Timeline for Elementary question: Intuition for equivariant cohomology
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 8, 2016 at 6:49 | vote | accept | HLC | ||
| Aug 4, 2016 at 12:18 | comment | added | Allen Knutson | $H^*_{S^1}(pt) := H^*(ES^1/S^1) = H^*(\mathbb{CP}^\infty) = \mathbb Z[t]$ where $\deg t = 2$. This is very, very basic to the study of equivariant cohomology. Note as a mnemonic that since these are cohomology rings, they are supercommutative not commutative, so will only give you polynomial rings if the generators are in even degree. | |
| Aug 2, 2016 at 17:12 | comment | added | HLC | What does it mean $\deg t_i=2$? Why is it not $1$? (I'm really sorry for burdening MO like this.) | |
| Aug 2, 2016 at 16:52 | vote | accept | HLC | ||
| Aug 2, 2016 at 16:52 | |||||
| Aug 2, 2016 at 13:32 | history | answered | Allen Knutson | CC BY-SA 3.0 |