Smith theory (part 2)

In Smith theory (part 1) we discussed the Smith Conjecture which stated that the fixed point set of a smooth cyclic action on the 3-sphere is always unknotted.  

Today, the goal is to give a hands on disproof of the 4D Smith Conjecture.  We'll cook up $\mathbb{Z}/2$-actions on $S^4$ with knotted fixed point set.  (The action we build will be smooth and the fixed point set will be knotted in the topological and smooth categories.)

To do this, we need a more interesting way to decompose $S^4$ into two pieces.  First, an observation.  Consider a Mazur manifold $W$ - a four-manifold built by attaching a single 1-handle to $B^4$ and a 2-handle that algebraically links it once.  (The framing can be anything, the 2-handle can be knotted in whatever way.)  Here's an example:

These manifolds will be contractible, but might still not be homeomorphic to $B^4$; in fact, if your 2-handle is attached along anything other than $pt \times S^1$, the resulting boundary will *not* be $S^3$.  Now, if you take the double of $W$, i.e. $W \cup -W$, then we do always get $S^4$.  The point is that we can get a Kirby diagram with one 1-handle and two 2-handles (and unique 0, 3-handle, and 4-handle attachments).  The first 2-handle is the original 2-handle $J$ and the second one $\mu$ links it like a little meridian and has 0-framing. 

So, we can do handleslides of $J$ over $\mu$ to perform self-crossing changes, until we turn $J$ into a copy of $pt \times S^1$ in the boundary of the 1-handle attachment.  Therefore, we can cancel the 1- and 2-handles and then with what's leftover is a 0-framed unknotted 2-handle which we can cancel with the 3-handle.  So, 0-handle + 4-handle makes $S^4$!  If this is all mystery talk, then head right on down to the library and get yourself a copy of Gompf and Stipsicz's "Kirby Calculus and 4-Manifolds" and work through Chapters 4 and 5.  You'll thank me later. 

 

Anyway, with this, we are ready for our strategy to disprove the Smith Conjecture.  Take a knotted sphere $S$ in $S^4$ such that $S$ is the double of a disk $D$ in $B^4$.  If the branched double cover of $D$ is a Mazur manifold, then the branched double cover of $S$ will be the double of the Mazur manifold, i.e. $S^4$.  Let $\widetilde{S}$ denote the preimage of $S$ in $\Sigma_2(S^4, S)$.  This is the fixed point set of the covering involution.  So, if we can show $\widetilde{S}$ is knotted then we've found our counterexample to the Smith Conjecture.  So, how do we do this?  With more observations!

 

- Suppose $K$ is a knot in $S^3$ obtained by a band surgery along a 2-component unlink (i.e. attach a band and look at the new one-component knot that results as the boundary).  Such a knot is called fusion number 1.  Here's a picture: 

- $K$ bounds a ribbon disk $D$ in $B^4$, which is obtained from taking two trivial disks to bound the 2-component unlink and then attaching the band.  (This disk will have self-intersections in $S^3$ but you can make this disk embedded in $B^4$.)  The complement of $D$ in $B^4$ has a handle decomposition with two 1-handles and one 2-handle.

 

- The double cover of $B^4$ branched over $D$ will be obtained from taking a 1-handle and attaching a 2-handle.  (If we did more bands to a bigger unlink, then the branched double cover would have more 1- and 2-handles.)  Now, the 2-handle may not be algebraic linking 1, but if $det(K) = 1$, then the branched double cover of $K$ will be a homology sphere and so the linking of the 2-handle will have to be 1!  (If the algebraic linking of the 2-handle is $n$, then the homology of the boundary of the four-manifold can be computed to have order $n^2$ using the linking matrix.)  

In summary: the branched double cover of the associated ribbon disk for a determinant 1, fusion number 1 knot is a Mazur manifold.

 

So, now we just need a knot $K$ with:

1) determinant 1         

2) fusion number 1 (i.e. it can be obtained from banding a two-component unlink)

3) the preimage of the double of this ribbon disk is not an unknotted 2-sphere in the branched double cover is knotted.

Probably it's reasonable to believe we can find a non-trivial knot satisfying 1) and 2).  The hard part is guaranteeing 3).  Here's how we will do it.  If the complement of the ribbon disk $D$ in $B^4$ has non-cyclic $\pi_1$ then so will the double of that disk, and hence it will be a knotted 2-sphere, $S$, downstairs.  On the other hand, the complement of $\widetilde{S}$ in $S^4$ is the double cover of $S^4 - S$, so $\pi_1(S^4 - \widetilde{S})$ is an index two subgroup of $\pi_1(S^4 - S)$ which abelianizes to $\mathbb{Z}$.  If $\pi_1(S^4 - \widetilde{S})$ was $\mathbb{Z}$, then this would imply $\pi_1(S^4 - S)$ would be $\mathbb{Z}$ as well.  (The only $\mathbb{Z}$-extensions of $\mathbb{Z}/2$, either central or non-central, are $\mathbb{Z}, \mathbb{Z} \oplus \mathbb{Z}/2$ and the infinite dihedral group.)  Therefore, we would get that $\pi_1(S^4 - \widetilde{S})$ must be non-cyclic.  TL; DR: if we can show that $\pi_1(B^4  - D)$ is non-cyclic, then $\widetilde{S}$ is knotted.

 

In more summary: We want a determinant 1, fusion number 1 knot whose ribbon disk exterior in $B^4$ has non-cyclic fundamental group.  So, we just need to find the right knot.  The knot $10_{153}$ will do!  Here's a picture of the knot and the band move to a 2-component unlink (in a weird diagram). 

 

You can check at KnotInfo, this knot has determinant 1.  But why does the complement of the ribbon disk $D$ in $B^4$ have non-cyclic fundamental group?  If it did, then all cyclic branched covers of $D$ would be simply-connected and built out of just 0-, 1-, and 2-handles. Even better, they are contractible four-manifolds.  (Proof: Since there's no 3-handles, this means there's no $H_n$ for $n \geq 3$ and $H_2$ has no torsion.  Smith Theory (pt 1) tells us the Euler characteristic of the $n$-fold branched cover agrees mod $n$ with the Euler characteristic of the fixed point set, which is a disk, and so the rank of $H_2$ must be 0.  Therefore, by Hurewicz + Whitehead, this is contractible.)  Now, the boundary of a contractible four-manifold is always a homology sphere.  However, the 3-fold branched cover of $10_{153}$ is $\mathbb{Z}/7 \oplus \mathbb{Z}/7$, and hence our disk has non-cyclic exterior.  (To verify this computation, you can compute this from a Seifert matrix or have KnotInfo tell you using their "Torsion Numbers" option.)  This was the last thing we needed to show that the 4D Smith Conjecture is false!