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Let $X$ be a surface in $\mathbb{P}^3$. We have a fibration $f: X \longrightarrow \mathbb{P}^1$, and $f^{-1}(s_1)$ and $f^{-1}(s_2)$ have the same singularity type. Let $\gamma_1$ and $\gamma_2$ be the loops in $\pi_1(\mathbb{P}^1-S,x_0)$ and going around $s_1$ and $s_2$ respectively exactly once and not going around any other point belongs to $S$ where $S$ is the singular locus of $f$ which does not contain $x_0$. Let $y_0 \in X$ s.t. $f(y_0)=x_0$.

My question is : Is it possible to find generators for $\pi_1(f^{-1}(x_0),y_0)$ such that the action of the loops $\gamma_1$ and $\gamma_2$ can be given by the same matrix since they have the same singularity type?

Note: The example I have is the Fermat's cubic surface and after blowing up this surface at three points with $x$ and $y$ coordinates are zero, fibrating over $\mathbb{P}^1$ with the hyperplanes $x=0$ and $y=0$.

Thanks,

Ozlem

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    $\begingroup$ the description of which loops you've chosen has an ambiguity the size of a braid group $\endgroup$
    – Vivek Shende
    Commented Nov 3, 2014 at 5:15
  • $\begingroup$ @VivekShende Sorry, I don't understand what you mean ? is it not clear what i have chosen ? $\endgroup$
    – ozheidi
    Commented Nov 3, 2014 at 5:18
  • $\begingroup$ An isotrivial elliptic surface cannot have semistable fibers. $\endgroup$
    – Will Sawin
    Commented May 3, 2015 at 18:01

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In general, not, as the monodromy group does not need to be abelian. Say, take a typical elliptic surface: the monodromy must kill all $H_1$ of the fiber, which is $\Bbb Z^2$; a cyclic group can only kill $\Bbb Z$.

I could not understand your particular example.

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  • $\begingroup$ My example(which is probably an elliptic surface) is where the generic fibers are elliptic curves and singular fibers are (3 of them) three lines intersecting at one point and sorry but why does it have to kill H_1 of the fiber in the elliptic surface case? $\endgroup$
    – ozheidi
    Commented Nov 3, 2014 at 6:44
  • $\begingroup$ OK, by "typical" I meant simply connected. As it is in your case. But yours is not quite "typical" as its fibers are of type $\tilde A_2^*$. For such a surface, the monodromy group is cyclic if and only if the fibration is isotrivial. $\endgroup$
    – Alex Degtyarev
    Commented Nov 3, 2014 at 6:55
  • $\begingroup$ thanks! and isotrivial means nonsingular fibers are isomorphic ? also what is a good reference for this result ? $\endgroup$
    – ozheidi
    Commented Nov 4, 2014 at 0:55
  • $\begingroup$ Yes, isotrivial means that. I'm not good at refs. The thing is that, the monodromy group of a non-isotrivial elliptic surface over $\Bbb P^1$ is a subgroup of the modular group of genus zero, hence of finite index, hence non-abelian. I can refer to my book MR2952675, but I'm sure that the statement is much older than that. $\endgroup$
    – Alex Degtyarev
    Commented Nov 4, 2014 at 13:43

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