the homomorphism lattice induced by a finite algebra...some examples i the case where c is the...

The homomorphism latticeinduced by a finite algebra

Brian A. Davey, Charles T. Gray and Jane G. Pitkethly

AAA94+NSAC 2017

The homomorphism order on a category

I Let C be a category.I Define a quasi-order→ on the objects of C by X→ Y if

there exists a morphism from X to Y.I Now set

X ≡ Y ⇐⇒ X→ Y and Y→ X.

Then→ induces an order on the class C/≡.

I To ensure that C/≡ is a set we take C to be a class of finitestructures with all homomorphisms between them.

I Then we can define the ordered set

PC := 〈C/≡;→〉,

which we refer to as the homomorphism order on C.

The homomorphism order on a category

I Let C be a category.I Define a quasi-order→ on the objects of C by X→ Y if

there exists a morphism from X to Y.I Now set

X ≡ Y ⇐⇒ X→ Y and Y→ X.

Then→ induces an order on the class C/≡.I To ensure that C/≡ is a set we take C to be a class of finite

structures with all homomorphisms between them.I Then we can define the ordered set

PC := 〈C/≡;→〉,

which we refer to as the homomorphism order on C.

Categories with product and coproduct

I PC := 〈C/≡;→〉, where X ≡ Y ⇐⇒ X→ Y and Y→ X.

I If the category C has pairwise products and coproducts,then PC is a lattice:

I the meet of X/≡ and Y/≡ is (X× Y)/≡,I the join of X/≡ and Y/≡ is (X t Y)/≡.

Some examples

I The case where C is the category of finite graphs orfinite digraphs has been studied extensively:

I in both cases, PC = 〈C/≡;→〉 is a distributive lattice(in fact, a Heyting algebra) into which every countableordered set can be embedded.

I We are interested in the case where C is a category offinite algebras.

I For many natural categories of finite algebras, the orderedset PC is trivial as there are homomorphisms betweenevery pair of objects in C. For example:

I groups, vector spaces over a finite field, rings, lattices,semigroups.

Some examples

Locally finite varieties

I Let A be a variety of algebras and let Afin denote theclass of finite members of A.

I If Afin is closed under forming finite coproducts,then PAfin is a lattice.

I Hence, it is natural to look at locally finite varieties.

LemmaLet A be a locally finite variety. Then the homomorphism orderPAfin = 〈Afin/≡;→〉 is a countable bounded lattice.

NoteI The top of PAfin is the equivalence class consisting of

algebras in Afin that have a trivial subalgebra.I The bottom of PAfin is the equivalence class that contains

all finitely generated free algebras in A.

The homomorphism lattice induced by a finite algebra

FactLet A be a finite algebra. Then the variety Var(A) generated byA is locally finite.

NotationGiven a finite algebra A, we define the lattice

LA := 〈Var(A)fin/≡;→〉

and refer to it as the homomorphism lattice induced by A.

QuestionI Which lattices arise as LA for some finite algebra A?

Warm-up exercisesExample 1. Representing finite chains

I The 2-element chain:

A = 〈{0,1}; f 〉

LA ∼= 2

A = 〈{0,1,2,3}; f 〉

0 1 2 3f

LA ∼= 3

I In general, if A = 〈{0,1, . . . ,2n−1 − 1}; f 〉, where f issuccessor modulo 2n−1, then LA ∼= n.

A = 〈{0,1}; f 〉

LA ∼= 2

A = 〈{0,1,2,3}; f 〉

0 1 2 3f

LA ∼= 3

A = 〈{0,1}; f 〉

LA ∼= 2

A = 〈{0,1,2,3}; f 〉

0 1 2 3f

LA ∼= 3

An interlude on cores

I A finite algebra C is a core if every endomorphism of C isan automorphism.

I For each finite algebra B, there is a subalgebra C (uniqueup to isomorphism) such that C is a core and there is aretraction ϕ : B→ C.

I If we let K consist of one copy (up to isomorphism) of eachcore in Var(A)fin, then K is a transversal of the equivalenceclasses of LA.

I Moreover, the cores are precisely the algebras that areminimal-sized within their equivalence classes.

I So the lattice LA is finite if and only if there is a finite boundon the sizes of the cores in Var(A)fin.

Warm-up exercisesExample 2. The homomorphism lattice induced by a finite monounary algebra

Questions1. Which lattices can be obtained as LA, for some finite

monounary algebra A?

2. Can we obtain the lattice 22 in this way?

LemmaLet A = 〈A; f 〉 be a finite monounary algebra. Then, for some `and some k1, . . . , k`,

LA ∼= (k1 t · · · t k`)⊕ 1,

where the coproduct is in the variety D of distributive lattices.(Let n be the least common multiple of the sizes of the cyclesof A. Then pk1

1 · · · pk`` is the prime decomposition of n.)

I So the answer to Question 2 is ‘No’.

monounary algebra A?2. Can we obtain the lattice 22 in this way?

LA ∼= (k1 t · · · t k`)⊕ 1,

Warm-up exercisesExample 3. Representing the lattice 22

A = 〈{0,a,b,1};∨,∧,0,1,a,b〉

LA ∼= 22

Warm-up exercisesExample 4. Representing finite partition lattices

I To represent the lattice M3, we can use the nullary algebraA = 〈{1,2,3};1,2,3〉:

2=31=2 3=1

LA ∼= M3

I If A = 〈{1,2, . . . ,n};1,2, . . . ,n〉, then LA ∼= Equiv(n).

QuestionsI Can we represent each finite distributive lattice as LA?I We have represented M3. What about N5?

We will answer both of these questions later in the talk.

2=31=2 3=1

LA ∼= M3

2=31=2 3=1

LA ∼= M3

How to find complicated examplesFinite-to-finite universal varieties

I A variety V of algebras is finite-to-finite universal if thecategory G of digraphs has a finiteness-preserving fullembedding into V.

I This is equivalent to requiring that every variety of algebrashas a finiteness-preserving full embedding into V.

I If Var(A) is finite-to-finite universal, thenI there is an order-embedding of PGfin into LA,I every countable ordered set order-embeds into LA,I in particular, LB order-embeds into LA, for all finite B.

Finite-to-finite universal varieties

Finite algebras A with Var(A) finite-to-finite universal:

1. The bounded lattice M3. [Goralcík, Koubek, Sichler 1990]

2. The algebra A = 〈A;∨,∧,0,1,a,b〉, where〈A;∨,∧,0,1〉 is the bounded distributive lattice 1⊕ 22 ⊕ 1freely generated by {a,b}.[Uses Adams, Koubek, Sichler 1985]

3. Let P = 〈{0,1,2}; ·〉 be the semigroupshown on the right.

Let A = 〈A; ·,a,b, c〉, where 〈A; ·〉 isfreely generated in Var(P)by {u, v ,a,b, c}.[Uses Demlová, Koubek 2003]

· 0 1 20 0 0 01 0 1 22 0 0 0

Even a small unary algebra can generate a finite-to-finiteuniversal variety.

4. Let A = 〈{0,1,2,u, v}; f1, f2〉 be the 5-element unaryalgebra shown below.

(a,b)(a,a) (b, c) (c,b)

The map G 7→ G∗ embeds the category G of digraphs intothe category Var(A). [Uses Hedrlín, Pultr 1966]

Distributive lattices via quasi-primal algebras

A finite algebra A is quasi-primal if the ternary discriminatoroperation τ is a term function of A, where

τ(x , y , z) :=

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

Theorem (Pixley 1970)A finite algebra A is quasi-primal if and only if every non-trivialsubalgebra of A is simple and the variety Var(A) is bothcongruence permutable and congruence distributive.

Theorem (Pixley 1970)Let A be a quasi-primal algebra. Then every finite algebra inVar(A) is isomorphic to a product of subalgebras of A.

NotationI We use Sub(A) to denote the set of all subalgebras of an

algebra A. Note that 〈Sub(A);6〉 is a join-semilattice.I For an ordered set P, we use O(P) to denote the lattice of

all down-sets of P, ordered by inclusion.

Theorem (Birkhoff 1933)For every finite distributive lattice L, there exists a finite orderedset P such that L ∼= O(P). Indeed,

L ∼= O(J (L)),

where J (L) is the ordered set of join-irreducible elements of L.

TheoremLet A be a quasi-primal algebra. Define the ordered set

Q := 〈Sub(A)/≡;→〉

and let Q denote Q without its top.a. If A has no trivial subalgebras, then LA ∼= O(Q).b. If A has a trivial subalgebra, then LA ∼= O(Q).

This result gives a strategy to show that every finite distributivelattice L arises as LA, for some finite algebra A:

I Let P = J (L). Hence, O(P) ∼= L.I Let P> denote P with a new top element > added.I Construct a quasi-primal algebra A, with a trivial

subalgebra, such that 〈Sub(A)/≡;→〉 ∼= P>. Then

LA ∼= O(P>) = O(P) ∼= L.

TheoremLet A be a quasi-primal algebra. Define the ordered set

Q := 〈Sub(A)/≡;→〉

and let Q denote Q without its top.a. If A has no trivial subalgebras, then LA ∼= O(Q).b. If A has a trivial subalgebra, then LA ∼= O(Q).

This result gives a strategy to show that every finite distributivelattice L arises as LA, for some finite algebra A:

I Let P = J (L). Hence, O(P) ∼= L.I Let P> denote P with a new top element > added.I Construct a quasi-primal algebra A, with a trivial

subalgebra, such that 〈Sub(A)/≡;→〉 ∼= P>. Then

LA ∼= O(P>) = O(P) ∼= L.

Interlude: a motivating example

Lemma (Birkhoff, Frink 1948)For each finite join-semilattice S, there exists a finite algebra Asuch that 〈Sub(A);6〉 is order-isomorphic to S.

Proof.Let S = 〈S;∨〉 be a finite semilattice.

I For each covering pair a ≺ b in S,define fba : S → S by

fba(x) :=

{a if x = b,x otherwise.

x 6= bfba

I Now define F := { fba | a ≺ b in S } and A := 〈S;∨,F 〉.I Let ↓x denote the subalgebra of A with universe ↓x .I Then α : S→ Sub(A), given by α(x) := ↓x , is an

order-isomorphism.

fba(x) :=

x 6= bfba

order-isomorphism.

fba(x) :=

x 6= bfba

order-isomorphism.

fba(x) :=

x 6= bfba

I Now define F := { fba | a ≺ b in S } and A := 〈S;∨,F 〉.

I Let ↓x denote the subalgebra of A with universe ↓x .I Then α : S→ Sub(A), given by α(x) := ↓x , is an

order-isomorphism.

fba(x) :=

x 6= bfba

I Now define F := { fba | a ≺ b in S } and A := 〈S;∨,F 〉.I Let ↓x denote the subalgebra of A with universe ↓x .

I Then α : S→ Sub(A), given by α(x) := ↓x , is anorder-isomorphism.

fba(x) :=

x 6= bfba

order-isomorphism.

From L to the quasi-primal algebra A

1. Let L be a finite distributive lattice.

2. Let P = J (L).3. Let P> denote P with a new top element > added.

1. Let L be a finite distributive lattice.

2. Let P = J (L).3. Let P> denote P with a new top element > added.

1. Let L be a finite distributive lattice.2. Let P = J (L).

3. Let P> denote P with a new top element > added.

1. Let L be a finite distributive lattice.2. Let P = J (L).3. Let P> denote P with a new top element > added.

4. Let S be the universal covering tree of the ordered set P>.

5310 5320 6310 6320 6410 6420

310 320 410 420

5. As a first approximation to A, let A1 = 〈S;∨,F1, τ〉, whereF1 := { fba | a ≺ b in S } and τ is the ternary discriminator.

5. A1 = 〈S;∨,F1, τ〉, where F1 := { fba | a ≺ b in S } and τ isthe ternary discriminator.

Bad: We have ↓x 6∼= ↓y for all x 6= y in S\Min(S),but want ↓x ∼= ↓y whenever ϕ(x) = ϕ(y).

Fix: Redefine fba and index via covers a ≺ b in P>.

6. A2 = 〈S;∨,F2, τ〉, where F2 := { fba | a ≺ b in P> } and τ isthe ternary discriminator.

Good: We now have ↓x ∼= ↓y whenever ϕ(x) = ϕ(y).Bad: All subalgebras of A2 are homomorphically equivalent.

Fix: Let S] be S with every minimal element doubled.Then add operations gm, indexed by Min(P>), that flip thenew and old minimal elements in an appropriate way.

Good: We now have ↓x ∼= ↓y whenever ϕ(x) = ϕ(y).Bad: All subalgebras of A2 are homomorphically equivalent.Fix: Let S] be S with every minimal element doubled.

Then add operations gm, indexed by Min(P>), that flip thenew and old minimal elements in an appropriate way.

Good: We now have ↓x ∼= ↓y whenever ϕ(x) = ϕ(y).Bad: All subalgebras of A2 are homomorphically equivalent.Fix: Let S] be S with every minimal element doubled.

Then add operations gm, indexed by Min(P>), that flip thenew and old minimal elements in an appropriate way.

7. A3 = 〈S];∨,F2,G, τ〉, where F2 := { fba | a ≺ b in P> },G = {gm | m ∈ Min(P>) }, τ is the ternary discriminator.

Good: Now every subalgebra of A3 is of the form ↓x , for somex ∈ S, and ↓x ≡ ↓y ⇐⇒ ↓x ∼= ↓y ⇐⇒ ϕ(x) = ϕ(y).

Bad: A3 has no trivial subalgebras.Fix: Remove all f>a from F2

and add an operation h so that(∗) {>} is the only new subuniverse.

Bad: A3 has no trivial subalgebras.

Fix: Remove all f>a from F2

Bad: A3 has no trivial subalgebras.Fix: Remove all f>a from F2

Bad: A3 has no trivial subalgebras.Fix: Remove all f>a from F2 and add an operation h so that

(∗) {>} is the only new subuniverse.

8. A = 〈S];∨,F3,G,h, τ〉, where F3 := { fba | a ≺ b in P },G = {gm | m ∈ Min(P>) }, h has property (∗), and τ is theternary discriminator.

Now every non-trivial subalgebra of A is of the form ↓x , forsome x ∈ S, and ↓x → ↓y if and only if ϕ(x) 6 ϕ(y).Thus A is a quasi-primal algebra, with a trivial subalgebra, suchthat 〈Sub(A)/≡;→〉 ∼= P>. Hence LA ∼= O(P>) = O(P) ∼= L.

Now every non-trivial subalgebra of A is of the form ↓x , forsome x ∈ S, and ↓x → ↓y if and only if ϕ(x) 6 ϕ(y).

Thus A is a quasi-primal algebra, with a trivial subalgebra, suchthat 〈Sub(A)/≡;→〉 ∼= P>. Hence LA ∼= O(P>) = O(P) ∼= L.

Now every non-trivial subalgebra of A is of the form ↓x , forsome x ∈ S, and ↓x → ↓y if and only if ϕ(x) 6 ϕ(y).Thus A is a quasi-primal algebra, with a trivial subalgebra, suchthat 〈Sub(A)/≡;→〉 ∼= P>. Hence LA ∼= O(P>) = O(P) ∼= L.

TheoremFor each finite distributive lattice L, there exists a quasi-primalalgebra A such that LA is isomorphic to L.

The class Lhom

NotationLet Lhom be the class of all lattices L such that L ∼= LA, for somefinite algebra A.

QuestionsI We wish to know which (finite) lattices belong to Lhom.I What general properties does the class Lhom have?

For example:I We know Lhom is closed under finite products.I It is not clear whether Lhom is closed under forming

homomorphic images or taking sublattices.

Congruence lattices

Finite Lattice Representation ProblemDoes every finite lattice arises as the congruence lattice of afinite algebra?

I This is one of the most famous unsolved problems inuniversal algebra.

I It is therefore natural to ask whether the congruence latticeof each finite algebra belongs to Lhom.

I That is, given a finite algebra A, does there exist a finitealgebra B such that LB ∼= Con(A)?

I We focus on the special case where LA ∼= Con(A).

Congruence lattices

I We can assume, without affecting the lattice Con(A), thatevery element of A is the value of a nullary term function.

I Then, for all B ∈ Var(A), the values of the nullary termfunctions of B form a subalgebra that we will denote by B0.

Congruence Lattice LemmaLet A be a finite algebra such that each element is the value ofa nullary term function. Then the following are equivalent:

1. LA ∼= Con(A);2. for every finite algebra B ∈ Var(A), we have B→ B0;3. a. B→ B0, for every finite subdirectly irreducible algebra

B ∈ Var(A), andb. (A/θ1)× (A/θ2)→ A/(θ1 ∩ θ2), for all θ1, θ2 ∈ Con(A).

Congruence lattices

I We can assume, without affecting the lattice Con(A), thatevery element of A is the value of a nullary term function.

I Then, for all B ∈ Var(A), the values of the nullary termfunctions of B form a subalgebra that we will denote by B0.

Congruence Lattice LemmaLet A be a finite algebra such that each element is the value ofa nullary term function. Then the following are equivalent:

1. LA ∼= Con(A);2. for every finite algebra B ∈ Var(A), we have B→ B0;3. a. B→ B0, for every finite subdirectly irreducible algebra

Subspace lattices and partition lattices

Corollary of (2)⇒ (1)Let A be a finite algebra.

I Assume that every finite algebra in Var(A) has a retractiononto each of its subalgebras.

I Let A+ be the algebra obtained from A by making eachelement of A the value of a nullary operation.

Then LA+ ∼= Con(A), whence Con(A) ∈ Lhom.

Applications of the corollaryI Let V be a finite dimensional vector space over a finite

field. The lattice of subspaces of V belongs to Lhom.I Let S be a finite set. The lattice of equivalences on S

belongs to Lhom. That is, the class Lhom contains all finitepartition lattices.

field. The lattice of subspaces of V belongs to Lhom.

I Let S be a finite set. The lattice of equivalences on Sbelongs to Lhom. That is, the class Lhom contains all finitepartition lattices.

field. The lattice of subspaces of V belongs to Lhom.I Let S be a finite set. The lattice of equivalences on S

belongs to Lhom. That is, the class Lhom contains all finitepartition lattices.

Partition lattices

Theorem (Pudlák, Tuma 1980)Every finite lattice embeds into a finite partition lattice.

ConsequencesI Every finite lattice embeds into a finite lattice in Lhom.I Thus, if we could show that Lhom were closed under

forming sublattices, then it would follow that every finitelattice belongs to Lhom.

I Since the variety of lattices is generated by its finitemembers, the only equations true in the class Lhom arethose true in all lattices.

I In fact, more is true. The only universal first-ordersentences true in the class Lhom are those true in alllattices.

Partition lattices

Subgroup lattices

The following are equivalent [Pálfy, Pudlák 1980]:

I every finite lattice arises as the congruence lattice of afinite algebra;

I every finite lattice arises as an interval [H,G] in thesubgroup lattice of a finite group G.

So it is natural to investigate which intervals in subgroup latticesbelong to Lhom.

Application of the lemma (Keith Kearnes)I Let G be a group and let A = 〈A; {λg | g ∈ G}〉 be a G-set,

with at least one singleton orbit, regarded as a unaryalgebra. Then LA+ ∼= Con(A).

I If A = G/H ∪ {∞} with the natural left action of G oncosets and with∞ fixed, then LA+ ∼= Con(A) ∼= [H,G]⊕ 1.

Subgroup lattices

The following are equivalent [Pálfy, Pudlák 1980]:

I every finite lattice arises as the congruence lattice of afinite algebra;

I every finite lattice arises as an interval [H,G] in thesubgroup lattice of a finite group G.

So it is natural to investigate which intervals in subgroup latticesbelong to Lhom.

Application of the lemma (Keith Kearnes)I Let G be a group and let A = 〈A; {λg | g ∈ G}〉 be a G-set,

with at least one singleton orbit, regarded as a unaryalgebra. Then LA+ ∼= Con(A).

I If A = G/H ∪ {∞} with the natural left action of G oncosets and with∞ fixed, then LA+ ∼= Con(A) ∼= [H,G]⊕ 1.

The lattice N5

I With the exception of N5, we now know that all lattices ofsize up to 5 belong to Lhom.

I Our aim now is to give an example of a finite algebra Asuch that LA ∼= N5.

I We shall apply (3)⇒ (1) of the Congruence LatticeLemma.

Congruence Lattice Lemma

1. LA ∼= Con(A);3. a. B→ B0, for every finite subdirectly irreducible algebra

The lattice N5

I With the exception of N5, we now know that all lattices ofsize up to 5 belong to Lhom.

I Our aim now is to give an example of a finite algebra Asuch that LA ∼= N5.

I We shall apply (3)⇒ (1) of the Congruence LatticeLemma.

Congruence Lattice Lemma

1. LA ∼= Con(A);3. a. B→ B0, for every finite subdirectly irreducible algebra

Distributive bisemilattices

I An algebra A = 〈A;∧,u〉 is a distributive bisemilattice ifboth ∧ and u are semilattice operations on A and eachdistributes over the other.

I Up to isomorphism, the subdirectly irreducible distributivebisemilattices are D, S and L. [Kalman 1971]

I S is term equivalent to the two-element semilattice.I L is the two-element lattice.

I S is term equivalent to the two-element semilattice.

I L is the two-element lattice.

The lattice N5

ExampleLet A = 〈{0,a,b,1};∧,u,0,a,b,1〉 be the distributivebisemilattice S× L with all four elements added as nullaries.Then LA ∼= N5.

(x , y)

(y , y)

(x , z)

(y , z)

∧(x , z)

(x , y)

(y , z)

(y , y)

uThe distributive bisemilattice S× L

I The bisemilattice S× L is a reduct of the 4-elementdistributive bilattice made famous by Belnap (1977).

The lattice N5

(x , y)

(y , y)

(x , z)

(y , z)

(x , z)

(x , y)

(y , z)

(y , y)

uThe algebra A

The lattice N5

(x , y)

(y , y)

(x , z)

(y , z)

(x , z)

(x , y)

(y , z)

(y , y)

uThe algebra A

Proof of the N5 example

I The algebra A:

I The congruences on A:

Con(A)

I The congruences on A:

Con(A)

I The corresponding quotients:

∧A/β 0b

∧A/γ b1

∧A/α 0b

Skip proof

∧A/β 0b

∧A/γ b1

∧A/α 0b

I Since α, β and γ are meet-irreducible in Con(A), it followsthat A/α, A/β and A/γ are subdirectly irreducible algebrasin Var(A).

I We claim that, up to isomorphism, these are the onlysubdirectly irreducible algebras in Var(A).

I Note that each of these algebras satisfies B = B0.

∧A/β 0b

∧A/γ b1

∧A/α 0b

∧A/β 0b

∧A/γ b1

∧A/α 0b

∧S x

∧L z

∧D x

I 0 is the bottom for ∧I 1 is the top for ∧I a is the top for uI b is the bottom for u 0

∧S x

∧L z

∧D x

∧S x

∧L z

∧D x

∧S x

∧L z

∧D x

∧S x

∧L z

∧D x

∧S x

∧L z

∧D x

∧S x

∧L z

∧D x

∧S x

∧L z

∧D x

I Hence every subdirectly irreducible algebra in Var(A) isisomorphic to one of A/α, A/β and A/γ.

I So (3a) of the Congruence Lattice Lemma holds:i.e., every subdirectly irreducible algebra B in Var(A)satisfies B→ B0 (indeed, B = B0).

I We now check that (3b) holds:i.e., for all θ1, θ2 ∈ Con(A),

(A/θ1)×(A/θ2)→ A/(θ1∧θ2).0A

Con(A)

I Since α 6 β and β ∧ γ = 0A, it is enough to show that(A/β)× (A/γ) ∼= A, which follows immediately from theeasily checked fact that β ∧ γ = 0A and β · γ = 1A.

I Hence LA ∼= Con(A) ∼= N5, by the Congruence LatticeLemma.

(A/θ1)×(A/θ2)→ A/(θ1∧θ2).0A

Con(A)

(A/θ1)×(A/θ2)→ A/(θ1∧θ2).0A

Con(A)

(A/θ1)×(A/θ2)→ A/(θ1∧θ2).0A

Con(A)

(A/θ1)×(A/θ2)→ A/(θ1∧θ2).0A

Con(A)

ProblemsRepresentation

I Does every countable bounded lattice belong to Lhom?I In particular, does Lhom contain:

I every finite lattice?I the congruence lattice of every finite algebra?I every interval in the subgroup lattice of a finite group?

I Is every finite homomorphism lattice LA representable asthe congruence lattice of a finite algebra?

FinitenessI For which finite algebras A is the lattice LA finite?

Is this decidable?

Classes of algebrasI Restrict to natural classes of algebras. For example:

unary algebras, semigroups with constants.

the homomorphism lattice induced by a finite algebra...some examples i the case where c is the...

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