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Constraints in Universal Algebra
Ross Willard
University of Waterloo, CAN
SSAOS 2014September 7, 2014
Lecture 1
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Outline
Lecture 1: Intersection problems and congruence SD(∧) varieties
Lecture 2: Constraint problems in ternary groups (and generalizations)
Lecture 3: Constraint problems in Taylor varieties
Almost all algebras will be finite.
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Quiz!
Fix an algebra A. Suppose
C,D ≤ An for some n ≥ 3
proji , j (C ) = proji , j (D) for all 1 ≤ i < j ≤ n.
Question: Does it follow that C ∩ D 6= ∅?
Answer: No, of course not!
Let A be the set {0, 1} (with no operations – ha ha!).
Let n = 3 and put
C = {x ∈ {0, 1}3 : x1 + x2 + x3 = 0 (mod 2)}D = {x ∈ {0, 1}3 : x1 + x2 + x3 = 1 (mod 2)}
C,D ≤ A3 and proji , j (C ) = proji , j (D) = {0, 1}2 for all i < j , yetC ∩ D = ∅.
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Apology
I’m sorry. Choosing A to be a set with no operations is pathetic.
Better example: A = ({0, 1}; x+y+z (mod 2)) with the same C,D ≤ A3.
More generally, for any R-module RA, take the associated affine R-module
A = (A ; x−y+z , {r x+(1−r)y : r ∈ R} )
and let C ,D be different cosets of {(x , y , z) : x + y + z = 0} ≤ A3.
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Harder Quiz
Bonus Problem: For which A is the answer “Yes”?
(The question: if C,D ≤ An and proji , j (C ) = proji , j (D) for all i < j ,does it follow that C ∩ D 6= ∅?)
Subproblem: are there any A for which the answer is “Yes”?
Answer: Of course!
Any algebra A having a constant term operation has this property.(So any group, ring, module, . . . )
I This is cheating.I We can forbid cheating by requiring that A be idempotent, i.e., all
1-element subsets must be subalgebras.
Problem: are there any idempotent A for which the the answer is “Yes”?
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The k-intersection property
Definition (Valeriote). Let A be an algebra.1 If C ,D ⊆ An and 0 < k ≤ n, we write C
k= D to mean
projJ(C ) = projJ(D) for all J ⊆ {1, . . . , n} satisfying |J| = k .
2 For example:
I C1= D iff proji (C ) = proji (D) for all i .
I C2= D iff proji, j (C ) = proji, j (D) for all i < j .
I Cn= D iff C = D.
3 We say that A has the k-intersection property (or k-IP) if for alln > k and every family {Ct ≤ An : t ∈ T},(
Csk= Ct for all s, t ∈ T
)⇒
⋂t∈T
Ct 6= ∅.
Note: 1-IP ⇒ 2-IP ⇒ 3-IP ⇒ 4-IP ⇒ · · ·
Problem (modified): are there any idempotent algebras with 2-IP?What about 1-IP? Or k-IP for some k?
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Theorem1 Every lattice (or lattice expansion) has 2-IP.
2 Every finite semilattice (or expansion) has 1-IP.
Proof.(1) Every lattice (or lattice expansion) L has a majority term
m(x , x , y) = m(x , y , x) = m(y , x , x) = x for all x , y ∈ L.
Hence for any C ≤ Ln, C is determined by (proji , j (C ) : 1 ≤ i < j ≤ n)(Baker-Pixley 1975).
Thus if {Ct ≤ Ln : t ∈ T} satisfies Cs2= Ct ∀s, t, then Cs = Ct ∀s, t.
So⋂Ct 6= ∅.
Generalization. If A has a (k + 1)-ary near unanimity (NU) term, thenA has k-IP. (Again by Baker-Pixley)
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Theorem1 Every lattice (or lattice expansion) has 2-IP.
2 Every finite semilattice (or expansion) has 1-IP.
Proof.(2) Suppose L is finite and has a semilattice term ∧.
For any C ≤ Ln, the ∧-least element of C is determined by(proji (C ) : 1 ≤ i ≤ n).
Thus if {Ct ≤ Ln : t ∈ T} satisfies Cs1= Ct ∀s, t, then all the Cs have
the same ∧-least element.
So⋂Ct 6= ∅.
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Summary
Idempotent algebras that have k-IP for some k:
Lattices
NU algebras
Finite semilattices
Idempotent algebras that do not have k-IP for any k :
Sets (pathetic, but true)
Affine R-modules
Question: What algebraic property separates “lattices, NU algebras andsemilattices” from “sets and affine R-modules”?
Answer: Congruence meet semi-distributivity
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Meet Semi-Distributivity (SD(∧))
Definition. A lattice L is meet semi-distributive (or SD(∧)) if itsatisfies the implication
x ∧ y = x ∧ z =: u ⇒ x ∧ (y ∨ z) = u.
Basic facts:
1 Every distributive lattice is SD(∧).
2 There exist SD(∧) lattices that are not modular. E.g.,
3 M3 is not SD(∧): x y z
u
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Congruence SD(∧)
Definition.
1 An algebra is congruence SD(∧) if its congruence lattice is SD(∧).
2 A variety is congruence SD(∧) if every algebra in the variety iscongruence SD(∧).
Theorem (Lipparini 1998; Kearnes, Szendrei 1998; Kearnes, Kiss 2013;cf. Hobby, McKenzie 1988)
For a variety V, the following are equivalent:
1 V is congruence SD(∧).
2 M3 does not embed into Con(B), for any B ∈ V.
If V is idempotent, then we can add
3 No nontrivial algebra B ∈ V is a term reduct of an affine R-module.
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Congruence SD(∧) varieties
groups
modules
affine modules
lattices semilattices
unary algebras
sets most semigroups
CD
NU
congruence distributive
CSD(∧)CM
congruence modular
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Valeriote’s observation
Theorem (Valeriote 2009)
Assume A is finite and idempotent.
If A has k-IP for some k , then HSP(A) is CSD(∧).
Proof.
Assume HSP(A) is not CSD(∧).
By the previous theorem, there exists a nontrivial B ∈ V which is aterm reduct of an affine R-module M.
We can assume B ∈ HSPfin(A).
Assume A has k-IP; then so does every algebra in HSPfin(A).
Hence B has k-IP.
Hence M has k-IP, but we can show this is impossible.
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Valeriote’s conjecture
Conjecture (Valeriote 2009)
And conversely.
That is, if A is finite, idempotent, and HSP(A) is CSD(∧), then A hask-IP for some k .
Theorem (Barto 2014 ms)
Valeriote’s conjecture is true.
In fact, if A is finite and HSP(A) is CSD(∧), then A has 2-IP.
This is a surprising result with a beautiful proof.
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Constraint Satisfaction Problem (CSP)
Let A be a finite algebra.
An instance of CSP(A) of degree n is a list
(s1,C1), (s2,C2), . . . , (sp,Cp)
of “specifications” of subalgebras of An (of a certain kind).
Each si is a non-empty subset of {1, 2, . . . , n}.Each Ci is a non-empty subuniverse of Asi .
(si ,Ci ) “specifies” the subalgebra {a ∈ An : projsi(a) ∈ Ci} of An.
I I denote this subalgebra by Jsi ,CiK.
Computer Science jargon:
{1, 2, . . . , n} is the set of variables.
Each (si ,Ci ) is a constraint.
si is the scope of (si ,Ci ). Ci is the constraint relation.
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Example
Let A = ({0, 1},∧). (The 2-element semilattice)
With n = 4, define
s1 = {2, 3, 4}C1 = {a ∈ A{2,3,4} : a2 ≤ a3 ≤ a4}
= {(0, 0, 0), (0, 0, 1), (0, 1, 1), (1, 1, 1)}.
The subalgebra of A4 “specified” by (s1,C1) is
Js1,C1K := {(a1, a2, a3, a4) ∈ A4 : a2 ≤ a3 ≤ a4}.
Similarly define (s2,C2), (s3,C3) by
s2 = {1, 3, 4}, C2 = {a ∈ A{1,3,4} : a3 ≤ a4 ≤ a1}s3 = {1, 2, 4}, C3 = {a ∈ A{1,2,4} : a4 ≤ a1 ≤ a2}.
(s1,C1), (s2,C2), (s3,C3) is an instance of CSP(A).
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SolutionsIn general, given a CSP(A) instance (s1,C1), . . . , (sp,Cp), we ask whether
Js1,C1K ∩ · · · ∩ Jsp,CpK 6= ∅.
The elements of Js1,C1K ∩ · · · ∩ Jsp,CpK (if any) are called solutions.
In the previous example, the subalgebras of ({0, 1},∧)4 “specified” by theconstraints are:
Js1,C1K := {a ∈ {0, 1}4 : a2 ≤ a3 ≤ a4}Js2,C2K := {a ∈ {0, 1}4 : a3 ≤ a4 ≤ a1}Js3,C3K := {a ∈ {0, 1}4 : a4 ≤ a1 ≤ a2}
This instance has two solutions, since
Js1,C1K ∩ Js2,C2K ∩ Js3,C3K = {a ∈ {0, 1}4 : a1 = a2 = a3 = a4}.
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(2,3)-minimal instances
Definition
An instance (s1,C1), . . . , (sp,Cp) of CSP(A) (say of degree n) is(2,3)-minimal if:
For any two constraints (si ,Ci ), (sj ,Cj ), if
J ⊆ si ∩ sj and 1 ≤ |J| ≤ 2
then projJ(Ci ) = projJ(Cj ).
For every 3-element subset J ⊆ {1, . . . , n} there exists a constraint(si ,Ci ) such that J ⊆ si .
Remark. The first requirement is like Jsi ,CiK2= Jsj ,CjK, but only requiring
it on coordinates in their common scopes.
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With A = ({0, 1},∧), recall the instance with three constraints:
s1 = {2, 3, 4}, C1 = {a ∈ A{2,3,4} : a2 ≤ a3 ≤ a4}s2 = {1, 3, 4}, C2 = {a ∈ A{1,3,4} : a3 ≤ a4 ≤ a1}s3 = {1, 2, 4}, C3 = {a ∈ A{1,2,4} : a4 ≤ a1 ≤ a2}.
Surprise Quiz: is this instance (2,3)-minimal?
For any two constraints (si ,Ci ), (sj ,Cj ), if J ⊆ si ∩ sj and |J| ≤ 2,then projJ(Ci ) = projJ(Cj ).
For every 3-element subset J ⊆ {1, . . . , n} there exists a constraint(si ,Ci ) such that J ⊆ si .
Answers
No: proj2,4(C1) 6= proj2,4(C3).
No: {1, 2, 3} is not contained in the scope of any constraint.
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Main Theorem
Theorem (Barto 2014 ms, improving Barto, Kozik 2009; Bulatov ms)
Suppose A is finite and HSP(A) is CSD(∧). Then every (2,3)-minimalinstance of CSP(A) has a solution.
Proof: Prague absorption.
Corollary (Barto)
If A is finite and HSP(A) is CSD(∧), then A has 2-IP.
Proof. Given {Ct ≤ An : 1 ≤ t ≤ p} with Cs2= Ct for all s, t, consider
the CSP(A) instance
({1, . . . , n},C1), ({1, . . . , n},C2), . . . , ({1, . . . , n},Cp).
It is (2,3)-minimal.
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Application (if time)
Definition. Let A be an algebra. A weak majority term for A is a termt(x , y , z) satisfying the idempotent law t(x , x , x) = x and
t(x , x , y) = t(x , y , x) = t(y , x , x) for all x , y ∈ A.
Similarly, for any k ≥ 2 we can define a k-ary weak NU term (WNU).
Examples of WNUs
For semilattices (or lattices) we can take t(x1, . . . , xn) = x1 ∧ · · · ∧ xn.
For (Z2,+) we have t(x1, . . . , xn) = x1 + · · ·+ xn (for any odd n ≥ 3).
Theorem (Kozik; in Kozik et al 2014?)
Let A be a finite algebra. The following are equivalent:
1 HSP(A) is congruence SD(∧).
2 A has 3-ary and 4-ary WNU terms t1(x , y , z) and t2(x , y , z ,w)satisfying t1(x , x , y) = t2(x , x , x , y) for all x , y ∈ A.
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Let A be a finite algebra. The following are equivalent:
1 HSP(A) is congruence SD(∧).
2 A has 3-ary and 4-ary WNU terms t1(x , y , z) and t2(x , y , z ,w)satisfying t1(x , x , y) = t2(x , x , x , y) for all x , y ∈ A.
Proof sketch. We can assume A is idempotent.
(2) ⇒ (1). Assume A has such WNUs but CSP(A) is not CSD(∧).
Then ∃B ∈ HSP(A) such that |B| > 1 and B is a term reduct of an affineR-module M.
B also has such WNUs, so M has such WNUs, contradiction.
(1) ⇒ (2). (Variation of an argument due to E. W. Kiss)
Let F2 be the free algebra of rank 2 in HSP(A). Let n = 3|F2|+ 1.
One can define a (2,3)-minimal instance of CSP(F2) of degree n, anysolution of which will give the desired WNUs. (Details in Kozik et al.)
By Barto’s theorem, a solution to the instance exists.
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A. A. Bulatov, Bounded relational width, manuscript, 2009.
L. Barto, The collapse of the bounded width hierarchy, manuscript, 2014.
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K. A. Kearnes and E. W. Kiss, The shape of congruence lattices, Mem. Amer.Math. Soc. 222 (2013), no. 1046.
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varieties, Canad. J. Math. 61 (2009), 451–464.R. Willard (Waterloo) Constraints in Universal Algebra SSAOS 2014 23 / 23