Questions tagged [self-distributivity]

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Is every critically subsimple Laver-like algebra a quotient of a critically simple Laver-like algebra on the same number of generators?

A finite reduced Laver-like algebra is a finite algebra $(X,*,1)$ that satisfies the identities $1*x=x,x*1=1,x*(y*z)=(x*y)*(x*z)$ and where there is a natural number $n$ and a function $\mathrm{crit}:...
2 votes
1 answer
105 views

Attraction in Laver tables

If $X$ is a self-distributive algebra, then define $x^{[n]}$ for all $n\geq 1$ by letting $x^{[1]}=x$ and $x^{[n+1]}=x*x^{[n]}$. The motivation for this question comes from the following fact about ...
3 votes
1 answer
313 views

Can Laver tables go extinct?

An algebra $(X,*)$ is said to be self-distributive if it satisfies the identity $x*(y*z)=(x*y)*(x*z)$ for all $x,y,z\in X$. If $(X,*)$ is an algebra, then a subset $L\subseteq X$ is said to be a left-...
11 votes
2 answers
705 views

Motivation for Laver's use of large cardinals to show finite combinatorial properties of Laver tables

Laver showed in 1995 that the period of the first row of certain Laver tables is unbounded, assuming that a rank-into-rank cardinal exists. The most accessible proof of his result that I was able to ...
4 votes
1 answer
227 views

Why do highly composite rows on the bad Laver tables have longer periods?

For all natural numbers $n$, let $(B_{n},*_{n})$ be the algebraic structure with underlying set $\{1,\dots,n\}$ where $x*_{n}1=x+1\mod n$, $n*_{n}y=y$, and $x*_{n}(y+1)=(x*_{n}y)*_{n}(x+1)$ for $x<...
1 vote
2 answers
210 views

Example of idempotent left quasigroups which are right-distributive but not left-distributive

I am looking for examples of the following algebraic structure: a set (X,.) which satisfy the axioms (idempotent) x.x = x (left quasigroup) the equation a.x = b has a unique solution denoted by x = ...
16 votes
2 answers
590 views

Formally undecidable problems on finitely presented quandles

In the literature, one sometimes sees the claim that finitely presented quandles (in particular, knot quandles) are "hard to deal with". Hence, a great deal of effort has gone into studying finite ...
7 votes
0 answers
329 views

Hemi-semi direct product of racks or quandles

In the category of racks (similarly quandles), instead of well-known semidirect product, we have the hemi-semi direct product construction as seen on Wagemann & Crans. As far as I know, semi ...
1 vote
0 answers
65 views

Posets with two partial (self-)distributive operations

Let $(X, {\sqsubset}, {\circ}, {\ast})$ be a set $X$ with a strict partial order $\sqsubset$ and two partial binary operations $\circ$ and $\ast$ such that for any $a, b, c \in X$: $a \circ b$ and $a ...
10 votes
4 answers
1k views

Conjugation Quandles and... "Quandle-Groups"? From quandles to Groups

This question is already asked MathSE A quandle $(Q,*,/ )$ is a idempotent right-distributive and right invertible structure. 1) $a*a=a$ 2) $(a*b)*c=(a*c)*(b*c)$ 3) $(a*b) /b=(a/b)*b=a$ ...
3 votes
1 answer
398 views

Shelves with "trichotomy"

A left shelf $(S, \rhd)$ is a magma with the left self-distributive law: $$ \forall x, y, z \in S: x \rhd (y \rhd z) = (x \rhd y) \rhd (x \rhd z). $$ Shelves are generalization of racks and quandles ...
3 votes
1 answer
239 views

Racks with "trichotomy"

(This is a follow-up question; the original question was about shelves.) A rack $(R, \rhd, \lhd)$ is a set $R$ with two binary operations $\rhd$ and $\lhd$ such that for all $x, y, z \in R$: $x \rhd ...
5 votes
1 answer
393 views

One question about the quandle

Given a finite quandle $Q$, for any knot $K$ one can associate an invariant, i.e. the number of proper colorings $p(K)$. Let us consider the inverse $K^{-1}$ and mirror image $K'$ of $K$. My queston ...
4 votes
1 answer
229 views

Quandle homomorphism does not always induces group homomorphism on inner automorphism groups of quandles

Let $X$ and $Y$ be two quandles and $f: X \rightarrow Y$ be a quandle homomorphism. Then we can define a map $\bar f: Inn(X) \rightarrow Inn(Y)$ as $\bar f(S_a)=S_{f(a)}$, where $a \in X$. Then $\bar ...
12 votes
0 answers
253 views

Higher homotopical information in racks and quandles

A quandle is defined to be a set $Q$ with two binary operations $\star,\bar\star\colon\ Q\times Q\to Q$ for which the following axioms hold. Q1. a $\star$ a = a Q2. (a $\star$ b) $\bar\star$ b = (a $\...
6 votes
0 answers
187 views

Can finite binary self-distributive algebras fit into small $n$-ary self-distributive algebras?

A binary operation $*$ is said to be self-distributive if it satisfies the identity $x*(y*z)=(x*y)*(x*z)$. An $n+1$-ary operation $t$ is said to be self-distributive if it satisfies the identity $$t(...
1 vote
0 answers
121 views

Basic questions about varieties of uniformly partially permutative algebras

Define the Fibonacci terms $t_{n}(x,y)$ for all $n\geq 1$ by letting $t_{1}(x,y)=y,t_{2}(x,y)=x,t_{n+2}(x,y)=t_{n+1}(x,y)*t_{n}(x,y)$. We say that an algebra $(X,*)$ is $N$-uniformly partially ...
4 votes
0 answers
113 views

How many compatible linear orders exist on the classical Laver tables?

Let $A_{n}$ be the unique algebra $(\{1,\dots,2^{n}\},*_{n})$ such that $x*_{n}1=x+1\mod 2^{n}$ and $x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$ for all $x,y,z$. We say that a linear ordering $\preceq$ ...
1 vote
0 answers
90 views

Why does $p_{n}(i,1)=1$ so often where the polynomials $p_{n}$ are obtained from the classical Laver tables

So I was doing some computer calculations with the classical Laver tables and I found polynomials $p_{n}(x,y)$ such that $p_{n}(i,1)=1$ for many $n$. The $n$-th classical Laver table is the unique ...
1 vote
0 answers
60 views

Growth rate of the critical points of the Fibonacci terms $t_{n}(x,y)$ vs $t_{n}(1,1)$ in the classical Laver tables

The classical Laver table $A_{n}$ is the unique algebra $(\{1,\dots,2^{n}\},*_{n})$ where $x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$ and $x*_{n}1=x+1\mod 2^{n}$ for all $x,y,z\in A_{n}$. Define the ...
1 vote
0 answers
75 views

Multiple roots in the classical Laver tables

The classical Laver table $A_{n}$ is the unique algebraic structure $$(\{1,\dots,2^{n}\},*_{n})$$ such that $x*_{n}1=x+1\mod 2^{n}$ and $$x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$$ for all $x,y,z\in\{1,...
1 vote
0 answers
59 views

Can we have $\sup\{\alpha\mid(x*x)^{\sharp}(\alpha)>x^{\sharp}(\alpha)\}=\infty$ in an algebra resembling the algebras of elementary embeddings?

A finite algebra $(X,*,1)$ is a reduced Laver-like algebra if it satisfies the identities $x*(y*z)=(x*y)*(x*z)$ and if there is a surjective function $\mathrm{crit}:X\rightarrow n+1$ where $\mathrm{...
1 vote
0 answers
60 views

In the classical Laver tables, do we have $o_{n}(1)<o_{n}(2)$ for any $n>8$?

The classical Laver table $A_{n}$ is the unique algebraic structure $(\{1,\dots,2^{n}\},*_{n})$ where $$x*_{n}(y*_{n}z)=(x*_{n}y)*_{n}(x*_{n}z)$$ and where $$x*_{n}1=x+1\mod 2^{n}$$ for $x,y,z\in\{1,\...
1 vote
0 answers
43 views

What possible order type can the critical points of these algebras with one generator achieve?

Suppose that $(X,*)$ is an algebra that satisfies the self-distributivity identity $x*(y*z)=(x*y)*(x*z)$. We say that an element $x\in X$ is a left-identity if $x*y=y$ for all $x\in X$. Let $\mathrm{...
2 votes
0 answers
81 views

For each $n$ is it possible to have $\mathrm{crit}(x^{[n]}*y)>\mathrm{crit}(x^{[n-1]}*y)>\dots>\mathrm{crit}(x*y)$?

Suppose that $(X,*,1)$ satisfies the following identities: $x*(y*z)=(x*y)*(x*z),1*x=x,x*1=1$. Define the Fibonacci terms $t_{n}(x,y)$ for $n\geq 1$ by letting $$t_{1}(x,y)=y,t_{2}(x,y)=x,t_{n+2}(x,y)=...
1 vote
0 answers
44 views

Vastness of inverse systems of Laver-like algebras

Suppose that $(X,*,1)$ satisfies the identities $x*(y*z)=(x*y)*(x*z),x*1=1,1*x=x$. Then we say that $(X,*,1)$ is a reduced Laver-like algebra if whenever $x_{n}\in X$ for all $n\in\omega$, there is ...
1 vote
0 answers
33 views

Can we always extend a finitely generated reduced Laver-like algebra to a vast inverse system of Laver-like algebras?

An $(X,*,1)$ that satisfies the identities $x*(y*z)=(x*y)*(x*z),1*x=x,x*1=1$ is said to be a reduced Laver-like algebra if whenever $x_{n}\in X$ for $n\in\omega$, there is some $N\in\omega$ where $x_{...
3 votes
0 answers
240 views

Ordering large cardinal axioms around the level of $n$-huge by consistency strength?

So the large cardinal axioms are for the most part considered to be linearly ordered by consistency strength. For the large cardinals between extendibility and rank-into-rank (i.e. the $n$-huge ...
2 votes
0 answers
53 views

Calibrating the strength of the quotients of subalgebras of the classical Laver tables

Define an algebraic structure $A_{n}$ by letting $$A_{n}=(\{1,\dots,2^{n}-1,2^{n}\},*_{n})$$ where $*_{n}$ is the unique operation such that $x*_{n}1=x+1\mod 2^{n}$ for $$x\in\{1,\dots,2^{n}-1,2^{n}\}$...
1 vote
0 answers
47 views

Density of different types of critical points in an algebra of elementary embeddings

Suppose that $j,k:V_{\lambda}\rightarrow V_{\lambda}$ are elementary embeddings. Let $\mathrm{crit}_{n}(j,k)$ denote the $n$-th element in $\{\mathrm{crit}(\ell)\mid\ell\in\langle j,k\rangle\}$. ...
1 vote
0 answers
42 views

Density of critical points subalgebras of the algebras of elementary embeddings

Let $j:V_{\lambda}\rightarrow V_{\lambda}$ be an elementary embedding. Then $\{\mathrm{crit}(k)\mid k\in\langle j\rangle\}$ has order type $\omega$, so let $\mathrm{crit}_{n}(j)$ denote the $n$-th ...
1 vote
0 answers
73 views

Which varieties are compatible with the classical Laver tables?

Let $$A_{n}=(\{1,\dots,2^{n}-1,2^{n}\},*_{n})$$ denote the $n$-th classical Laver table. The operation $*_{n}$ is the unique binary operation on $\{1,\dots,2^{n}\}$ such that $$x*_{n}(y*_{n}z)=(x*_{n}...
5 votes
0 answers
93 views

The descriptive complexity and definiteness of the space of all elementary embeddings $j:V_{\lambda+1}\rightarrow V_{\lambda+1}$

Let $\mathcal{E}_{\lambda}$ be the set of all elementary embeddings $j:V_{\lambda}\rightarrow V_{\lambda}$. Suppose that $(\alpha_{n})_{n}$ is an increasing cofinal sequence in $\lambda$. Give $\...
2 votes
0 answers
80 views

Can a finitely generated algebra of rank-into-rank embeddings grow at rate $O(n\cdot\log(n))$?

Let $\mathcal{E}_{\lambda}$ be the set of all elementary embeddings from $V_{\lambda}$ to $V_{\lambda}$. If $j\in\mathcal{E}_{\lambda}$ is a non-trivial elementary embedding, then define $\mathrm{crit}...
2 votes
0 answers
107 views

Is the action of free self-distributive algebras on racks computable in polynomial time?

Let $B_{\infty}$ denote the infinite strand braid group. Let $\mathrm{sh}:B_{\infty}\rightarrow B_{\infty}$ be the mapping where $\mathrm{sh}(\sigma_{i})=\sigma_{i+1}$ whenever $i\geq 1$. Then $B_{\...
2 votes
0 answers
102 views

Does shifted conjugacy still give you free self-distributive algebras on one generator for quotient groups of the braid groups?

Let $B_{\infty}$ denote the infinite strand braid group. Let $\mathrm{sh}:B_{\infty}\rightarrow B_{\infty}$ be the group homomorphism where $\mathrm{sh}(\sigma_{i})=\sigma_{i+1}$ for all $i>0$. ...