Timeline for Generalization of winding number to higher dimensions
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
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| S Mar 6, 2019 at 8:03 | history | bounty ended | Ali Taghavi | ||
| S Mar 6, 2019 at 8:03 | history | notice removed | Ali Taghavi | ||
| S Mar 5, 2019 at 15:33 | history | suggested | Ali Taghavi |
I add two tags.
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| Mar 5, 2019 at 15:20 | review | Suggested edits | |||
| S Mar 5, 2019 at 15:33 | |||||
| Mar 5, 2019 at 9:05 | answer | added | Ali Taghavi | timeline score: 2 | |
| S Mar 5, 2019 at 7:25 | history | bounty started | Ali Taghavi | ||
| S Mar 5, 2019 at 7:25 | history | notice added | Ali Taghavi | Reward existing answer | |
| Apr 10, 2018 at 15:23 | answer | added | Piotr Hajlasz | timeline score: 4 | |
| Apr 10, 2018 at 15:01 | answer | added | M.G. | timeline score: 3 | |
| Jan 8, 2017 at 12:44 | comment | added | Thomas Rot | Interesting. But if I equip both manifolds $M$ (The domain and codomain) with different orientations, our notion of degree will differ. Probably it is sensible that the identity map has degree one. But the degree I defined above is also defined for maps between different manifolds. | |
| Jan 8, 2017 at 12:39 | comment | added | Denis Nardin | @ThomasRot I probably was just channeling my inner pedant. If $M$ is an orientable $n$-manifold an orientation is the same thing as an isomorphism $H_n(M)\cong \mathbb{Z}$. But even without choosing an orientation there is a canonical isomorphism $End(H_n(M))\cong \mathbb{Z}$, so the degree of $f:M\to M$ does not depend on the orientation. | |
| Jan 8, 2017 at 12:34 | comment | added | Thomas Rot | @denis Nardin: I don't think I understand that. | |
| Jan 8, 2017 at 11:20 | comment | added | Denis Nardin | @ThomasRot You don't even need oriented, orientable is enough (since the endomorphism ring of a free abelian group of rank 1 is still canonically $\mathbb{Z}$). | |
| Jan 8, 2017 at 2:42 | vote | accept | Joseph O'Rourke | ||
| Jan 8, 2017 at 2:24 | answer | added | T. Amdeberhan | timeline score: 14 | |
| Jan 8, 2017 at 1:49 | comment | added | Thomas Rot | The top homology group of a closed oriented manifold is canonically isomorphic to $\mathbb Z$. The induced map on the top homology groups can then be identified with the multiplication with a number, which is the degree. For non-oriented manifolds there is a notion of degree modulo two. Any algebraic topology book ought to explain this, it is for sure in Hatcher's book | |
| Jan 8, 2017 at 1:48 | answer | added | user103319 | timeline score: 28 | |
| Jan 8, 2017 at 1:44 | comment | added | Thomas Rot | For maps $S^n\rightarrow S^n$, or more generally between closed oriented manifolds of the same dimension, there is the notion of the degree. | |
| Jan 8, 2017 at 1:35 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |