Happy Birthday Cameron! (part 1)

There is an upcoming conference celebrating Cameron Gordon's 80th birthday.  Cameron has had a big influence on my career (both directly and through his mathematics), so I wanted to write some posts giving quick summaries of his influential work.  These will be short samples instead of giving full details on anything.  This post was inspired by a comment of Josh Wang.

I thought it might be kind of cliche to start with the most famous results, like the Knot Complement Theorem, so I'll instead start with one of my personal favorites: "Only integral surgeries can yield reducible manifolds" by Gordon-Luecke.  While the title is the result, we'll see some more background and at the end, I'll give a really nice application to the unknotting number of knots.  

First, we need to have a little chat about Dehn surgeries.  (As an aside, Cameron is also a historian of topology.  Since Dehn came up, I'll point out Cameron wrote a historical article on Dehn's work in topology.)  Given a knot $K$ in the three-sphere, we can construct $p/q$-Dehn surgery, $S^3_{p/q}(K)$, which means to remove a tubular neighborhood of $K$ and reglue so that the meridian of the tubular neighborhood gets sent to a curve on the boundary of the exterior of $K$ which represents $p$ times a meridian + $q$ times a longitude.  Lots of interesting three-manifolds can be built in this way and a major study in low-dimensional topology is to try to understand surgery descriptions of three-manifolds.  For example, $p/q$-surgery on the unknot gives the lens space $L(p,q)$, while $+1/1$-surgery on the right-handed trefoil gives the Poincare homology sphere.  A more surprising example is the following: Let $K$ be the $(p,q)$-torus knot.  Then $S^3_{pq/1}(K) = L(p,q) \# L(q,p)$.  If you want more details, there's some proofs at the end of this post.  In fact, more generally, if $K$ is the $(p,q)$-cable of $J$, then $S^3_{pq}(K) = L(p,q) \# S^3_{q/p}(J)$.  Here's a picture of the $(2,1)$-cable of the figure-eight knot to give a sense of what a cable knot is.

One of the biggest problems in Dehn surgery theory is to characterize when surgery on a knot can give a connected sum of non-trivial three-manifolds.  The Cabling Conjecture predicts that the only time this happens is this case of $pq/1$-surgery on a $(p,q)$-cable.  This conjecture is big-time: I'd say it's a guaranteed Annals paper if you solve it.  In "Only integral surgeries can yield reducible manifolds", they show that if surgery on a knot gives a reducible manifold, this surgery is integral, i.e. $p/q  \in \mathbb{Z}$.  (Reducible just means $S^2 \times S^1$ or a non-trivial connected sum.  The case of $S^2 \times S^1$ is not relevant to us since that can only arise as 0-surgery on a knot, and in fact, only the unknot.  So I'll just say reducible in the context of the cabling conjecture.)  This may not seem like the most exciting result if you're not into Dehn surgery problems.  However, rather than sketch the proof, I wanted to give some motivation for how these types of results can be useful.  I'll explain how we can quickly apply the Gordon-Luecke theorem to prove the following result:

Theorem:  A connected sum of non-trivial knots has unknotting number at least 2.  

Note that until about a month ago, it was unknown if the unknotting number was additive under connected sum.  (Now it's known not to be by Brittenham-Hermiller.)  This result was originally proved by Scharlemann using some more hands-on three-manifold topology.  However, there is a quick proof using the Gordon-Luecke result due to Xingru Zhang.  

Proof:  Let $K$ be a knot in $S^3$ described as $K_1 \# K_2$, where the $K_i$ are non-trivial.  Then $\Sigma_2(K) = \Sigma_2(K_1) \# \Sigma_2(K_2)$, where $\Sigma_2$ denotes the branched double cover.  Since the $K_i$ are non-trivial, their branched double covers are not $S^3$.  (This is a very special case of the Smith Conjecture, as discussed slightly in this older post.  The case of involutions long predates the full proof though, so this is not as big of a hammer as it seems.)  Consequently, $\Sigma_2(K)$ is a reducible manifold.  If $K$ had unknotting number one, by the Montesinos trick $\Sigma_2(K) = S^3_{d/2}(J)$, where $J$ is a knot in $S^3$ and $d \in \mathbb{Z}$.  However, this means that we have found a non-integral reducible surgery on a knot in $S^3$.  That is a contradiction! 

(I don't think I've done a post yet on the Montesinos trick.  Hopefully sometime in the near future I'll write a post on this.  If you haven't seen this before, a nice explanation can be found here.  The key idea is that if two knots are related by removing a rational tangle and replacing it with another one, the branched double covers are related by a Dehn surgery.)    

For anyone who wants more details on reducible surgeries on cabled knots, here's one way I think about them.  To see why $T_{p,q}$ has a reducible surgery, think of this knot as sitting on a standardly embedded torus $T$ in $S^3$.  To do the Dehn surgery, we want to remove a neighborhood of $T_{p,q}$.  A neighborhood of $T$ can be thought of as $A \times I$ in $T \times I$, where $A$ is an annular neighborhood of $T_{p,q}$ on $T$.  The boundary of this annulus, say $A \times \{1/2\}$, determines a slope $\gamma$ on the boundary of the exterior of $T_{p,q}$.  You can check that this is in fact $pq/1$.  (Note that it's guaranteed to be integral because it's coming from a framing of the knot.)  To find $S^3_{pq/1}(T_{p,q})$ we attach a solid torus so that meridional disks have boundary parallel to $\gamma$.  That means that in the surgered manifold we can build a surface by $T \times \{1/2\} - A \times \{1/2\}$ together with two disks from the surgery solid torus: one for each of the boundary components of $A \times \{1/2\}$.  Since $T - A$ is an annulus, that means we've built a sphere!  What is the rest of the manifold?  I'll leave it as an exercise to see why this sphere separates the surgered manifold into two punctured lens spaces.  Hint: Each of the two summands can be built in the following way: take a solid torus in the genus 1 Heegaard splitting of $S^3$ together with "half" of the surgery solid torus, so this is adding a three-dimensional 2-handle to a solid torus. 

The case of a cabled knot can actually be handled similarly.  Here's a description of the $(p,q)$-cable of a knot $J$.  On the boundary of a neighborhood of $J$, look at a $(p,q)$-curve, where we parametrize that torus as (longitude) x (meridian).  That is the $(p,q)$-cable of $J$.  Now, we can run the same argument as before, where instead of the standard genus 1 Heegaard splitting for $S^3$, the torus $T$ is from the boundary of a neighborhood of $J$.  In this case, when we do $S^3_{pq/1}(C_{p,q}(J))$ we will get two pieces: one is the exterior of $J$ with a 2-handle attached and the other is a solid torus with a 2-handle attached.  This gives the connected sum of a surgery on $J$ with a lens space!