Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 23G (Riemann Surfaces)

An elliptic function for a lattice $\Lambda$ is a meromorphic function on $\mathbb{C}$ satisfying $f(z+\lambda )=f(z)$ for all $\lambda \in \Lambda$. Its order is the number of poles, counted with multiplicity, in a fundamental parallelogram, equivalently the degree of its map from the torus to the Riemann sphere. Differentiation preserves the periods. If the distinct poles have orders $r_1,\dots ,r_s$, the derivative has exactly the same pole locations with orders $r_j+1$. ThereforeThe bounds use $1\le s\le m$, valid for a nonconstant elliptic function.

For the Weierstrass elliptic function $\wp$, $f=\wp {\wp }^{\prime }$ has a single pole of order five, so its derivative has order six. For the second example choose $c$ not a critical value of $\wp$ and choose $a$ away from zero and the two points where $\wp =c$. Thenhas a double pole at $a$ and three simple poles: the quotient has residue $-2$ at zero and residue 1 at each of its two other poles. No locations overlap. Hence its order is five and its derivative has order $3+2+2+2=9$, as required.

The monodromy theorem states that a germ admitting analytic continuation along every path in a simply connected domain extends to a single-valued holomorphic function there. Equivalently continuations along homotopic paths with fixed endpoints agree. For a biholomorphism between the two tori, lift its composition with $\mathbb{C}\to \mathbb{C}/{\Lambda }_1$ through the second covering. Simple connectivity, or monodromy applied to local inverse charts, yields a holomorphic lift $F:\mathbb{C}\to \mathbb{C}$.

Lift the inverse similarly. The compositions of the lifts differ from the identity by lattice translations; adjusting the lifts shows that $F$ is bijective. By the allowed classification, $F(z)=az+b$ with $a\ne 0$. For $\lambda \in {\Lambda }_1$, its two arguments $z,z+\lambda$ represent the same torus point, so $a\lambda =F(z+\lambda )-F(z)\in {\Lambda }_2$. Thus $a{\Lambda }_1\subseteq {\Lambda }_2$. Apply the inverse lift to obtain the reverse inclusion. ThereforeThe translation $b$ affects the chosen origin, not the period lattice.