# Are link (non-splittable) quandles complete invariant up to orientation?

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.

Let me review here a definition of quandles and how the proof goes that they are a complete invariant of unoriented knots up to 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 $$\partial\nu(L)$$ (where $$\nu(L)$$ is a tubular neighborhood, and where the homotopies allow the end to slide along $$\partial\nu(L)$$), 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 a quandle (sometimes called 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 $$p\triangleright p=p$$.

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 reversal and 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 $$S^3-\nu (K)\cong S^3-\nu (K')$$. This map might be orientation reversing, and this can affect the orientations of both longitudes and meridians. In any case, the meridians allow us to glue solid tori into the knot complements uniquely and in a way that the homeomorphism extends to all of $$S^3$$, and it follows that either $$K$$ and $$\pm K'$$ or $$K$$ and $$\pm mK'$$ are isotopic since there are only two isotopy classes of homeomorphisms $$S^3\to S^3$$. Q.E.D.

Theorem 2. For a nonsplit oriented link $$L$$, $$Q(L)$$ is a complete invariant of the link, up to total orientation reversal and 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, and again by Waldhausen 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 $$\pm L'$$ or $$L$$ and $$\pm mL'$$ are isotopic. Q.E.D.

• @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? – eyp Jan 21 '19 at 8:36
• @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. – Kyle Miller Jan 21 '19 at 21:40