Timeline for Open immersions of open manifolds
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 13, 2016 at 17:15 | comment | added | Tom Goodwillie | I don't know what I was thinking! | |
| Jun 22, 2012 at 21:55 | comment | added | Misha | @Tom: Could you write more details as an "answer": I tried, unsuccessfully, inductive arguments and it would be nice to know why exactly they do no work. What is the least dimension where the relative version fails? | |
| Jun 22, 2012 at 0:38 | comment | added | Misha | @Ben: Thanks for the comments, I will take a look at the B-P papers. | |
| Jun 21, 2012 at 21:52 | comment | added | Tom Goodwillie | Note that an obvious relative version is false. You might hope to construct an immersion $M\to N$ by induction over a handle structure of $M$ (without any index $n$ handles, since $M$ is open)). But there can be no $N$ such that immersions of $n$-manifolds in $N$ can always be extended across a handle of index less than $n$: by Smale-Hirsch, that would mean that the relative homotopy groups of the tangent bundle map $N\to BO_n$ vanish up to $\pi_{n-1}$. But that would force the mod $2$ Betti numbers of $N$ to violate Poincare duality. | |
| Jun 20, 2012 at 22:12 | answer | added | Daniele Zuddas | timeline score: 8 | |
| Jun 20, 2012 at 19:24 | comment | added | Ben Wieland | Roughly, you are trying to approximate the $n-1$-skeleton of $BO(n)$ by a closed (compact?) $n$-manifold and its canonical map. I think that the finiteness of the skeleton means that there is a best $n$-manifold, so that all open $n$-manifolds that immerse in a closed $n$-manifold immerse in this one. The Wu formula restricts tangent bundles, so we cannot fully approximate the $n-1$-skeleton. Key question: do all restrictions on tangent bundles of closed manifolds apply to open manifolds? (Brown and Peterson may be relevant. They probably built something more relevant than BO(n).) | |
| Jun 20, 2012 at 18:53 | comment | added | Misha | @Daniele: Could you provide details as an "answer" since there is probably not enough room for a proof in the comments? | |
| Jun 20, 2012 at 16:52 | comment | added | Misha | @Daniele: This is nice and goes in the right direction. How do you construct a bundle map in the case of an open 4-manifold like this? | |
| Jun 20, 2012 at 15:25 | comment | added | Daniele Zuddas | I modified my comment with a further remark. | |
| Jun 20, 2012 at 15:22 | comment | added | Daniele Zuddas | May be this paper can help you: A. Phillips, "Submersions of open manifolds", Topology 6 (1967), 171-206. The main result is that, for $M$ open and $\dim M \geq \dim N$, there is a submersion $M \to N$ iff there is a bundle epimorphism $TM \to TN$. From this you get the following: every oriented 4-manifold with the homotopy type of a (not necessarily finite) 2-complex can be immersed in $CP^2$. | |
| Jun 20, 2012 at 15:15 | comment | added | Misha | @Daniele: I know this paper and used it in the past in a work which was a partial motivation for my question. Philips' result is a variation on H-S theory and gives the same answer: Reduction of the problem to a homotopy-theoretic question. | |
| Jun 20, 2012 at 4:31 | history | edited | John Pardon | CC BY-SA 3.0 |
added italics for clarity
|
| Jun 20, 2012 at 3:57 | comment | added | Misha | @Scott: Torus trick is based on the fact that if $M$ is a closed $n$-manifold with trivial tangent bundle (say, a torus), then $M-p$ admits an immersion in ${\mathbb R}^n$, which is a direct corollary of Hirsch's theorem. This does not really help (at least in an obvious way) with general smooth manifolds which have nontrivial tangent bundles. They will not immerse in ${\mathbb R}^n$, so one needs more general targets, which you can see already in dimension $2$. | |
| Jun 20, 2012 at 3:53 | history | edited | Misha | CC BY-SA 3.0 |
added 334 characters in body
|
| Jun 20, 2012 at 3:49 | comment | added | Misha | @Igor: You are right, I have to include complex-projective spaces as possible factors. | |
| Jun 19, 2012 at 23:51 | comment | added | Scott Carter | What does the torus trick, as given in Kirby Siebenmann do for you in regards to this question? | |
| Jun 19, 2012 at 23:21 | comment | added | Igor Belegradek | Doesn't the connected sum of products of projective spaces have zero rational Pontryagin classes? If so, the same would be true for any immersed submanifold of the same dimension. | |
| Jun 19, 2012 at 22:57 | history | asked | Misha | CC BY-SA 3.0 |