Meridional almost normal surfaces in knot complements

The proofs of the following Claims 17, 18, 19, and 20 can be found in [16]. Let $(M_{i},\Sigma_{i},\Gamma_{i})$ be as described above, where $\mathcal{T}^{1}_{i}$ is in thin position with respect to the foliation $F_{i}$ of $M_{i}$. Since there is a thick region of $\mathcal{T}^{1}_{i}$ in $F_{i}$ we can apply Claim 4.5 of [16].

Let $L$ be a leaf of $F_{i}$ in a thick region intersecting the 2-skeleton in normal arcs and simple closed curves disjoint from $\mathcal{T}^{1}$ as is guaranteed by Claim 17. Then we can apply the following Claims 18, 19, and 20 to the leaf $L$.

The above claims together imply that this leaf $L$ of the foliation $F_{i}$ intersects the 2-skeleton only in simple closed curves disjoint from the 1-skeleton and normal curves of lengths three, four, and at most one of length eight. Compressing the simple closed curves in $\mathcal{T}^{2}$ and the surfaces in the interior of the tetrahedra gives a collection of normal surfaces with at most one almost normal surface. The almost normal surface, if it exists, must be a normal octagon since we have compressed all annuli in the interior of the tetrahedra. We can think of the leaf $L$ as this collection of normal and almost normal surfaces with tubes attached. Since our triangulation has no normal 2-spheres we can conclude that this collection will contain no almost normal 2-spheres as well since any almost normal 2-sphere can be isotoped to give a normal one by Lemma 14. Also notice that in this collection we will not have a normal surface tubed to the opposite side of another normal surface or we get a contradiction to the leaf being weakly compressible. To see that all of the compressions of $L$ are to one side, observe that as long as there are no normal 2-spheres in the collection then we know that compressions of $L\cap H$ are compressions of $L$ for any tetrahedron $H$ of $\mathcal{T}$. Lemma 15 implies that after compressing $L$ to one side the remaining surface is incompressible to the opposite side, thus there cannot be compressions on opposite sides of $L$. This completes the proof of Lemma 16. ∎

This claim follows from a standard innermost disk and outermost arc argument. ∎

We will prove the Lemma for arcs of $\mathcal{T}^{1}_{i}$ with both endpoints on $\Gamma_{i}^{top}$. The argument for arcs of $\mathcal{T}^{1}_{i}$ with both endpoints on $\Gamma_{i}^{bot}$ is symmetric. Let $L$ be a leaf of the foliation near the top singular leaf, so that $L$ consists of the normal surface $\Gamma_{i}^{top}$ punctured and attached to the tubes $t_{i}^{top}=\partial N(\Sigma_{i}^{top})$. See Figure 1 for the case $i=0$ and $M=S^{3}$. Choose an arc $\beta$, a subarc of $\mathcal{T}^{1}_{i}$. Since there is no thick region for $\mathcal{T}^{1}_{i}$, $\beta$ has only a single minimum and it is parallel to an arc on $L$, so there is a lower disk $E$ whose boundary is the union of $\beta$ and an arc $\alpha$ in $L$. We will show that after some edge slides and isotopies of $\Sigma_{i}^{top}$, $\alpha$ runs once over exactly one tube of $t_{i}^{top}$, and that this tube connects two normal disks in a tetrahedron, therefore is part of an almost normal surface.

Define the complexity of $E$ to be $(a,b)$, lexicographically ordered, where $a$ is the number of points of $\Sigma_{i}^{top}\cap\mathcal{T}^{2}$ to which $\alpha$ is also incident, and $b$ is the number of components in which $E$ meets the 2-skeleton of $\mathcal{T}$. We will assume that the complexity of $E$ has been minimized over all choices of $E$.

Observe that the arc $\alpha$ of $\partial E$ can’t lie entirely in $\Gamma_{i}^{top}$. Otherwise $E$ would be a boundary compressing disk in the complement of the 1-skeleton of the normal surface $\Gamma_{i}^{top}$ which contradicts Claim 22. Our strategy will be to show that there is a sequence of proper isotopies of $\Sigma_{i}^{top}$ and slides of ends of arcs of $\Sigma_{i}^{top}$ over each other and over $\Gamma_{i}^{top}$ (neither of which affect the isotopy class of $\Gamma_{i}^{top}\cup\Sigma_{i}^{top}$) so that afterwards $\alpha$ is incident to a single edge $z$ in $\Sigma_{i}^{top}$, $\alpha$ runs along this arc once, and $E$ lies entirely inside a single tetrahedron. Then $\Gamma_{i}^{top}\cup\partial(N(z))$ is the required almost normal surface obtained from $L$ by compressing all other tubes of $\Sigma_{i}^{top}$. Now that we have established some notation, for the rest of the proof we will consider the intersections of the disk $E$ with the 2-skeleton of $\mathcal{T}$ and we will show that any intersection violates the minimality of $(a,b)$

When we consider the arcs of intersection between $E$ and the 2-skeleton, we can get four types of components of $E\cap\mathcal{T}^{2}$ in $E$. Components of Type I are simple closed curves in $E$. Components of Type II are arcs with both endpoints on $\alpha$. Components of Type III are arcs with both endpoints on $\beta$, and components of Type IV are arcs with one endpoint on $\alpha$ and the other endpoint on $\beta$. See Figure 3. Next we will describe how each type of component of intersection of $E\cap\mathcal{T}^{2}$ can be removed, violating minimality of $(a,b)$.

Components of Type I are simple closed curves in $E$. A component of Type I that is innermost in a 2-simplex of $\mathcal{T}^{2}$ can be removed by substituting the disk it bounds in $\mathcal{T}^{2}$ for the disk it bounds in $E$. This reduces the number of times that $E$ meets the 2-skeleton, thus reducing $b$ and contradicting the minimality of $E$.

A component of Type II corresponds to an arc in a face $\sigma$ of some tetrahedron of $\mathcal{T}$ that either 1) has both endpoints on distinct components of $(\Sigma_{i}^{top}\cup\Gamma_{i}^{top})\cap\sigma$, or 2) has both endpoints on the same component of $(\Sigma_{i}^{top}\cup\Gamma_{i}^{top})\cap\sigma$. Suppose $\gamma$ is an arc of intersection between $\mathcal{T}^{2}$ and $E$ that is outermost in $E$ and is of Type II. Let $\delta$ be a subarc of $\alpha$ such that $\gamma\cup\delta$ is the boundary of a disk $D$ in $E-\mathcal{T}^{2}$. See Figure 4b). Then the arc $\gamma$ is either of type 1) or 2) above. In what follows we will use edge slides of $\Sigma_{i}^{top}$ to remove components of $\mathcal{T}^{2}\cap E$ and $\mathcal{T}^{2}\cap\Sigma_{i}^{top}$. Recall that $t_{i}^{top}$ is the boundary of a neighborhood of $\Sigma_{i}^{top}$, and we will abuse notation and consider $\delta\subset\alpha$ as an arc on $\Sigma_{i}^{top}$ when we really mean that $\delta$ is an arc on $t_{i}^{top}$.

Any two ends of edges of $\Sigma_{i}^{top}$ that meet the same normal disk in $\Gamma_{i}^{top}$ can be isotoped together so that there is at most one edge incident to each normal disk. Since $\Sigma_{i}^{top}$ can be extended to give a spine of $W$, any cycle in $\Sigma_{i}^{top}$ gives a cycle in $\Sigma$. Since $W\cup_{S}W^{\prime}$ is an irreducible Heegaard splitting of $M$ it follows from [4] (also see Proposition 2.5 of [14]) that no cycle of $\Sigma_{(W,K)}$ lies in a 3-ball. Hence for any tetrahedron $H$ of $\mathcal{T}$, $\Sigma_{i}^{top}\cap H$ cannot contain a cycle. Thus $\Sigma_{i}^{top}\cap H$ is a union of trees for each tetrahedron $H$ in $\mathcal{T}$. Each component of $\Sigma_{i}^{top}\cap H$ is a tree and each component of $(\Sigma_{i}^{top}\cup\Gamma_{i}^{top})\cap H$ is a tree with disks attached and so is simply connected.

Arcs of type 1) fall into the following subcases:

Case a) Let $H$ be the tetrahedron in $\mathcal{T}$ that contains $\delta$ and let $q$ denote the component of $\Sigma_{i}^{top}\cap H$ that contains $\delta$. If $q$ is a single arc with both endpoints on the same face of $H$ then $D$ describes an isotopy that removes two points of intersection of $q$ with $\mathcal{T}^{2}$. This reduces the number of points of $\Sigma_{i}^{top}\cap\mathcal{T}^{2}$ to which $\alpha$ is incident, thus reducing $a$, which is a contradiction.

Case b) Now suppose that $q$ is not an arc and $\delta$ is a path in $q$ that begins at point $x$ in $q$ that is not in a normal disk but is in some face $\sigma$ of $H$. Let $z$ be the edge of $\Sigma_{i}^{top}$ containing $x$. Then $\delta$ describes a series of edge slides of $z$ which culminate by introducing an extra point of intersection between $\Sigma_{i}^{top}$ and $\sigma$. However, after the edge slides the disk $D$ runs only over the edge $z$. Hence we can reduce the number of intersections of edges of $\Sigma_{i}^{top}$ incident to $\alpha$ that meet $\mathcal{T}^{2}$ by two as in Case a) reducing $a$ by at least one and contradicting the minimality assumptions. See Figure 5.

Case c) If $\delta$ is an arc on $L$ that has both endpoints of $\mathcal{T}^{2}\cap E$ on normal disks of $L$, then either $\delta$ must run over some edges of $\Sigma_{i}^{top}$ or it lies in a normal disk. If $\delta$ lies in a normal disk then as an outermost arc of the normal disk it cuts off a subdisk $D$ in the normal disk. Together $\gamma$ and $\delta$ bound a disk $D^{\prime}$ so that $D\cup D^{\prime}$ bounds a 3-ball that can be used to isotope $\delta$ into the next tetrahedron removing $\gamma$ and thus reducing complexity.

So we can assume that $\delta$ runs over some edges of $\Sigma_{i}^{top}$. Say that $\delta$ runs from normal disk $D_{1}$ to normal disk $D_{2}$. Since $\Sigma_{i}^{top}$ is incident to $D_{1}$ in only a single point, $\delta$ is incident to $\partial D_{1}$ in a single point. It follows that $D_{1}-N(\Sigma_{i}^{top})$ is an annulus and $\delta$ intersects the annulus in a single spanning arc. Thus $\delta$ runs precisely once along the edge $z$ that is incident to $D_{1}$. Then, as above, $D$ describes a slide and isotopy of $z$ that carries it to the arc $\gamma$ in a simplex of $\mathcal{T}^{2}$. But then a subdisk of that face describes a parallelism between $z$ and a subarc of $\mathcal{T}^{1}$. In particular, attaching a tube to $\Gamma_{i}^{top}$ along $z$ gives an almost normal surface.

Arcs of type 2) have endpoints on the same component of $q\cap\partial H$. Let $x$ denote the endpoint of $q$ in the face $\sigma$ of tetrahedron $H$. We assume that $x\in\Sigma_{i}^{top}$, so $\gamma$ forms a loop based at $x$ in $\sigma$ bounding a disk $A$ in $\sigma$; the case where both ends of $\gamma$ lie on a normal disk is similar:

Case a) If $interior(A)\cap\Sigma_{i}^{top}=\emptyset$, then construct new disks $E^{\prime}$ and $E^{\prime\prime}$ by cutting the disk $D$ along $\gamma$ and attaching a copy of $A$ to each piece. One of the disks $E^{\prime}$ or $E^{\prime\prime}$ will still be a lower disk and it will meet $\mathcal{T}^{2}$ in fewer components than $E$, contradicting the minimality of $E$.

Case b) Now suppose that $A\cap\Sigma_{i}^{top}\neq\emptyset$. Since $\Sigma_{i}^{top}$ is a union of trees in $H$, we know that a neighborhood of each component $q$ of $\Sigma_{i}^{top}\cap H$ is a 3-ball. So there is a disk $D^{\prime}$ in $N(q)$ whose boundary is the union of $\delta$ and a diameter $\delta^{\prime}$ of a small disk $\epsilon$ with which $N(q)$ meets $\partial H$ at $x$.

Isotope the leaf $L$ by compressing $\delta$ to $\delta^{\prime}$ in $N(q)$, splitting the disk $\epsilon$ in two. The effect on the spine is a possibly complicated series of edge slides. The overall effect is that the number of components of $\Sigma_{i}^{top}\cap\sigma$ increases by one when $\epsilon$ splits, and $D\cup D^{\prime}$ becomes a disk disjoint from $\Sigma_{i}^{top}$ and parallel to $A$. The disk $A$ now contains at least two points of $\Sigma_{i}^{top}\cap\sigma$. Now push $D\cup D^{\prime}$ across $A$ to remove $\gamma$, thus reducing $b$ which is a contradiction. See Figure 6.

To see how to remove components of Types III and IV it will be helpful to view the arc $\beta$ that runs along the edge $e$ of $\mathcal{T}^{1}$ as an arc that lies on $\partial N(e)$. As an arc on $\partial N(e)$, $\beta$ may wind around the edge $e$. If the winding is not monotone a priori then we can reduce the number of components in which the disk $E$ meets the faces of $\mathcal{T}^{2}$, contradicting minimality. Thus we may assume that the curve $\beta$ winds monotonically around the edge $e$. This implies that there are no curves of Type III since the existence a curve of Type III means that the arc $\beta$ must ‘double back’ as it winds around $e$, contradicting monotonicity.

Let $\gamma$ be an outermost arc component of Type IV, and let $D$ be the corresponding outermost sub-disk of $E$. Let $\sigma$ be the face of $\mathcal{T}^{2}$ that contains $\gamma$. The disk $D$ is co-bounded by a sub-arc $\rho$ of $\alpha$, a sub-arc $\lambda$ of $\beta$, and $\gamma$. See Figure 4a). Let $H$ denote the tetrahedron containing $D$ in its interior and with $\sigma$ as a face.

There are two cases that we will consider separately. The first case is when the arc $\gamma$ of $E\cap\sigma$ that runs from the edge $e$ to $L$ ends on a normal disk $\eta$ of $L$. The second case is when $\gamma$ ends on a tube (neighborhood of $\Sigma_{i}^{top}$) of $L$.

Case a): Suppose first that $\gamma$ ends on a normal disk $\eta$. In this situation there are two subcases. Either $\Sigma_{i}^{top}\cap\rho=\emptyset$ or $\Sigma_{i}^{top}\cap\rho\neq\emptyset$. See Figure 7a) and 7b).

Subcase 1): Suppose that $\Sigma_{i}^{top}\cap\rho=\emptyset$. In this case the arc $\rho$ runs over only normal disks. Observe that there is a disk $D^{\prime}$ in $\partial N(e)$ that is bounded by $\lambda$, a copy of part of the edge $e$ that bounds the face $\sigma$, and a copy of a meridian of $\partial N(e)$. In this case $D^{\prime}\cup\sigma\cup\eta$ bounds a 3-ball in $H$ that we can use to isotope $D$ across $\sigma$ and into the next tetrahedron, removing $\gamma$ and reducing $b$, thus reducing the complexity of $E$, a contradiction. See Figure 7a).

Subcase 2): Suppose now that $\Sigma_{i}^{top}\cap\rho\neq\emptyset$. Since $\Sigma_{i}^{top}$ is a union of trees, each component of $N(\Sigma_{i}^{top})$ is a 3-ball. In particular, there is a disk $\Delta$ in $N(\Sigma_{i}^{top})$ whose boundary is the union of a sub-arc of $\rho$ and a diameter $d$ of the disk $\epsilon$ with which $N(\Sigma_{i}^{top})$ intersects the normal disk $\eta$. Isotope $N(\Sigma_{i}^{top})$ by compressing $d$ to $\rho$ in $N(\Sigma_{i}^{top})$, splitting the disk $\epsilon$ in two. The effect on $\Sigma_{i}^{top}$ is a series of edge slides that results in a new component of $\Sigma_{i}^{top}\cap H$ that is on the same side of $\rho$ as $e$. Now proceed as in Subcase 1). See Figure 7b).

Case b): Now suppose that $\gamma$ ends on a tube of $L$. The core of this tube is an edge $\tau$ that may connect to other edges of $\Sigma_{i}^{top}$ in $\Sigma_{i}^{top}\cap H$, and $\Sigma_{i}^{top}$ connects to a normal disk $\eta$. See Figure 8. We will describe in two steps a slide of $\tau$ and an isotopy of $D$ that will remove a component of $\Sigma_{i}^{top}$, reducing $a$, and thereby reducing the complexity of $E$.

First, since $\tau$ connects to other edges of $\Sigma_{i}$ in $\Sigma_{i}\cap H$, $\rho$ describes an edge slide of $\tau$ that keeps $\tau\cap\sigma$ fixed but slides the opposite end of the edge $\tau$ off of $\Sigma_{i}$ and onto the normal disk $\eta$. We continue to slide $\tau$ along $\eta$ following $\rho$ until it almost meets $\partial N(e)$. See Figure 8b). Now we can use the disk $D$ to isotope all of $\tau$ until it lies close to $\lambda\cup\gamma$. At this point the entire disk $D$ and tube $\tau$ lie very close to $\beta\cup\gamma$ in $H$. See Figure 8c).

For the second step recall the disk $D^{\prime}$ in $\partial N(e)$ that is bounded by $\lambda$, a copy of part of the edge $e$ that bounds the face $\sigma$, and a copy of a meridian of $\partial N(e)$. The disks $D$ and $D^{\prime}$ describe an isotopy of $\tau$ across the face $\sigma$ and into the next tetrahedron, removing the component $\gamma$ from $\sigma\cap E$ and, in particular, removing the component of intersection between $\tau$ and $\sigma\in\mathcal{T}^{2}$, reducing $a$, which is a contradiction. See Figure 8d). Thus there can be no arcs of Type IV. Therefore the arc $\beta$ of $\mathcal{T}^{1}_{i}$, the edges of $\Sigma_{i}^{top}$ that $\alpha$ runs along, and the disk $E$ are all contained in one tetrahedron. The proof now follows as in Case 1c) above. ∎

Recall that $\partial M_{i}=\Gamma_{i}^{top}\cup\Gamma_{i}^{bot}$. Since $\Gamma_{i}^{top}$ and $\Gamma_{i}^{bot}$ are normal with respect to $\mathcal{T}$ it follows that there are two possibilities for how the region $M_{i}$ between $\Gamma_{i}^{top}$ and $\Gamma_{i}^{bot}$ can intersect a face of the 2-skeleton. Either the region bounded by $\Gamma_{i}^{top}$ and $\Gamma_{i}^{bot}$ is a trapezoid region or a hexagon region.

Suppose that there is a hexagon region of $M_{i}\cap\mathcal{T}^{2}$. Then three edges of the hexagon are arcs of $(\Gamma_{i}^{top}\cup\Gamma_{i}^{bot})\cap\mathcal{T}^{2}$ and that the other three edges are arcs of $\mathcal{T}^{1}_{i}$ connecting the three components of $\Gamma_{i}\cap\mathcal{T}^{2}$. But this implies that some arc of $\mathcal{T}^{1}_{i}$ connects either $\Gamma_{i}^{top}$ to $\Gamma_{i}^{top}$ or $\Gamma_{i}^{bot}$ to $\Gamma_{i}^{bot}$ which is a contradiction. Therefore there cannot be any hexagonal components and all regions of intersection between $M_{i}$ and $\mathcal{T}^{2}$ are trapezoids.