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Let $(M,J)$ be a complex manifold. Suppose that $X$ is a real vector field such that the flow of $X$ is by biholomorphisms.Question Show the flow of $JX$ is by biholomorphisms.

I know one reference to show the answer is yes (where it is stated that: flow of $X$ is by biholomorphisms $\iff$ $X-iJX$ is a holomorphic vector field) but I am interested to find a simple explanation (if indeed one exists).

I would also be interested if anyone knows about the corresponding question for an almost complex manifold $(M,J)$ with a vector field with flow by J-isomorphism's? - for this question I do not have a reference (or clue if it is true).

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A vector field $X$ has flow preserving $J$ just when $L_X J=0$, i.e. just when $[X,JY]=J[X,Y]$ for any vector field $Y$. Take $Y$ to also have such flow. Many such $Y$ exist locally, spanning the tangent bundle, because $(M,J)$ is a complex manifold. Vanishing of the Nijenhuis tensor gives $[JX,JY]=[X,Y]+J([JX,Y]+[X,JY])$. But then we apply the flows of $X$ and $Y$: $[JX,JY]=[X,Y]+J(J[X,Y]+J[X,Y])=-[X,Y]=J^2[X,Y]=J[X,JY]=J[JX,Y]$. So $[JX,JY]=J[JX,Y]$. So $L_{JX} J=0$.

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As Ben's argument suggests, the proof that, if the flow of $X$ preserves $J$ then the flow of $JX$ preserves $J$ does depend on the integrability of $J$.

As a concrete example of an almost-complex manifold for which this property does not hold, consider $S^6 = \mathrm{G}_2/\mathrm{SU}(3)$ with its standard (non-integrable) $\mathrm{G}_2$-invariant almost-complex structure $J$.

It is known (Submanifolds and special structures on the octonians, J. Differential Geom. 17 (1982), 185–232) that the vector fields on $S^6$ whose flows preserve $J$ consist exactly of the vector fields associated to $1$-parameter subgroups of $\mathrm{G}_2$, and hence they form a $14$-dimensional Lie algebra ${\frak{g}}_2$. Moreover, any vector field $X$ on a connected open subset $U\subset S^6$ whose flow preserves $J$ is the restriction to $U$ of a (unique) global vector field on $S^6$ that preserves $J$.

However, any such global $X$ that is not identically zero must vanish at some point $p\in S^6$, and its linearization at $p$ is a nonzero linear transformation $X'(p):T_pS^6\to T_pS^6$ with purely imaginary eigenvalues (since the flow of $X$ must also preserve the metric on $S^6$). It follows that the linearization of $JX$ at $p$ is a nonzero linear transformation with real eigenvalues. Consequently, $JX$ cannot belong to ${\frak{g}}_2$.

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Without lose of generality we may work locally in $\mathbb{R}^{2n}$ with its standard complex structure $J$. Let $\phi$ be the flow of the vector field $x'= f(x)$. So the statement in the question is a consequence of the variational equation $$ \partial_t \partial_x \phi_t = Df(\phi). \partial_x \phi_t$$ If the flow is holomorphic then $\partial_x \phi_t$ and its time derivative $\partial_t \partial _x \phi_t $ commute with $J$. This implies that $Df.J= J.Df$. Q.E.D

In the same manner one can shows the following:

Fact: If the flow of a vector field $x'=f(x)$ on $\mathbb{R}^n$ is harmonic then $f$ is harmonic, too.

Proof:

The Laplacian, the gradient and the Hessian of a vector valued function is defined naturally by componentwise corresponding operators. Hence the Hessian of a vector valued function is a $3$ dimensional matrix whose trace is a "vector".

With such notation we have the following formula:

$$\partial_t \Delta \phi= trace( {(\partial _x \phi_t)}^{tr}Hess(f)\partial_x \phi_t)+ \Delta \phi . \nabla f$$

This shows (using the time independence of $f$ to evaluate at $t = 0$) that if the flow of a vector field is harmonic then the vector field is a harmonic vector valued function.

But what about the converse? Is the flow of a harmonic vector field, a harmonic function? Moreover, how can we rephrase the above Euclidean fact in an abstract Riemannian manifold?

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  • $\begingroup$ @WillieWong If the flow is standard harmonic then $\partial_t \Delta \phi= trace( {(\partial _x \phi_t)}^{tr}Hess(f)\partial_x \phi_t)+ \Delta \phi . \nabla f$ implies that $ trace( {(\partial _x \phi_t)}^{tr}Hess(f)\partial_x \phi_t)=0$ for all t. Now put t=0 then f is harmonic. Right? $\endgroup$ May 19, 2017 at 7:13
  • $\begingroup$ Ah, okay. I see what you are doing there. $\endgroup$ May 19, 2017 at 13:44

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