Let $(M,J)$ be a complex manifold, where $J$ is the integrable complex structure. Let $X$ be a holomorphic vector field on $M$ and let $\varphi_{t} : M\rightarrow M $ be its flow. Question: Is $\varphi_{t}$ a biholomorphism? It is a diffeomorphism but is it holomorphic?
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4$\begingroup$ Yes. The solutions of a holomorphic differential equation are holomorphic. The question would be more appropriate on MSE. $\endgroup$– abxCommented Jan 2, 2018 at 11:36
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$\begingroup$ Can you provide some details or reference? $\endgroup$– DanielCommented Jan 2, 2018 at 11:40
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1$\begingroup$ Henri Cartan, Elementary Theory of Analytic Functions of One or Several Complex Variables, Chap. 7 (Dover Publications). This is the translation of the notes of one of Cartan's undergraduate courses. $\endgroup$– abxCommented Jan 2, 2018 at 13:59
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The question is local, so assume that $M$ is open in $\mathbb C^n$, so that we do not have to deal with the second tangent bundle. Then: \begin{align*} \partial_t \phi_t &= X\circ \phi_t \\ T(\partial_t \phi_t) &= \partial_t T(\phi_t) = TX\circ T(\phi_t) \\ \partial_t J\circ T(\phi_t) &= J\circ\partial_t T(\phi_t) = J\circ TX\circ T(\phi_t) = TX\circ J\circ T(\phi_t) \\ \partial_t T(\phi_t)\circ J &= TX\circ T(\phi_t)\circ J \end{align*} so $J\circ T(\phi_t)$ and $T(\phi_t)\circ J$ solve the same equation with the same initial value $\phi_0= Id$, thus they are equal.
answered Jan 2, 2018 at 13:06
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$\begingroup$ If the vector field is "complete" (meaning its solution curves and hence its flow) exist for all complex times t), then for each time t the flow at that time is a biholomorphism. $\endgroup$ Commented Oct 20, 2019 at 5:36
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3$\begingroup$ On a meta level, I find it tremendously unfortunate that a perfectly valid question like this one, by someone who perhaps has not studied this area very much, needs to be closed off. It smacks of snobbery and elitism in the worst way. I cannot imagine how allowing such questions does any harm to MO in any way. $\endgroup$ Commented Oct 20, 2019 at 5:41