Timeline for A map of spaces implementing the Pontryagin Thom collapse map? (collapse maps in families)
Current License: CC BY-SA 4.0
14 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| May 9, 2020 at 21:02 | vote | accept | Saal Hardali | ||
| May 22, 2018 at 20:54 | answer | added | John Klein | timeline score: 4 | |
| May 22, 2018 at 16:44 | answer | added | Neil Strickland | timeline score: 10 | |
| May 22, 2018 at 16:09 | comment | added | Oscar Randal-Williams | @SaalHardali Let me be more direct: I don't believe the function you have in mind is well-defined. Let $j' \in [j]$. It has a collapse map in $\pi_m(Th(N_{j'}))$, and you want to associate to it an element of $\pi_m(Th(N_{j}))$. You want to do so by saying that there exists a homotopy equivalence from $Th(N_{j'})$ to $Th(N_{j})$, which there does: in fact, there exist many; which do you choose? | |
| May 22, 2018 at 14:47 | comment | added | Saal Hardali | @OscarRandal-Williams I agree that it is suspicious from that point of view. As far as I understand though the part of the proof youre reffering to (of the PT theorem) is strictly needed only at the step where you prove the invariance under cobordisms (which is really only true after composing with the map to the thom space of the taotological bundle over the grassmanian). The invarianve under isotopy is more elementary and i think this is not neeeded for it. Seems to me like a reasonable thing to expect that such a map exists in any case. | |
| May 22, 2018 at 14:45 | comment | added | Oscar Randal-Williams | @SaalHardali: There is something strange about this set-up. Often in the P-T construction one obtains a map to the Thom space of the tautological bundle over a Grassmannian, by composing the map you discussed with the classifying map for the normal bundle. It is only after doing this that it really makes sense to compare the P-T maps of two different embeddings: otherwise isotopic embeddings give classes in the homotopy of different spaces $Th(N_j)$ and $Th(N_{j'})$, and while these are homotopy equivalent they are not canonically so. | |
| May 22, 2018 at 13:32 | comment | added | Saal Hardali | @ThomasRot I don't think so. For example I would expect a path in the space of embeddings i.e. an isotopy to go to a homotopy between collapse maps i.e. a path in the RHS (which is what happens in practice when you prove that the first function is well defined). If the right hand side is discrete then you don't get this. (as I was trying to allude to in the question, a "map of spaces" in this context is a homotopy class of maps between homotopy types, otherwise nothing makes sense). | |
| May 22, 2018 at 13:24 | comment | added | Thomas Rot | don't take it as a complaint, but as a lack of my understanding. If you declare space to be a homotopy type, a map should be a homotopy class right? Isn't your proposed map then the composition $Emb(M,\mathbb{R}^m)\rightarrow E\rightarrow \coprod_{[j]}[S^m,Th(N_j)]$? | |
| May 22, 2018 at 11:56 | history | edited | Saal Hardali | CC BY-SA 4.0 |
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| May 22, 2018 at 11:55 | comment | added | Saal Hardali | @ThomasRot Perhaps I don't understand your complaint but If you declare that space means "homotopy type" then I don't see a problem. As I said the homotopy type of the Thom space only depends on the isotopy class of the embedding. | |
| May 22, 2018 at 11:53 | comment | added | Thomas Rot | The boxed equation does not make sense to me: You take the coproduct over equivalence classes, and then take a representative for each one. How do you choose the representative? To me this sound like an evil thing to do: You want to pick a preferred representative out of an equivalence class. | |
| May 22, 2018 at 10:57 | history | edited | Saal Hardali | CC BY-SA 4.0 |
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| May 22, 2018 at 10:52 | history | asked | Saal Hardali | CC BY-SA 4.0 |