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Oct 8, 2010 at 21:01 comment added Ryan Budney Mazur manifolds (at least, ones that can be represented as $S^1 \times D^3$ union a $2$-handle) are all smooth submanifolds of $S^4$. On the Wikipedia page the reason is sketched in the section titled "Mazur's Observation". When you refer to configuration spaces, are you trying to go after a Salvatore-Longoni type problem, looking at the homotopy-type of configuration spaces of a Mazur manifold, that kind of thing?
Oct 8, 2010 at 16:44 comment added Victor Thank you Ryan. As usual you give a very interesting answer. In relation with that, don't you know what is known about the configuration spaces of these manifolds? And what about their embeddings into $R^3$, $R^4$, $R^5$...?
Oct 7, 2010 at 16:28 comment added Ryan Budney In full generality there's likely a natural Whitehead-torsion obstruction to doing what you want to do, given by the s-cobordism theorem. That would require more thought than I'm prepared for this morning.
Oct 7, 2010 at 16:12 comment added Ryan Budney I believe it's unknown whether or not there are exotic smooth structures on open 4-balls that extend to smooth structures on a compact 4-ball so that would be a potential counter-example for $k=0$. Less hypothetically, Mazur manifolds are counter-examples. In high dimensions if your manifold and its boundary are simply-connected the minimal handle presentation theorem would kick in and say there are no counter-examples. But once you have $\pi_1$ in either the manifold or the boundary there are potential obstructions. I wrote up a little thing on Mazur manifolds on Wikipedia.
Oct 7, 2010 at 14:40 comment added Victor Can one find a counterexample among interiors of compact manifolds with boundary?
Oct 5, 2010 at 18:59 vote accept Victor
Oct 5, 2010 at 18:15 history edited Ryan Budney CC BY-SA 2.5
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Oct 5, 2010 at 17:45 history edited Ryan Budney CC BY-SA 2.5
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Oct 5, 2010 at 17:37 history answered Ryan Budney CC BY-SA 2.5