SMALL CONGRUENCE LATTICES
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1 SMALL CONGRUENCE LATTICES William DeMeo University of South Carolina joint work with Ralph Freese, Peter Jipsen, Bill Lampe, J.B. Nation BLAST Conference Chapman University August 5 9, 2013
2 THE PROBLEM CHARACTERIZE CONGRUENCE LATTICES OF FINITE ALGEBRAS. For an arbitrary algebra, there is essentially no restriction on the shape of its congruence lattice. THEOREM (GRÄTZER-SCHMIDT, 1963) Every algebraic lattice is isomorphic to the congruence lattice of an algebra.
3 THE PROBLEM CHARACTERIZE CONGRUENCE LATTICES OF FINITE ALGEBRAS. For an arbitrary algebra, there is essentially no restriction on the shape of its congruence lattice. THEOREM (GRÄTZER-SCHMIDT, 1963) Every algebraic lattice is isomorphic to the congruence lattice of an algebra. If an algebra is finite, then its congruence lattice is...
4 THE PROBLEM CHARACTERIZE CONGRUENCE LATTICES OF FINITE ALGEBRAS. For an arbitrary algebra, there is essentially no restriction on the shape of its congruence lattice. THEOREM (GRÄTZER-SCHMIDT, 1963) Every algebraic lattice is isomorphic to the congruence lattice of an algebra. If an algebra is finite, then its congruence lattice is... finite.
5 THE PROBLEM CHARACTERIZE CONGRUENCE LATTICES OF FINITE ALGEBRAS. For an arbitrary algebra, there is essentially no restriction on the shape of its congruence lattice. THEOREM (GRÄTZER-SCHMIDT, 1963) Every algebraic lattice is isomorphic to the congruence lattice of an algebra. If an algebra is finite, then its congruence lattice is... finite. Problem: Given an arbitrary finite lattice L, does there exist finite algebra A such that Con A = L? DEFINITION We call a finite lattice representable if it is (isomorphic to) the congruence lattice of a finite algebra.
6 A FEW IMPORTANT THEOREMS THEOREM (PUDLÁK AND TŮMA, 1980) Every finite lattice can be embedded in Eq(X), with X finite.
7 A FEW IMPORTANT THEOREMS THEOREM (PUDLÁK AND TŮMA, 1980) Every finite lattice can be embedded in Eq(X), with X finite. THEOREM (PÉTER PÁL PÁLFY AND PAVEL PUDLÁK, 1980) The following statements are equivalent: (I) Every finite lattice is representable. (II) Every finite lattice is isomorphic to an interval in the subgroup lattice of a finite group.
8 A FEW IMPORTANT THEOREMS THEOREM (PUDLÁK AND TŮMA, 1980) Every finite lattice can be embedded in Eq(X), with X finite. THEOREM (PÉTER PÁL PÁLFY AND PAVEL PUDLÁK, 1980) The following statements are equivalent: (I) Every finite lattice is representable. (II) Every finite lattice is isomorphic to an interval in the subgroup lattice of a finite group. THEOREM (BERMAN, QUACKENBUSH & WOLK, 1970) Every finite distributive lattice is representable.
9 PART 1: METHODS Given a finite lattice L, find a finite algebra A with Con A = L.
10 HOW TO FIND A FINITE ALGEBRA WITH A GIVEN CONGRUENCE LATTICE?
11 HOW TO FIND A FINITE ALGEBRA WITH A GIVEN CONGRUENCE LATTICE? 1. USE CLOSURE PROPERTIES Relate the given lattice to other lattices known to be representable. If L is representable, so is A. the dual of L (Kurzweil 1985, Netter) B. any interval sublattice of L (follows from A.) C. any sublattice that is the union of a principal filter and principal idea of L (Snow, 2000) If L 1 and L 2 are representable, so is 1. the direct product of L 1 and L 2 (Tůma1989) 2. the ordinal sum of L 1 and L 2 (McKenzie 1984, Snow 2000) 3. the parallel sum of L 1 and L 2 (Snow 2000)
12 HOW TO FIND A FINITE ALGEBRA WITH A GIVEN CONGRUENCE LATTICE? 2. THE CLOSURE METHOD Find a closed representation of L in Eq(X). YAGC For L Eq(X) define For F X X define λ(l) = {f X X : ( θ L) f (θ) θ} ρ(f) = {θ Eq(X) : ( f F) f (θ) θ} The map ρλ is a closure operator on Sub[Eq(X)]. (idempotent, extensive, order preserving)
13 HOW TO FIND A FINITE ALGEBRA WITH A GIVEN CONGRUENCE LATTICE? 2. THE CLOSURE METHOD Find a closed representation of L in Eq(X). YAGC For L Eq(X) define For F X X define λ(l) = {f X X : ( θ L) f (θ) θ} ρ(f) = {θ Eq(X) : ( f F) f (θ) θ} The map ρλ is a closure operator on Sub[Eq(X)]. (idempotent, extensive, order preserving) THEOREM A lattice L Eq(X) is a congruence lattice if and only if it is closed, i.e. ρλ(l) = L, in which case L = Con X, λ(l).
14 HOW TO FIND A FINITE ALGEBRA WITH A GIVEN CONGRUENCE LATTICE? 2. THE CLOSURE METHOD Find a closed representation of L in Eq(X). YAGC For L Eq(X) define For F X X define λ(l) = {f X X : ( θ L) f (θ) θ} ρ(f) = {θ Eq(X) : ( f F) f (θ) θ} The map ρλ is a closure operator on Sub[Eq(X)]. (idempotent, extensive, order preserving) THEOREM A lattice L Eq(X) is a congruence lattice if and only if it is closed, i.e. ρλ(l) = L, in which case L = Con X, λ(l). Example: M 3 = L Eq(5)
15 HOW TO FIND A FINITE ALGEBRA WITH A GIVEN CONGRUENCE LATTICE? 3. THE G-SET METHOD Find L as an interval in a subgroup lattice of a finite group. If H G are finite groups, then the filter above H in Sub(G), H, G := {K : H K G}, is isomorphic to Con G/H, Ḡ.
16 HOW TO FIND A FINITE ALGEBRA WITH A GIVEN CONGRUENCE LATTICE? 3. THE G-SET METHOD Find L as an interval in a subgroup lattice of a finite group. If H G are finite groups, then the filter above H in Sub(G), H, G := {K : H K G}, is isomorphic to Con G/H, Ḡ. 4. THE RABBIT EARS METHOD (AKA OVERALGEBRAS, AKA EXPANSION-EXTENSION) Build the required algebra by gluing together isomorphic copies of an algebra and adding new operations.
17 THE G-SET METHOD: DETAILS For groups H G, let A = H\G, Ḡ denote the algebra with universe: the right cosets H\G = {Hx : x G} operations: Ḡ = {g A : g G}, where g A (Hx) = Hxg.
18 THE G-SET METHOD: DETAILS For groups H G, let A = H\G, Ḡ denote the algebra with universe: the right cosets H\G = {Hx : x G} operations: Ḡ = {g A : g G}, where g A (Hx) = Hxg. THEOREM Con A = H, G := {K : H K G}. The isomorphism H, G K θ K Con A is given by θ K = {(Hx, Hy) : xy 1 K}. The inverse isomorphism Con A θ K θ H, G is K θ = {g G : (H, Hg) θ}.
19 THE G-SET METHOD: DETAILS For groups H G, let A = H\G, Ḡ denote the algebra with universe: the right cosets H\G = {Hx : x G} operations: Ḡ = {g A : g G}, where g A (Hx) = Hxg. THEOREM Con A = H, G := {K : H K G}. The isomorphism H, G K θ K Con A is given by θ K = {(Hx, Hy) : xy 1 K}. The inverse isomorphism Con A θ K θ H, G is K θ = {g G : (H, Hg) θ}. Aside: properties of such congruence lattices correspond to properties of subgroup lattices. For example, LEMMA In Con H\G, Ḡ, two congruences, θ K1 and θ K2, n-permute if and only if the corresponding subgroups, K 1 and K 2, n-permute.
20 FILTER+IDEAL METHOD: DETAILS LEMMA Suppose L 0 = Con A, F, and α, β L0 \ {0, 1}. L 0 β α
21 FILTER+IDEAL METHOD: DETAILS LEMMA Suppose L 0 = Con A, F, and α, β L0 \ {0, 1}. Consider L = α β. L L 0 β α
22 FILTER+IDEAL METHOD: DETAILS LEMMA Suppose L 0 = Con A, F, and α, β L0 \ {0, 1}. Consider L = α β. There exists a set F A A such that L = Con A, F F. L L 0 β α
23 FILTER+IDEAL METHOD: DETAILS LEMMA Suppose L 0 = Con A, F, and α, β L0 \ {0, 1}. Consider L = α β. There exists a set F A A such that L = Con A, F F. Proof: Fix θ L 0 \ L. Then α θ β, so (a, b) α \ θ, (u, v) θ \ β. L L 0 Define f θ : A A by f θ (x) = { a b x u/β, x / u/β. Then (f θ (u), f θ (v)) = (a, b) / θ, so f θ (θ) θ, ker f θ β, so f θ (γ) γ for all γ β, f θ (A) {a, b}, so f θ (γ) γ for all γ α. α θ β Let F = {f θ : θ L 0 \ L}.
24 PART 2: ATLAS Which finite lattices are known to be representable?
25 LATTICES WITH AT MOST 6 ELEMENTS ARE REPRESENTABLE. Watatani (1996) J. Funct. Anal. Aschbacher (2008) JAMS
26 LATTICES WITH AT MOST 6 ELEMENTS ARE REPRESENTABLE. Watatani (1996) J. Funct. Anal. Aschbacher (2008) JAMS Theorem: Lattices with at most 6 elements are intervals in subgroup lattices of finite groups.
27 ARE ALL LATTICES WITH AT MOST 7 ELEMENTS REPRESENTABLE? As of Spring Figure courtesy of Peter Jipsen.
28 ARE ALL LATTICES WITH AT MOST 7 ELEMENTS REPRESENTABLE? L 19 L 20 L 17 L 13 L 11 L 9 L 10
29 ARE ALL LATTICES WITH AT MOST 7 ELEMENTS REPRESENTABLE? L 19 L 20 L 17 L 13 L 11 L 9 L 10
30 FINDING REPRESENTATIONS......AS INTERVALS IN SUBGROUP LATTICES L 17 L 13
31 FINDING REPRESENTATIONS......AS INTERVALS IN SUBGROUP LATTICES G L 17 H SmallGroup(288,1025) G : H = 48 The group G = (A 4 A 4) C 2 has a subgroup H = S 3 such that H, G = L so the dual L 16 is also representable. L 13
32 FINDING REPRESENTATIONS......AS INTERVALS IN SUBGROUP LATTICES G L 17 H SmallGroup(288,1025) G : H = 48 The group G = (A 4 A 4) C 2 has a subgroup H = S 3 such that H, G = L so the dual L 16 is also representable. G L 13 The group G = (C 2 C 2 C 2 C 2) A 5 has a subgroup H = A 4 such that H, G = L 13. H SmallGroup(960,11358) G : H = 80
33 ARE ALL LATTICES WITH AT MOST 7 ELEMENTS REPRESENTABLE? L 19 L 20 L 17 L 13 L 11 L 9 L 10
34 FINDING REPRESENTATIONS......USING SUBGROUP LATTICE INTERVALS AND THE FILTER+IDEAL LEMMA. L 11
35 FINDING REPRESENTATIONS......USING SUBGROUP LATTICE INTERVALS AND THE FILTER+IDEAL LEMMA. SmallGroup(288,1025) G L 11 β γ H α Let G = (A 4 A 4) C 2. G has a subgroup H = C 6 with H, G = N 5. Let H, G = {H, α, β, γ, G} = N 5. 1 G : H = 48
36 FINDING REPRESENTATIONS......USING SUBGROUP LATTICE INTERVALS AND THE FILTER+IDEAL LEMMA. SmallGroup(288,1025) G L 11 β γ K H α Let G = (A 4 A 4) C 2. G has a subgroup H = C 6 with H, G = N 5. Let H, G = {H, α, β, γ, G} = N 5. Sub(G) is a congruence lattice, so if there exists a subgroup K 1, below β and not below γ, then 1 G : H = 48 L 11 = K H.
37 FINDING REPRESENTATIONS......USING SUBGROUP LATTICE INTERVALS AND THE FILTER+IDEAL LEMMA. SmallGroup(288,1025) G L 11 β γ K H α Let G = (A 4 A 4) C 2. G has a subgroup H = C 6 with H, G = N 5. Let H, G = {H, α, β, γ, G} = N 5. Sub(G) is a congruence lattice, so if there exists a subgroup K 1, below β and not below γ, then 1 G : H = 48 L 11 = K H. L 17 A 4 P V 4 Sub(A 4) is a congruence lattice (of A 4 acting regularly on itself). Therefore, L 17 = V 4 P is a congruence lattice.
38 ARE ALL LATTICES WITH AT MOST 7 ELEMENTS REPRESENTABLE? L 19 L 20 L 17 L 13 L 11 L 9 L 10
39 ARE ALL LATTICES WITH AT MOST 7 ELEMENTS REPRESENTABLE? L 19 L 20 L 17 L 13 L 11 L 9 L 10
40 CONSTRUCTION OF AN ALGEBRA A WITH Con A = L 9. M 4 L 9
41 CONSTRUCTION OF AN ALGEBRA A WITH Con A = L 9. STEP 1 Take a permutational algebra B = B, F with congruence lattice Con B = M 4. 1 B α β γ δ Con B 0 B L 9
42 CONSTRUCTION OF AN ALGEBRA A WITH Con A = L 9. 1 B α β γ δ Con B 0 B STEP 1 Take a permutational algebra B = B, F with congruence lattice Con B = M 4. Example: Let B = {0, 1,..., 5} index the elements of S 3 and consider the right regular action of S 3 on itself. g 0 = (0, 4)(1, 3)(2, 5) and g 1 = (0, 1, 2)(3, 4, 5) generate this action group, the image of S 3 S 6. Con B, {g 0, g 1 } = M 4 with congruences α = , β = , γ = , δ = L 9
43 CONSTRUCTION OF AN ALGEBRA A WITH Con A = L 9. 1 B α β γ δ Con B 0 B STEP 1 Take a permutational algebra B = B, F with congruence lattice Con B = M 4. Example: Let B = {0, 1,..., 5} index the elements of S 3 and consider the right regular action of S 3 on itself. g 0 = (0, 4)(1, 3)(2, 5) and g 1 = (0, 1, 2)(3, 4, 5) generate this action group, the image of S 3 S 6. Con B, {g 0, g 1 } = M 4 with congruences α = , β = , γ = , δ = Goal: expand B to an algebra A that has α doubled in Con A. α α Con A
44 CONSTRUCTION OF AN ALGEBRA A WITH Con A = L 9. 1 B α β γ δ Con B 0 B STEP 1 Take a permutational algebra B = B, F with congruence lattice Con B = M 4. Example: Let B = {0, 1,..., 5} index the elements of S 3 and consider the right regular action of S 3 on itself. g 0 = (0, 4)(1, 3)(2, 5) and g 1 = (0, 1, 2)(3, 4, 5) generate this action group, the image of S 3 S 6. Con B, {g 0, g 1 } = M 4 with congruences α = , β = , γ = , δ = Goal: expand B to an algebra A that has α doubled in Con A. α α Con A STEP 2 Since α = Cg B (0, 2), we let A = B 0 B 1 B 2 where B 0 = {0, 1, 2, 3, 4, 5} = B B 1 = {0, 6, 7, 8, 9, 10} B 2 = {11, 12, 2, 13, 14, 15}.
45 CONSTRUCTION OF AN ALGEBRA A WITH Con A = L 9. 1 B α β γ δ Con B 0 B STEP 1 Take a permutational algebra B = B, F with congruence lattice Con B = M 4. Example: Let B = {0, 1,..., 5} index the elements of S 3 and consider the right regular action of S 3 on itself. g 0 = (0, 4)(1, 3)(2, 5) and g 1 = (0, 1, 2)(3, 4, 5) generate this action group, the image of S 3 S 6. Con B, {g 0, g 1 } = M 4 with congruences α = , β = , γ = , δ = Goal: expand B to an algebra A that has α doubled in Con A. α α Con A STEP 2 Since α = Cg B (0, 2), we let A = B 0 B 1 B 2 where B 0 = {0, 1, 2, 3, 4, 5} = B B 1 = {0, 6, 7, 8, 9, 10} B 2 = {11, 12, 2, 13, 14, 15}. STEP 3 Define unary operations e 0, e 1, e 2, s, g 0e 0, and g 1e 0.
46 CONSTRUCTION OF AN ALGEBRA A WITH Con A = L 9. 1 B 1 A α β γ δ α α β γ δ 0 B 0 A Con A, F A α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10, 13, 14, 15 α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10 13, 14, 15 β = 0, 3, 8 1, 4 2, 5, 15 6, 9 7, 10 11, 13 12, 14 γ = 0, 4, 9 1, 5 2, 3, 13 6, 10 7, 8 11, 14 12, 15 δ = 0, 5, 10 1, 3 2, 4, 14 6, 8 7, 9, 11, 15 12, 13
47 CONSTRUCTION OF AN ALGEBRA A WITH Con A = L 9. 1 B 1 A α β γ δ α α β γ δ 0 B 0 A Con A, F A α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10, 13, 14, 15 α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10 13, 14, 15 β = 0, 3, 8 1, 4 2, 5, 15 6, 9 7, 10 11, 13 12, 14 γ = 0, 4, 9 1, 5 2, 3, 13 6, 10 7, 8 11, 14 12, 15 δ = 0, 5, 10 1, 3 2, 4, 14 6, 8 7, 9, 11, 15 12, 13 α = α B 2 = α B 2, β = β B 2,...
48 B 0
49 α B 0 2 5
50 β B 0 2 5
51 γ B 0 2 5
52 δ B 0 2 5
53 α B B 0 = { }
54 B 1 B B 0 A = B 0 B 1 B 2 B 1 = { } B 0 = { } B 2 = { }
55 B 1 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
56 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
57 10 9 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
58 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
59 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
60 B 0 B A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
61 7 10 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
62 B 0 B 2 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
63 B 0 B 2 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
64 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
65 B 0 B 2 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
66 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
67 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
68 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
69 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
70 B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
71 B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
72 B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
73 B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
74 B A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
75 B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
76 B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
77 B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
78 B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
79 B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
80 B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
81 B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
82 B 1 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
83 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
84 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
85 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
86 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
87 B B A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
88 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
89 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
90 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
91 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
92 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
93 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
94 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
95 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
96 B A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
97 B A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
98 B A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
99 B A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
100 B A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
101 B A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
102 B A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
103 B A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
104 B A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
105 B A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
106 B A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
107 B A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
108 B A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
109 B 1 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
110 B 1 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
111 B 1 B B 0 A = B 0 B 1 B 2 Unary operations e 0: A B 0 e 1: A B 1 e 2: A B 2 s: A B 0 B 1 = { } B 0 = { } B 2 = { } ge 0: A e 0 g B 0 B 0
112 B 1 B B 0 Con A, F A α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10, 13, 14, 15 α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10 13, 14, 15 β = 0, 3, 8 1, 4 2, 5, 15 6, 9 7, 10 11, 13 12, 14 γ = 0, 4, 9 1, 5 2, 3, 13 6, 10 7, 8 11, 14 12, 15 δ = 0, 5, 10 1, 3 2, 4, 14 6, 8 7, 9, 11, 15 12, 13
113 B 1 B α B 0 Con A, F A α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10, 13, 14, 15 α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10 13, 14, 15 β = 0, 3, 8 1, 4 2, 5, 15 6, 9 7, 10 11, 13 12, 14 γ = 0, 4, 9 1, 5 2, 3, 13 6, 10 7, 8 11, 14 12, 15 δ = 0, 5, 10 1, 3 2, 4, 14 6, 8 7, 9, 11, 15 12, 13
114 B 1 B α 1 4 B Con A, F A α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10, 13, 14, 15 α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10 13, 14, 15 β = 0, 3, 8 1, 4 2, 5, 15 6, 9 7, 10 11, 13 12, 14 γ = 0, 4, 9 1, 5 2, 3, 13 6, 10 7, 8 11, 14 12, 15 δ = 0, 5, 10 1, 3 2, 4, 14 6, 8 7, 9, 11, 15 12, 13
115 B 1 B α 1 4 B Con A, F A α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10, 13, 14, 15 α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10 13, 14, 15 β = 0, 3, 8 1, 4 2, 5, 15 6, 9 7, 10 11, 13 12, 14 γ = 0, 4, 9 1, 5 2, 3, 13 6, 10 7, 8 11, 14 12, 15 δ = 0, 5, 10 1, 3 2, 4, 14 6, 8 7, 9, 11, 15 12, 13
116 B 1 B β B Con A, F A α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10, 13, 14, 15 α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10 13, 14, 15 β = 0, 3, 8 1, 4 2, 5, 15 6, 9 7, 10 11, 13 12, 14 γ = 0, 4, 9 1, 5 2, 3, 13 6, 10 7, 8 11, 14 12, 15 δ = 0, 5, 10 1, 3 2, 4, 14 6, 8 7, 9, 11, 15 12, 13
117 B 1 B β 1 4 B Con A, F A α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10, 13, 14, 15 α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10 13, 14, 15 β = 0, 3, 8 1, 4 2, 5, 15 6, 9 7, 10 11, 13 12, 14 γ = 0, 4, 9 1, 5 2, 3, 13 6, 10 7, 8 11, 14 12, 15 δ = 0, 5, 10 1, 3 2, 4, 14 6, 8 7, 9, 11, 15 12, 13
118 B 1 B β 1 4 B Con A, F A Why don t the β classes of B 1 and B 2 mix? α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10, 13, 14, 15 α = 0, 1, 2, 6, 7, 11, 12 3, 4, 5 8, 9, 10 13, 14, 15 β = 0, 3, 8 1, 4 2, 5, 15 6, 9 7, 10 11, 13 12, 14 γ = 0, 4, 9 1, 5 2, 3, 13 6, 10 7, 8 11, 14 12, 15 δ = 0, 5, 10 1, 3 2, 4, 14 6, 8 7, 9, 11, 15 12, 13
119 VARIATIONS ON THE SAME EXAMPLE... Suppose we want β = Cg B (0, 3) = 0, 3 2, 5 1, 4 to have non-trivial inverse image β 1 B = [β, β]. 1 B α β γ δ 0 B β B 0
120 VARIATIONS ON THE SAME EXAMPLE... Suppose we want β = Cg B (0, 3) = 0, 3 2, 5 1, 4 to have non-trivial inverse image β 1 B = [β, β]. Select elements 0 and 3 as intersection points: 1 B A = B 0 B 1 B 2 where α β γ δ B 0 = {0, 1, 2, 3, 4, 5} B 1 = {0, 6, 7, 8, 9, 10} B 2 = {11, 12, 13, 3, 14, 15}. 0 B β B 0
121 VARIATIONS ON THE SAME EXAMPLE... Suppose we want β = Cg B (0, 3) = 0, 3 2, 5 1, 4 to have non-trivial inverse image β 1 B = [β, β]. Select elements 0 and 3 as intersection points: 1 B A = B 0 B 1 B 2 where α β γ δ B 0 = {0, 1, 2, 3, 4, 5} B 1 = {0, 6, 7, 8, 9, 10} B 2 = {11, 12, 13, 3, 14, 15}. 0 B B β B B 0
122 VARIATIONS ON THE SAME EXAMPLE... Suppose we want β = Cg B (0, 3) = 0, 3 2, 5 1, 4 to have non-trivial inverse image β 1 B = [β, β]. Select elements 0 and 3 as intersection points: 1 B A = B 0 B 1 B 2 where α β γ δ B 0 = {0, 1, 2, 3, 4, 5} B 1 = {0, 6, 7, 8, 9, 10} B 2 = {11, 12, 13, 3, 14, 15}. 0 B B β α β β 1 A γ δ B B 0 0 A
123 VARIATIONS ON THE SAME EXAMPLE... Suppose we want β = Cg B (0, 3) = 0, 3 2, 5 1, 4 to have non-trivial inverse image β 1 B = [β, β]. Select elements 0 and 3 as intersection points: 1 B A = B 0 B 1 B 2 where α β γ δ B 0 = {0, 1, 2, 3, 4, 5} B 1 = {0, 6, 7, 8, 9, 10} B 2 = {11, 12, 13, 3, 14, 15}. 0 B B β α β β 1 A γ δ B B 0 0 A
124 VARIATIONS ON THE SAME EXAMPLE... Suppose we want β = Cg B (0, 3) = 0, 3 2, 5 1, 4 to have non-trivial inverse image β 1 B = [β, β]. Select elements 0 and 3 as intersection points: 1 B A = B 0 B 1 B 2 where α β γ δ B 0 = {0, 1, 2, 3, 4, 5} B 1 = {0, 6, 7, 8, 9, 10} B 2 = {11, 12, 13, 3, 14, 15}. 0 B B β ε α β 1 A β ε β γ δ B B 0 0 A
125 VARIATIONS ON THE SAME EXAMPLE... Suppose we want β = Cg B (0, 3) = 0, 3 2, 5 1, 4 to have non-trivial inverse image β 1 B = [β, β]. Select elements 0 and 3 as intersection points: 1 B A = B 0 B 1 B 2 where α β γ δ B 0 = {0, 1, 2, 3, 4, 5} B 1 = {0, 6, 7, 8, 9, 10} B 2 = {11, 12, 13, 3, 14, 15}. 0 B B β ε α β β ε β ε β 1 A γ δ B B 0 0 A
126 VARIATIONS ON THE SAME EXAMPLE... Suppose we want β = Cg B (0, 3) = 0, 3 2, 5 1, 4 to have non-trivial inverse image β 1 B = [β, β]. Select elements 0 and 3 as intersection points: 1 B A = B 0 B 1 B 2 where α β γ δ B 0 = {0, 1, 2, 3, 4, 5} B 1 = {0, 6, 7, 8, 9, 10} B 2 = {11, 12, 13, 3, 14, 15}. 0 B B β ε α β β ε β ε β 1 A γ δ B B 0 0 A
127 SEVEN ELEMENT LATTICES: SUMMARY L 19 L 20 L 17 L 13 L 11 L 9 L 7
128 SEVEN ELEMENT LATTICES: SUMMARY L 19 L 20 L 17 L 13 L 11 L 9 L 7
129 INTERVAL ENFORCEABLE PROPERTIES A minimal representation of L 7 must come from a transitive G-set. L 7
130 INTERVAL ENFORCEABLE PROPERTIES A minimal representation of L 7 must come from a transitive G-set. Suppose L 7 = H, G for some finite groups H < G. What can we say about the group G? G H
131 INTERVAL ENFORCEABLE PROPERTIES A minimal representation of L 7 must come from a transitive G-set. Suppose L 7 = H, G for some finite groups H < G. What can we say about the group G? If we prove G must have certain properties, then FLRP has a positive answer iff every finite lattice is an interval in the subgroup lattice of a group satisfying all of these properties. H G
132 INTERVAL ENFORCEABLE PROPERTIES A minimal representation of L 7 must come from a transitive G-set. Suppose L 7 = H, G for some finite groups H < G. What can we say about the group G? If we prove G must have certain properties, then FLRP has a positive answer iff every finite lattice is an interval in the subgroup lattice of a group satisfying all of these properties. PROPOSITION H G Suppose H < G, core G(H) = 1, L 7 = H, G. (I) G is a primitive permutation group. (II) If N G, then C G(N) = 1. (III) G contains no non-trivial abelian normal subgroup. (IV) G is not solvable. (V) G is subdirectly irreducible. (VI) With the possible exception of at most one maximal subgroup, all proper subgroups in the interval H, G are core-free.
133 OPEN PROBLEMS 1. Are homomorphic images of a representable lattices representable? 2. Are subdirect products of representable lattices representable? 3. Does representable imply group representable? i.e., is the congruence lattice of a finite algebra isomorphic to an interval in the subgroup lattice of a finite group? 4. Is the class of representable lattices recursive?
134 REMARKS ON THE PROBLEMS Are homomorphic images of representable lattices representable? If L = Con A, F and L is the homomorphic image {θ B 2 θ L}, where B = e(a) for some operation e 2 = e F, then L = Con B, F B.
135 REMARKS ON THE PROBLEMS Are homomorphic images of representable lattices representable? If L = Con A, F and L is the homomorphic image {θ B 2 θ L}, where B = e(a) for some operation e 2 = e F, then L = Con B, F B. Is the class of representable lattices recursive? In other words, is the membership problem for this class decidable? (Note that the class is recursively enumerable by the closure method.)
136 REMARKS ON THE PROBLEMS Are homomorphic images of representable lattices representable? If L = Con A, F and L is the homomorphic image {θ B 2 θ L}, where B = e(a) for some operation e 2 = e F, then L = Con B, F B. Is the class of representable lattices recursive? In other words, is the membership problem for this class decidable? (Note that the class is recursively enumerable by the closure method.) This question asks whether it s possible to write a program that, when given a finite lattice L, halts with output True if L is representable and False otherwise. A negative answer would solve the finite lattice representation problem.
137 OPEN PROBLEMS One approach: try to find a computable function f such that, if L is a representable lattice of size n, then L is representable as the congruence lattice of an algebra of cardinality f (n) or smaller. This would answer the decidability question, as follows: IsRepresentable( L ) { } n:= Size( L ) N:= f( n ) for each L in Sub[Eq(N)] { } if ( L isomorphic to L ) and ( L is closed ): return True return False
138 <ADVERTISEMENT> Workshop on Computational Universal Algebra Friday, October 4, 2013 University of Louisville, KY universalalgebra.wordpress.com
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