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Let $f:X\to B$ be a family of curves of genus $g$ over a smooth curve $B$. Let $F_0$ be a singular fiber.

If $F_0$ is a semistable fiber, the monodromy matrix can be gotten by the classical Picard-Lefschetz formula.

If $F_0$ is non-semistable, I don't know how to compute its monodromy matrix. For example, in Namikawa and Ueno's paper[1], they can compute the Picard-Lefschetz monodromy matrix for each type of singular fiber of genus 2. It's not clear to me how they did that.

[1] Namikawa, Y. and Ueno, K., The complete classification of fibres in pencils of curves of genus two, Manuscripta math., Vol. 9 (1973), 143-186.

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up vote 4 down vote accepted

One approach (I don't know how effective it is in the genus 2 case you asked about) is to explicitly apply the semi-stable reduction theorem, and so reduce to the semi-stable case.

To achieve semi-stable reduction, you have to alternately blow-up singular points in the special fibre, and then make ramified base-changes. The latter operation just extracts a root of the monodromy operator (i.e. if $\gamma$ is a generator of $\pi_1$ of the punctured $t$-disk, and we set $t = s^n$, then $\gamma = \tau^n,$ where $\tau$ is a generator of $\pi_1$ of the punctured $s$-disk), so it is easy to see how the monodromy matrix changes. And blowing up a point in the special fibre doesn't change the monodromy action around the puncture at all.

So using this process, one can relate the original (unknown) monodromy matrix to the corresponding matrix in the semi-stable context, where it is known thanks to the Picard--Lefshcetz formula.

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If you want to do semi-stable reduction in practice, Section 3.C of Moduli of Curves, by Harris and Morrison, has many very helpful worked examples. –  David Speyer Apr 2 '10 at 11:24
    
Thanks to Emerton and Speyer. I'm very familiar with semistable reduction of a family of curves. Could you tell me how to see the change of the monodromy matrix in a base change? I think it may be a key of my question. –  Jun Lu Apr 3 '10 at 10:25
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This response is over $\mathbb{C}$, because I'm not sure what the monodromy matrix means in other settings: Shrink $B$ to a disc around $b_0$. Let $b_0$ be the point of $B$ under $F_0$ and let $B' \to B$ be your branched cover, of degree $k$. The monodromy of $B'$ is the $k$-th power of the monodromy of $B$. So you can compute the monodromy of $B$ up to any ambiguity about taking the $k$-th root of a matrix. –  David Speyer Apr 20 '10 at 13:26
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