2-bridge knots are small

Here's a cool fact due to Hatcher and W. Thurston (article here): 2-bridge knots are small.  I never realized the proof was so simple (and clever), so I wanted to write about it.  I'll briefly explain what small means and then sketch the proof (which is actually fairly short).  

Remark:  They prove something much more general.  They are characterizing incompressible surfaces in the complement of 2-bridge knots, but I won't discuss that here.   

Background:  An embedded surface in a three-manifold is incompressible if the inclusion map induces an injection on $\pi_1$.  For example, the boundary of a handlebody is not incompressible but $S^1 \times \Sigma_g$ has many incompressible surfaces: $pt \times \Sigma_g$ or $S^1 \times \gamma$ for $\gamma$ non-trivial in $\pi_1(\Sigma_g)$.  A neat fact is that a knot in $S^3$ is non-trivial if and only if the boundary of the exterior is incompressible.  A knot is small if there are no closed incompressible surfaces in the exterior other than the boundary torus.  A good example of a knot which is not small is a satellite knot, like a connected sum of two non-trivial knots or the Whitehead double of the trefoil.  For some intuition, note that the exterior of $P(K)$, where $P$ is the pattern knot in the solid torus and $K$ is the companion, can be built by gluing the exterior of $K$ to the exterior of $P$ in $D^2 \times S^1$.  The boundary torus of the exterior of $K$ is not boundary-parallel  and is incompressible (except some degenerate cases).  On the other hand, all torus knots are small.  (The torus knots $T(2,n)$ are all 2-bridge knots, so the proof below will cover at least this case.)

One way to think of 2-bridge knots is as follows.  A knot $K$ embedded in $S^3$ is an $n$-bridge position if the usual height function on $S^3$ (or $\mathbb{R}^3$) restricts to have $n$ maxima on $K$ (and necessarily $n$ minima).  Alternatively, think of two copies of $B^3$ with $n$ trivial arcs inside (hitting the boundary sphere in $2n$ points), and glue these together in some way - the arcs join up to become a link (or knot if glued together appropriately).  For some pictures and discussion about bridge number, see this note by Jennifer Schultens.  When we have 2 trivial arcs in $B^3$, this is called a rational tangle.  This can be recorded on a 4-punctured sphere by the choice of curve which bounds the disk in $B^3$ separating the two tangle arcs.  The choices of such curves are parameterized by rational numbers (and $1/0$), explaining the term rational.  So, 2-bridge knots can be built by gluing together two rational tangles.  

Proof that 2-bridge knots are small:  Let $K$ be a 2-bridge knot and suppose that $F$ is a closed incompressible surface in the exterior of $K$.  Our strategy is to show that it is isotopic to the boundary of the exterior of $K$.  We will do this by looking at the height function on $S^3$ and using it to understand the level sets of $F$.  Let's assume that the max and min for $K$ occur at $+1/2$ and $-1/2$ respectively.  

Let's wiggle $F$ a little so that the height function on $S^3$ is also a Morse function on $F$ and there's no critical points at $\pm 1/2$.  For notation, let $S^2_t$ denote the sphere in $S^3$ at height $t$, and write $F_t$ for $F \cap S^2_t$, which will be a collection of (disjoint) circles on $S^2_t$.  

Let's look at $F_t$ for some $t \in (-1/2,1/2)$.  Since $F$ is in the exterior of $K$, these circles miss the 4 points where $K$ intersects $S^2_t$.  There could be a few types of circles: those that split the 4 points into two pairs, those that split into 1 point and 3 points, or those that split into 0 and 4 points.  Let's call them 2/2, 1/3, and 0/4 circles.  To make my life a little easier, let's just say there are no 0/4 circles.  (They're not really a big deal, because I say they're not really a big deal.)  Now, note that all of the 2/2-curves at a given level have to be parallel - if they weren't, they would intersect, giving a self-intersection of $F$.  Near $\pm 1/2$, if there are 2/2 curves, their slopes have to be exactly the slopes of the corresponding rational tangle.  (If not, we couldn't keep building the surface $F$ past $t = \pm 1/2$ without creating intersections with $K$.)  If there are 2/2 circles near $t = -1/2$ and $t = +1/2$, their slopes have to be different, because otherwise we would be getting a 2-component link.  This is because these slopes are describing the type of rational tangle!    

The only time these circles change in topology as $t$ varies is when we pass through a critical level of $F$.  Since we have no 0/4 circles, this means for $t \in (-1/2, 1/2)$, we can merge/split circles (i.e. pass index 1 critical points).  Since the slopes at $t = -1/2$ and $t = 1/2$ are different, at some regular value $t$, we have to see no 2/2 circles.  (You can't merge/split 2/2 circles to get 2/2 circles of a different slope.  You have to pass through 1/3 circles.)         

Consider $S^2_t$ and look at the two balls $B_1$ and $B_2$ to each side.  We can now think of $K$ being split into two rational tangles instead by $B_1$ and $B_2$.   There's a disk $D_1$ in $B_1$ with boundary on $S^2_t$ which splits the rational tangle into two strands and $D_1$ is disjoint from $F_r$.  (This is because the 1/3 circles can be made small and encircle one puncture and $D_1$ misses the punctures.)  We can do the same with a disk $D_2$ in $B_2$.  Now, after isotoping $F$, we can choose these disks so that they miss our surface.  (Basically, if the surface intersected the disk in an essential loop, you could use the disk to compress the surface.  These are the obstruction to being able to avoid the disk.)  That means that in $F \cap B_1$, we have two pieces - each piece lives in half of $B_1$ and encapsulates one strand of the rational tangle (i.e. one strand of $K \cap B_1$).  In fact, each of these pieces has to be a tube around the strands of $K$, because if the pieces had more topology, then we could compress the extra loops.  (Basically, we are splitting a rational tangle along $D_1$ into two (1,1)-tangles, and $F$ has a component in each half.  There's not room for anything interesting to happen.)  

Now, do the same for the other half $B_2$, to see that $F \cap B_2$ is just two tubes along the strands of $K \cap B_2$.  Putting these together, we see that $F$ is just isotopic to the boundary of a tubular neighborhood of $K$!  That means that $F$ was boundary parallel in the exterior of $K$.  This is what we wanted to show.

Pretty neat!  I'd pick an exact triangle argument over this any day of the week, but I know these "real" topology arguments are valuable and so I'll keep working on them.