Timeline for Invariant differential forms on commutative group schemes are closed!?
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 23, 2011 at 15:33 | comment | added | Philipp Hartwig | Thanks a lot, this is a very neat argument. So we get the result whenever 2 is invertible on G. I'd still be very interested in an argument that also works in characteristic 2 though. | |
| May 22, 2011 at 5:55 | comment | added | naf | Thanks; I realized this later. It's annoying though that the argument doesn't work in characteristic 2. | |
| May 21, 2011 at 18:37 | comment | added | Keerthi Madapusi | The fact that you quote should be true for invariant forms on any commutative group scheme. After all, multiplication by $n$ is just composition of $n$-fold multiplication with the $n^{\text{th}}$ diagonal map. You just have to pick an $n$ that's invertible, and your argument should go through. | |
| May 21, 2011 at 17:21 | history | answered | naf | CC BY-SA 3.0 |