an introduction to relational complexity: …an introduction to relational complexity: background,...
TRANSCRIPT
![Page 1: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/1.jpg)
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
![Page 2: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/2.jpg)
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
![Page 3: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/3.jpg)
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
![Page 4: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/4.jpg)
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
![Page 5: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/5.jpg)
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
![Page 6: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/6.jpg)
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
![Page 7: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/7.jpg)
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
![Page 8: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/8.jpg)
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
![Page 9: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/9.jpg)
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
![Page 10: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/10.jpg)
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
![Page 11: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/11.jpg)
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
![Page 12: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/12.jpg)
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
![Page 13: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/13.jpg)
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
![Page 14: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/14.jpg)
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
![Page 15: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/15.jpg)
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
![Page 16: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/16.jpg)
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
![Page 17: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/17.jpg)
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
![Page 18: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/18.jpg)
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
![Page 19: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/19.jpg)
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
![Page 20: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/20.jpg)
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
![Page 21: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/21.jpg)
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
![Page 22: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/22.jpg)
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
![Page 23: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/23.jpg)
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
![Page 24: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/24.jpg)
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
![Page 25: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/25.jpg)
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
![Page 26: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/26.jpg)
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
![Page 27: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/27.jpg)
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
![Page 28: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/28.jpg)
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
![Page 29: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/29.jpg)
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
![Page 30: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/30.jpg)
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
![Page 31: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/31.jpg)
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
![Page 32: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/32.jpg)
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
![Page 33: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/33.jpg)
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
![Page 34: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/34.jpg)
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
![Page 35: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/35.jpg)
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
![Page 36: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/36.jpg)
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
![Page 37: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/37.jpg)
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
![Page 38: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/38.jpg)
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
![Page 39: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/39.jpg)
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
![Page 40: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/40.jpg)
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
![Page 41: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/41.jpg)
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
![Page 42: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/42.jpg)
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
![Page 43: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/43.jpg)
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
![Page 44: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/44.jpg)
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
![Page 45: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/45.jpg)
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
![Page 46: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/46.jpg)
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
![Page 47: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/47.jpg)
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
![Page 48: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/48.jpg)
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
![Page 49: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/49.jpg)
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
![Page 50: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/50.jpg)
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
![Page 51: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/51.jpg)
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
![Page 52: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/52.jpg)
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
![Page 53: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/53.jpg)
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
![Page 54: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/54.jpg)
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
![Page 55: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/55.jpg)
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
![Page 56: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/56.jpg)
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
![Page 57: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/57.jpg)
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
![Page 58: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/58.jpg)
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
![Page 59: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/59.jpg)
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
![Page 60: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/60.jpg)
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
![Page 61: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/61.jpg)
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
![Page 62: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/62.jpg)
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
![Page 63: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/63.jpg)
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
![Page 64: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/64.jpg)
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
![Page 65: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/65.jpg)
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
![Page 66: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/66.jpg)
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
![Page 67: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/67.jpg)
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
![Page 68: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/68.jpg)
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
![Page 69: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/69.jpg)
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
![Page 70: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/70.jpg)
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
![Page 71: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/71.jpg)
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
![Page 72: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/72.jpg)
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
![Page 73: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/73.jpg)
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
![Page 74: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/74.jpg)
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
![Page 75: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/75.jpg)
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
![Page 76: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/76.jpg)
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
![Page 77: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/77.jpg)
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
![Page 78: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/78.jpg)
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
![Page 79: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/79.jpg)
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
![Page 80: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/80.jpg)
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
![Page 81: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/81.jpg)
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
![Page 82: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/82.jpg)
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
![Page 83: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/83.jpg)
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
![Page 84: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/84.jpg)
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
![Page 85: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/85.jpg)
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
![Page 86: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/86.jpg)
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
![Page 87: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/87.jpg)
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
![Page 88: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/88.jpg)
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
![Page 89: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/89.jpg)
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
![Page 90: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/90.jpg)
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
![Page 91: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/91.jpg)
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
![Page 92: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/92.jpg)
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
![Page 93: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/93.jpg)
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
![Page 94: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/94.jpg)
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
![Page 95: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/95.jpg)
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
![Page 96: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/96.jpg)
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
![Page 97: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/97.jpg)
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
![Page 98: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/98.jpg)
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
![Page 99: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/99.jpg)
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
![Page 100: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/100.jpg)
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
![Page 101: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/101.jpg)
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
![Page 102: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/102.jpg)
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
![Page 103: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/103.jpg)
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
![Page 104: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/104.jpg)
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
![Page 105: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/105.jpg)
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
![Page 106: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/106.jpg)
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
![Page 107: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/107.jpg)
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
![Page 108: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/108.jpg)
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
![Page 109: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/109.jpg)
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
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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
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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
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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
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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
![Page 114: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/114.jpg)
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
![Page 115: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/115.jpg)
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
![Page 116: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/116.jpg)
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
![Page 117: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/117.jpg)
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
![Page 118: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/118.jpg)
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
![Page 119: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/119.jpg)
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
![Page 120: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/120.jpg)
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
![Page 121: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/121.jpg)
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
![Page 122: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/122.jpg)
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
![Page 123: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/123.jpg)
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
![Page 124: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/124.jpg)
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
![Page 125: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/125.jpg)
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
![Page 126: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/126.jpg)
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
![Page 127: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/127.jpg)
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
![Page 128: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/128.jpg)
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
![Page 129: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/129.jpg)
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
![Page 130: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/130.jpg)
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
![Page 131: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/131.jpg)
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
![Page 132: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/132.jpg)
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
![Page 133: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/133.jpg)
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
![Page 134: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/134.jpg)
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
![Page 135: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/135.jpg)
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
![Page 136: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/136.jpg)
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
![Page 137: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/137.jpg)
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
![Page 138: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/138.jpg)
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
![Page 139: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/139.jpg)
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
![Page 140: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/140.jpg)
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
![Page 141: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/141.jpg)
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
![Page 142: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/142.jpg)
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
![Page 143: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/143.jpg)
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
![Page 144: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/144.jpg)
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
![Page 145: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/145.jpg)
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
![Page 146: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/146.jpg)
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
![Page 147: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/147.jpg)
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
![Page 148: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/148.jpg)
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
![Page 149: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/149.jpg)
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
![Page 150: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/150.jpg)
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
![Page 151: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/151.jpg)
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
![Page 152: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/152.jpg)
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
![Page 153: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/153.jpg)
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
![Page 154: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/154.jpg)
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
![Page 155: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/155.jpg)
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
![Page 156: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/156.jpg)
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
![Page 157: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/157.jpg)
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
![Page 158: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/158.jpg)
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
![Page 159: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/159.jpg)
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
![Page 160: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/160.jpg)
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
![Page 161: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/161.jpg)
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
![Page 162: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/162.jpg)
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
![Page 163: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/163.jpg)
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
![Page 164: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/164.jpg)
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
![Page 165: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/165.jpg)
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
![Page 166: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/166.jpg)
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
![Page 167: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/167.jpg)
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
![Page 168: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/168.jpg)
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
![Page 169: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/169.jpg)
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
![Page 170: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/170.jpg)
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
![Page 171: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/171.jpg)
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
![Page 172: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/172.jpg)
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
![Page 173: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/173.jpg)
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
![Page 174: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/174.jpg)
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
![Page 175: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/175.jpg)
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
![Page 176: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/176.jpg)
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
![Page 177: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/177.jpg)
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
![Page 178: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/178.jpg)
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
![Page 179: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/179.jpg)
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
![Page 180: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/180.jpg)
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
![Page 181: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/181.jpg)
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
![Page 182: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/182.jpg)
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
![Page 183: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/183.jpg)
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
![Page 184: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/184.jpg)
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
![Page 185: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/185.jpg)
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
![Page 186: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/186.jpg)
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
![Page 187: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/187.jpg)
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
![Page 188: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/188.jpg)
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
![Page 189: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/189.jpg)
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
![Page 190: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/190.jpg)
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
![Page 191: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/191.jpg)
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
![Page 192: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/192.jpg)
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
![Page 193: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/193.jpg)
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
![Page 194: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/194.jpg)
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
![Page 195: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/195.jpg)
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
![Page 196: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/196.jpg)
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
![Page 197: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/197.jpg)
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
![Page 198: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/198.jpg)
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
![Page 199: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/199.jpg)
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
![Page 200: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/200.jpg)
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
![Page 201: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/201.jpg)
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
![Page 202: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/202.jpg)
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
![Page 203: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/203.jpg)
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
![Page 204: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/204.jpg)
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
![Page 205: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/205.jpg)
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
![Page 206: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/206.jpg)
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
![Page 207: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/207.jpg)
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
![Page 208: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/208.jpg)
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
![Page 209: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/209.jpg)
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
![Page 210: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/210.jpg)
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
![Page 211: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/211.jpg)
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
![Page 212: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/212.jpg)
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
![Page 213: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/213.jpg)
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
![Page 214: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/214.jpg)
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
![Page 215: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/215.jpg)
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
![Page 216: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/216.jpg)
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
![Page 217: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/217.jpg)
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
![Page 218: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/218.jpg)
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
![Page 219: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/219.jpg)
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
![Page 220: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/220.jpg)
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
![Page 221: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/221.jpg)
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
![Page 222: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/222.jpg)
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
![Page 223: An introduction to relational complexity: …An introduction to relational complexity: background, questions, and a few answers Joshua Wiscons California State University, Sacramento](https://reader036.vdocuments.mx/reader036/viewer/2022081405/5f0a6ffd7e708231d42ba166/html5/thumbnails/223.jpg)
Thank You
Joshua Wiscons Relational complexity