Alternating links on surfaces and volume bounds

Part (1) is a special case of Proposition 4.10 of [31]. We note that if $S^{3}-K$ contains an essential torus, then $F-\pi(K)$ must contain some regions that are not discs. Part (2) is a special case of Theorem 4.6 of [31]. ∎

The best way to see both results is by inspection. The two diagrams are explicit. It is a somewhat soothing exercise to sketch the diagrams and colour the checkerboard surfaces by hand.

Alternatively, $\pi(\hat{K})$ can be projected onto the usual plane of projection in $S^{3}$, giving an ordinary alternating link, which must be checkerboard coloured. Note that in this ordinary alternating link, the region on the outside has the same colouring as the two regions inside the loops that ran around the handles before the projection. Therefore the colouring transfers to the genus-2 surface, giving a consistent colouring on the single 3-punctured sphere region under the plane of projection. Adding four parallel curves to this, as in $\pi(\hat{K}^{*})$, does not affect most faces. One can check, using the fact that there are exactly four new strands, that the faces that are changed can have colouring modified to give a unified checkerboard colouring of the diagram.

As for twist reduced, suppose there is a disc $D$ in either $\pi(\hat{K})$ or $\pi(\hat{K}^{*})$ with $\partial D$ meeting the diagram exactly four times, adajcent to exactly two crossings, as in Figure 1, left. Isotope $\partial D$ to run through the adjacent crossings. This splits $\partial D$ into two arcs between crossings, with each such arc lying in a single region of the diagram. Moreover, these two regions have the same colour. Thus such a disc will correspond to two regions of the same colour meeting at two distinct crossings. By inspection, most regions of the diagram are quads with four distinct regions meeting the four vertices (crossings) of the quad, so such a disc cannot run there. In the first template there is the 3-holed sphere region, but all similar coloured regions adjacent to the 3-holed sphere across crossings are distinct. Similarly in the second template there is another large disc region, but it meets only distinct regions at each of its crossings. Finally, in both templates there are exactly two bigon faces, and thus a disc meeting exactly two crossings encircling the bigon. This satisfies the definition of twist reduced. ∎

The diagrams of $\pi(\hat{K})$ and $\pi(\hat{K}^{*})$ are checkerboard coloured and twist reduced by Lemma 3.1. When adding bigons to these diagrams to form $\pi(K)$ and $\pi(K^{*})$, the diagram remains checkerboard coloured and twist reduced, because new bigons are added only to existing twist regions.

Showing weakly prime requires a little more work. Consider the three closed curves of intersection of $F$ with the usual plane of projection in $S^{3}$; call them $x$, $y$, $z$. These divide $F$ into two pairs of pants, or 3-punctured spheres, one above the plane of projection and one below. The entire knot $K$ lies above the plane of projection, and is disjoint from the 3-punctured sphere below the plane of projection. Most of the knot $K^{*}$ lies above the plane of projection, but eight unknotted arcs of the diagram lie below.

Now, suppose there is an embedded disc $D$ on $F$ whose boundary $\partial D$ intersects the diagram $\pi(K)$ or $\pi(K^{*})$ in exactly two points. The boundary $\partial D$ must intersect each of the curves $x$, $y$, and $z$ in an even number of intersection points. If it intersects any of $x$, $y$, $z$ in more than zero points, then there exists an outermost arc of $\partial D$ co-bounding a disc with an arc of $x$, $y$, or $z$, forming a bigon $B$.

Suppose first that the boundary of the bigon $\partial B$ is disjoint from the knot diagram. Then we may use $B$ to isotope $D$ through $x$, $y$, or $z$, to reduce the number of intersections with $x$, $y$, and $z$. Repeat this a finite number of times to remove all bigons that are disjoint from the diagram.

Next suppose that the boundary of $B$ intersects the diagram. Again $B$ is a disc, so $\partial B$ intersects the knot diagram twice. Note that $B$ is embedded either above or below the plane of projection. If below, there are no crossings that $B$ could enclose, hence it bounds a single unknotted arc. If above, $B$ is a disc in a 3-punctured sphere containing a graph whose only vertices come from bigons replacing vertices in a grid of 72 or 120 points, with edges running from the outside of this grid. A grid is well known to form a prime diagram, with primeness unchanged by replacing a vertex with a twist region, so $B$ bounds a single unknotted arc of the diagram.

The boundary $\partial B$ is made up of two arcs, one on $\partial D$ and one on $x$, $y$, or $z$. The diagram either intersects $\partial D$ twice, or $\partial D$ once and $x$, $y$, or $z$ once (if the diagram is that of $K^{*}$). In the first case, the diagram intersects $\partial D$ twice in $\partial B$; these must be the only intersections of $\partial D$ with the diagram. Hence since $B$ bounds an unknotted arc, so does $D$. In the second case, use $B$ to isotope $D$ along the unknotted arc of the diagram, pushing it past $x$, $y$, or $z$, and removing two intersections with these curves. After a finite number of steps, there are no bigons $B$.

Then $D$ lies either completely above or completely below the plane of projection. As before, it follows that it meets the diagram in a single unknotted arc. Thus the diagram is weakly prime. ∎

The nontrivial curve $x$ on $F$ intersects $\pi(K^{*})$ four times, so edge-representativity is at most four. To show it is equal to four, we need to show that any other essential curve on $F$ intersects $\pi(K^{*})$ at least four times.

Let $\ell\subset F$ be an essential curve. Consider how $\ell$ intersects the 3-holed spheres obtained by cutting $F$ along the curves $u$, $v$, and $w$ of Figure 4. If $\ell$ is completely contained in one of the two 3-holed spheres, then the fact that $\ell$ is essential on $F$ means it is parallel to one of the three boundary components. But in either 3-holed sphere, the graph of the template includes six strands of the diagram running between each pair of boundary components, separated by a large grid. A closed curve must intersect this graph at least six times; it follows that $\ell$ intersects $\pi(K^{*})$ at least six times.

Next suppose $\ell$ intersects both 3-holed spheres. Then it must meet each in a collection of arcs. If any of these arcs co-bounds a bigon with one of $u$, $v$, $w$, then $\ell$ can be simplified by isotoping through the bigon. Such an isotopy will at worst, only decrease the number of intersections of $\pi(K^{*})$ with $\ell$, so we assume now that there are no bigons.

Because $\ell$ must alternate running between the two 3-holed spheres, it must have an essential arc on the left side. This is an essential arc on the 3-holed sphere, running from one boundary component to another, possibly the same component. If it runs from one boundary component to a distinct one, it is a seam. Otherwise it is a wave. Each seam must run through the four parallel strands appearing in $K^{*}$ and not $K$. Each wave either runs through these strands, or through the six parallel strands or the grid portion of the graph appearing in both $K$ and $K^{*}$. Thus in all cases, the arc meets $\pi(K^{*})$ at least four times. It follows that the edge-representativity is four. ∎

Twisting along any curve $u$, $v$ or $w$ restricts to a power of a Dehn twist on $F$; such a homeomorphism is the identity away from an annulus neighbourhood of the curve. That annulus intersects the knot diagram in six essential arcs running from one boundary component to the other. The arcs subdivide the annulus into six discs that are coloured in a checkerboard fashion. After twisting, the arcs run into the annulus, around its core some number of times, and then out, but no new crossings are added. Because the knot was alternating on $F$ before twisting, the new knot remains alternating on $F$. The checkerboard coloured discs between arcs of the knot diagram are mapped to new discs, with checkerboard colouring preserved under twisting.

To see that the diagram is still weakly prime, suppose a disc $D$ on $F$ meets the diagram $\pi(K_{T})$ or $\pi(K_{T}^{*})$ twice transversally in edges. Apply the inverse Dehn twist to the picture. This preserves $F$ and takes diagram to $\pi(K)$ or $\pi(K^{*})$, respectively, and takes $D$ to a disc now meeting $\pi(K)$ or $\pi(K^{*})$ twice. Because $\pi(K)$ and $\pi(K^{*})$ are weakly prime by Lemma 3.2, the disc must contain no crossings, both before and after homeomorphism, and $\pi(K_{T})$ and $\pi(K_{T}^{*})$ are weakly prime. A similar argument shows the diagram is still twist reduced after twisting.

As for edge-representativity, recall that by Lemma 3.3 any essential curve on $F$ meets $\pi(K^{*})$ at least four times. Since surface homeomorphisms map essential curves to essential curves and preserve intersection numbers of curves, it follows that any essential curve on $F$ meets $\pi(K_{T}^{*})$ at least four times. Thus the edge-representativity of $\pi(K_{T}^{*})$ is four. ∎

The surface $F$ divides $S^{3}$ into two handlebodies. Denote the bounded handlebody by $H_{1}$, and the unbounded one (i.e. that contains infinity) by $H_{2}$. We must argue that every curve $C$ that bounds a compressing disc in either $H_{1}$ or in $H_{2}$ will intersect the diagram in at least six points. Observe first that $u$, $v$, and $w$ bound meridian disc of $H_{1}$ meeting the diagram at least six times. We need to show a similar result for any choice of a curve $C$ bounding a compressing disc $E$.

Suppose first that $C$ bounds a compressing disc for $H_{1}$. Note that the curves $u,v,w$ cut the surface $F$ into two pairs of pants, $P_{1}$, $P_{2}$, and the discs bounded by $u,v,w$ cut $H_{1}$ into 3-balls. This is shown on the right of Figure 5. Isotope $C$ so that its intersection with $u\cup v\cup w$ is minimal on $F$. By further isotopy we can arrange so that the disc $E$ bounded by $C$ intersects those bounded by $u,v,w$ in arcs (not simple closed curves). Now an arc of this intersection that is innermost on $E$ will define a wave on one of $P_{1},P_{2}$. That is, it forms an essential arc on $P_{i}$ with both of its boundary points on the same component of $\partial P_{i}$. Such a wave will separate the two boundary components of $P_{i}$ that is disjoint from. Thus it either runs through at least six parallel strands of the diagram, or it intersects the diagram in the alternating tangle. Within the alternating tangle, the number of intersection points must be at least six. Hence in either case, the wave will intersect the diagram in at least six points. This finishes the claim for $H_{1}$.

The proof for $H_{2}$ is similar, only now cut $H_{2}$ into balls bounded by pants along the plane of projection. This cuts $F$ into two pants, one above and one below the plane of projection, as shown in Figure 6.

Denote the boundary curves of the pants by $x$, $y$, and $z$. A compressing disc $E$ for $H_{2}$ can be isotoped to intersect $x$, $y$, and $z$ minimally, and to intersect the discs bounded by these curves only in arcs. Then an outermost arc of $E$ has boundary forming a wave. This wave separates boundary components, so again it either runs through the alternating tangle, meeting the knot diagram at least six times, or runs through a region with a high number of parallel strands, again meeting the knot diagram at least six times. ∎

Consider the original template. Each curve $C_{i}$ bounds a disc in $S^{3}$ that is disjoint from all the curves of the original template, except meets each of the six parallel strands running over the handles of $H_{1}$ exactly twice. Orient these strands in the same direction, as in the construction of $K$ and $K^{*}$. Note each strand runs through the disc in one direction, then runs through again in the opposite direction, so each such strand has linking number zero with $C_{i}$. When we modified the template above to create the knot, in Subsection 3.1, we first inserted crossings at vertices. This did not affect orientation of strands through the $C_{i}$. We then added bigons to crossings on $F$, chosen in such a way that the orientation of the six strands around handles is not affected, as in Figure 3. Finally we twisted along $u$, $v$, and $w$. Again this had no effect on orientation of strands. Thus the linking number of $K_{T}$ or $K_{T}^{*}$ with any $C_{i}$ is zero. ∎

We have shown that $\pi(K_{T})$ and $\pi(K_{T}^{*})$ are alternating on $F$, checkerboard coloured, weakly prime, and each component projects with at least one crossing by Lemma 3.4. The representativity of each is at least six by Lemma 3.5. Thus $\pi(K_{T})$ and $\pi(K_{T}^{*})$ are weakly generalized alternating. They are twist reduced with edge-representativity at least four by Lemma 3.4. We now show these properties for $\pi(K_{n})$ and $\pi(K_{n}^{*})$.

By choice, $H_{1}$ is constructed as a 3-ball $B$ with with two unknotted and unlinked 2-handles. The co-cores of these 2-handles are meridians of $H_{1}$. The cores can be joined to a wedge of two circles on which $H_{1}$ deformation retracts. Twisting along the curves $C_{i}$ leaves the co-cores of the 2-handles of $H_{1}$ fixed and inserts a 4-string pure braid to the wedge of two circles on which $H_{1}$ retracts. The complement $H_{2}$ is still a handlebody after twisting along the curves $C_{i}$. Indeed it is a complement of a 2-bridge link with an unknotting tunnel drilled out; see for example [34]. Thus after the twisting operation we still get a Heegaard splitting of $S^{3}$. We can isotope this Heegaard splitting to the original one by sliding around the discs on $B$, on which the 2-handles of $H_{1}$ are attached, until the 4-string braid on the spine is trivialized. This implies that the effect of twisting along the curves $C_{i}$ is also achieved by a homeomorphism of $S^{3}$ that preserves $H_{1}$ and $H_{2}$. The restriction of the homeomorphism on $F$ fixes a set of meridian curves of $H_{1}$ and restricts to a pure mapping class of the 4-holed sphere.

Since the property of bounding a meridian disc in $H_{1}$ (resp. $H_{2}$) is preserved under homeomorphisms of $H_{1}$ (resp. $H_{2}$) and intersection numbers of curves are preserved under homeomorphisms of $F$, it follows that $\pi(K_{n})$ and $\pi(K_{n}^{*})$ have representativity at least six. Similarly, $\pi(K_{n})$ and $\pi(K_{n}^{*})$ remain alternating on $F$ after twisting along $C_{i}$, with the same number of crossings on $F$ as before, and the same twist number, and they remain checkerboard colored, twist reduced, and weakly prime on $F$.

Thus the representativity of $\pi(K_{n})$ is identical to the representativity of $\pi(K)$, hence it is at least six. Similarly for $\pi(K_{n}^{*})$. Similarly, edge-representativity is identical in both cases, hence it is at least four. ∎

For $K_{T}^{*}$ and $K_{n}^{*}$, the result follows immediately from Theorem 2.6: These knots are WGA in $S^{3}$, all regions on $F-\pi(K_{T}^{*})$ and $F-\pi(K_{n}^{*})$ are discs, and the representativity is at least six, by Lemma 3.7. Thus by that theorem, these knots are hyperbolic.

The result is not quite as straightforward for $K_{T}$ and $K_{n}$, because these knots have one region $E$ that is a 3-holed sphere, not a disc. For these knots, we show their complements are irreducible, boundary irreducible, anannular and atoroidal. For notational simplicity we will use $S^{3}-J$ to denote any of these manifolds.

We recall that  [31, Corollary 3.16] states that if $\pi(L)$ is a reduced alternating link projection on a surface $F$ in a 3-manifold $Y$, them $Y-L$ is is irreducible and boundary irreducible. Hence the claim that $S^{3}-J$ is irreducible and boundary irreducible follows from the fact that the knot is WGA.

By construction, $J$ has a WGA knot diagram $\pi(J)$ that has more than one twist region. Thus part (2) of Theorem 2.7 implies that $S^{3}-J$ cannot contain an essential annulus.

Next we claim that $S^{3}-J$ is atoroidal. To prove it, suppose that $T$ is an essential torus in $S^{3}-J$. By part (1) of Theorem 2.7, the checkerboard surfaces of $\pi(J)$ cut $T$ into annuli $A_{1},\dots,A_{j}$, such that each boundary component of each $A_{i}$ lies entirely in a single region of $F-\pi(J)$. Such a component cannot bound a disc (else use the fact that $T$ is incompressible and $S^{3}-J$ is irreducible to remove such an intersection), thus it must lie on a non-disc face of $F-J$. In our case there is only one non-disc face, namely the 3-holed sphere $E$. The three components of $\partial E$ represent distinct free homotopy classes in $H_{1}$. Thus no two of these components can co-bound an annulus in $H_{1}$.

It follows that the components $\partial A_{i}$ must be parallel to a single component of $\partial E$ and parallel to each other. Then $A_{i}$ is an annulus in a handlebody with parallel boundary components on the boundary of the handlebody. Thus $A_{i}$ co-bounds a solid torus in the handlebody with an annulus on the boundary, and either $A_{i}$ can be isotoped away from $F$, or $J$ lies inside that solid torus. However by Lemma 3.7, $J$ has edge-representativity at least four. If $J$ were contained in a solid torus parallel to an annulus on the boundary of $H_{1}$, then there would be an essential curve on $\partial H_{1}$ disjoint from that annulus, implying the edge-representativity is zero. This is impossible. Thus $J$ cannot lie entirely within a solid torus defined by $A_{i}$, and so we may isotope $T$ to avoid the intersections of $\partial A_{i}$. Repeating this argument, all the intersections $T\cap F$ can be eliminated by isotoping $T$. But then $T$ is inessential in $S^{3}-J$, a contradiction. ∎