Timeline for Can all n-manifolds be obtained by gluing finitely many blocks?
Current License: CC BY-SA 2.5
12 events
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| Jul 12, 2011 at 16:25 | comment | added | Daniel Moskovich | Slightly tangential, but: if the blocks allowed to have corners, then there's an interesting "yes" answer (the boring "yes" answer is "chop into simplexes) where the pieces behave like generators of a Hopf algebra. See ldtopology.wordpress.com/2011/06/26/… | |
| Nov 11, 2010 at 1:48 | answer | added | michael freedman | timeline score: 12 | |
| Nov 4, 2010 at 8:46 | vote | accept | Bruno Martelli | ||
| Nov 2, 2010 at 1:05 | answer | added | michael freedman | timeline score: 23 | |
| Oct 22, 2010 at 12:47 | history | edited | Bruno Martelli |
added "four-manifolds" tag
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| Oct 18, 2010 at 20:32 | comment | added | Bruno Martelli | Yes, I am convinced it works. One only needs to fix once for all the same Morse function (or splitting) on each block and then glue them together (see also Greg Kuperberg below). | |
| Oct 18, 2010 at 19:52 | answer | added | Greg Kuperberg | timeline score: 8 | |
| Oct 18, 2010 at 19:46 | comment | added | Ian Agol | Bruno - you're right, scratch the TQFT comment (I hadn't thought it through carefully). However, I'm sure that the argument in the other question works. The examples are a sequence of small 3-manifolds with Heegaard genus going to infinity. Then any decomposition into manifolds has to have a dual graph a tree, since $b_1=0$. So you can make a Morse function (or generalized Heegaard splitting) out of each piece, and apply the argument in that answer. | |
| Oct 18, 2010 at 17:07 | comment | added | Bruno Martelli | Thanks! On using TQFT: the number of blocks is however not bounded because of repetitions. For instance, graph manifolds have arbitrarily large TQFT invariants (for instance, they contain the connected sums of $S^2\times S^1$). Moreover, I think it is still unknown whether the invariants of graph manifolds cover all possible invariants of 3-manifolds. | |
| Oct 18, 2010 at 16:38 | comment | added | Ian Agol | Bruno: this is closely related to this question: mathoverflow.net/questions/30567/level-sets-of-morse-functions/… In particular, I think the same argument shows that there is not such set of finitely many compact 3-manifolds generating all hyperbolic 3-manifolds. One could likely make another argument using TQFT's. The value of a unitary TQFT invariant made from finitely many manifolds is bounded by the norm of the manifolds in the vector space associated to the boundary. So if one had manifolds with arbitrarily large TQFT invts., then this would be impossible. | |
| Oct 18, 2010 at 15:19 | answer | added | HJRW | timeline score: 4 | |
| Oct 18, 2010 at 13:15 | history | asked | Bruno Martelli | CC BY-SA 2.5 |