I ask for some hint/help about finding the monodromy homomorphism of a holomorphic map $g: \Bbb{P}^1_{\mathbb{C}}\rightarrow \Bbb{P}^1_{\mathbb{C}}$ locally defined by $$ f(z) = \frac{27z^2(z-1)^2}{4(z^2-z+1)^3} $$
I see that $deg(g) = 6$ and that we have two ramification points $\infty$ and $0$ with three points in the fiber of $0$, each of multiplicity 2 and two point in the fiber of $\infty$, each of multiplicity 3. Call this branch locus $B$ and consider $p\notin B$.
Hence, by Riemann Existence theorem, we get a monodromy homomorphism
$$ \rho:\pi_{1}(\mathbb{P}^1\smallsetminus B, p)\simeq \Bbb{Z}*\Bbb{Z}*\Bbb{Z}*\Bbb{Z}\rightarrow \mathcal{S}_6 $$ with transitive image. I know that the loop around multiplicity 2 -points should be 2-cycle, and similarly loops around multiplicity 3-points should be mapped to 3-cycle. But how can I determine the exact permutation for each homotopy class generator?
Thank you for any help you can provide.
It seems to me that the derivative of $f(z)$ vanishes at the points $-1$, $0$, $1/2$, $1$, $2$, and $\infty$. The images of these points are the values $0$ and $1$.
It also seems to me that the derivative of $1/f(1/w)$ vanishes at $-1$, $1/2$, $2$, and the points $\exp(\pm \pi i / 3)$. The images of these are again $0$ and $1$ (but in the $w$-plane).
So I believe that the branch locus $B$ (in the $z$-plane) consists of the points $0$, $1$, and $\infty$. The fundamental group of you want is that of $\mathbb{CP}^1 - B = \mathbb{C} - \{0, 1\}$, namely a free group of rank two.
The easiest monodromy to compute is the one about zero in the $z$-plane. So, take a circle of radius $r$ (with $r$ very small, positive, and real) and find the roots of $f(z) = r\exp(i\theta)$, as a function of $\theta$. As we walk around the circle, these six roots permute in the desired fashion.