Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 3 / 23F (Riemann Surfaces)

A point is ramified with multiplicity $m_i$ exactly when it is a zero of $f^{\prime }$ of multiplicity $m_i-1$. Since $f^{\prime }$ has degree $d-1$, the fundamental theorem of algebra gives ${\sum }_i(m_i-1)=d-1$.

Conversely, choose distinct complex numbers $a_i$ and put $g(z)=c{\prod }_i(z-a_i{)}^{m_i-1}$ with $c\ne 0$. The assumption makes $g$ a polynomial of degree $d-1$. Any antiderivative $f$ of $g$ has degree $d$ and precisely the prescribed ramification multiplicities.

Outside the finite set of branch points, the derivative is nonzero. The inverse function theorem supplies evenly covered neighbourhoods, and compactness of the Riemann sphere makes every fibre finite, so the restriction is a covering map.

Lift a loop $\gamma$ from each point of the fibre over $z_0$; its endpoints define a permutation ${\sigma }_{\gamma }$. The homotopy lifting property shows that this permutation depends only on the based homotopy class. The punctured source is connected, so a path between any two points in the fibre projects to a loop whose lift joins them. Hence the monodromy group acts transitively.

The monodromy around both $0$ and $1$ must be a product of two disjoint transpositions, hence a $(2,2)$ cycle type in $S_4$. The product relation for loops around the three punctures says ${\sigma }_0{\sigma }_1{\sigma }_{\infty }=1$. By the stated fact, ${\sigma }_0{\sigma }_1$ is either the identity or another $(2,2)$ permutation. Consequently ${\sigma }_{\infty }=({\sigma }_0{\sigma }_1{)}^{-1}$ cannot contain a $3$-cycle. But ramification index $3$ over infinity would force exactly such a cycle, a contradiction.