SNACKs and slice disks

Here's a short post on a construction I really enjoy.  

Theorem: (Kawauchi)  Let $K \subset S^3$ be a strongly negative amphichiral knot (SNACK).  Then $K$ bounds a smoothly embedded disk in a rational homology ball.  

Let me first explain the statement and then give a proof.  A knot $K$ is strongly negative amphichiral if there is an orientation-reversing diffeomorphism of $(S^3,K)$ which is also orientation-reversing on $K$ and the fixed point set is two points which sit on $K$.  Sometimes $K$ can be slice, e.g. the square knot is a slice SNACK, but often it is not, e.g. the figure-eight knot.  (The Fox-Milnor condition shows it is not even topologically slice.)  

Here is the proof, which is very quick.  First, $X_0(K)$ denote a 0-framed 2-handle attachment to $B^4$ along $K$.  Then $K$ is smoothly slice in this four-manifold - the slice disk is just the core of the 2-handle!  (This applies even if $K$ is not SNACK.)  Now, this four-manifold isn't a rational homology ball (it has the wrong homology and an extra boundary component homeomorphic to $S^3_0(K)$).  But we'll cap this off to get our rational ball.  

Now, suppose that $K$ is a SNACK.  Then, consider the orientation-reversing involution $\phi : S^3 \to S^3$.  Consequently, $\phi|_{E(K)}$ is a free orientation-reversing involution sending $\mu$ to $\mu$, like a rotation by $\pi$, and $\lambda$ to $-\lambda$.  We can extend this over a solid torus to get a free orientation-reversing involution $\widehat{\phi}$ on $S^3_0(K)$.  (Think of the involution on $S^1 \times S^1 = \mu \times \lambda$ as $\phi(x,y) = (-x, \overline{y})$.  Then, this formula extends over $S^1 \times D^2$ and there are no fixed points.  This is how the solid torus gets in glued for the 0-surgery.)  Now $S^3_0(K) \to S^3_0(K)/ \widehat{\phi}$ is a 2-sheeted covering (with the quotient being non-orientable), so let $W$ denote the mapping cylinder of the covering map $S^3_0(K)  \to S^3_0(K) / \widehat{\phi}$, or equivalently, $W$ is the twisted $I$-bundle over $S^3_0(K) / \widehat{\phi}$ corresponding to this 2-sheeted cover.  Then $W$ will have boundary $S^3_0(K)$ and the rational homology of a circle.  Gluing $W$ to $X_0(K)$ now gives a rational ball (the $H_1$ from $W$ cancels the $H_2$ from $X_0(K)$ rationally).  In fact, you can compute that $H_1(W \cup X_0(K)) = H_2(W \cup X_0(K)) = \mathbb{Z}/2$.  It turns out this manifold is spin (yey!) but the slice disk is homologically non-trivial rel boundary (boo!).   

I'll point out that in a more recent paper of Adam Levine, he shows that this rational homology ball is actually the same for every SNACK!  (In particular, it is the one you get for the unknot, which can be computed rather explicitly.  This has a simple handle decomposition with one 1-handle, two 2-handles, and one 3-handle.)   

In general, I think there's a useful takeaway from this that applies in broad settings.  Suppose you have a manifold with a boundary component you're trying to close off in a simple way.  If there's an orientation-reversing free involution of the boundary, then you can close off by simply quotienting by that involution.  Pretty neat!