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Let $K_1$ and $K_2$ be two knots such that for all finite quandles $X$, the number of colorings of $K_1$ by $X$ is the same as the number of colorings of $K_2$ by $X$. Then my question is, must $K_1$ and $K_2$ either be the same knot or mirror images of each other?

If not, does anyone know of a counterexample?

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The short answer is that I think this is an open question. This is stated as Conjecture 3.4 here, proved for knots up to 12 crossings:

Clark, W. Edwin; Elhamdadi, Mohamed; Saito, Masahico; Yeatman, Timothy, Quandle colorings of knots and applications, J. Knot Theory Ramifications 23, No. 6, Article ID 1450035, 29 p. (2014). ZBL1302.57016. MR3253967.

Looking at articles citing this one, there doesn't seem to have been subsequent direct progress on this problem, making it appear to be still open.

On the other hand, this question is related to the question of "profinite rigidity" of compact 3-manifold fundamental groups, including knot groups. A knot $K$ is said to be profinitely rigid if for any compact 3-manifold $N$ with $\pi_1N$ and $\pi_1(S^3-K)$ having the same finite quotients, then $\pi_1 N=\pi_1 (S^3-K)$. I.e., the groups have the same profinite completion iff they are isomorphic.

Profinite rigidity is known for some knot groups, including graph knots and the figure eight knot; see Bridson-Reid, Boileau-Friedl, and Wilkes.

Prime knots are determined by their fundamental groups up to mirror image, whereas composite knots can have the same fundamental group but different quandles. So one may restrict to prime knots to get examples of knots which are determined by the profinite completion of the fundamental groups of their complements by the above cited theorems.

How does this relate to your question? Consider the class of finite quandles obtained by taking conjugacy in finite groups $G$. A quandle homomorphism from the fundamental quandle of a knot is thus equivalent to a map $\varphi$ of the conjugacy class of the meridian to the finite group quandle (mapping to some conjugacy class in $G$). So one has $\varphi(a\lhd b) = \varphi(aba^{-1})=\varphi(a)\varphi(b)\varphi(a)^{-1} = \varphi(a)\lhd \varphi(b)$. In turn, this induces a homomorphism $\Phi: \pi_1(S^3-K) \to Inn(G)$. To see this, note that the Wirtinger presentation involves only generators in the meridian conjugacy class, and relators of the form $aba^{-1}= a\lhd b = c$. Hence a quandle homomorphism induces a group homomorphism to $Inn(G)$, where $\Phi(ab)(c)=abcb^{-1}a^{-1}= a \lhd (b\lhd c)=\Phi(a)\Phi(b)(c)$, and extending by induction to all elements of $\pi_1(S^3-K)$. If $G$ is center-free, then $Inn(G)\cong G$, and hence we get an induced homomorphism $\pi_1(S^3-K)\to G$.

So it remains to show that if all homomorphisms of $\pi_1(S^3-K)$ to finite groups $Inn(G)$ are the same, then all homomorphisms to finite groups are the same. But any finite group $H$ is a subgroup of $Inn(G)$ for some finite group $G$. So one can "see" homomorphisms to $H$ via the conjugation quandle associated to $G$. A fortiori, if all quandle homomorphisms to finite quandles are the same, then all homomorphisms of the fundamental quandle to conjugacy quandles are the same, and hence all homomorphisms of the fundamental group to finite groups are the same. Hence we get classes of knots which are determined by homomorphisms to finite quandles. Moreover, if profinite rigidity were known for knot groups, it would imply this property for prime knots. It should also be true for composite knots (following from profinite rigidity) using the fact that their quandle is determined by the fundamental group and the peripheral subgroup, using e.g. subgroup separability results.

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  • $\begingroup$ Just to be clear, are you saying my question is an open problem, or what? $\endgroup$ – Keshav Srinivasan Aug 30 '19 at 23:22
  • $\begingroup$ I believe it's an open problem in general, but I haven't done a literature search. $\endgroup$ – Ian Agol Aug 30 '19 at 23:34
  • $\begingroup$ So two different knots have never been found which have have this property? $\endgroup$ – Keshav Srinivasan Aug 31 '19 at 0:24
  • $\begingroup$ Not that I know of. The point of my answer is that if two knots had this property, then they would have the same profinite completion of the fundamental group. And no such pair of knots is known. $\endgroup$ – Ian Agol Aug 31 '19 at 1:15
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Not a direct answer but a related one:

The granny knot and the square knot are known to have isomorphic $\pi_1$, but they can be detected by quandle coloring.

An application of that is the following: fix a knot $K_0$, and then for a general knot $K$ consider the quandle colorings of $K\#K_0$.

If you take $K_0$ to be the treefoil, and $K$ equal to the same treefoil or it mirror image, then you get the square and the granny knot. So, quandle coloring of $(-)\#K_0$ can detect mirror image. The reference of this idea is "Quandle coloring and cocycle invariants of composite knots and abelian extensions", by W. Edwin Clark, Masahico Saito, and Leandro Vendramin, J Knot Theory Ramif. 2016 Apr; 25(5): 1650024. Also in https://arxiv.org/abs/1407.5803

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