Timeline for Is Cohen immersion conjecture (theorem) known for vector bundles?
Current License: CC BY-SA 4.0
8 events
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| May 22, 2019 at 23:36 | comment | added | Ryan Budney | Have you looked at Massey's paper where he formulated the immersion conjecture? The idea is to look at the generators of the cobordism ring. Massey proved his conjecture on those generators. That led him to the conjecture for arbitrary manifolds. | |
| May 22, 2019 at 14:14 | comment | added | Luis A. Florit | @Ryan: I don't see how to do what you propose, since the base is any compact manifold of any dimension... | |
| May 22, 2019 at 10:53 | comment | added | Ryan Budney | I don't think anyone has written anything up. The Massey conjecture bound is "any bound" but as noted above, it's not sharp. I suppose if you were looking for sharp bounds, the place to start would be to do what Massey did: enumerate bundles over all cobordism class representatives, and find the best immersion dimension for those bundles. | |
| May 21, 2019 at 20:14 | comment | added | Luis A. Florit | Sorry, maybe I should have stated my question more explicitly. By "this kind" I did not meant to say that "2n-a(n)" is sharp. Any bound, as long as it is better than Whitney's. | |
| May 21, 2019 at 17:20 | comment | added | mme | Therefore $di'$ extends to a bundle monomorphism of $\lambda \to S$ and hence $i'$ extends to an immersion $\lambda \to \Bbb R^4$; of course because we can make the underlying manifold embedded, we can make this line bundle embedded. | |
| May 21, 2019 at 17:18 | comment | added | mme | An obnoxiously fancy way to phrase your final paragraph: rank 2 vector bundles are classified by their first Stiefel-Whitney class and their twisted Euler class. Given any immersion $i: S \to \Bbb R^3$, the normal bundle of $i': S \to \Bbb R^4$ is isomorphic to $\det(S) \oplus \Bbb R$, and hence has trivial twisted Euler class; but given any line bundle $\lambda$ over $S$, $\lambda \oplus (\det(S) \otimes \lambda^{-1})$ has $w_1 = w_1(S)$ and has $w_2 = w_1(\lambda)^2 - w_1(S) w_1(\lambda) = 0$, using the Wu relations or an explicit calculation of $w_1(S)$ and the cohomology of a surface $S$. | |
| May 21, 2019 at 12:01 | history | edited | Ryan Budney | CC BY-SA 4.0 |
clean up statement -- although it's maybe a little vague
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| May 21, 2019 at 11:41 | history | answered | Ryan Budney | CC BY-SA 4.0 |