An introduction to relational complexity: …An introduction to relational complexity: background,...
Transcript of An introduction to relational complexity: …An introduction to relational complexity: background,...
An introduction to relational complexity: background,questions, and a few answers
Joshua Wiscons
California State University, Sacramento
Workshop on the Model Theory of Finite and PseudofiniteStructures
Joshua Wiscons Relational complexity
Relational Complexity
Let X = (X,R1, . . . ,Rm) be a finite relational structure.
Let L(X) be the language obtained by naming the definable subsets of allfinite powers of X.
Let Lk(X) be the language obtained by considering only the definablesubsets of Xi for i ≤ k.
DefinitionThe relational complexity of X is defined to be the least k such that for everyrelation R in L(X) there exists a quantifier-free formula ϕ in Lk(X) withX |= ∀x (R(x)↔ ϕ(x)).
Definition (via Homogeneity)The relational complexity of X is the least k such that X is equivalent to ahomogeneous structure (X, S1, . . . , Sn) with every Si of arity at most k.
Joshua Wiscons Relational complexity
Relational Complexity
Let X = (X,R1, . . . ,Rm) be a finite relational structure.
Let L(X) be the language obtained by naming the definable subsets of allfinite powers of X.
Let Lk(X) be the language obtained by considering only the definablesubsets of Xi for i ≤ k.
DefinitionThe relational complexity of X is defined to be the least k such that for everyrelation R in L(X) there exists a quantifier-free formula ϕ in Lk(X) withX |= ∀x (R(x)↔ ϕ(x)).
Definition (via Homogeneity)The relational complexity of X is the least k such that X is equivalent to ahomogeneous structure (X, S1, . . . , Sn) with every Si of arity at most k.
Joshua Wiscons Relational complexity
Relational Complexity
Let X = (X,R1, . . . ,Rm) be a finite relational structure.
Let L(X) be the language obtained by naming the definable subsets of allfinite powers of X.
Let Lk(X) be the language obtained by considering only the definablesubsets of Xi for i ≤ k.
DefinitionThe relational complexity of X is defined to be the least k such that for everyrelation R in L(X) there exists a quantifier-free formula ϕ in Lk(X) withX |= ∀x (R(x)↔ ϕ(x)).
Definition (via Homogeneity)The relational complexity of X is the least k such that X is equivalent to ahomogeneous structure (X, S1, . . . , Sn) with every Si of arity at most k.
Joshua Wiscons Relational complexity
Relational Complexity
Let X = (X,R1, . . . ,Rm) be a finite relational structure.
Let L(X) be the language obtained by naming the definable subsets of allfinite powers of X.
Let Lk(X) be the language obtained by considering only the definablesubsets of Xi for i ≤ k.
DefinitionThe relational complexity of X is defined to be the least k such that for everyrelation R in L(X) there exists a quantifier-free formula ϕ in Lk(X) withX |= ∀x (R(x)↔ ϕ(x)).
Definition (via Homogeneity)The relational complexity of X is the least k such that X is equivalent to ahomogeneous structure (X, S1, . . . , Sn) with every Si of arity at most k.
Joshua Wiscons Relational complexity
Relational Complexity
Let X = (X,R1, . . . ,Rm) be a finite relational structure.
Let L(X) be the language obtained by naming the definable subsets of allfinite powers of X.
Let Lk(X) be the language obtained by considering only the definablesubsets of Xi for i ≤ k.
DefinitionThe relational complexity of X is defined to be the least k such that for everyrelation R in L(X) there exists a quantifier-free formula ϕ in Lk(X) withX |= ∀x (R(x)↔ ϕ(x)).
Definition (via Homogeneity)The relational complexity of X is the least k such that X is equivalent to ahomogeneous structure (X, S1, . . . , Sn) with every Si of arity at most k.
Joshua Wiscons Relational complexity
Relational Complexity
Let X = (X,R1, . . . ,Rm) be a finite relational structure.
Let L(X) be the language obtained by naming the definable subsets of allfinite powers of X.
Let Lk(X) be the language obtained by considering only the definablesubsets of Xi for i ≤ k.
DefinitionThe relational complexity of X is defined to be the least k such that for everyrelation R in L(X) there exists a quantifier-free formula ϕ in Lk(X) withX |= ∀x (R(x)↔ ϕ(x)).
Definition (via Homogeneity)The relational complexity of X is the least k such that X is equivalent to ahomogeneous structure (X, S1, . . . , Sn) with every Si of arity at most k.
Joshua Wiscons Relational complexity
Relational Complexity
Let X = (X,R1, . . . ,Rm) be a finite relational structure.
Let L(X) be the language obtained by naming the definable subsets of allfinite powers of X.
Let Lk(X) be the language obtained by considering only the definablesubsets of Xi for i ≤ k.
DefinitionThe relational complexity of X is defined to be the least k such that for everyrelation R in L(X) there exists a quantifier-free formula ϕ in Lk(X) withX |= ∀x (R(x)↔ ϕ(x)).
Definition (via Homogeneity)The relational complexity of X is the least k such that X is equivalent to ahomogeneous structure (X, S1, . . . , Sn) with every Si of arity at most k.
Joshua Wiscons Relational complexity
Relational Complexity
Let X = (X,R1, . . . ,Rm) be a finite relational structure.
DefinitionThe relational complexity of X is defined to be the least k such that for everyrelation R in L(X) there exists a quantifier-free formula ϕ in Lk(X) withX |= ∀x (R(x)↔ ϕ(x)).
1 rc(X) is really an invariant of X, hence of Aut(X) = Aut(X).2 The definable subsets of Xk are unions of orbits of Aut(X) (acting
diagonally) on Xk.
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Joshua Wiscons Relational complexity
Relational Complexity
Let X = (X,R1, . . . ,Rm) be a finite relational structure.
DefinitionThe relational complexity of X is defined to be the least k such that for everyrelation R in L(X) there exists a quantifier-free formula ϕ in Lk(X) withX |= ∀x (R(x)↔ ϕ(x)).
1 rc(X) is really an invariant of X, hence of Aut(X) = Aut(X).
2 The definable subsets of Xk are unions of orbits of Aut(X) (actingdiagonally) on Xk.
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Joshua Wiscons Relational complexity
Relational Complexity
Let X = (X,R1, . . . ,Rm) be a finite relational structure.
DefinitionThe relational complexity of X is defined to be the least k such that for everyrelation R in L(X) there exists a quantifier-free formula ϕ in Lk(X) withX |= ∀x (R(x)↔ ϕ(x)).
1 rc(X) is really an invariant of X, hence of Aut(X) = Aut(X).2 The definable subsets of Xk are unions of orbits of Aut(X) (acting
diagonally) on Xk.
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Joshua Wiscons Relational complexity
Relational Complexity
Let X = (X,R1, . . . ,Rm) be a finite relational structure.
DefinitionThe relational complexity of X is defined to be the least k such that for everyrelation R in L(X) there exists a quantifier-free formula ϕ in Lk(X) withX |= ∀x (R(x)↔ ϕ(x)).
1 rc(X) is really an invariant of X, hence of Aut(X) = Aut(X).2 The definable subsets of Xk are unions of orbits of Aut(X) (acting
diagonally) on Xk.
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Joshua Wiscons Relational complexity
Relational Complexity
Let X = (X,R1, . . . ,Rm) be a finite relational structure.
DefinitionThe relational complexity of X is defined to be the least k such that for everyrelation R in L(X) there exists a quantifier-free formula ϕ in Lk(X) withX |= ∀x (R(x)↔ ϕ(x)).
1 rc(X) is really an invariant of X, hence of Aut(X) = Aut(X).2 The definable subsets of Xk are unions of orbits of Aut(X) (acting
diagonally) on Xk.
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Joshua Wiscons Relational complexity
Relational Complexity
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Terminology
Let (X,G) be a permutation group.
1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).2 The orbits of G on Xn are called n-types.3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are in
the same orbit as the corresponding k elements from y.
(x1, x2, x3)
(y1, y2, y3)
g
(3-)equivalence
(x1, x2, x3)
(y1, y2, y3)
g1 g2
g3
2-equivalence
Note: 3-equivalenceimplies 2-equivalence
Joshua Wiscons Relational complexity
Relational Complexity
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Terminology
Let (X,G) be a permutation group.
1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).2 The orbits of G on Xn are called n-types.3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are in
the same orbit as the corresponding k elements from y.
(x1, x2, x3)
(y1, y2, y3)
g
(3-)equivalence
(x1, x2, x3)
(y1, y2, y3)
g1 g2
g3
2-equivalence
Note: 3-equivalenceimplies 2-equivalence
Joshua Wiscons Relational complexity
Relational Complexity
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Terminology
Let (X,G) be a permutation group.1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).
2 The orbits of G on Xn are called n-types.3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are in
the same orbit as the corresponding k elements from y.
(x1, x2, x3)
(y1, y2, y3)
g
(3-)equivalence
(x1, x2, x3)
(y1, y2, y3)
g1 g2
g3
2-equivalence
Note: 3-equivalenceimplies 2-equivalence
Joshua Wiscons Relational complexity
Relational Complexity
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Terminology
Let (X,G) be a permutation group.1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).2 The orbits of G on Xn are called n-types.
3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are inthe same orbit as the corresponding k elements from y.
(x1, x2, x3)
(y1, y2, y3)
g
(3-)equivalence
(x1, x2, x3)
(y1, y2, y3)
g1 g2
g3
2-equivalence
Note: 3-equivalenceimplies 2-equivalence
Joshua Wiscons Relational complexity
Relational Complexity
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Terminology
Let (X,G) be a permutation group.1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).2 The orbits of G on Xn are called n-types.3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are in
the same orbit as the corresponding k elements from y.
(x1, x2, x3)
(y1, y2, y3)
g
(3-)equivalence
(x1, x2, x3)
(y1, y2, y3)
g1 g2
g3
2-equivalence
Note: 3-equivalenceimplies 2-equivalence
Joshua Wiscons Relational complexity
Relational Complexity
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Terminology
Let (X,G) be a permutation group.1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).2 The orbits of G on Xn are called n-types.3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are in
the same orbit as the corresponding k elements from y.
(x1, x2, x3)
(y1, y2, y3)
g
(3-)equivalence
(x1, x2, x3)
(y1, y2, y3)
g1 g2
g3
2-equivalence
Note: 3-equivalenceimplies 2-equivalence
Joshua Wiscons Relational complexity
Relational Complexity
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Terminology
Let (X,G) be a permutation group.1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).2 The orbits of G on Xn are called n-types.3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are in
the same orbit as the corresponding k elements from y.
(x1, x2, x3)
(y1, y2, y3)
g
(3-)equivalence
(x1, x2, x3)
(y1, y2, y3)
g1 g2
g3
2-equivalence
Note: 3-equivalenceimplies 2-equivalence
Joshua Wiscons Relational complexity
Relational Complexity
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Terminology
Let (X,G) be a permutation group.1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).2 The orbits of G on Xn are called n-types.3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are in
the same orbit as the corresponding k elements from y.
(x1, x2, x3)
(y1, y2, y3)
g
(3-)equivalence
(x1, x2, x3)
(y1, y2, y3)
g1 g2
g3
2-equivalence
Note: 3-equivalenceimplies 2-equivalence
Joshua Wiscons Relational complexity
Relational Complexity
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Terminology
Let (X,G) be a permutation group.1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).2 The orbits of G on Xn are called n-types.3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are in
the same orbit as the corresponding k elements from y.
(x1, x2, x3)
(y1, y2, y3)
g
(3-)equivalence
(x1, x2, x3)
(y1, y2, y3)
g1
g2
g3
2-equivalence
Note: 3-equivalenceimplies 2-equivalence
Joshua Wiscons Relational complexity
Relational Complexity
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Terminology
Let (X,G) be a permutation group.1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).2 The orbits of G on Xn are called n-types.3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are in
the same orbit as the corresponding k elements from y.
(x1, x2, x3)
(y1, y2, y3)
g
(3-)equivalence
(x1, x2, x3)
(y1, y2, y3)
g1 g2
g3
2-equivalence
Note: 3-equivalenceimplies 2-equivalence
Joshua Wiscons Relational complexity
Relational Complexity
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Terminology
Let (X,G) be a permutation group.1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).2 The orbits of G on Xn are called n-types.3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are in
the same orbit as the corresponding k elements from y.
(x1, x2, x3)
(y1, y2, y3)
g
(3-)equivalence
(x1, x2, x3)
(y1, y2, y3)
g1 g2
g3
2-equivalence
Note: 3-equivalenceimplies 2-equivalence
Joshua Wiscons Relational complexity
Relational Complexity
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Terminology
Let (X,G) be a permutation group.1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).2 The orbits of G on Xn are called n-types.3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are in
the same orbit as the corresponding k elements from y.
(x1, x2, x3)
(y1, y2, y3)
g
(3-)equivalence
(x1, x2, x3)
(y1, y2, y3)
g1 g2
g3
2-equivalence
Note: 3-equivalenceimplies 2-equivalence
Joshua Wiscons Relational complexity
Relational Complexity
Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.
Terminology
Let (X,G) be a permutation group.1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).2 The orbits of G on Xn are called n-types.3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are in
the same orbit as the corresponding k elements from y.
DefinitionThe relational complexity of a permutation group (X,G) is the smallest k suchthat for all n ≥ k, k-equivalence of n-tuples implies equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.
Question: (1, 2, 3) ∼ (1, 5, 4)? Yes.(1, 2, 3)
(1, 5, 4)
?reflection: (25)(34)(7 10)(89)
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2.
It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.Question: (1, 2, 3) ∼ (1, 5, 4)?
Yes.
(1, 2, 3)
(1, 5, 4)
?
reflection: (25)(34)(7 10)(89)
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2.
It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.Question: (1, 2, 3) ∼ (1, 5, 4)? Yes.
(1, 2, 3)
(1, 5, 4)
?
reflection: (25)(34)(7 10)(89)
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2.
It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.
Question: (1, 3, 7) ∼ (1, 3, 9)?(1, 3, 7)
(1, 3, 9)
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2.
It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼ (1, 3, 9)?
(1, 3, 7)
(1, 3, 9)
?
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2.
It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼ (1, 3, 9)? No.
(1, 3, 7)
(1, 3, 9)
?
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2.
It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼2 (1, 3, 9)?
(1, 3, 7)
(1, 3, 9)
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2.
It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼2 (1, 3, 9)?
(1, 3, 7)
(1, 3, 9)
id
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2.
It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼2 (1, 3, 9)?
(1, 3, 7)
(1, 3, 9)
id
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2.
It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼2 (1, 3, 9)?
(1, 3, 7)
(1, 3, 9)
reflection
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2.
It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼2 (1, 3, 9)?
(1, 3, 7)
(1, 3, 9)
reflection
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2.
It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼2 (1, 3, 9)?
(1, 3, 7)
(1, 3, 9)
it works . . . but hard to see
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2.
It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼2 (1, 3, 9)?
(1, 3, 7)
(1, 3, 9)
it works . . . but hard to see
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2.
It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼2 (1, 3, 9)? Yes!
(1, 3, 7)
(1, 3, 9)
it works . . . but hard to see
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2.
It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼2 (1, 3, 9)? Yes!
(1, 3, 7)
(1, 3, 9)
it works . . . but hard to see
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2.
It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼2 (1, 3, 9)? Yes!
(1, 3, 7)
(1, 3, 9)
it works . . . but hard to see
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2.
It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼2 (1, 3, 9)? Yes!
(1, 3, 7)
(1, 3, 9)
it works . . . but hard to see
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2. It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
A First Example
Example (Petersen Graph)
1
62
7
3
8
4
9
510
Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼2 (1, 3, 9)? Yes!
(1, 3, 7)
(1, 3, 9)
it works . . . but hard to see
Thus, 2-equivalence does not imply equivalence.
Thus, rc(G) > 2. It turns out that rc(G) = 3.
RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries
3 rc(X,G) ≤ max(2, |X| − 1)
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 2
2 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries
3 rc(X,G) ≤ max(2, |X| − 1)
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 2
2 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries
3 rc(X,G) ≤ max(2, |X| − 1)
Proof.
Choose a nontrivial g ∈ G.
Choose distinct x, y ∈ X with xg = y.
Then (x, x) and (x, y) are 1-equivalent but not 2-equivalent.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 2
2 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries
3 rc(X,G) ≤ max(2, |X| − 1)
Proof.
Choose a nontrivial g ∈ G.
Choose distinct x, y ∈ X with xg = y.
Then (x, x) and (x, y) are 1-equivalent but not 2-equivalent.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 2
2 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries
3 rc(X,G) ≤ max(2, |X| − 1)
Proof.
Choose a nontrivial g ∈ G.
Choose distinct x, y ∈ X with xg = y.
Then (x, x) and (x, y) are 1-equivalent but not 2-equivalent.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 2
2 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries
3 rc(X,G) ≤ max(2, |X| − 1)
Proof.
Choose a nontrivial g ∈ G.
Choose distinct x, y ∈ X with xg = y.
Then (x, x) and (x, y) are 1-equivalent but not 2-equivalent.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 2
2 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries
3 rc(X,G) ≤ max(2, |X| − 1)
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries
3 rc(X,G) ≤ max(2, |X| − 1)
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Proof.
Let x and y be (|X| − 1)-equivalent.
We may assume that neither tuple has repeated entries, so
x and y areenumerations of X.
If g ∈ G takes the first (|X| − 1) entries of x to the first (|X| − 1) entriesy, it must take the remaining entry of x to the remaining entry of y.
Thus, x and y are |X|-equivalent.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Proof.
Let x and y be (|X| − 1)-equivalent.
We may assume that neither tuple has repeated entries, so
x and y areenumerations of X.
If g ∈ G takes the first (|X| − 1) entries of x to the first (|X| − 1) entriesy, it must take the remaining entry of x to the remaining entry of y.
Thus, x and y are |X|-equivalent.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Proof.
Let x and y be (|X| − 1)-equivalent.
We may assume that neither tuple has repeated entries, so x and y areenumerations of X.
If g ∈ G takes the first (|X| − 1) entries of x to the first (|X| − 1) entriesy, it must take the remaining entry of x to the remaining entry of y.
Thus, x and y are |X|-equivalent.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Proof.
Let x and y be (|X| − 1)-equivalent.
We may assume that neither tuple has repeated entries, so x and y areenumerations of X.
If g ∈ G takes the first (|X| − 1) entries of x to the first (|X| − 1) entriesy, it must take the remaining entry of x to the remaining entry of y.
Thus, x and y are |X|-equivalent.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Proof.
Let x and y be (|X| − 1)-equivalent.
We may assume that neither tuple has repeated entries, so x and y areenumerations of X.
If g ∈ G takes the first (|X| − 1) entries of x to the first (|X| − 1) entriesy, it must take the remaining entry of x to the remaining entry of y.
Thus, x and y are |X|-equivalent.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) =
22 rc(X,Alt(n)) =
n−1
Assume x and y are 2-equivalent.
x1 x2 · · · xn
y1 y2 · · · yn
· · ·
all distinct
all distinct
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) =
22 rc(X,Alt(n)) =
n−1
Assume x and y are 2-equivalent.
x1 x2 · · · xn
y1 y2 · · · yn
· · ·
all distinct
all distinct
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 2
2 rc(X,Alt(n)) =
n−1
Assume x and y are 2-equivalent.
x1 x2 · · · xn
y1 y2 · · · yn
· · ·
all distinct
all distinct
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 2
2 rc(X,Alt(n)) =
n−1
Assume x and y are 2-equivalent.
x1 x2 · · · xn
y1 y2 · · · yn
· · ·
all distinct
all distinct
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 2
2 rc(X,Alt(n)) =
n−1
Assume x and y are 2-equivalent.
x1 x2 · · · xn
y1 y2 · · · yn
· · ·
all distinct
all distinct
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 2
2 rc(X,Alt(n)) =
n−1
Assume x and y are 2-equivalent.
x1 x2 · · · xn
y1 y2 · · · yn
· · ·
all distinct
all distinct
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 2
2 rc(X,Alt(n)) =
n−1
Assume x and y are 2-equivalent.
x1 x2 · · · xn
y1 y2 · · · yn
?· · ·
all distinct
all distinct
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 2
2 rc(X,Alt(n)) =
n−1
Assume x and y are 2-equivalent.
x1 x2 · · · xn
y1 y2 · · · yn
Of Course!· · ·
all distinct
all distinct
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 2
2 rc(X,Alt(n)) =
n−1
Assume rc(X,Alt(n)) = r with 2 ≤ r ≤ n− 2.
Let x, y be any two enumerations of X.
Key point: (X,Alt(n)) is (n− 2)-transitive.
Thus, x and y are (n− 2)-equivalent.
So, if r ≤ n− 2, x and y are equivalent.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 22 rc(X,Alt(n)) =
n−1
Assume rc(X,Alt(n)) = r with 2 ≤ r ≤ n− 2.
Let x, y be any two enumerations of X.
Key point: (X,Alt(n)) is (n− 2)-transitive.
Thus, x and y are (n− 2)-equivalent.
So, if r ≤ n− 2, x and y are equivalent.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 22 rc(X,Alt(n)) = n−1
Assume rc(X,Alt(n)) = r with 2 ≤ r ≤ n− 2.
Let x, y be any two enumerations of X.
Key point: (X,Alt(n)) is (n− 2)-transitive.
Thus, x and y are (n− 2)-equivalent.
So, if r ≤ n− 2, x and y are equivalent.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 22 rc(X,Alt(n)) = n−1
Assume rc(X,Alt(n)) = r with 2 ≤ r ≤ n− 2.
Let x, y be any two enumerations of X.
Key point: (X,Alt(n)) is (n− 2)-transitive.
Thus, x and y are (n− 2)-equivalent.
So, if r ≤ n− 2, x and y are equivalent.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 22 rc(X,Alt(n)) = n−1
Assume rc(X,Alt(n)) = r with 2 ≤ r ≤ n− 2.
Let x, y be any two enumerations of X.
Key point: (X,Alt(n)) is (n− 2)-transitive.
Thus, x and y are (n− 2)-equivalent.
So, if r ≤ n− 2, x and y are equivalent.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 22 rc(X,Alt(n)) = n−1
Assume rc(X,Alt(n)) = r with 2 ≤ r ≤ n− 2.
Let x, y be any two enumerations of X.
Key point: (X,Alt(n)) is (n− 2)-transitive.
Thus, x and y are (n− 2)-equivalent.
So, if r ≤ n− 2, x and y are equivalent.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 22 rc(X,Alt(n)) = n−1
Assume rc(X,Alt(n)) = r with 2 ≤ r ≤ n− 2.
Let x, y be any two enumerations of X.
Key point: (X,Alt(n)) is (n− 2)-transitive.
Thus, x and y are (n− 2)-equivalent.
So, if r ≤ n− 2, x and y are equivalent.
Joshua Wiscons Relational complexity
Remarks And More Examples
RemarkLet (X,G) be a nontrivial permutation group.
1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))
where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)
Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 22 rc(X,Alt(n)) = n−1
Assume rc(X,Alt(n)) = r with 2 ≤ r ≤ n− 2.
Let x, y be any two enumerations of X.
Key point: (X,Alt(n)) is (n− 2)-transitive.
Thus, x and y are (n− 2)-equivalent.
So, if r ≤ n− 2, x and y are equivalent.
Joshua Wiscons Relational complexity
Some Problems (posed by Cherlin)
Problems
1 Gather data: determine the relational complexities of natural permutationgroups.
E.g. study the various natural actions of classical groups.E.g, study the various natural actions of Sn (and An).
2 Classify the finite permutation groups of relational complexity at most r.
Need to restrict to so-called primitive groups.E.g, what are the finite primitive permutation groups of relationalcomplexity 2?
3 Classify the finite permutation groups of relational complexity at least rwhen r is “big” compared to |X|.
Joshua Wiscons Relational complexity
Some Problems (posed by Cherlin)
Problems1 Gather data: determine the relational complexities of natural permutation
groups.
E.g. study the various natural actions of classical groups.E.g, study the various natural actions of Sn (and An).
2 Classify the finite permutation groups of relational complexity at most r.
Need to restrict to so-called primitive groups.E.g, what are the finite primitive permutation groups of relationalcomplexity 2?
3 Classify the finite permutation groups of relational complexity at least rwhen r is “big” compared to |X|.
Joshua Wiscons Relational complexity
Some Problems (posed by Cherlin)
Problems1 Gather data: determine the relational complexities of natural permutation
groups.E.g. study the various natural actions of classical groups.
E.g, study the various natural actions of Sn (and An).2 Classify the finite permutation groups of relational complexity at most r.
Need to restrict to so-called primitive groups.E.g, what are the finite primitive permutation groups of relationalcomplexity 2?
3 Classify the finite permutation groups of relational complexity at least rwhen r is “big” compared to |X|.
Joshua Wiscons Relational complexity
Some Problems (posed by Cherlin)
Problems1 Gather data: determine the relational complexities of natural permutation
groups.E.g. study the various natural actions of classical groups.E.g, study the various natural actions of Sn (and An).
2 Classify the finite permutation groups of relational complexity at most r.
Need to restrict to so-called primitive groups.E.g, what are the finite primitive permutation groups of relationalcomplexity 2?
3 Classify the finite permutation groups of relational complexity at least rwhen r is “big” compared to |X|.
Joshua Wiscons Relational complexity
Some Problems (posed by Cherlin)
Problems1 Gather data: determine the relational complexities of natural permutation
groups.E.g. study the various natural actions of classical groups.E.g, study the various natural actions of Sn (and An).
2 Classify the finite permutation groups of relational complexity at most r.
Need to restrict to so-called primitive groups.E.g, what are the finite primitive permutation groups of relationalcomplexity 2?
3 Classify the finite permutation groups of relational complexity at least rwhen r is “big” compared to |X|.
Joshua Wiscons Relational complexity
Some Problems (posed by Cherlin)
Problems1 Gather data: determine the relational complexities of natural permutation
groups.E.g. study the various natural actions of classical groups.E.g, study the various natural actions of Sn (and An).
2 Classify the finite permutation groups of relational complexity at most r.Need to restrict to so-called primitive groups.
E.g, what are the finite primitive permutation groups of relationalcomplexity 2?
3 Classify the finite permutation groups of relational complexity at least rwhen r is “big” compared to |X|.
Joshua Wiscons Relational complexity
Some Problems (posed by Cherlin)
Problems1 Gather data: determine the relational complexities of natural permutation
groups.E.g. study the various natural actions of classical groups.E.g, study the various natural actions of Sn (and An).
2 Classify the finite permutation groups of relational complexity at most r.Need to restrict to so-called primitive groups.E.g, what are the finite primitive permutation groups of relationalcomplexity 2?
3 Classify the finite permutation groups of relational complexity at least rwhen r is “big” compared to |X|.
Joshua Wiscons Relational complexity
Some Problems (posed by Cherlin)
Problems1 Gather data: determine the relational complexities of natural permutation
groups.E.g. study the various natural actions of classical groups.E.g, study the various natural actions of Sn (and An).
2 Classify the finite permutation groups of relational complexity at most r.Need to restrict to so-called primitive groups.E.g, what are the finite primitive permutation groups of relationalcomplexity 2?
3 Classify the finite permutation groups of relational complexity at least rwhen r is “big” compared to |X|.
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2.
Then rc(V,GL(V)) = ?1 Lower bound:
rc(V,GL(V)) ≥ d + 1Let x = (e1, . . . , ed,
∑ei) and y = (e1, . . . , ed,
∑2ei).
Key point: GL(V) is transitive on the set of bases.Thus, x and y are d-equivalent.But, x and y are not equivalent.
2 Upper bound:
rc(V,GL(V)) ≤ d + 1
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?
1 Lower bound:
rc(V,GL(V)) ≥ d + 1Let x = (e1, . . . , ed,
∑ei) and y = (e1, . . . , ed,
∑2ei).
Key point: GL(V) is transitive on the set of bases.Thus, x and y are d-equivalent.But, x and y are not equivalent.
2 Upper bound:
rc(V,GL(V)) ≤ d + 1
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound:
rc(V,GL(V)) ≥ d + 1Let x = (e1, . . . , ed,
∑ei) and y = (e1, . . . , ed,
∑2ei).
Key point: GL(V) is transitive on the set of bases.Thus, x and y are d-equivalent.But, x and y are not equivalent.
2 Upper bound:
rc(V,GL(V)) ≤ d + 1
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound:
rc(V,GL(V)) ≥ d + 1
Let x = (e1, . . . , ed,∑
ei) and y = (e1, . . . , ed,∑
2ei).
Key point: GL(V) is transitive on the set of bases.Thus, x and y are d-equivalent.But, x and y are not equivalent.
2 Upper bound:
rc(V,GL(V)) ≤ d + 1
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound:
rc(V,GL(V)) ≥ d + 1
Let x = (e1, . . . , ed,∑
ei) and y = (e1, . . . , ed,∑
2ei).Key point: GL(V) is transitive on the set of bases.
Thus, x and y are d-equivalent.But, x and y are not equivalent.
2 Upper bound:
rc(V,GL(V)) ≤ d + 1
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound:
rc(V,GL(V)) ≥ d + 1
Let x = (e1, . . . , ed,∑
ei) and y = (e1, . . . , ed,∑
2ei).Key point: GL(V) is transitive on the set of bases.Thus, x and y are d-equivalent.
But, x and y are not equivalent.2 Upper bound:
rc(V,GL(V)) ≤ d + 1
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound:
rc(V,GL(V)) ≥ d + 1
Let x = (e1, . . . , ed,∑
ei) and y = (e1, . . . , ed,∑
2ei).Key point: GL(V) is transitive on the set of bases.Thus, x and y are d-equivalent.But, x and y are not equivalent.
2 Upper bound:
rc(V,GL(V)) ≤ d + 1
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 1
Let x = (e1, . . . , ed,∑
ei) and y = (e1, . . . , ed,∑
2ei).Key point: GL(V) is transitive on the set of bases.Thus, x and y are d-equivalent.But, x and y are not equivalent.
2 Upper bound:
rc(V,GL(V)) ≤ d + 1
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 1
Let x = (e1, . . . , ed,∑
ei) and y = (e1, . . . , ed,∑
2ei).Key point: GL(V) is transitive on the set of bases.Thus, x and y are d-equivalent.But, x and y are not equivalent.
2 Upper bound:
rc(V,GL(V)) ≤ d + 1
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 1
Let x = (e1, . . . , ed,∑
ei) and y = (e1, . . . , ed,∑
2ei).Key point: GL(V) is transitive on the set of bases.Thus, x and y are d-equivalent.But, x and y are not equivalent.
2 Upper bound: rc(V,GL(V)) ≤ d + 1
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 1
Let x = (e1, . . . , ed,∑
ei) and y = (e1, . . . , ed,∑
2ei).Key point: GL(V) is transitive on the set of bases.Thus, x and y are d-equivalent.But, x and y are not equivalent.
2 Upper bound: rc(V,GL(V)) ≤ d + 1
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1
Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that
- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to y
m + 1 equivalence implies that
- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym
=⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1
Repeating, we find that x′i = yi for all i ≥ mThus, x ∼ (y1, . . . , ym, x′m+1, . . . , x
′n) = y
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1
Want to show that (d + 1)-equivalence implies equivalence
Assume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that
- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to y
m + 1 equivalence implies that
- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym
=⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1
Repeating, we find that x′i = yi for all i ≥ mThus, x ∼ (y1, . . . , ym, x′m+1, . . . , x
′n) = y
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1
Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalent
Let m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that
- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to y
m + 1 equivalence implies that
- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym
=⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1
Repeating, we find that x′i = yi for all i ≥ mThus, x ∼ (y1, . . . , ym, x′m+1, . . . , x
′n) = y
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1
Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)
Without loss, assume that x1, . . . , xm are independentm equivalence implies that
- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to y
m + 1 equivalence implies that
- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym
=⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1
Repeating, we find that x′i = yi for all i ≥ mThus, x ∼ (y1, . . . , ym, x′m+1, . . . , x
′n) = y
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1
Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independent
m equivalence implies that
- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to y
m + 1 equivalence implies that
- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym
=⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1
Repeating, we find that x′i = yi for all i ≥ mThus, x ∼ (y1, . . . , ym, x′m+1, . . . , x
′n) = y
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1
Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that
- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to y
m + 1 equivalence implies that
- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym
=⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1
Repeating, we find that x′i = yi for all i ≥ mThus, x ∼ (y1, . . . , ym, x′m+1, . . . , x
′n) = y
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1
Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that
- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and
- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to ym + 1 equivalence implies that
- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym
=⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1
Repeating, we find that x′i = yi for all i ≥ mThus, x ∼ (y1, . . . , ym, x′m+1, . . . , x
′n) = y
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1
Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that
- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to y
m + 1 equivalence implies that
- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym
=⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1
Repeating, we find that x′i = yi for all i ≥ mThus, x ∼ (y1, . . . , ym, x′m+1, . . . , x
′n) = y
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1
Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that
- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to y
m + 1 equivalence implies that
- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym
=⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1
Repeating, we find that x′i = yi for all i ≥ mThus, x ∼ (y1, . . . , ym, x′m+1, . . . , x
′n) = y
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1
Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that
- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to y
m + 1 equivalence implies that- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so
- g fixes y1, . . . , ym
=⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1
Repeating, we find that x′i = yi for all i ≥ mThus, x ∼ (y1, . . . , ym, x′m+1, . . . , x
′n) = y
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1
Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that
- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to y
m + 1 equivalence implies that- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym
=⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1
Repeating, we find that x′i = yi for all i ≥ mThus, x ∼ (y1, . . . , ym, x′m+1, . . . , x
′n) = y
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1
Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that
- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to y
m + 1 equivalence implies that- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym =⇒ g fixes span{y1, . . . , ym}
=⇒ x′m+1 = ym+1
Repeating, we find that x′i = yi for all i ≥ mThus, x ∼ (y1, . . . , ym, x′m+1, . . . , x
′n) = y
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1
Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that
- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to y
m + 1 equivalence implies that- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym =⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1
Repeating, we find that x′i = yi for all i ≥ mThus, x ∼ (y1, . . . , ym, x′m+1, . . . , x
′n) = y
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1
Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that
- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to y
m + 1 equivalence implies that- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym =⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1
Repeating, we find that x′i = yi for all i ≥ m
Thus, x ∼ (y1, . . . , ym, x′m+1, . . . , x′n) = y
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1
Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that
- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to y
m + 1 equivalence implies that- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym =⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1
Repeating, we find that x′i = yi for all i ≥ mThus, x ∼ (y1, . . . , ym, x′m+1, . . . , x
′n) = y
Joshua Wiscons Relational complexity
An easy, natural example
Example (GL(V))
Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = d + 11 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1
Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that
- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to y
m + 1 equivalence implies that- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym =⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1
Repeating, we find that x′i = yi for all i ≥ mThus, x ∼ (y1, . . . , ym, x′m+1, . . . , x
′n) = y
Joshua Wiscons Relational complexity
A harder, natural example
DefinitionPm(n) is the set of all partitions of {1, . . . , n} into subsets of (equal) size m.
Example
P4(8) :
[1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]
P2(8) :
[12 | 34 | 56 | 78], . . .
P3(9) :
[147 | 258 | 369], . . .
Of course, there are many partitions of a given set, e.g. |P3(9)| = 280.
Definition (Action on Partitions)
Every σ ∈ Sn can be naturally regarded as a permutation of Pm(n) via the rule
σ ([x1 · · · xm | y1 · · · ym | · · · ]) = [σ(x1) · · ·σ(xm) | σ(y1) · · ·σ(ym) | · · · ].
Joshua Wiscons Relational complexity
A harder, natural example
DefinitionPm(n) is the set of all partitions of {1, . . . , n} into subsets of (equal) size m.
Example
P4(8) :
[1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]
P2(8) :
[12 | 34 | 56 | 78], . . .
P3(9) :
[147 | 258 | 369], . . .
Of course, there are many partitions of a given set, e.g. |P3(9)| = 280.
Definition (Action on Partitions)
Every σ ∈ Sn can be naturally regarded as a permutation of Pm(n) via the rule
σ ([x1 · · · xm | y1 · · · ym | · · · ]) = [σ(x1) · · ·σ(xm) | σ(y1) · · ·σ(ym) | · · · ].
Joshua Wiscons Relational complexity
A harder, natural example
DefinitionPm(n) is the set of all partitions of {1, . . . , n} into subsets of (equal) size m.
ExampleP4(8) :
[1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]
P2(8) :
[12 | 34 | 56 | 78], . . .
P3(9) :
[147 | 258 | 369], . . .
Of course, there are many partitions of a given set, e.g. |P3(9)| = 280.
Definition (Action on Partitions)
Every σ ∈ Sn can be naturally regarded as a permutation of Pm(n) via the rule
σ ([x1 · · · xm | y1 · · · ym | · · · ]) = [σ(x1) · · ·σ(xm) | σ(y1) · · ·σ(ym) | · · · ].
Joshua Wiscons Relational complexity
A harder, natural example
DefinitionPm(n) is the set of all partitions of {1, . . . , n} into subsets of (equal) size m.
ExampleP4(8) : [1234 | 5678],
[2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]
P2(8) :
[12 | 34 | 56 | 78], . . .
P3(9) :
[147 | 258 | 369], . . .
Of course, there are many partitions of a given set, e.g. |P3(9)| = 280.
Definition (Action on Partitions)
Every σ ∈ Sn can be naturally regarded as a permutation of Pm(n) via the rule
σ ([x1 · · · xm | y1 · · · ym | · · · ]) = [σ(x1) · · ·σ(xm) | σ(y1) · · ·σ(ym) | · · · ].
Joshua Wiscons Relational complexity
A harder, natural example
DefinitionPm(n) is the set of all partitions of {1, . . . , n} into subsets of (equal) size m.
ExampleP4(8) : [1234 | 5678], [2468 | 1357],
. . . note: [1234 | 5678] = [8765 | 4321]
P2(8) :
[12 | 34 | 56 | 78], . . .
P3(9) :
[147 | 258 | 369], . . .
Of course, there are many partitions of a given set, e.g. |P3(9)| = 280.
Definition (Action on Partitions)
Every σ ∈ Sn can be naturally regarded as a permutation of Pm(n) via the rule
σ ([x1 · · · xm | y1 · · · ym | · · · ]) = [σ(x1) · · ·σ(xm) | σ(y1) · · ·σ(ym) | · · · ].
Joshua Wiscons Relational complexity
A harder, natural example
DefinitionPm(n) is the set of all partitions of {1, . . . , n} into subsets of (equal) size m.
ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . .
note: [1234 | 5678] = [8765 | 4321]
P2(8) :
[12 | 34 | 56 | 78], . . .
P3(9) :
[147 | 258 | 369], . . .
Of course, there are many partitions of a given set, e.g. |P3(9)| = 280.
Definition (Action on Partitions)
Every σ ∈ Sn can be naturally regarded as a permutation of Pm(n) via the rule
σ ([x1 · · · xm | y1 · · · ym | · · · ]) = [σ(x1) · · ·σ(xm) | σ(y1) · · ·σ(ym) | · · · ].
Joshua Wiscons Relational complexity
A harder, natural example
DefinitionPm(n) is the set of all partitions of {1, . . . , n} into subsets of (equal) size m.
ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]
P2(8) :
[12 | 34 | 56 | 78], . . .
P3(9) :
[147 | 258 | 369], . . .
Of course, there are many partitions of a given set, e.g. |P3(9)| = 280.
Definition (Action on Partitions)
Every σ ∈ Sn can be naturally regarded as a permutation of Pm(n) via the rule
σ ([x1 · · · xm | y1 · · · ym | · · · ]) = [σ(x1) · · ·σ(xm) | σ(y1) · · ·σ(ym) | · · · ].
Joshua Wiscons Relational complexity
A harder, natural example
DefinitionPm(n) is the set of all partitions of {1, . . . , n} into subsets of (equal) size m.
ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]
P2(8) :
[12 | 34 | 56 | 78], . . .
P3(9) :
[147 | 258 | 369], . . .
Of course, there are many partitions of a given set, e.g. |P3(9)| = 280.
Definition (Action on Partitions)
Every σ ∈ Sn can be naturally regarded as a permutation of Pm(n) via the rule
σ ([x1 · · · xm | y1 · · · ym | · · · ]) = [σ(x1) · · ·σ(xm) | σ(y1) · · ·σ(ym) | · · · ].
Joshua Wiscons Relational complexity
A harder, natural example
DefinitionPm(n) is the set of all partitions of {1, . . . , n} into subsets of (equal) size m.
ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]
P2(8) : [12 | 34 | 56 | 78], . . .
P3(9) :
[147 | 258 | 369], . . .
Of course, there are many partitions of a given set, e.g. |P3(9)| = 280.
Definition (Action on Partitions)
Every σ ∈ Sn can be naturally regarded as a permutation of Pm(n) via the rule
σ ([x1 · · · xm | y1 · · · ym | · · · ]) = [σ(x1) · · ·σ(xm) | σ(y1) · · ·σ(ym) | · · · ].
Joshua Wiscons Relational complexity
A harder, natural example
DefinitionPm(n) is the set of all partitions of {1, . . . , n} into subsets of (equal) size m.
ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]
P2(8) : [12 | 34 | 56 | 78], . . .
P3(9) :
[147 | 258 | 369], . . .
Of course, there are many partitions of a given set, e.g. |P3(9)| = 280.
Definition (Action on Partitions)
Every σ ∈ Sn can be naturally regarded as a permutation of Pm(n) via the rule
σ ([x1 · · · xm | y1 · · · ym | · · · ]) = [σ(x1) · · ·σ(xm) | σ(y1) · · ·σ(ym) | · · · ].
Joshua Wiscons Relational complexity
A harder, natural example
DefinitionPm(n) is the set of all partitions of {1, . . . , n} into subsets of (equal) size m.
ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]
P2(8) : [12 | 34 | 56 | 78], . . .
P3(9) : [147 | 258 | 369], . . .
Of course, there are many partitions of a given set, e.g. |P3(9)| = 280.
Definition (Action on Partitions)
Every σ ∈ Sn can be naturally regarded as a permutation of Pm(n) via the rule
σ ([x1 · · · xm | y1 · · · ym | · · · ]) = [σ(x1) · · ·σ(xm) | σ(y1) · · ·σ(ym) | · · · ].
Joshua Wiscons Relational complexity
A harder, natural example
DefinitionPm(n) is the set of all partitions of {1, . . . , n} into subsets of (equal) size m.
ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]
P2(8) : [12 | 34 | 56 | 78], . . .
P3(9) : [147 | 258 | 369], . . .
Of course, there are many partitions of a given set, e.g. |P3(9)| = 280.
Definition (Action on Partitions)
Every σ ∈ Sn can be naturally regarded as a permutation of Pm(n) via the rule
σ ([x1 · · · xm | y1 · · · ym | · · · ]) = [σ(x1) · · ·σ(xm) | σ(y1) · · ·σ(ym) | · · · ].
Joshua Wiscons Relational complexity
A harder, natural example
DefinitionPm(n) is the set of all partitions of {1, . . . , n} into subsets of (equal) size m.
ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]
P2(8) : [12 | 34 | 56 | 78], . . .
P3(9) : [147 | 258 | 369], . . .
Of course, there are many partitions of a given set, e.g. |P3(9)| = 280.
Definition (Action on Partitions)
Every σ ∈ Sn can be naturally regarded as a permutation of Pm(n) via the rule
σ ([x1 · · · xm | y1 · · · ym | · · · ]) = [σ(x1) · · ·σ(xm) | σ(y1) · · ·σ(ym) | · · · ].
Joshua Wiscons Relational complexity
A harder, natural example
DefinitionPm(n) is the set of all partitions of {1, . . . , n} into subsets of (equal) size m.
ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]
P2(8) : [12 | 34 | 56 | 78], . . .
P3(9) : [147 | 258 | 369], . . .
Of course, there are many partitions of a given set, e.g. |P3(9)| = 280.
Definition (Action on Partitions)
Every σ ∈ Sn can be naturally regarded as a permutation of Pm(n) via the rule
σ ([x1 · · · xm | y1 · · · ym | · · · ]) = [σ(x1) · · ·σ(xm) | σ(y1) · · ·σ(ym) | · · · ].
Joshua Wiscons Relational complexity
A harder, natural example
ProblemDetermine the relational complexity of Sn acting on Pm(n).
Some answers are known, but the reasons why aren’t so unclear.
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P2(6) : [∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S6) = 3P2(8) : [∗ ∗ | ∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S8) = 4
P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P3(9) : [∗ ∗ ∗ | ∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S9) = 5
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P4(8) : [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗] =⇒ rc(S8) = 5P5(10) : [∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗] =⇒ rc(S10) = 4P6(12) : [∗ ∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗ ∗] =⇒ rc(S12) ≥ 6
Joshua Wiscons Relational complexity
A harder, natural example
ProblemDetermine the relational complexity of Sn acting on Pm(n).
Some answers are known, but the reasons why aren’t so unclear.
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P2(6) : [∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S6) = 3P2(8) : [∗ ∗ | ∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S8) = 4
P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P3(9) : [∗ ∗ ∗ | ∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S9) = 5
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P4(8) : [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗] =⇒ rc(S8) = 5P5(10) : [∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗] =⇒ rc(S10) = 4P6(12) : [∗ ∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗ ∗] =⇒ rc(S12) ≥ 6
Joshua Wiscons Relational complexity
A harder, natural example
ProblemDetermine the relational complexity of Sn acting on Pm(n).
Some answers are known, but the reasons why aren’t so unclear.
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2
P2(6) : [∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S6) = 3P2(8) : [∗ ∗ | ∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S8) = 4
P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P3(9) : [∗ ∗ ∗ | ∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S9) = 5
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P4(8) : [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗] =⇒ rc(S8) = 5P5(10) : [∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗] =⇒ rc(S10) = 4P6(12) : [∗ ∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗ ∗] =⇒ rc(S12) ≥ 6
Joshua Wiscons Relational complexity
A harder, natural example
ProblemDetermine the relational complexity of Sn acting on Pm(n).
Some answers are known, but the reasons why aren’t so unclear.
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P2(6) : [∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S6) = 3
P2(8) : [∗ ∗ | ∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S8) = 4
P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P3(9) : [∗ ∗ ∗ | ∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S9) = 5
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P4(8) : [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗] =⇒ rc(S8) = 5P5(10) : [∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗] =⇒ rc(S10) = 4P6(12) : [∗ ∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗ ∗] =⇒ rc(S12) ≥ 6
Joshua Wiscons Relational complexity
A harder, natural example
ProblemDetermine the relational complexity of Sn acting on Pm(n).
Some answers are known, but the reasons why aren’t so unclear.
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P2(6) : [∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S6) = 3P2(8) : [∗ ∗ | ∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S8) = 4
P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P3(9) : [∗ ∗ ∗ | ∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S9) = 5
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P4(8) : [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗] =⇒ rc(S8) = 5P5(10) : [∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗] =⇒ rc(S10) = 4P6(12) : [∗ ∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗ ∗] =⇒ rc(S12) ≥ 6
Joshua Wiscons Relational complexity
A harder, natural example
ProblemDetermine the relational complexity of Sn acting on Pm(n).
Some answers are known, but the reasons why aren’t so unclear.
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P2(6) : [∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S6) = 3P2(8) : [∗ ∗ | ∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S8) = 4
P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3
P3(9) : [∗ ∗ ∗ | ∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S9) = 5
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P4(8) : [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗] =⇒ rc(S8) = 5P5(10) : [∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗] =⇒ rc(S10) = 4P6(12) : [∗ ∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗ ∗] =⇒ rc(S12) ≥ 6
Joshua Wiscons Relational complexity
A harder, natural example
ProblemDetermine the relational complexity of Sn acting on Pm(n).
Some answers are known, but the reasons why aren’t so unclear.
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P2(6) : [∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S6) = 3P2(8) : [∗ ∗ | ∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S8) = 4
P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P3(9) : [∗ ∗ ∗ | ∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S9) = 5
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P4(8) : [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗] =⇒ rc(S8) = 5P5(10) : [∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗] =⇒ rc(S10) = 4P6(12) : [∗ ∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗ ∗] =⇒ rc(S12) ≥ 6
Joshua Wiscons Relational complexity
A harder, natural example
ProblemDetermine the relational complexity of Sn acting on Pm(n).
Some answers are known, but the reasons why aren’t so unclear.
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P2(6) : [∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S6) = 3P2(8) : [∗ ∗ | ∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S8) = 4
P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P3(9) : [∗ ∗ ∗ | ∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S9) = 5
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2
P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P4(8) : [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗] =⇒ rc(S8) = 5P5(10) : [∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗] =⇒ rc(S10) = 4P6(12) : [∗ ∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗ ∗] =⇒ rc(S12) ≥ 6
Joshua Wiscons Relational complexity
A harder, natural example
ProblemDetermine the relational complexity of Sn acting on Pm(n).
Some answers are known, but the reasons why aren’t so unclear.
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P2(6) : [∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S6) = 3P2(8) : [∗ ∗ | ∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S8) = 4
P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P3(9) : [∗ ∗ ∗ | ∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S9) = 5
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3
P4(8) : [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗] =⇒ rc(S8) = 5P5(10) : [∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗] =⇒ rc(S10) = 4P6(12) : [∗ ∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗ ∗] =⇒ rc(S12) ≥ 6
Joshua Wiscons Relational complexity
A harder, natural example
ProblemDetermine the relational complexity of Sn acting on Pm(n).
Some answers are known, but the reasons why aren’t so unclear.
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P2(6) : [∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S6) = 3P2(8) : [∗ ∗ | ∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S8) = 4
P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P3(9) : [∗ ∗ ∗ | ∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S9) = 5
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P4(8) : [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗] =⇒ rc(S8) = 5
P5(10) : [∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗] =⇒ rc(S10) = 4P6(12) : [∗ ∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗ ∗] =⇒ rc(S12) ≥ 6
Joshua Wiscons Relational complexity
A harder, natural example
ProblemDetermine the relational complexity of Sn acting on Pm(n).
Some answers are known, but the reasons why aren’t so unclear.
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P2(6) : [∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S6) = 3P2(8) : [∗ ∗ | ∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S8) = 4
P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P3(9) : [∗ ∗ ∗ | ∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S9) = 5
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P4(8) : [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗] =⇒ rc(S8) = 5P5(10) : [∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗] =⇒ rc(S10) = 4
P6(12) : [∗ ∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗ ∗] =⇒ rc(S12) ≥ 6
Joshua Wiscons Relational complexity
A harder, natural example
ProblemDetermine the relational complexity of Sn acting on Pm(n).
Some answers are known, but the reasons why aren’t so unclear.
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P2(6) : [∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S6) = 3P2(8) : [∗ ∗ | ∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S8) = 4
P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P3(9) : [∗ ∗ ∗ | ∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S9) = 5
P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P3(6) : [∗ ∗ ∗ | ∗ ∗ ∗] =⇒ rc(S6) = 3P4(8) : [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗] =⇒ rc(S8) = 5P5(10) : [∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗] =⇒ rc(S10) = 4P6(12) : [∗ ∗ ∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗ ∗ ∗] =⇒ rc(S12) ≥ 6
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
[1234 | 5678]
(1, 3)-pattern
[1234 | 5678]
[5634 | 1278]
(2, 2)-pattern
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
[1234 | 5678]
(1, 3)-pattern
[1234 | 5678]
[5634 | 1278]
(2, 2)-pattern
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
[1234 | 5678]
(1, 3)-pattern
[1234 | 5678]
[5634 | 1278]
(2, 2)-pattern
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
[1234 | 5678]
[5234 | 1678]
(1, 3)-pattern
[1234 | 5678]
[5634 | 1278]
(2, 2)-pattern
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
[1234 | 5678]
[5234 | 1678]
(1, 3)-pattern
[1234 | 5678]
[5634 | 1278]
(2, 2)-pattern
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
[1234 | 5678]
[1678 | 3245]
(1, 3)-pattern
[1234 | 5678]
[5634 | 1278]
(2, 2)-pattern
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
[1234 | 5678]
[5234 | 1678]
(1, 3)-pattern
[1234 | 5678]
[5634 | 1278]
(2, 2)-pattern
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
[1234 | 5678]
[5234 | 1678]
(1, 3)-pattern
[1234 | 5678]
[5634 | 1278]
(2, 2)-pattern
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
[1234 | 5678]
[5234 | 1678]
(1, 3)-pattern
[1234 | 5678]
[5634 | 1278]
(2, 2)-pattern
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
[1234 | 5678]
[5234 | 1678]
(1, 5)
[1234 | 5678]
[6234 | 5178]
(1, 6)
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
[1234 | 5678]
[5234 | 1678]
(1, 5)
[1234 | 5678]
[6234 | 5178]
(1, 6)
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
[1234 | 5678]
[5234 | 1678]
(1, 5)
[1234 | 5678]
[6234 | 5178]
(1, 6)
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
[1234 | 5678]
[5234 | 1678]
(1, 5)
[1234 | 5678]
[6234 | 5178]
(1, 6)
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
[1234 | 5678]
[5234 | 1678]
(1, 5)
[1234 | 5678]
[6234 | 5178]
(1, 6)
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
1
5
2
6
3
7
4
8
y1
y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
Set x = [1234 | 5678].
Every x′ 6= x can have one of two relationships with x.
Let Y ⊂ P4(8) be the partitions that have a (1, 3)-pattern with x.
Every y ∈ Y is determined by two numbers.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
No! The stabilizer of (x, y1, y2, y3) is〈(34)〉.
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
No!
The stabilizer of (x, y1, y2, y3) is〈(34)〉.
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
No! The stabilizer of (x, y1, y2, y3) is〈(34)〉.
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
Yes!
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
Yes!
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
Yes: (23)(78) (“rotation”)
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
Yes: (23)(78) (“rotation”)
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
Yes: (15)(27)(38)(46) (“reflection”)
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
Yes: (15)(27)(38)(46) (“reflection”)
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
Yes: “reflection”
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
Yes: “reflection”
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
Yes!
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
Yes!
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
Yes!
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
A harder, natural (concrete) example
Let’s look at P4(8), i.e. [∗ ∗ ∗ ∗ | ∗ ∗ ∗ ∗], and try to see why rc(S8) = 5.
1
5
2
6
3
7
4
8
y1 y2
y3 u
v
x = [1234 | 5678]
y1 = [5234 | 1678]
y2 = [6234 | 5178]
y3 = [1734 | 5628]
u = [1834 | 5672]
v = [1274 | 5638]
x y1 y2 y3 u
x y1 y2 y3 v
Yes!
So these tuples are 4-equivalentbut not equivalent.
rc(S8) ≥ 5
Joshua Wiscons Relational complexity
Some Problems (posed by Cherlin)
Problems1 Gather data: determine the relational complexities of natural permutation
groups.E.g. study the various natural actions of classical groups.E.g, study the various natural actions of Sn (and An).
2 Classify the finite permutation groups of relational complexity at most r.Need to restrict to so-called primitive groups.E.g, what are the finite primitive permutation groups of relationalcomplexity 2?
3 Classify the finite permutation groups of relational complexity at least rwhen r is “big” compared to |X|.
Joshua Wiscons Relational complexity
Groups of relational complexity 2
Conjecture (Cherlin, 2000)A finite primitive permutation group of relational complexity 2 is either
Sn acting naturally on {1, . . . , n},a cyclic group of prime order acting regularly, or
an affine orthogonal group V o O(V) where V is a vector spaceequipped with an anisotropic quadratic form.
Fact (O’Nan–Scott)A finite primitive permutation group falls into one of five classes: (1) affine,(2) regular nonabelian, (3) almost simple, (4) diagonal, or (5) product.
Theorem
Cherlin, ’15: The conjecture holds for affine groups.
W, ’16: The conjecture reduces to the almost simple case.
Joshua Wiscons Relational complexity
Groups of relational complexity 2
Conjecture (Cherlin, 2000)A finite primitive permutation group of relational complexity 2 is either
Sn acting naturally on {1, . . . , n},
a cyclic group of prime order acting regularly, or
an affine orthogonal group V o O(V) where V is a vector spaceequipped with an anisotropic quadratic form.
Fact (O’Nan–Scott)A finite primitive permutation group falls into one of five classes: (1) affine,(2) regular nonabelian, (3) almost simple, (4) diagonal, or (5) product.
Theorem
Cherlin, ’15: The conjecture holds for affine groups.
W, ’16: The conjecture reduces to the almost simple case.
Joshua Wiscons Relational complexity
Groups of relational complexity 2
Conjecture (Cherlin, 2000)A finite primitive permutation group of relational complexity 2 is either
Sn acting naturally on {1, . . . , n},a cyclic group of prime order acting regularly, or
an affine orthogonal group V o O(V) where V is a vector spaceequipped with an anisotropic quadratic form.
Fact (O’Nan–Scott)A finite primitive permutation group falls into one of five classes: (1) affine,(2) regular nonabelian, (3) almost simple, (4) diagonal, or (5) product.
Theorem
Cherlin, ’15: The conjecture holds for affine groups.
W, ’16: The conjecture reduces to the almost simple case.
Joshua Wiscons Relational complexity
Groups of relational complexity 2
Conjecture (Cherlin, 2000)A finite primitive permutation group of relational complexity 2 is either
Sn acting naturally on {1, . . . , n},a cyclic group of prime order acting regularly, or
an affine orthogonal group V o O(V) where V is a vector spaceequipped with an anisotropic quadratic form.
Fact (O’Nan–Scott)A finite primitive permutation group falls into one of five classes: (1) affine,(2) regular nonabelian, (3) almost simple, (4) diagonal, or (5) product.
Theorem
Cherlin, ’15: The conjecture holds for affine groups.
W, ’16: The conjecture reduces to the almost simple case.
Joshua Wiscons Relational complexity
Groups of relational complexity 2
Conjecture (Cherlin, 2000)A finite primitive permutation group of relational complexity 2 is either
Sn acting naturally on {1, . . . , n},a cyclic group of prime order acting regularly, or
an affine orthogonal group V o O(V) where V is a vector spaceequipped with an anisotropic quadratic form.
Fact (O’Nan–Scott)A finite primitive permutation group falls into one of five classes: (1) affine,(2) regular nonabelian, (3) almost simple, (4) diagonal, or (5) product.
Theorem
Cherlin, ’15: The conjecture holds for affine groups.
W, ’16: The conjecture reduces to the almost simple case.
Joshua Wiscons Relational complexity
Groups of relational complexity 2
Conjecture (Cherlin, 2000)A finite primitive permutation group of relational complexity 2 is either
Sn acting naturally on {1, . . . , n},a cyclic group of prime order acting regularly, or
an affine orthogonal group V o O(V) where V is a vector spaceequipped with an anisotropic quadratic form.
Fact (O’Nan–Scott)A finite primitive permutation group falls into one of five classes: (1) affine,(2) regular nonabelian, (3) almost simple, (4) diagonal, or (5) product.
Theorem
Cherlin, ’15: The conjecture holds for affine groups.
W, ’16: The conjecture reduces to the almost simple case.
Joshua Wiscons Relational complexity
Groups of relational complexity 2
Conjecture (Cherlin, 2000)A finite primitive permutation group of relational complexity 2 is either
Sn acting naturally on {1, . . . , n},a cyclic group of prime order acting regularly, or
an affine orthogonal group V o O(V) where V is a vector spaceequipped with an anisotropic quadratic form.
Fact (O’Nan–Scott)A finite primitive permutation group falls into one of five classes: (1) affine,(2) regular nonabelian, (3) almost simple, (4) diagonal, or (5) product.
TheoremCherlin, ’15: The conjecture holds for affine groups.
W, ’16: The conjecture reduces to the almost simple case.
Joshua Wiscons Relational complexity
Groups of relational complexity 2
Conjecture (Cherlin, 2000)A finite primitive permutation group of relational complexity 2 is either
Sn acting naturally on {1, . . . , n},a cyclic group of prime order acting regularly, or
an affine orthogonal group V o O(V) where V is a vector spaceequipped with an anisotropic quadratic form.
Fact (O’Nan–Scott)A finite primitive permutation group falls into one of five classes: (1) affine,(2) regular nonabelian, (3) almost simple, (4) diagonal, or (5) product.
TheoremCherlin, ’15: The conjecture holds for affine groups.
W, ’16: The conjecture reduces to the almost simple case.
Joshua Wiscons Relational complexity
Some Problems (posed by Cherlin)
Problems1 Gather data: determine the relational complexities of natural permutation
groups.E.g. study the various natural actions of classical groups.E.g, study the various natural actions of Sn (and An).
2 Classify the finite permutation groups of relational complexity at most r.Need to restrict to so-called primitive groups.E.g, what are the finite primitive permutation groups of relationalcomplexity 2?
3 Classify the finite permutation groups of relational complexity at least rwhen r is “big” compared to |X|.
Joshua Wiscons Relational complexity
Upper bounds on rc(X,G) in terms of |X|
Let (X,G) be a finite primitive permutation group.
Remark1 If (X,G) is of affine type, then rc(X,G) ≤ log2(|X|) + 2.
2 Generically, rc(X,G) is likely not to be much bigger than log(|X|)3 But, we have already seen that big examples exist (Alt(n) on {1, . . . , n})
Proof.
In this type, (X,G) ∼= (V,H) where
V is a vector space, say of characteristic p and dimension d, andH ≤ AGL(V) = V o GL(V)
Our method for bounding rc(V,GL(V)) above by d + 1 can be used toshow that rc(V,AGL(V)) ≤ d + 2
More generally, rc(V,H) ≤ d + 2
Thus rc(X,G) = rc(V,H) ≤ logp(|V|) + 2
Joshua Wiscons Relational complexity
Upper bounds on rc(X,G) in terms of |X|
Let (X,G) be a finite primitive permutation group.
Remark1 If (X,G) is of affine type, then rc(X,G) ≤ log2(|X|) + 2.
2 Generically, rc(X,G) is likely not to be much bigger than log(|X|)3 But, we have already seen that big examples exist (Alt(n) on {1, . . . , n})
Proof.
In this type, (X,G) ∼= (V,H) where
V is a vector space, say of characteristic p and dimension d, andH ≤ AGL(V) = V o GL(V)
Our method for bounding rc(V,GL(V)) above by d + 1 can be used toshow that rc(V,AGL(V)) ≤ d + 2
More generally, rc(V,H) ≤ d + 2
Thus rc(X,G) = rc(V,H) ≤ logp(|V|) + 2
Joshua Wiscons Relational complexity
Upper bounds on rc(X,G) in terms of |X|
Let (X,G) be a finite primitive permutation group.
Remark1 If (X,G) is of affine type, then rc(X,G) ≤ log2(|X|) + 2.
2 Generically, rc(X,G) is likely not to be much bigger than log(|X|)3 But, we have already seen that big examples exist (Alt(n) on {1, . . . , n})
Proof.
In this type, (X,G) ∼= (V,H) where
V is a vector space, say of characteristic p and dimension d, andH ≤ AGL(V) = V o GL(V)
Our method for bounding rc(V,GL(V)) above by d + 1 can be used toshow that rc(V,AGL(V)) ≤ d + 2
More generally, rc(V,H) ≤ d + 2
Thus rc(X,G) = rc(V,H) ≤ logp(|V|) + 2
Joshua Wiscons Relational complexity
Upper bounds on rc(X,G) in terms of |X|
Let (X,G) be a finite primitive permutation group.
Remark1 If (X,G) is of affine type, then rc(X,G) ≤ log2(|X|) + 2.
2 Generically, rc(X,G) is likely not to be much bigger than log(|X|)3 But, we have already seen that big examples exist (Alt(n) on {1, . . . , n})
Proof.In this type, (X,G) ∼= (V,H) where
V is a vector space, say of characteristic p and dimension d, andH ≤ AGL(V) = V o GL(V)
Our method for bounding rc(V,GL(V)) above by d + 1 can be used toshow that rc(V,AGL(V)) ≤ d + 2
More generally, rc(V,H) ≤ d + 2
Thus rc(X,G) = rc(V,H) ≤ logp(|V|) + 2
Joshua Wiscons Relational complexity
Upper bounds on rc(X,G) in terms of |X|
Let (X,G) be a finite primitive permutation group.
Remark1 If (X,G) is of affine type, then rc(X,G) ≤ log2(|X|) + 2.
2 Generically, rc(X,G) is likely not to be much bigger than log(|X|)3 But, we have already seen that big examples exist (Alt(n) on {1, . . . , n})
Proof.In this type, (X,G) ∼= (V,H) where
V is a vector space, say of characteristic p and dimension d, and
H ≤ AGL(V) = V o GL(V)
Our method for bounding rc(V,GL(V)) above by d + 1 can be used toshow that rc(V,AGL(V)) ≤ d + 2
More generally, rc(V,H) ≤ d + 2
Thus rc(X,G) = rc(V,H) ≤ logp(|V|) + 2
Joshua Wiscons Relational complexity
Upper bounds on rc(X,G) in terms of |X|
Let (X,G) be a finite primitive permutation group.
Remark1 If (X,G) is of affine type, then rc(X,G) ≤ log2(|X|) + 2.
2 Generically, rc(X,G) is likely not to be much bigger than log(|X|)3 But, we have already seen that big examples exist (Alt(n) on {1, . . . , n})
Proof.In this type, (X,G) ∼= (V,H) where
V is a vector space, say of characteristic p and dimension d, andH ≤ AGL(V) = V o GL(V)
Our method for bounding rc(V,GL(V)) above by d + 1 can be used toshow that rc(V,AGL(V)) ≤ d + 2
More generally, rc(V,H) ≤ d + 2
Thus rc(X,G) = rc(V,H) ≤ logp(|V|) + 2
Joshua Wiscons Relational complexity
Upper bounds on rc(X,G) in terms of |X|
Let (X,G) be a finite primitive permutation group.
Remark1 If (X,G) is of affine type, then rc(X,G) ≤ log2(|X|) + 2.
2 Generically, rc(X,G) is likely not to be much bigger than log(|X|)3 But, we have already seen that big examples exist (Alt(n) on {1, . . . , n})
Proof.In this type, (X,G) ∼= (V,H) where
V is a vector space, say of characteristic p and dimension d, andH ≤ AGL(V) = V o GL(V)
Our method for bounding rc(V,GL(V)) above by d + 1 can be used toshow that rc(V,AGL(V)) ≤ d + 2
More generally, rc(V,H) ≤ d + 2
Thus rc(X,G) = rc(V,H) ≤ logp(|V|) + 2
Joshua Wiscons Relational complexity
Upper bounds on rc(X,G) in terms of |X|
Let (X,G) be a finite primitive permutation group.
Remark1 If (X,G) is of affine type, then rc(X,G) ≤ log2(|X|) + 2.
2 Generically, rc(X,G) is likely not to be much bigger than log(|X|)3 But, we have already seen that big examples exist (Alt(n) on {1, . . . , n})
Proof.In this type, (X,G) ∼= (V,H) where
V is a vector space, say of characteristic p and dimension d, andH ≤ AGL(V) = V o GL(V)
Our method for bounding rc(V,GL(V)) above by d + 1 can be used toshow that rc(V,AGL(V)) ≤ d + 2
More generally, rc(V,H) ≤ d + 2
Thus rc(X,G) = rc(V,H) ≤ logp(|V|) + 2
Joshua Wiscons Relational complexity
Upper bounds on rc(X,G) in terms of |X|
Let (X,G) be a finite primitive permutation group.
Remark1 If (X,G) is of affine type, then rc(X,G) ≤ log2(|X|) + 2.
2 Generically, rc(X,G) is likely not to be much bigger than log(|X|)3 But, we have already seen that big examples exist (Alt(n) on {1, . . . , n})
Proof.In this type, (X,G) ∼= (V,H) where
V is a vector space, say of characteristic p and dimension d, andH ≤ AGL(V) = V o GL(V)
Our method for bounding rc(V,GL(V)) above by d + 1 can be used toshow that rc(V,AGL(V)) ≤ d + 2
More generally, rc(V,H) ≤ d + 2
Thus rc(X,G) = rc(V,H) ≤ logp(|V|) + 2
Joshua Wiscons Relational complexity
Upper bounds on rc(X,G) in terms of |X|
Let (X,G) be a finite primitive permutation group.
Remark1 If (X,G) is of affine type, then rc(X,G) ≤ log2(|X|) + 2.2 Generically, rc(X,G) is likely not to be much bigger than log(|X|)
3 But, we have already seen that big examples exist (Alt(n) on {1, . . . , n})
Proof.In this type, (X,G) ∼= (V,H) where
V is a vector space, say of characteristic p and dimension d, andH ≤ AGL(V) = V o GL(V)
Our method for bounding rc(V,GL(V)) above by d + 1 can be used toshow that rc(V,AGL(V)) ≤ d + 2
More generally, rc(V,H) ≤ d + 2
Thus rc(X,G) = rc(V,H) ≤ logp(|V|) + 2
Joshua Wiscons Relational complexity
Upper bounds on rc(X,G) in terms of |X|
Let (X,G) be a finite primitive permutation group.
Remark1 If (X,G) is of affine type, then rc(X,G) ≤ log2(|X|) + 2.2 Generically, rc(X,G) is likely not to be much bigger than log(|X|)3 But, we have already seen that big examples exist (Alt(n) on {1, . . . , n})
Proof.In this type, (X,G) ∼= (V,H) where
V is a vector space, say of characteristic p and dimension d, andH ≤ AGL(V) = V o GL(V)
Our method for bounding rc(V,GL(V)) above by d + 1 can be used toshow that rc(V,AGL(V)) ≤ d + 2
More generally, rc(V,H) ≤ d + 2
Thus rc(X,G) = rc(V,H) ≤ logp(|V|) + 2
Joshua Wiscons Relational complexity
Some Problems (posed by Cherlin)
Problems1 Gather data: determine the relational complexities of natural permutation
groups.E.g. study the various natural actions of classical groups.E.g, study the various natural actions of Sn (and An).
2 Classify the finite permutation groups of relational complexity at most r.Need to restrict to so-called primitive groups.E.g, what are the finite primitive permutation groups of relationalcomplexity 2?
3 Classify the finite permutation groups of relational complexity at least rwhen r is “big” compared to |X|.
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.
What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type.
This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type.
This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type.
This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Proof.
Want to show (1, a, ah) ∼ (1, ah, a)
This is the same as (1, a, a−h) ∼ (1, ah, a−1)
Suffices to show (1, a, a−h) ∼2 (1, ah, a−1)
a · (aha−1)
= (aah)a−1 = ahaa−1 = ah
1 a a−h
1 ah a−1
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Proof.
Want to show (1, a, ah) ∼ (1, ah, a)
This is the same as (1, a, a−h) ∼ (1, ah, a−1)
Suffices to show (1, a, a−h) ∼2 (1, ah, a−1)
a · (aha−1)
= (aah)a−1 = ahaa−1 = ah
1 a a−h
1 ah a−1
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Proof.
Want to show (1, a, ah) ∼ (1, ah, a)
This is the same as (1, a, a−h) ∼ (1, ah, a−1)
Suffices to show (1, a, a−h) ∼2 (1, ah, a−1)
a · (aha−1)
= (aah)a−1 = ahaa−1 = ah
1 a a−h
1 ah a−1
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Proof.
Want to show (1, a, ah) ∼ (1, ah, a)
This is the same as (1, a, a−h) ∼ (1, ah, a−1)
Suffices to show (1, a, a−h) ∼2 (1, ah, a−1)
a · (aha−1)
= (aah)a−1 = ahaa−1 = ah
1 a a−h
1 ah a−1
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Proof.
Want to show (1, a, ah) ∼ (1, ah, a)
This is the same as (1, a, a−h) ∼ (1, ah, a−1)
Suffices to show (1, a, a−h) ∼2 (1, ah, a−1)
a · (aha−1)
= (aah)a−1 = ahaa−1 = ah
1 a a−h
1 ah a−1
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Proof.
Want to show (1, a, ah) ∼ (1, ah, a)
This is the same as (1, a, a−h) ∼ (1, ah, a−1)
Suffices to show (1, a, a−h) ∼2 (1, ah, a−1)
a · (aha−1)
= (aah)a−1 = ahaa−1 = ah
1 a a−h
1 ah a−1
h
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Proof.
Want to show (1, a, ah) ∼ (1, ah, a)
This is the same as (1, a, a−h) ∼ (1, ah, a−1)
Suffices to show (1, a, a−h) ∼2 (1, ah, a−1)
a · (aha−1)
= (aah)a−1 = ahaa−1 = ah
1 a a−h
1 ah a−1
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Proof.
Want to show (1, a, ah) ∼ (1, ah, a)
This is the same as (1, a, a−h) ∼ (1, ah, a−1)
Suffices to show (1, a, a−h) ∼2 (1, ah, a−1)
a · (aha−1)
= (aah)a−1 = ahaa−1 = ah
1 a a−h
1 ah a−1
h−1
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Proof.
Want to show (1, a, ah) ∼ (1, ah, a)
This is the same as (1, a, a−h) ∼ (1, ah, a−1)
Suffices to show (1, a, a−h) ∼2 (1, ah, a−1)
a · (aha−1)
= (aah)a−1 = ahaa−1 = ah
1 a a−h
1 ah a−1
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Proof.
Want to show (1, a, ah) ∼ (1, ah, a)
This is the same as (1, a, a−h) ∼ (1, ah, a−1)
Suffices to show (1, a, a−h) ∼2 (1, ah, a−1)
a · (aha−1)
= (aah)a−1 = ahaa−1 = ah
1 a a−h
1 ah a−1
aha−1
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Proof.
Want to show (1, a, ah) ∼ (1, ah, a)
This is the same as (1, a, a−h) ∼ (1, ah, a−1)
Suffices to show (1, a, a−h) ∼2 (1, ah, a−1)
a · (aha−1)
= (aah)a−1 = ahaa−1 = ah
1 a a−h
1 ah a−1
aha−1
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Proof.
Want to show (1, a, ah) ∼ (1, ah, a)
This is the same as (1, a, a−h) ∼ (1, ah, a−1)
Suffices to show (1, a, a−h) ∼2 (1, ah, a−1)
a · (aha−1) = (aah)a−1
= ahaa−1 = ah
1 a a−h
1 ah a−1
aha−1
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Proof.
Want to show (1, a, ah) ∼ (1, ah, a)
This is the same as (1, a, a−h) ∼ (1, ah, a−1)
Suffices to show (1, a, a−h) ∼2 (1, ah, a−1)
a · (aha−1) = (aah)a−1 = ahaa−1
= ah
1 a a−h
1 ah a−1
aha−1
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Proof.
Want to show (1, a, ah) ∼ (1, ah, a)
This is the same as (1, a, a−h) ∼ (1, ah, a−1)
Suffices to show (1, a, a−h) ∼2 (1, ah, a−1)
a · (aha−1) = (aah)a−1 = ahaa−1 = ah
1 a a−h
1 ah a−1
aha−1
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Punchline:
in these cases, complexity equal to 2 can be used to createinvolutions in a point stabilizer. (This is very useful.)
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Punchline: in these cases, complexity equal to 2 can be used to createinvolutions in a point stabilizer.
(This is very useful.)
Joshua Wiscons Relational complexity
Back to binary groups
Let (X,G) be a finite primitive permutation group of relational complexity 2.What does this tell us about (X,G)?
Example
Assume that (X,G) is of affine or nonabelian regular type. This implies that
G ∼= M o H with H the stabilizer of a point,
X can be identified with M in such a way that (X,G) ∼= (M,MH) where
a · mh = (am)h
Claim: if a ∈ X, h ∈ H, and [a, ah] = 1, then ∃h′ ∈ H swapping a and ah.
Punchline: in these cases, complexity equal to 2 can be used to createinvolutions in a point stabilizer. (This is very useful.)
Joshua Wiscons Relational complexity
Thank You
Joshua Wiscons Relational complexity