1841 ?=> 203

Stream: Equational

[Bernhard Reinke]

I thought a bit about the implication 1841 => 203. At the moment it is open, right?

One thing I think I can prove it that 1841 => 203 holds for magmas that are (left) translation invariant for a (not necessarily commutative) group. For this, suppose $G$ is a group and we have an operation x ◇ y that is of the form $x ◇ y = x f(x^{-1} y)$ for some function $f \colon G \rightarrow G$. Then the RHS of 1841 can be written as

$$(x ◇ (x ◇ y)) ◇ (y ◇ y) = (x ◇ (x f(x^{-1} y)) ◇ (y f(1)) = \\ (x f(f(x^{-1} y)) ◇ (y f(1)) = x f(f(x^{-1} y)) f( f(f(x^{-1} y))^{-1} x^{-1} y f(1))$$

writing $z = x^{-1} y$, we get that 1841 is equivalent to

$$f(f(z)) f( f(f(z))^{-1} z f(1)) = 1$$

on the other hand, the RHS of 203 can be written as

$$(x ◇ (x ◇ x)) ◇ x = (x ◇ (x f(1))) ◇ x = (x f(f(1))) ◇ x = x f(f(1))f ( f(f(1))^{-1})$$

so 203 is equivalent to $f(f(1)) f (f(f(1))^{-1}) = 1$.

As a shorthand, let $a = f(1)^{-1}, b = f(f(a))^{-1}$. If we plug in $a$ into the equation we get $1 = f(f(a)) f( f(f(a))^{-1} a f(1)) = b^{-1} f(b)$, so $f(b) = b$. Plugging in $b$ we get $1 = f(f(b)) f( f(f(b))^{-1} b f(1)) = b f(f(1))$, so $b = f(f(1))^{-1}$. But then 203 holds: indeed, we have $f(f(1)) f (f(f(1))^{-1}) = b^{-1} f(b) = 1$.

I haven't checked yet what would happen with right transitive instead of left transitive.

[Bernhard Reinke]

I am not sure about one really needs to additionally consider right translation invariant magmas, if $x ◇ y$ is left translation invariant, isn't $(x^{-1} ◇ y^{-1})^{-1}$ right translation invariant and satisfies the same laws?

[Daniel Weber]

Interestingly, assuming 99 together with 1841 still doesn't seem to imply 203

[Bernhard Reinke]

Related to this argument, I am curious whether there a term t(x) only in x (maybe even t(x) = (x ◇ x) ◇ ((x ◇ x) ◇ x) works ), such that 203 holds for x substituted by t(x). Then the statement about translation invariance follows as $x \mapsto t(x)$ is always surjective for magmas that have a transitive automorphism group.

[Bernhard Reinke]

Daniel Weber said:

Interestingly, assuming 99 together with 1841 still doesn't seem to imply 203

How did you compute this? I would interested to know if 1841 + ((x ◇ x) ◇ x) ◇ ((x ◇ x) ◇ x) = (x ◇ x)imply 203

[Bernhard Reinke]

Not sure if this has an equation number, might be too big

[Daniel Weber]

Bernhard Reinke said:

Daniel Weber said:

Interestingly, assuming 99 together with 1841 still doesn't seem to imply 203

How did you compute this? I would interested to know if 1841 + ((x ◇ x) ◇ x) ◇ ((x ◇ x) ◇ x) = (x ◇ x)imply 203

I just fed it to Vampire. It doesn't seem to imply 203, no

[Bernhard Reinke]

Hm, actually I want to understand 1841 + ( x ◇ (x ◇ x)) ◇ (x ◇ (x ◇ x) ) = (x ◇ x), but this might be a good change to play around with Vampire myself

[Bernhard Reinke]

Ok, I think I have a geometric construction. It is related to the previous tree constructions.

We start of again with a planar 3-regular tree, and consider a 1-factor F (so if we construct the tree as the Cayley graph of the free product of three copies of C_2, one can take the edges corresponding to a fixed generator). The magma is now defined on the vertex set of the tree together with the edges in the 1-factor F.
The rules are as follows: if e is in F, then it is left absorbent: e ◇ y = e. Now consider a vertex v. We set v ◇ v to be the adjacent edge to v in F. If w is a neighbor of v, then v ◇ w is the next neighbor of v in the cyclical ordering. If e = (v,v') is the adjacent edge in F to v, then v ◇ e = v' . Otherwise v ◇ y is the neighbor in the direction of y (this also makes sense for nonadjacent edges in F).

Let us see that (x ◇ (x ◇ y)) ◇ (y ◇ y) = x holds: If xis in F, it is clear as x is left absorbent. Otherwise suppose x is a vertex with neighbors (in cyclic order) a,b,c, and suppose e = (x,a) is the adjacent edge of F to x,. Let us deal with the special cases $y = \{x, a, e\}$:

$$(x ◇ (x ◇ x)) ◇ (x ◇ x) = (x ◇ e) ◇ e = a ◇ e = x \\ (x ◇ (x ◇ a)) ◇ (a ◇ a) = (x ◇ b) ◇ e = c ◇ e = x \\ (x ◇ (x ◇ e)) ◇ (e ◇ e) = (x ◇ a) ◇ e = b ◇ e = x \\$$

In the other cases, we always see that (x ◇ (x ◇ y)) is a neighbor of x that is away from (y ◇ y), so (x ◇ (x ◇ y)) ◇ (y ◇ y) = x holds as well.

Finally, we have (for x a vertex as above)
$(x ◇ (x ◇ x)) ◇ x = (x ◇ e) ◇ x = a ◇ x$, this is a neighbor of a different to x, so 203 does not hold.

[Bernhard Reinke]

Hm, I think it should also work if we work with the half-edges in F instead of the edges. So we can take as the underlying set the disjoint union of two copies of the vertex set of our graph. This might make the implementation easier

[Joachim Breitner]

Bernhard Reinke said:

Hm, I think it should also work if we work with the half-edges in F instead of the edges. So we can take as the underlying set the disjoint union of two copies of the vertex set of our graph. This might make the implementation easier

Likely, given that we have G = C_2×C_2×C_2 more or less already constructed.

Can you reformulate your construction in terms of G with generators $a,b,c$?

[Bernhard Reinke]

Here is my attempt: we keep the notation of this construction. In particular, we let $G = C_2 * C_2 * C_2$ with generating set $\{a, b, c\}$. We let $M = G \times \{0, 1\}$. Let $f : G \rightarrow \{1, a, b, c \}$ as in the previous construction ( so f(1)=1,f(a)=b,f(b)=c,f(c)=a, f(aw)=a, etc.)

We define $x ◇ y$ as follows

(g,1) ◇ y = (g, 1)
(g,0) ◇ (g, 0) = (g, 1)
(g,0) ◇ (h, 0) = (g f(g^{-1}h), 0)  for h != g
(g,0) ◇ (g, 1) = (ga, 0)
(g,0) ◇ (ga, 1) = (ga, 0)
(g,0) ◇ (gb, 1) = (gb, 0)
(g,0) ◇ (gc, 1) = (gc, 0)
(g,0) ◇ (h, 1) =  (g f(g^{-1}h), 0)  for h != g, ga, gb, gc

[Terence Tao]

Created equational#621 for this, and will mark it as a conjecture as well.

[Bernhard Reinke]

I think this is slightly different in the handling of edges that are distance 1 to x compared to my original description, but this should also work

[Bernhard Reinke]

Here is my attempt:

Bernhard Reinke said:

I think this is slightly different in the handling of edges that are distance 1 to x compared to my original description, but this should also work

actually, there is an issue, i will update the formula to reflect my original construction

[Bernhard Reinke]

@Joachim Breitner , have you started implementing this? If not, I would give it a try

[Joachim Breitner]

no, all yours

[Bernhard Reinke]

There is now a PR