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.


1 Answer 1


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 $S^1\times B^3$.

  • $\begingroup$ 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
    Dec 26, 2020 at 21:35
  • 1
    $\begingroup$ 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. $\endgroup$ 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
    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. $\endgroup$ 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
    Dec 29, 2020 at 11:19

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