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Feb 9, 2023 at 17:58 history edited LSpice CC BY-SA 4.0
Names of references, while this is on the front page
Jan 8, 2017 at 3:33 comment added Tim Perutz @Ashley: No, this approach gives us no control over that, so far as I can see.
Jan 6, 2017 at 23:47 comment added Ashley @TimPerutz Hope this isn't a silly question, but regarding "via the Lagrangian neighbourhood theorem, you can take the symplectic form to be the canonical one if you'll settle for a Kaehler structure only near the zero-section." Do you know if this Kahler form take on a nice form in holomorphic coordinates? I'm wondering because the usual form as $\sum_j dx_j \wedge dy_j$ will not come from holomorphic coordinates $z_j = x_j +i y_j$ unless the cotangent bundle is flat.
Mar 7, 2013 at 22:27 comment added Tim Perutz Right, but Whitney proved that every smooth manifold is diffeomorphic to a real analytic manifold. Kai's answer is indeed a good one (I didn't know of the work he cited).
Mar 7, 2013 at 19:05 vote accept Mohammad Farajzadeh-Tehrani
Mar 7, 2013 at 19:04 comment added Mohammad Farajzadeh-Tehrani does not your first argument need M to be an analytic manifold? I found the papers mentioned by Zehmisch and the one in the comment below that interesting as well.
Jun 2, 2010 at 13:17 vote accept Mohammad Farajzadeh-Tehrani
Mar 7, 2013 at 19:04
Jun 2, 2010 at 4:18 history edited Tim Perutz CC BY-SA 2.5
Typo fixed, slight rephrasing.
Jun 2, 2010 at 4:10 history edited Tim Perutz CC BY-SA 2.5
Fixed arxiv links
Jun 2, 2010 at 4:01 history answered Tim Perutz CC BY-SA 2.5