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If we have a complex Morse function on a complex four-manifold, $f: X\to \mathbb{C} $, can we tell from the function how the genus of inverse images $f^{-1}(z)$ (for regular values) may change? under what conditions can we asssure that the genus does not change?

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There are various notions of being Morse for a function $X\to\mathbb C=\mathbb R^2$. Could you say specifically which you mean? – John Pardon May 18 '13 at 18:29
I didn't know there many. I meant a non-degenerate complex-analytic function. Can tell me about the other notions? – nikita May 18 '13 at 18:41
kaavek: Then $f$ is a fibration on open and dense subset with constant genus of the fiber on this subset (I am assuming your function is also proper, otherwise, genus is not even defined). Is this what you wanted to know? – Misha May 18 '13 at 19:42
In this case $f$ is equivalent, at singular points, to the function $f(z_1, \dots, z_n) = z_1^2 + \cdots + z_n^2$ (in your case, $n=4$). This holds at least with respect to smooth coordinates, I don't know if you can assume holomorphic coordinates. In other words, $f$ is a Lefschetz fibration (assuming proper), and the regular fiber doesn't change in topology. In the case of real manifolds of even real dimension, this is the starting of the topological Lefschetz fibrations (maps which can be expressed in that way at singular points). – Daniele Zuddas May 18 '13 at 21:59
I am thinking about a Morse theory proof that every Stein 4-manifold admits a Lefschetz fibration over $D^2$. So according to the last comment, we need to show that every Stein admits a non-degenerate complex morse function.Is it possible to use the fact that it can be embedded in $C^n$? – nikita May 19 '13 at 5:48

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