Abstract clones

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Correspondences between interpretations of formulas

Formulas can be expressed to relate elements in an abstract clone as if they operated on an unknown module M, using universal quantifiers on the variables in M . Namely, a formula R of equality «t = t'» between 2 n-ary terms t,t', can be written (for example with n=2):
x,y, R(x,y)
to actually mean the relation in C2 that is R0,21,2) because this indeed implies the above universal formula in any representation. In particular, it implies R0,31,3), i.e. ∀x,y,z, R(x,y).
In fact these are equivalent, but the reason for the converse implication (R0,31,3)⇒R0,21,2)) has a subtlety, that is the same already faced to represent an operation symbol in an algebra with higher arity. In other words, to deduce (∀x,y, R(x,y)) from (∀x,y,z, R(x,y)) we need to choose a value for z. We may set it as equal to x or y, except if n=0 when they are not available, but then we can use any available symbol of constant instead (there will be, since terms not using variables must use constants).

Centralizers in abstract clones

We can generalize the concept of centralizer (or commutant), which was defined on monoids, to the case of abstract clones. It is directly obtained by abstracting with the above trick, the definition of the Galois connection Pol-Pol between sets of operations, where the condition of commutation between xCi and yCj is an equation expressed in Cij .
So, the centralizer Pol X of any subset X of an abstract clone C, is a sub-clone of C.
The center of C, is the centralizer of C itself, Pol C = ∩xC Pol {x}. We have Pol Pol C = C.

For any p<n, Pol Cn ⊂ Pol Cp because all p-ary operations are represented among n-ary operations, with equivalence of formulas between p-ary and n-ary views for the above reason.
kC0, Pol{k} ∩ C0 = {k}.

The multiplication monoid in an abstract clone

Elements of C1 will be called scalars. The unit element of the clone, is the scalar π0,1 that will be simply denoted 1. The operation that applies any scalar a to any element x of any representation, (a,x)↦ax will be called multiplication. The multiplication between scalars forms a monoid, with 1 as identity element:
1⋅x = x
a,bC1,∀x  (ab)⋅x = a⋅(bx)

The multiplication is right distributive over any operation : for example if #∈C2,
a,bC1,∀x, (a#b)⋅x =(ax)#(bx)
However, the left distributivity formula (∀x,y, a⋅(x#y) = (ax)#(ay)) means that # commutes with a.
Similarly for constants (nullary operations) : as any constant symbol kC0 can be interpreted both in C1 and in any other representation, both are related by
x, kx = k
The same works for any constant, i.e. defined by a term that does not depend on any "variable". For example, if #∈C2 and k,k'∈C0, we have k#k'∈C0, thus
x, (k#k')⋅x = k#k'.
Conversely, if C0≠Ø and an element kC1 satisfies ∀xC1, kx = k then for any k'∈C0 , the element kk' ∈ C0 satisfies kk' = k in C1, so that k is the image in C1 of kk'∈C0.
As already argued above on the correspondence between interpretations of formulas, if two constants k,k'∈C0 commute by multiplication in C1 then they are equal in C1 (k = kk' = k'⋅k = k'∈C1) and thus also in C0 (k = kk = k'⋅k = k'∈C0).
Every term in the language C (i.e. without ⋅ nor 1) with only one variable, is reducible to a scalar multiplying the variable : for example, ∀#∈C2, ∀x,
x#k = (1⋅x)#(kx) = (1#k)⋅ x
  x#x = (1⋅x)#(1⋅x) = (1#1)⋅ x

Proposition.  C0 ∩ Pol C1 ⊂ Pol C.

Proof. Let kC0∩ Pol C1 , i.e. ∀aC1, ak=ka=k.
Let n∈ℕ and xCn. The commutation relation between k and x is verified as
x(k,...,k) = x(1⋅k,...,1⋅k) = x(1,...,1)⋅k = k

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