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Let $(X,J,\omega)$ be a Kähler manifold. Let $\dim_{\mathbb{R}}(X)=2n$ and we also know that $X$ splits as $X=M\times \mathbb{R}$, where $\dim_{\mathbb{M}}=2n-1$. My question is now: does there exists coordinates $(x_{1},...,x_{n},y_{1},...,y_{n-1},t)$, where $t$ is the global coordinate on $\mathbb{R}$, such that $(z_{1},...,z_{n})$, with $z_{j}=x_{j}+iy_{j},j=1,...,n-1$ and $z_{n}=x_{n}+it$, are holomorphic coordinates?

I considered the global defined vector field $\frac{\partial}{\partial t}$, which never vanishes on $X$ and then look at $J\frac{\partial}{\partial t}$ which also never vanishes on $X$. Then for some point in $X$ choose a complement complex frame i.e. $\{ X_{1},...,X_{n-1},Y_{1},...,Y_{n-1}\}$, with the property that $JX_{i}=Y_{i}$ and $JY_{i}=-X_{i}$. Does this somehow give raise to my desired coordinates somehow? If yes, how does the argument go on? If no, is it even possible to do this?

Greetings Nina

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No in general, the function $t$ on $X$ needs to be pluriharmonic for this to be true even locally. – Misha Jul 18 '13 at 5:00
you mean $t$ as a projection $t : X \rightarrow \mathbb{R}$? If the function $t$ is real-analytic, is this true then ? – Nina Jul 18 '13 at 5:08
I am looking for such kind of coordinates in a neighbourhood of the "zero section" $M\times \{ 0 \}$? can one do this here ? – Nina Jul 18 '13 at 5:12
Have a look at You will see that real analytic is not sufficient for what Misha is telling you. It's similar to how being real analytic is nowhere near sufficient to guarantee that you are the real part of a holomorphic function locally. – Daniel Pomerleano Jul 18 '13 at 7:43
What role does the Kähler metric play in your question? When you say that $X$ "splits", do you mean that it is isometric to a Riemannian product, or just that it is diffeomorphic to $M \times \mathbb{R}$? – Johannes Nordström Jul 18 '13 at 8:41

The question makes sense only locally, in which case you are effectively asking the following

Question Q: Let $h: {\mathbb C}^n\to {\mathbb R}$ be a smooth function defined near $0$, which has $0$ as its regular point. Under what conditions there exists a complex-analytic function $f: {\mathbb C}^n\to {\mathbb C}$ (again, defined near $0$), so that $h$ is the real part of $f$. Note that such function $f$ is automatically regular at $0$, i.e., $df(0)\ne 0$. Given this, one can find a locally (near $0$) defined biholomorphic function $F: {\mathbb C}^n\to {\mathbb C}^n$ so that $f$ is the last component of $F$. (This is just the implicit function theorem for holomorphic functions.)

The answer to Question Q is classical and could be found in any book on several complex variables:

Such $f$ exists if and only if $h$ is pluriharmonic.

A randomly chose real analytic function, of course, is not going to be pluriharmonic. Just think about complex 1-dimensional case, where pluriharmonicity is equivalent to harmonicity.

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