarrow}}\mathbf{L}\stackrel{{\scriptstyle\varphi}}{{\twoheadrightarrow}}\mathbf{M}_{3}\quad\text{with}\quad\mathbf{L}^{\prime}\leq\mathbf{Con}(\mathbf{A}).$

Let $a$, $b$ and $c$ be the atoms of $\mathbf{M}_{3}$. Choose $\alpha$,
$\beta$ and $\gamma\in\mathbf{L}^{\prime}$ to be pre-images of $a$, $b$ and
$c$ under $\psi\mathbin{\circ}\varphi$.

By Theorem 2.5(ii) (and the symmetry between $\beta$ and $\gamma$) there is an
$m$ such that $\beta^{m+1}=\beta^{m}$ and $\gamma^{m+1}=\gamma^{m}$, where
$\beta^{n}$ and $\gamma^{n}$ are defined by (2.2). Then

$\beta^{m}=\beta^{m+1}=\beta\wedge(\alpha\vee\gamma^{m})\leq\alpha\vee\gamma^{m},$

and so $\alpha\vee\beta^{m}\leq\alpha\vee\gamma^{m}$, and by symmetry,
$\alpha\vee\beta^{m}=\alpha\vee\gamma^{m}$. Now an easy induction shows
$(\phi\mathbin{\circ}\varphi)(\beta^{n})=b$ and
$(\phi\mathbin{\circ}\varphi)(\gamma^{n})=c$ for all $n$. So, changing
notation, we may assume $\alpha$, $\beta$ and $\gamma$ are pre-images of $a$,
$b$ and $c$ and that $\alpha\vee\beta=\alpha\vee\gamma$. Now let
$\alpha^{\prime}=\alpha\vee(\beta\wedge\gamma)$ and note that
$\alpha^{\prime}$ is a pre-image of $a$ and that
$\alpha^{\prime}\vee\beta=\alpha^{\prime}\vee\gamma$ so, with another change
in notation, we may assume our pre-images satisfy

$\alpha\vee\beta=\alpha\vee\gamma\text{ \ and \ }\beta\wedge\gamma\leq\alpha.$

By Theorem 2.5(iii) the interval $I[\alpha,\alpha\vee\beta\vee\gamma]$ is
abelian.

By Theorem 2.5(i) $\mathscr{V}$ has a weak difference term so by Theorem
2.4(i) meeting with $\beta$ we have that
$I[\alpha\wedge\beta,(\alpha\vee\beta\vee\gamma)\wedge\beta]=I[\alpha\wedge\beta,\beta]$
is abelian. Similarly, $I[\alpha\wedge\gamma,\gamma]$ is abelian. Joining the
first of these with $\alpha\wedge\gamma$ yields that
$I[(\alpha\wedge\beta)\vee(\alpha\wedge\gamma),\beta\vee(\alpha\wedge\gamma)]$
is abelian. Joining the second with $\beta$ gives
$I[\beta\vee(\alpha\wedge\gamma),\beta\vee\gamma]$ is abelian. So we have the
chain

$(\alpha\wedge\beta)\vee(\alpha\wedge\gamma)\leq\beta\vee(\alpha\wedge\gamma)\leq\beta\vee\gamma,$

and it follows that the interval
$I[(\alpha\wedge\beta)\vee(\alpha\wedge\gamma),\beta\vee\gamma]$ is solvable
and so by Theorem 2.4(ii) it is modular. Let

$\displaystyle\alpha^{\prime}$ $\displaystyle=\alpha\wedge(\beta\vee\gamma),$
$\displaystyle\beta^{\prime}$ $\displaystyle=\beta\vee(\alpha\wedge\gamma),$
$\displaystyle\gamma^{\prime}$ $\displaystyle=\gamma\vee(\alpha\wedge\beta),$

and note these lie in the interval and are pre-images of $a$, $b$ and $c$,
respectively. Now the desired result follows from the projectivity of
$\mathbf{M}_{3}$ in the variety of modular lattices. Explicitly, defining

$\displaystyle\alpha^{\prime\prime}$
$\displaystyle=(\alpha^{\prime}\wedge(\beta^{\prime}\vee\gamma^{\prime}))\vee(\beta^{\prime}\wedge\gamma^{\prime})=\alpha^{\prime}\vee(\beta^{\prime}\wedge\gamma^{\prime})$
$\displaystyle\beta^{\prime\prime}$
$\displaystyle=(\beta^{\prime}\wedge(\alpha^{\prime}\vee\gamma^{\prime}))\vee(\alpha^{\prime}\wedge\gamma^{\prime})=\beta^{\prime}\wedge(\alpha^{\prime}\vee\gamma^{\prime})$
$\displaystyle\gamma^{\prime\prime}$
$\displaystyle=(\gamma^{\prime}\wedge(\alpha^{\prime}\vee\beta^{\prime}))\vee(\alpha^{\prime}\wedge\beta^{\prime})=\gamma^{\prime}\wedge(\alpha^{\prime}\vee\beta^{\prime})$

we have that $\alpha^{\prime\prime}$, $\beta^{\prime\prime}$ and
$\gamma^{\prime\prime}$ generate a sublattice which is isomorphic to
$\mathbf{M}_{3}$ and maps onto $\mathbf{M}_{3}$ under
$\psi\mathbin{\circ}\varphi$. ∎

##  5\. Proof of Theorem 1.3($\operatorname{1}^{\prime}$)

Let $\mathbf{L}=\mathbf{H}_{n}(\mathbf{F})$ be one of Haiman’s lattices and
suppose $\mathbf{L}$ lies in the congruence variety $\mathscr{K}$ associated
with a variety $\mathscr{V}$ and that $\mathscr{K}$ is not the variety of all
lattices. By Lemma 2.1(ii) $\mathbf{L}$ is a homomorphic image of a lattice
$\mathbf{L}^{\prime}$ which is a sublattice of $\mathbf{Con}(\mathbf{A})$ for
some $\mathbf{A}\in\mathscr{V}$. Let
$\psi:\mathbf{L}^{\prime}\twoheadrightarrow\mathbf{L}$ be this epimorphism. By
Lemma 3.6(3) there are elements $x_{i}$, $x_{i}^{\prime}$ and
$x_{i}^{\prime\prime}$, $i<n$, of $\mathbf{L}$ which are atoms of a sublattice
isomorphic to $\mathbf{M}_{3}$ such that $x_{i}$ and $x_{i}^{\prime}$ generate
$\mathbf{L}$ and witness a failure of ($\text{D}^{*}_{n}$) in $\mathbf{L}$.
Since $\mathbf{M}_{3}$ is projective for $\mathcal{K}$ by Theorem 4.1 there
are elements $\overline{x_{i}}$, $\overline{x_{i}}^{\prime}$ and
$\overline{x_{i}}^{\prime\prime}$ of $\mathbf{L}^{\prime}$ which map under
$\psi$ to $x_{i}$, $x_{i}^{\prime}$ and $x_{i}^{\prime\prime}$ and are the
atoms of a sublattice isomorphic to $\mathbf{M}_{3}$. Corollary 3.5 implies
that ($\text{D}^{*}_{n}$) holds for these $\overline{x_{i}}$ and
$\overline{x_{i}}^{\prime}$, $i<n$. Applying $\psi$ we get that
($\text{D}^{*}_{n}$) holds in $\mathbf{L}$ for $x_{i}$ and $x_{i}^{\prime}$.
This contraction completes the proof. ∎

As mentioned just before Theorem 3.2, ($\text{D}^{*}_{3})$ is equivalent to
the arguesian law so, in a nondesaguesian projective plane, there are points
$x_{i}$ and $x_{i}^{\prime}$, $i=0,1,2$, which witness the failure of
($\text{D}^{*}_{3})$. Since every line in a projective plane has at least 3
points, there are points $x_{i}^{\prime\prime}$, $i=0,1,2$, such that $x_{i}$,
$x_{i}^{\prime}$ and $x_{i}^{\prime\prime}$ are the atoms of sublattice
isomorphic to $\mathbf{M}_{3}$ in the lattice of subspaces of the plane. Now
using the arguments just above we get the following theorem.

######  Theorem 5.1.

Let $\mathbf{L}$ be the lattice of subspaces of a nonarguesian projective
plane. Then $\mathbf{L}$ lies in no congruence variety other than the variety
of all lattices.∎

The arguments above also give the following theorem which has a weaker
hypothesis but also a weaker conclusion. It should be compared to the concept
of omitted lattices of §4.3 of [27].

######  Theorem 5.2.

Let $\mathbf{L}$ be one of Haiman’s lattices or the lattice of subspaces of a
nonarguesian projective plane. Let $\mathscr{V}$ be a variety with a weak
difference term. Then $\mathbf{L}\notin\SS\,\mathbf{Con}(\mathscr{V})$.∎

##  6\. Lattices in $\SS\,\mathbf{Con}(\mathscr{V})$

Before getting to the proof of Theorem 1.3($\text{2}^{\prime}$) we prove some
interesting embedding theorems. For example we show there is a large class of
modular lattices, $\mathscr{K}_{\infty}$, all of which can be embedded into a
member of $\mathbf{Con}(\mathscr{V})$ as long as $\mathscr{V}$ is not
congruence meet semidistributive and has a weak difference term.

######  Theorem 6.1.

Let $\mathscr{V}$ be a variety having a weak difference term, and assume that
$\mathscr{V}$ is not congruence meet semidistributive. Then there is a prime
field $\mathbf{P}$ such that, for every finite $n$, the lattice of subspaces
of a vector space over $\mathbf{P}$ of dimension $n$ lies in
$\SS\,\mathbf{Con}(\mathscr{V})$.

###### Proof.

A variety is called _congruence neutral_ if $[\alpha,\beta]=\alpha\wedge\beta$
holds in every algebra in the variety. Equivalently $[\alpha,\alpha]=\alpha$.
By [29, Corollary 4.7], see also [16, Theorem 11.37], a variety is congruence
neutral if and only if it is congruence meet semidistributive. Since we are
assuming $\mathscr{V}$ is not congruence meet semidistributive, there is a
congruence $\alpha$ on an algebra $\mathbf{A}\in\mathscr{V}$ with
$[\alpha,\alpha]<\alpha$. By [5, Theorem 6.2] there is an element $\psi$ of
$\operatorname{Con}(\mathbf{A})$ with $[\alpha,\alpha]\leq\psi$ but
$\alpha\not\leq\psi$, and such that $\psi$ is completely meet irreducible with
unique cover $\psi\vee\alpha$. By Theorem 2.4(i) the interval
$I[\psi,\psi\vee\alpha]$ is abelian. So, using the basic properties of the
commutator ([16, Lemma 11.4(viii)]), $\mathbf{A}/\psi$ is subdirectly
irreducible with an abelian monolith.

Changing notation, we may assume there is a subdirectly irreducible algebra
$\mathbf{A}\in\mathscr{V}$ with an abelian monolith $\alpha$. Let
$\mathbf{A}^{n}(\alpha)$ be the subalgebra of $\mathbf{A}^{n}$ with universe

$\\{\langle a_{0},a_{1},\ldots,a_{n-1}\rangle\in
A^{n}:a_{i}\mathrel{\alpha}a_{j}\text{ for all $i$ and $j$}\\}.$

When $n=2$, $\mathbf{A}^{2}(\alpha)=\mathbf{A}(\alpha)$ which is described on
pages 96–97 in [15]. Let
$\overline{\alpha}\in\operatorname{Con}(\mathbf{A}^{n}(\alpha))$ be such that
$\langle a_{0},\ldots,a_{n-1}\rangle\mathrel{\overline{\alpha}}\langle
b_{0},\ldots,b_{n-1}\rangle$ provided all these elements are $\alpha$ related.
Let $\eta_{i}$ be the kernel of the $i^{\text{th}}$ projection of
$\mathbf{A}^{n}(\alpha)$ onto $\mathbf{A}$. Of course
$\mathbf{A}^{n}(\alpha)/\eta_{i}\cong\mathbf{A}$ and under this isomorphism
$\overline{\alpha}/\eta_{i}$ corresponds to $\alpha$. Since
$[\alpha,\alpha]=0$ in $\mathbf{A}$, it follows using Lemma 11.4(viii) of [16]
that $C(\overline{\alpha},\overline{\alpha};\eta_{i})$ holds for each $i$.
Part (v) of that Lemma gives
$C(\overline{\alpha},\overline{\alpha};\bigwedge_{i}\eta_{i})=C(\overline{\alpha},\overline{\alpha};0)$.
Hence $\overline{\alpha}$ is an abelian congruence of
$\mathbf{A}^{n}(\alpha)$. Lemma 2.4(ii) implies the interval
$I[0,\overline{\alpha}]$ in $\mathbf{Con}(\mathbf{A}^{n}(\alpha))$ is a
modular lattice. Let $\mathbf{L}_{n}$ denote this lattice. Since
$\eta_{i}\prec\overline{\alpha}$, $\mathbf{L}_{n}$ has length $n$ and its
least element is the meet of its coatoms. By (4.3) of [5], each
$\mathbf{L}_{n}$ is a complemented modular lattice.

For $i\neq j$ the interval $I[\eta_{i}\wedge\eta_{j},\overline{\alpha}]$ has
length 2. We claim this interval contains an element which is a complement of
$\eta_{i}$ and of $\eta_{j}$ and so contains a sublattice isomorphic to
$\mathbf{M}_{3}$. We can prove this with $n=2$ (so
$\mathbf{A}^{2}(\alpha)=\mathbf{A}(\alpha)$) and then apply the Correspondence
Theorem, [32, Theorem 4.12]. Let $\Delta$ be the congruence on
$\mathbf{A}(\alpha)$ generated by

$\\{\langle\langle a,a\rangle,\langle
b,b\rangle\rangle:a\mathrel{\alpha}b\\}.$

Easy element-wise calculations show $\Delta\vee\eta_{i}=\overline{\alpha}$,
$i=0,1$. Using the weak difference term it is not hard to show
$\Delta\wedge\eta_{i}=\eta_{0}\wedge\eta_{1}=0$. But actually these properties
of $\Delta$ can be proved with the weaker assumption that $\mathscr{V}$ has a
Taylor term, as was shown by Kearnes and Kiss; see [27, Claim 3.25].

By elementary modular lattice theory the existence of these $\mathbf{M}_{3}$’s
force $\mathbf{L}_{n}$ to be simple. Classical coordination theorems of Artin
[2], Birkhoff [4] and Frink [17], see Chapter 13 of [5] and also [25], show
that, for $n\geq 4$, $\mathbf{L}_{n}$ is isomorphic to the lattice of
subspaces of a vector space over a skew field $\mathbf{F}$. Moreover, since
$\mathbf{L}_{n-1}$ is isomorphic to an interval (actually to a principal
filter with generator $\eta_{0}\wedge\eta_{1}\wedge\cdots\wedge\eta_{n-2}$) in
$\mathbf{L}_{n}$ by the Correspondence Theorem, the vector space for
$\mathbf{L}_{n-1}$ is a subspace of the one for $\mathbf{L}_{n}$. This implies
the skew field $\mathbf{F}$ is the same for all $n$.

If $\mathbf{P}$ is the prime subfield of $\mathbf{F}$ then the lattice of
subspaces of an $n$-dimensional vector space over $\mathbf{P}$ can be embedded
via a cover-preserving map into the lattice of subspaces of an $n$-dimensional
vector space over $\mathbf{F}$. This completes the proof of Theorem 6.1. ∎

######  Remark 6.2.

  1. (1)

The embedding of Theorem 6.1 can be assumed to be cover-preserving, as the
proof shows.

  2. (2)

In [11] it is shown that the congruence varieties
$\textit{\emphslb{V}}\,\mathbf{Con}(\mathscr{M}_{p})$, $p$ a prime or $0$,
associated with the variety of vector spaces over a prime field are minimal
modular congruence varieties and every nondistributive modular congruence
variety contains one of these. Theorem 6.1 gives a clearer understanding of
the situation. The minimal nondistributive congruence varieties are these and
$\textit{\emphslb{V}}\,\mathbf{Con}(\mathscr{P})$, the congruence variety of
Polin’s variety; see [6]. If $\mathscr{V}$ is not congruence meet
semidistributive, then its congruence variety contains
$\textit{\emphslb{V}}\,\mathbf{Con}(\mathscr{M}_{p})$ for some $p$. If
$\mathscr{V}$ is not congruence modular, then it contains
$\textit{\emphslb{V}}\,\mathbf{Con}(\mathscr{P})$. By Remark 6.5 below the
intersection of any two of the
$\textit{\emphslb{V}}\,\mathbf{Con}(\mathscr{M}_{p})$’s is the variety of
modular, 2-distributive lattices. By the remarks of this paragraph, this
variety is not a congruence variety, so the intersection of two congruence
varieties need not be a congruence variety. On the othe hand, the join of two
congruence varieties is a congruence variety.

Recall that $\mathscr{M}_{\mathbf{F}}$ is the variety of vector spaces over
$\mathbf{F}$ and $\mathscr{M}_{\mathbf{F}}^{\text{fd}}$ is the class of all
finite dimensional vector spaces over $\mathbf{F}$.

If $\mathbf{P}$ is the prime field of characteristic $p$, a prime or 0, we let
$\mathscr{M}_{p}=\mathscr{M}_{\mathbf{P}}$ and
$\mathscr{M}_{p}^{\text{fd}}=\mathscr{M}_{\mathbf{P}}^{\text{fd}}$. Let

$\mathscr{K}_{\infty}=\bigcap_{\text{$p$ a prime or
$0$}}\mathscr{\SS}\,\mathbf{Con}(\mathscr{M}_{p}^{\text{fd}}).$

be the class of all modular lattices that, for each $p$ a prime or $0$, can be
embedded into the lattice of subspaces of some finite dimensional vector space
over the prime field of characteristic $p$.

######  Corollary 6.3.

If $\mathscr{V}$ is a variety with a weak difference term but which is not
congruence meet semidistributive, then
$\mathscr{K}_{\infty}\subseteq\SS\,\mathbf{Con}(\mathscr{V})$.∎

$\mathscr{K}_{\infty}$ is a broad class that includes most modular lattices
that have ever been pictured. Two members of this class are drawn in Figure 2.
We present a nice characterization of this class due primarily to C. Herrmann,
A. Huhn, J. B. Nation and B. Jónsson. A lattice is 2-distributive if is
satisfies the equation:

$u\wedge(x\vee y\vee z)\approx(u\wedge(x\vee y))\vee(u\wedge(x\vee
z))\vee(u\wedge(y\vee z)).$ (6.1)

Figure 2. Two members of $\mathscr{K}_{\infty}$

######  Theorem 6.4.

$\mathscr{K}_{\infty}$ is the class of all finite modular 2-distributive
lattices.

###### Proof.

Suppose that $\mathbf{L}$ is a finite, modular 2-distributive lattice. In [26]
Jónsson and Nation show that, for every field $\mathbf{F}$, $\mathbf{L}$ can
be embedded via a covering preserving embedding into the lattice of subspaces
of an $\mathbf{F}$-vector space, provided $|\mathbf{F}|>|\mathbf{L}|$. Since
the embedding is cover preserving and $\mathbf{L}$ is finite, we may assume
these vector spaces are finite dimensional. This shows
$\mathbf{L}\in\SS\,\mathbf{Con}(\mathscr{M}_{p}^{\text{fd}})$ if $p=0$ (or if
$p>|\mathbf{L}|$). For arbitrary $p>0$ we can choose a finite field
$\mathbf{F}$ of characteristic $p$ with $|\mathbf{F}|>|\mathbf{L}|$. Let
$\mathbf{P}$ be the prime subfield of $\mathbf{F}$. Then an
$\mathbf{F}$-vector space ${}_{\mathbf{F}}\mathbf{V}$ has a reduct to a
$\mathbf{P}$-vector space ${}_{\mathbf{P}}\mathbf{V}$ by restricting to scalar
multiplication by elements of $\mathbf{P}$. Of course
$\dim{}_{\mathbf{P}}\mathbf{V}=\dim{}_{\mathbf{F}}\mathbf{V}\cdot\dim{}_{\mathbf{P}}\mathbf{V}$
and so is finite. This proves the reverse containment.

To see the opposite containment, let $\mathbf{P}$ and $\mathbf{Q}$ be prime
fields of characteristics $p$ and $q$, where $p\neq q$ are prime or $0$.
Suppose $\mathbf{L}\in\mathscr{K}_{\infty}$. Then there are vector spaces
${}_{\mathbf{P}}\mathbf{V}$ and ${}_{\mathbf{Q}}\mathbf{U}$ such that
$\mathbf{L}$ can be embedded into the lattice of subspaces of each. If
$\mathbf{L}$ is not $2$-distributive it contains a $2$-diamond (equivalently a
$3$-frame) by Huhn’s Theorem [22]. Since $\mathbf{L}$ embeds into the subspace
lattice of ${}_{\mathbf{P}}\mathbf{V}$, this $2$-diamond ($3$-frame) has
characteristic $p$, as defined in [9] and [19]. Since this holds for
${}_{\mathbf{Q}}\mathbf{U}$ as well, the frame has characteristic $q$ as well.
But, as is shown in [9] and [19], this is impossible. This contradiction
completes the proof. ∎

######  Remark 6.5.

In [20] Herrmann, Pickering and Roddy generalized the result of Jónsson and
Nation [26] by showing that, for every field $\mathbf{F}$, every modular
$2$-distributive lattice can be embedded into a vector space lattice over
$\mathbf{F}$. This, together with the projectivity of $n$-frames of
characteristic $p$ [9, Theorem 1.6], shows that the congruence varieties
associated with $\mathscr{M}_{p}$ and with $\mathscr{M}_{q}$, $p\neq q$,
intersect to the variety of modular, $2$-distributive lattices. Of course the
intersection of any modular variety of lattices with a semidistributive
variety of lattices is distributive. In particular
$\textit{\emphslb{V}}\,\mathbf{Con}(\mathscr{M}_{p})\cap\textit{\emphslb{V}}\,\mathbf{Con}(\mathscr{P})$,
$p$ a prime or $0$, is the variety of distributive lattices.

######  Remark 6.6.

In a recent paper [1] Agliano, Bartali and Fioravanti show in their Theorem
1.4 that, if $\mathscr{V}$ is a variety having a member whose congruence
lattice has $\mathbf{M}_{3}$ as a sublattice such that the congruences
permute, then $\SS\,\mathbf{Con}(\mathscr{V})$ contains rods and snakes, which
roughly are finite stacks of $\mathbf{M}_{3}$’s glued over one-dimensional
intervals. $\mathbf{M}_{3,3}$ is an example. Moreover, in their concluding
remarks the authors refer to a private communication by Keith Kearnes that the
hypothesis of their Theorem 1.4 is satisfied if $\mathscr{V}$ is not
congruence meet semidistributive. This result follows from Corollary 6.3 but
under the additional assumption that $\mathscr{V}$ has a weak difference term,
and so it is a strengthening at least regarding rods and snakes. On the other
hand both the lattices of Figure 2 lie in $\mathscr{K}_{\infty}$. One wonders
if they lie in $\SS\,\mathbf{Con}(\mathscr{V})$ whenever $\mathscr{V}$ is not
congruence meet semidistributive.

##  7\. Proofs of Theorem 1.3($\operatorname{2}^{\prime}$) and Theorem 1.2

We begin with a folklore result connecting the equational theory of algebraic
lattices to that of their compact elements.

######  Lemma 7.1.

An algebraic lattice satisfies a lattice equation if and only this equation
holds whenever compact elements are substituted for the variables.

###### Proof.

This lemma implies that the ideal lattice of a lattice satisfies the same
identities as the lattice and its proof is essentially the same. But
nevertheless we give an outline. Let $\mathbf{L}$ be an algebraic lattice and
let $\mathbf{L}^{\text{c}}$ be the join subsemilattice of compact elements of
$\mathbf{L}$. It is well known that $\mathbf{L}$ is isomorphic to the lattice
of ideals of $\mathbf{L}^{c}$ (join closed down-sets). It follows that if
$c\leq a_{0}\vee a_{1}$, with $c$ compact, then $c\leq c_{0}\vee c_{1}$ with
$c_{i}$ compact and $c_{i}\leq a_{i}$, $i=0,1$. More generally, an inductive
argument on the complexity of a term $t$, shows that

$\text{if $a_{i}\in L$, $c$ is compact and }c\leq
t(a_{0},\ldots,a_{n-1}),\text{ then }c\leq t(c_{0},\ldots,c_{n-1})$

for some compact elements $c_{i}\leq a_{i}$.

To prove the lemma suppose $s(x_{0},\ldots,x_{n-1})\approx
t(x_{0},\ldots,x_{n-1})$ holds for compact elements but fails in general.
Then, by symmetry, we may assume there exist $a_{i}$’s such that
$s(a_{0},\ldots,a_{n-1})\nleq t(a_{0},\ldots,a_{n-1})$ in $\mathbf{L}$.
Consequently there is a compact element $c$ with

$c\leq s(a_{0},\ldots,a_{n-1}),\text{ but }c\nleq t(a_{0},\ldots,a_{n-1})$
(7.1)

Now by the previous paragraph there are compact elements $c_{i}\leq a_{i}$
with $c\leq s(c_{0},\ldots,c_{n-1})$. But then

$c\leq s(c_{0},\ldots,c_{n-1})=t(c_{0},\ldots,c_{n-1})\leq
t(a_{0},\ldots,a_{n-1}),$

since $s=t$ holds for compact elements and lattice term functions are order-
preserving. This contradicts (7.1) and so proves the lemma. ∎

######  Lemma 7.2.

Let $\mathbf{F}$ and $\mathbf{K}$ be skew fields.

  1. (1)

The variety of lattices generated by
$\mathbf{Con}(\mathscr{M}^{\text{{fd}}}_{\mathbf{F}})$ equals the congruence
variety of $\mathscr{M}_{\mathbf{F}}$. That is,

$\textit{\emphslb{V}}\,\mathbf{Con}(\mathscr{M}^{\text{{fd}}}_{\mathbf{F}})=\textit{\emphslb{V}}\,\mathbf{Con}(\mathscr{M}_{\mathbf{F}}).$

  2. (2)

$\mathbf{F}$ and $\mathbf{K}$ have the same characteristic if and only if

$\SS\,\mathbf{Con}(\mathscr{M}_{\mathbf{F}})=\SS\,\mathbf{Con}(\mathscr{M}_{\mathbf{K}}).$

Moreover, if $\mathbf{F}$ and $\mathbf{K}$ have different characteristics then

$\textit{\emphslb{V}}\,\mathbf{Con}(\mathscr{M}_{\mathbf{F}})\nsubseteq\textit{\emphslb{V}}\,\mathbf{Con}(\mathscr{M}_{\mathbf{K}}).$

###### Proof.

Let $\mathbf{V}$ be a vector space over $\mathbf{F}$ and let
$\mathbf{L}=\mathbf{Sub}(\mathbf{V})$. Of course,
$\mathbf{Con}(\mathbf{V})\cong\mathbf{Sub}(\mathbf{V})$ so
$\mathbf{L}\in\mathbf{Con}(\mathscr{M}_{\mathbf{F}})$. Since any finite set of
compact elements of $\mathbf{L}$ is contained in $\mathbf{W}$ for some finite
dimensional vector space $\mathbf{W}$ of $\mathbf{V}$, (1) follows from the
previous lemma.

For (2) first assume $\mathbf{F}\leq\mathbf{K}$. Then

$\SS\,\mathbf{Con}(\mathscr{M}_{\mathbf{K}})\subseteq\SS\,\mathbf{Con}(\mathscr{M}_{\mathbf{F}}).$
(7.2)

Indeed, if ${}_{\mathbf{K}}\\!\mathbf{V}$ is a vector space over $\mathbf{K}$
then its reduct to $\mathbf{F}$, ${}_{\mathbf{F}}\\!\mathbf{V}$, is a vector
space over $\mathbf{F}$. By Lemma 6.8 of [15],
$\mathbf{Con}({}_{\mathbf{K}}\\!\mathbf{V})\leq\mathbf{Con}({}_{\mathbf{F}}\\!\mathbf{V})$,
and hence
$\mathbf{Con}({}_{\mathbf{K}}\\!\mathbf{V})\in\SS\,\mathbf{Con}(\mathscr{M}_{\mathbf{F}})$.
So (7.2) holds.

Since (7.2) is the only part of statement (2) of the lemma required in this
paper, we will only sketch the rest of the proof. The opposite containment of
(7.2), still assuming $\mathbf{F}\leq\mathbf{K}$, can be proved using tensor
products: if $\mathbf{V}$ is a vector space over $\mathbf{F}$, then
$\mathbf{V}\otimes_{\mathbf{F}}\mathbf{K}$ is a vector space over $\mathbf{K}$
and, since ${}_{\mathbf{F}}\\!\mathbf{K}$ is flat, the map
$\mathbf{U}\mapsto\mathbf{U}\otimes_{\mathbf{F}}\mathbf{K}$ embeds the lattice
of subspaces of $\mathbf{V}$, $\mathbf{Sub}(\mathbf{V})$, into that of
$\mathbf{V}\otimes_{\mathbf{F}}\mathbf{K}$. Thus, if $\mathbf{P}$ is the prime
subfield of $\mathbf{K}$, we have
$\SS\,\mathbf{Con}(\mathscr{M}_{\mathbf{P}})=\SS\,\mathbf{Con}(\mathscr{M}_{\mathbf{K}})$,
and (2) easily follows.

The last statement of (2) follows from [23, Theorems 4 and 5]. ∎

Let $\mathscr{K}$ be the congruence variety of a variety $\mathscr{V}$ and
assume $\mathscr{K}$ is not join semidistributive. To prove Theorem
1.3($\operatorname{2}^{\prime}$) we need to show that there is a field
$\mathbf{F}$ such that any nonprincipal ultraproduct of
$\\{\mathbf{H}_{n}(\mathbf{F}):n\geq 3\\}$ lies in $\mathscr{K}$. This is
trivial if $\mathscr{K}$ is the variety of all lattices so we may assume
$\mathscr{V}$ has a nontrivial (pure lattice) congruence identity. We know
$\mathscr{K}$ is not join semidistributive which implies $\mathscr{V}$ is not
congruence join semidistributive, as was shown by Kearnes and Kiss in [27,
Theorem 8.14 (1) $\Leftrightarrow$ (8)]. Now (1) $\Leftrightarrow$ (6) of that
same theorem implies $\mathscr{V}$ is not congruence meet semidistributive.
Also, by the Kearnes-Szendrei result Theorem 2.5(i) above, $\mathscr{V}$ has a
weak difference term.

Theorem 6.1 can be interpreted to say that there is a prime field $\mathbf{P}$
such that
$\mathbf{Con}(\mathscr{M}^{\text{{fd}}}_{\mathbf{P}})\subseteq\SS\,\mathbf{Con}(\mathscr{V})$.
Now by Lemma 7.2(1) we have that the congruence variety of $\mathscr{V}$,
namely $\mathscr{K}$, contains the congruence variety of
$\mathscr{M}_{\mathbf{P}}$. Let $\mathbf{F}$ be a field whose prime subfield
is $\mathbf{P}$. (If $\mathbf{P}$ is the two element field, $\mathbf{F}$
should have at least 4 elements.) By Lemma 3.6(4) any nonprincipal
ultraproduct of $\\{\mathbf{H}_{n}(\mathbf{F}):n\geq 3\\}$ lies in
$\SS\,\mathbf{Con}(\mathscr{M}_{\mathbf{P}})$ and so in $\mathscr{K}$.

This completes the proof of Theorem 1.3($\operatorname{2}^{\prime}$) and hence
of Theorem 1.3. ∎

To see Theorem 1.2 let $\mathscr{V}$ be a variety of algebras such that
$\mathbf{Con}\;\mathbf{(}\mathscr{V})$ satisfies a nontrivial lattice identity
and suppose $\mathbf{Con}(\mathscr{V})$ is not semidistributive. By the
Kearnes-Kiss result cited above, this implies $\mathbf{Con}(\mathscr{V})$ is
not join semidistributive. By Theorem 1.3($\operatorname{1}^{\prime}$) and
($\operatorname{2}^{\prime}$) there is a collection of lattices not in
$\textit{\emphslb{V}}\,\mathbf{Con}(\mathscr{V})$ whose ultraproduct is in.
Consequently the congruence variety of $\mathscr{V}$ is not finitely based by
[15, Theorem 8.52]. ∎

### Data availability

Data sharing not applicable to this article as datasets were neither generated
nor analyzed.

### Compliance with ethical standards

The authors declare that they have no conflict of interest. The first author
belongs to the Editorial Board of Algebra Universalis.

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