2
$\begingroup$

I am reading knot quandles. I read that knot quandles are complete knot invariant upto orientation from thesis of David joyce, An Algebraic Approach to Symmetry with Applications to Knot Theory, page no. $53$. In the proof it is used that in case of knot, the knot group act transitively on the knot quandle. While reading the book "Surface-Knots in 4-Space: An Introduction" by Seiichi Kamada, on page no.$152$ it is mentioned in the comment $22$: "A similar result holds when $K$ and $K'$ are non-splittable oriented links.

I am not able to prove it, because in case of links with more that one component, the action of link group on link quandle is not transitive.

Can someone help me? Any suggestions or hints would be helpful.

$\endgroup$
1
  • $\begingroup$ A point of clarification: "up to orientation" means "up to orientation-reversed mirror image." The fundamental quandle can depend on the orientation of an oriented knot/link. $\endgroup$ May 25, 2020 at 20:06

1 Answer 1

2
$\begingroup$

Edit: Many thanks to Lorenzo Traldi for helping me understand the correct statement for the way in which the fundamental quandle is a complete invariant.

Let me review here a definition of quandles and how the proof goes that they are a complete invariant of oriented knots up to orientation-reversed mirror image.

The link quandle $Q(L,*)$ of an oriented link $L\subset S^3$ with basepoint $*\in S^3-\nu(L)$ is the collection of all homotopy classes of paths from $*$ to $\partial(S^3-\nu(L))$ (where $\nu(L)$ is a tubular neighborhood of $L$, and where the homotopies allow the end to slide along $\partial(S^3-\nu(L))$), along with

  1. A function $\lambda:Q(L,*)\to \pi_1(S^3-\nu(L),*)$ given by $p\mapsto p\; \overline{\mu_{p(1)}}\; \overline{p}$, where $\mu_{p(1)}$ is some meridian loop in $\nu(L)$ at $p(1)$ having $+1$ linking number with $L$. (Note: this requires knowing $L\subset S^3$ and cannot in general be recovered from $S^3-\nu (L)$). This is the loop corresponding to a path, hence $\lambda$.

  2. A group action of $\pi_1(S^3-\nu(L),*)$ on $Q(L,*)$ by $g\cdot p=gp$ (concatenation).

These form an enriched quandle in the following sense:

  1. $g\lambda_p g^{-1}=\lambda_{gp}$ for all $g\in \pi_1(S^3-\nu(L))$ and $p\in Q(L)$.
  2. $\lambda_pp=p$ for all $p\in Q(L)$.

If you substitute in $g=\lambda_q$, you can get $\lambda_q\lambda_pr=\lambda_{\lambda_q p}\lambda_q r$, which with $y\triangleright x:=\lambda_xy$ renders as $(r\triangleright p)\triangleright q=(r\triangleright q)\triangleright(p\triangleright q)$. The second renders as $p\triangleright p=p$.

The non-enriched version of the quandle is from forgetting $\pi_1(S^3-L)$ and instead considering the map as $\lambda:Q(L,*)\to\operatorname{Aut}(Q(L,*))$, which sends each element of $Q(L,*)$ to its action on $Q(L,*)$. (Quandles are also known as automorphic sets.) Up to an automorphism of a peripheral group system for $\pi_1(S^3-L)$, the enriched quandle structure can be recovered from the quandle.

A note about $\lambda$: the image $\lambda(Q(L))$ is a union of conjugacy classes of $\pi_1(S^3-\nu(L))$, due to $g\lambda_p g^{-1}=\lambda_{gp}$. In particular, through the construction each conjugacy class corresponds to the free homotopy class of some meridian --- one class per link component --- so the quandle is certainly not transitive for links of more than one component. It's worth knowing that $\lambda(Q(L))$ generates $\pi_1(S^3-\nu(L))$, with a quick reason being from the Wirtinger presentation.

Another note about it: $\lambda$ is functorial in the sense that if there is a quandle homomorphism $f:Q(L)\to Q(L')$, then there is an induced homomorphism $f_*:\pi_1(S^3-\nu(L))\to \pi_1(S^3-\nu(L'))$ compatible with $\lambda$. (In a way, this induced homomorphism is part of the definition of the (enriched) quandle homomorphism, so this is just putting different terminology on the concept.)

Theorem 1. For an oriented knot $K$, $Q(K)$ is a complete invariant of the knot, up to orientation-reversed mirror image.

Proof. Fix an element $p\in Q(K)$, which represents a particular choice of meridian for $K$. Let $G=\pi_1(S^3-\nu(K))$ and $H=\operatorname{Stab}_G(p)$.

  • For $g\in \pi_1(\partial\nu(K),p(1))$, $(pg\overline{p})p\sim p$ in $Q(K)$, hence $p\pi_1(\partial\nu(K),p(1))\overline{p}\subseteq H$.

  • For $g\in H$, since $gp\sim p$, we can take the edge of this homotopy to get a path $h$ from $gp(1)$ to $p(1)$ in $\partial\nu(K)$. We can write $g=p\overline{h}\overline{p}$, which is in $p\pi_1(\partial\nu(K),p(1))\overline{p}$.

Thus, $H=p\pi_1(\partial\nu(K),p(1))\overline{p}$. That is, we can recover a peripheral subgroup and a meridian. Suppose there is an oriented knot $K'$ with $Q(K)\cong Q(K')$, and let $p'$ be the image of $p$ in $Q(K')$. By functoriality, there is an isomorphism $G\cong G'=\pi_1(S^3-\nu K')$, and it carries $H$ to $H'=\operatorname{Stab}_{G'}(p')$, which by the above argument is a peripheral subgroup. Since knot complements are Haken manifolds with connected boundaries, Waldhausen '68 applies and there is a homeomorphism $S^3-\nu (K)\cong S^3-\nu (K')$ carrying oriented meridian to oriented meridian. The meridians allow us to glue solid tori into the knot complements in a way that the homeomorphism extends to all of $S^3$. If the homeomomorphism is orientation-preserving, then $K$ and $K'$ are isotopic. Otherwise, if it is orientation-reversing, then $K$ and the mirror image $mK'$ as unoriented knots are isotopic. The orientation of a meridian determines the orientation of a knot, so since the mirror image reverses the orientation of a meridian, it is that $K$ and $-mK'$ are isotopic as oriented knots. Q.E.D.

Theorem 2. For a nonsplit oriented link $L$, $Q(L)$ is a complete invariant of the link, up to orientation-reversed mirror image.

Proof. Let $L$ be a nonsplit link having components $L_1,\dots,L_k$. Define an equivalence relation on $Q(L)$ generated by $p_1\sim p_2$ if there exists a $q\in Q(L)$ such that $p_1=\lambda_qp_2$. Equivalence classes are in one-to-one correspondence with components of $L$, so choose a representative element $p_i\in Q(L)$ for each component, with $p_i(1)\in\partial\nu(L_i)$ (though we might not know which corresponds to which component). With $H_i=\operatorname{Stab}_{G}(p_i)$, as before we get peripheral subgroups $H_i=p_i\pi_1(\partial\nu(L_i),p_i(1))\overline{p_i}$. Since $L$ is nonsplit, then $S^3-\nu(L)$ is a Haken manifold, so by Lemmas 13.8 and 13.7 in Hempel we get from the isomorphism $G\cong G'$ that carries each $H_i$ to $H_i'$ a homeomorphism $S^3-\nu(L)\cong S^3-\nu(L')$ when $L'$ is another nonsplit link with $Q(L)\cong Q(L')$. By a similar argument as before using the meridians implied by the $p_i$'s, either $L$ and $L'$ or $L$ and $-mL'$ are isotopic. Q.E.D.

This result is a consequence of Theorem 5.2 and Corollary 5.3 of

Fenn, Roger; Rourke, Colin, Racks and links in codimension two, J. Knot Theory Ramifications 1, No. 4, 343-406 (1992). ZBL0787.57003.

For them, a link $L$ in $S^3$ is an oriented link, and they view the orientation in terms of consistently orienting the transverse disks of a tubular neighborhood for $L$ (the link is transversely oriented). Stated for quandles, they prove that if $Q(L)$ and $Q(L')$ are isomorphic for non-split links $L$ and $L'$ in $S^3$, then there is a homeomorphism $h : S^3 \to S^3$ carrying $L$ to $L'$ and transverse orientations to transverse orientations (equivalently, oriented meridians to oriented meridians). If $h$ is orientation-preserving, then $L \sim L'$, and if $h$ is orientation-reversing, $L \sim -mL'$.

One can show that $Q(L)\cong Q(-mL)$ and $Q(mL)\cong Q(-L)$, so this is the strongest possible statement.

$\endgroup$
10
  • $\begingroup$ @Miller: So the proof of Theorem $2$ implies that a non-splittable link is completely classified by its link group and peripheral systems. Am I right? $\endgroup$
    – eyp
    Jan 21, 2019 at 8:36
  • 1
    $\begingroup$ @eyp Yes, up to mirror image. See chapter 13 of Hempel's 3-Manifolds ($\pi_2(S^3-L)=0$ for non-splittable links, so all homomorphisms are induced by continuous maps, and 13.8 says the map can be boundary preserving. 13.7 says this map is homotopic to a homeomorphism, where the special case for link complements is just a Hopf link, I believe.) With split links, there are homomorphisms corresponding to mirror imaging only part of the link. $\endgroup$ Jan 21, 2019 at 21:40
  • 1
    $\begingroup$ @eyp A correction: the link group, peripheral subgroups, and oriented meridians together completely classify a non-split link up to orientation-reversed mirror image. If you don't know the orientations of the meridians, then you only know it up to orientation-reversal and mirror image. The quandle provides orientations of the meridians. (I've corrected the answer.) $\endgroup$ May 25, 2020 at 19:57
  • 1
    $\begingroup$ @eyp For the first question, you can think of it like a stronger version of Eilenberg-MacLane spaces; the group homomorphism already maps the peripheral subgroup and the oriented meridian to the corresponding peripheral subgroup and oriented meridian since the homomorphism is induced from the quandle homomorphism. Hempel lemma 13.8 then says this peripheral group system is induced by a continuous map. $\endgroup$ May 26, 2020 at 20:34
  • 1
    $\begingroup$ @ChoMedit It's a peripheral subgroup for the knot, with $p$ giving a path from the basepoint to the basepoint on the boundary (this is what the first line after the bullets in the proof of Theorem 1 is indicating). $\endgroup$ Aug 13, 2022 at 17:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .