Timeline for Compute the rational cohomology ring $H^*(M;\mathbb{Q})$ of a product of Lie groups (and their quotients)
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 25, 2018 at 0:51 | vote | accept | Sergio Charles | ||
| Jul 28, 2018 at 5:25 | comment | added | Venkataramana | If $G/K$ is a compact symmetric space, the real cohomology is given by the K invariant in the exterior algebra of the cotangent space at identity: $(\wedge ^* {\mathfrak p})^K$ . From this the rational cohomology is easy to compute. | |
| Jul 28, 2018 at 4:34 | history | edited | Sergio Charles | CC BY-SA 4.0 |
edited title
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| Jul 28, 2018 at 4:09 | comment | added | Sergio Charles | @ArunDebray No, you are right! I should have said and quotients of Lie groups. | |
| Jul 28, 2018 at 4:01 | answer | added | Igor Rivin | timeline score: 4 | |
| Jul 28, 2018 at 4:01 | comment | added | Arun Debray | Forgive me for being pedantic, but not all of the spaces in your product are Lie groups. For example, $\mathrm{Sp}(n_8+1)/(\mathrm{Sp}(1)\times\mathrm{Sp}(n_8))$ is the quaternionic projective space $\mathbb{HP}^{n_8}$. I think you'll have that problem with all of the quotients in that formula. | |
| Jul 28, 2018 at 3:53 | history | asked | Sergio Charles | CC BY-SA 4.0 |