Timeline for Integrable compatible complex structures
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 17, 2014 at 20:02 | answer | added | Richard Montgomery | timeline score: 1 | |
| Jan 16, 2014 at 23:05 | vote | accept | David P | ||
| Jan 16, 2014 at 22:38 | comment | added | Jonny Evans | @David Petrecca: This $\mathcal{C}$ will be at least as big as the $Ham(M,\omega_0)$-orbit of $J_0$ (i.e. infinite-dimensional). A very nice study of $\mathcal{J}_{int}$ in some special cases (which in particular discusses your set $\mathcal{C}$) can be found in the papers: arxiv.org/abs/math/0610436 (in particular section 2) and arxiv.org/abs/math/0507382 by Abreu-Granja-Kitchloo (who prove $\mathcal{J}_{int}$ is contractible if $X$ is a rational ruled complex surface). | |
| Jan 16, 2014 at 22:00 | answer | added | Ben McKay | timeline score: 3 | |
| Jan 16, 2014 at 16:49 | comment | added | David P | Gauduchon also introduces another space, when you start with a Kaehler structure $(J_0, \omega_0)$. Namely he defines the space $\mathcal C$ of the $J$'s which are of the form $\phi J_0$ for some $\phi \in Diffeo(M)^0$. Do you mean that this $\mathcal C$ is a point for the Kodaira-Siu examples? | |
| Jan 16, 2014 at 16:18 | comment | added | BS. | @Gunnar: uniqueness is up to isomorphism. And in the case of $P^1$, $\mathcal{I}_int=\mathcal{I}$. In general, $\mathcal{I}_int$ is invariant by the whole symplectomorphism group of $(M,\omega)$, so it is infinite dimensional once it is nonempty (the stabilisers are finite dimensional -- riemannian isometry groups). | |
| Jan 16, 2014 at 16:03 | comment | added | Gunnar Þór Magnússon | I'm pretty sure it can. Kodaira and Siu have results proving the uniqueness of the complex Kahler structure on projective space, so your $\mathcal I_{\mathrm{int}}$ should be a point for the underlying smooth $M$ with the standard Fubini-Study symplectic form $\omega_0$. For a very elementary example, take the projective line $\mathbb P^1$. | |
| Jan 16, 2014 at 15:51 | history | asked | David P | CC BY-SA 3.0 |