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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$\begingroup$ There are various notions of being Morse for a function $X\to\mathbb C=\mathbb R^2$. Could you say specifically which you mean? $\endgroup$– John PardonCommented May 18, 2013 at 18:29
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$\begingroup$ I didn't know there many. I meant a non-degenerate complex-analytic function. Can tell me about the other notions? $\endgroup$– nikitaCommented May 18, 2013 at 18:41
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$\begingroup$ 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? $\endgroup$– MishaCommented May 18, 2013 at 19:42
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$\begingroup$ 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). $\endgroup$– Daniele ZuddasCommented May 18, 2013 at 21:59
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$\begingroup$ 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 $C^n$? $\endgroup$– nikitaCommented May 19, 2013 at 5:48
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