Timeline for Measure in $\mathbb {C} ^p$ [closed]
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Aug 28, 2021 at 16:17 | history | closed |
abx Steven Landsburg Ben McKay Ryan Budney Mark Wildon |
Not suitable for this site | |
| Aug 19, 2021 at 20:04 | review | Close votes | |||
| Aug 28, 2021 at 16:17 | |||||
| Aug 19, 2021 at 19:07 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 21, 2021 at 19:41 | review | Close votes | |||
| Apr 26, 2021 at 3:06 | |||||
| Apr 21, 2021 at 18:09 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Dec 22, 2020 at 17:07 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 22, 2020 at 16:10 | answer | added | Incnis Mrsi | timeline score: 0 | |
| Aug 9, 2020 at 18:54 | comment | added | Kamel | @Ben Mackay thank you | |
| Aug 9, 2020 at 17:36 | comment | added | Ben McKay | @Kamel: The pullback $f^*\omega^p$ vanishes just when $f$ has rank less than $p$ at every point of $\mathbb{C}^p$ mapping to a point of $X$ where $\omega\ne 0$.This is just determinant of a product is the product of determinants in local coordinates, or in a holomorphic basis of local sections of the cotangent bundle. | |
| Aug 9, 2020 at 17:27 | comment | added | Kamel | @Ben Mackay, Okay, if $f^*(\omega ^p) =0$, it implies that $f$ is a constant? Or just $f$ is with rank $<p$? | |
| Aug 9, 2020 at 7:23 | comment | added | Ben McKay | Zero is a measure. | |
| Aug 9, 2020 at 4:52 | comment | added | Kamel | The fact that $f$ is not constant is sufficient or it is also necessary either of rank $=p$ | |
| Aug 9, 2020 at 1:46 | comment | added | Amir Sagiv | If $f$ is measurable, why not? | |
| Aug 9, 2020 at 1:22 | comment | added | LSpice |
The grammar of what you wrote didn't make sense to me, so I made what seemed like a small change that hopefully preserved meaning. I think that I didn't change the meaning, but please revert if I did. (In that case, though, you might want to make sure to use $\mathbb C$ $\mathbb C$ rather than $\ C$ $\ C$ (a $C$ with a forced space beforehand) as you originally had.)
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| Aug 9, 2020 at 1:21 | history | edited | LSpice | CC BY-SA 4.0 |
Proofreading
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| Aug 8, 2020 at 21:55 | review | First posts | |||
| Aug 8, 2020 at 22:19 | |||||
| Aug 8, 2020 at 21:52 | history | asked | Kamel | CC BY-SA 4.0 |