Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 24H / Solution

The valency theorem states that for a nonconstant analytic map between compact connected Riemann surfaces,for every target point $w$.

For a rational map $f=p/q$ on the Riemann sphere, first cancel common factors; then the degree of a rational map of the Riemann sphere isThe analytic isomorphisms are exactly the degree-one maps, namely the Möbius transformations.

The required transformation isIt sends $\infty \leftrightarrow 1$ and $0\leftrightarrow -1$. Its fixed points satisfy

The octahedral rotation orbits on the Riemann sphere have possible sizescorresponding respectively to vertices, face centres, edge centres, and generic points.

In the displayed $F$, no numerator factor vanishes at $0$ or at a fourth root of one. Hence $0,\pm 1,\pm i$ are poles of order four. The numerator and denominator have degrees $24$ and $20$, so $F(z)\sim z^4$ at infinity; infinity is also a pole of order four. Thusand

Finally let $z$ have stabilizer of order $e$. A local coordinate turns its stabilizer action into rotation by $e$th roots of unity. Since $F$ is invariant, the first nonconstant term of its local expansion has exponent divisible by $e$, so $m_F(z)\ge e$. The orbit has $24/e$ points and therefore contributes at least $24$ to the fibre through $F(z)$. The valency theorem and $\mathrm{deg}F=24$ leave no room for any further point. The degree-sized invariant separates finite-group orbits, so $F(z)=F(w)$ implies that $z$ and $w$ lie in the same orbit.