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We know that if we attach $4$-dimensional $2$-handle $D^2 \times D^2$ to $S^1 \times S^2$, then we produce a contractible $4$-manifold. In this case, $S^1 \times S^2$ is $0$-surgery on the unknot.

If we replace the unknot with a slice knot, can we still have a contractible manifold? Is there an easy argument for this?

Note: Here, slice knots bound smoothly properly embedded disks in $4$-balls.

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2 Answers 2

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Yes, this can be done, but requires a little care with the fundamental group. First, let me tighten up your description; one is attaching the 2-handle to $S^1 \times B^3$ along a curve $\gamma$ in its boundary $S^1 \times S^2$. In order to get a contractible manifold, $\gamma$ should generate $\pi_1(S^1\times B^3) = \pi_1(S^1\times S^2)$. The point of this is to make sure that the resulting fundamental group is trivial. Such manifolds are often called Mazur manifolds (although the terminology is not universal.)

The key observation is that $S^1 \times B^3$ is the exterior of an unknotted disk in $B^4$. You could replace that with any disk D (with a slice knot K as boundary) but now we have to be more careful about the fundamental group. You want to require that $\gamma$ normally generate the fundamental group of $B^4 -nhd(D)$. If $\gamma$ is freely homotopic to a meridian of $S^3 - K$, then this will hold. In this case the manifold $W$ obtained by adding a 2-handle to $B^4 - nhd(D)$ is simply connected, and it's easy to check that its homology vanishes, so it will be contractible.

The boundary of the manifold built this way is obtained by surgery on $0$-surgery on $K$. Presumably one can find examples of this sort that aren't obtained by adding a 2-handle to $S^1\times B^3$.

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  • $\begingroup$ Thus, if $\gamma$ is a knot in the $3$-manifold obtained by $0$-surgery on a slice knot, and if we attach $2$-handle along $\gamma$, we produce a contractible manifold, right? $\endgroup$
    – user160180
    Commented Dec 26, 2020 at 21:35
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    $\begingroup$ Yes, providing that the condition I mentioned is satisfied: the homotopy class of $\gamma$ must normally generate the fundamental group of the 0-surgered manifold. $\endgroup$ Commented Dec 26, 2020 at 21:49
  • $\begingroup$ Attaching $2$-handle corresponds to doing integral surgery. Let $Y$ be the $3$-manifold which is obtained by $0$-surgery on a slice knot. We also need to assume that the resulting $3$-manifold which is obtained by surgery on a knot in $Y$ is a homology sphere, right? Or it is a consequence of having contractible $4$-manifold? $\endgroup$
    – user160180
    Commented Dec 27, 2020 at 11:22
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    $\begingroup$ There are two knots in question. The slice knot K and the other curve $\gamma$. If you do 0-framed surgery on K, and $\gamma$ normally generates the fundamental group, then you get a homology sphere no matter what the framing is on $\gamma$. As you say, the boundary of a contractible manifold must be a homology sphere. $\endgroup$ Commented Dec 27, 2020 at 19:51
  • $\begingroup$ I checked the algebraic details so that everything holds. I have two more questions: 1. Why we can replace the slice disk freely? 2. These contractible manifolds are all built from one 0-handle, one 1-handle and one 2-handle, i.e., all of them are Mazur? $\endgroup$
    – user160180
    Commented Dec 29, 2020 at 11:19
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I'm not sure if I should create a new post or this answer is fine.

Danny's answer is really great, and it helped me understand a really nice trick I found in Piccirillo-Hayden paper "New curiosities in the menagerie of corks" at https://arxiv.org/abs/2005.08928.

As he said, the key observation is that the exterior of any properly embedded unknotted disk is $S^1\times B^3$. In Gompf-Stipsicz, 6.2, the 1-handle notation is extended so that a dotted ribbon knot refers to the manifold obtained by removing the 4d neighborhood of the ribbon disk. This extends indeed because, in dotted circle notation, attaching a 1-handle is the same as removing the 4d neighborhood of the obvious disk the dotted circle bounds (you have to push it to the interior of the 4-ball so it's properly embedded).

Now, here's the trick: say we start with a diagram with a 1-handle and a 0-framed 2-handle which is a ribbon knot. Let their linking number be one. As per Danny's answer, this means the 4-manifold is contractible, since linking number one means the 2-handle's homotopy class is that of the meridian of the 1-handle. Now "switch" their roles, dotting the ribbon knot instead (and 0-frame on the other). The result will be contractible for the same reason. But more than that, they will have the same integer homology sphere for boundary!

The boundary in both cases is the same because as far as the 3d boundary is concerned, the effect of removing the (thickened) ribbon disk of a ribbon knot or of attaching a 0-framed 2-handle along it is the same: Dehn surgery with slope $\frac{0}{1}$. One just checks that the removal of the ribbon disk removes an $S^1\times D^2$ from $S^3$ and glues another one sending meridian to meridian.

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