Timeline for Stiefel-Whitney class of unordered configuration space
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 10, 2015 at 2:02 | comment | added | QSR | The question is on mathoverflow.net/questions/217408/… | |
| Sep 10, 2015 at 2:01 | comment | added | QSR | Thanks! I totally understand the answer. Moreover, if we change $S^m$ to other manifolds, for example, projective spaces, Grassmannians, then how to compute the corresponding Stiefel-Whitney class? | |
| Sep 8, 2015 at 11:49 | comment | added | Yury Ustinovskiy | Given a short exact sequence of vector bundles $0\to V\to W\to U\to 0$, you can always introduce a metric on $W$ and take the orthogonal complement to $V\subset W$. This gives you a splitting $W=V\oplus U$. | |
| Sep 8, 2015 at 7:00 | comment | added | QSR | Why the Stiefel-Whitney class's Cartan formula for Whitney sum is also true for short exact sequence? | |
| Sep 8, 2015 at 5:09 | comment | added | QSR | Thanks! Why the exact sequence of vector bundles can imply that the Stiefel-Whitney class satisfy the product formula? I am only clear with the special case when the short exact sequence split, then we have the Whitney sum. | |
| Sep 8, 2015 at 2:33 | comment | added | Yury Ustinovskiy | The second map is just the map induced by $\pi_*\colon TTot(E)\to T\mathbb RP^m$. The kernel of this map is the vertical component of $TTot(E)$, which is precisely $\pi^*E$. | |
| Sep 8, 2015 at 2:26 | comment | added | QSR | Thanks! How to obtain the (short?) exact sequence? I am not clear.... | |
| Sep 8, 2015 at 1:15 | vote | accept | QSR | ||
| Sep 8, 2015 at 0:08 | history | edited | Yury Ustinovskiy | CC BY-SA 3.0 |
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| Sep 8, 2015 at 0:02 | history | answered | Yury Ustinovskiy | CC BY-SA 3.0 |