Timeline for Hodge isometry sending the Kahler class to its opposite
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 14, 2014 at 15:22 | vote | accept | Scott Koster | ||
| May 14, 2014 at 12:24 | answer | added | Misha Verbitsky | timeline score: 4 | |
| May 14, 2014 at 7:45 | comment | added | Henri | The cohomology class of $f^*\omega'$ should be both pseudoeffective (even big) and anti-Kähler, so it is impossible. | |
| May 13, 2014 at 16:37 | comment | added | abx | @Gunnar: the question is whether $f^*(\omega')=-\omega$ in $H^2(Y)$. How does your argument apply? | |
| May 13, 2014 at 16:23 | comment | added | Gunnar Þór Magnússon | Just look at what happens to the pullback in local coordinates. You'll find that the pullback of a positive $(1,1)$-form by a holomorphic map is always semipositive. Your morphism is only bimeromorphic, but this is enough to see that we can't get $f^*(\omega') = -\omega$ on an open dense set. | |
| May 13, 2014 at 15:17 | comment | added | Scott Koster | yes, i'm sorry, so let's say $X$ and $Y$ are compact Kahler surfaces and $f$ is an isometry with respect to the intersection form which also respect the Hodge decomposition (this is what i mean by Hodge isometry) | |
| May 13, 2014 at 14:59 | comment | added | abx | You probably want $X$ and $Y$ compact. And could you explain what you call a Hodge isometry? | |
| May 13, 2014 at 14:51 | review | First posts | |||
| May 13, 2014 at 15:05 | |||||
| May 13, 2014 at 14:31 | history | asked | Scott Koster | CC BY-SA 3.0 |