text:
\begin{document}

   \title[Adding 2--handles to sutured manifolds]{Adding 2--handles to sutured manifolds}
   \author{Scott A Taylor}
   \email{staylor@math.ucsb.edu}
   \thanks{This research was partially supported by a grant from the National Science Foundation.}

\begin{abstract}
Combinatorial sutured manifold theory is used to study the effects of attaching a 2--handle to an essential simple closed curve on a genus two boundary component of a compact, orientable 3--manifold. The main results concern degenerating handle additions to a simple 3--manifold and essential surfaces in the exterior of a knot or link obtained by ``boring'' a split link or unknot. (Boring is an operation on knots and two-component links which generalizes rational tangle replacement.) Generalizations and new proofs of several well known theorems from classical knot theory are obtained. These include superadditivity of genus under band connect sum and the fact that unknotting number one knots are prime.
\end{abstract}

   \date{\today}

 \maketitle

\section{Introduction}
Sutured manifold theory has produced many stunning results in the study of Dehn surgery on knots, including the solution to the property R and Poenaru conjectures \cite{G}, the result that satellite knots have property P \cite{S2}, and an understanding of how Dehn surgery on a linking number zero knot in a link complement restricts the intersections between essential surfaces and the knot \cite{L1}. Adding a 2--handle to a boundary component of genus greater than 1 is a natural generalization of Dehn surgery, but sutured manifold theory has not often been exploited in this study. Typically, 2--handle addition has been studied using purely combinatorial techniques. For example, Scharlemann and Wu \cite{SW} show that if a genus 2 or greater boundary component of a hyperbolic manifold contains a reducing slope and a boundary-reducing slope, either the slopes are disjoint or they can be isotoped into a common once-punctured torus. Zhang, Qiu, and Li \cite{ZQL} have shown that if $M$ is simple and if the genus of a boundary component $F$ is at least 2, then any two reducing slopes on $F$ must intersect minimally no more than 4 times. The same authors \cite{LQZ} show that if $F$ has genus 2, then there is at most one separating slope which will create a boundary-reducible manifold.

Using combinatorial sutured manifold theory, the paper \cite{T1} studied a limited form of 2-handle attachment, called ``refilling meridians of a genus 2 handlebody''. This paper continues the project of that paper, although the sutured manifold techniques in this paper are easier (in spirit) and apply to a wider variety of situations. In particular, we are able to draw conclusions about 2--handle additions to a genus 2 boundary component $F$ of a simple 3--manifold $M$ which produce a non-simple manifold. In general, there is not much overlap between the work of this paper and earlier work on 2--handle additions. Roughly speaking, this is because although sutured manifold theory is a very powerful technique, it is limited by the need to place sutures on non-torus boundary components. If, in a particular situation, there were knowledge about the placement of sutures the main theorem of this paper would allow a fairly thorough study of 2--handle addition along even non-separating curves.

The main theorem of this paper applies to sutured manifolds with any number of boundary components of any positive genus. For the applications, we focus attention on a genus two boundary component. The advantage of such a restriction is that, without knowing much about the manifold, we can be specific about suitable sutured manifold structures. This allows us to prove theorems that don't use the language of sutured manifold theory in their statements. Here are some of the more interesting results. A 3--manifold $N$ is \defn{simple} if it does not contain an essential sphere, disc, annulus, or torus. For a curve $a \subset \boundary N$, let $N[a]$ denote the result of attaching a 2--handle to $a$.

\begin{ThmSimpleHandle}
Suppose that $N$ is a compact, orientable, simple 3--manifold and that $F\subset \boundary N$ is a genus 2 component. Suppose that $a$ and $b$ are non-isotopic separating curves on $F$. Suppose that $N[a]$ is reducible and that $N[b]$ is non-simple. Then $N[b]$ contains an essential annulus with boundary on non-torus boundary components and $a$ and $b$ can be isotoped to intersect exactly four times.
\end{ThmSimpleHandle}

The remaining applications all concern the operation of ``refilling meridians''.
Let $W$ be a genus two handlebody embedded in 3--manifold $M$. For an essential disc $\alpha \subset W$, let $M[\alpha]$ denote the result of attaching a regular neighborhood of $\alpha$ to $M - \inter{W}$. Let $\alpha$ and $\beta$ be essential discs in $W$ which have been isotoped so that their boundaries intersect minimally in $\Delta > 0$ points.

The first application solves a conjecture of Scharlemann for a large class of 3--manifolds $M$ and embeddings of $W \subset M$. Here is the statement for $M = S^3$.

\begin{ThmMSC}
Suppose that $M = S^3$ and that $S^3 - \inter{W}$ is boundary-irreducible. Then one of $S^3[\alpha]$ or $S^3[\beta]$ is irreducible and boundary\hyph irreducible.
\end{ThmMSC}

The next result can be used to give new proofs of both the superadditivity of genus under band-sum and the fact that tunnel number one knots and links have minimal genus Seifert surfaces disjoint from a given tunnel. Let $L_\alpha$ be the core curves of the solid torus or tori $W - \inter{\eta}(\alpha)$. Let $\ob{\alpha}$ be the cocore of the 2--handle $\eta(\alpha)$. Similarly define $L_\beta$. We say that $L_\alpha$ and $L_\beta$ are related by \defn{boring}.

\begin{ThmDisjointSS}
Suppose that $L_\alpha$ is a knot or link in $S^3$ obtained by boring a knot or link $L_\beta$ using handlebody $W$. If $\alpha$ is separating, assume that $\boundary W - \boundary \alpha$ is incompressible in $S^3 - \inter{W}$. If $L_\beta$ is a split link or unknot, then $L_\alpha$ is not a split link and there is a minimal genus Seifert surface for $L_\alpha$ which is disjoint from $\ob{\alpha}$.
\end{ThmDisjointSS}

In addition to studying the presence of essential spheres and discs and minimal genus Seifert surfaces in the exterior of a knot or link $L_\beta$ obtained by boring a split link or unknot, the techniques of this paper are useful for studying many other essential surfaces. It is natural to ask about conditions that guarantee that $L_\beta$ is hyperbolic. The next theorem shows that under some mild assumptions it is enough to require that $\Delta \geq 4$. The non-simplified version of the theorem weakens the assumption that $N$ be simple.

\begin{ThmEssentialTorus}
Suppose that $L_\beta$ is a knot or link obtained by boring the link $L_\alpha$ using a handlebody $W \subset S^3$ with $N = S^3 - \inter{W}$ simple. Suppose that $L_\alpha$ is a split link or that there is no minimal genus Seifert surface for $L_\alpha$ disjoint from $\ob{\alpha}$. If the exterior of $L_\beta$ contains an essential annulus or torus then $\Delta = 2$.
\end{ThmEssentialTorus}

Rational tangle replacement is a particularly interesting type of boring. The next two theorems concern essential planar surfaces and (punctured) tori in the exterior of a knot or link obtained by a rational tangle replacement operation. The first is a generalization of, and gives a new proof of, the fact that unknotting number one knots are prime. For each of them suppose that $L_\beta$ is a knot or link in $S^3$ and that $B' \subset S^3$ is a 3--ball which intersects $L_\beta$ in two strands $r_\beta$ such that $(B',r_\beta)$ is a rational tangle. Suppose that $(S^3 - \inter{\eta}(B'),L_\beta - \inter{\eta}(B'))$ is a prime tangle. Suppose that $(B',r_\alpha)$ is a rational tangle of distance $d \geq 1$ from $(B',r_\beta)$. (Here, $d = \Delta/2$ and $\alpha$ and $\beta$ are the trivializing discs for the tangles $(B',r_\alpha)$ and $(B',r_\beta)$ respectively.) Suppose also that $L_\alpha = (L_\beta - B') \cup r_\alpha$ is a split link or does not have a minimal genus Seifert surface with interior disjoint from $B'$.

\begin{ThmMeridionalPlanar}
If $L_\beta$ has an essential properly embedded meridional planar surface with $m$ boundary components, it contains such a surface $\ob{Q}$ with $|\boundary \ob{Q}| \leq m$ such that either $\ob{Q}$ is disjoint from $B'$ or
\[
p(d - 1) \leq |\boundary \ob{Q}| - 2
\]
where $p$ is the number of times that the interior of $\ob{Q}$ intersects $B'$.
\end{ThmMeridionalPlanar}

The second theorem may be useful for studying non-hyperbolic surgeries on hyperbolic knots and links.

\begin{ThmRTPlanarTori}
If $L_\beta$ contains an essential planar surface or punctured torus $\ob{R}$ in its exterior, then one of the following holds:
\begin{enumerate}
\item $L_\beta$ is a link and $\boundary \ob{R}$ is disjoint from some component of $L_\beta$.
\item $\ob{R}$ has meridional boundary on some component of $L_\beta$
\item The interior of $\ob{R}$ is disjoint from $B'$ and $\ob{\beta}$, the cocore of the trivializing disc for $r_\beta$, can be slid and isotoped to lie in $\ob{R}$.
\item $d \leq 3$.
\end{enumerate}
\end{ThmRTPlanarTori}
In the non-simplified version of the theorem, much more detail is given concerning the last possibility.

\subsection*{Notation and Conventions} We assume a working knowledge of combinatorial sutured manifold theory as developed in \cite{S2}. All (co)homology groups have integer coefficients. $S^3$ denotes the 3--sphere, $\inter{X}$ denotes the interior of $X$, and $\eta(X)$ denotes a closed regular neighborhood of $X$. $N$ will always denote a compact, orientable 3--manifold with boundary component $F$ of genus at least two. $N[a]$ denotes the 3--manifold obtained by attaching a 2--handle with core $\alpha$ to an essential simple closed curve $a \subset F$. The cocore of the 2--handle $\alpha \times I$ is denoted by $\ob{\alpha}$. Similarly $b$ will always be an essential simple closed curve curve on $F$ and $\beta$ and $\ob{\beta}$ will be the core and co-core of a 2--handle attached to it. $M$ will be a compact orientable 3--manifold containing a genus two handlebody $W$. For essential discs $\alpha$ and $\beta$ in $W$, $M[\alpha]$ and $M[\beta]$ will denote the manifolds obtained by refilling the meridians $\alpha$ and $\beta$ respectively. $L_\alpha$ and $L_\beta$ will denote the core of the solid torus/tori $W - \inter{\eta}(\alpha)$ and $W - \inter{\eta}(\beta)$ respectively.

\subsection*{Acknowledgements} The work in this paper forms part of my Ph.D. dissertation at University of California, Santa Barbara. I am grateful to my advisor, Martin Scharlemann, for many helpful conversations. This work was partially supported by a grant from the National Science Foundation.

\section{Sutured manifolds and 2--handle addition}
Suppose that in $\boundary N$ a collection of pairwise disjoint essential simple closed curves $\gamma$, all disjoint from the simple closed curve $a \subset F$, has been chosen so that $(N,\gamma \cup a)$ is a taut sutured manifold. Notice that $(N[a],\gamma)$ is a sutured manifold.

Suppose that $\mc{B} = \{b_1, \hdots, b_{|\mc{B}|}\}$ are pairwise disjoint, pairwise nonparallel essential curves in $F$ each of which intersects $a \cup \gamma$ minimally. Suppose that $Q \subset N$ is a surface with $q_i$ boundary components parallel to $b_i$. Let $\boundary_0 Q$ denote the boundary components of $Q$ which are not parallel to any curve in $\mc{B}$ and suppose that $\boundary_0 Q \cap b_i = \nil$ for all $i$. Assume that $\boundary Q$ intersects $\gamma \cup a$ minimally with $|\boundary Q \cap a| > 0$ and that no component of $Q$ is a sphere or disc disjoint from $a \cup \gamma$. We think of $Q$ as being $\ob{Q} \cap N$ where $\ob{Q}$ is a surface in the manifold $N[\mc{B}]$ obtained from attaching 2--handles along the curves of $\mc{B}$ and filling in any 2--sphere boundary components with 3--balls. $\ob{Q}$ is obtained from $Q$ by attaching discs to the components of $\boundary Q$ parallel to curves in $\mc{B}$. We then have $\boundary_0 Q= \boundary \ob{Q}$. Define $\Delta_i = |b_i \cap a|$, $\Delta_\boundary = |\boundary_0 Q \cap a|$, $\nu_i = |b_i \cap \gamma|$, $\nu_\boundary = |\boundary_0 Q \cap \gamma|$ and
\[
K(\ob{Q}) = \sum_{i = 1}^{n} q_i(\Delta_i - \nu_i - 2) + (\Delta_\boundary - \nu_\boundary).
\]

The presence of such a surface can lead to information on the sutured manifold $(N[a],\gamma)$ since $Q$ is a parameterizing surface. Before stating and proving the Main Theorem, we define two important notions.

First, define an \defn{$a$--boundary compressing disc} for $Q$ to be a boundary compressing disc $D$ for $Q$ such that $\boundary D = \delta \cup \epsilon$ where $\delta$ and $\epsilon$ are arcs with $\boundary \delta = \boundary \epsilon = \delta \cap \epsilon$. The arc $\delta$ is an essential arc in $Q$. The arc $\epsilon$ is a subarc of some essential circle in $\eta(a) \subset F$.

Second, recall that a \defn{conditioned surface} $S \subset N$ is a surface such that
\begin{enumerate}
\item all components of $\boundary S$ on $T(\gamma)$ are coherently oriented parallel essential circles,
\item all components of $\boundary S$ in a component of $\eta(\gamma \cup a)$ are either parallel coherently oriented circles or essential coherently oriented arcs,
\item and no collection of simple closed curves of $\boundary S \cap R(\gamma)$ is trivial in $H_1(R(\gamma),\boundary R(\gamma))$.
\end{enumerate}

\begin{theorem}[Main Theorem]\label{Thm: Main}
Let $(N,\gamma \cup a)$ and $Q$ be as described above. If either of the following holds:
\begin{itemize}
\item $(N[a],\gamma)$ is not taut
\item There exists a surface $S$ in $N[a]$ which is disjoint from $\ob{\alpha}$, is a conditioned surface in $N[a]$, and is taut in $N$ but is not taut in $N[a]$.
\end{itemize}
then one of the following holds:
\begin{enumerate}
\item $N[a]$ contains an essential separating sphere intersecting $\ob{\alpha}$ exactly twice and which cannot be isotoped to intersect that arc any fewer times. Furthermore, this sphere bounds a non-trivial homology ball in $N[a]$
\item There is an $a$--boundary compressing disc for $Q$
\item $-2\chi(\ob{Q}) \geq K(\ob{Q}).$
\end{enumerate}
\end{theorem}

The remainder of this section proves the theorem. The proof was inspired by Lackenby's work \cite{L1} on Dehn surgery on linking number zero knots in sutured manifolds.

We begin by creating a sequence of taut sutured manifold decompositions of $(N,\gamma \cup a)$. In order to effectively apply the main theorems of combinatorial sutured manifold theory, this sequence will need to be constructed in a particular fashion.

It will be useful to be able to convert sutures into arcs and \textit{vice versa}. This can be done with the following lemma. For the statement, we introduce some terminology. Suppose that $a'$ is a suture on the boundary of a sutured 3--manifold $(X,\Gamma \cup a',\psi)$. Let $\ob{\alpha}'$ be the cocore of a 2--handle attached to $a'$. We say that $(X,\Gamma \cup a',\psi)$ is obtained by \defn{converting} the arc $\ob{\alpha}'$ in $(X[a'],\Gamma,\psi \cup \ob{\alpha}')$ into a suture. Similarly, we say that $(X[a'],\Gamma,\psi \cup \ob{\alpha}')$ is obtained by converting the suture $a'$ into an arc. The next lemma shows that converting an arc to a suture or \textit{vice versa} does not affect tautness.

\begin{lemma}[{\cite[Lemma 2.3]{S4}}]\label{Lem: Suture Conversion}
The sutured manifold $(X[a'],\Gamma,\psi \cup \ob{\alpha}')$ is $(\psi \cup \ob{\alpha}')$--taut if and only if $(X,\Gamma \cup a',\psi)$ is $\psi$-taut.
\end{lemma}

The next lemma will provide the surfaces that are essential for creating a useful sutured manifold hierarchy.

\begin{lemma}\label{Lem: Conditioned Surfaces}
Suppose that $(X,\Gamma \cup a)$ is a taut sutured manifold and that $H_2(X[a],\boundary X[a]) \neq 0$. Then, in $X[a]$, there is a conditioned surface $S$ which is disjoint from $\ob{\alpha}$. $S$ is a taut conditioned surface in $X$ and $\boundary S \cap \eta(a) = 0$.
\end{lemma}
\begin{proof}
By the proofs of Theorems 2.5 and 2.6 of \cite{S2}, given a non-trivial homology class $z \in H_2(X[a],\boundary X[a])$ there exists a taut conditioned surface $S'$ in the $\ob{\alpha}$--taut sutured manifold $(X[a],\Gamma,\ob{\alpha})$. After possibly replacing $z$ with $-z$ we may assume that $\ob{\alpha}$ has algebraic intersection number $i \geq 0$ with $S'$. By the choice of orientation of the arc $\ob{\alpha}$, $\ob{\alpha}$ has algebraic intersection number $-1$ with $R_+(\Gamma)$. The surface $S''$ which is the the double curve sum of $S'$ with $i$ copies of $R_+(\Gamma)$ has algebraic intersection number zero with $\ob{\alpha}$. Notice that $\boundary S''$ satisfies the necessary criteria for $S''$ to be conditioned in $(X,\Gamma \cup a)$. Tube together points of opposite intersection number to create from $S''$ a surface $S$ which is disjoint from $\ob{\alpha}$ and for which $\boundary S = \boundary S''$. The surface $S$ is a conditioned surface in $(X,\Gamma \cup a)$. After replacing $S$ with a taut surface in $(X,\Gamma \cup a)$ having the same boundary.
\end{proof}

To create a taut sutured manifold decomposition that is adapted to the parameterizing surface $Q$, we may need to take the double curve sum of our favorite conditioned surface $S$ in a sutured manifold $(X,\Gamma \cup a)$ with some number $k$ of copies of $R_+(\Gamma \cup a)$ and some number $l$ of copies of $R_-(\Gamma \cup a)$, creating the surface $S_{k,l}$. We then decompose using the surface $S_{k,l}$ instead of $S$. The conditioned surfaces that we use will be the ones provided by Lemma \ref{Lem: Conditioned Surfaces}. Performing the double curve sums creates boundary components of $S_{k,l}$ which are located in $\eta(a)$. Attaching discs to each of those boundary components creates a surface $S^a_{k,l} \subset X[a]$. The surface $S^a_{k,l}$ can also be created by taking the double curve sum of $S$ with $k$ copies of $R_+(\Gamma) \subset X[a]$ and $l$ copies of $R_-(\Gamma) \subset X[a]$.

The next lemma guarantees that if we use such a surface to perform a decomposition of the sutured manifold $(X[a],\Gamma,\ob{\alpha})$ then all but one arc of $\ob{\alpha} - \eta(S^a_{k,l})$ can be cancelled. Let $*_a$ denote the point on $\ob{\alpha}$ to which the curve $a$ retracts under the standard retraction of $\eta(\ob{\alpha})$ to $\ob{\alpha}$. If $(X[a],\Gamma,\ob{\alpha})$ is a sutured manifold which is decomposed along a surface $\Sigma$ which is disjoint from $*_a$, call any component of $\ob{\alpha} - \inter{\eta}(\Sigma)$ which doesn't contain $*_a$ a \defn{residual arc}. If such an arc is converted into a suture, call the resulting suture a \defn{residual suture}.

\begin{lemma}\label{Lem: Cancelling All Arcs But One}
Let $S^a_{k,l} \subset X[a]$ be a surface created from the $S$ provided by Lemma \ref{Lem: Conditioned Surfaces}. Then after decomposing $X[a]$ along $S^a_{k,l}$ there exists a cancelling disc or a self-amalgamating disc for each residual arc.
\end{lemma}
\begin{proof}
Suppose that $\ob{\alpha}'$ is a residual arc. Since $S$ is disjoint from $a$, each endpoint of $\ob{\alpha}'$ is on $R_\pm(\Gamma)$ or on a pushed off copy of $R_\pm(\Gamma)$. In other words, since $\ob{\alpha}'$ is a residual arc both endpoints are on different copies of $R_+(\Gamma)$ or on different copies of $R_-(\Gamma)$. Let $P$ be the product region $R_\pm \times I$ between these copies. Then each component of $S \cap P$ is an $\text{(arc)} \times I$ or
an $S^1 \times I$. The arc $\ob{\alpha}' = \ob{\alpha} \cap P$ is a copy of $\text{(point)} \times I$.

If the component of $(R_\pm \times \{0\}) - S$ containing the endpoint of $\ob{\alpha}'$ has any part of its boundary intersecting $\boundary (R_\pm \times \{0\})$ choose a path $p$ in that component from $\boundary \ob{\alpha}'$ to $\boundary (R_\pm \times \{0\})$. If not, then there is an essential closed curve $p$ in that component which passes through the endpoint of $\ob{\alpha}'$ and is isotopic to a component of $\boundary S$. In the first case, $p \times I$ is a cancelling disc for $\ob{\alpha}'$ and in the second case, $p\times I$ is a self-amalgamating disc for $\ob{\alpha}'$.
\end{proof}

In creating a hierarchy of $(N,\gamma \cup a)$ it may be necessary to eliminate index zero discs. Certain index-zero discs need to be treated carefully. To that end, suppose that $(X[a],\Gamma,\ob{\alpha})$ is a sutured manifold with $\ob{\alpha}$ a collection of arcs. Suppose that $D \subset X[a]$ is a cancelling disc for a component $\ob{\alpha}'$ of $\ob{\alpha}$. A regular neighborhood of $D$ is a 3--ball $B$ containing $\ob{\alpha}'$. Cutting open $X[a]$ along the disc $E = \cls(\boundary B \cap \inter{X})$ produces a sutured manifold, one component of which is a 3--ball containing $\ob{\alpha}'$. The 3--ball has a single suture in its boundary. We may then remove the arc $\ob{\alpha}'$ without affecting $\ob{\alpha}$-tautness. The other component is the sutured manifold we would obtain by cancelling the arc $\ob{\alpha}'$ in $X[a]$. By converting all arcs to sutures we obtain a decomposition of $(X,\Gamma \cup a)$ which eliminates the index zero disc $D$. Indeed, by decomposing along $E$ but not the disc $D$ we can eliminate an index-zero disc in $X$ without cutting along $a$ or a residual suture. This is at the cost of introducing a component which is a solid torus having two longitudinal sutures in its boundary. Exactly one of those sutures is either $a$ or a residual suture. If it is a residual suture call the component a \defn{residual torus}.

Suppose that $D \subset X[a]$ is a self-amalgamating disc for a component $\ob{\alpha}'$ of $\ob{\alpha}$. Then slightly enlarging it produces a non-trivial product annulus $A$ in $X$. There is a parallelism of $\ob{\alpha}$ in $X[a]$ into $A$. After decomposing $X[a]$ along $A$ there is a cancelling disc for $\ob{\alpha}'$ which may then be eliminated as above. Notice that since each component of $\boundary S$ is essential in $\boundary X[a]$ (by the construction of $S$ in Lemma \ref{Lem: Conditioned Surfaces}) the product annuli created by the self-amalgamating discs of Lemma \ref{Lem: Cancelling All Arcs But One} have ends which are essential in $\boundary X[a]$. Thus, if a product annulus created from a self-amalgamating disc has both ends inessential in $\boundary X[a]$ it must have arisen from a self-amalgamating disc for the suture $a$. But it is easy to see that in this case all such product annuli must have both ends essential in $\boundary X[a]$. This observation will be useful in the proof of Lemma \ref{Lem: N' Not Taut} below.

\begin{lemma}\label{Lem: Hier Exists}
There is a taut sequence of sutured manifold decompositions
\begin{equation*}\tag{$\dagger$}\label{Hier}
(N,\gamma \cup a) = (N_0,\gamma_0 \cup a) \stackrel{S_1}{\to} (N_1,\gamma_1 \cup a) \stackrel{S_2}{\to} \hdots \stackrel{S_n}{\to} (N_n,\gamma_n \cup a)
\end{equation*}
adapted to the parameterizing surface $Q$ such that
\begin{enumerate}
\item each decomposition is either a decomposition along a product disc or product annulus or along a surface $S_{k,l}$ given by Lemma \ref{Lem: Cancelling All Arcs But One}. If the product disc intersects a residual suture then the decomposition is performed as described above. All decompositions along product annuli arise from this method of eliminating product discs, as described above.
\item $H_2(N_n[\mathfrak{a}],\boundary N_n[\mathfrak{a}]) = 0$ where $\mathfrak{a}$ is the curve $a$ together with all the residual sutures.
\item if a component of $N_n$ does not contain $a$, it is either a residual torus or a 3--ball containing a single suture in its boundary.
\end{enumerate}
\end{lemma}

Another formulation of (2) is that if we convert $a$ and all residual sutures to arcs, the resulting manifold has trivial homology relative to its boundary.

\begin{proof}
This is essentially the proof that taut sutured manifold hierarchies exist \cite[Theorem 4.19]{S2}. The proof of that theorem makes the hierarchy stop when $H_2(N_n,\boundary N_n) = 0$. By Lemma \ref{Lem: Conditioned Surfaces}, we can instead stop the hierarchy when $H_2(N_n[\mathfrak{a}],\boundary N_n[\mathfrak{a}]) = 0$. If it is necessary to eliminate a product disc which intersects twice a residual suture or the suture $a$ then the decomposition should be performed as described previously. By Lemma \ref{Lem: Cancelling All Arcs But One}, there exists such a product disc for all residual sutures. Hence, all residual sutures end up in residual tori. Any component of $N_n$ which does not contain a residual suture or $a$ must be a 3--ball with a single suture in its boundary since $(N_n,\gamma_n \cup a)$ is taut and $H_2(N_n[\mathfrak{a}],\boundary N_n[\mathfrak{a}]) = 0$.
\end{proof}

Let $N'$ denote the component of $N_n$ which contains $a$. We can now use the hypotheses of the theorem we are trying to prove to conclude that $(N'[a],\gamma_n \cap N')$ is not taut.

\begin{lemma}\label{Lem: N' Not Taut}
$(N'[a],\gamma_n \cap N')$ is not taut.
\end{lemma}
\begin{proof}
Since a component of $N_n - N'$ is either a 3--ball with a single suture in its boundary or a residual torus, all components of $N_n[\mathfrak{a}] - N'[a]$ are $\nil$--taut. Thus, if $(N'[a],\gamma_n \cap N')$ is taut, so is $(N_n[\mathfrak{a}],\gamma_n - \mathfrak{a})$.

Convert the hierarchy ($\dagger$) into a sequence of sutured manifold decompositions of the sutured manifold $(N[a],\gamma,\ob{\alpha})$ by converting the suture $a$ into an arc $\ob{\alpha}$ and any surface $S_{k,l}$ into $S^a_{k,l}$ as described previously. Let $S^a_1$ denote the result of applying this conversion to $S_1$. Then each surface in the hierarchy is either a product disc, non-trivial product annulus (by the remarks preceding Lemma \ref{Lem: Hier Exists}), or conditioned surface. Thus, if $(N_n[\mathfrak{a}],\gamma_n - \mathfrak{a})$ is $\nil$-taut, by  Corollary 3.9 of \cite{S2}, $(N[a],\gamma)$ is taut and $S^a_1$ is taut. The surface $S^a_1$ is obtained by taking the double curve sum of $S$ with $k$ copies of $R_+(\gamma)$ and $l$ copies of $R_-(\gamma)$. If $S$ is not $\nil$--taut, then it does not minimize the Thurston norm (in $H_2(N[a],\boundary S)$). But in this case, the double curve sum of $S$ with $k$ copies of $R_+$ and $l$ copies of $R_-$ is not Thurston norm minimizing either, implying that $S^a_1$ is not taut, a contradiction. Thus, if $(N'[a],\gamma_n \cap N')$ is taut so is $(N_n[\mathfrak{a}],\gamma_n - \mathfrak{a})$. In which case, we can also conclude that $(N[a],\gamma)$ and $S$ are taut. But this contradicts the hypotheses of our theorem.
\end{proof}

\begin{remark}
Here is a brief aside to explain the route taken for the proof up until this point. Psychologically, it would be easier to have taken an $\ob{\alpha}$--taut hierarchy of $(N[a],\gamma)$. However, would need that at the end of the hierarchy there is at most one arc which cannot be cancelled. This requires that the conditioned surfaces be taken to be disjoint from $\ob{\alpha}$ (except for the result of double curve summing with $R_\pm$. \textit{A priori} decompositions along such surfaces may not be $\ob{\alpha}$--taut. There is then no clear way to guarantee that the sutured manifold at the end is $\ob{\alpha}$--taut, in other words that it has the structure that we will now be making use of. Furthermore, it is unclear whether or not such a sequence of decompositions can be guaranteed to terminate. The proof given here avoids these difficulties by constructing taut decompositions of $(N,\gamma \cup a)$.
\end{remark}

Carefully examining $N'$ will enable us to conclude the proof of the theorem.

\begin{lemma}\label{Lem: Boundary of N'}
$\boundary N'$ is a torus and $N'[a]$ is an integer homology ball.
\end{lemma}
\begin{proof}
The proof is similar to \cite[Lemma A.4]{L2}. Let $A = \boundary N' - \inter{\eta}(a)$. By construction of the hierarchy, $H_2(N',A) = 0$. Thus, by duality for manifolds with boundary $H^1(N',\eta(a)) = 0$. By the Universal Coefficient Theorem, $H_1(N',\eta(a)) = 0$. From the exact sequence for the homology of the pair $(N',\eta(a))$, $H_1(\eta(a))$ surjects onto $H_1(N')$. Thus, $H_1(N')$ is cyclic. Since $H_2(N',A) = 0$, by the long exact sequence for the pair $(N',A)$, $H_1(A)$ injects into $H_1(N')$. Since $A$ is a surface and $\boundary \eta(a)$ has two components, $A$ is a collection of spheres and either an annulus or two discs. Since $a$ does not compress in $N$, $A$ does not contain a disc. The existence of a sphere would contradict tautness of $N'$, and so $A$ is an annulus.

Since $H_1(A)$ is isomorphic to $\Z$ and it injects into the cyclic group $H_1(N')$, $H_1(N')$ is also isomorphic to $\Z$. Since $\eta(a)$ is an annulus and since $H_1(\eta(a))$ surjects $H_1(N')$, the inclusion of $\eta(a)$ into $N'$ induces an isomorphism on first homology. Since $A$ is an annulus and $H_2(N',A) = 0$, the exact sequence for the pair $(N',A)$ shows that $H_2(N') = 0$. It is then easy to see that $N'[a]$ is a homology ball.
\end{proof}

Since $(N',(\gamma_n \cup a) \cap N')$ is a sutured manifold and $\boundary N'$ is a torus containing the suture $a$ there must be an odd number $r$ of other sutures. The proof of the theorem concludes by examining two cases. The first case is when $r = 1$ and the second case is when $r \geq 3$.

Suppose that $r = 1$. Then $\boundary N'[a]$ is a sphere containing a single suture. Since $(N'[a],\gamma_n \cap N')$ is an integer homology ball (Lemma \ref{Lem: Boundary of N'}) which is not taut, the integer homology ball is not a 3--ball. Push $\boundary N'[a]$ slightly into $N[a]$. Then, $\boundary N'[a]$ must be a reducing sphere for $N[a]$ which is intersected exactly twice by $\ob{\alpha}$ and which bounds a non-trivial integer homology ball. If $\ob{\alpha}$ could be isotoped to intersect the sphere $\boundary N'[a]$ fewer times, it could be isotoped to be disjoint from that sphere and $N'$ would be reducible, contrary to the hypothesis that $(N,\gamma \cup a)$ is taut. Hence, conclusion (1) holds.

Suppose, therefore, that $r \geq 3$. Let $Q_n$ be the parameterizing surface in $N_n$ obtained from $Q$.  Since index does not increase during a hierarchy \cite[Definition 7.7]{S2}, the index of $Q_n$ is no more than the index of $Q$. No component of $Q_n$ is a sphere or a disc disjoint from $\gamma_n$, hence each component of $Q_n$ has non-negative index. Suppose that $\zeta$ is a component of $\boundary Q_n$ which crosses $a$ at least once. Let $A = \boundary N' - \inter{\eta}(a)$. If $\zeta \cap A$ contains an arc inessential in $A$ then either there is an isotopy of $Q$ reducing $|\boundary Q \cap \eta(a)|$ or an outermost such arc in $A$ bounds an $a$--boundary compressing disc $D$ for $Q$ in $N$. The former is forbidden by our hypothesis that $\boundary Q$ intersects $\eta(a)$ minimally and the latter is the second of our conclusions. Assume this doesn't happen. Then $\zeta$ is an essential loop in the torus $\boundary N'$ which intersects $\eta(a)$ minimally a positive number of times. Hence, $\zeta$ intersects all $r + 1$ sutures on $\boundary N'$.

Let $Q'$ be a component of $Q_n$ such that at least one component of $\boundary Q'$ intersects $\eta(a)$. Notice that $-2\chi(Q') \geq -2$. Let $z_{Q'} = |\boundary Q' \cap \eta(a)|$. Then $\boundary Q'$ has at least $z_{Q'}(r+1)$ intersections with the sutures $\gamma_n$. Hence,
\[
I(Q') \geq z_{Q'}(r+1) - 2\chi(Q') \geq z_{Q'}(r+1) - 2 \geq z_{Q'}(r-1).
\]

Then,
\[
I(Q_n) \geq \sum I(Q') \geq (r-1)\sum z_{Q'}
\]
where the sums are taken over all components $Q'$ of $Q$ which have at least one boundary component intersecting $\eta(a)$. By the construction of $Q_n$ from $Q$, we have that $\sum z_{Q'} = |\boundary Q \cap a|$. Thus,
\[
|\boundary Q \cap \gamma| + |\boundary Q \cap a| - 2\chi(Q) = I(Q) \geq I(Q_n) \geq (r-1)|\boundary Q \cap a|.
\]

Consequently,
\[
|\boundary Q \cap \gamma| - 2\chi(Q) \geq (r-2)|\boundary Q \cap a| \geq |\boundary Q \cap a|
\]
since $r \geq 3$.

Recalling that $Q$ is obtained from $\ob{Q}$ by removing $q_i$ discs with boundary parallel to $b_i$, that $|\boundary Q \cap \gamma| = \sum q_i \nu_i + \nu_\boundary$, that $|\boundary Q \cap a| = \sum q_i \Delta_i + \Delta_\boundary$, and that $\chi(Q) = \chi(\ob{Q}) - \sum q_i$ we obtain:
\[
\sum q_i \nu_i + \nu_\boundary - 2\chi(\ob{Q}) + 2\sum q_i \geq \sum q_i\Delta_i + \Delta_\boundary.
\]

It is easy to rearrange this to obtain
\[
-2\chi(\ob{Q}) \geq K(\ob{Q})
\]
as desired.\qed

\section{Sutures and $a$--boundary compressions}\label{Sutures}

For the main theorem to be useful, we must place sutures $\gamma$ on $\boundary N$ so that $(N,\gamma \cup a)$ is taut. We now describe how to do this, beginning with the sutures $\hat{\gamma}$ on $F$. For the remainder of the paper we require $F$ to have genus two.

If $a$ is separating, define $\hat{\gamma} = \nil$. If $a$ is non-separating, define $\hat{\gamma}$ to be a pair of essential disjoint simple closed curves on $F - \inter{\eta}(a)$ which separate $\boundary \eta(a)$. The next lemma shows how to define sutures $\tild{\gamma}$ on non-torus components of $\boundary N - F$. If $\boundary N - F$ consists of tori, then $\tild{\gamma} = \nil$. The next lemma also shows how to tell if $(N,\gamma \cup a)$ is taut, where $\gamma = \hat{\gamma} \cup \tild{\gamma}$.

\begin{lemma}[Lemma 4.1 of \cite{T1}]\label{Lem: Choosing Sutures}
Suppose that $F - (\gamma \cup a)$ is incompressible in $N$ and that $N$ is irreducible. Suppose also that if $\boundary N - F$ contains a non-torus component then there is no essential annulus in $N$ with boundary on $\hat{\gamma} \cup a$. Then $\tild{\gamma}$ can be chosen so that $(N,\gamma \cup a)$ is taut.  Furthermore, if $c \subset \boundary N - F$ is a collection of disjoint, non-parallel curves such that:
\begin{itemize}
\item $|c| \leq 2$
\item All components of $c$ are on the same component of $\boundary N - F$
\item No curve of $c$ cobounds an essential annulus in $N$ with a curve of $\hat{\gamma} \cup a$
\item If $|c| = 2$ then there is no essential annulus in $N$ with boundary $c$
\item If $|c| = 2$ and $a$ is separating, there is no essential thrice-punctured sphere in $N$ with boundary $c \cup a$.
\end{itemize}
then $\tild{\gamma}$ can be chosen to be disjoint from $c$.
\end{lemma}

We also need to understand the presence of surfaces $Q$. To that end, suppose that $b \subset F$ is a simple closed curve. If $b$ is non-separating, there are multiple ways to obtain a manifold homeomorphic to $N[b]$. Certainly attaching a 2--handle to $b$ is one such way. If $b^*$ is any curve in $F$ which cobounds in $F$ with $\boundary \eta(b)$ a thrice-punctured sphere, then attaching 2--handles to both $b^*$ and $b$ creates a manifold with a spherical boundary component. Filling in that sphere with a 3--ball creates a manifold homeomorphic to $N[b]$. We will often think of $N[b]$ as obtained in the fashion. Say that a surface $\ob{Q} \subset N[b]$ is \defn{suitably embedded} if each component of $\boundary Q - \boundary \ob{Q}$ is a curve parallel to $b$ or to some $b^*$. (Recall that $Q = \ob{Q} \cap N$.) We denote the number of components of $\boundary Q - \boundary \ob{Q}$ parallel to $b$ by $q = q(\ob{Q})$ and the number parallel to $b^*$ by $q^* = q^*(\ob{Q})$. If $b$ is separating, define $b^* = \nil$. Define $\Delta = |b \cap a|$, $\Delta^* = |b^* \cap a|$, $\nu = |b \cap \gamma|$, and $\nu^* = |b^* \cap \gamma|$. Also define $\tild{q} = q + q^*$, $\boundary_1 N[b] = \boundary N - F$, and $\boundary_0 N[b] = \boundary N[b] - \boundary_1 N[b]$. A surface in $N[b]$ is \defn{essential} if it is incompressible, boundary-incompressible, and no component is boundary-parallel, a sphere bounding a ball, or a disc with inessential boundary.

Letting $\mc{B} = \{b,b^*\}$ we have:
\[
K(\ob{Q}) = (\Delta - \nu - 2)q + (\Delta^* - \nu^* - 2)q^* + \Delta_\boundary - \nu_\boundary.
\]

To get the most use out of the main theorem, we would like to know that there are useful $a$--boundary incompressible surfaces. Theorem 5.1 of \cite{T1} allows us to find some. We state here the version of the theorem which will be useful for our applications.

\begin{theorem}\label{Thm: Constructing Q}
\begin{enumerate}
\item[]
\item[(I)] Suppose that $N[b]$ contains a non-empty collection $\ob{R}$ of essential spheres and discs. Then there is a suitably embedded essential disc or sphere $\ob{Q} \subset N[b]$ such that $\tild{q}(\ob{Q}) \leq \tild{q}(\ob{R})$ and
\begin{itemize}
\item If $\ob{R}$ is a sphere or disc with boundary on $\boundary_0 N[b]$ then $\ob{Q}$ is a sphere or disc with boundary on $\boundary_0 N[b]$.
\item There is no $a$--boundary compressing disc for $Q = \ob{Q} \cap N$.
\end{itemize}
\item[(II)] Suppose that $N[b]$ does not contain an essential disc or sphere and that $\ob{R} \subset N[b]$ is an essential surface. Then there is a suitably embedded essential connected surface $\ob{Q} \subset N[b]$ such that the following hold:
\begin{itemize}
\item $-\chi(\ob{Q}) \leq -\chi(\ob{R})$ and the genus of $\ob{Q}$ is no more than the total of the genera of the components of $\ob{R}$.
\item $\tild{q}(\ob{Q}) \leq \tild{q}(\ob{R})$
\item If $\ob{R}$ is non-separating, so is $\ob{Q}$.
\item Each curve of $\boundary \ob{Q} \cap \boundary_1 N[b]$ is isotopic to a curve of $\boundary \ob{R} \cap \boundary_1 N[b]$.
\item Either $\ob{Q}$ is contained in $N$ or there is no $a$--boundary compressing disc for $Q$.
\end{itemize}
\end{enumerate}
\end{theorem}
\begin{proof}
This easily follows from Theorem 5.1 and Corollary 5.2 of \cite{T1}. The only point that is not immediately obvious is how to guarantee that $\tild{q}(\ob{Q}) \leq \tild{q}(\ob{R})$ in case (II). The statement of the theorem and corollary only guarantee that $(-\chi(\ob{Q}),\tild{q}(\ob{Q})) \leq (-\chi(\ob{R}), \tild{q}(\ob{R}))$ in lexicographic order. To achieve the stronger statement here, notice that under the operations described in the proof of \cite[Theorem 5.1]{T1} the only time $\tild{q}$ increases is when (B1) of that theorem is invoked or when $a$--torsion $2g$--gons (with $g \geq 2$) are being eliminated. For the theorem here do not invoke (B1) and do not attempt to eliminate $a$--torsion $2g$--gons with $g \geq 2$.
\end{proof}

It is worth noting that the surface $\ob{Q}$ is created from $\ob{R}$ by boundary compressing $R = \ob{R} \cap N$ using $a$--boundary compressing discs. this observation will be useful later. We now turn to applications.

\section{2--Handle Addition}\label{Section: 2-HandleAddition}
Our first application is to the study of degenerating handle additions to the boundary of a simple manifold. We will assume that $F \subset \boundary N$ has genus 2 and that both $a$ and $b$ are separating. This means that $\hat{\gamma} = \gamma \cap F$ and $b^*$ can both be taken to be the empty set.

\begin{theorem}\label{Thm: hyperbolic}
Suppose that $F$ has genus 2, and that $N$ is simple. Suppose that $a$ and $b$ are non-isotopic separating curves on $F$. Suppose that $N[a]$ is reducible. Then if $N[b]$ is non-simple, it contains an essential annulus with boundary on non-torus components of $\boundary N[b]$ and $\Delta = 4$.
\end{theorem}

\begin{proof}
Notice, first, that since $a$ and $b$ are separating, $\Delta$ is even and positive. If $\Delta$ were equal to two then $a - b$ would have a single arc on each once-punctured torus component of $F - b$, implying that $a$ was non-separating. Thus, $\Delta \geq 4$.

Now suppose that $N[b]$ contains a surface $\ob{R}$ which is an essential sphere, disc, torus, or annulus. Let $\ob{Q}$ be the surface obtained by applying Theorem \ref{Thm: Constructing Q} to $\ob{R}$. The surface $\ob{Q}$ is still an essential sphere, disc, annulus, or torus. It is not disjoint from $a$ since $N$ is simple and $\Delta > 0$. Furthermore, there is no $a$--boundary compressing disc for $Q$. We may assume that out of all such $\ob{Q}$, $q$ has been minimized. If $\ob{Q}$ is an annulus, suppose for the time being that $\boundary \ob{Q}$ does not have all its boundary components on non-torus components of $\boundary N - F$. Since $a$ and $b$ are separating, $\hat{\gamma} = \nil$ and $b^* = \nil$.

If $\ob{Q}$ has a boundary component on a non-torus component of $\boundary N - F$, let $c$ be that component of $\boundary \ob{Q}$. Since $N$ is simple, $c$ satisfies the requirements for an application of Lemma \ref{Lem: Choosing Sutures}. Let $\gamma$ be the sutures provided by that lemma.

Notice that $q > 0$ since $N$ is simple. We may now apply the Main Theorem. Since $N$ does not contain an essential annulus, conclusion (1) does not occur. By the construction of $\ob{Q}$, there is no $a$--boundary compressing disc for $Q$. Thus,
\[
(\Delta - 2) q + \Delta_\boundary \leq -2\chi(\ob{Q}).
\]
Hence,
\[
\Delta \leq 2 + (-\Delta_\boundary + -2\chi(\ob{Q}))/q \leq 2 - \Delta_\boundary/q.
\]

Since $\Delta_\boundary$ is non-negative, we have $\Delta = 2$. This contradicts our initial observation that $\Delta \geq 4$.

We may, therefore, assume that $\ob{Q}$ is an annulus with both boundary components on non-torus components of $\boundary N - F$. Let $G$ be the components of $\boundary N - F$ containing $\boundary \ob{Q}$. Let $N'$ be the manifold obtained by doubling $N$ along $G$. That is, $N'$ is formed by gluing a copy $N_2$ of $N$ to $N_1 = N$ along $G$. Let $F_i$, $a_i$, $b_i$, $Q_i$ be the copy of $F$, $a$, $b$, and $Q$ lying in $N_i$. The gluing should be performed so that $\ob{Q}' = \ob{Q} \cup Q_2$ is a punctured torus in $N'[b_1]$ with punctures on $F_2$ parallel to $b_2$. It is easy to show that $N'$ is simple. Notice that $N'[a_1]$ is reducible.

Let $Q' = \ob{Q'} \cap N'$. Suppose that $D$ is an $a_1$--boundary compressing disc for $Q'$ with $\epsilon = \boundary D \cap F_1$. Since $N_1$ and $N_2$ are simple, we may assume that $D \cap G$ consists of arcs which are essential in $Q'$. Since there is no $a$--boundary compressing disc for $Q$ in $N$, this collection of arcs is non-empty. Since $G$ is disjoint from $F_1$, there is some arc of $D \cap G$ which is outermost on $D$ and does not contain $\epsilon$ in the outermost disc it bounds. Let $E$ be the outermost disc bounded by that arc. Then $E$ is a boundary compressing disc for $Q_1$ or $Q_2$. Without loss of generality, suppose it to be $Q_1$. Since $\ob{Q}$ is essential in $N_1[b_1]$, the arc $\boundary E \cap \ob{Q}$ must be inessential in $\ob{Q}$. Cutting $\ob{Q}$ along $\boundary E$ produces a surface with an annulus component and a disc component. Since $N_1[b_1]$ contains no essential discs, the disc component must be inessential. But this implies that there is an isotopy of $\ob{Q}$ reducing $q$, contradicting our choice of $\ob{Q}$. Hence, there is no $a_1$--boundary compressing disc for $Q'$.

Let $c = b_2$ and apply Lemma \ref{Lem: Choosing Sutures} to construct sutures $\gamma$ on $\boundary N'$ which are disjoint from $c$ so that $(N',\gamma \cup a_1)$ is a taut sutured manifold. Since all boundary components of $\ob{Q}'$ are parallel to $b_2$, $\Delta_\boundary = \nu_\boundary = 0$. Also, $-2\chi(\ob{Q}') = 2q$ since $\ob{Q}'$ is a punctured torus with $q$ boundary components. If $-2\chi(\ob{Q}) < K(\ob{Q})$ then the main theorem shows that $N'[a_1]$ is irreducible, a contradiction. Hence, $-2\chi(\ob{Q}) \geq K(\ob{Q})$. Thus,
\[
2q \geq q(\Delta - 2).
\]

Solving for $\Delta$, we observe $\Delta \leq 4$. Since $\Delta \geq 4$, we conclude $\Delta = 4$.
\end{proof}

\begin{remark}
A separating curve in $\boundary N$ is an example of what Scharlemann and Wu \cite{SW} call a \defn{basic} curve. They prove that if $N$ is simple and one of  $a$ and $b$ is basic, then if $N[a]$ is reducible and $N[b]$ is boundary-reducible then $a$ and $b$ can be isotoped to be disjoint. They conjecture that if both $a$ and $b$ are basic and neither $N[a]$ nor $N[b]$ is simple then $\Delta \leq 5$. Theorem \ref{Thm: hyperbolic} gives some evidence for their conjecture.
\end{remark}

\section{Refilling Meridians}
The remaining applications concern the following situation. Suppose that $W$ is a genus two handlebody embedded in a compact orientable 3--manifold $M$ and take $N = M - \inter{W}$. Assume that $N$ is irreducible. Suppose that $a$ and $b$ are essential simple closed curves in $F = \boundary W$ bounding discs $\alpha$ and $\beta$ in $W$ intersecting minimally and non-trivially. Notice that each $b^*$ curve will also bound a disc $\beta^*$ in $W$. We say that $M[\alpha] = N[a]$ is obtained by \defn{refilling} the meridian $\alpha$ \cite{S3}. The solid torus or tori $W - \inter{\eta}(\alpha)$ is the regular neighborhood of a knot or 2-component link $L_\alpha$ in $M$. We say that $L_\alpha$ is obtained by \defn{boring} $L_\beta$ (and vice versa). $W$ will occasionally be called the \defn{boring handlebody}.

We will need an understanding of how $a$ and $b$ lie on $F$. The assertions of this paragraph are easy to prove. More detail can be found in \cite{S3} and \cite{T1}. Each arc of $\alpha \cap \beta$ which is outermost on $\beta$ bounds a subdisc of $\beta$ which is a meridian of a solid torus component of $W - \inter{\eta}(\alpha)$. The arc lying in $b$ which forms part of the boundary of such a subdisc of $\beta$ has both endpoints lying on the same component of $\boundary \eta(a) \subset F$. Call an arc in $b$ lying in $F - \inter{\eta}(a)$ which, together with an arc in $\alpha$, bounds a meridian disc of $W - \inter{\eta}(\alpha)$ a \defn{meridional arc} of $b - a$. Let $\mc{M}_a(b)$ denote the total number of meridional arcs of $b-a$. If $a$ is non-separating then any arc of $b - \inter{\eta}(a)$ with endpoints on the same component of $\boundary \eta(a)$ is a meridional arc. In this case, there are the same number of meridional arcs based at each component of $\boundary \eta(a)$. Thus, when $a$ is non-separating, $\mc{M}_a(b)$ is even.

Notice that if a surface $\ob{Q} \subset N[b]$ has boundary disjoint from $a$ then its boundary is disjoint from the meridional arcs of $a - b$. Suppose that $\ob{Q}$ has a boundary component on $\boundary N[b] - \boundary M$ which is not a meridian. If $b$ is separating, this means that all of the meridional arcs of $a - b$ lie on a single component of $\boundary N[b]$ and the non-meridional boundary of $\ob{Q}$ is disjoint from that component of $\boundary N[b]$. Similarly, if $b$ is non-separating, the non-meridional boundary component of $\ob{Q}$ must be disjoint from $\boundary N[b] - \boundary M$.

We desire to choose sutures so that $(N,\gamma \cup a)$ is a taut sutured manifold. We will, of course, use Lemma \ref{Lem: Choosing Sutures}, but we also need to choose the sutures $\hat{\gamma}$ more carefully than we did previously. If $\alpha$ is separating, choose $\gamma = \nil$. Otherwise, in the boundary of the solid torus $W - \inter{\eta}(a)$ choose two meridians $\gamma$, separating the components of $\boundary \eta(a)$, intersecting $b$ minimally, and disjoint from the meridional arcs of $b - a$. Also, if $\alpha$ is non-separating, $\Delta - \nu = \mc{M}_a(b) \geq 2$ and $\Delta^* - \nu^* \geq \mc{M}_a(b^*) \geq 2$ (if $b^* \neq \nil$). Since $\boundary \ob{Q}$ is disjoint from $b$, in particular from the meridional arcs of $b - a$, \[
|\boundary \ob{Q} \cap a| - |\boundary \ob{Q} \cap \hat{\gamma}| \geq 0.
\]
Hence, if $\boundary \ob{Q}$ is disjoint from $\gamma \cap \boundary M$, $K(\ob{Q})$ is non-negative. We will also need to consider the cocores $\ob{\alpha}$ and $\ob{\beta}$ of the 2--handles $\eta(\alpha)$ and $\eta(\beta)$ respectively.

Before proceeding to more interesting results, we need to know that there are taut conditioned Seifert surfaces. The next lemma is a hands-on version of Lemma \ref{Lem: Conditioned Surfaces}.

\begin{lemma}\label{Lem: Taut Conditioned SS}
Suppose that $L_\alpha$ is null-homologous in $M$. Then there is a Seifert surface $S$ for $L_\alpha$ which is disjoint from $\ob{\alpha}$ (i.e. lies in $N$) and is a taut conditioned surface in $N$ with boundary disjoint from $a$.
\end{lemma}
\begin{proof}
First we show that $L_\alpha$ does contain a conditioned Seifert surface disjoint from $\ob{\alpha}$. Choose a Seifert surface $\Sigma_0 \subset N[a]$ for $L_\alpha$. Since $L_\alpha$ is null-homologous in $M$ such a surface exists. If $L_\alpha$ is a link, $\Sigma_0$ may not be connected. Since $\boundary \Sigma_0$ is longitudinal on $\boundary N[a]$, we may assume (when $\alpha$ is non-separating) that it intersects $\gamma$ exactly twice.

Calculate the algebraic intersection number between $\ob{\alpha}$ and each component of $\Sigma_0$. If it is $n \neq 0$, an endpoint of $\ob{\alpha}$ may be isotoped around $\boundary S^3[\alpha]$ creating $n$ intersections of sign $-n/|n|$. Perform the isotopy so that $\boundary \ob{\alpha}$ is always disjoint from $\gamma$. Rather than isotoping $\ob{\alpha}$, we may instead isotope $\Sigma_0$. We take this latter viewpoint. The requirement, from the former viewpoint, that $\boundary \ob{\alpha}$ be disjoint from $\gamma$ guarantees that, from the latter viewpoint, if $\alpha$ is non-separating then $\boundary \Sigma_0$ still intersects each component of $\gamma$ exactly once.

We may, therefore, assume that the intersection number of $\ob{\alpha}$ with each component of $\Sigma_0$ is zero. Choosing an arc $\sigma$ of $\ob{\alpha} - \Sigma_0$ with endpoints creating intersections of opposite sign on the same component of $\Sigma_0$, we attach a tube containing $\sigma$ to $\Sigma_0$, decreasing $|\Sigma_0 \cap \ob{\alpha}|$ (but increasing the genus of $\Sigma_0$). The algebraic intersection number of $\ob{\alpha}$ and $\Sigma_0$ is still zero. Continuing in this manner, we may construct a conditioned Seifert surface $\Sigma$ for $L_\alpha$ which is disjoint from $\ob{\alpha}$. Out of all Seifert surfaces for $L_\alpha$ which are disjoint from $\ob{\alpha}$ and which have boundary $\boundary \Sigma$ choose one of minimal genus and call it $S$. Then $S$ is a taut conditioned surface in $N$.
\end{proof}

\begin{remark}
Notice that even though $\boundary S$ (where $S$ is the surface created by the previous lemma) is a longitude on $\boundary_0 N[a]$ (when $\alpha$ is separating) it may intersect meridional arcs of $b - a$ more than once. It must, however, intersect them at least once. See Figure \ref{Fig: SpiralMeridians}, for a depiction of the ``spiralling $\boundary \ob{\alpha}$'' viewpoint.
\end{remark}

\begin{figure}[ht]
\labellist
\hair 2pt
\pinlabel $\gamma$ [br] at 32 197
\pinlabel {\tiny $\boundary \eta(a)$} at 435 65
\pinlabel {\tiny $\boundary \eta(a)$} at 435 284
\pinlabel $\gamma$ [t] at 403 135
\pinlabel $\boundary S$ [b] at 220 245
\endlabellist
\center
\includegraphics[scale=0.4]{Figures/SpiralMeridians}

\caption{The result of spiralling $\boundary \ob{\alpha}$ around $\boundary_0 N[a]$}
\label{Fig: SpiralMeridians}
\end{figure}

\subsection{Scharlemann's Conjecture}
In \cite{S3}, Scharlemann considered refilling meridians of a genus two handlebody $W$ in a wide variety of compact, orientable 3--manifolds $M$. He showed in a number of situations that if neither $M[\alpha]$ or $M[\beta]$ is irreducible and boundary-irreducible (with minor assumptions on the embedding of $W$ in $M$) then either $W \subset S^3$ is unknotted or $\alpha$ and $\beta$ are ``aligned'' in $W$. He conjectured that this is always the case (given his hypotheses on the pair $(M,W)$). In \cite{T1}, under slightly different hypotheses, his conjecture was mostly completed. The only interesting case left was when $M = S^3$, $\boundary W$ is incompressible in $N$ and both $S^3[\alpha]$ and $S^3[\beta]$ are solid tori. Here is a fairly general theorem which completes the last interesting case.

\begin{theorem}\label{Thm: MSC 2}
Suppose that any two curves of $\boundary M$ which compress in $M$ are on the same component of $\boundary M$. Suppose that $W$ is a genus two handlebody embedded in $M$ such that $W$ intersects every essential sphere in $M$ at least three times and every essential disc at least two times. Suppose also that $N = M - \inter{W}$ is irreducible. Let $\alpha$ and $\beta$ be essential discs in $W$ which cannot be isotoped to be disjoint. Assume that $M[\alpha]$ and $M[\beta]$ contain no essential disc which is contained in $N$ and that $\boundary \alpha$ and $\boundary \beta$ do not compress in $N$. Then the following hold:
\begin{itemize}
\item One of $M[\alpha]$ and $M[\beta]$ is irreducible and is not a solid torus
\item If one of them is reducible the other is boundary-irreducible
\item If $c_a \subset \boundary M$ is a curve which compresses in $M[\alpha]$ and if $c_b \subset \boundary M$ is a curve which compresses in $M[\beta]$ then $c_a$ and $c_b$ cannot be isotoped in $\boundary M$ to be disjoint.
\end{itemize}
\end{theorem}

\begin{proof}
Without loss of generality, assume that $M[\beta]$ is reducible or boundary-reducible and let $\ob{Q}$ be an essential sphere or disc obtained by applying Theorem \ref{Thm: Constructing Q}. If $\boundary \ob{Q}$ is on $\boundary M$ then we may assume that $\boundary \ob{Q} = c_b$. Let $T = T(\gamma)$ be the torus components of $\boundary M$.

We need to place sutures on $\boundary M$.  We will use Lemma \ref{Lem: Choosing Sutures} and we start by defining curves $c$ for use in that lemma. If $\ob{Q}$ is a sphere or disc with boundary on $\boundary_0 M[\beta] \cup T$, define $c_\beta = \nil$. Otherwise, let $c_\beta = c_b$. If no curve of $\boundary M$ disjoint from $c_b$ compresses in $M[\alpha]$, then let $c_\alpha = \nil$. If $c_b$ compresses in $M[\alpha]$, let $c_\alpha = c_b$. If $c_b$ does not compress in $M[\alpha]$ but a curve $c_a$ disjoint from $c_b$ does, let $c_\alpha = c_a$. Define $c = c_\alpha \cup c_\beta$ and notice that if $|c| = 2$, there is no essential annulus in $N$ with boundary equal to $c$. Also, if a component of $c$ bounds an essential annulus with a curve of $\hat{\gamma} \cup a$ then, because, the components of $\hat{\gamma} \cup a$ bound discs in $W$, $W$ would intersect a compressing disc for $\boundary M$ exactly once. This is forbidden by our hypotheses. Furthermore, if $|c| = 2$ then one component, $c_\alpha$, of $c$ bounds a disc in $M[\alpha]$ and the other component $c_\beta = c_b$ does not. If $a$ is separating, $|c| = 2$, and $c \cup a$ bounds an essential thrice-punctured sphere in $M[\alpha]$, then attaching discs to $c_\alpha$ and to $a$ shows that $c_\beta = c_b = c - c_\alpha$ compresses in $M[\alpha]$, but this contradicts the construction of $c_\alpha$. Thus $c$ satisfies the criteria for an application of Lemma \ref{Lem: Choosing Sutures}. Let $\gamma = \tild{\gamma} \cup \hat{\gamma}$ be the sutures on $\boundary M[\alpha]$ provided by that Lemma.

If $M[\alpha]$ is reducible or if $c_\alpha \neq \nil$ then $(M[\alpha],\gamma)$ is not taut. If $M[\alpha]$ is a solid torus, then, by our hypotheses, every taut conditioned surface with boundary on $\boundary_0 M[\alpha]$, of which there is one (Lemma \ref{Lem: Taut Conditioned SS}), intersects $\ob{\alpha}$. We can, therefore, apply the main theorem. Since $\ob{Q}$ is a disc or sphere, $-2\chi(\ob{Q}) < 0 \leq K(\ob{Q})$. By the construction of $\ob{Q}$, there is no $a$--boundary compressing disc for $Q$ in $N = M - \inter{W}$. Thus, $M[\alpha]$ contains an essential separating sphere $S$ intersecting $\ob{\alpha}$ twice and which cannot be isotoped to intersect $\ob{\alpha}$ fewer times. Furthermore, this sphere bounds a non-trivial homology ball in $M[\alpha]$, and therefore in $M$.

Because $W$ intersects every essential sphere in $M$ at least three times, $S$ cannot be an essential sphere for $M$. Let $B$ be the ball in $M$ which $S$ bounds. Notice that this implies that $M$ is a non-trivial homology sphere. Since $S$ is separating and $B$ is not contained in $M[\alpha]$, $(\boundary M[\alpha] - \boundary M) \subset B$. Attaching $\eta(\ob{\alpha})$ to $B$ produces a solid torus $V$ containing $W$, with $\boundary V$ compressible in $V[\alpha]$ and $V - \inter{W}$ irreducible. Since $V$ does not contain a non-trivial homology sphere connected summand, we may apply the work of the previous paragraphs to conclude that $V[\beta]$ is irreducible and that $\boundary V[\beta]$ does not compress in $V[\beta]$. Thus, $\ob{Q}$ intersects $\boundary V$ and (perhaps after an isotopy to eliminate inessential curves of intersection) an innermost disc of intersection $D$ on $\ob{Q}$ is a compressing disc for $\boundary V$ contained outside $V$. If $\boundary D$ intersected a meridian curve on $\boundary V$ exactly once, $\boundary D$ would run exactly once along a regular neighborhood of $\ob{\alpha}$. $D$ then guides an isotopy of $\ob{\alpha}$ into $B$, contradicting the construction of $S$. If $\boundary D$ is a meridional curve of $\boundary V$, then $W$ is contained in an $S^1 \times S^2$ summand of $M$. If $\boundary D$ intersects every meridional curve of $\boundary V$ more than once then $W$ is contained in a lens space connected summand of $M$. By hypothesis, $W$ intersects every reducing sphere in $M$, so $M$ is $S^1 \times S^2$ or a lens space. Both possibilities contradict our previous conclusion that $M$ was a homology sphere. Hence, $M[\alpha]$ is irreducible, the torus components of $\boundary M[\alpha]$ are incompressible in $M[\alpha]$, and $c_\alpha = \nil$.
\end{proof}

\subsection{Essential surfaces and boring knots and links}
For the remainder, we restrict to $M = S^3$ for simplicity. For knots or links $L_\alpha$ and $L_\beta$ which are related by boring, we desire to understand essential surfaces in the exterior of one by placing hypotheses on the other and on the embedding of $W$ in $S^3$.

\begin{theorem}\label{Thm: Disjt SS}
Suppose that $L_\alpha$ is a knot or link in $S^3$ obtained by boring a knot or link $L_\beta$ using handlebody $W$. If $\alpha$ is separating, assume that $\boundary W - \boundary \alpha$ is incompressible in $N$. If $L_\beta$ is a split link or unknot, then $L_\alpha$ is not a split link and there is a minimal genus Seifert surface for $L_\alpha$ which is disjoint from $\ob{\alpha}$.
\end{theorem}

\begin{proof}
If $\alpha$ is non-separating, $\gamma \neq \nil$. Since $\boundary W - (a \cup \gamma)$ consists of two thrice-punctured spheres each with meridional boundary, it is incompressible in $N$. If $\alpha$ is separating, by hypothesis $\boundary W - \boundary \alpha$ is incompressible in $N$. Thus, in either case, by Lemma \ref{Lem: Choosing Sutures}, $(N,\gamma \cup a)$ is taut. Let $\ob{R}$ be an essential disc or sphere in $S^3[\beta]$ and let $\ob{Q}$ be the disc or sphere provided by Theorem \ref{Thm: Constructing Q}. If $\tild{q} = 0$ then, since $N$ is irreducible, $\ob{Q}$ must be a disc. Discs are boundary\hyph incompressible and so in no case is there an $a$--boundary compressing disc for $Q$. $\ob{Q}$ cannot be disjoint from $a$, since either $\tild{q} > 0$ or $\boundary \ob{Q}$ intersects the meridional arcs of $a - b$. Hence, $-2\chi(\ob{Q}) < 0 \leq K(\ob{Q})$.

By the main theorem and Lemma \ref{Lem: Taut Conditioned SS}, $S^3[\alpha]$ is irreducible and $L_\alpha$ has a minimal genus Seifert surface disjoint from $\ob{\alpha}$ (that is, contained in $N$).
\end{proof}

\begin{corollary}[{\cite[Proposition 4.2]{ST2}}]
If $\ob{\alpha}$ is a tunnel for a tunnel number one knot or link $L_\alpha$, $L_\alpha$ has a minimal genus Seifert surface disjoint from $\ob{\alpha}$.
\end{corollary}

\begin{proof}
Every tunnel number one knot or link can be obtained by boring an unknot $L_\beta$ using the standard unknotted genus two handlebody in $S^3$. A tunnel for a non-trivial tunnel number one knot or link is a boring arc for converting the knot or link into the unknot $L_\beta$. Thus, unless $\alpha$ is separating and $\boundary W - \boundary \alpha$ is compressible in $N$, the corollary follows immediately from Theorem \ref{Thm: Disjt SS}.

We may, therefore, assume that $L_\alpha$ is a split link. The surface $\boundary W$ is a genus two Heegaard surface for $S^3[\alpha]$. If $L_\alpha$ is a split link, $S^3[\alpha]$ contains an essential sphere, so by Haken's Lemma for Heegaard splittings there is an essential sphere $P$ intersecting the Heegaard surface in a single loop. One side of the Heegaard surface is a compressionbody with two boundary components, each a torus. Thus, $P$ must intersect that compressionbody in the unique (up to isotopy) essential disc. That disc is parallel to $\alpha$. $\boundary W - \inter{\eta}(P)$ has two components each of which is a genus one Heegaard splitting for the exterior of a component of $L_\alpha$. The only knot with a genus one Heegaard splitting for its exterior is the unknot and so $L_\alpha$ is the unlink of two components. Since the connected sum of Heegaard splittings is well-defined $\ob{\alpha} \cap (S^3 - \inter{\eta}(P))$ consists of two unknotted arcs. Thus, each component of $L_\alpha$ bounds a disc disjoint from $\ob{\alpha}$ and the corollary is proved when $L_\alpha$ is a split link.
\end{proof}

\begin{remark}
The proof of the previous corollary is not any better than Scharlemann and Thompson's proof. Indeed, their proof is certainly easier to understand than the arguments of this paper. However, it is interesting to note that they do rely on a theorem of Gabai which was proved using sutured manifold theory. The point of Theorem \ref{Thm: Disjt SS} is that a rather significant property of tunnel number one knots has a natural generalization to knots and links obtained by boring an unknot. Indeed, this theorem also generalizes the fact (due to Gabai and Scharlemann) that genus is superadditive under band sum (see Corollary \ref{Cor: Band Sum}).
\end{remark}

Using Theorem \ref{Thm: Disjt SS}, we can reverse the roles of $\alpha$ and $\beta$ to obtain:

\begin{theorem}\label{Thm: Boring Genus}
Suppose that $L_\beta \subset S^3$ is obtained by boring a split link or unknot $L_\alpha$. If $L_\alpha$ is a split link, assume that $\boundary W - \boundary \alpha$ is incompressible in $N$. If $L_\alpha$ is an unknot, assume that there does not exist an essential disc in $S^3[\alpha]$ disjoint from $\ob{\alpha}$. Then $L_\beta$ is not a split link or unknot and $L_\beta$ has a minimal genus Seifert surface $\ob{Q}$ properly embedded in $S^3[\beta]$, which is disjoint from $\ob{\beta}$ and for which one of the following is true:
\begin{itemize}
\item $-2 \chi(\ob{Q}) \geq \Delta_\boundary - \nu_\boundary$
\item There is an $a$--boundary compressing disc for $\ob{Q}$ in $N$
\end{itemize}
\end{theorem}
\begin{remark}
Corollary \ref{Cor: RT Genus} rephrases this theorem for rational tangle replacements. Following that theorem, there is an example which shows that the possibility that there is an $a$--boundary compressing disc for $\ob{Q}$ cannot be eliminated. Notice that if $\ob{\beta}$ is isotopic to a non-trivial arc in $\ob{Q}$ then there is an $a$--boundary compressing disc for $\ob{Q}$ in $S^3$.
\end{remark}

\begin{proof}
If $L_\beta$ is a split link or unknot then by Theorem \ref{Thm: Disjt SS}, $L_\alpha$ would not be a split link and would have a minimal genus Seifert surface which was disjoint from $\ob{\alpha}$. These conclusions contradict the assumption that $L_\alpha$ is a split link or is an unknot without an essential disc disjoint from $\ob{\alpha}$. Hence, $L_\beta$ is not a split link or unknot. If $\beta$ is separating and $\boundary W - b$ is compressible in $N$ then it is easy to see that $b$ compresses in $N$. This, however, implies that $L_\beta$ is a split link, a contradiction. Hence, if $\beta$ is separating $\boundary W - b$ is incompressible in $N$.

By Theorem \ref{Thm: Disjt SS}, applied with $\alpha$ and $\beta$ reversed, there is a minimal genus Seifert surface $\ob{Q}$ for $L_\beta$ which is disjoint from $\ob{\beta}$; that is, it is contained in $N$. The only way in which $\ob{Q}$ could be disjoint from the meridional arcs of $a - b$ is if $\beta$ were separating and $\ob{Q}$ had boundary on a single component of $L_\beta$. This contradicts the definition of Seifert surface for $L_\beta$, so $\ob{Q}$ is not disjoint from $\eta(a)$.

If there is an $a$--boundary compressing disc for $\ob{Q}$ in $N$, we are done, so suppose that no such disc exists. If $-2\chi(\ob{Q}) < K(\ob{Q})$ the first sutured manifold theorem and Lemma \ref{Lem: Taut Conditioned SS} imply that $S^3[\alpha]$ is irreducible and that there is a minimal genus Seifert surface for $L_\alpha$ which is disjoint from $\ob{\alpha}$. The first option means that $L_\alpha$ isn't a split link and the second that $L_\alpha$ isn't an unknot with no unknotting disc contained in $N$. Hence, $-2\chi(\ob{Q}) \geq K(\ob{Q})$. Since $\ob{Q}$ is disjoint from $\ob{\beta}$, $q = q^* = 0$. The given inequality follows from the definition of $K(\ob{Q})$.
\end{proof}

With the stronger assumption that $\boundary W$ is incompressible in $N$, we can restrict the possibilities for obtaining a non-hyperbolic knot or link from a split link by boring.

\begin{theorem}\label{Thm: Essential Torus}
Suppose that $L_\beta$ is a knot or link obtained by boring the link $L_\alpha$ using a handlebody $W \subset S^3$ with $N = S^3 - \inter{W}$ boundary-irreducible. Suppose that $L_\alpha$ is a split link or that there is no minimal genus Seifert surface for $L_\alpha$ disjoint from $\ob{\alpha}$. If the exterior of $L_\beta$ contains an essential annulus or torus then one of the following holds:
\begin{enumerate}
\item $\Delta = 2$ and if there is an essential annulus then there is one which is either disjoint from or has meridional boundary on some component of $L_\beta$.
\item There is an essential annulus in the exterior of $L_\beta$ disjoint from $\ob{\beta}$ and which is either disjoint from or has meridional boundary on some component of $L_\beta$.
\item There is an essential torus in $N$.
\end{enumerate}
\end{theorem}
\begin{example}
Figure \ref{Fig: BandSum} shows that a composite knot can be obtained from a split link by a band sum. Thickening the band and the unknot gives us $W$, and the exterior of $W$ is boundary-irreducible. This shows that the first case can arise.

\begin{figure}[ht]
\center
\includegraphics[scale=0.4]{Figures/BandSum}

\caption{A band sum creating the granny knot}
\label{Fig: BandSum}
\end{figure}

Both versions of the second conclusion are possible. Figure \ref{Fig: MeridianAnnulus} shows a spine for a genus two handlebody. The ``S''-shaped arc is disjoint from an essential meridional annulus $A$. Refilling the meridian of that arc creates a split link with one component a trefoil and the other component an unknot. It is not hard to show that the exterior of the handlebody is boundary-irreducible. Using the ``S'' shaped arc to perform a band-sum creates a knot $L_\beta$ which is the connected sum of a trefoil and a $6_1$ knot. (Jiho Kim's KnotSketcher and Charles Livingston's KnotFinder helped with this calculation.)

\begin{figure}[ht]
\center
\includegraphics[scale=0.4]{Figures/MeridianAnnulus}

\caption{Performing a rational tangle replacement on the ``S'' shaped arc leaves the meridional annulus untouched.}
\label{Fig: MeridianAnnulus}
\end{figure}

Figure \ref{Fig: DisjointAnnulus} shows a split link $L_\alpha$ consisting of a trefoil (drawn so the ``cabling'' annulus is visible) and an unknot. There is an ``S'' shaped arc joining them. On the trefoil the annulus has boundary slope $\pm 6$. Use the ``S''-shaped arc to perform a Kirby band move of the unknot over the trefoil (giving the trefoil a framing of $\pm 6$). We now have a new link $L_\beta$ with one component the trefoil. By construction the cabling annulus for the trefoil persists into $L_\beta$. It is not difficult to show that the band move can be obtained by a boring operation with the exterior of the boring handlebody boundary-irreducible.

It is easy to use a ``satellite construction'' to concoct an example of the last possibility. Figure \ref{Fig: EssentialTorus} shows a spine for a genus two handlebody $W$ inside a knotted solid torus $\boundary V$. Cutting the edge of the spine containing the local trefoil produces the unlink $L_\alpha$ in $S^3$. By Theorem \ref{Thm: MSC 2}, $\boundary V$ remains essential in the exterior of any knot or link $L_\beta$ obtained from $L_\alpha$ by boring using $W$. It is easy to show that $\boundary W$ is incompressible in both $V - \inter{W}$ and $S^3 - \inter{W}$.
\end{example}

\begin{figure}[ht]
\center
\includegraphics[scale=0.4]{Figures/DisjointAnnulus}

\caption{Performing a Kirby band move using the ``S'' shaped arc leaves the trefoil's essential annulus untouched.}
\label{Fig: DisjointAnnulus}
\end{figure}
\begin{figure}[ht]
\center
\includegraphics[scale=0.4]{Figures/EssentialTorus}

\caption{An essential torus in the exterior of $W$.}
\label{Fig: EssentialTorus}
\end{figure}

\begin{proof}[Proof of Theorem \ref{Thm: Essential Torus}]
Suppose that there is no essential torus in $N$. Theorem \ref{Thm: Disjt SS} shows that $L_\beta$ is not an unknot or split link; consequently, there is no essential disc or sphere in $S^3[\beta]$. Let $\ob{R}$ be an essential annulus or torus in $S^3[\beta]$ and apply Theorem \ref{Thm: Constructing Q}, obtaining a connected surface $\ob{Q}$. Since $\ob{Q}$ is not a sphere or disc and since $-\chi(\ob{Q}) \leq - \chi(\ob{R})$, $\ob{Q}$ is an annulus or torus. Since the genus of $\ob{Q}$ is no higher than the genus of $\ob{R}$, if $\ob{R}$ was an annulus, then $\ob{Q}$ is an annulus. If $\ob{Q}$ is disjoint from $\ob{\beta}$ then it is contained in $N$ and must be an annulus by our initial assumption that $N$ contains no essential torus. In this case, if there is an $a$--boundary compression for $\ob{Q}$, $N$ would contain an essential disc, contradicting the assumption that $\boundary W$ is incompressible in $N$.

We may, therefore, assume that there is no $a$--boundary compressing disc for $\ob{Q}$. If $\ob{Q}$ is completely disjoint from $a$, then $\ob{Q}$ is an annulus which is disjoint from the meridional arcs of $a-b$. From our observations about meridional arcs, this means that $\ob{Q} \subset N$ is an annulus which is either disjoint from or has meridional boundary on one component of $\boundary S^3[\beta]$.

Suppose, therefore, that $\ob{Q}$ is not completely disjoint from $a$. By the main theorem, $-2\chi(\ob{Q}) \geq K(\ob{Q})$. Since $\chi(\ob{Q}) = 0$ and since $K(\ob{Q}) \geq 0$ we have $K(\ob{Q}) = 0$. That is,
\[
q(\Delta - 2) + q^*(\Delta^* - 2) + \Delta_\boundary = 0.
\]

Since each term is non-negative, each term must be zero. Hence $\Delta_\boundary = 0$, implying that either $\ob{Q}$ is a torus or it is an annulus with boundary disjoint from or consisting of meridians on some component of $\boundary S^3[\beta]$. If $q^* \neq 0$, then $\beta$ is non-separating and we must have $\Delta^* = 2$. Since $b^*$ intersects each meridional arc of $a - b$ at least twice, this means that there is exactly one such meridional arc. The number of meridional arcs is even, so this is a contradiction. If $q \neq 0$ then we have $\Delta = 2$. If both $q$ and $q^*$ are equal to zero, then since $\Delta_\boundary = 0$, $\ob{Q}$ is an annulus disjoint from $a$, a possibility we have already considered.
\end{proof}

\subsection{Rational Tangle Replacements}
By restricting attention to borings which are actually rational tangle replacements, we can obtain more results, including new proofs of some classical theorems. Suppose that $K$ is a knot or link in $S^3$ and $B' \subset S^3$ is a 3-ball which intersects $K$ in two strands which form a rational tangle in $B'$. Let $K'$ be a knot or link obtained by replacing $B' \cap K$ with a different rational tangle. See \cite{EM1} for definitions relating to rational tangles. It is easy to see that there are discs $\alpha$ and $\beta$ in $W = \eta(K) \cup B'$ such that $L_\alpha = K'$ and $L_\beta = K$. The distance $d$ between the two rational tangles is defined to be $\Delta(\alpha,\beta)/2$. Define $\gamma \subset \boundary W$ as before. We let $(B,\tau)$ be the tangle with $B = S^3 - \inter{B'}$ and $\tau = L_\beta \cap B$. A tangle $(B,\tau)$ is \defn{prime} if there are no local knots and if there is no disc separating the strands of $\tau$.

The following lemmas are easy to prove, or see \cite{T1}.
\begin{TangleCalcI}[$\beta$ separating]\label{Lem: Tangle Calcs, Sep}
Suppose that $L_\beta$ is a link obtained from $L_\alpha$ by a rational tangle replacement of distance $d$ using $W$. Let $\ob{Q}$ be a suitably embedded surface in the exterior $S^3[\beta]$ of $L_\beta$. Let $\boundary_1 \ob{Q}$ be the components of $\boundary \ob{Q}$ on one component of $\boundary S^3[\beta]$ and $\boundary_2 \ob{Q}$ be the components on the other. Let $n_i$ be the minimum number of times a component of $\boundary_i \ob{Q}$ intersects a meridian of $\boundary S^3[\beta]$.
\begin{itemize}
\item If $L_\alpha$ is a link then
\[
K(\ob{Q}) \geq 2q(d - 1) + d(|\boundary_1 \ob{Q}|n_1 + |\boundary_2 \ob{Q}|n_2).
\]

\item If $L_\alpha$ is a knot then
\[K(\ob{Q}) \geq 2q(d - 1) +  (d-1)(|\boundary_1 \ob{Q}|n_1 + |\boundary_2 \ob{Q}|n_2).
\]
\end{itemize}
\end{TangleCalcI}
\begin{TangleCalcII}[$\beta$ non-separating]
Suppose that $L_\beta$ is a knot obtained from $L_\alpha$ by a rational tangle replacement of distance $d$ using $W$. Let $\ob{Q}$ be a suitably embedded surface in the exterior $S^3[\beta]$ of $L_\beta$. Suppose that each component of $\boundary \ob{Q}$ intersects a meridian of $\boundary S^3[\beta]$ $n$ times.
\begin{itemize}
\item If $L_\alpha$ is a link then
\[
K(\ob{Q}) \geq 2q(d - 1) + 2q^*(2d - 1) + 2d|\boundary \ob{Q}|n.
\]
\item If $L_\beta$ is a knot then
\[
K(\ob{Q}) \geq 2(d-1)(q + 2q^*) + 2(d-1)|\boundary \ob{Q}|n.
\]
\end{itemize}
\end{TangleCalcII}

It will be useful to understand the relationship between the tangle $(B,\tau)$ and the required assumptions about the incompressibility of $\boundary N$.

\begin{lemma}[\cite{T1}]\label{Lem: Tangle Compressing}
Suppose that $(B,\tau)$ and $(B',r_\alpha)$ are tangles embedded in $S^3$ with $(B',r_\alpha)$ a rational tangle so that $\boundary B = \boundary B'$ and $\boundary \tau = \boundary r_\alpha$. Suppose that $(B',r_\beta)$ is rational tangle of distance at least one from $(B',r_\alpha)$. Define the sutures $\gamma \cup a$ on $\boundary N$ as before. If one of the following holds
\begin{itemize}
\item $\alpha$ is non-separating in the handlebody $W = B' \cup \eta(\tau)$, or
\item if $(B,\tau)$ is a prime tangle, or
\item if $(B,\tau)$ is a rational tangle and $\boundary \alpha$ does not bound a trivializing disc for $(B,\tau)$, or
\item if $\boundary \alpha$ does not compress in $(B,\tau)$
\end{itemize}
then $\boundary W - (\gamma \cup a)$ is incompressible in $N$. Consequently, $(N,\gamma \cup a)$ is $\nil$--taut and $(N[a],\gamma)$ is $\ob{\alpha}$--taut.
\end{lemma}

Our next observation concerns the implications of an $a$--boundary compressing disc.

\begin{lemma}\label{Lem: Isotopic}
Suppose that $\ob{Q}$ is an essential surface in $S^3[\beta]$ disjoint from $\ob{\beta}$. If all components of $\boundary \ob{Q}$ are meridians then there does not exist an $a$--boundary compressing disc joining two components of $\boundary \ob{Q}$. If $\boundary \ob{Q}$ has components on all components of $\boundary S^3[\beta]$ and no component is a meridian, then if there is an $a$--boundary compression for $\ob{Q}$ in $S^3 - \inter{W}$ the arc $\ob{\beta}$ is properly isotopic into $\ob{Q}$.
\end{lemma}

\begin{proof}
Notice, first, that if all components of $\boundary \ob{Q}$ are meridians then $\boundary \ob{Q} \cap a = \nil$, since all arcs of $a-b$ are meridional. Suppose therefore that $\boundary \ob{Q}$ intersects each component of $\boundary S^3[\beta]$ and that no component of $\boundary \ob{Q}$ is a meridian. Let $D$ be an $a$--boundary compression. Let $\epsilon = \boundary D \cap \boundary W$. It is a component of $a - \boundary \ob{Q}$. Since $\ob{Q}$ is essential in $S^3[\beta]$, the arc runs at least once across $\eta(b)$. Since no component of $\boundary \ob{Q}$ is a meridian and since it intersects each component of $\boundary S^3[\beta]$, each arc of $a - \boundary \ob{Q}$ which runs across $\eta(b)$ does so exactly once. Hence, after pushing $\epsilon$ into $W$ slightly, $\eta(\ob{\beta})$ can be viewed as a regular neighborhood of $\epsilon$. Then $D$ guides an isotopy of $\ob{\beta}$ into $\ob{Q}$. See Figure \ref{Fig: IsotopicArc}.
\end{proof}

\begin{figure}[ht]
\labellist
\hair 2pt
\pinlabel $\boundary \ob{Q}$ [b] at 13 142
\pinlabel $\ob{\beta}$ [t] at 162 18
\pinlabel $a$ [bl] at 297 153
\pinlabel $\epsilon$ [l] at 215 195
\endlabellist
\center
\includegraphics[scale=0.9]{Figures/IsotopicArc}

\caption{The arc $\ob{\beta}$ is parallel to $\epsilon$.}
\label{Fig: IsotopicArc}
\end{figure}

The final preliminary lemma strengthens Theorem \ref{Thm: Constructing Q}.

\begin{lemma}\label{Lem: Isotopic Surfaces}
Suppose that $L_\beta$ and $L_\alpha$ are related by a rational tangle replacement of distance $d \geq 1$ and that $\ob{R} \subset S^3[\beta]$ is an essential surface. Assume that $\boundary \ob{R}$ is not disjoint from any component of $\boundary S^3[\beta]$ and that no component of $\boundary \ob{R}$ is a meridian. Then the surface $\ob{Q}$ constructed by Theorem \ref{Thm: Constructing Q} is isotopic to $\ob{R}$.
\end{lemma}

\begin{proof}
The surface $\ob{Q}$ is constructed by boundary compressing $R = \ob{R} \cap N$ using $a$--boundary compressing discs. Suppose that $D$ is an $a$--boundary compressing disc for $R$. Let $\delta = \boundary D \cap R$ and $\epsilon = \boundary D \cap \boundary W$. If $\epsilon$ lies in an annulus component of $\boundary W - \interior \ob{R}$ then, as in the proof of \cite[Theorem 5.1]{T1}, there is an isotopy of $\ob{R}$ which reduces $\tild{q}$. If $\epsilon$ joins different components of $\boundary \eta(b) \subset \boundary W$ then there is an isotopy of $\ob{R}$ to a suitably embedded surface, such that $q$ has been decreased by two and $q^*$ has been increased by one. If $\epsilon$ joins a component of $\boundary R$ which is parallel to $b^*$ to one which is parallel to $b$ then there is an isotopy which reduces $q^*$ by one and leaves $q$ unchanged. If $\epsilon$ joins $\boundary \ob{R}$ to $\boundary R - \boundary \ob{R}$ there is an isotopy of $\ob{R}$ which moves $\boundary \ob{R}$ and which reduces $\tild{q}$. This isotopy has the same effect as performing the boundary compression.

Thus, it remains to examine what happens when $\epsilon$ joins $\boundary \ob{R}$ to itself or when it joins a component of $\boundary R - \boundary \ob{R}$ to itself. Since $\boundary S^3[\beta]$ consists of tori and since $\ob{R}$ is essential, $\ob{R}$ is boundary\hyph incompressible in $S^3[\beta]$. Thus, if $\epsilon$ joins $\boundary \ob{R}$ to itself, the arc $\delta$ must be inessential in $\ob{R}$. There is, therefore, an isotopy of $\ob{R}$ which reduces $\tild{q}$ and accomplishes the boundary compression.

Suppose, therefore, that $\epsilon$ joins a component of $\boundary R - \boundary \ob{R}$ to itself. Since $L_\beta$ is obtained from $L_\alpha$ by a rational tangle replacement the arc $\epsilon$ is a subarc of a meridional arc of $a - b$. (In fact, all arcs of $a - b$ are meridional.) The arc $\epsilon$ is disjoint from $\boundary \ob{R}$ and so is a meridional arc of $a - b$, and not just a subset of one. Let $T$ be the component of $\boundary S^3[\beta]$ containing $\epsilon$. Since $\epsilon \cap \boundary \ob{R} = \nil$, either $\boundary \ob{R}$ is disjoint from $T$ or $\boundary \ob{R} \cap T$ consists of meridians. Thus, by hypothesis, this case cannot occur and $\ob{Q}$ is isotopic to $\ob{R}$.
\end{proof}

The work of this paper can be used to give new proofs of several classical theorems about rational tangle replacements, including five out of the six theorems of \cite{EM1}. This is done in \cite{T2}. Here, we prefer to prove some new results; however, we do not shy away from showing how these results can give new proofs of some well-known theorems.

We begin by restating Theorem \ref{Thm: Boring Genus} for rational tangles:

\begin{corollary}\label{Cor: RT Genus}
Suppose that $L_\beta \subset S^3$ is obtained by a rational tangle replacement of distance $d\geq1$ on a split link or unknot $L_\alpha$. If $L_\alpha$ is a split link, assume that $a$ is incompressible in $B - \tau$. If $L_\alpha$ is an unknot, assume that there does not exist an essential disc in $S^3[\alpha]$ disjoint from $\ob{\alpha}$. Then $L_\beta$ is not a split link and it has a minimal genus Seifert surface $\ob{Q}$ disjoint from $\ob{\beta}$ such that one of the following holds:
\begin{itemize}
\item $\ob{\beta}$ is properly isotopic into $\ob{Q}$
\item $-\chi(\ob{Q}) \geq d$ and $L_\alpha$ is a split link
\item $-\chi(\ob{Q}) \geq d - 1$ and $L_\alpha$ is an unknot.
\end{itemize}
\end{corollary}
\begin{proof}
The assumption that if $\alpha$ is separating then $a$ is incompressible in $B - \tau$ implies (Lemma \ref{Lem: Tangle Compressing}) that $\boundary N - (a \cup \gamma)$ is incompressible in $N$. Applying Theorem \ref{Thm: Boring Genus}, we produce the Seifert surface $\ob{Q}$ and either there is an $a$--boundary compressing disc for $\ob{Q} \subset N$ or $-2\chi(\ob{Q}) \geq K(\ob{Q})$. If the former happens, by Lemma \ref{Lem: Isotopic}, we conclude that $\ob{\beta}$ is properly isotopic into $\ob{Q}$. Suppose, therefore, that $-2\chi(\ob{Q}) \geq K(\ob{Q})$. Using the Tangle Calculations and the fact that $q = q^* = 0$ we see that if $L_\alpha$ is a link, then $-2\chi(\ob{Q}) \geq 2d$. If $L_\alpha$ is a knot, then $-2\chi(\ob{Q}) \geq 2(d-1)$. The given inequalities follow immediately.
\end{proof}

A pleasing corollary is Gabai and Scharlemann's result that genus is superadditive under band sum. A band sum is a rational tangle replacement of distance 1 on a split link.

\begin{corollary}[Gabai \cite{G2}, Scharlemann \cite{S3}]\label{Cor: Band Sum}
Suppose that $K_1 \#_b K_2$ is the band sum of knots $K_1$ and $K_2$. Then
\[
\genus(K_1 \#_b K_2) \geq \genus(K_1) + \genus(K_2)
\]
with equality only if $K_1$ and $K_2$ have minimal genus Seifert surfaces disjoint from the band.
\end{corollary}
\begin{proof}
The statement holds if the band sum is a connected sum (i.e. if the band intersects a splitting sphere exactly once), so we may assume that the band intersects every essential sphere in the exterior of $L_\alpha = K_1 \cup K_2$ more than once. Let $W = \eta(K_1 \cup K_2 \cup b)$ where $b$ is the band. (Note the ambiguity associated with the letter `$b$' in this context.) Let $\alpha$ be a disc in $\eta(b)$ intersected once transversally by the core of $b$. Let $\beta$ be a disc intersecting $\alpha$ once and which is ``parallel'' to the cocore of the band so that $L_\beta = K_1 \#_b K_2$. Since the band sum is not a connected sum, $\boundary W - \boundary \alpha$ is incompressible in $S^3 - \inter{W}$ (Lemma \ref{Lem: Tangle Compressing}). Applying Corollary \ref{Cor: RT Genus}, we produce a minimal genus Seifert surface $\ob{Q}$ for $L_\beta$ which is disjoint from $\ob{\beta}$, the cocore of the band. The proof now proceeds as in \cite{G2} and \cite{S3}.
\end{proof}

Superadditivity of genus under band sum provides a more interesting estimate of the genus of a knot $L_\beta$ obtained by a rational tangle replacement on a split link than does Corollary \ref{Cor: RT Genus}. To see this, notice that the rational tangle replacement on a split link can be seen as a band sum of knots $K_1$ and $K_2$ with a 2--bridge knot $K_3$ inserted in the middle of the band. By moving the 2--bridge knot along the band so that it is close to $K_2$, we see that $L_\beta = K_1 \#_b (K_3 \# K_2)$. Thus, by supperadditivity of genus under band sum,
\[
\genus(L_\beta) \geq \genus(K_1) + \genus(K_3) + \genus(K_2).
\]
I believe that the result of Corollary \ref{Cor: RT Genus} for $L_\beta$ an unknot is genuinely new. Similar to the previous case, this result can be interpreted as a result about attaching a band to a 2--bridge knot or link. However, not every such band attachment can be described as a rational tangle replacement on the unknot. The superadditivity of genus under band sum is used in the next example to show that the possibility that $\ob{\beta}$ is isotopic into $\ob{Q}$ cannot be removed from Corollary \ref{Cor: RT Genus}.

\begin{example}
Figure \ref{fig: example} depicts the diagram of a $9_{37}$ knot $L_\beta$. (Jiho Kim's KnotSketcher and Charles Livingston's KnotFinder were used to determine the name of this knot.) The indicated rational tangle replacement converts $L_\beta$ into a split link $L_\alpha$. The rational tangle replacement has distance $d = 5$. In the diagram, it is not difficult to find a Seifert surface $S$ for $L_\beta$ consisting of an annulus and three twisted bands. Two of the bands have one half twist each and the third has three half twists. Thus, $-\chi(S) = 3$ and $\genus(S) = 2$. $L_\beta$ is the band sum of the unknot with a figure eight knot. The band is not disjoint from Seifert surfaces for the unknot and the figure eight knot. Hence, by Corollary \ref{Cor: Band Sum}, $\ob{Q}$ is a minimal genus Seifert surface for $L_\beta$. It is easy to see that $\ob{\beta}$ is isotopic into $\ob{Q}$.
\end{example}

\begin{figure}[ht]
\center
\includegraphics[scale=0.4]{Figures/Example2}

\caption{The knot $L_\beta$ and a rational tangle replacement.}
\label{fig: example}
\end{figure}
\begin{remark}
Scharlemann and Thompson \cite{ST2} have shown that, in many cases, a tunnel for a tunnel number 1 knot can be isotoped and slid to lie in a minimal genus Seifert surface for the knot. Since tunnel number 1 knots are those knots which are obtained by boring the unknot or unlink using an unknotted handlebody, perhaps the first possible conclusion of Corollary \ref{Cor: RT Genus} points to a more general phenomenon.
\end{remark}

We now turn to an examination of genus zero and one surfaces in the exterior of knots and links obtained by a rational tangle replacement on a split link or unknot. The next theorem is a more sophisticated version of \cite[Theorem 7.3]{T1}.

\begin{theorem}\label{Thm: Meridional Planar}
Suppose that $L_\beta$ is a knot or link obtained by a rational tangle replacement of distance $d \geq 1$ on the knot or link $L_\alpha$. Suppose that either $L_\alpha$ is a knot or that $\boundary W - \boundary \alpha$ does not compress in $N$. Suppose also that $L_\alpha$ is a split link or does not contain a minimal genus Seifert surface disjoint from $\ob{\alpha}$. Then, if $L_\beta$ has an essential properly embedded meridional planar surface with $m$ boundary components, it contains such a surface $\ob{Q}$ with $|\boundary \ob{Q}| \leq m$ such that either $\ob{Q}$ is disjoint from $\ob{\beta}$ or
\[
|\ob{Q} \cap \ob{\beta}|(d - 1) \leq |\boundary \ob{Q}| - 2
\]
\end{theorem}
\begin{proof}
The proof is a fun exercise using the machinery developed so far. Alternatively, adapt the proof of \cite[Theorem 7.3]{T1} by replacing the appeal to main theorem of that paper with an appeal to the main theorem of this paper.
\end{proof}

A crossing change or generalized crossing change of a knot $K$ is achieved by choosing a disc $D \subset S^3$ which is pierced twice by $K$ with opposite sign and by performing a $\pm 1/n$ Dehn surgery on $\boundary D$ with $n \in \N$. If $n = 1$, the new knot is obtained by changing the crossing of $K$. It is easy to see that a generalized crossing change can be achieved by rational tangle replacement of distance $d = 2n$.

\begin{corollary}[Scharlemann \cite{S1}, Scharlemann and Thompson \cite{ST1}]\label{Cor: Composite Unknotting}
No generalized crossing change on a composite knot will produce the unknot.
\end{corollary}

\begin{proof}
Suppose that $L_\beta = K_1 \# K_2$ is an unknotting number one knot with $K_1$ and $K_2$ non-trivial knots. Let $D$ be a crossing disc for $L_\beta$ such that $\pm 1/n$ surgery on $\boundary D$ converts $L_\beta$ to the unknot $L_\alpha$. Let $W = \eta(L_\beta \cup D)$ and notice that $L_\alpha$ can be obtained from $L_\beta$ by a rational tangle replacement of distance $d = 2n$. Notice that $\alpha$ is non-separating.

Apply Theorem \ref{Thm: Meridional Planar} beginning with an essential meridional annulus in $S^3[\beta]$. The surface $\ob{Q}$ is then either an essential annulus or an essential disc. Since $L_\beta$ is not the unknot, $\ob{Q}$ is an annulus. If it were disjoint from $\ob{\beta}$, the crossing change would be a crossing change on either $K_1$ or $K_2$ and so would not convert $L_\beta$ into the unknot. The inequality $|\ob{Q} \cap \ob{\beta}|(d - 1) \leq |\boundary \ob{Q}| - 2$ becomes
\[
0 \leq |\ob{Q} \cap \ob{\beta}|(d - 1) \leq 0,
\]
implying that $\ob{Q}$ is disjoint from $\eta(b)$ after all. This contradiction shows that $L_\beta$ cannot be a composite unknotting number one knot.
\end{proof}

If a non-trivial surgery on a hyperbolic knot or link $L_\beta \subset S^3$ produces a manifold containing an essential sphere or torus, it is easy to show that the exterior of $L_\beta$ contains an essential planar surface or punctured torus. The final theorem examines the possibilities for such surfaces in the exterior of a knot $L_\beta$ obtained by rational tangle replacement on a split link or knot without a minimal genus Seifert surface disjoint from the boring arc.

\begin{theorem}\label{Thm: RT Planar/Tori}
Suppose that $L_\beta$ is a knot or link obtained by rational tangle replacement of distance $d \geq 1$ on a knot or link $L_\alpha$ using handlebody $W$. Suppose either that $\alpha$ is non-separating or that $\boundary W - \boundary \alpha$ is incompressible in $N$. Assume also that $L_\alpha$ is a split link or does not have a minimal genus Seifert surface disjoint from $\ob{\alpha}$. Suppose that $L_\beta$ contains an essential planar surface or punctured torus $\ob{R}$ in its exterior. Then $\ob{R}$ can be properly isotoped so that one of the following is true:
\begin{enumerate}
\item $L_\beta$ is a link and $\boundary \ob{R}$ is disjoint from some component of $L_\beta$.
\item $\ob{R}$ has meridional boundary on some component of $L_\beta$.
\item $\ob{R}$ is disjoint from $\ob{\beta}$ and $\ob{\beta}$ can be slid and isotoped to lie in $\ob{R}$.
\item $L_\beta$ and $L_\alpha$ are both links, $d = 2$, and $\ob{R}$ is a punctured torus disjoint from $\ob{\beta}$ with integer slope on both components of $\boundary S^3[\beta]$.
\item $L_\beta$ is a link, $L_\alpha$ is a knot, $d \leq 2$, and $\ob{R}$ is a planar surface.
\item $L_\beta$ is a link, $L_\alpha$ is a knot, $d \leq 3$, and $\ob{R}$ is a punctured torus.
\item $L_\beta$ is a knot, $L_\alpha$ is a link, $d = 1$, $\ob{R}$ is a punctured torus with $\boundary \ob{R}$ having integer slope.
\item $L_\beta$ and $L_\alpha$ are both knots, $d = 1$ and $\ob{R}$ is a planar surface.
\item $L_\beta$ and $L_\alpha$ are both knots, $d \leq 2$ and $\ob{R}$ is a punctured torus.
\end{enumerate}
\end{theorem}
\begin{proof}
By Corollary \ref{Cor: RT Genus}, there is no essential sphere or disc in the exterior of $L_\beta$. Apply Theorem \ref{Thm: Constructing Q} version II to $\ob{R}$ obtaining a connected suitably embedded surface $\ob{Q}$. Suppose that neither conclusion (1) nor conclusion (2) holds. By Lemma \ref{Lem: Isotopic Surfaces}, $\ob{Q}$ is isotopic to $\ob{R}$. In particular, it is also a planar surface or punctured torus and $\boundary \ob{Q}$ has the same slope as $\boundary \ob{R}$.

If $\ob{Q}$ is disjoint from $\ob{\beta}$ and if there is an $a$--boundary compressing disc for $\ob{Q}$ then, by Lemma \ref{Lem: Isotopic}, $\ob{\beta}$ is isotopic into $\ob{Q}$. Examining the isotopy of $\ob{R}$ to $\ob{Q}$ shows that this is equivalent to conclusion (3). By the construction of $\ob{Q}$, we may assume that there is no $a$--boundary compressing disc for $Q$.

If $\ob{Q}$ is disjoint from $a$ then it is disjoint from all meridional arcs of $a - b$ and so must have meridional boundary or must be disjoint from some component of $\boundary S^3[\beta]$, contradicting our denial of (1) and (2). Hence, we may apply the main theorem to conclude that $-2\chi(\ob{Q}) \geq K(\ob{Q})$. Let $s = 2$ if $\ob{Q}$ is a planar surface and let $s = 0$ if $\ob{Q}$ is a punctured torus. We now consider the possibilites for $\alpha$ and $\beta$. We use the notation and results of the Tangle Calculation Lemmas.

\subsubsection*{Case 1: $\beta$ and $\alpha$ are both separating}
In this case, notice that $d \geq 2$. Since $-2\chi(\ob{Q}) = -2s + 2(|\boundary_1 \ob{Q}| + |\boundary_2 \ob{Q}|)$ we have
\[
-2s + 2(|\boundary_1 \ob{Q}| + |\boundary_2 \ob{Q}|) \geq 2q(d-1) + d(|\boundary_1 \ob{Q}|n_1 + |\boundary_2 \ob{Q}|n_2).
\]
Rearrange this to obtain
\[
-2s \geq 2q(d-1) + |\boundary_1 \ob{Q}|(dn_1 - 2) + |\boundary_2 \ob{Q}|(dn_2 - 2).
\]
If $\ob{Q}$ is a planar surface, then we must have either $dn_1 < 2$ or $dn_2 < 2$. Since $d$, $n_1$, and $n_2$ are all non-zero by hypothesis, we contradict the observation that $d \geq 2$. Hence $\ob{Q}$ is not a planar surface.

If $\ob{Q}$ is a punctured torus, then we must have $dn_1 \leq 2$ and $dn_2 \leq 2$. Since $d \geq 2$, we must have $d = 2$ and $n_1 = n_2 = 1$. This is conclusion (4).

\subsubsection*{Case 2: $\beta$ is separating and $\alpha$ is non-separating}
We have
\[
-2s + 2(|\boundary_1 \ob{Q}| + |\boundary_2 \ob{Q}|) \geq 2q(d-1) + (d-1)(|\boundary_1 \ob{Q}|n_1 + |\boundary_2 \ob{Q}|n_2).
\]

Rearranging, we obtain
\[
-2s \geq 2q(d-1) + |\boundary_1 \ob{Q}|((d-1)n_1 - 2) + |\boundary_2 \ob{Q}|((d-1)n_2 - 2).
\]

Therefore, we have $(d-1)n_1 \leq 2$ or $(d-1)n_2 \leq 2$. If $\ob{Q}$ is a planar surface the inequalities are strict. This produces conclusion (5). Otherwise, we obtain conclusion (6).

\subsubsection*{Case 3: $\beta$ is non-separating and $\alpha$ is separating}
Now we have,
\[
-2s + 2|\boundary \ob{Q}| \geq 2q(d-1) + 2q^*(2d-1) + 2d|\boundary \ob{Q}|n.
\]
Rearranging we find
\[
-2s \geq 2q(d-1) + 2q^*(2d-1) + |\boundary \ob{Q}|(2dn - 2).
\]

Since $d$, $n$, and $|\boundary \ob{Q}|$ are all positive, $s = 0$ and $d = n = 1$. This is conclusion (7).

\subsubsection*{Case 4: $\beta$ and $\alpha$ are both non-separating}
Finally, we have
\[
-2s \geq 2(q + 2q^*)(d-1) + 2|\boundary \ob{Q}|((d-1)n - 1)
\]

If $s = 2$, then $(d-1)n < 1$ implying $d = 1$. If $s = 0$, then $(d-1)n \leq 1$. This implies $d \leq 2$. These are conclusions (8) and (9).
\end{proof}