Satellite knots and JSJ tori — Examples and Counterexamples

Take the core pattern $P=S^1\times \{0\}\subset V$ and choose its tubular neighbourhood $N(P)$ to be a smaller concentric solid torus. The companion embedding sends this core to $K$, so $P(K)=K$ with the smaller chosen tube, while the companion torus is the boundary of the larger tube.

Radial interpolation in the $D^2$ coordinate identifies the shell between the two concentric tubes with $T^2\times [0,1]$. It lies in the satellite exterior and has the companion torus and the satellite exterior boundary as its ends. Thus the companion torus is boundary-parallel and is not essential by [F2]. This refutes the unrestricted statement.

The circle action $t\cdot (z,w)=(e^{4\pi it}z,e^{2\pi it}w)$ on $S^1\times D^2$ has generic orbit slope $2{\lambda }_V+{\mu }_V$. The given pattern is one such orbit. The core $S^1\times \{0\}$ has stabilizer of order $2$, while generic points have trivial stabilizer. Removing a tube about the regular orbit $P_{2,1}$ therefore gives a Seifert fibration of $X_{P_{2,1}}$ over an annulus with one exceptional point of multiplicity $2$. On the outer torus its regular-fiber slope is $2{\lambda }_V+{\mu }_V$.

Remove a small neighborhood of the exceptional point from the annulus base. The remaining pair of pants has fundamental group freely generated by a loop $a$ around the outer boundary and a loop $c$ around the removed cone point; its third boundary loop is $(ac{)}^{-1}$. The circle bundle over this pair of pants has a section: it retracts to a graph, and a section over that graph is obtained by choosing one point in the fiber at the vertex and extending over each edge. Thus its group has generators $a,c,h$, with $h$ the fiber and $[a,h]=[c,h]=1$. The order-two local model $(z,w)\mapsto (-z,-w)$ around the exceptional core identifies the filled meridian with $c^{2}h$ after replacing $c$ by $c{h}^k$ if necessary: one circuit of the cone-point boundary lifts from $(z,w)$ to $(-z,-w)$, so two circuits equal one inverse fiber turn, and the twisting coefficient is odd. Filling that neighborhood kills this meridian. By [F3],

Its quotient by $⟨h⟩$ is $\mathbb{Z}\ast C_2$, which is nonabelian by reduced normal forms [F4]; hence $X_{P_{2,1}}$ is not $T^2\times I$. Its outer boundary subgroup is generated by $a,h$. The abelianization is free on $[a],[c]$ with $[h]=-2[c]$, so $a^m{h}^n=1$ implies $m=0$ and $n=0$. Thus the outer torus group injects into ${\pi }_1(X_{P_{2,1}})$ and has no compressing disk.

For the trefoil, [F2] gives ${\pi }_1(E_K)=⟨x,y\mid x^2=y^3⟩$ and regular fiber $h_K=x^2=y^3$. Its quotient by $⟨h_K⟩$ is $C_2\ast C_3$, nonabelian by [F4], and $h_K$ is infinite because its abelianization is $6$ times the generator of $H_1(E_K)\cong \mathbb{Z}$. The group maps onto $S_3$ by $x\mapsto (12)$ and $y\mapsto (123)$, so it is nonabelian and $K$ is not the unknot, whose exterior group is $\mathbb{Z}$.

The peripheral formula in [F2] gives $h_K={\lambda }_K+6{\mu }_K$. The companion embedding identifies $({\mu }_V,{\lambda }_V)$ with $({\mu }_K,{\lambda }_K)$. Therefore the two Seifert fibrations on the cut sides induce slopes $2\lambda +\mu$ and $\lambda +6\mu$ on the companion torus. Their determinant in the ordered basis $(\mu ,\lambda )$ is $\det \left(\begin{array}{cc}1& 6\\ 2& 1\end{array}\right)=-11$, so the slopes are not parallel.

By [F7], the companion torus $T_c$ is essential. Hence the whole exterior $M=E_{P_{2,1}(K)}$ is not atoroidal. Both pieces cut along $T_c$ are Seifert-fibered by steps 1.1 and 1.3.

The horizontal case is impossible. The interval-bundle bases inherit smooth transverse charts from the horizontal surface and the fiber charts. For any such compact connected surface with nonempty boundary, [F10] gives a finite triangulation. Root a spanning tree of its triangle-adjacency graph at a boundary edge. Delete each triangle across its exposed parent edge, starting at the root; this collapses the surface to a finite connected graph. Its group is therefore free, with rank equal to the rank of its abelianization, by collapsing a graph spanning tree. For a closed surface, the polygon models of [F5] give abelianization ${\mathbb{Z}}^{2g}$ in the orientable genus $g$ case and ${\mathbb{Z}}^{k-1}\oplus \mathbb{Z}\mathrm{/}2$ in the nonorientable genus $k$ case, by adding the signed edge occurrences in its single face word. Hence a compact surface group with torsion-free rank-two abelianization is either the torus group ${\mathbb{Z}}^2$ or the free group of rank two; with torsion-free rank-one abelianization it is infinite cyclic. The cable group has abelianization ${\mathbb{Z}}^2$, is nonabelian because its quotient is $\mathbb{Z}\ast C_2$, and has the nonidentity central element $h$ since $[h]=-2[c]$ in its abelianization. It is neither ${\mathbb{Z}}^2$ nor the free group of rank two, whose center is trivial by [F11]. The trefoil group has abelianization $\mathbb{Z}$ and is nonabelian by step 1.3, so it too is not a surface group with that abelianization. Thus neither cut piece can be an interval bundle over a surface.

Suppose $M$ had a Seifert fibration. By [A2], countable choice holds, so [F13] makes $M$ irreducible. Then [F6] makes $T_c$ ambient isotopic to a vertical or horizontal torus. Isotopy preserves the homeomorphism types of the cut pieces. In the horizontal case, cutting along $T_c$ gives interval bundles over compact connected surfaces, so each side group is a surface group.

Suppose instead that $T_c$ is vertical after isotopy. Pull the hypothetical fibration back by that ambient isotopy, so the original $T_c$ is vertical and the original cut-piece markings are retained. Restriction gives a Seifert fibration on each cut piece with one common regular-fiber slope on $T_c$. Their boundary groups inject independently of any general knot-boundary incompressibility theorem. For the cable side this was proved in step 1.2. For the trefoil side, the peripheral formula of [F2], with $2(-1)+3(1)=1$, gives $\mu =y^{-1}x$ and $\lambda =h_K{\mu }^{-6}$. Thus $(\mu ,h_K)$ is also a basis for its boundary-torus group. In the quotient $C_2\ast C_3$ the word $y^{-1}x$ is cyclically reduced of length two, so all its nonzero powers are nonidentity by [F4]. If ${\mu }^a{h}_K^b=1$, quotienting forces $a=0$, and the infinite order of $h_K$ from step 1.3 forces $b=0$. This proves injectivity of that boundary group. The two original fiber quotients are $\mathbb{Z}\ast C_2$ and $C_2\ast C_3$, respectively, by steps 1.2 and 1.3; neither is the two-order-two exception. Apply [F9] separately to the restricted hypothetical fibration on each side. Its common boundary-fiber slope must equal both original slopes, contradicting their determinant $-11$ in step 2.1. This rules out the vertical case; together with step 2.3 it proves that $M$ is not Seifert-fibered.

Steps 2.2 and 4.1 verify the nonamalgamation hypotheses of [F8], so the companion torus is a JSJ torus and the component outside the companion solid torus is $E_K$. The smooth ambient uniqueness in [F8] carries the original marked outside component to the outside component for any isotopic minimal JSJ torus.