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Suppose that $f\colon X\to \mathbb P^N$ is a finite morphism, where $X$ is a smooth projective variety over $\mathbb C$. Then one may consider monodromy of the (singular) cohomology of the subvariety $f^{-1}(H)$, where $H\subset\mathbb P^N$ is a general hyperplane (monodromy as $H$ varies). I strongly suspect that properties of this monodromy are to a great extent parallel to those of monodromy of the hyperplane section (i.e., to the Picard-Lefschetz theory). In particular, I believe that the monodromy group is generated by reflexions in "vanishing cycles" admitting a more or less explicit description.

My question is whether there exists a text in which this theory is written up (may be somebody's Ph.D. thesis?).

Thanks in advance, Serge

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If you take a divisor of bidegree $(1,1)$ in $\mathbb P^n \times \mathbb P^n$ and take the fiber product over the first $\mathbb P^n$ with $X$, you get a family of varieties over the second $\mathbb P^n$, the fibers of which are the $f^{-1}(H)$. Essentially, you are asking what the monodromy of this family is. The singular fibers will be a codimension $1$ subset. I think the fundamental group of the complement of the singular locus will always be generated by small loops around it. For those small loops, simply choose a $\mathbb P^1$ containing them and apply regular Picard-Lefschetz theory.

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  • $\begingroup$ Will, thank you! So you suggest to choose a general pencil of hyperplanes in $\mathbb P^N$ and to blow up $X$ at the preimage of this pencil's axis. Thus, one obtains a family over $\mathbb P^1$ (with fibers of the form $f^{-1}(H)$) which is going to play the role of L:efschetz pencil. That's right, but this is not a particular case of the standard P-L theory, so one should do the local analysis of monodromy and degenerations: exactly what I hoped that somebody had already done for me:)) $\endgroup$
    – Serge Lvovski
    Commented Mar 26, 2013 at 7:10
  • $\begingroup$ @Will: On second thought, you are absolutely right, the standard Picard-Lefshcetz theory is applicable in this situation. So thank you again, I am accepting the answer. $\endgroup$
    – Serge Lvovski
    Commented Mar 26, 2013 at 10:34

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