# Mazur homology spheres

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?

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 $$S$$).

• How about the case $L_2$ is $0$-surgery on the unknot as well?
– user160180
Dec 30, 2020 at 23:39
• 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?
– user160180
Dec 31, 2020 at 9:29
• 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. Dec 31, 2020 at 14:46
• If you attach two 2-handles to $B^4$, you never get a contractible manifold (that's the answer to your second question). Dec 31, 2020 at 16:46
• 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. Jan 1, 2021 at 21:52