Contractible four-manifolds that are not Mazur-type (part 2)

Back for more!  Recall that we are interested in proving the following proposition: 

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.  

In the last post, we showed that it suffices to show that there is a prime homology 3-sphere that bounds a contractible 4-manifold that is *not* surgery on a knot in $S^2 \times S^1$.  That was because being surgery on a knot in $S^2 \times S^1$ is equivalent to bounding a Mazur manifold, and those are the only contractible 4-manifolds with a handle decomposition with fewer than five handles.  That's our goal.  If you don't care about a proof of the proposition, here's a surgery picture for an example (and if you do care, use this picture as reference for the constructive proof below):  

We'll start the proof by giving an obstruction to a homology sphere being surgery on a knot in $S^2 \times S^1$.  (A more general version of this result is described here.)  

Lemma: If $Y$ is a homology sphere that is surgery on a knot in $S^2 \times S^1$, then $\pi_1(Y)$ is weight one, i.e. it is normally generated by one element.  

Proof:  Suppose that $J \subset Y$ is a knot for which surgery produces $S^2 \times S^1$.  For homology reasons, the surgery coefficient for $J$ must be 0.  Let $M = Y - J$.  Then, we can see that $\pi_1(M)/ \ll \lambda_J \gg = \mathbb{Z}$ and $\pi_1(M)/\ll \mu_J \gg = \pi_1(Y)$.  Therefore, $\pi_1(Y)/ \ll \lambda_J\gg$ is the quotient of a cyclic group and a perfect group.  It follows that the resulting group is trivial.  In other words, the image of $\lambda_J$ in $\pi_1(Y)$ normally generates.  

The problem with this lemma is that until very recently, we knew no finitely generated perfect groups with weight bigger than one.  So this lemma was useless!  In fact, the famous Wiegold problem asked if every finitely generated perfect group was weight one.  We now have the following awesome theorem: 

Theorem (Chen-Lodha): If $G_1, G_2$ are non-trivial left-orderable groups, then $G_1 * G_2$ is not weight one.  

Here left-orderable means that the group admits a left-invariant order, e.g. < on $\mathbb{Z}$.  Lots of homology spheres have non-trivial left-orderable fundamental groups, such as any Seifert fibered homology sphere other than $S^3$ or the Poincare homology sphere.  (Conjecturally, having non-trivial left-orderable fundamental group is equivalent to being an L-space for prime rational homology spheres.)  Now, the unfortunate thing is that if we want a 3-manifold with $\pi_1$ a free product, it's going to be a connected sum.  But we wanted prime examples!  Nathan Dunfield made the following observation for getting around this issue.  If $\pi_1(Y_1), \pi_1(Y_2)$ are non-trivial, left-orderable and $K$ is a nullhomotopic knot in $Y_1 \# Y_2$, then any surgery on $K$ produces another 3-manifold whose fundamental group surjects $\pi_1(Y_1 \# Y_2)$ and hence will also fail to be weight 1.  But if we choose $K$ and the surgery coefficient carefully, then surgeries will no longer be prime, and still can't be surgery on a knot in $S^2 \times S^1$.  (For example, if $K$ is hyperbolic in $Y_1 \# Y_2$, then all but finitely many surgeries will be prime.)  

In case you got lost, here's our new goal: find a prime homology sphere which bounds a contractible 4-manifold that is obtained as surgery on a nullhomotopic knot in a connected sum of homology spheres with non-trivial left-orderable fundamental group.  Let's build one!!! 

The building block will be the Brieskorn sphere $Y = \Sigma(2,3,13)$, which can be described as $+1$-surgery on the pretzel knot $P(3,-3,6)$ in $S^3$, which is slice. (For example, see this paper of Jeff Meier for more on surgery on the family of slice pretzel knots $P(3,-3,k)$.)  As discussed in part 1, we can surger the slice disk for $P(3,-3,6)$ to get a contractible manifold $W$ with boundary $\Sigma(2,3,13)$ and disk in $W$ with boundary the core of the $+1$-surgery.  For notation, let $K$ in $\Sigma(2,3,13)$ denote the core of the $+1$-surgery on $P(3,-3,6)$ and $D$ the associated slice disk.  Note now that $Y \# -Y$ bounds a contractible 4-manifold: the boundary sum of $W$ and $-W$.  (The orientation reversal on the second summand is not necessary, but it will make the relevant surgery diagrams for our final manifolds a little simpler.)  Suppose we can find a nullhomotopic knot $J$ in $Y \# - Y$ which is slice in $W \natural -W$.  Then we could surger the slice disk to get a contractible 4-manifold whose boundary cannot have weight 1 fundamental group by the Chen-Lodha theorem + Dunfield observation!  This cannot be surgery on a knot in $S^2 \times S^1$.  If this homology sphere is prime, then we win!

In case you got lost again, here's our final goal: Find a nullhomotopic knot $J$ in $\Sigma(2,3,13) \# -\Sigma(2,3,13)$ which is slice in $W \natural -W$ for which some $1/n$-surgery on $J$ is prime.  

This might seem hard to find such a knot, but it's actually easy with a little trick.  If $Q \subset M^3$ is a slice knot in a 4-manifold $V$, then so is the Whitehead double of $Q$ - you can essentially Whitehead double the disk.  Here's how to do this if you care: punching out a neighborhood of a point in the slice disk in $V$ gives a concordance from $(S^3,U)$ to $(M,Q)$ in $V - B^4$.  A neighborhood of the concordance is $S^1 \times D^2 \times I$, where the concordance is $S^1 \times 0 \times I$.  Now Whitehead double the concordance at each $t \in I$.  This gives a concordance from $(S^3, Wh(U))$ to $(M, Wh(Q))$.  But the Whitehead double of $U$ is $U$, so we can cap off this concordance to get a slice disk for $Wh(Q)$ in $V$.  (See this paper for example for a more general principle but applied in $B^4$.)  Who cares?  We just took a knot and made it more complicated... But the Whitehead double of any knot in any 3-manifold is always nullhomotopic!!!  

So, now here's what we do.  Take the knot $K$ in $\Sigma(2,3,13)$ which is slice in the contractible $W$.  Take $Wh(K)$, which is nullhomotopic in $\Sigma(2,3,13)$ and still slice in $W$.  So take $Wh(K) \# - Wh(K)$ which is nullhomotopic in $Y \# - Y$ and slice in $W \natural -W$ because we just take the boundary sum of the two slice disks.  (It's important to make the distinction here that in general being slice in the 4-manifold doesn't mean you are nullhomotopic in the bounding 3-manifold.  So we really did need this Whitehead doubling trick.  To apply the Chen-Lodha theorem we really need the relevant knot to be nullhomotopic in the 3-manifold.)  

A little 3-manifold topology shows that every surgery on $Wh(K) \# -Wh(K)$ in $Y \# - Y$ will be prime.  (See below for a proof.)  Thus, a $1/n$-surgery on $J$ will produce a homology sphere that is prime, bounds contractible, but can't be surgery on $S^2 \times S^1$, so we have completed our proof!!!  If you were following along with the picture of the surgery description at the top, the curve in the center is the knot $J = Wh(K) \# - Wh(K)$.    

Appendix: Here's a sketch of the proof that surgery on $J = Wh(K) \# - Wh(K)$ in $Y \# - Y$ will always be irreducible. First, notice that $Y - K = S^3 - P(3,-3,6)$.  Since this pretzel knot is hyperbolic, e.g. you can check this in Snappy, we have that $Y - K$ is hyperbolic. The JSJ decomposition of the exterior of $Wh(K) \# - Wh(K)$ can then be described in the following way: there are two copies of $Y - K$ (hyperbolic), two copies of the exterior of the Whitehead link (hyperbolic), and one the exterior of the 3-component keychain link, pictured below.  

The keychain link exterior is the piece of the JSJ decomposition that contains the boundary of the exterior of $J$ in $Y \# - Y$, and the boundary corresponds to the middle component.  Any Dehn filling of the middle component (other than the trivial filling) can be shown to produce a prime Seifert fibered 3-manifold whose boundary is incompressible.  It follows that surgery on $J$ is prime with JSJ decomposition having four hyperbolic pieces (the four from the exterior of $J$) and one Seifert fibered piece (the corresponding Dehn filling of the keychain link).