This is Example 6.47 in Saveliev's book Invariants for homology $3$-spheres:

Let us consider a two-component link $\mathcal L = L_1 \cup L_2$ in $S^3$ such that $\mathrm{lk}(L_1,L_2) = \pm 1$ and the component $L_1$ is an unknot. Let $\Sigma_p (\mathcal L)$ be the integral homology sphere obtained by surgery on the link $\mathcal L$ with $L_1$ framed by $0$ and $L_2$ framed by $p$. The manifolds $\Sigma_p (\mathcal L)$ are referred as Mazur homology spheres because they generalize the original Mazur's example of a homology sphere bounding a smooth acyclic $4$-manifold.

  1. How we generalize Mazur's construction? Originally, we attach a $2$-handle $B^2 \times B^2$ to $S^1 \times B^3$ along a knot on the boundary $\partial S^1 \times B^3 = S^1 \times S^2$.
  2. Do we attach two $2$-handles to obtain an acyclic $4$-manifold?

1 Answer 1


Saveliev writes:

they generalize the original Mazur's example

and not "exampleS". The original example he's referring to is a single contractible 4-manifold with a boundary that's not the 3-sphere. Then there is a wealth of Mazur (4-)manifolds which are obtained, as you say, by attaching a 2-handle to $S^1\times B^3$. Saveliev talks about Mazur homology (3-)spheres, which are the boundaries of such objects. Now that we have cleared this, I'll get to your questions.

  1. To me, the key point is that we can skip the 1-handle attachment in the construction of a Mazur manifold, and work directly in $S^1\times S^2$ (which we can then fill in uniquely with $S^1 \times B^3$, thanks to Laudenbach and Poenaru). In this sense, to get a Mazur manifold, we just need a knot in $S^1\times S^2$ that generates the homology but such that doing surgery along it doesn't produce $S^3$ (essentially, thanks to Gabai, we just need a knot that's not isotopic to a fibre). In Saveliev's example, once we do the 0-surgery along $L_1$ we are in business, since $L_1$ is unknotted and doing 0-surgery along it this yields $S^1\times S^2$. (Then we need to make sure that $L_2$ does not become a fibre, but this might be harder to do.)

  2. No, attaching the two 2-handles does not yield a homology ball. For instance, because the Euler characteristic is 3 instead of 1. What you need to do is surger out the 2-sphere $S$ of self-intersection 0 that you get by capping off a Seifert surface for $L_1$ and replace it with an $S^1\times D^3$ (this corresponds to attaching a 5-dimensional 3-handle to $S$).

  • $\begingroup$ How about the case $L_2$ is $0$-surgery on the unknot as well? $\endgroup$
    – user160180
    Dec 30, 2020 at 23:39
  • $\begingroup$ I shall clarify my previous comment. We may attach two $2$-handles to two $S^1 \times S^2$. If these unknots are not knotted, then we produce an acyclic manifold rather than a contractible manifold. How can we produce a contractible manifold by generalizing Mazur's example? $\endgroup$
    – user160180
    Dec 31, 2020 at 9:29
  • $\begingroup$ If I understand correctly, you'll just obtain two (a priori non-diffeomorphic) contractible 4-manifolds with the same boundary. Even if they're diffeomorphic, you might gain a so-called cork. $\endgroup$ Dec 31, 2020 at 14:46
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    $\begingroup$ If you attach two 2-handles to $B^4$, you never get a contractible manifold (that's the answer to your second question). $\endgroup$ Dec 31, 2020 at 16:46
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    $\begingroup$ There is no "right" Kirby diagram. But you can read off a presentation of $\pi_1$ of the 4-manifold from the surgery diagram. If you can show that the fundamental group is trivial, you're in business. $\endgroup$ Jan 1, 2021 at 21:52

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