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Dec 6, 2011 at 3:30 comment added Lucas Culler You can also use an embedding in S^6 (a bit easier to construct). If nM = normal bundle, then H^2(S^6-nM) = Z by Mayer Vietoris. A geometric representative of the generating class will be a 4-manifold X with dX = M. More precisely, dX will be cobordant to a nonvanishing section of nM.
Apr 29, 2011 at 23:37 comment added Dylan Thurston @Greg: It is related, though in a different direction; see my answer. @Daniel: I disagree that Whitney immersion is hard, and Hirsch's theorem is not necessary, as Bruno sketches.
Apr 29, 2011 at 16:47 comment added Daniel Moskovich This is beautifully intuitive! But it's definitely massive overkill- each step is in itself more difficult that the fact being proven. Statement (1) is the easiest statement among: 1) Omega_3=0 2)Whitney immersion theorem 3)Hirsch's theorem 4) Alexander duality in this context. This is highlighted by Rourke's short elementary proof (which isn't half as conceptually satisfying for me, though).
Apr 29, 2011 at 16:34 comment added Bruno Martelli You can avoid Hirsch's theorem by using Whitney's immersion theorem of $M^n$ in $\mathbb R^{2n-1}$, which implies that every 3-manifold $M$ immerses in $\mathbb R^5$. Such an immersion can be perturbed so that self-intersections have dimension 3+3-5 = 1, i.e. are circles. You can then surger the manifold around the circles to get an embedding, and such a surgery can easily be realized by a cobordism. Then you use Alexander duality. This was Rohlin's original argument.
Apr 29, 2011 at 13:52 comment added Autumn Kent It's funny, I sorta thought this was how everybody thought about it. I guess having Cameron Gordon as an advisor colors your world view.
Apr 29, 2011 at 13:24 comment added Greg Kuperberg That's quite interesting! And potentially related to the paper of Costantino and Thurston on complexity of 4-manifolds bounded by a 3-manifold.
Apr 29, 2011 at 13:01 history edited Autumn Kent CC BY-SA 3.0
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Apr 29, 2011 at 12:36 history answered Autumn Kent CC BY-SA 3.0