Contractible four-manifolds that are not Mazur-type (part 1)

This sequence of posts was inspired by some conversations with Mike Miller Eismeier

In this post, I want to talk about some contractible four-manifolds and connections with Dehn surgery.  Recall that lots of homology 3-spheres bound contractible four-manifolds.  For example, $S^3_{1/n}(K)$ bounds a contractible four-manifold if $K$ is a smoothly slice knot, as do the Brieskorn spheres $\Sigma(3,4,5)$ and $\Sigma(2,3,13)$.  By a quick Euler characteristic computation, it is easy to see that every contractible four-manifold other than the four-ball requires at least three handles.  In this sequence of posts I want to give a proof of the following goofy result (and give it some context). 

Proposition: There exists an irreducible, i.e. prime, homology 3-sphere which bounds a contractible 4-manifold and no bounding contractible four-manifold can be built with fewer than five handles.  Alternatively, there are prime homology 3-spheres that bound contractible 4-manifolds which are not surgery on a knot in $S^2 \times S^1$.  

My guess is someone knows a more elementary proof of this, but I wanted to write a proof of this to highlight a recent breakthrough result in group theory due to Lvzhou Chen and Yash Lodha, called the Wiegold problem.  If I was a more thoughtful blogger I would tell you some interesting background on weight one groups, the Wiegold problem, and its rich connection with Dehn surgery in $S^3$.  Unfortunately, I am not such a person, so I will refer you to their paper for a great treatise on this.  Instead, I am going to start from the naive assumption that you think contractible four-manifolds are interesting and go from there.  

If you are a 4-manifold and/or Dehn surgery pro, you may want to skip to Part 2 where I'll actually prove the Proposition and talk about Chen-Lodha's result.  The rest of this Part 1 post is background on contractible four-manifolds, their handle decompositions, and connections with Dehn surgery.   

To begin, we need to analyze the handle decompositions of contractible 4-manifolds.  Suppose that $X$ is a contractible 4-manifold and it has one 0-handle, $k$ 1-handles, $\ell$ 2-handles, and $m$ 3-handles.  Let $Y$ denote the boundary of $X$, which is necessarily a homology 3-sphere.  The following result really re-shaped how I think about 3.5-dimensional topology, so I want to include it here: 

Proposition (Casson): If the boundary of $X$ is not $S^3$, then $k>0$.  

Proof: Suppose for contradiction that $k = 0$.  Then flipping $X$ upside down, $X$ is built from $Y$ by adding $m$ 1-handles and $\ell$ 2-handles and a 4-handle.  By contractibility, $\ell = m$.  We can then get a presentation for $\pi_1(X)$, i.e. the trivial group, by taking the free product of $\pi_1(Y)$ with a free group on $m$ generators and then adding $m$ relations.  In the abelianization of $\pi_1(Y) * F_m$, which is $\mathbb{Z}^m$, the $m$ relations are linearly independent.  Here's where the magic comes in! The fundamental groups of 3-manifolds have a property called residually finite, which means that for every non-trivial element in $\pi_1(Y)$, we can find a homomorphism to a finite group that is non-vanishing on this element.  Gerstenhaber-Rothaus proved that residually finite groups have the amazing property that when you take a non-trivial group and free product with a rank $m$ free group and add exactly $m$ relations which are linearly independent after abelianizing and projecting to the $\mathbb{Z}^m$ factor coming from the free group, the new group is still non-trivial!  This means that in our purported contractible 4-manifold, $\pi_1(X)$ is in fact non-trivial if $\pi_1(Y)$ is non-trivial.  Since $S^3$ is the only simply-connected homology 3-sphere, $\pi_1(Y)$ is non-trivial, and that's our contradiction.    

Cool!  This is super old school, and this general idea has been around in many other places, such as Cameron Gordon's classical work on ribbon concordance.  

Ok, so what does this mean for us?  Suppose we have a contractible four-manifold whose boundary is not $S^3$.  Then, we know our contractible four-manifold has at least one 1-handle.  If $\pi_1 = 0$, there better be a 2-handle killing it off.  So, together with the 0-handle, we are up to at least 3 handles.  Actually, there are lots of interesting homology spheres that bound contractible four-manifolds built out of exactly 3 handles.  Here how to build one of them (in fact all of them): 1) Attach a single 1-handle to $B^4$, so the boundary is now $S^2 \times S^1$, 2) attach a single 2-handle along a knot in $S^2 \times S^1$ that is homologous but *not* isotopic to $pt \times S^1$.  Here's a sample picture of such a knot: 

This manifold has $\pi_1 = 0$ (the 2-handle kills $\pi_1$ from the 1-handle) and all reduced homology trivial.  Hurewicz + Whitehead now tell us our manifold is contractible! These contractible manifolds are called Mazur or Mazur-type manifolds.  The boundary can be described as surgery on the chosen knot in $S^2 \times S^1$, and the surgery coefficient depends on the framing of the 2-handle.  

Gabai's Property R theorem says that the boundary of this contractible four-manifold is never $S^3$.  (Recall that Property R says that only the unknot in $S^3$ can surger to $S^2 \times S^1$; dually, only $pt \times S^1$ in $S^2 \times S^1$ can surger to $S^3$.)  As a funny remark, Gabai's Property R theorem actually implies that any contractible 4-manifold with boundary $S^3$ that is not $B^4$ necessarily has at least 5 handles as well!

So, how can we build interesting homology spheres that don't bound contractible four-manifolds with the minimal number of handles (3)?  The most naive thing is the following: let $X_1, X_2$ be two Mazur 4-manifolds with boundary $Y_1, Y_2$.  Then take the boundary sum of $X_1$ and $X_2$, which has boundary $Y_1 \# Y_2$.  This 4-manifold now naturally has a handle decomposition with 5 handles and seems intrinsically more complicated.  But maybe it could have a simpler handle decomposition, or $Y_1 \# Y_2$ could bound a different contractible manifold which is Mazur-type.  However, we can now use Dehn surgery magic: Gordon-Luecke tell us that a non-prime homology sphere can never arise as surgery on a knot in $S^2 \times S^1$.  So, this means we can build lots of (non-prime) homology spheres bounding contractible 4-manifolds which can't bound a Mazur manifold.  This means that any contractibles that they bound have to have at least four handles.  As a cheat, since the Euler characteristic is 1, they need at least five handles!  So, now the next natural (i.e. contrived) question to ask is whether one can cook up prime examples.  

In the next part, we'll construct the promised prime homology spheres with the same property - they bound contractible, but any bounding contractible needs at least five handles total.