Equational Theories

17 Equation 1516

In this chapter we study magmas that obey equation 1516,

x=(yy)(x(xy))
1

for all x,y. Using the squaring operator Sy:=yy and the left and right multiplication operators Lyx:=yx and Ryx=xy, this law can be written as

LSyLxLxy=x.

We begin by studying a greedily constructed translation invariant model

xy=x+f(yx)
2

on the carrier Z with some function f:ZZ with f(0)=0. This ensures that Sx=x. If y=x+h, then Lxy=x+f(h), Lx2y=x+f2(h), and LyLx2y=y+f(f2(h)h), so the law Equation 1 simplifies to

f(f2(h)h)=h.
3

Thus, if we let E={(h,f(h))} be the graph of f, then we have the following property: if (a,b),(b,c)E, then (ca,a)E. This helps motivate the following definition.

Definition 17.1 1516 seed
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A 1516 seed is a finite collection E of pairs (a,b) with a,bZ obeying the following axioms:

  • Axiom 1: (0,0)E.

  • Axiom 2: If (a,b)E and a0, then b0,a.

  • Axiom 3: If (a,b),(a,b)E, then b=b.

  • Axiom 4: If (a,b),(b,c)E, then (ca,a)E.

  • Axiom 5: If (b,a),(b,a),(b,d),(b,d)E with bb, then b+dd,b+d.

An extension of a 1516 seed E is a 1516 seed E that contains E.

This definition has an extension property:

Lemma 17.2 1516 extension

Let E be a 1516 seed, and let a0Z. Then there exists an extension E of E that contains (a0,c0) for some c0.

Proof

For technical reasons (which will be helpful later when we expand the magma) we also give a variant that will ensure a useful “double surjectivity” property:

Lemma 17.3 1516 extension variant

Let E be a 1516 seed, and let hZ be non-zero. Then there exists an extension E of E that contains (ai,ai+h) for i=1,2,3,4, for some distinct a1,a2,a3,a4.

Proof
Corollary 17.4 Base magma

There exists a 1516 magma with carrier Z with the properties that

  • (i) Sa=a for all a.

  • (ii) For any distinct a,bZ, the equation Rac=b has at least four solutions c.

  • (iii) For each aZ, there exist at least two ba such that LaRab=b.

Note from Equation 1 and the hypothesis Sa=a that Rac=a if and only if c=a. Thus, the requirement that a,b be distinct in (ii) is necessary.

Proof

Let us denote by A the free group over N. Note that a magma with the same properties as in Corollary 17.4 but having carrier A can be constructed similarly. The same thing is true for the following results. In fact, the Lean implementation has A whenever Z is; this has been done to reuse some preexisting code.

Now we construct a more complex 1516 magma, whose carrier G is ZG, where

G:={(a,c,n)Z×Z×N:ac}.

The first component Z will represent the squares, and the second component G will represent the non-squares. The most important coordinate of an element (a,c,n) of G is a; the other two components c,n are technical, with n being needed to ensure a certain infinite surjectivity property, and c being such that Lc(a,c,n) remains at one’s disposal to select at certain stages of the inductive process.

The 1516 magma constructed in Corollary 17.4 will be the restriction of G to Z; thus ab is already defined in G for a,bZ, but the rest of the multiplication table is not currently defined. By construction, S is already defined and equal to the identity on Z, and we have the 1516 equation

LSaLbLba=b

for a,bZ, with the left multiplication operators Lb currently only defined as maps from Z to Z. Among other things, this means that La=LSa is surjective as a map from Z to Z for any aZ.

We extend the squaring map S to all of G by declaring S(a,c,n):=a, thus S maps G to Z. To create a 1516 magma structure on all of G, we need to extend the left multiplication operators Lb, bZ as maps from G to G, and also introduce additional maps Lx:GG for xG, obeying the following axioms:

  • Axiom A: For any xG, Lxx=Sx.

  • Axiom B: For any xG and bZ, LSxLbLbx=b.

  • Axiom C: For any xG and yG, LSxLyLyx=y.

We first address Axiom B, which purely concerns how to extend the Lb operators for bZ, and also impose an additional technical “infinitely surjective” requirement that will help us satisfy Axiom C later. We first need some preparatory lemmas. We begin by locating some useful elements cy,b of Z for all yG and bZ.

Lemma 17.5 Useful elements

One can assign cy,bZ for each y=(a,c,n)G and bZ with the following properties:

  • (i) For all y=(a,c,n)G and bZ, cy,ba,b,c and LaLcy,bb=cy,b.

  • (ii) For all yG, the map bcy,b is injective.

Proof

Let cy,b be as above. Define a partial solution to be a partial assignment of LcyG for cZ and yG that obeys the following properties:

  • (a) If yZ, then Lcy agrees with the operation defined in Corollary 17.4.

  • (b) If y=(a,c,0)G, then Lay=a.

  • (c) If y=(a,c,n)G and n0, then Lay=(a,c,0).

  • (d) If y=(a,c,n)G and bZ, then Lcy,by=b.

  • (e) For any a,c, Lc(a,c,n) is only defined for finitely many n.

  • (f) If y=(a,c,n)G and cZ are such that Lcy is defined, then LcLcy and LaLcLcy are defined, with LaLcLcy equal to c.

  • (g) If y=(a,c,n)G and cZ are such that ca and Lcy is defined, then Lcyc.

Lemma 17.6 Existence of partial solution

A partial solution exists.

Proof

We can make Lcy defined in a partial solution:

Lemma 17.7 First extension

Suppose we have a partial solution for which Lcy is currently undefined for some cZ and yG. Then one can extend this partial solution in such a manner that Lcy is now defined.

Proof

We can also insert inverses of Lc:

Lemma 17.8 Second extension

Suppose we have a partial solution, and let cZ and y=(a,c,n)G. Then, unless one is in the case when c=a and n=0, then one can find zG such that Lcz is currently undefined, but for which one can find an extension of the partial solution for which Lcz is now equal to y.

Proof

Now we can establish Axiom B.

Proposition 17.9 Obtaining Axiom B

There exists a way to extend Lb:ZZ to Lb:GG for all bZ, in such a way that Axiom B holds, and furthermore for each bZ and xG, the set {yG:Lby=x} is infinite. Also, we can ensure that Lbxx for any bZ and xG.

Proof

Finally, we construct the remaining Lx.

Proposition 17.10 Obtaining Axioms A, C

One can find maps Lx:GG for each “non-square” xG, such that Axioms A, C hold.

Proof
Corollary 17.11

There exists a 1516 magma that does not obey the 255 equation

x=((xx)x)x.
Proof