Stone-topological/profinite equivalence for quandles

Question

A quandle $(Q,\triangleleft,\triangleleft^{-1})$ is a set $Q$ with two binary operations $\triangleleft,\triangleleft^{-1}:Q\times Q\to Q$ such that the following hold for all $x,y,z\in Q$:

(Q1) Idempotency: $x\triangleleft x=x$.

(Q2) Right-invertibility: $(x\triangleleft y)\triangleleft^{-1}y=x=(x\triangleleft^{-1}y)\triangleleft y$.

(Q3) Right-distributivity: $(x\triangleleft y)\triangleleft z=(x\triangleleft z)\triangleleft(y\triangleleft z)$.

These axioms may seem somewhat arbitrary, but I claim they aren't: They correspond to the Reidemeister moves in knot theory. Moreover, if $G$ is a group, then defining the operation $x\triangleleft y=y^{-1}xy$ on $G$ gives a quandle structure, and as shown by Joyce in 1982, these three axioms are sufficient to capture the equational content of conjugation.

A profinite quandle is an inverse limit $\varprojlim_{\lambda\in\Lambda}Q_\lambda$ of an inverse system of finite quandles. Those familiar with the theory of profinite groups will recognize this construction.

It is known for profinite groups the following result: Every profinite group can be endowed with a Stone topology, i.e., compact, Hausdorff, and totally disconnected. On the other hand, every Stone-topological group is profinite. This result can be found in Johnstone's book on Stone spaces, though it is presented in much more generality such that any finitary algebraic category with a "complete word set" in some variable admits this Stone-topological/profinite equivalence. Johnstone also gives sufficient requirements for having a complete word set. However, one of these requirements (distributivity) is not met by quandles.

Is there reason to believe that quandles admit such an equivalence? Perhaps there is some reason to believe this in relation to quandles characterizing group conjugation, but I haven't been able to work it out. I know there is a direct proof for Stone-topological groups being profinite, see my previous question. However, this relies on having an identity element, which quandles don't have. Any insight would be appreciated.

Answer 1

Here is how the proof works for monoids. This doesn't really answer the question so I made this communitywiki, but maybe the OP will find it useful.

Let $X$ be a compact Hausdorff space. Call an equivalence relation $R$ on $X$ open if it has open equivalence classes. By compactness, an open equivalence relation $R$ has finitely many classes and $X/R$ is a finite discrete space. Notice that the intersection of two open equivalence relations is open.

Note that a compact Hausdorff space is totally disconnected if and only if the intersection of all the open equivalence relations (in $X\times X$) is the diagonal. Moreover, $X\cong \varprojlim X/R$ where $R$ is any cofinal set downward directed set of open equivalence relations.

Numakura's Theorem: Let $M$ be a compact and totally disconnected topological monoid (the monoid is done just to make the congruence slightly less complicated to describe). Then $M$ is profinite.

Proof. It suffices by the above discussion to show that if $R$ is an open equivalence relation on $M$ then there is an open congruence contained in $R$ (note the open congruences are closed under intersection and hence downward directed).

Define $\sim$ by $m\sim n$ iff $umv\mathrel{R} unv$ for all $u,v\in M$ (for semigroups, one has to adjust to compensate for not having the possibility $u,v$ are the identity).

It is an easy exercise to verify that $\sim$ is a congruence contained in $R$. We claim it is open. If $m\in M$, then the $\sim$-class of $m$ can be described as $\bigcap_{Y\in M/R}(\bigcup_{u,v\in M,umv\in Y}u^{-1}Yv^{-1})$ where $u^{-1} Yv^{-1} =\{n\in M\mid unv\in Y\}$. Since each $Y\in M/R$ is open and multiplication is continuous, $u^{-1}Yv^{-1}$ is open. Since $M/R$ is finite, we conclude that the $\sim$-class of $m$ is open.