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Consider the family of proper complex genus two curves with affine equation $y^2 = x(x-1)(x-a)(x-b)(x-c)$, defined over an open subset $U$ of $\mathbb{C}^3$. Here, $U$ consists of all triples $(a,b,c)$ such that the corresponding curve is smooth.

Fix a point $x \in U$, and let $C$ denote the corresponding curve. The fundamental group $\pi_1(U,x)$ acts on the set of isotopy classes of simple closed curves on $C$. Is this action transitive on the set of isotopy classes of non-separating simple closed curves? Is this action transitive on the set of isotopy classes of separating simple closed curves? (The answer in both cases is yes if $\pi_1(U,x)$ is replaced by the mapping class group of $C$)

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    $\begingroup$ Can you spell put the action you are asking about? $\endgroup$
    – Andy Putman
    Commented Sep 19, 2016 at 1:06
  • $\begingroup$ Parallel transport gives a map from $\pi_1(U,x)$ to the mapping class group. I think this is an inclusion, and gives a subgroup of index either 6 or 12, and if the index is 6, I think the quotient is $S_6$ (something similar if the index is 12). $\endgroup$
    – user98640
    Commented Sep 19, 2016 at 1:13
  • $\begingroup$ @user98640: What parallel transport do you mean? In order to make this work you would need a flat bundle over this surface whose structure group is the mapping class group. There is no natural way to make this work. Please, revise your question to make the construction which you have in mind clear. $\endgroup$
    – Misha
    Commented Sep 19, 2016 at 1:39
  • $\begingroup$ If $X \rightarrow B$ is a family of smooth projective varieties, the Gauss-Manin connection on the relative de Rham cohomology gives a notion of parallel transport. $\endgroup$
    – user98640
    Commented Sep 19, 2016 at 1:42
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    $\begingroup$ Any loop based at $x$ gives a homeomorphism from $C$ to $C$. The isotopy class of this homeomorphism depends only on the class of the loop in $\pi_1(U,x)$. Therefore, there's a map from $\pi_1(U,x)$ to the mapping class group of $C$. The mapping class group of $C$ acts on the set of isotopy classes of simple closed curves of C. Therefore the fundamental group also acts on the set of isotopy classes of simple closed curves of C. $\endgroup$
    – user98640
    Commented Sep 19, 2016 at 1:49

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The action is not transitive on simple closed curves. For instance, some of the simple closed curves lift to the following unbranched, $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ cover, yet others do not: the projective smooth model of the affine curve $\text{Zero}(y^2-x(x-1)(x-a)(x-b)(x-c), 4xz^2-(z^2+1)^2) \subset \mathbb{C}^3$. Since your parameterization of the curves "fixes" the branch points over $x=0$, $x=1$, and $x=\infty$, you are also "fixing" the quotient fundamental group of the orbifold with underlying manifold $\mathbb{CP}^1$ (the "$x$-line") and $\mathbb{Z}/2\mathbb{Z}$-orbifold points at $x=0$, $x=1$, and $x=\infty$. This orbifold fundamental group is isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. So the image of a free homotopy class in this quotient group is fixed by the monodromy action.

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