Timeline for Turning cocycles in cobordism into an inclusion or a fibering
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 9, 2023 at 8:02 | comment | added | Mark Grant | Are you familiar with Thom's work on cobordism? He shows that there are integral cohomology classes that aren't realized by submanifolds. Together with the map from complex cobordism to integral cohomology, I think this gives a no answer to your first question. | |
| Sep 9, 2023 at 5:14 | comment | added | მამუკა ჯიბლაძე | Well you can represent homotopy type of any finite CW-complex by a manifold with boundary (as Quillen does), what I want to say is that in some cases, in particular for cobordism cocycles, it is sometimes convenient not to do so. | |
| Sep 9, 2023 at 5:11 | comment | added | timaeus | @მამუკაჯიბლაძე Typo fixed, and I was referring to this note, where (on page 447) the representatives are considered as manifolds. | |
| Sep 9, 2023 at 5:04 | history | edited | timaeus | CC BY-SA 4.0 |
edited body
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| Sep 9, 2023 at 4:47 | comment | added | მამუკა ჯიბლაძე | $M$ need not be a manifold. See Section 1 in Elementary proofs of some results of cobordism theory using Steenrod operations (Quillen, AIM 7 (1971) 29-56). Also, if you speak of cobordisms, usually you put $n$ into superscript and if of bordisms, then into subscript. | |
| Sep 9, 2023 at 3:55 | comment | added | LSpice |
TeX note: $\operatorname{dim} X$ \operatorname{dim} X spaces better than $\mathrm{dim} X$ \mathrm{dim} X. Actually, in this case, it's pre-defined, so you can just use $\dim X$ \dim X. I edited accordingly.
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| Sep 9, 2023 at 3:54 | history | edited | LSpice | CC BY-SA 4.0 |
`\dim`; deleted "thanks"
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| Sep 9, 2023 at 3:19 | history | asked | timaeus | CC BY-SA 4.0 |