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Jan 15, 2012 at 5:35 comment added Matt Brin @Chris. Lickorish's proof is a triumph of finding generators for the mapping class groups of surfaces and using the information to get conclusions about Heegard splittings. I believe that Wallace's proof uses Morse theory, but I have not read the proof. Lickorish's paper is quite readable.
Feb 10, 2010 at 19:58 comment added Igor Belegradek Maybe instead of saying that the empty set is simply-connected it is better to say that since a closed oriented 3-manifold bounds a compact 4-manifold, it is cobordant to the 3-sphere; the cobordism is obtained by removing an open disk from the interior of the 4-manifold.
Feb 10, 2010 at 19:15 comment added José Figueroa-O'Farrill On the subject of references for surgery theory, there are a number of "goodies" at my colleague Andrew Ranicki's web page: maths.ed.ac.uk/~aar/surgery
Feb 10, 2010 at 18:55 comment added Chris Schommer-Pries @Tim: You're right! I was implicitly thinking high dimensions. I edited to reflect the problem in dimension 3, and suggested one way to deal with it. Presumably there are more direct arguments from low-dimensional topology, rather then homotopy theory. For example it also follows from the Lickorish-Wallace theorem. How difficult is that to prove?
Feb 10, 2010 at 18:40 history edited Chris Schommer-Pries CC BY-SA 2.5
corrected spelling.
Feb 10, 2010 at 17:14 comment added Jason DeVito - on hiatus I see. When I think of a "generating set", I seem to (incorrectly) think of a "minimal generating set" instead. Thanks for the clarification.
Feb 10, 2010 at 17:12 comment added Tim Perutz @Jason. It can be a generator if it's killed by a relation.
Feb 10, 2010 at 17:09 comment added Jason DeVito - on hiatus I don't understand. If it's contractible, how can it be a generator?
Feb 10, 2010 at 16:29 comment added Tim Perutz Chris, what happens if your generator happens to be a contractible loop in a 3-manifold?
Feb 10, 2010 at 15:50 comment added Justin Curry Yeah, I think Hopkins was supposed to be at Princeton today (snow storm might have delayed this), explaining recent work with Mike Hill and Doug Ravenel that resolves the Kervaire Invariant problem in every dimension except 126.
Feb 10, 2010 at 15:36 comment added Chris Schommer-Pries btw, if you try to do this same thing but using framed manifolds you can't. Even in dimension 2 there is an obstruction known as the Arf invariant. This lead to a higher dimensional obstruction known as the Kervaire invariant, which has been the subject of some very exciting recent research.
Feb 10, 2010 at 15:32 comment added Chris Schommer-Pries No, I don't. It's rather disgraceful actually. I never had a proper class in surgery theory and I'm not at all familiar with the references. I just picked up bits and pieces along the way. I do hear that good references exist, though. Hopefully someone will chime in with some.
Feb 10, 2010 at 15:16 vote accept Justin Curry
Feb 10, 2010 at 15:16 comment added Justin Curry Thanks! As I am just learning some surgery theory, it is a new vocabulary that requires some practice. Do you have references that you recommend?
Feb 10, 2010 at 15:10 history answered Chris Schommer-Pries CC BY-SA 2.5