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10 votes
4 answers
2k 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) $(a*b) /b=(a/b)*b=a$ ...
MphLee's user avatar
  • 233
16 votes
2 answers
602 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 ...
James's user avatar
  • 1,889
4 votes
1 answer
244 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 ...
eyp's user avatar
  • 163
12 votes
0 answers
259 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 $\...
Nicola Di Vittorio's user avatar
2 votes
0 answers
108 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_{\...
Joseph Van Name's user avatar
7 votes
0 answers
342 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 ...
Kadir Emir's user avatar
5 votes
1 answer
403 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 ...
Zhiyun Cheng's user avatar