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May 8, 2020 at 15:40 history edited Martin Sleziak CC BY-SA 4.0
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May 8, 2020 at 14:08 answer added Bananeen timeline score: 1
Apr 23, 2017 at 20:20 vote accept Saal Hardali
Apr 23, 2017 at 14:02 comment added nfdc23 Much as the holomorphic Poincare Lemma follows from the smooth one because the Cauchy-Riemann equations imply holomorphicity of output when input is holomorphic, why so skeptical for Frobenius' theorem? Multivariable differential calculus works in the holomorphic setting as in the smooth setting by the same proofs (e.g., inverse function theorem), so a $C^{\infty}$-submanifold of a complex manifold is a complex submanifold iff its tangent space is a $\mathbf{C}$-subspace at all points (see para. 2 of p. 10 of math.stanford.edu/~conrad/papers/rhtalk.pdf); that is what you're overlooking.
Apr 23, 2017 at 11:12 history edited Saal Hardali CC BY-SA 3.0
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Apr 23, 2017 at 10:44 answer added Ben McKay timeline score: 6
Apr 23, 2017 at 10:32 history edited Saal Hardali CC BY-SA 3.0
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Apr 23, 2017 at 10:19 history edited Saal Hardali CC BY-SA 3.0
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Apr 23, 2017 at 10:03 comment added Saal Hardali @Omar I don't think this can be done without using some form of holomorphic Poincare lemma (ensuring that solutions to holomorphic first order systems are holomorphic). In particular I'm not sure that one can prove holomorphic frobenius from real frobenius alone as the latter doesn't know about holomorphic Poincare lemma.
Apr 23, 2017 at 10:00 history edited Saal Hardali CC BY-SA 3.0
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Apr 23, 2017 at 10:00 comment added Omar One idea that might help you understand tht theorem is simply reproving Frobenius theorem while checking that if you start with holomorphic vector bundles the local submersions you get are holomorphic.
Apr 23, 2017 at 9:57 history edited Saal Hardali CC BY-SA 3.0
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Apr 23, 2017 at 9:45 history edited Saal Hardali CC BY-SA 3.0
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Apr 23, 2017 at 9:37 history asked Saal Hardali CC BY-SA 3.0