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Some examples of irreducible homology 3-spheres that bound smooth contractible 4-manifolds are listed in the comment to problem 4.2 in Kirby's problem list, and all of them happen to occur among the Brieskorn spheres $\Sigma(p,q,r)$ modelled on $\widetilde{SL}_2(\mathbb R)$, i.e. such that $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}<1$ (except for the standard $S^3$, of course).

Question. Are there any known homology 3-spheres that bound smooth contractible 4-manifolds and are modelled on other geometries, e.g. NIL or the hyperbolic space?

EDIT: Glazner in the paper "Uncountably many contractible 4-manifolds" constructed some other examples but I cannot recognize the geometry. (Glazner's six page paper is easily googlable by title, and it gives an explicit representation for the fundamental group, denoted $G_n$ on page 40).

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3 Answers 3

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Yes, there's loads of other contractible 4-manifolds bounding homology spheres with other geometries.

For example $1$-surgery on the Stevedore knot is a hyperbolic manifold with volume $1.3985\cdots$. This bounds a contractible manifold by the same kind of arguments Casson and Harer used to cook up their big list of Mazur manifolds. There are other examples like this that occur in my arXiv preprint: http://front.math.ucdavis.edu/0810.2346 Some are geometric, some have incompressible tori.

I don't think any Nil manifolds bound contractible 4-manifolds. Crisp and Hillman determined all the Nil manifolds that embed smoothly (or topologically) in the $4$-sphere. The only Nil manifolds on that list have non-trivial homology. Crisp & Hillman "Embedding Seifert fibred 3-manifolds and Sol-manifolds in 4-space".

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The paper is structured sort of like a "workbook" with techniques building up in the beginning, computations in the middle and appendices in the back. This example is manifold 30, on page 22 of the current arXiv version (v4). See also Proposition 2.10 on page 11 for context and details of the argument. Towards the end of the summer I'll upload v5 of the paper -- the paper will be much closer to being complete. –  Ryan Budney Aug 11 '12 at 2:07
    
Thank you! I see manifold 30, but where are the "other examples" you mention? Where do you prove 30 is hyperbolic? Also am I correct that this example is due to yourself (and was not worked out by Casson-Harer)? –  Igor Belegradek Aug 11 '12 at 2:20
    
As far as I know it's due to me. If you look in the credits for the paper I certainly thank a lot of people for their insights. Danny Ruberman and I talked about these types of manifolds quite a bit so he certainly deserves some credit. Manifold 31 also bounds a contractible 4-manifold. I think there's others in the list -- when I wrote the paper determining whether or not the bounding 4-manifolds was contractible was not a priority, it was at most a step towards finding embeddings in $S^4$. The proof of hyperbolicity is via SnapPea (which is rigorous with the Harriet Moser criterion). –  Ryan Budney Aug 11 '12 at 2:32
    
I only asked about Casson-Harer because you mentioned them. Also many thanks for bringing to my attention Harriet Moser's work front.math.ucdavis.edu/0809.1203 which I was not aware of. –  Igor Belegradek Aug 11 '12 at 2:48
    
Incidentally, Glazner's manifolds in my EDIT embed into $S^4$ but you do not refer to that paper. –  Igor Belegradek Aug 11 '12 at 3:05

If you have a contractible 2-complex, then you can construct many different 4-manifolds with homology sphere boundary, and I think it is flexible enough that you can make them hyperbolic. Take the 1-skeleton of the complex, and thicken it to a 4-dim. handlebody, whose boundary is $\sharp _k S^2\times S^1$. The boundaries of the 2-cells in the complex may be realized by loops in $\sharp_k S^2\times S^1$ realized by a link, and one attaches a 2-handle to each such link. The resulting manifold is obtained by $0$-framed surgery on this link. By choosing different such links representing the same homotopy class in the 4-handlebody, I think you can arrange a manifold to be hyperbolic.

For example, if the 2-complex is a disk, then the 1-skeleton is $S^1$, so you want to take a knot in $S^2\times S^1$ which meets $S^2$ algebraically once. I think 0-framed surgery on a sufficiently complicated such knot will usually give a hyperbolic 3-manifold bounding a contractible 4-manifold.

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Thank you, this clarifies things for me quite a bit. The power of hyperbolic Dehn surgery allows one not to be very particular about what complex to take, or how to thicken it. –  Igor Belegradek Aug 11 '12 at 15:24
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I'm not sure why I specified 0-framed surgery - as Bruno points out in his answer, one may perform n-surgery on the link components, and the resulting manifold is still contractible. –  Ian Agol Aug 15 '12 at 6:54

This is a variation of Agol's answer. Take any contractible special polyhedron $X$, that is a compact polyhedron such that every point has a neighborhood of one of these three types:

Neighborhoods

and such that the natural stratification is a cellularization, i.e. all the points as in the center (right) form disjoint open 1-cells (2-cells). A famous example is Bing's house: Bings house

Now thicken 4-dimensionally the polyhedron $X$. You do this first by thickening the 1-skeleton, and this can be done in a unique way (we only construct orientable 4-manifolds). The rest can be thickened in various ways, but for simplicity we restrict ourselves only to locally flat thickenings, i.e. thickenings that locally lie in a (smooth) 3-dimensional slice (a very reasonable assumption). As shown by Turaev, the locally flat thickenings are parametrized by assigning a (half-)integer at each 2-cell of the polyhedron (these numbers are called gleams), which measures the "Euler number" of the thickening. This is very similar to the description of the 4-manifold via a Kirby diagram. For instance in the Bing's house there are three 2-cells and each needs to be coloured by an integer.

The boundary 3-manifolds of all these thickenings are all obtained by Dehn surgery on a fixed link $L$ in a connected sum $\sharp_h(S^2\times S^1)$ of some $h$ copies of $S^2\times S^1$. Different gleams yield different surgeries on the same link $L$. The nice thing here is the following: Costantino and Thurston have shown that the link $L$ is always hyperbolic; the complement $\sharp_h(S^2\times S^1)\setminus L$ can be constructed by gluing some regular ideal octahedra. You need two regular octahedra for each vertex of $X$, so that the volume of the hyperbolic manifold is precisely twice the volume of the regular ideal hyperbolic octahedron times the number of vertices of $X$.

Now as Agol said you only need to use Thurston Dehn filling theorem that ensures you that by avoiding a finite set of gleams on each face the resulting 3-manifold is hyperbolic. That is, the boundary of the resulting thickened 4-manifold is hyperbolic. Moreover the 6 theorem tells you that if all the gleams are bigger than 6 you are certainly done (because we know very well how the cusps are). Hence you can decorate Bing's house with a triple of integers bigger than 6 and you get plenty of hyperbolic 3-manifolds bounding contractible 4-manifolds.

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Thank you! I am confused why "the boundary 3-manifolds of all these thickenings are all obtained by Dehn surgery on a fixed link $L$ in a connected sum of $S^2\times S^1$'s. Does this follow by construction of the thickening? –  Igor Belegradek Aug 12 '12 at 14:00
    
Yes, the thickening is done in two steps: first thicken a regular neighborhood $N$ of the 1-skeleton of $X$ to a 4-dimensional handlebody $H$ made of $0$- and $1$-handles (corresponding to vertices and edges of $X$). You can think of $N$ as properly embedded in $H$ so that $\partial N\subset \partial H$ is a link in the closed 3-manifold $\partial H$ which is homeomorphic to $\sharp_k (S^2\times S^1)$. Such a link is fully determined by $X$ and can be drawn from it, there should be an example in the paper of Costantino and Thurston. –  Bruno Martelli Aug 12 '12 at 16:20
    
The second step is thickening the remaining 2-cells. For every 2-cell there is a $\mathbb Z$-freedom, like in Kirby diagrams. The thickening is determined by an integer (the gleam). The boundary 3-manifold changes by surgerying along the link $\partial N$ via a longitudinal surgery, which also depends on such integer. –  Bruno Martelli Aug 12 '12 at 16:22
    
That's just an extension to the well-known 4-dimensional thickenings on a closed surface $\Sigma$: the thickening on a 1-skeleton is determined, but there is a $\mathbb Z$-freedom on each 2-cell, giving globally the Euler number of the thickening. –  Bruno Martelli Aug 12 '12 at 16:23
    
I really aprreciate that you took the time to spell things out. –  Igor Belegradek Aug 12 '12 at 17:08

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