We know, if $π : X → S$ is a generically smooth family of complex projective varieties, such that $X_0 := π^{−1}(0)$ is an snc divisor in $X$, then the monodromy representation is unipotent. Now assume the monodromy representation is unipotent, then is the central fibre $X_0 := π^{−1}(0)$ simple normal crossing?
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5$\begingroup$ The answer to the question as stated is no: you could take the constant family $X=S\times X_0$ and then blow up along a bad subscheme in the special fiber. The monodromy would still be the identity. A better question could be: suppose that the monodromy is unipotent, does there exist a regular model of the family whose special fiber is snc? Note that by Nagata and Hironaka, you can always find a regular model such that $\pi^{-1}(0)_{\rm red}$ is snc. This is key in the geometric proof of the fact that the monodromy is quasi-unipotent (cf. Illusie "Autour de theoreme de monodromie locale"). $\endgroup$– Piotr AchingerFeb 10, 2017 at 15:05
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$\begingroup$ I suspect that the answer would still be no, but I don't know of a counterexample. $\endgroup$– Piotr AchingerFeb 10, 2017 at 15:06
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Just to be clear, the local monodromy is unipotent if $X_0$ is reduced with simple normal crossings (that's probably what you meant). As Piotr pointed out, your question, as originally formulated, has an easy negative answer. However, as he suggested, the question can be modified to a something more reasonable:
If the local monodromy is unipotent, does there exist a birational model with $X_0$ reduced snc?
I think this is also no. Take a Lefschetz pencil of sextics $Y\to S=\mathbb{P}^1$ in $\mathbb{P}^4$, so that $Y_0$ has a node $p$ and no other singularities. By the Picard-Lefschetz formula, the local monodromy is unipotent. Let $X\to S$ be a birational model with $X_0$ snc. We have a morphism $X\to Y$ given by the relative canonical map. This will dominate the blow up $Bl_pY$. The exceptional divisor of $Bl_pY\to Y$ will have multiplicity $2$. So that the components of $X_0$ lying over this also have multiplicity at least $2$.
answered Feb 10, 2017 at 16:04
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$\begingroup$ what about when we use semi-stable reduction? then can $X_0$ can be snc? $\endgroup$– HelenFeb 10, 2017 at 22:43
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$\begingroup$ Yes, the semistable reduction theorem says if we allow binational modifications plus base change, then we can make $X_0$ reduced snc. $\endgroup$ Feb 10, 2017 at 23:03
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2$\begingroup$ The semistable reduction theorem is only known in dimension 1. $\endgroup$ Feb 11, 2017 at 0:55