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

Question

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) $(a*b) /b=(a/b)*b=a$

If we have a group $(G, \cdot, e,^{-1})$ and $*$ is the cojugation operation on $G$

$$a*b:=bab^{-1}$$

and

$$a/b:=a*b^{-1}$$

then $(G,*,/)$ is denoted with $Conj(G)$ and is a quandle because it satisfies the quandles axioms

1) $a*a=a=aaa^{-1}$

2) $(a*b)*c=(a*c)*(b*c)$

because $c(bab^{-1})c^{-1}=(cbc^{-1})cac^{-1}(cb^{-1}c^{-1})=cbab^{-1}c^{-1}$

3) $(a*b) *b^{-1}=(a*b^{-1})*b=a$

because $b^{-1}(bab^{-1})b=b(b^{-1}ab)b^{-1}=a$

I read that we have too that a group homomorphism between two groups $G$ and $G'$ is a quandle homomorphism between theire cojugation quandles $Conj(G)$ and $Conj(G')$ and that makes $Conj$ a functor betwen the category of Groups and the category of quandles...

---

I wanted to know more about this functor $Conj$ that "maps" Groups to Quandles and since I'm not expert of category theory I apologize if I use a wrong terminology

$Q1a$ - I learnt that not every Quandle is a conjugation Quandle or in other words $conj$ is not ""surjective"" on the "set" of all quandles (that is not a set but a class i think) so how can I prove that a Quandle is a Conjugation Quandle too?

$Q1b$ -Wich extra "axioms" must hold in a Quandle that is a Conjugation Quandle?

This has something to do with the inverse construction of $Conj$ so my next question is

$Q2$ - There is a way to define a group operation starting with a quandle operation? Like an inverse $Conj$ construction that build a "Quandle-Group" $Conj^{-1}(Q)$ from a conjugation quandle $Q$.

$Q3$ Is this process not unique?

With not unique I mean that is possible to have two different groups $G=(G,\cdot,\phi)$ and $G'=(G,\circ,\varphi)$ and $$Conj (G)= Conj(G')$$

this should mean that is possible to have

$$b\cdot a\cdot \phi(b)=b \circ a \circ \varphi(b)$$

where $\phi(b)$and $\varphi(b)$ are the inverse functions and

$$a \cdot b \neq a \circ b$$

$Q4$ My last question is if possible to generalize the conjugation operation of groups for monoids and semigruops in a way that these "Monoid-conjugations" and "Semigroup-conjugations" are quandles. If is possible I would like to read more about.

Answer 1

Q1b Joyce in A classifying invariant of knots, the knot quandle. J. Pure Appl. Algebra 23 (1982) proves that the equational theory of quandles and that of groups endowed with a quandle structure via conjugation is the same. So conjugation quandles are not a subvariety of quandles. There is however one more axiom which is necessary (I follow your convention of right quandle action)

$$ y * x = y \iff x *y =x $$

Q1a I'm not aware of any sufficient condition for a quandle to be a conjugation subquandle of a group.

Q2 Given a quandle you can construct its Enveloping Group, or Adjoint Group, which is the free group on the elements modulo the relations $$y *x = x^{-1}y x $$ Joyce shows that this construction is left adjoint to that of taking the conjugation quandle. This is more explicitly stated in Andruskiewitsch, Nicolás; Graña, Matías From racks to pointed Hopf algebras. Adv. Math. 178 (2003)

Q3 Consider that every abelian group with the same number of elements give rise to the same (trivial) quandle

Answer 2

You have two excellent expert answers. Here is a very simple answer to some of your questions.

First a slight overview. Given a group one has the conjugation operation $a*b=bab^{-1}$ This is just $a*b=a$ when $G$ is an abelian group. A group with $n$ elements gives a quandle with $n$ elements which satisfies the axioms you listed.

A finite quandle can be thought of as an $n$ by $n$ "multiplication" table which satisfies certain rules including every symbol appears once in each column and : in the $x$ column, the symbol $x$ occurs in the $x$ row. I'm not sure how interesting quandles are without a group or perhaps another structure to motivate them.

Anyway, there is only one group of order $5$ and it is abelian so it has quandle

$$\left[ \begin {array}{ccccc} 0&0&0&0&0\\ 1&1&1&1&1 \\ 2&2&2&2&2\\3&3&3&3&3 \\ 4&4&4&4&4\end {array} \right] .$$ This is the only quandle with $5$ elements which can come from a group.

Here are two other quandles with $5$ elements $$\left[ \begin {array}{ccccc} 0&4&3&2&1\\ 2&1&0&4&3 \\ 4&3&2&1&0\\ 1&0&4&3&2 \\ 3&2&1&0&4\end {array} \right] $$

$$ \left[ \begin {array}{ccccc} 0&3&1&4&2\\ 3&1&4&2&0 \\ 1&4&2&0&3\\ 4&2&0&3&1 \\ 2&0&3&1&4\end {array} \right] $$

We can see that they are not isomorphic since only the second has $a*b=b*a$. It might not be hard to find all quandles (up to isomorphism) with $5$ elements, but I did not do that.

The axioms could be checked by brute force but it is easier if one knows that the first is $a*b=2a-b \mod 5$ and the second $a*b=3a-2b \mod 5$. The general construction for $m$ elements is $a*b=ka-(k-1)b \mod m$ where we need that $k$ is relatively prime to $m.$

In answer to a later question: The construction is actually very general. Suppose we have a ring (maybe not commutative) with an identity, and that $k$ is an element with a multiplicative inverse $k^{-1}$. Then the operation $a*b=ka+(1-k)b$ certainly satisfies $a*a=a.$ Also $$(a*c)*(b*c)=$$ $$k(ka+(1-k)c)+(1-k)(kb+(1-k)c)=k^2a+(1-k)kb+(1-k)c$$ while $$(a*b)*c=k(ka+(1-k)b)+(1-k)c$$ so these are equal, since in all cases $(1-k)k=k(1-k).$ So it remains to see what is needed to have a solution $c=a/b$ to $c*b=a.$ That would have to be $a/b=k^{-1}(a+(k-1)b)$ in which case $(a*b)/b=k^{-1}(ka+(1-k)b+(k-1)b)=a.$

Answer 3

Let me point out a sort of characterization of conjugation quandles. (This is too long for a comment.)

Let $X$ be a finite quandle. The enveloping group $G_X$ is the group with generators $x\in X$ and relations $x(x*y)=yx$ for all $x,y\in X$.

The enveloping group has the following universal property:

For any group $G$ and any map $f:X\to G$ satisfying $f(x*y)=f(x)^{-1}f(y)f(x)$ there exists a unique group homomorphism $g:G_X\to G$ such that $f=g\circ\partial$, where $\partial:X\to G_X$, $i\mapsto x_i$.

See for example:

• Andruskiewitsch, Nicolás; Graña, Matías. From racks to pointed Hopf algebras. Adv. Math. 178 (2003), no. 2, 177--243. MR1994219 (2004i:16046)
• Fenn, Roger; Rourke, Colin. Racks and links in codimension two. J. Knot Theory Ramifications 1 (1992), no. 4, 343--406. MR1194995 (94e:57006), link

Now suppose that $X$ is connected. (This means that the group generated by the permutations $\varphi_x\colon y\mapsto y*x$, $x\in X$, acts transitively on $X$.) Then it follows from the definitions that all the permutations $\varphi_x$ have the same order. Suppose that the order of the permutations $\varphi_x$ is $d$. Then:

1. $\delta(x)^d=\delta(y)^d$ for all $x,y\in X$,
2. $\delta(x)^d$ is central in $G_X$, and
3. the quotient group $F_X=G_X/\langle \delta(x)^d\rangle$ is finite.

Since there is a group homomorphism $\deg:G_X\to\mathbb{Z}$ given by $\deg(x)=1$ for all $x\in X$, the group $G_X$ is $\mathbb{Z}$-graded. Now consider the quotient group $F_X$ and let $p:G_X\to F_X$ be the canonical surjection. Write the generators of $G_X$ as $x_i$ for all $i\in X$.

The following is useful to check whether a connected quandle is a conjugacy class:

The map $\partial:X\to G_X$ is injective if and only if the map $X\to F_X$, $i\mapsto p(x_i)$, is injective.

This follows from the following observation: $p$ restricted to the conjugacy class of $x_1$ is bijective. To prove this, let $u,v$ be conjugate elements of $G_X$. Then $\deg(u)=\deg(v)$. If $p(u)=p(v)$ then $u=vx_1^m$ for some $m$. Taking degrees, $m=0$ and hence $u=v$.

Application. This website contains computational results on small connected quandles and their knot colorings obtained by W. Edwin Clark and Timothy Yeatman. In this subpage you can find a list of 16 quandles which are not known to be conjugation quandles. Apparently it is not easy to use the computer and the universal property of $G_X$ to check whether these quandles are conjugation quandles. The group $F_X$ does the trick and one can easily decide which of these 16 quandles are conjugation quandles!

Answer 4

The notion of a semigroup with conjugation can be formalized by the notions of an LD-monoid, LRD-monoid, and LRDQ-monoid. The information in this answer can be found in the book Braids and Self-Distributivity by Patrick Dehornoy.

$\textbf{LD-monoids}$

An LD-monoid (LD stands for left-distributive) is an algebra $(M,\cdot,1,\wedge )$ such that $(M,\cdot,1)$ is a monoid and the following identities are satisfied: $$x\cdot y=(x\wedge y)\cdot x,(x\cdot y)\wedge z=x\wedge (y\wedge z), x\wedge (y\cdot z)=(x\wedge y)\cdot(x\wedge z),1\wedge x=x,x\wedge 1=1.$$

If $(M,\cdot,1,\wedge )$ is an LD-monoid, then $(M,\wedge )$ is generally not the quandle. However, $(M,\wedge )$ still satisfies the left-distributivity law $x\wedge (y\wedge z)=(x\wedge y)\wedge (x\wedge z)$. Algebras that satisfy the law $x\wedge (y\wedge z)=(x\wedge y)\wedge (x\wedge z)$ are commonly called LD-systems and LD-systems are still of interest to at least some knot theorists even though they are usually not quandles..

The operation $\wedge$ acts as a sort of conjugation operation on monoids.

For instance, suppose that $(M,\cdot,1)$ is a monoid where for all $x,y\in M$ there exists a unique $z$ where $z\cdot x=x\cdot y$ (in particular, if $M$ is a group). Then $(M,\cdot,1)$ becomes an LD-monoid when we attach the operation $x\wedge y$ defined by $(x\wedge y)\cdot x=x\cdot y.$ Furthermore, if $(G,\cdot,1,\wedge)$ is an LD-monoid and $G$ is a group, then $x\wedge y=x\cdot y\cdot x^{-1}$.

One can even construct LD-monoids by taking a collection of functions and a sort of conjugacy operation. For example, let $\mathfrak{I}_{\infty}$ be the collection of all injective functions from $\mathbb{N}$ to $\mathbb{N}$. Then $(\mathfrak{I}_{\infty},\cdot,1,\wedge)$ is an LD-monoid where $\cdot$ is the composition function, $1$ is the identity mapping, and $(f\wedge g)(x)=f\circ g\circ f^{-1}(x)$ whenever $x\in Im(f)$ and where $(f\wedge g)(x)=x$ otherwise.

$\textbf{LRDQ-monoids and quandles}$

Even though LD-monoids do not give us quandles, one can generalize the notion of an LD-monoid to LRDQ-monoids and LRDQ-monoids do give us quandles.

An LRD-monoid is an algebra $(M,\cdot,1,\wedge,\vee)$ that satisfies the following identities. $$1\wedge x=1\vee x=x,x\wedge 1=x\vee 1=1;x\cdot y=(x\wedge y)\cdot x=y\cdot(y\vee x)$$

$$(x\cdot y)\wedge z=x\wedge(y\wedge z),(x\cdot y)\vee z=y\vee(x\vee z)$$ $$x\wedge(y\cdot z)=(x\wedge y)\cdot(x\wedge z),x\vee(y\cdot z)=(x\vee y)\cdot(x\vee z).$$

An LRDQ-monoid is an LRD-monoid that satisfies the identity $x\wedge(x\vee y)=x\vee(x\wedge y)=x$. Every group becomes an LRDQ-monoid with operations $\wedge$ and $\vee$ defined by $x\wedge y=xyx^{-1},x\vee y=x^{-1}yx$.

If $(M,\cdot,1,\wedge,\vee)$ is an LRDQ-monoid, then $\wedge$ is idempotent if and only if $\vee$ is idempotent. If $(M,\cdot,1,\wedge,\vee)$ is an LRDQ-monoid such that $\wedge$ and $\vee$ are idempotent, then $(X,\wedge,\vee)$ is a quandle.

For example, suppose that $(M,\cdot,1)$ is a monoid such that for all $x,y$ there are unique $z_{L},z_{R}$ such that $xy=z_{L}x=xz_{R}$. Then one may define unique operations $\wedge$ and $\vee$ on $(M,\cdot,1)$ such that $x\cdot y=(x\wedge y)\cdot x$ and $x\cdot y=y\cdot(y\vee x)$. Then $(M,\cdot,1,\vee,\wedge)$ is an LRDQ-monoid, and the operations $\vee$ and $\wedge$ are idempotent, so $(M,\vee,\wedge)$ is a quandle.

LD-monoids originally to axiomatize the identities satisfied by the algebra of elementary embeddings obtained from the I3 axiom which states that there exists some non-trivial elementary embedding $V_{\lambda}\rightarrow V_{\lambda}$. The collection of elementary embeddings from $V_{\lambda}$ to $V_{\lambda}$ becomes an LD-monoid. The I3 axiom is one of the strongest large cardinal axioms that far extends the axioms of set theory.