standard topological algebras and syntactic congruences

Standard topological algebras and syntactic congruences

David M. Clark, Brian A. Davey, Ralph S. Freese, and Marcel Jackson

Abstract. A topological quasi-variety X generated by a finite algebra M∼ with the discrete topology is said tobe standard if it admits a canonical axiomatic description. Drawing on the formal language notion of syntacticcongruences, we prove that X is standard provided that the algebraic quasi-variety generated by M∼ is a variety, andthat syntactic congruences in that variety are determined by a finite set of terms. We give equivalent semantic andsyntactic conditions for a variety to have Finitely Determined Syntactic Congruences, show that FDSC generalizesthe notion of Definable Principle Congruences and exhibit many familiar algebras M∼ that our method reveals tobe standard.

1. Background and motivation

An algebra M = 〈M ;F 〉 with finite underlying set M and operations F generates an (algebraic)quasi-variety A := ISPM consisting of all isomorphic copies of subalgebras of direct powers of M.Similarly a structure M∼ = 〈M ;G,H,R,T 〉 with finite underlying set M , operations G, partial oper-ations H , relations R and discrete topology T generates a topological quasi-variety X := IScP

+ M∼consisting of all isomorphic copies of topologically closed substructures of non-zero direct powers, withthe product topology, of M∼ . Interest in topological quasi-varieties stems from the fact that they ariseas the duals to algebraic quasi-varieties under natural dualities. The general theory of natural dualitiesprovides methods to produce, from the algebra M, a structure M∼ that will yield a natural duality on A.(See Clark and Davey [4].)

To maximize the usefulness of a natural duality, it is necessary to find a simple description of themembers of the topological quasi-variety X. Attempts to do this have drawn on Mal’cev’s description ofthe members of a finitely generated algebraic quasi-variety A. Mal’cev [17] discovered that A consistsexactly of the models of the quasi-equations (equations and implications) that hold in M; in symbols,

A = Mod(Thqe(M)).

Often a small set Σ ⊆ Thqe(M) of axioms for Thqe(M) provides a simple description of the membersof A as A = Mod(Σ).

The presence of partial operations, relations and topology in X makes an analogous descriptionconsiderably more illusive. By a Boolean topological structure (of type 〈G,H,R〉) we mean a structureX∼ = 〈X ;GX∼, HX∼, RX∼,TX∼ 〉 such that

(i) 〈X ; TX∼ 〉 is a Boolean (i.e. compact totally disconnected) space,(ii) if h ∈ G∪H is n–ary, then the domain dom(hX∼) is a closed subset of Xn and hX∼ : dom(hX∼) → X

is continuous, and(iii) if r ∈ R is n–ary, then rX∼ is a closed subset of Xn.A quasi-atomic formula in the first-order language of M∼ is an expression of one of the forms

χ or∨i∈I

¬ψi or∧i∈I

ψi ⇒ χ (∗)

Date: October 21, 2002.2000 Mathematics Subject Classification: 08C15, 54E99, 03C99, 06D50, 68Q45, 06B20 .Key words and phrases: standard topological quasi-variety, syntactic congruences, definable principal congruences.

1

2 DAVID M. CLARK, BRIAN A. DAVEY, RALPH S. FREESE, AND MARCEL JACKSON

where χ and each ψi are atomic formulæ and I is a finite set. We denote by Thqa(M∼ ) the set of allquasi-atomic formulæ that hold in M∼ . If Σ is a set of quasi-atomic formulæ, ModT(Σ) denotes theclass of all Boolean structures that satisfy each quasi-atomic formula in Σ. Necessary conditions fora structure X∼ to be in X := IScP

+ M∼ are given by the Preservation Theorem (Clark and Davey [4,1.4.3]), which states that

X ⊆ ModT(Thqa(M∼ )).

In [5], Clark, Davey, Haviar, Pitkethly and Talukder define X := IScP+ M∼ to be a standard topological

quasi-variety, or M∼ to be standard, if X is exactly the class of all Boolean models of the quasi-atomictheory of M∼ ; in symbols,

X = ModT(Thqa(M∼ )).

This is not the case for every choice of M∼ . A classic example of Stralka [26] shows that the orderedspace M∼ = 〈{0, 1},≤,T〉, which generates the category of Priestley spaces, is not standard.

A subset Σ ⊆ Thqa(M∼ ) is said to axiomatize X if X = ModT(Σ). In [5] we find many examples inwhich X has a simple finite axiomatization, and is therefore standard. We also find that knowing inadvance that X is standard greatly simplifies the process of finding an axiomatization by making it anessentially finitary process:

Proposition 1.1. ([5, 1.4]) Assume that X := IScP+ M∼ is standard. Then Σ ⊆ Thqa(M∼ ) axiomatizes

X provided every model of Σ is locally finite and each finite model of Σ is in X.

In this paper we restrict our attention to the case in which M∼ has neither partial operations norrelations. Thus we assume that M∼ = 〈M ;G,T〉 is a Boolean topological algebra with the discretetopology T. Algebras M whose generated quasi-variety admits a strong duality via a total algebraare described by the Two-for-One Strong Duality Theorem ([4, 3.3.2]). However, we note here thatthe theory of natural dualities will play no further role in this study. We mention it only to providemotivation for the investigation of the question as to when the topological quasi-variety X is standard.

2. Finitely Determined Syntactic Congruences (FDSC)

Serendipitously it turns out that a notion from the theory of formal languages leads to an algebraiccondition that provides a simple criterion for a topological algebra M∼ = 〈M ;G,T 〉 to be standard. Inthis and the next section we will develop that algebraic condition, and in subsequent sections we willapply it to the standardness problem.

Given a finite alphabet A, let A∗ denote the free monoid of words over A. A subset L ⊆ A∗ iscalled a language over A. An important class of languages, the regular languages, have a numberof equivalent characterizations. Primarily, they are the languages accepted by finite state acceptors,the languages generated by regular grammars, and the languages denoted by regular expressions. Onecharacterization is purely algebraic. A language L over A determines an equivalence relation θL onA∗ with the two classes, L and A∗\L. Let Syn(L) denote the set of all (a, b) ∈ A∗ × A∗ such that(cad, cbd) ∈ θL for all c, d ∈ A∗. Then Syn(L) is a congruence on A∗ called the syntactic congruenceof L, and the language L is regular if and only if Syn(L) has finite index in A∗. (See, for example,Eilenberg [10].)

The notion of syntactic congruences has a natural extension to arbitrary algebras. To see this, letX = 〈X ;GX〉 be an algebra and let θ be an equivalence relation on X . Let Ty be the set of all termswith operation symbols G in the countable sequence of variables y, x1, x2, x3, . . . , and let F be a subsetof Ty. We define a relation θF on X by (a, b) ∈ θF if and only if, for all f(y, x1, x2, . . . ) ∈ F ,

(f(a, c1, c2, . . . ), f(b, c1, c2, . . . )) ∈ θ for all c1, c2, · · · ∈ X. (∗)

Finally, we define Syn(θ) = θTy . The following facts are easy to check.

STANDARDNESS AND SYNTACTIC CONGRUENCES 3

Lemma 2.1. For every algebra X and every equivalence relation θ on X,(i) Syn(θ) is a congruence on X,(ii) Syn(θ) ⊆ θ (since y ∈ Ty), and(iii) ψ ⊆ Syn(θ) if ψ is a congruence on X and ψ ⊆ θ.Thus Syn(θ) is the largest congruence on X that is contained in θ.

We shall refer to Syn(θ) as the syntactic congruence of θ. This formulation of syntactic congruencesappears to have first been observed by S lominski [25]. For a detailed discussion of the role of syntacticcongruences in the study of pseudo-varieties, see Chapter 3 of [1].

Our present study begins with the observation that it is never necessary to test all terms t ∈ Ty inorder to establish membership in Syn(θ). For example, if X is a ring and (∗) is true for f = x1y + x2,then (∗) is certainly also true for f = x3y + x4 and for f = (x2 − x3x5)y. Thus Syn(θ) may be moreefficiently defined as being θF for some proper subset F ⊆ Ty. We will say that a subset F ⊆ Ty ofterms determines syntactic congruences in a class M of algebras if Syn(θ) = θF for every X ∈ M andevery equivalence relation θ on X . In Section 4 we will see that Syn(θ) will be useful to us provided thatit is determined by some finite set of terms. Accordingly we say that a class M of algebras has finitelydetermined syntactic congruences, abbreviated FDSC, if there is a finite set F ⊆ Ty that determinessyntactic congruences in M.

In the next theorem we show that one particular subset of Ty is always sufficient to determinesyntactic congruences. Let T1y denote the set of all terms in Ty that have exactly one occurrence of thevariable y. For example, y occurs three times in the group term z−1y3x but exactly once in y−1z3x.

In the remainder of this section we will develop a practical method to verify that a set of termsdetermines syntactic congruences in a class of algebras. For this purpose we fix a type G and a classM of algebras of type G. For a set F ⊆ Ty of terms and a single term t ∈ Ty, we write

F |= t in M (F semantically shadows t in M)

if, for each X ∈ M and each equivalence θ on X , we have θF ⊆ θt. Thus F |= t in M if t is redundantin the presence of F for the determination of syntactic congruences on members of M. We say thatF |= F ′ in M if F |= t in M for each t ∈ F ′. Equivalently, F |= F ′ in M if θF ⊆ θF ′ for every X ∈ M

and every equivalence θ on X. If t(y, x1, x2, . . . ) does not depend upon y in M, that is, if M satisfiesthe identity

t(x, x1, x2, . . . ) ≈ t(y, x1, x2, . . . )(in particular, if t is a nullary term), then θt = X ×X , for each X ∈ M, and hence F |= t, for everysubset F of Ty. Notice that |= is a transitive relation on the subsets of Ty.

The following theorem gives five criteria for a set F to determine syntactic congruences. Criterion (ii)shows that it is equivalent to semantic shadowing. Criterion (iii) will be needed later in our applicationsof the Shadowing Theorems 3.3 and 3.4. Criterion (iv) implies that shadowing can be established bychecking finitely many instances in case F and G are both finite. The two examples that follow willillustrate criteria (iv) and (v). Note that F |= y in M if and only if θF ⊆ θ for every X ∈ M and everyequivalence θ on X , since we always have θy = θ.

Theorem 2.2. For a set F ⊆ Ty of terms and a class M of algebras, the following are equivalent:(i) F determines syntactic congruences in M.(ii) F |= Ty in M;(iii) F |= T1y in M;(iv) F |= y and, for each f(y, xm+1, . . . , xm+n) ∈ F , each m–ary fundamental operation g ∈ G and

each i with 1 ≤ i ≤ m, we have

F |= f(g(x1, . . . , xi−1, y, xi+1, . . . , xm), xm+1, . . . , xm+n) in M;

(v) F |= y and θF is a congruence for each X ∈ M and each equivalence θ on X.


Proof (i)→(ii) Let t ∈ Ty and let θ be an equivalence on X where X ∈ M. Then θF = Syn(θ) =θTy ⊆ θt so F |= t.

(ii)→(iv), trivially.(iv)→(v) Let θ be an equivalence on X with X ∈ M. Let (a, b) ∈ θF , let g ∈ G be m–ary, let

1 ≤ i ≤ m and let c1, . . . , cm ∈ X . To see that θF is a congruence it suffices to show that

(g(c1, . . . , ci−1, a, . . . , cm), g(c1, . . . , ci−1, b, . . . , cm)) ∈ θF .

Assume that f(y, xm+1, . . . , xm+n) ∈ F and cm+1, . . . , cm+n ∈ X . We define

t(y, x1, . . . , xm+n) = f(g(x1, . . . , xi−1, y, xi+1, . . . , xm), xm+1, . . . , xm+n).

By (iv) we have F |= t, so (a, b) ∈ θt. Thus,

(t(a, c1, . . . , cn+m), t(b, c1, . . . , cn+m)) ∈ θ,

and hence (g(c1, . . . , ci−1, a, . . . , cm), g(c1, . . . , ci−1, b, . . . , cm)) ∈ θF , as required.(v)→(i) Take X ∈ M and let θ be an equivalence on X . Then θF ⊆ θy = θ so θF ⊆ Syn(θ), the

largest congruence contained in θ, and Syn(θ) = θTy ⊆ θF since F ⊆ Ty. Thus θF = Syn(θ).(ii)→(iii)→(i) The implication (ii)→(iii) is trivial, so assume (iii) to prove (i). Let X ∈ M and let θ

be an equivalence on X . By (iii) we have Syn(θ) ⊆ θF ⊆ θT1y ⊆ θ. Now it is straightforward to checkthat θT1y is a congruence. By Lemma 2.1(iii) we have θT1y ⊆ Syn(θ). Thus θF = Syn(θ), showing thatF determines syntactic congruences in M.

Our first example shows that our notion of syntactic congruence is consistent with the formal languagetheoretic notion.

Example 2.3. The single term F = {x1yx2} determines syntactic congruences in the variety ofmonoids, which therefore has FDSC.

Proof We verify Theorem 2.2(v). Let X be a monoid, let θ be an equivalence on X and assume(a, b) ∈ θF . Then (a, b) = (1a1, 1b1) ∈ θ = θy, so F |= y. Now choose c, d, e ∈ X . Then

(d · ac · e, d · bc · e) = (d · a · ce, d · b · ce) ∈ θ,

so (ac, bc) ∈ θF . Similarly, (ca, cb) ∈ θF , showing that θF is a congruence.

The algebra M is unary if each of its operations is unary. Natural dualities given by unary topologicalalgebras M∼ are of particular interest since they tend to give accessible dual categories. Algebras Mstrongly dualised by a unary topological algebra are characterized in [4, 6.4.5]. In [5] it is shown thatevery finite unary topological algebra with a single unary operation is standard. Our second examplewill be used in Proposition 5.8 to exhibit many more unary algebras that generate a standard topologicalquasi-variety.

Example 2.4. Every variety generated by a finite unary algebra M = 〈M ;G〉 has FDSC.

Proof Since M is finite, there is a finite set F of unary terms in the variable y such that y ∈ F and{fM | f(y) ∈ F} is the submonoid of MM generated by G. We verify condition (iv) of Theorem 2.2.Let f(y) ∈ F and g ∈ G. Then there is a term t(y) ∈ F such that fMgM = tM, so f(g(y)) ≈ t(y) holdsin M. Thus F |= f(g(y)).

3. Syntactic shadowing and the Completeness Theorem

Condition (ii) of Theorem 2.2 suggests a quite different method for establishing that F determinessyntactic congruences. We would like a purely syntactic procedure, that refers only to terms and not tomodels and congruences, to produce t from F when F |= t. A simple example suggests how this might


be done. According to Example 2.3, the term x1yx2 determines syntactic congruences in the variety ofmonoids. By Theorem 2.2(ii) it must be true that x1yx2 semantically shadows, for example, the termy3. To verify this fact directly, let θ be an equivalence on a monoid X, let a, b ∈ X with (a, b) ∈ θx1yx2 .To see that (a, b) ∈ θy3 , we have

a3 = 1 · a · aaθ 1 · b · aa = b · a · a

θ b · b · a = bb · a · 1θ bb · b · 1 = b3,

showing that (a3, b3) ∈ θ. Notice that, starting from a3, we move right using monoid identities and wemove down using the fact that (a, b) ∈ θx1yx2 . Based on this and many other examples, we propose thefollowing definition relative to the fixed type G and class M of algebras of type G. For a set F ⊆ Ty ofterms and a single term t ∈ Ty, we write

F ` t in M (F syntactically shadows t) in M

provided that there exists k ≥ 0, terms f1(y, x1, . . . , xm), . . . , fk(y, x1, . . . , xm) in F and maps ν1, . . . , νk

from {x1, . . . , xm} to the set Txy of all terms in the expanded set of variables {x, y, x1, x2, . . . } suchthat each of the following identities is satisfied by each member of M:

t(x, x1, . . . , xn) ≈ f1(z1, ν1(x1), . . . , ν1(xm))fi(z′i, νi(x1), . . . , νi(xm)) ≈ fi+1(zi+1, νi+1(x1), . . . , νi+1(xm))fk(z′k, νk(x1), . . . , νk(xm)) ≈ t(y, x1, . . . , xn),

where {zi, z′i} = {x, y}. (When k = 0, this family of identities reduces to the single identity

t(x, x1, . . . , xn) ≈ t(y, x1, . . . , xn),

that is, to the statement that, in M, the term t(y, x1, . . . , xn) does not depend upon the variable y.)For example, x1yx2 syntactically shadows y3 in the variety of monoids as we have

x3 ≈ 1 · x · xx1 · y · xx ≈ y · x · x

y · y · x ≈ yy · x · 1yy · y · 1 ≈ y3.

In order to justify our definition of syntactic shadowing, we need to know that F syntacticallyshadows only what it should and everything that it should.

Soundness Theorem 3.1. If F ` t in M, then F |= t in M for every class M of algebras.

Proof Assume that F ` t as described above. Let θ be an equivalence on an algebra X ∈ M and leta, b ∈ X with (a, b) ∈ θF . To see that (a, b) ∈ θt, choose c1, . . . , cn ∈ X . Then we have

t(a, c1, . . . , cn) = f1(d1, ν1(c1), . . . , ν1(cm))θ f1(d′1, ν1(c1), . . . , ν1(cm)) = f2(d2, ν2(c1), . . . , ν2(cm))θ f2(d′2, ν2(c1), . . . , ν2(cm)) = f3(d3, ν3(c1), . . . , ν3(cm))

...θ fk(d′k, νk(c1), . . . , νk(cm)) = t(b, c1, . . . , cn),

where (di, d′i) = (a, b) if (zi, z

′i) = (x, y) and (di, d

′i) = (b, a) if (zi, z

′i) = (y, x) for i = 1, 2, . . . , k. Thus

(t(a, c1, . . . , cn), t(b, c1, . . . , cn)) ∈ θ and therefore (a, b) ∈ θt.


Completeness Theorem 3.2. Assume that M contains a free countably generated algebra. If F |= tin M, then F ` t in M.

Proof Suppose that F |= t in M. Let Fxy be the free countably generated algebra in M with itsgenerators labelled x,y,x1,x2, . . . . Let S be the set of all

(f(x, z1, z2, . . . ), f(y, z1, z2, . . . ))

and(f(y, z1, z2, . . . ), f(x, z1, z2, . . . )),

where f(y, x1, x2, . . . ) ∈ F and z1, z2, . . . ∈ Fxy, together with the identity relation on Fxy. Then thetransitive closure of S is an equivalence relation θ on Fxy for which we have (x,y) ∈ θF . Consequently(x,y) ∈ θt. In particular,

(t(x,x1,x2, . . . ), t(y,x1,x2, . . . )) ∈ θ.

This says that t(x,x1,x2, . . . ) and t(y,x1,x2, . . . ) are connected via the transitive closure of S. Hencethere exists k ≥ 0, terms fi(y, x1, x2, . . . ) in F for i = 1, . . . , k, choices of zi, z

′i so that {zi, z

′i} = {x, y}

and maps νj : {x1,x2, . . . } → Fxy such that

t(x,x1,x2, . . . ) = f1(z1, ν1(x1), ν1(x2), . . . )fi(z′i, νi(x1), νi(x2), . . . ) = fi+1(zi+1, νi+1(x1), νi+1(x2), . . . )fk(z′k, νk(x1), νk(x2), . . . ) = t(y,x1,x2, . . . ).

Since Fxy is M-freely generated by {x,y,x1,x2, . . . }, we see that the equations

t(x, x1, x2, . . . ) ≈ f1(z1, ν1(x1), ν1(x2), . . . )fi(z′i, νi(x1), νi(x2), . . . ) ≈ fi+1(zi+1, νi+1(x1), νi+1(x2), . . . )fk(z′k, νk(x1), νk(x2), . . . ) ≈ t(y, x1, x2, . . . )

all hold in M. Thus F ` t in M.

Together these two theorems tell us that in an ISP-closed class, and in particular in a variety M,the two notions of shadowing are identical. Thus we say that F shadows t in an ISP-closed class M ifF |= t and/or F ` t in M, and F shadows F ′ in M provided that F shadows t in M for each t ∈ F ′.It will be useful to reiterate the equivalence of (i) and (iii) of Theorem 2.2 in this context.

General Shadowing Theorem 3.3. A subset F ⊆ Ty determines syntactic congruences in an ISP-closed class M of algebras if and only if F shadows T1y in M.

It turns out that a very simple version of shadowing is sufficient for a large number of applications.Relative to a class M of algebras, we say that F simply shadows a term t ∈ Ty if there are termsw1, w2, w3, . . . in variables {x1, x2, x3, . . . }, and a term f(y, x1, x2, . . . ) ∈ F and such that the identity

t(y, x1, x2, . . . ) ≈ f(y, w1, w2, . . . ) (∗)

holds in M. We say that F simply shadows F ′ in M if F simply shadows each term of F ′ in M.

Simple Shadowing Theorem 3.4. A subset F ⊆ Ty determines syntactic congruences in a class M

of algebras provided that F simply shadows T1y in M.

Proof Let t ∈ T1y. Since F simply shadows t, there are terms w1, w2, w3, . . . in variables x1, x2, x3, . . .and a term f(y, x1, x2, . . . ) in F such that the identity (∗) holds in M. Taking k = 1, f1 = f , z1 = x,z1

′ = y, and ν1(xi) = wi, we obtain F ` t. Thus F ` T1y and so F |= T1y in M. The conclusion followsfrom (iii)→(i) of Theorem 2.2.

This theorem would be true if T1y were replaced by Ty, but this formulation would not be very useful.For example, the reader can check that Examples 2.3 and 2.4 can both be established as immediateconsequences of the Simple Shadowing Theorem.


Since T1y determines syntactic congruences in any class M of algebras, we know that T1y |= t in M,for all t ∈ Ty. The following lemma is a local version of the corresponding result for syntactic shadowing.This, along with the Soundness Theorem 3.1, provides a direct proof of (ii)→(i) in Theorem 2.2. If wereplace some or all occurrences of variables in a term t by other variables, then we say that the resultingterm t′ is obtained from t by variable replacement. For example, the term t′ = ((y∧x1)∨(x2∧x3))∧x4

may be obtained by variable replacement from t = ((y ∧x2)∨ (y ∧ x3))∧ y. The significance of variablereplacement is that it does not alter the shape and therefore the complexity of a term.

Lemma 3.5. Let t be a term in which the variable y occurs k > 0 times. Then there are termst1, . . . , tk ∈ T1y such that each ti is obtained from t by variable replacement and {t1, . . . , tk} ` t.

Proof Let t′ be obtained from t be replacing the k occurrences of y with new variables y1, . . . , yk.Let ti be obtained from t′ by replacing yi with y. It is straightforward to show that {t1, . . . , tk} ` t.

4. FDSC-HSP Theorem

We will now apply the work of the previous two sections to give an efficient method to verify that atopological quasi-variety of algebras is standard. Given a topological algebra X∼ := 〈X ;G,T 〉, we denotethe underlying (discrete) algebra by X := 〈X ;G〉. Our central result, the FDSC-HSP Theorem, will saythat M∼ = 〈M ;G,T 〉 generates a standard topological quasi-variety provided that the algebraic quasi-variety generated by M is a variety and that this variety has FDSC. We use the basic characterizationof topological quasi-varieties, which we restate here for total algebras.

Separation Theorem 4.1. ([4, 1.4.4]) Let M∼ = 〈M ;G,T 〉 be a finite algebra with T discrete, letX := IScP

+ M∼ , and let X∼ be a Boolean topological algebra of the same type as M∼ . Then X∼ ∈ X if andonly if either X∼ has only one element and M∼ has a one-element subalgebra, or X∼ has more than oneelement and for each x, y ∈ X where x 6= y, there is an α : X∼ → M∼ such that α(x) 6= α(y).

It will be instructive to prove our result first in the case that M∼ is unary since that argument ismuch simpler but still illustrates the method we use in the general case.

Proposition 4.2. Let M∼ = 〈M ;G,T 〉 be a finite unary algebra with T discrete for which ISPM =HSPM. Then X = IScP

+ M∼ is standard.

Proof Let X∼ ∈ ModT(Thqa(M∼ )). To show that X∼ ∈ X, we consider two cases. First assume that X∼is a one-element algebra. As M is finite we may assume, without loss of generality, that G is finite, sayG = {g1, . . . , gk}. If M∼ has no one-element subalgebra, then

M∼ |= (g1(x) 6≈ x) ∨ · · · ∨ (gk(x) 6≈ x)

and hence ModT(Thqa(M∼ )) contains no one-element algebras. Since |X | = 1 and X∼ belongs toModT(Thqa(M∼ )), it follows that M∼ has a one-element subalgebra and consequently X∼ ∈ X = IScP

+ M∼ .Otherwise, we choose a, b ∈ X with a 6= b and apply the Separation Theorem 4.1. Let Y be a clopen

subset of X containing a but not b, and consider the equivalence relation θ with the two equivalenceclasses Y and X\Y . By Example 2.4 there is a finite set F of unary terms f(y) that determine syntacticcongruences in HSPM.

Consider the congruence θF on X. We claim that each θF class is clopen. To prove it, let c ∈ X .The equivalence class c/θF is equal to⋂

{f−1(Y ) | f ∈ F, f(c) ∈ Y } ∩⋂

{f−1(X\Y ) | f ∈ F, f(c) /∈ Y }.

Since F is finite, this set is clopen.Since X∼ is compact, θF must have finite index in X. It follows that the quotient homomorphism

α : X → X/θF is continuous, and it separates a and b since θF ⊆ θ. Finally X∼/θF (with the discrete


topology) is in X, since HSPM = ISPM. Consequently there is a homomorphism β : X/θF → M thatseparates α(a) and α(b), and β is continuous since these algebras are finite. The map α ◦ β : X∼ → M∼separates a and b.

Let X∼ = 〈X ;G,T 〉 be a Boolean topological algebra and let θ be an equivalence relation on X . Wesay that θ is a clopen equivalence relation on X∼ if each θ class is clopen; equivalently, if θ is a clopensubset of X×X . As we observed in the proof of Proposition 4.2, a clopen equivalence relation must havefinite index since X∼ is compact. The general, non-unary, case requires a more substantive applicationof compactness which we prove as a separate lemma. It is the application of this lemma that leads usto require syntactic congruences be finitely determined.

Lemma 4.3. Let X∼ = 〈X ;G,T 〉 be a Boolean topological algebra and let θ be a clopen equivalencerelation on X. If F ⊆ Ty is a finite set of terms of X∼, then θF is also a clopen equivalence relationon X.

Proof Let a ∈ X . We first argue that a/θF is closed. If the term f(y, x1, . . . , xn) belongs to F andc := (c1, . . . , cn) ∈ Xn, then the map f(−, c) : X → X is continuous. Thus, for b ∈ X , we have

b ∈ a/θF ⇐⇒ (f(b, c), f(b, c)) ∈ θ for all f ∈ F and c ∈ Xn

⇐⇒ b ∈⋂

{f(−, c)−1(f(a, c)/θ) | f ∈ F, c ∈ Xn}.

Thus a/θF =⋂{f(−, c)−1(f(a, c)/θ) | f ∈ F, c ∈ Xn}, which is closed since each f(−, c)−1(f(a, c)/θ)

is clopen.We now prove that a/θF is open as well by constructing a clopen subset W of a/θF containing a.

Again fix a term f(y, x1, . . . , xn) ∈ F and c := (c1, . . . , cn) ∈ Xn. Since f is continuous at (a, c) andf(a, c) is in the clopen set f(a, c)/θ, there are clopen sets Uc ⊆ X containing a and Vc ⊆ Xn containingc such that f(Uc, Vc) ⊆ f(a, c)/θ. Since X∼

n is compact, there is a finite set {c1, c2, . . . , ck} ⊆ Xn suchthat Vc1 , Vc2 , . . . , Vck

cover Xn. Define

Wf := Uc1 ∩ · · · ∩ Uck.

Then Wf is a clopen subset of X containing a. We now define

W :=⋂

{Wf | f(y, x1, . . . , xn) ∈ F}.

Then W contains a, and W is clopen since F is finite.To see that W ⊆ a/θF , choose b ∈ W . Let f(y, x1, . . . , xn) ∈ F and c ∈ Xn. Then c ∈ Vcj for

some j ≤ k. Since b ∈ W ⊆ Wf ⊆ Ucj , we have f(b, c) ∈ f(a, cj)/θ. Since a ∈ Ucj , we also havef(a, c) ∈ f(a, cj)/θ. Consequently (f(a, c), f(b, c)) ∈ θ. Since f(y, x1, . . . , xn) ∈ F and c ∈ Xn werearbitrary, we conclude that (a, b) ∈ θF . Thus W ⊆ a/θF .

We can now put the pieces together, giving a general condition which shows how FDSC leads tostandardness.

FDSC-HSP Theorem 4.4. Let M∼ = 〈M ;G,T 〉 be a finite algebra with T discrete and assume thatISPM = HSP M. If HSPM has FDSC, then X := IScP

+ M∼ is standard.

Proof Let X∼ ∈ ModT(Thqa(M∼ )). To show that X∼ ∈ X, we consider two cases. First assume that X∼is a one-element algebra. In this case, to show that X∼ ∈ X it suffices to show that M∼ has a one-elementsubalgebra. Suppose that M∼ has no one-element subalgebra. Then, for all a ∈ M , there exists ga ∈ Gsuch that ga(a, . . . , a) 6= a. Thus

M∼ |=∨

a∈M

ga(x, . . . , x) 6≈ x


and hence ModT(Thqa(M∼ )) contains no one-element algebras. Since |X | = 1 and X∼ belongs toModT(Thqa(M∼ )), it follows that M∼ has a one-element subalgebra, as required.

Now assume that X∼ has more than one element. We choose a, b ∈ X with a 6= b and again usethe Separation Theorem 4.1. Let Y be a clopen subset of X containing a but not b, and let θ bethe equivalence relation with two classes Y and X\Y . Choose a finite set F of terms that determinesyntactic congruences in HSPM. By Lemma 4.3, each θF class is clopen. Since X∼ is compact, θF musthave finite index in X∼. It follows that the quotient homomorphism α : X → X/θF is also continuous,and it separates a and b since θF ⊆ θ. Finally X∼/θF , endowed with the discrete topology, is in X sinceHSPM = ISPM. Consequently there is a homomorphism β : X/θF → M that separates α(a) andα(b), and β is continuous since these algebras are finite. The map α ◦ β : X∼ → M∼ separates a and b.

5. Examples of simple shadowing

We will now exhibit a large assortment of varieties of algebras that can be shown to have FDSCusing the Simple Shadowing Theorem 3.4. Many of our examples generate a quasi-variety that is also avariety. The FDSC-HSP Theorem tells us that each of these examples, when endowed with the discretetopology, generates a standard topological quasi-variety.

The condition of the Simple Shadowing Theorem 3.4 says, loosely speaking, that F determinessyntactic congruences on M if every term with a single y can be rewritten, modulo the equationaltheory of M, in the form of a member of F . For example, let M be the variety of distributive latticesL = 〈L;∨,∧〉 and define

F := {y, y ∨ x2, x1 ∧ y, (x1 ∧ y) ∨ x2}.If t(y, x1, x2, . . . ) ∈ T1y, then it can be rewritten as (w1 ∧ y) ∨w2, (or as y ∨w2 or w1 ∧ y or y if eitherw1 or w2 or both are missing) where w1 and w2 are terms in x1, x2, . . . .

Corollary 5.1. Let D be the category of Boolean distributive lattices, that is, Boolean spaces withcontinuous distributive lattice operations and let M∼ = 〈M ;∨,∧,T 〉 be any non-trivial finite memberof D. Let M be the variety of distributive lattices.

(i) F = {y, y ∨ x2, x1 ∧ y, (x1 ∧ y) ∨ x2} determines syntactic congruences in M.(ii) D = IScP

+ M∼ is standard.

Proof The Simple Shadowing Theorem 3.4 implies (i) by the argument above. Since ISPM =HSPM = M, the FDSC-HSP Theorem tell us that IScP

+ M∼ is standard. We know that Thqe(M∼ )is the quasi-equational theory of distributive lattices. Since IScP

+ M∼ is standard, we have D :=ModT(Thqe(M∼ )) = IScP

+ M∼ .

For bounded distributive lattices L = 〈L;∨,∧, 0, 1〉, a similar assertion is true: the first three termsof F are now redundant in the presence of the fourth. This illustrates a recurrent theme. Havingdistinguished identity elements for binary operations generally serves to reduce them number of termsneeded to define syntactic congruences.

We give several more examples whose proofs, which we omit, are easy and similar to the aboveargument for distributive lattices.

Corollary 5.2. Let K be a finite field. Let V be the category of Boolean vector spaces over K, thatis, Boolean spaces with continuous vector space operations, and let M∼ = 〈M ; +, α, 0,T 〉α∈K be anynon-trivial finite member of V. Let M be the variety of vector spaces V = 〈V ; +, α, 0〉α∈K over K.

(i) F = {x1 + αy | α ∈ K} determines syntactic congruences in M.(ii) V = IScP

+ M∼ is standard.


Corollary 5.3. Let B be the category of non-trivial Boolean topological Boolean algebras, that is,Boolean spaces with continuous Boolean algebra operations, and let M∼ = 〈M ;∨,∧,′ , 0, 1,T 〉 be anynon-trivial finite member of B. Let M be the variety of Boolean algebras B = 〈B;∨,∧,′ , 0, 1〉.

(i) F = {(x1 ∧ y) ∨ x2, (x1 ∧ y′) ∨ x2} determines syntactic congruences in M.(ii) B = IScP

+ M∼ is standard.

Corollary 5.4. Let A be the category of Boolean abelian groups, that is, Boolean spaces with continuousabelian group operations, and let M∼ = 〈M ; +,−, 0,T 〉 be any non-trivial finite member of A. Let M

be the variety of abelian groups A = 〈A; +,−, 0〉.(i) F = {x1 + y, x1 − y} determines syntactic congruences in M.(ii) A = IScP

+ M∼ is standard.

Part (ii) of the previous three corollaries was established in Davey and Werner [9], where it is shownthat V is dual to the variety of all vector spaces over K, that B is dual to the variety of non-emptysets, and that A is dual to the variety of all abelian groups satisfying mx = 0 for some integer m.

We exhibit several other familiar varieties that can be easily shown to have FDSC using the SimpleShadowing Theorem 3.4. In these examples it is not true that every finite member generates a quasi-variety that is a variety, but many finite members do. In those cases we can apply the FDSC-HSPTheorem to show that they generate a standard topological quasi-variety.

Corollary 5.5. Syntactic congruences in the variety of groups G = 〈G; ·,−1, 1〉 are determined by thefinite set

F = {x1yx2, x1y−1x2}.

Corollary 5.6. Syntactic congruences in the variety of semigroups S = 〈S; ·〉 are determined by thefinite set

F = {y, x1y, yx2, x1yx2}.

Comparing semigroups and monoids, we again see the need for additional terms when the identityelement is missing.

Corollary 5.7. Syntactic congruences in the variety of rings R = 〈R; +, ·,−, 0〉 are determined by thefinite set

F = {x1 + y, x1 − y, x1 + x2y, x1 + yx3, x1 + x2yx3}.

From Example 2.4 we know that every finite unary algebra generates a variety with FDSC. Thefollowing observation, kindly contributed to this effort by Jane Pitkethly, exhibits many finite unaryalgebras that generate a quasi-variety that is a variety.

Proposition 5.8. Let M be a unary algebra that has at most one constant term function. Then there isa unary algebra M′ such that HSP M = HSPM′ = ISPM′, and consequently M∼

′ generates a standardtopological quasi-variety.

Proof Higgs [15] gave a short proof that the variety generated by a finite unary algebra has onlyfinitely many isomorphism types of subdirectly irreducibles, all of which are finite. Let S be a set ofdisjoint representatives of these types. We take M′ to be the coproduct of S, that is, M′ is the union ofS if M has no constant term function and is the union of S with the values of the constant term functionamalgamated otherwise. Because of the restriction on constant term functions, M′ must satisfy everyequation satisfied by M. Thus M and M′ generate the same variety, which is also ISPM′.

Recall that an Ockham algebra is an algebra O = 〈O;∨,∧, c, 0, 1〉 where 〈O;∨,∧, 0, 1〉 is a boundeddistributive lattice and the identities

c(0) ≈ 1, c(1) ≈ 0, c(x ∨ y) ≈ c(x) ∧ c(y), c(x ∧ y) ≈ c(x) ∨ c(y)


are satisfied. The variety of Ockham algebras includes Boolean algebras, Kleene algebras and De Morganalgebras as subvarieties.

Corollary 5.9. The variety generated by a finite Ockham algebra has FDSC.

Proof Let O = 〈O;∨,∧, c, 0, 1〉 be a finite Ockham algebra. Since O is finite, there is a positiveinteger m such that, for all n > m there is an i ≤ m where cn(x) ≈ ci(x) holds in O. Define

F := {fi(y, x1, x2) := (x1 ∧ ci(y)) ∨ x2 | i ≤ m}.

We use the Simple Shadowing Theorem to show that F determines syntactic congruences in HSPO. Ift(y, x1, x2, . . . ) ∈ T1y, we can find terms w1(x1, x2, . . . ) and w2(x1, x2, . . . ) and a non-negative integern such that t ≈ (w1 ∧ cn(y)) ∨w2 holds for all Ockham algebras. Let i ≤ m be such that cn(x) ≈ ci(x)holds in O. Then O satisfies the identity t(y, x1, x2, . . . ) ≈ fi(y, w1, w2).

Goldberg [13] gives an elementary way of deciding if the quasi-variety generated by a finite subdirectlyirreducible Ockham algebra is a variety. If it is, then we can conclude that the finite Ockham algebragenerates a standard topological quasi-variety.

6. Examples of general shadowing

The reader may well wonder if every instance of shadowing in familiar varieties can be achieved bysimple shadowing. In this section we present examples of varieties that we prove to have FDSC usingapplications of general shadowing. We begin with a familiar variety in which it is easy to see that simpleshadowing is not adequate.

Corollary 6.1. Syntactic congruences in the variety of groups G = 〈G; ·,−1, 1〉 are determined bythe finite set

F = {x1yx2},but this fact can not be established by simple shadowing.

Proof To apply the General Shadowing Theorem 3.3, let t(y, x1, x2, . . . ) ∈ T1y be a group term. Lett′(y, x1, x2, . . . ) be obtained from t by successively replacing inverses of products with reversed productsof inverses. Then t ≈ t′ holds in every group, and there are terms w1(x1, x2, . . . ) and w2(x1, x2, . . . ) sothat t′ is either w1yw2 or w1y

−1w2. We now show that x1yx2 syntactically shadows t.If t′ = w1yw2, then x1yx2 simply shadows t′ and therefore also t. Assume t′ = w1y

−1w2. Then wehave

t(x, x1, x2, . . . ) ≈ w1x−1w2 ≈w1x

−1yy−1w2

w1x−1xy−1w2 ≈ w1y

−1w2 ≈ t(y, x1, x2, . . . )

The fact that x1yx2 shadows the term y−1 can not be established by simple shadowing, as there do notexist terms w1(x1, x2, . . . ) and w2(x1, x2, . . . ) for which y−1 ≈ w1yw2 holds in every group.

Hidden in this syntactic shadowing argument is the exploitation of a peculiar property of groups:every equivalence on a group that is preserved by the binary operation is also preserved by the inverseoperation. The fact that x1yx2 determines syntactic congruences in groups can be established directlyfrom this property as well. Consider an equivalence θ on a group G = 〈G; ·,−1, 1〉. According toExample 2.3, θx1yx2 is a congruence on the monoid 〈G; ·, 1〉 and is therefore also a congruence on G.Thus F := {x1yx2} determines syntactic congruences in the variety of groups.


We saw in Example 2.4 that the variety generated by a finite unary algebra has FDSC. In order tocomplement this result, consider a fixed finite type of algebras with at least one non-unary operation.Murskiı [24] showed that, in a reasonable statistical sense, almost all finite algebras of this type arequasi-primal. Recall that M is quasi-primal if it has the ternary discriminator operation

t3(x, y, z) :=

{z if x = y;x if x 6= y.

as a term function, and therefore also has the normal transform

n(x, y, u, v) :=

{u if x = y;v if x 6= y.

}= t3(t3(x, y, u), t3(x, y, v), v)

as a term function.

Corollary 6.2. Syntactic congruences in the variety generated by a quasi-primal algebra are determinedby F = {n(x1, y, x2, x3)}. Consequently,

(i) the topological quasi-variety generated by a quasi-primal algebra with the discrete topology is stan-dard, and

(ii) almost every finite algebra of finite type generates a variety with FDSC.

Proof For an arbitrary term t(y, x1, x2, . . . ) we have

t(x, x1, x2, . . . ) ≈n(x, x, t(x, x1, x2, . . . ), t(y, x1, x2, . . . ))

n(x, y, t(x, x1, x2, . . . ), t(y, x1, x2, . . . )) ≈ t(y, x1, x2, . . . ).

Thus F determines syntactic congruences and we apply the FDSC-HSP Theorem to obtain (i). Then(ii) follows from (i), Example 2.4 and Murskiı [24].

Our notion of syntactic shadowing is reminiscent of a much more restrictive notion that has played animportant role in the study of certain varieties. (See Blok and Pigozzi [2], Burris and Lawrence [3] andFried, Gratzer and Quackenbush [12].) The variety V is said to have definable principal congruences(DPC) if there is a finite collection Π of formulæ π(x, y, u, v) of the form

∃−→w{x ≈ p1(z1,

−→w ) &

[&

1≤i≤kpi(z

′i,−→w ) ≈ pi+1(zi+1,

−→w )

]& pk(zk,−→w ) ≈ y

},

where {zi, z′i} = {u, v}, such that for all A ∈ V and a, b, c, d ∈ A we have (a, b) is in the principal

congruence Θ(c, d) if and only if A |= π(a, b, c, d) for some π ∈ Π.

Lemma 6.3. If a variety has definable principle congruences (DPC), then it also has finitely determinedsyntactic congruences (FDSC).

Proof Let V be a variety with DPC witnessed by the finite collection Π of formulæ. Let F consistof all terms pi(y, x1, x2, . . . ) that are used in the formulæ of Π, and let t(y, x1, x2, . . . ) ∈ Ty. We showthat F shadows t and then apply the General Shadowing Theorem 3.3. In the V-free algebra Fxy freelygenerated by x,y,x1,x2, . . . it is certainly true that

(t(x,x1,x2, . . . ), t(y,x1,x2, . . . )) ∈ Θ(x,y).

Therefore there is a π(x, y, u, v) ∈ Π such that Fxy satisfies

π(t(x,x1,x2, . . . ), t(y,x1,x2, . . . ),x,y).

This means there is a shadowing of t(y, x1, x2, . . . ) using F .

The converse of Lemma 6.3 is not true; for example, in [3], Burris and Lawrence found finite groupsand rings that generate varieties without DPC, whereas we have seen that all varieties of groups andrings have FDSC.


Lemma 6.3 gives us many new examples of varieties with FDSC. Burris and Lawrence [3] showedthat the variety I of implication algebras I = 〈I;→, 1〉 has DPC and therefore also has FDSC. SinceI = ISP2 ([23, 9]), we conclude that 2∼ = 〈2;→, 1,T 〉 is standard. McKenzie [18] found that everyvariety that is directly representable by a finite set of finite algebras has DPC. This fact applies to thevariety generated by any para primal algebra (Clark and Krauss [6]). While not all para primalalgebras generate a quasi-variety that is a variety, a method to recognize those that do is given in [6].Among those that do are all quasi-primal algebras of Corollary 6.2.

We round out this section by showing that finitely generated lattice varieties have FDSC. We beginby defining lattice terms f0 = y and fm = fm−1 ∧ xm if m is odd, and fm = fm−1 ∨ xm if m is even.The first few of these are

f0 = y

f1 = y ∧ x1

f2 = (y ∧ x1) ∨ x2

f3 = ((y ∧ x1) ∨ x2) ∧ x3

The dual of a lattice term t will be denoted by t∂ .

Lemma 6.4. In the variety of all lattices, fm+1 shadows both fm and f∂m.

Proof We first show that fm+1 ` fm.

fm(x, x1, . . . , xm) ≈ fm+1(x, x1, . . . , xm, fm(x, x1, . . . , xm))

fm+1(y, x1, . . . , xm, fm(x, x1, . . . , xm)) ≈ fm+1(x, x1, . . . , xm, fm(y, x1, . . . , xm))

fm+1(y, x1, . . . , xm, fm(y, x1, . . . , xm)) ≈ fm(y, x1, . . . , xm)

Thus fm+1 syntactically shadows fm. Now consider f∂m.

f∂m(x, x1, . . . , xm) ≈ fm+1(x, x, x1, . . . , xm)

fm+1(y, x, x1, . . . , xm) ≈ fm+1(x, y, x1, . . . , xm)

fm+1(y, y, x1, . . . , xm) ≈ f∂m(y, x1, . . . , xm)

Hence fm+1 syntactically shadows f∂m.

We wish to induct on the complexity of bracket-reduced lattice terms in which all parentheses whichare unnecessary by associativity have been omitted. We now make this precise. Formal joins and formalmeets are expressions obtained by a finite number of applications of the following rules:

(i) each variable is both a formal join and a formal meet,(ii) if k ≥ 2 and t1, . . . , tk are formal meets, then (t1 ∨ . . . ∨ tk) is a formal join,(iii) if k ≥ 2 and t1, . . . , tk are formal joins, then (t1 ∧ . . . ∧ tk) is a formal meet.An expression t is a bracket-reduced lattice term if it is either a formal join or a formal meet. As usual,we always omit the outer pair of parentheses. The a-height (short for associative height) of a bracket-reduced lattice term is defined inductively by declaring that variables have a-height 0, if t = t1∨ . . .∨ tkis a formal join and the maximum a-height of t1, . . . , tk is n, then the a-height of t is n+1, and similarlyfor formal meets. The a-height of a usual lattice term is defined to be the a-height of the correspondingbracket-reduced lattice term. Thus, for example, the a-height of the lattice term

x1 ∧ ((x1 ∨ (x2 ∨ x3)) ∧ (x2 ∨ x4))

is 2 since the corresponding bracket-reduced lattice term is

x1 ∧ (x1 ∨ x2 ∨ x3) ∧ (x2 ∨ x4).


Lemma 6.5. If t ∈ Ty is a lattice term of a-height d then, for all m > d, the term fm shadows t in thevariety of all lattices.

Proof We may assume that t involves the variable y as otherwise every term shadows t. Sincevariable replacement does not alter the a-height of a lattice term, by Lemma 3.5 we may assume thatt ∈ T1y. Let t′ be the bracket-reduced lattice term corresponding to t. If t′ is a formal join, then usingcommutativity and associativity we can write t′ = g ∨ h, where g is a formal meet involving y and his free of y. Let t′′ = g ∨ xk, where xk does not occur in t. Clearly t′′ shadows t. A dual reductionprocess produces t′′ = g ∧ xk in the case that t′ is a formal meet. Now we apply the same procedureto g. Inducting backwards in this way we obtain, up to variable substitution, either fk or f∂

k for somek ≤ d. By Lemma 6.4 it follows that fm shadows t for all m > d.

Corollary 6.6. For any class L of lattices { fm | m = 0, 1, 2, . . .} determines syntactic congruencesin L. If L has FDSC, then there exists m such that {fm} determines syntactic congruences in L.

The key to showing that the variety V(L) generated by L has FDSC is to prove there is a bound dsuch that for every term t there is a term t′ of a-height at most d such that V(L) satisfies t ≈ t′. LetFV(X) denote the free algebra in V generated by X .

Lemma 6.7. Let V be a variety of lattices, let X and Y be finite sets, and let γ : FV(Y ) � FV(X)be a surjective homomorphism such that γ(Y ) = X. Let βγ : FV(X) � FV(Y ) be the homomorphismdetermined by βγ(x) :=

∧{y ∈ Y | γ(y) = x}, for all x ∈ X. Then βγ(w) is the least element of FV(Y )

mapping to w under γ, for each w ∈ FV(X).

Proof It is easy to check that each of the following hold on the generators and hence on all elements:

γ(βγ(w)) = w for all w ∈ FV(X), (1)

βγ(γ(u)) ≤ u for all u ∈ FV(Y ). (2)

By (1), the element βγ(w) is a preimage of w. Assume that w ≤ γ(u) for some w in FV(X) and someu in FV(Y ). Then βγ(w) ≤ βγ(γ(u)) ≤ u, by (2). Thus βγ(w) is the least preimage of w.

Theorem 6.8. Let L be a finite lattice. Then there is a number d such that for each lattice term tthere is a term t′ of a-height at most d such that t ≈ t′ holds in L.

Proof By Jonsson’s Theorem [16] we can choose n so that every subdirectly irreducible lattice inV = V(L) has at most n elements: indeed, n := |L| will suffice. Let X be a set with n elements. Ofcourse each element of FV(X) can be represented by some term with variables from X . Since FV(X)is finite we can choose k such that every one of these representatives has a-height at most k. We shallprove that we can choose d := k + 2.

Now let Y be an arbitrary finite set. Let u be a join irreducible element in FV(Y ) and let u∗ beits unique lower cover. There is a surjective homomorphism ϕ : FV(Y ) � K, for some subdirectlyirreducible lattice K in V, with ϕ(u) 6= ϕ(u∗). Clearly u is the least preimage of ϕ(u). Define anequivalence relation ∼ on Y by y1 ∼ y2 if ϕ(y1) = ϕ(y2) and let X = Y/∼. Since K has at most nelements, X has at most n elements. Define γ : FV(Y ) � FV(X) to be the homomorphism determinedby γ(y) = y/∼, for all y ∈ Y . Of course ϕ can be factored through γ; that is, there is a homomorphismη : FV(X) � K with ϕ = ηγ. The least preimage maps satisfy βϕ = βγβη. By the above remarksβη(ϕ(u)) has a term representing it of a-height at most k. By the previous lemma, u = βϕ(ϕ(u)) =βγ(βη(ϕ(u))) has a-height at most k+ 1 as βγ increases the a-height by at most 1. Since every elementis a join of join-irreducibles, every element of FV(Y ) can be represented by a term of a-height at mostd := k + 2.


Theorem 6.9. Every finitely generated lattice variety has FDSC.

Proof Let V = V(L), where L is a finite lattice and let m be one greater than the number d fromTheorem 6.8. We claim fm shadows every t ∈ Ty. By Theorem 6.8 we may assume that t has a-heightat most d and so the result follows from Lemma 6.5.

7. Varieties without FDSC

Not every variety has FDSC. We will present three examples of varieties that do not have FDSC,each significant in a different way.

In Corollary 5.9 we saw that every finite Ockham algebra generates a variety with FDSC, but that weneeded a larger finite set F to determine congruences when the sequence c, c2, c3, . . . continued longerbefore it began to repeat. This suggests that no finite set F is large enough to determine syntacticcongruences on all Ockham algebras.

Proposition 7.1. The variety O of all Ockham algebras does not have FDSC.

Proof Assume that a finite subset F ⊆ Ty determines syntactic congruences on O. Since T1y |= Ty,by Theorem 2.2, we have T1y |= F and hence T1y ` F . Clearly ` induces an algebraic closure operationon Ty, so there is a finite F0 ⊆ T1y such that F0 ` F . Applying De Morgan’s laws, we observe thatevery term in T1y is simply shadowed by a term of the form (x1 ∧ ck(y))∨ x2. Thus there is a finite setF1 of terms of the form (x1 ∧ ck(y))∨ x2 which shadows F0. This gives us F1 ` F0 ` F ` Ty, so that F1

is also a finite subset of Ty that determines syntactic congruences on O. It follows that we may assumethat

F = {(x1 ∧ ck(y)) ∨ x2 | k < n} ∪ {y}for some positive integer n.

Let m ∈ N where n < 2m. Let X := {0, 1, . . . , 2m} and define γ : X → X by γ(k) = k − 1 if k > 0and γ(0) = 0. Let O be the power set of X . For each P ∈ O we define c(P ) = X \ γ−1(P ). Then

O := 〈O;∪,∩, c,∅, X〉

is an Ockham algebra. Let θ be the equivalence given by (P,Q) ∈ θ if P ∩{2m} = Q∩{2m}. Beginningwith the fact that (0,∅) ∈ θ, we see that θ is preserved by the first 2m− 1 applications of c:

Z : {0} θ ∅

c(Z) : {2, 3, . . . , 2m− 2, 2m− 1, 2m} θ X

c2(Z) : {0, 1, 2} θ ∅

c3(Z) : {4, . . . , 2m− 2, 2m− 1, 2m} θ X

...

c2m−3(Z) : {2m− 2, 2m− 1, 2m} θ X

c2m−2(Z) : {0, 1, . . . , 2m− 2} θ ∅

c2m−1(Z) : {2m} θ X

In particular, we have (ck({0}), ck(∅)) ∈ θ for k < n < 2m. Since θ is a distributive-lattice congruence,it follows that

(P1 ∧ ck({0}) ∨ P2, P1 ∧ ck(∅) ∨ P2) ∈ θ


for every P1, P2 ∈ O. Thus ({0},∅) ∈ θF . Applying (i)→(v) of Theorem 2.2, we conclude that(c2m({0}), c2m(∅)) ∈ θF ⊆ θ. However, if we continue the above computation one more step, we obtainc2m({0}) = X and c2m(∅) = ∅. This is a contradiction, as (c2m({0}), c2m(∅)) /∈ θ.

0

a

b

...

Figure 1.

Proposition 7.2. The variety of all modular lattices, and therefore the variety of all lattices, does nothave FDSC.

Proof Suppose that the variety M of all modular lattices did have FDSC. Then, by Corollary 6.6,there is an m such that {fm} determines syntactic congruences in M. Let L be the finite modularlattice of Figure 1 and let θ be the equivalence relation on L with 0 in a block by itself and all the otherelements in a second block. Since L is simple, Syn(θ) = 0. Since lattice polynomials preserve order and{fm} determines syntactic congruences, there are c1, . . . , cm ∈ L such that fm(b, c1, . . . , cm) = 0 whilefm(a, c1, . . . , cm) 6= 0. Let ak := fk(a, c1, . . . , ck) and bk := fk(b, c1, . . . , ck), for k = 0, . . . ,m. Note thatak 6= bk and hence ak > bk, for all k.

If k is odd, then ak = ak−1 ∧ ck and bk = bk−1 ∧ ck. Since ak 6= bk, the element ck cannot be belowbk−1 and this easily implies that the height in L of bk can be at most two less than the height of bk−1.Of course when k is even the height of bk is at least that of bk−1. So if we choose L long enough, bmcannot be 0, a contradiction.

The application of FDSC given in the FDSC-HSP Theorem 4.4, which provides the motivation forthis notion, concerns only varieties generated by a single finite algebra. It is therefore important to askif there is a variety with a single finite generator that does not have FDSC. The answer is affirmative,as we show with two closely related examples.

We no longer require the concept of a-height; we require the usual concept instead. The height ht(t)of a term t is the height of the term tree of t. More formally, constant symbols and variables have


height 0 and if f is an k–ary fundamental operation symbol and u1, u2, . . . , uk are terms of maximumheight n, then the term f(u1, u2, . . . , uk) has height n+ 1.

We adopt the notation of R. McKenzie [21] for reasons that will soon be apparent. Let Mω be thealgebra

Mω := 〈{0, an, bn | n ∈ N};∧, ·〉with distinct elements 0, an and bn, for each n ∈ N, and binary operations defined by

x ∧ y = 0 if x 6= y

x ∧ x = x

an · bn+1 = bn for all n ∈ N

x · y = 0 otherwise.

It is easy to check that Mω is subdirectly irreducible: every proper congruence includes the pair (0, b0).Let θ be the equivalence on the set Mω with two classes, {b0} and Mω \ {b0}. Then Syn(θ) is the

identity congruence and consequently, for each n ∈ N, we have (0, bn) /∈ Syn(θ). This fact is witnessedby the height-n term

fn(y, x0, x1, . . . , xn−1) := x0 · (x1 · (x2 · (. . . (xn−1 · y) . . . )))

since

fn(0, a0, a1, . . . , an−1) =a0(a1(a2(. . . (an−10) . . . ))) = 0

6≡θ b0 =a0(a1(a2(. . . (an−1bn) . . . ))) = fn(bn, a0, a1, . . . , an−1)

We prove that no term of lower height witnesses (0, bn) /∈ Syn(θ).

Lemma 7.3. Let f(y, x0, . . . , xj−1) ∈ Ty have height k < n; let c0, . . . , cj−1 ∈ Mω and assume thatf(0, c0, . . . , cj−1) 6= f(bn, c0, . . . , cj−1). Then f(0, c0, . . . , cj−1) = 0 and f(bn, c0, . . . , cj−1) = bm forsome m with m ≥ n− k. In particular, (0, bn) ∈ θf .

Proof The statement is clearly true if f is the height 0 term y. Assume that f has height k+ 1 < n.We consider two possibilities.

Assume that f = p ∧ q where the maximum of the heights of p and q is k. Since

f(0, c0, . . . , cj−1) 6= f(bn, c0, . . . , cj−1)

and ∧ is commutative, we may as well assume that

p(0, c0, . . . , cj−1) 6= p(bn, c0, . . . , cj−1).

By induction we conclude that

0 = p(0, c0, . . . , cj−1) = f(0, c0, . . . , cj−1)

and thatp(bn, c0, . . . , cj−1) = bm = q(bn, c0, . . . , cj−1) = f(bn, c0, . . . , cj−1)

for some m with m ≥ n− k > n− (k + 1).Now assume that f = p · q where again the maximum of the heights of p and q is k. Either

p(0, c0, . . . , cj−1) 6= p(bn, c0, . . . , cj−1) or q(0, c0, . . . , cj−1) 6= q(bn, c0, . . . , cj−1).

By induction f(0, c0, . . . , cj−1) = 0 in either case. Since f(bn, c0, . . . , cj−1) 6= 0 we have, for some m,that p(bn, c0, . . . , cj−1) = am, that q(bn, c0, . . . , cj−1) = bm+1 and that f(bn, c0, . . . , cj−1) = bm. As am

is neither a product nor a proper meet, p(y, x0, . . . , xj−1) must be a variable or meet of variables notinvolving y. Thus p(0, c0, . . . , cj−1) = am, so

q(0, c0, . . . , cj−1) 6= q(bn, c0, . . . , cj−1).

By induction, m+ 1 ≥ n− ht(q) ≥ n− k giving us m ≥ n− (k + 1).


Corollary 7.4. No finite set of terms determines the syntactic congruence of the equivalence θ on thealgebra Mω.

Proof If F is a finite set of terms, then we can choose an upper bound n for the heights of the termsin F . By Lemma 7.3 we have (0, bn) ∈ ∩{θf | f ∈ F} = θF , and therefore θF is not Syn(θ), the identitycongruence.

A significant class of finite algebras of current interest were constructed by McKenzie [21] for thepurpose of resolving certain algebraic decidability questions. McKenzie gave an effective procedure toconstruct, for each Turing machine T , a finite algebra A(T ) of finite type for which he established thefollowing facts.

McKenzie’s Theorem 7.5. [21] Let T be a Turing machine.

(i) If T eventually halts when started on a blank tape, then there is a finite bound on the sizes of thesubdirectly irreducibles in the variety V(A(T )) generated by A(T ).

(ii) If T does not halt when started on a blank tape, then the variety V(A(T )) contains an infinitesubdirectly irreducible.

(iii) The algebra A(T ) has a distinguished element 0 which determines a one-element subalgebraof A(T ).

Thus, for example, it is recursively undecidable as to whether or not the variety generated by a finitealgebra of finite type is residually finite.

Theorem 7.6. If the Turing machine T does not halt, then the variety generated by A(T ) does nothave FDSC.

Proof In [21, Lemma 4.1] McKenzie showed that V(A(T )) contains an algebra SZ which has asubalgebra Sω that is term equivalent to Mω. The result then follows form Corollary 7.4.

We will return to A(T ) and several applications of McKenzie’s Theorem 7.5 in the next section.Meanwhile, we give a self contained proof that a related finitely generated variety does not have FDSC.Let Jω be the algebra

Jω := 〈{0, 1, an, bn | n ∈ N}; ·〉obtained from Mω by deleting ∧ and adding a two-sided identity element 1. Then Jω is also subdirectlyirreducible as every proper congruence contains (0, b0). Let ψ be the equivalence on Jω with the twoclasses {b0} and Jω \ {b0}. Then Syn(ψ) is the identity congruence.

Lemma 7.7. No finite set of terms determines the syntactic congruence of the equivalence ψ on thealgebra Jω.

Proof Let F ⊆ Ty be a finite set of terms, and let n be an upper bound for the heights of the terms in F .We claim that (0, bn) ∈ ψF . Let f(y, x0, . . . , xl−1) ∈ F and let c0, . . . , cl−1 ∈ Mω. If f(0, c0, . . . , cl−1)and f(bn, c0, . . . , cl−1) are equal, they are related by ψ. Assume f(0, c0, . . . , cl−1) 6= f(b0, c0, . . . , cl−1).Let g(y, x0, . . . , xl−1) be obtained from f(y, x0, . . . , xl−1) by deleting xi if ci = 1, and then removingthe extraneous parentheses. For i < l, let di := ci if ci 6= 1 and let di := 0 if ci = 1. Then

g(0, d0, . . . , dl−1) = f(0, c0, . . . , cl−1) 6= f(0, c0, . . . , cl−1) = g(0, d0, . . . , dl−1).

Since g is also an Mω-term with 0, d0, . . . , dl−1 ∈Mω and ht(g) ≤ ht(f) ≤ n, we can apply Lemma 7.3to g to conclude that f(0, c0, . . . , cl−1) = 0 and f(bn, c0, . . . , cl−1) = bm with m ≥ n− ht(g) > 0. Thus(0, bn) ∈ ψF and consequently ψF 6= Syn(ψ).


For 1 < n < ω we define the finite subalgebra

Jn := 〈{0, 1, a0, . . . , an, b0, . . . , bn+1}; ·〉of Jω. Notice that each Jn is also subdirectly irreducible and that Jω is the nested union

Jω =⋃

{Jn | 1 < n < ω}.

Lemma 7.8. The algebras Jκ, where 1 < κ ≤ ω, each generate the same variety VJ.

Proof This will follow immediately if we can show that, for 1 ≤ n < ω, we have Jn+1 ∈ HS(Jn ×Jn).We define

1 := {(1, 1)}, 0 := {(a0, bn)} ∪ (Jn × {0}) ∪ ({0} × Jn)

A0 := {(a0, 1)}, A1 := {(a1, a0)}, . . . , An−1 := {(an−1, an−2)}, An := {(bn, an−1)},B0 := {(b0, b0)}, B1 := {(b1, b0)}, . . . , Bn−1 := {(bn−1, bn−2)}, Bn := {(bn, bn−1)},Bn+1 := {(1, bn)}.

It is then easy to check that• Kn := 1 ∪A0 ∪A1 ∪ · · · ∪An ∪B0 ∪B1 ∪ · · · ∪Bn+1 ∪ 0 forms a subalgebra Kn of Jn × Jn,• 0 is an ideal in Kn and• the quotient Kn/0 = 〈{1, A0, A1, . . . , An, B0, B1, . . . , Bn+1,0}; ·〉 is isomorphic to Jn+1 under

that map 0 7→ 0, 1 7→ 1, Ai 7→ ai, Bj 7→ bj.

Theorem 7.9. The variety VJ generated by each algebra Jκ, where 1 < κ ≤ ω, does not have FDSC.

The algebra Jn was in fact constructed independently of McKenzie’s work. It arose as the algebra ofall subterms of the term tn := x0(x1(. . . (xn−2(xn−1y)) . . . )) together with a two-sided identity 1 and a0 as the value of all products that do not form a subterm.

8. Lingering questions

The work presented in this paper is motivated by the following question.

Problem 8.1. Is there an algorithm to decide if a given finite algebra of finite type generates a standardtopological quasi-variety?

Our central FDSC-HSP Theorem 4.4 imposes two hypotheses on the finite algebra M,(i) the quasi-variety generated by M is a variety and(ii) the variety generated by M has FDSC,

which together imply that X = IScP+ M∼ is standard. We conclude by raising several questions suggested

by this theorem.Are conditions (i) and (ii) necessary for X to be standard? Davey and Talukder [8] show that

they are not by constructing a three element unary algebra M for which (i) fails but X is neverthelessstandard. This example shows that there is more to understand about standardness of algebras than theFDSC-HSP Theorem reveals. It seems unlikely that Problem 8.1 will be answered without first gainingan understanding of the extent to which standardness extends beyond applications of the FDSC-HSPTheorem.

Are conditions (i) and (ii) independent? By Corollary 5.6 we know that every variety of semigroupshas FDSC. Yet the studies of residually finite semigroups in Golubov and Sapir [14] and residually smallvarieties of semigroups in McKenzie [19] provide many examples of finite semigroups M generatingvarieties that are not even residually small, and therefore certainly do not satisfy HSPM = ISPM. Onthe other hand we do not at this time have an example of a finite algebra M satisfying (i) but not (ii).


Problem 8.2. If the quasi-variety generated by a finite algebra M is a variety, must this variety haveFDSC?

Our Theorem 6.9 for finite lattices suggests that perhaps (ii) follows from a condition weaker than (i).

Problem 8.3. Let M be a finite algebra. If M generates a congruence-distributive variety, or moregenerally, if the variety generated by M has only finitely many subdirectly irreducibles all of which arefinite, must this variety have FDSC?

We can also inquire as to whether or not we can effectively determine if the FDSC-HSP Theorem isapplicable.

Problem 8.4. Is there an algorithm to decide if the quasi-variety generated by a finite algebra of finitetype is a variety?

Note that HSPM = ISPM if and only if each subdirectly irreducible of HSPM embeds into M. Thuswe can decide this equality for the algebras M in a given class K if, within K, we have

(1) an effective method to determine if HSPM is residually finite and, in case it is,(2) an effective method to compute a number f(M) ∈ N such that HSPM is residually less than f(M).

Indeed, if HSP M is residually less than f(M), then every subdirectly irreducible of HSP M is isomorphicto a quotient of the free algebra in HSPM on f(M) generators. We can then effectively (althoughlaboriously) check if each subdirectly irreducible quotient of this finite free algebra embeds into M.

McKenzie and Freese [11, Theorem 10.15] exhibited (1) and (2) for those finite algebras M generatinga congruence modular variety, giving f as f(M) = |M |+ |M ||M||M|+3

. Higgs [15] gave a short argumentthat if M = 〈M ;F 〉 is a unary algebra, then HSPM is residually less than 2n+1, where n is the sizeof the one-generated free algebra in HSPM. Algorithms (1) and (2) can be extracted for semigroupsfrom the papers Golubov and Sapir [14] and McKenzie [19], and for rings from McKenzie [20].

In spite of these far reaching results, a positive answer to Problem 8.4 can not be established for allfinite algebras by means of (1) and (2) alone. Indeed, McKenzie’s Theorem 7.5 shows that (1) does nothold in the class of all finite algebras.

Problem 8.5. Is there an algorithm to decide if the variety generated by a finite algebra of finite typehas FDSC?

In view of Theorem 7.6 and McKenzie’s Theorem 7.5, we could conclude that there was no such algorithmif we could show that the variety generated by A(T ) has FDSC when T does halt. An alternate proofthat there is no such algorithm might come from a link between Problems 8.4 and 8.5.

Proposition 8.6. At least one of the following statements is true.

(i) Some finite algebra generates a quasi-variety that is a variety, but that variety does not haveFDSC.

(ii) There is no algorithm to decide if an arbitrary finite algebra of finite type generates a variety withFDSC.

Proof Assume (i) fails, that is, HSP M = ISPM implies that HSPM has FDSC for every finitealgebra M. We prove (ii) by arguing that, for a Turing machine T , the algebra A(T ) generates avariety with FDSC if and only if T halts.

If T does not halt, then the variety generated by A(T ) does not have FDSC by Theorem 7.6. Assumethat T halts. Applying McKenzie’s Theorem 7.5(i) and (iii), we take A(T )′ to be the direct productof the non-isomorphic subdirectly irreducibles in HSPA(T ). Then A(T )′ is a finite algebra containingcopies of all the subdirectly irreducibles in HSPA(T ), and HSPA(T ) = HSP A(T )′ = ISPA(T )′ hasFDSC by our assumption.


On two different occasions we have had to construct, from a finite algebra M, a finite algebra M′

for which HSPM = HSPM′ = ISPM′: in the proofs of Propositions 5.8 and 8.6. This constructionsuggests a variant of Problem 8.4 which is also answered by McKenzie’s Theorem.

Proposition 8.7. There is no algorithm to decide, given a finite algebra M of finite type, if there is afinite algebra M′ such that HSPM = HSPM′ = ISPM′.

Proof From McKenzie’s Theorem 7.5 we see that, for a Turing machine T , there is a finite algebraA(T )′ such that HSPA(T ) = HSPA(T )′ = ISPA(T )′ if and only if T halts; namely, take A(T )′ tobe the direct product of the non-isomorphic subdirectly irreducibles in HSPA(T ).

Yet another variant of Problem 8.4 is to ask for an algorithm to decide when a given finite set ofquasi-identities determines a variety. McNulty [22, Theorem 18] shows that this also is undecidable.

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SUNY, New Paltz, NY 12561, U.S.A.

E-mail address: [email protected]

La Trobe University, Victoria 3086, Australia


University of Hawaii, Honolulu, HI 96822, U. S. A


La Trobe University, Victoria 3086, Australia


standard topological algebras and syntactic congruences

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