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in our lectures we have seen the following:

Let $(M, J)$ be a complex manifold, i.e., $J$ is an integrable almost complex structure. Then a real vector field $X$ is an infinitesimal automorphism of $J$ if and only if its component $X^{1,0}\in \Gamma(T^{1,0}M)$ is holomorphic.

Now let $(M, J)$ be an almost complex manifold and $X \in \chi(M)$ be an infinitesimal automorphism of $J$, i.e., $[X, JY]=J[X, Y]$ for all $Y \in \chi(M)$.

Can we then conclude that the mapping $X^{1,0}f$ with $X^{1,0}:=\frac{1}{2}(X-iJX)$ is pseudoholomorphic for every pseudoholomorphic mapping $f: M \to \mathbb{C}$?

Thank you and best regards,


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Can someone edit the question a bit? It needs some clean-up and I don't have the power for that. – Spiro Karigiannis Nov 8 '10 at 21:53
Done. I hope it reads better now. – José Figueroa-O'Farrill Nov 8 '10 at 22:09
If $J$ is sufficiently non-integrable (meaning that the values of $N_J$ generate the tangent space), there are non non-constant $J$-holomorphic functions $f:M\to\mathbb{C}$, since $N_J(X,Y)f=0$ for all vector fields $X,Y$. – BS. Nov 9 '10 at 13:07
... but note that $N_J$ has at most a one (complex) dimensional image if $M$ has complex dimension 2, so maybe there are a few non-constant $J$-holomorphic functions in this case ("of one variable", so to speak). – BS. Nov 9 '10 at 13:12
up vote 6 down vote accepted

I can give you a partial answer- a reformulation of your question. Perhaps others will then be able to give a counterexample. (I think it is not true in general.) Let's start from the beginning.

A map $h : M \to \mathbb C$ is $J$-holomorphic iff $h_* (J Y) = i h_* (Y)$ for every vector field $Y$ on $M$. Equivalently, $dh(JY) = i dh(Y)$, or $(JY) (h) = i Y(h)$, or $(Y + i JY)h = 0$, or $(Y - iJY)h = 2Yh$ for any $Y$.

The condition that $X$ is an infinitesmal automorphism of the almost complex structure $J$ is that $[X,JY] = J[X,Y]$ for all $Y$. We want to investigate the question of under what conditions on $J$, for such an $X$, is it true that $(X- i JX)f = 2 Xf$ will be $J$-holomorphic for any $J$-holomorphic $f$.

Ignoring the factor of $2$, we see that \begin{equation*} Xf \text{ is $J$-holomorphic} \Leftrightarrow (Y + iJY)(Xf) = 0 \text{ for all $Y$} \end{equation*} \begin{equation*} \Leftrightarrow Y(Xf) = -i(JY)(Xf) \text{ for all $Y$} \end{equation*}

Recall the definition of the Nijenhuis tensor $N_J(X,Y) = [JX, JY] - J[JX, Y] - J[X,JY] - [X,Y]$. Since $X$ is an infinitesmal automorphism of $J$, the last two terms cancel. Applying both sides of this equation to a $J$-holomorphic map $f$, we get \begin{equation*} N_J(X,Y) f = (JX)((JY)f) - (JY)((JX)f) - (J[JX,Y])f = (JX)(iYf) - (JY)(iXf) - i [JX,Y]f \end{equation*} \begin{equation*} = i \left( (JX) (Yf) - (JY)(Xf) - (JX)(Yf) + Y((JX)f) \right) \end{equation*} \begin{equation*} = i \left( - (JY)(Xf) + i Y(Xf) \right) \end{equation*} which is exactly the quantity that we need to vanish.

Therefore, $Xf$ will also be $J$-holomorphic iff $N_J(X,Y)f = 0$ for all vector fields $Y$. In particular, this always holds if $J$ is integrable, since then $N_J$ vanishes identically. I don't know an example where $J$ is not integrable but where such an $X$ and $f$ exist. If there is one (which I would bet there is), then in that case $Xf$ would not be $J$-holomorphic.

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Okay, thanks... I assume that there is one, too... Now we only need someone you can construct an example :) – Differentialgeometer Nov 9 '10 at 11:49

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