Timeline for Counterexample showing that G-invariant de Rham cohomology different from cohomology of G-invariant sub-complex?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 12, 2018 at 11:50 | history | edited | Michael Albanese | CC BY-SA 3.0 |
deleted 1 character in body
|
| Mar 11, 2018 at 17:19 | history | edited | David E Speyer | CC BY-SA 3.0 |
added 676 characters in body
|
| Mar 11, 2018 at 16:05 | history | edited | David E Speyer | CC BY-SA 3.0 |
added 1515 characters in body
|
| Mar 11, 2018 at 15:47 | history | edited | David E Speyer | CC BY-SA 3.0 |
deleted 201 characters in body
|
| Mar 11, 2018 at 15:10 | comment | added | David E Speyer | Expand $\phi^{\ast} (g(\theta) d \theta) = g( \theta) d \theta$ to give $g(\phi(\theta)) \phi'(\theta) = g(\theta)$. Then, inductively, $g(\phi^n(\theta)) \prod_{k=0}^{n-1} \phi'(\phi^k(\theta)) = g(\theta)$. | |
| Mar 11, 2018 at 15:06 | vote | accept | ychemama | ||
| Mar 11, 2018 at 15:05 | comment | added | ychemama | And I don't really understand where your formula for g(\phi^n(\theta_0)) come from... | |
| Mar 11, 2018 at 14:57 | comment | added | David E Speyer | Yes, that is right. | |
| Mar 11, 2018 at 14:56 | comment | added | ychemama | yes it's good enough, thx ! Just to be sure, your Z action on S^1 is p.x = \phi \circ ... \circ \phi(x) = \phi^p (x) ? | |
| Mar 11, 2018 at 14:31 | comment | added | David E Speyer | See my edit. I have an example of failure of surjectivity, but it uses an action which isn't free (and isn't even proper!) so I don't know if it is good enough for you. | |
| Mar 11, 2018 at 14:30 | history | edited | David E Speyer | CC BY-SA 3.0 |
added 1179 characters in body
|
| Mar 11, 2018 at 14:24 | comment | added | ychemama | thx for that quick answer, but what I really need is an example where surjectivité fails, I have edited my question. | |
| Mar 11, 2018 at 13:57 | history | answered | David E Speyer | CC BY-SA 3.0 |