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?

  • $\begingroup$ 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

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