3
$\begingroup$

I would like to ask if there is a holomorphic version of Darboux's theorem. More concretely, given a holomorphic symplectic manifold $(X, \omega)$ is there a local holomorphic symplectomorphism from $(X, \omega)$ to $(\mathbb{C}^{2n}, \omega_0)$ where $\omega_0$ is the holomorphic equivalent of the standard symplectic form in $\mathbb C^{2n}$. To put it differently, is it true that $X$ locally looks like a cotangent bundle? Do you have a reference?

$\endgroup$
  • $\begingroup$ google lead to this paper: arxiv.org/pdf/0707.4253.pdf $\endgroup$ – Aknazar Kazhymurat Feb 4 at 17:03
  • $\begingroup$ Just follow the usual proofs from the real case. $\endgroup$ – Ben McKay Feb 4 at 20:16
  • $\begingroup$ @Aknazar Kazhymurat: Thank you for the link. It is an interesting paper. They mention the "holomorphic Darboux theorem" which is what I am looking for. But the theorem itself is not stated. I guess that is because it is the exact holomorphic version of the classical Darboux theorem? $\endgroup$ – Flavius Aetius Feb 5 at 16:44
  • $\begingroup$ @Ben McKay: So there is a holomorphic Dardoux's theorem and it is exactly as the real one except that we replace "smooth" by "holomorphic". Is that right? No hidden traps there? Do you have a standard reference? $\endgroup$ – Flavius Aetius Feb 5 at 16:46
  • $\begingroup$ @FlaviusAetius If you mean "exact" as in "exact symplectic manifold", then no I do not think that that is the case. There is, I believe, a formulation of the theorem in that paper. Look at page 8, "The holomorphic Darboux theorem asserts that, in a neighborhood of each point,..." $\endgroup$ – Aknazar Kazhymurat Feb 5 at 16:48

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.