Timeline for Is the degree a sufficient condition such that a measure is the pullback of another one?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 3, 2020 at 16:08 | comment | added | Willie Wong | a note on terminology: usually measures are only pushed forward, and differential forms are pulled back. Defining pullback measures is tricky mathoverflow.net/q/122704/3948 . | |
| S Sep 3, 2020 at 15:03 | history | edited | Yemon Choi |
corrected spelling in title
|
|
| S Sep 3, 2020 at 15:03 | history | suggested | RobPratt | CC BY-SA 4.0 |
corrected spelling in title and added a tag
|
| Sep 3, 2020 at 14:54 | review | Suggested edits | |||
| S Sep 3, 2020 at 15:03 | |||||
| Sep 3, 2020 at 13:31 | review | Close votes | |||
| Sep 18, 2020 at 3:06 | |||||
| Sep 3, 2020 at 13:16 | history | edited | RaphaelB4 | CC BY-SA 4.0 |
added 3 characters in body
|
| Sep 3, 2020 at 13:11 | answer | added | Sebastian | timeline score: 3 | |
| Sep 3, 2020 at 13:09 | comment | added | mme | $S^n$ does not have a Haar measure unless $n=0,1,3$. Maybe $7$ if you stretch the meaning of Haar measure a bit. Ignoring this, most manifolds do not have self-maps of arbitrary degree; and if you want $\nu$ to also be nonvanishing then $f$ should be a covering map (even less common). It is a theorem of Moser that if $\int \omega = \int \nu$ for two nonvanishing top forms, then there's an oriented diffeomorphism $f$ with $f^*\nu = \omega$. The proof is what's usually called the Moser trick. | |
| Sep 3, 2020 at 12:52 | history | asked | RaphaelB4 | CC BY-SA 4.0 |