Timeline for Thom Class of tensor bundles
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Sep 9, 2015 at 10:55 | history | bounty ended | CommunityBot | ||
| S Sep 9, 2015 at 10:55 | history | notice removed | CommunityBot | ||
| S Sep 1, 2015 at 8:51 | history | bounty started | user43326 | ||
| S Sep 1, 2015 at 8:51 | history | notice added | user43326 | Draw attention | |
| Sep 1, 2015 at 8:46 | comment | added | user43326 | Basically the fact that one can use the splitting principle is a consequence of the fact that the map $BO(1)^n\rightarrow BO(n)$ induces injection in mod $2$ cohomology, thus to compute the induced map in cohomology by $BO(n)\times BO(m)\rightarrow BO(mn)$, one can replace left hand side by $(BO(1)^m)\times (BO(1)^n)$. Of course, the same kind of arguments work over integers if we place $O$ by $U$, which means that if your bundles happen to be complex, then you will have a formula. As to the your original question, as $H^*(BO(n);Z)$ is known, there might be a way to get a formula. | |
| Jul 7, 2015 at 10:35 | comment | added | Panagiotis Konstantis | @Tom: Right! And I think in that case I know the answer: Since every vector bundle over a point is trivial, the Thom space of a trivial bundle $\varepsilon^n$ over a point is the $n$-sphere. Hence the Thom space of $\varepsilon^n \otimes \varepsilon^m$ is $S^{nm}$ and the Thom class may be represented as follows: take a generator of $H^n(S^n;\mathbb Z)$ say $u$, which represents the Thom class of $\varepsilon^n$ over a point. Then the Thom class of $\varepsilon^n \otimes \varepsilon^m$ is $\Sigma^{(m-1)n} u$, the $(m-1)n$-th suspension of $u$ (the roles of $n$ and $m$ can be interchanged). | |
| Jul 7, 2015 at 9:09 | comment | added | Panagiotis Konstantis | @Achim: Unfortunately I don't know the (exact) answer for ${\rm mod}\, 2$ coefficients but I would expect (but now the more I think about the less I am convinced about it) that the Thom class $u(L\otimes L')$ would be something like $u(L)+u(L')$ in some reasonable way (since this work for the Euler and Stiefel-Whitney class). But I see some problemes here: In case of the Thom class of $\xi\otimes\eta$ there could be an expression like $u(L)\cup u(L)$ which should make the Thom class of the tensor product zero?! | |
| Jul 6, 2015 at 23:11 | comment | added | Achim Krause | @James: You sound like you're aware of a corresponding formula for $Z/2$ coefficients (the splitting principle would work there), care to explain? In particular for two line bundles, I am not even sure how to relate the thom spaces to the Thom space of the tensor product (which would be necessary for any meaningful formula on cohomology) | |
| Jul 6, 2015 at 21:51 | comment | added | Tom Goodwillie | I don't have an answer, but I have some advice: First think of the case where $B$ is a point. This may help you to clarify the question, and what form of answer you are looking for. | |
| Jul 6, 2015 at 19:53 | history | asked | Panagiotis Konstantis | CC BY-SA 3.0 |