Satellite knots and trivializing bands

The winding number of $K^{\prime}$ is $V$ is $\pm 1$ or $\pm 3$, so the wrapping number is 1 or 3. Suppose that it is 1. Then there is a meridian disk $D$ of $V$ which intersects $K^{\prime}$ in one point. We can assume that $\partial D$ is disjoint from $\partial D_{i}$ for all $i$, hence $D\cap D_{i}$ consists only of simple closed curves. Let $\gamma$ be an innermost such curve in $D$. Then $\gamma$ bounds a disk $D^{\prime}$ in $D$ with interior disjoint from the $D_{i}^{\prime}s$, and it is either disjoint from $K^{\prime}$ or intersects it in one point. The disk $D^{\prime}$ must be in a region $H_{i}$. Then in fact $D^{\prime}$ is isotopic to a disk in $D_{i}$, and after an isotopy the intersection is removed. We continue until there is no intersection left between $D$ and the $D_{i}^{\prime}s$. Then $D$ would be contained in a region $H_{i}$, which clearly is not possible. ∎

The proof is similar to that of Lemma 2.1, by looking at the intersections between a sphere with the disks $D_{i}$. ∎

Suppose that the 3-tangle $(H_{i},t_{i})$ is not a braid, for in this case the conclusion is obvious. Let $E^{\prime}$ be a disk in $H_{i}$, such that $\partial E^{\prime}=\alpha\cup\beta$, where $\alpha$ is an arc in $D_{i}$ and $\beta$ is an arc in $D_{i1}$. Suppose that the interior of $E^{\prime}$ is disjoint from $D_{i1}$, and that $E^{\prime}$ is disjoint from $K^{\prime}$ or intersects it transversely in one point. The arc $\beta$ in $D_{i1}$ cuts off a disk $D^{\prime}$, which is disjoint from $K^{\prime}$ or intersects it in one point. If $E^{\prime}$ is disjoint from $K^{\prime}$ and $D^{\prime}$ intersects $K^{\prime}$ in one point, then the disk $E^{\prime}$ would separate the strings of the tangle $(B,t)$ determined by $D_{i1}$, which is not possible. In all other cases, $E^{\prime}$ and $D^{\prime}$ are parallel, that is, cobound with part of $D_{i}$ a 3-ball disjoint from $K^{\prime}$, or intersecting it in an unknotted spanning arc.

Suppose $E$ is a disk as in the statement of the lemma, whose boundary lies on $D_{i}$. Look at the intersections between $D_{i1}$ and $E$. Let $\gamma$ be a simple closed curve of intersection which is innermost in $E$. If $\gamma$ bounds a disk disjoint from $K^{\prime}$ or intersecting it in one point, then this is easily removed, for there are no local knots in $K^{\prime}$. Now suppose that $\gamma$ is an outermost arc of intersection in $E$, bounding a disk $E^{\prime}$. Then, by the previous paragraph, this disk would be parallel to a disk in $D_{i1}$, and by an isotopy the intersection could be removed. The only remaining case is that $\gamma$ bounds a disk intersecting $K^{\prime}$ twice, and that there are no arcs of intersection. In that case $\partial E$ must be parallel to $\partial D_{i1}$, and then $E$ is parallel to a disk in $D_{i}$, or it is parallel to $D_{i1}$. ∎

Again, look at the intersections between $E$ and $D_{i1}$, $D_{i2}$ and $A_{i}$. ∎

Suppose that $S$ is a Conway sphere for $K^{\prime}$ in $V$, that is, $S$ is a sphere in $V$ that intersects $K^{\prime}$ transversely in four points, and $S-K^{\prime}$ is incompressible in $V-K^{\prime}$. Look at the intersections between $S$ and the disks $D_{i}$. If this intersection is empty, then $S$ would be contained in some $H_{i}$, but this is not possible for $(H_{i},t_{i})$ is a trivial tangle. Let $\gamma$ be a simple closed curve of intersection which is innermost in $S$. If $\gamma$ bounds a disk disjoint from $K^{\prime}$ or intersecting it in one point, then this is easily removed. So $\gamma$ bounds a disk intersecting $K^{\prime}$ twice. It follows that the intersection consists of concentric curves, with two of them, say $\gamma_{1}$ and $\gamma_{2}$, bounding disks $E_{1}$ and $E_{2}$ which intersect $K^{\prime}$ in two points. Between $E_{1}$ and $E_{2}$ there is a collection of annuli $F_{1},\dots,F_{r}$, which may be empty. If one of $E_{i}$ or $F_{j}$ is parallel to a disk or annulus in some $D_{i}$, then the number of curves of intersection could be reduced. Then, say, $E_{1}$ is in some $H_{i}$ and by Lemma 2.3 is parallel to the disk $D_{i2}$, and similarly $E_{2}$ is in some $H_{j}$ and is parallel to the disk $D_{jt}$. The annuli $F_{j}$ are then spanning annuli in some $H_{i}$. Note that $\partial D_{i2}$ and $\partial D_{(i+1)1}$ have non-empty intersection, and also $\partial D_{i2}$ and $\partial A_{i+1}$ have non-empty intersection, except perhaps if $H_{i+1}$ is a braid. By inspection it follows that is not possible to assemble the pieces to get the desired sphere.∎

Let $T$ be an incompressible torus in $V-K^{\prime}$. Look at the intersection between $T$ and the disks $D_{i}$. This intersection consists of curves that divide $T$ into annuli, which we can suppose are not annuli parallel to an annulus lying in some $D_{i}$, that is, all are spanning annuli in the $H_{i}$. If in some $H_{i}$ there is an annulus which is parallel to an annulus $A_{i}$, then it can be seen that the torus cannot be assembled. Then each such annuli is either an annulus running along an arc of $t_{i}$, or is parallel to $\partial V$. We conclude that the torus is parallel to $\partial V$, or parallel to $\partial N(K^{\prime})$. ∎

Let $A$ be an incompressible annulus in $V-K^{\prime}$. Look at the intersection between $A$ and the disks $D_{i}$. This intersection consists of arcs, which we can assume divide the annulus into rectangles. That is, the intersection of $A$ with each $H_{i}$ consists of one of more rectangles. But these rectangles cannot be assembled to form an annulus, except perhaps if each tangle $H_{i}$ is a braid, that is, if $K^{\prime}$ is a closed braid in $V$ and that $K^{\prime}$ is parallel to a curve on $\partial V$. $K^{\prime}$ is a closed braid only when $m=\pm 1$, $n=\pm 1$ and $p=0$, or when $m=\pm 1$, and $p=1,q=3,n=1$, or when $m=\pm 1$, and $p=-1,q=-3,n=-1$. This gives a total of 8 closed braids, and it can be checked that none of them is parallel to a curve contained in $\partial V$. If there is a Möbius band disjoint from $K^{\prime}$, then there is also an incompressible torus disjoint from $K^{\prime}$, which is not possible.∎

To see this in an easy way, note that there is a Möbius band $\mathcal{M}$ properly embedded in the exterior of $J$ with slope $-2q$. Let $A$ be an annulus embedded in $V$ whose boundary consists of preferred longitudes of $V$, such that a diagram of $K^{\prime}$ can be drawn on $A$. Consider now $h(A)$ and note that one of its boundary components coincide with $\partial\mathcal{M}$. Then $\mathcal{M}^{\prime}=\mathcal{M}\cup h(A)$ is a Möbius band which contains a diagram of $K(m,n,p;q)$. Consider the band $b$ shown in Figure 3 and note that it is contained in $\mathcal{M}^{\prime}$, and then we have a diagram of $K_{b}$ on $\mathcal{M}^{\prime}$. Following Figure 4, do isotopies of $K_{b}$ along $\mathcal{M}^{\prime}$ to undo the crossing in the boxes $m$, $-m$, then undo the crossings in the boxes $n$ and $-n$, until we get to the penultimate diagram of Figure 4. There the crossings in the box labelled with $q$ cancel with the crossings introduced by the Möbius band, and then just undo the crossings in the box labeled by $p$. ∎

Suppose that $b$ is a band for $K=K(m,n,p;q)$ which produces by banding a trivial knot $U$ and that $b$ is disjoint from $Q$. Give an orientation to $K$. Remember that the wrapping number of $K$ in the solid torus $h(V)$ is $3$. Then $b$ intersects a meridian disk of $h(V)$ in a collection of $r$ arcs. So $U$ intersects a meridian disk of $h(V)$ in $3+2r$ points. Then the winding number of $U$ in $h(V)$ is an odd number. and then $U$ cannot be inside a $3$-ball contained in $h(V)$. It follows that $Q$ is incompressible in the exterior of $U$, then $U$ is not trivial. ∎

Suppose that $P$ is a Conway sphere for $K$. Consider first the intersections between $P$ and $Q$. We can assume that the intersection consists of curves that are essential on both $P$ and $Q$. Let $\gamma$ be an innermost curve of intersection in $P$. Then $\gamma$ bounds a disk $D$ in $P$ containing one or two points of intersection with $K$. Because the wrapping number of $K$ in $h(V)$ is $3$, this is not possible. Then $P$ is disjoint from $Q$, and it must be contained in $h(V)$. By Lemma 2.5 this is not possible.

Suppose now that $P$ is an essential torus not isotopic to $Q$. Again look at the intersection between $P$ and $Q$, which then consists of curves which are essential in both $P$ and $Q$. Let $\gamma_{1}$ and $\gamma_{2}$ be a pair of intersection curves that bound an annulus $A$ in $P$ with interior disjoint from $Q$ and disjoint from $N(J)$. If $A$ is a boundary parallel annulus in the exterior of $J$, $E(J)$, an isotopy reduces the number of arcs of intersection. If $A$ is essential in $E(J)$, then its boundary consists of curves of slope $-2q$ on $Q$. Then the annulus $A^{\prime}$ in $P$ adjacent to $A$ is boundary parallel in $h(V)$. By an isotopy the number of arcs of intersection are reduced. Then $P$ is disjoint from $Q$. As $J$ is a torus knot, there are not essential tori in its complement. Then $P$ is contained in $V$. By Lemma 2.6 this is not possible. ∎

The manifold $\tilde{M}$ is the double cyclic cover of the exterior of the $(2,-q)$ torus knot $J$, and then by [11] is the Seifert fibered manifold $(D^{2};\frac{(q-1)/2}{q},\frac{(q-1)/2}{q})$.

If $\tilde{V}$ is not an hyperbolic manifold, then by Thurston Hyperbolicity Theorem, $\tilde{V}$ will have a compressing disk, a reducing sphere, an essential annulus or an essential torus. If there is any of these surfaces, by one of the several equivariant results [9], [10], [8], there is a surface $S$ which is equivariant, that is, $\tau(S)=S$, or $\tau(S)\cap S=\emptyset$, where $\tau$ is the involution defined on $\tilde{V}$ by the double branched cover. By taking the projection $\tilde{\pi}(S)$, there will be a compression disk for $\partial V$ disjoint from $K^{\prime}$, a meridian disk intersecting $K^{\prime}$ in one point, a decomposing sphere for $K^{\prime}$, a meridian disk intersecting $K^{\prime}$ in two points, an essential annulus or Möbius band, or an essential torus or Conway sphere for $K^{\prime}$. None of these can exist by Lemmas 2.1, 2.2, 2.5, 2.6 or 2.7,. ∎

First note that $\tilde{K}$ cannot be a torus knot, for it has a Dehn surgery producing a 3-manifold containing a separating incompressible torus. Suppose $\tilde{K}$ is a satellite knot, and let $R$ be an incompressible torus in its exterior. As $\tilde{K}$ is strongly invertible, by the Equivariant Torus Theorem [8], there is an incompressible torus $\tilde{R}$, which is equivariant under the involution defined on $\tilde{K}$. $\tilde{R}$ bounds a solid torus $W$ which contains $\tilde{K}$. The complement of $W$ may contain incompressible tori other than $\tilde{R}$, but applying again the Equivariant Torus Theorem, there will be another equivariant torus in the complement of $W$. Therefore we can assume that there is an equivariant torus $R$, which defines then a companion knot $\overline{K}$ for $\tilde{K}$ which is hyperbolic or a torus knot.

Suppose first that $R$ compresses after performing $r$-Dehn surgery. Then by [5], $\tilde{K}$ is a 0 or 1-bridge braid in $W$, with winding number $w\geq 2$. It follows from [6] that the manifold $\tilde{K}(r)$ is obtained by $n^{2}$-Dehn surgery on $\overline{K}$, for some integer $n$. This is not possible if $\overline{K}$ is a torus knot, for the surgery produces a 3-manifold containing a separating incompressible torus. Also, this is not possible if $\overline{K}$ is an hyperbolic knot, because by [7], we should have $w^{2}\leq 2$.

Suppose now that $R$ is incompressible in $\tilde{K}(r)$. As $R$ is equivariant, $\pi(R)$, is a torus or a Conway sphere. But, by Lemma 2.10 there are no Conway spheres for $K$. If $\pi(R)$ is a satellite torus for $K$, it would have to be isotopic to $Q$ by Lemma 2.10. But, as $\pi(R)$ is disjoint from the band, this is not possible by Proposition 2.9. ∎