Counting the relations compatible with an algebra Brian Davey and - - PowerPoint PPT Presentation

Counting the relations compatible with an algebra

Brian Davey and Jane Pitkethly (La Trobe, Australia) AMS Hawaii, 4 March 2012

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Four finiteness conditions Proof-by-picture: An easy example Proof-by-picture: Two general conditions for ‘infiniteness’ A family of ‘finite’ examples

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Compatible relations

A compatible relation on a finite algebra A is a non-empty subuniverse of An, for some n ∈ N. There are several natural finiteness conditions on A that are based on ‘how many’ compatible relations it has.

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A weak finiteness condition

Condition (1): A is finitely related

There is a finite set of compatible relations on A from which all other compatible relations on A can be defined, via primitive positive formulæ.

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A weak finiteness condition

Condition (1): A is finitely related

There is a finite set of compatible relations on A from which all other compatible relations on A can be defined, via primitive positive formulæ.

◮ Equivalent: Clo(A) is determined by a finite set of relations.

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A weak finiteness condition

Condition (1): A is finitely related

There is a finite set of compatible relations on A from which all other compatible relations on A can be defined, via primitive positive formulæ.

◮ Equivalent: Clo(A) is determined by a finite set of relations. ◮ All finite lattices,1 groups,2 semilattices and unary algebras

are finitely related.

1Bergman 2Aichinger, Mayr, McKenzie 4 / 26

A weak finiteness condition

Condition (1): A is finitely related

There is a finite set of compatible relations on A from which all other compatible relations on A can be defined, via primitive positive formulæ.

◮ Equivalent: Clo(A) is determined by a finite set of relations. ◮ All finite lattices,1 groups,2 semilattices and unary algebras

are finitely related.

◮ Every finite commutative semigroup is finitely related,3

but not every finite semigroup.4

1Bergman 2Aichinger, Mayr, McKenzie 3Davey, Jackson, Pitkethly, Szabo 4Mayr 4 / 26

A weak finiteness condition

Condition (1): A is finitely related

There is a finite set of compatible relations on A from which all other compatible relations on A can be defined, via primitive positive formulæ.

◮ Equivalent: Clo(A) is determined by a finite set of relations. ◮ All finite lattices,1 groups,2 semilattices and unary algebras

are finitely related.

◮ Every finite commutative semigroup is finitely related,3

but not every finite semigroup.4

◮ The finite relatedness of A only depends on Var(A).3

1Bergman 2Aichinger, Mayr, McKenzie 3Davey, Jackson, Pitkethly, Szabo 4Mayr 4 / 26

A stronger finiteness condition

Condition (2): A has few subpowers

The logarithm of the number of n-ary compatible relations on A grows polynomially in n.

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A stronger finiteness condition

Condition (2): A has few subpowers

The logarithm of the number of n-ary compatible relations on A grows polynomially in n.

◮ Equivalent to A having an edge term.5

5Berman, Idziak, Markovi´

c, McKenzie, Valeriote, Willard

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A stronger finiteness condition

Condition (2): A has few subpowers

The logarithm of the number of n-ary compatible relations on A grows polynomially in n.

◮ Equivalent to A having an edge term.5 ◮ All finite lattices and groups have few subpowers,

but not semilattices or unary algebras.

5Berman, Idziak, Markovi´

c, McKenzie, Valeriote, Willard

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A stronger finiteness condition

Condition (2): A has few subpowers

The logarithm of the number of n-ary compatible relations on A grows polynomially in n.

◮ Equivalent to A having an edge term.5 ◮ All finite lattices and groups have few subpowers,

but not semilattices or unary algebras.

◮ (2) ⇒ (1): Few subpowers implies finitely related.6

5Berman, Idziak, Markovi´

c, McKenzie, Valeriote, Willard

6Aichinger, Mayr, McKenzie 5 / 26

An even stronger finiteness condition

Condition (3): Baker–Pixley

There is a finite set of compatible relations on A from which all other compatible relations on A can be defined, via conjunctions of atomic formulæ.

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An even stronger finiteness condition

Condition (3): Baker–Pixley

There is a finite set of compatible relations on A from which all other compatible relations on A can be defined, via conjunctions of atomic formulæ.

◮ Equivalent to A having a near-unanimity term.

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An even stronger finiteness condition

Condition (3): Baker–Pixley

There is a finite set of compatible relations on A from which all other compatible relations on A can be defined, via conjunctions of atomic formulæ.

◮ Equivalent to A having a near-unanimity term. ◮ All finite lattices satisfy Baker–Pixley. But not groups,

semilattices or unary algebras.

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An even stronger finiteness condition

Condition (3): Baker–Pixley

There is a finite set of compatible relations on A from which all other compatible relations on A can be defined, via conjunctions of atomic formulæ.

◮ Equivalent to A having a near-unanimity term. ◮ All finite lattices satisfy Baker–Pixley. But not groups,

semilattices or unary algebras.

◮ (3) ⇒ (2): Baker–Pixley implies few subpowers.

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An even stronger finiteness condition

Condition (3): Baker–Pixley

There is a finite set of compatible relations on A from which all other compatible relations on A can be defined, via conjunctions of atomic formulæ.

◮ Equivalent to A having a near-unanimity term. ◮ All finite lattices satisfy Baker–Pixley. But not groups,

semilattices or unary algebras.

◮ (3) ⇒ (2): Baker–Pixley implies few subpowers.

Condition (4): Finitely many relations

There is a finite set of compatible relations on A such that every

• ther compatible relation on A is interdefinable with one of

these relations, via conjunctions of atomic formulæ.

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An example

Two compatible relations on the 2-element bounded lattice L = {0, 1}; ∨, ∧, 0, 1:

⊆ L2 ❝ 00 ❝ 01 ❝ 11 ρ ⊆ L4 ❝

• ❅

❅ 0000 ❝ ❅ ❅ 0011 ❝

• 0100

❝ 0111 ❝ 1111

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An example

Two compatible relations on the 2-element bounded lattice L = {0, 1}; ∨, ∧, 0, 1:

⊆ L2 ❝ 00 ❝ 01 ❝ 11 ρ ⊆ L4 ❝

• ❅

❅ 0000 ❝ ❅ ❅ 0011 ❝

• 0100

❝ 0111 ❝ 1111

=

• (x, y) ∈ L2

(x, x, y, y) ∈ ρ

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An example

Two compatible relations on the 2-element bounded lattice L = {0, 1}; ∨, ∧, 0, 1:

⊆ L2 ❝ 00 ❝ 01 ❝ 11 ρ ⊆ L4 ❝

• ❅

❅ 0000 ❝ ❅ ❅ 0011 ❝

• 0100

❝ 0111 ❝ 1111

=

• (x, y) ∈ L2

(x, x, y, y) ∈ ρ

• ρ =
• (w, x, y, z) ∈ L4

w x & w y & y = z

• .

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An example

Two compatible relations on the 2-element bounded lattice L = {0, 1}; ∨, ∧, 0, 1:

⊆ L2 ❝ 00 ❝ 01 ❝ 11 ρ ⊆ L4 ❝

• ❅

❅ 0000 ❝ ❅ ❅ 0011 ❝

• 0100

❝ 0111 ❝ 1111

=

• (x, y) ∈ L2

(x, x, y, y) ∈ ρ

• ρ =
• (w, x, y, z) ∈ L4

w x & w y & y = z

• .

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An example

Two compatible relations on the 2-element bounded lattice L = {0, 1}; ∨, ∧, 0, 1:

⊆ L2 ❝ 00 ❝ 01 ❝ 11 ρ ⊆ L4 ❝

• ❅

❅ 0000 ❝ ❅ ❅ 0011 ❝

• 0100

❝ 0111 ❝ 1111

=

• (x, y) ∈ L2

(x, x, y, y) ∈ ρ

• ρ =
• (w, x, y, z) ∈ L4

w x & w y & y = z

• .

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An example

Two compatible relations on the 2-element bounded lattice L = {0, 1}; ∨, ∧, 0, 1:

⊆ L2 ❝ 00 ❝ 01 ❝ 11 ρ ⊆ L4 ❝

• ❅

❅ 0000 ❝ ❅ ❅ 0011 ❝

• 0100

❝ 0111 ❝ 1111

=

• (x, y) ∈ L2

(x, x, y, y) ∈ ρ

• ρ =
• (w, x, y, z) ∈ L4

w x & w y & y = z

• .

The relations and ρ are conjunct-atomic interdefinable, and so we will regard them as equivalent.

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An example

Two compatible relations on the 2-element bounded lattice L = {0, 1}; ∨, ∧, 0, 1:

⊆ L2 ❝ 00 ❝ 01 ❝ 11 ρ ⊆ L4 ❝

• ❅

❅ 0000 ❝ ❅ ❅ 0011 ❝

• 0100

❝ 0111 ❝ 1111

=

• (x, y) ∈ L2

(x, x, y, y) ∈ ρ

• ρ =
• (w, x, y, z) ∈ L4

w x & w y & y = z

• .

The relations and ρ are conjunct-atomic interdefinable, and so we will regard them as equivalent. Every compatible relation on L is equivalent to either or ∆L.

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Basic definitions

Two compatible relations on A are equivalent if each is conjunct-definable from the other. If the set of all compatible relations on A has only a finite number of equivalence classes, then we say that A admits only finitely many relations.

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Basic definitions

Two compatible relations on A are equivalent if each is conjunct-definable from the other. If the set of all compatible relations on A has only a finite number of equivalence classes, then we say that A admits only finitely many relations. In this case, the algebra A satisfies the Baker–Pixley condition and so A has a near-unanimity term.

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Basic definitions

Two compatible relations on A are equivalent if each is conjunct-definable from the other. If the set of all compatible relations on A has only a finite number of equivalence classes, then we say that A admits only finitely many relations. In this case, the algebra A satisfies the Baker–Pixley condition and so A has a near-unanimity term.

Question

Which finite algebras admit only finitely many relations?

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Examples

2-element Boolean algebra B = {0, 1}; ∧, ¬

One compatible relation: ∆B.

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Examples

2-element Boolean algebra B = {0, 1}; ∧, ¬

One compatible relation: ∆B.

2-element bounded lattice L = {0, 1}; ∨, ∧, 0, 1

Two compatible relations: ∆L, .

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Examples

2-element Boolean algebra B = {0, 1}; ∧, ¬

One compatible relation: ∆B.

2-element bounded lattice L = {0, 1}; ∨, ∧, 0, 1

Two compatible relations: ∆L, .

2-element lattice 2 = {0, 1}; ∨, ∧

Eight compatible relations: ∆2, , {0}, {1}, {(0, 1)}, ×{0}, ×{1}, ×{(0, 1)}.

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More examples

Finitely many relations:

◮ 3-element p-algebra S = {0, d, 1}; ∨, ∧, ∗, 0, 1, ◮ 4-element Boolean algebra with constants, ◮ ring with unity Zpq, for distinct primes p, q, ◮ every quasi-primal algebra (with R. Willard), ◮ each finite Heyting chain (Nguyen, Pitkethly).

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More examples

Finitely many relations:

◮ 3-element p-algebra S = {0, d, 1}; ∨, ∧, ∗, 0, 1, ◮ 4-element Boolean algebra with constants, ◮ ring with unity Zpq, for distinct primes p, q, ◮ every quasi-primal algebra (with R. Willard), ◮ each finite Heyting chain (Nguyen, Pitkethly).

Infinitely many relations:

◮ Boolean algebras of size 4, lattices of size 3,

p-algebras of size 4, non-chain Heyting algebras, . . .

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More examples

Finitely many relations:

◮ 3-element p-algebra S = {0, d, 1}; ∨, ∧, ∗, 0, 1, ◮ 4-element Boolean algebra with constants, ◮ ring with unity Zpq, for distinct primes p, q, ◮ every quasi-primal algebra (with R. Willard), ◮ each finite Heyting chain (Nguyen, Pitkethly).

Infinitely many relations:

◮ Boolean algebras of size 4, lattices of size 3,

p-algebras of size 4, non-chain Heyting algebras, . . . (If A admits infinitely many relations and A ∈ HS(B), then B admits infinitely many relations.)

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Proof-by-picture: An easy example

Consider the 2-element algebra M = {0, 1}; m, where m: {0, 1}3 → {0, 1} is the majority operation.

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Proof-by-picture: An easy example

Consider the 2-element algebra M = {0, 1}; m, where m: {0, 1}3 → {0, 1} is the majority operation. This picture proves that M admits infinitely many relations: M X2 X3 X X4 X . . . X

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Easy example: Steps 1 and 2

Step 1

The relation r := {(0, 0), (0, 1), (1, 0)} is compatible with M.

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Easy example: Steps 1 and 2

Step 1

The relation r := {(0, 0), (0, 1), (1, 0)} is compatible with M. Define M = {0, 1}; r. 1

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Easy example: Steps 1 and 2

Step 1

The relation r := {(0, 0), (0, 1), (1, 0)} is compatible with M. Define M = {0, 1}; r. 1

Step 2

Each structure Xn defines a compatible relation on M: rn := hom(Xn, M) MXn.

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Easy example: Steps 1 and 2

Step 1

The relation r := {(0, 0), (0, 1), (1, 0)} is compatible with M. Define M = {0, 1}; r. 1

Step 2

Each structure Xn defines a compatible relation on M: rn := hom(Xn, M) MXn. For example, 1 M 3 1 2 X3 r3 = hom(X3, M)

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Easy example: Steps 1 and 2

Step 1

The relation r := {(0, 0), (0, 1), (1, 0)} is compatible with M. Define M = {0, 1}; r. 1

Step 2

Each structure Xn defines a compatible relation on M: rn := hom(Xn, M) MXn. For example, 1 M 3 1 2 X3 r3 = hom(X3, M) = { (0, 0, 0),

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Easy example: Steps 1 and 2

Step 1

The relation r := {(0, 0), (0, 1), (1, 0)} is compatible with M. Define M = {0, 1}; r. 1

Step 2

Each structure Xn defines a compatible relation on M: rn := hom(Xn, M) MXn. For example, 1 M 3 1 2 X3 r3 = hom(X3, M) = { (0, 0, 0), (0, 0, 1),

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Easy example: Steps 1 and 2

Step 1

The relation r := {(0, 0), (0, 1), (1, 0)} is compatible with M. Define M = {0, 1}; r. 1

Step 2

Each structure Xn defines a compatible relation on M: rn := hom(Xn, M) MXn. For example, 1 M 3 1 2 X3 r3 = hom(X3, M) = { (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 0, 0) }.

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Easy example: Step 3

M 1 r2 = hom(X2, M) r3 = hom(X3, M) r4 = hom(X4, M) . . .

◮ We want to show that the relations r2, r3, r4, . . . are

pairwise non-equivalent.

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Easy example: Step 3

M 1 r2 = hom(X2, M) r3 = hom(X3, M) r4 = hom(X4, M) . . .

◮ We want to show that the relations r2, r3, r4, . . . are

pairwise non-equivalent.

◮ We will just check that r3 is not ca-definable from r4.

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Easy example: Step 3

M 1 r2 = hom(X2, M) r3 = hom(X3, M) r4 = hom(X4, M) . . .

◮ We want to show that the relations r2, r3, r4, . . . are

pairwise non-equivalent.

◮ We will just check that r3 is not ca-definable from r4. ◮ It suffices to find a map ϕ: Z → {0, 1}, for some

Z ⊆ {0, 1}n, such that ϕ preserves r4 but not r3.

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Step 3: Choosing ϕ: Z → {0, 1}

We want ϕ to preserve r4 but not r3.

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Step 3: Choosing ϕ: Z → {0, 1}

We want ϕ to preserve r4 but not r3. There’s always a canonical choice for dom(ϕ):

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Step 3: Choosing ϕ: Z → {0, 1}

We want ϕ to preserve r4 but not r3. There’s always a canonical choice for dom(ϕ): r3 = hom(X3, M) = { (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 0, 0) },

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Step 3: Choosing ϕ: Z → {0, 1}

We want ϕ to preserve r4 but not r3. There’s always a canonical choice for dom(ϕ): r3 = hom(X3, M) Z = r T

3 = { (0, 0, 0, 1),

= { (0, 0, 0), (0, 0, 1, 0), (0, 0, 1), (0, 1, 0, 0) } (0, 1, 0), r3 r3 r3 r3 (1, 0, 0) },

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Step 3: Choosing ϕ: Z → {0, 1}

We want ϕ to preserve r4 but not r3. There’s always a canonical choice for dom(ϕ): r3 = hom(X3, M) Z = r T

3 = { (0, 0, 0, 1),

ϕ → 1 = { (0, 0, 0), (0, 0, 1, 0), → 1 (0, 0, 1), (0, 1, 0, 0) } → 1 (0, 1, 0), r3 r3 r3 r3 r3

• (1, 0, 0) },

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Step 3: Choosing ϕ: Z → {0, 1}

We want ϕ to preserve r4 but not r3. There’s always a canonical choice for dom(ϕ): r3 = hom(X3, M) Z = r T

3 = { (0, 0, 0, 1),

ϕ → 1 = { (0, 0, 0), (0, 0, 1, 0), → 1 (0, 0, 1), (0, 1, 0, 0) } → 1 (0, 1, 0), r3 r3 r3 r3 r3

• (1, 0, 0) },

So ϕ doesn’t preserve r3.

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Step 3: Choosing ϕ: Z → {0, 1}

We want ϕ to preserve r4 but not r3. There’s always a canonical choice for dom(ϕ): r3 = hom(X3, M) Z = r T

3 = { (0, 0, 0, 1),

ϕ → 1 = { (0, 0, 0), (0, 0, 1, 0), → 1 (0, 0, 1), (0, 1, 0, 0) } → 1 (0, 1, 0), r3 r3 r3 r3 r3

• (1, 0, 0) },

So ϕ doesn’t preserve r3. 1 M X3

From the picture

◮ In fact, we have X3 ∈ ISP(M). Therefore X3 ∼

= Z M4.

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Step 3: Choosing ϕ: Z → {0, 1}

We want ϕ to preserve r4 but not r3. There’s always a canonical choice for dom(ϕ): r3 = hom(X3, M) Z = r T

3 = { (0, 0, 0, 1),

ϕ → 1 = { (0, 0, 0), (0, 0, 1, 0), → 1 (0, 0, 1), (0, 1, 0, 0) } → 1 (0, 1, 0), r3 r3 r3 r3 r3

• (1, 0, 0) },

So ϕ doesn’t preserve r3. 1 M X3

From the picture

◮ In fact, we have X3 ∈ ISP(M). Therefore X3 ∼

= Z M4.

◮ The map ϕ: X3 → {0, 1} does not preserve r3, because

it is not a graph homomorphism, i.e., ϕ / ∈ hom(X3, M).

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Step 3: Checking ϕ: Z → {0, 1} preserves r4

r4 = hom(X4, M) {0, 1}4 1 M X4

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Step 3: Checking ϕ: Z → {0, 1} preserves r4

r4 = hom(X4, M) {0, 1}4 1 M X4 Pick w, x, y, z ∈ Z. For example, w = (0, 0, 0, 1) ϕ → 1 x = (0, 0, 1, 0) → 1 y = (0, 1, 0, 0) → 1 z = (0, 0, 1, 0) → 1

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Step 3: Checking ϕ: Z → {0, 1} preserves r4

r4 = hom(X4, M) {0, 1}4 1 M X4 Pick w, x, y, z ∈ Z. For example, w = (0, 0, 0, 1) ϕ → 1 x = (0, 0, 1, 0) → 1 y = (0, 1, 0, 0) → 1 z = (0, 0, 1, 0) → 1 r4

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Step 3: Checking ϕ: Z → {0, 1} preserves r4

r4 = hom(X4, M) {0, 1}4 1 M X4 Pick w, x, y, z ∈ Z. For example, w = (0, 0, 0, 1) ϕ → 1 x = (0, 0, 1, 0) → 1 y = (0, 1, 0, 0) → 1 z = (0, 0, 1, 0) → 1 r4

• So ϕ preserves r4, since r Z

4 = ∅.

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Step 3: Checking ϕ: Z → {0, 1} preserves r4

r4 = hom(X4, M) {0, 1}4 1 M X4 Pick w, x, y, z ∈ Z. For example, w = (0, 0, 0, 1) ϕ → 1 x = (0, 0, 1, 0) → 1 y = (0, 1, 0, 0) → 1 z = (0, 0, 1, 0) → 1 r4

• So ϕ preserves r4, since r Z

4 = ∅.

From the picture

◮ Since Z = X3, we are picking a map ω: X4 → X3.

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Step 3: Checking ϕ: Z → {0, 1} preserves r4

r4 = hom(X4, M) {0, 1}4 1 M X4 Pick w, x, y, z ∈ Z. For example, w = (0, 0, 0, 1) ϕ → 1 x = (0, 0, 1, 0) → 1 y = (0, 1, 0, 0) → 1 z = (0, 0, 1, 0) → 1 r4

• So ϕ preserves r4, since r Z

4 = ∅.

From the picture

◮ Since Z = X3, we are picking a map ω: X4 → X3. ◮ Want: If each πi ◦ ω: X4 → {0, 1} is in r4, then

ϕ ◦ ω: X4 → {0, 1} is in r4.

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Step 3: Checking ϕ: Z → {0, 1} preserves r4

r4 = hom(X4, M) {0, 1}4 1 M X4 Pick w, x, y, z ∈ Z. For example, w = (0, 0, 0, 1) ϕ → 1 x = (0, 0, 1, 0) → 1 y = (0, 1, 0, 0) → 1 z = (0, 0, 1, 0) → 1 r4

• So ϕ preserves r4, since r Z

4 = ∅.

From the picture

◮ Since Z = X3, we are picking a map ω: X4 → X3. ◮ Want: If each πi ◦ ω: X4 → {0, 1} is in r4, then

ϕ ◦ ω: X4 → {0, 1} is in r4.

◮ But X3 M4, so we really want: If ω: X4 → X3, then

ϕ ◦ ω: X4 → M is a graph homomorphism. True vacuously.

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Proof-by-picture: Recap

Consider the 2-element algebra M = {0, 1}; m, where m: {0, 1}3 → {0, 1} is the majority operation. This picture below proves that M admits infinitely many relations. The red labels on Xn yield a map ϕn : Xn → {0, 1} that is not a graph homomorphism. 1 M 1 1 X2 1 1 1 X3 X 1 1 1 1 X4 X . . . X

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Proof-by-picture: General approach

Lemma

To show that A admits infinitely many relations, it suffices to find

◮ a structure A = A; R that is compatible with A, and

skip

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Proof-by-picture: General approach

Lemma

To show that A admits infinitely many relations, it suffices to find

◮ a structure A = A; R that is compatible with A, and

skip

◮ for all n ∈ N,

◮ a finite structure Xn ∈ ISP(A), and ◮ a map ϕn : Xn → A that is not a morphism from Xn to A 17 / 26

Proof-by-picture: General approach

Lemma

To show that A admits infinitely many relations, it suffices to find

◮ a structure A = A; R that is compatible with A, and

skip

◮ for all n ∈ N,

◮ a finite structure Xn ∈ ISP(A), and ◮ a map ϕn : Xn → A that is not a morphism from Xn to A

such that, either for all k < ℓ or for all k > ℓ, the following condition holds:

◮ for every morphism ω: Xℓ → Xk, the map

ϕk ◦ ω: Xℓ → A is a morphism from Xℓ to A

(i.e., the relation rk is not ca-definable from rℓ).

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A general condition for ‘infiniteness’

A bad relation

If a finite algebra A either

◮ has a pair of non-permuting congruences, or ◮ is of the form B2, for non-trivial B,

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A general condition for ‘infiniteness’

A bad relation

If a finite algebra A either

◮ has a pair of non-permuting congruences, or ◮ is of the form B2, for non-trivial B,

then A has elements a, b, c and a compatible binary relation r such that in r c a b not in r c a b

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A general condition for ‘infiniteness’

A bad relation

If a finite algebra A either

◮ has a pair of non-permuting congruences, or ◮ is of the form B2, for non-trivial B,

then A has elements a, b, c and a compatible binary relation r such that in r c a b not in r c a b

◮ For non-permuting congruences:

r = θ1 · θ2.

◮ For A = B2:

r = (x1, x2), (y1, y2)

• ∈ A2

x2 = y1

• .

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A general condition for ‘infiniteness’

The theorem

Theorem

Let A be a finite algebra with elements a, b, c and a compatible binary relation r such that in r c a b not in r c a b Then A admits infinitely many relations.

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A general condition for ‘infiniteness’

The theorem

Theorem

Let A be a finite algebra with elements a, b, c and a compatible binary relation r such that in r c a b not in r c a b Then A admits infinitely many relations. There is an obvious choice for A = A; R,

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A general condition for ‘infiniteness’

The theorem

Theorem

Let A be a finite algebra with elements a, b, c and a compatible binary relation r such that in r c a b not in r c a b Then A admits infinitely many relations. There is an obvious choice for A = A; R, namely, A := A; r.

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A general condition for ‘infiniteness’

The theorem

Theorem

Let A be a finite algebra with elements a, b, c and a compatible binary relation r such that in r c a b not in r c a b Then A admits infinitely many relations.

Proof of Case 1: (b, b) / ∈ r or (c, c) / ∈ r

X3 c a c X X4 c a c c X X5 c c a c c . . .

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A general condition for ‘infiniteness’

The theorem

Theorem

Let A be a finite algebra with elements a, b, c and a compatible binary relation r such that in r c a b not in r c a b Then A admits infinitely many relations.

Proof of Case 2: (b, b) ∈ r and (c, c) ∈ r

X3 c a c X4 c a c c X5 c c a c c . . .

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A general condition for ‘infiniteness’

Applications

Corollary

If one of the following holds, then A admits infinitely many relations:

• 1. A has a pair of non-permuting congruences;
• 2. A is isomorphic to B2, with B non-trivial;
• 3. A is a subalgebra of B2 such that

{(0, 0), (0, 1), (1, 0)} ⊆ A, for some 0 = 1 in B. This result covers:

◮ Boolean algebras of size 4, ◮ lattices of size 3, ◮ p-algebras of size 4, ◮ non-chain Heyting algebras, . . .

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Another general condition for ‘infiniteness’

Theorem and example

Theorem

If A admits only finitely many relations, then Con(A) is an N-free distributive lattice. α β γ δ N An N-free distributive lattice

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Another general condition for ‘infiniteness’

Theorem and example

Theorem

If A admits only finitely many relations, then Con(A) is an N-free distributive lattice. α β γ δ N An N-free distributive lattice

Example

Assume M = A × B × C, for non-trivial finite algebras A, B, C. Then M admits infinitely many relations.

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Another general condition for ‘infiniteness’

Proof-by-picture

Proof

Assume Con(A) contains α β δ .

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Another general condition for ‘infiniteness’

Proof-by-picture

Proof

Assume Con(A) contains α β δ . Take A = A; α, β, δ and choose (a, b) ∈ δ \ β.

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Another general condition for ‘infiniteness’

Proof-by-picture

Proof

Assume Con(A) contains α β δ . Take A = A; α, β, δ and choose (a, b) ∈ δ \ β. X4 b a b b δ β δ β α X6 b b a b b b α β δ β δ β α δ X8 b b b a b b b b δ α β δ β δ β α δ α β

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Another general condition for ‘infiniteness’

Proof-by-picture

Proof

Assume Con(A) contains α β δ . Take A = A; α, β, δ and choose (a, b) ∈ δ \ β. X4 b a b b δ β δ β α X6 b b a b b b α β δ β δ β α δ X8 b b b a b b b b δ α β δ β δ β α δ α β So A admits infinitely many relations.

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Finding examples of ‘finiteness’

A definition

A finite algebra A is strictly affine complete if: for all n ∈ N and all X ⊆ An, every function f : X → A that preserves each θ ∈ Con(A) extends to a polynomial of A.

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Finding examples of ‘finiteness’

A definition

A finite algebra A is strictly affine complete if: for all n ∈ N and all X ⊆ An, every function f : X → A that preserves each θ ∈ Con(A) extends to a polynomial of A.

Example

Every finite algebra that generates an arithmetical variety is strictly affine complete (Pixley).

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Finding examples of ‘finiteness’

A theorem

Theorem

Let A be a finite algebra such that

• 1. A is strictly affine complete,
• 2. each a ∈ A is the value of a constant term function of A,

and

• 3. Con(A) is an N-free lattice.

Then A admits only finitely many relations.

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Finding examples of ‘finiteness’

A theorem

Theorem

Let A be a finite algebra such that

• 1. A is strictly affine complete,
• 2. each a ∈ A is the value of a constant term function of A,

and

• 3. Con(A) is an N-free lattice.

Then A admits only finitely many relations.

Proof

We use the duality given by the alter ego A := A; Con(A), T to represent all the compatible relations on A.

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Finding examples of ‘finiteness’

Application of the theorem

Example

Let L be a finite N-free distributive lattice. There exists a finite algebra A with Con(A) ∼ = L such that A admits only finitely many relations.

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Finding examples of ‘finiteness’

Application of the theorem

Example

Let L be a finite N-free distributive lattice. There exists a finite algebra A with Con(A) ∼ = L such that A admits only finitely many relations.

Proof

L

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Finding examples of ‘finiteness’

Application of the theorem

Example

Let L be a finite N-free distributive lattice. There exists a finite algebra A with Con(A) ∼ = L such that A admits only finitely many relations.

Proof

L A = A; ∨, ∧, →, a1, a2, . . . , an

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Finding examples of ‘finiteness’

Another application of the theorem

Example

Assume that M = A × B is the independent product of two primal algebras. Then M admits only finitely many relations.

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Finding examples of ‘finiteness’

Another application of the theorem

Example

Assume that M = A × B is the independent product of two primal algebras. Then M admits only finitely many relations. For example, the following algebras admit only finitely many relations:

◮ the four-element Boolean algebra enriched with all

elements as constants;

◮ the ring with unity Zpq, for all distinct primes p and q.

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Finding examples of ‘finiteness’

Another application of the theorem

Example

Assume that M = A × B is the independent product of two primal algebras. Then M admits only finitely many relations. For example, the following algebras admit only finitely many relations:

◮ the four-element Boolean algebra enriched with all

elements as constants;

◮ the ring with unity Zpq, for all distinct primes p and q.

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Finding examples of ‘finiteness’

Another application of the theorem

Example

Assume that M = A × B is the independent product of two primal algebras. Then M admits only finitely many relations. For example, the following algebras admit only finitely many relations:

◮ the four-element Boolean algebra enriched with all

elements as constants;

◮ the ring with unity Zpq, for all distinct primes p and q.

Mahalo!

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