Math 2.0

General algebraic systems Definition

Polymorphism of a relation

Builds on Finitary operationCartesian productBinary relation

Definition

Let AA be a set, let m,n≥1m, n \ge 1, and let R⊆AmR \subseteq A^m be an mm-ary relation on AA. An nn-ary operation f ⁣:An→Af \colon A^n \to A is a polymorphism of RR if for all tuples (a11,…,am1),  …,  (a1n,…,amn)∈R(a_{1 1}, \dots, a_{m 1}), \; \dots, \; (a_{1 n}, \dots, a_{m n}) \in R the tuple (f(a11,…,a1n),…,f(am1,…,amn))\bigl(f(a_{1 1}, \dots, a_{1 n}), \dots, f(a_{m 1}, \dots, a_{m n})\bigr) is also in RR. For a set Γ\Gamma of finitary relations on AA, Pol⁡(Γ)\operatorname{Pol}(\Gamma) denotes the set of operations that are polymorphisms of every R∈ΓR \in \Gamma. For a set FF of finitary operations on AA, Inv⁡(F)\operatorname{Inv}(F) denotes the set of finitary relations on AA of which every f∈Ff \in F is a polymorphism.

Remarks

Example: on {0,1}\set{0, 1}, the majority operation mm is a polymorphism of the binary relation R={(0,1),(1,0)}R = \set{(0, 1), (1, 0)}. Check: given three pairs (aj,1−aj)(a_j, 1 - a_j) in RR, the second coordinates are the complements of the first, so m(1−a1,1−a2,1−a3)=1−m(a1,a2,a3)m(1 - a_1, 1 - a_2, 1 - a_3) = 1 - m(a_1, a_2, a_3) and the result lies in RR. Non-example: the operation ∧\wedge on {0,1}\set{0, 1} is not a polymorphism of the same RR, since applying it to (0,1)(0, 1) and (1,0)(1, 0) gives (0,0)∉R(0, 0) \notin R. For a unary relation B⊆AB \subseteq A, the polymorphisms are the operations under which BB is closed, so BB is a subuniverse. For an equivalence relation, the polymorphisms are the operations compatible with it, as in the definition of a congruence. If AA carries a relational structure, its polymorphisms are the homomorphisms from its nn-th power to itself.

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  1. Binary relation concept
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Further reading

  1. Web https://en.wikipedia.org/wiki/Clone_(algebra)
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Kinddefinition
WritingW0 stub
MSC08A40, 08A02
DefinesPolymorphism of a relation
UsesFinitary operation, Cartesian product, Binary relation

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