How do I compute the monodromy action of rotating three punctures of $\mathbb{P}^1$ on the fundamental group.

Question

Let $M\subset\mathbb{C}^3$ be the orbit of the circle group $S^1\subset\mathbb{C}^\times$ on the tuple $(1,e^{2\pi i/3},e^{4\pi i/3})$ (by multiplication). Thus, $M$ is homeomorphic to $S^1$, and so has fundamental group $\mathbb{Z}$.

Now consider the fiber bundle $X\subset\mathbb{P}^1_{\mathbb{C}}\times M$ defined as the subspace obtained by removing the points $\{a,b,c\}$ from each fiber $\mathbb{P}^1\times\{(a,b,c)\}$. Thus, $X$ is a fiber bundle over $M$ with fibers which are projective lines minus 3 points. This bundle admits the constant section "$\infty$".

Let $(a,b,c)\in M$, then we get an exact sequence of fundamental groups $$1\rightarrow \pi_1(\mathbb{P}^1-\{a,b,c\})\rightarrow\pi_1(X)\rightarrow\pi_1(M)\rightarrow 1$$ which is split by the section "$\infty$", from which we get an action $$\pi_1(M)\rightarrow\text{Aut}(\pi_1(\mathbb{P}^1-\{a,b,c\}))$$ By "drawing pictures", I'm pretty confident this action is trivial. How can we prove this?

Answer 1

There is an isomorphism of bundles over $M$: $$ \begin{align*} \left(\mathbb{P}^1\backslash\{1,e^{2\pi i/3},e^{4\pi i/3}\}\right)\times M&\to X,\\ (t,(z_1,z_2,z_3))&\mapsto (z_1t,(z_1,z_2,z_3)), \end{align*} $$ so $X$ is a trivial bundle. It follows that the action of $\pi_1(M)$ on the fundamental group of the fibers of $X$ is trivial.