Irreducible homology 3-spheres that bound smooth contractible manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T19:44:08Z http://mathoverflow.net/feeds/question/104451 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104451/irreducible-homology-3-spheres-that-bound-smooth-contractible-manifolds Irreducible homology 3-spheres that bound smooth contractible manifolds Igor Belegradek 2012-08-11T00:24:51Z 2012-08-14T07:55:35Z <p>Some examples of irreducible homology 3-spheres that bound <b> smooth</b> contractible 4-manifolds are listed in the comment to problem 4.2 in <a href="http://math.berkeley.edu/~kirby/problems.ps.gz" rel="nofollow"> Kirby's problem list</a>, and all of them happen to occur among the <a href="http://www.maths.ed.ac.uk/~aar/papers/milnbries.pdf" rel="nofollow"> Brieskorn spheres</a> $\Sigma(p,q,r)$ modelled on $\widetilde{SL}_2(\mathbb R)$, i.e. such that $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}&lt;1$ (except for the standard $S^3$, of course). </p> <p><b> Question.</b> 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?</p> <p>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). </p> http://mathoverflow.net/questions/104451/irreducible-homology-3-spheres-that-bound-smooth-contractible-manifolds/104452#104452 Answer by Ryan Budney for Irreducible homology 3-spheres that bound smooth contractible manifolds Ryan Budney 2012-08-11T01:09:44Z 2012-08-11T01:09:44Z <p>Yes, there's loads of other contractible 4-manifolds bounding homology spheres with other geometries. </p> <p>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: <a href="http://front.math.ucdavis.edu/0810.2346" rel="nofollow">http://front.math.ucdavis.edu/0810.2346</a> Some are geometric, some have incompressible tori. </p> <p>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 &amp; Hillman "Embedding Seifert fibred 3-manifolds and Sol-manifolds in 4-space". </p> http://mathoverflow.net/questions/104451/irreducible-homology-3-spheres-that-bound-smooth-contractible-manifolds/104491#104491 Answer by Agol for Irreducible homology 3-spheres that bound smooth contractible manifolds Agol 2012-08-11T14:54:15Z 2012-08-11T14:54:15Z <p>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. </p> <p>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. </p> http://mathoverflow.net/questions/104451/irreducible-homology-3-spheres-that-bound-smooth-contractible-manifolds/104526#104526 Answer by Bruno Martelli for Irreducible homology 3-spheres that bound smooth contractible manifolds Bruno Martelli 2012-08-11T22:55:58Z 2012-08-14T07:55:35Z <p>This is a variation of Agol's answer. Take any contractible <i>special polyhedron</i> $X$, that is a compact polyhedron such that every point has a neighborhood of one of these three types:</p> <p><img src="http://www.dm.unipi.it/~martelli/images/models.png" alt="Neighborhoods"></p> <p>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 <a href="http://sketchesoftopology.wordpress.com/2010/03/25/bings-house/" rel="nofollow">Bing's house</a>: <img src="http://farm3.static.flickr.com/2724/4462178049_3ac33f0414.jpg" alt="Bings house"></p> <p>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 <i>locally flat</i> 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 <i>gleams</i>), 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.</p> <p>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: <a href="http://arxiv.org/abs/math/0506577" rel="nofollow">Costantino and Thurston</a> 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 $X$.</p> <p>Now as Agol said you only need to use <a href="http://en.wikipedia.org/wiki/Hyperbolic_Dehn_surgery" rel="nofollow">Thurston Dehn filling theorem</a> 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 <a href="http://en.wikipedia.org/wiki/2%CF%80_theorem" rel="nofollow">6 theorem</a> 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.</p>