Timeline for Can a non-Kähler complex manifold be rationally connected?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 23, 2022 at 11:54 | answer | added | Misha Verbitsky | timeline score: 3 | |
| May 23, 2022 at 6:48 | vote | accept | ABBC | ||
| May 22, 2022 at 3:16 | comment | added | David E Speyer | @ABBC No. Let $H$ be a Hopf surface; it's universal cover is $\mathbb{C}^2 \setminus \{ 0 \}$. Since $\mathbb{CP}^1$ is simply connected, any map from $\mathbb{CP}^1$ to $H$ would lift to $\mathbb{C}^2 \setminus \{ 0 \}$. But there are no nonconstant maps $\mathbb{CP}^1 \longrightarrow \mathbb{C}^2$. | |
| May 21, 2022 at 11:35 | answer | added | Jason Starr | timeline score: 3 | |
| May 21, 2022 at 3:18 | comment | added | ABBC | @JasonStarr This is a separate question, but: Are there rational curves on a Hopf surface? | |
| May 21, 2022 at 1:20 | comment | added | Jason Starr | There are plenty of examples of non-Moishezon rationally connected manifolds, e.g., a threefold fibered over $\mathbb{CP}^1$ with Hopf surface fibers. | |
| May 21, 2022 at 0:08 | comment | added | R. van Dobben de Bruyn | You could take a rational smooth proper algebraic variety that is not projective, e.g. a variant of Hironaka's example (starting from $\mathbf P^3$). "Not Kähler" does not imply "not algebraic". In the presence of "Kähler", "projective" is equivalent to "Moishezon", so a more natural variant would be if there exist non-Moishezon examples (necessarily non-Kähler, as all Kähler examples are projective). | |
| May 20, 2022 at 23:54 | history | asked | ABBC | CC BY-SA 4.0 |