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Jason Starr
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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, degree $2$$\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ cover, yet others do not: the projective smooth model of the affine curve $\text{Zero}(y^2-x(x-1)(x-a)(x-b)(x-c), x-z^2) \subset \mathbb{C}^3$$\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 orbifold$\mathbb{Z}/2\mathbb{Z}$-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.

The action is not transitive on simple closed curves. For instance, some of the simple closed curves lift to the following unbranched, degree $2$ cover, yet others do not: projective smooth model of the affine curve $\text{Zero}(y^2-x(x-1)(x-a)(x-b)(x-c), x-z^2) \subset \mathbb{C}^3$. 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 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.

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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Jason Starr
  • 4.1k
  • 1
  • 93
  • 111

The action is not transitive on simple closed curves. For instance, some of the simple closed curves lift to the following unbranched, degree $2$ cover, yet others do not: projective smooth model of the affine curve $\text{Zero}(y^2-x(x-1)(x-a)(x-b)(x-c), x-z^2) \subset \mathbb{C}^3$. 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 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.

The action is not transitive on simple closed curves. For instance, some of the simple closed curves lift to the following unbranched, degree $2$ cover, yet others do not: $\text{Zero}(y^2-x(x-1)(x-a)(x-b)(x-c), x-z^2) \subset \mathbb{C}^3$. 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 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.

The action is not transitive on simple closed curves. For instance, some of the simple closed curves lift to the following unbranched, degree $2$ cover, yet others do not: projective smooth model of the affine curve $\text{Zero}(y^2-x(x-1)(x-a)(x-b)(x-c), x-z^2) \subset \mathbb{C}^3$. 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 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.

Source Link
Jason Starr
  • 4.1k
  • 1
  • 93
  • 111

The action is not transitive on simple closed curves. For instance, some of the simple closed curves lift to the following unbranched, degree $2$ cover, yet others do not: $\text{Zero}(y^2-x(x-1)(x-a)(x-b)(x-c), x-z^2) \subset \mathbb{C}^3$. 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 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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