Timeline for Can all n-manifolds be obtained by gluing finitely many blocks?
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Oct 20, 2010 at 12:15 | comment | added | Dylan Thurston | Hmm, interesting. I wonder if the block we described is the "simplest" in some sense. In 3-D, the elementary block needs to have positive hyperbolic volume in some suitable sense. The paper was motivated by 4-manifolds, and that block can be filled in to give a block for 4-manifolds that can be used to avoid 0-faces. (Also, note the spelling of Costantino's name.) | |
| Oct 19, 2010 at 7:41 | comment | added | Greg Kuperberg | Constantino and D. Thurston, to be precise. It seems that there is a lot left in this question. Even if one found an argument in higher dimensions, you could still ask for which $k$ could the blocks avoid $\le k$-dimensional faces. | |
| Oct 19, 2010 at 7:09 | comment | added | Bruno Martelli | I think you can do this in dimension 3. In jtopol.oxfordjournals.org/content/1/3/703.refs Costantino and Thurston define a universal set of cusped hypebolic manifolds that are constructed from copies of a single block $B$ with ridges. The block $B$ (also used in other papers by Agol, Minsky and others) is a 3-handlebody having 6 closed embedded curves as ridges. It is a hyperbolic manifold with geodesic boundary consisting of 4 pair of pants and having 6 annular cusps (the ridges). It is obtained by mirroring a checkerboard-coloured regular ideal octahedron along black triangles. | |
| Oct 18, 2010 at 20:39 | comment | added | Greg Kuperberg | I think that it's an interesting question whether you can make finitely many blocks with ridges but not corners. | |
| Oct 18, 2010 at 20:04 | comment | added | Greg Kuperberg | On the other hand, I don't see a complete argument using TQFTs, because even if you have finitely many block types you could still get large values for a TQFT invariant. | |
| Oct 18, 2010 at 19:52 | history | answered | Greg Kuperberg | CC BY-SA 2.5 |