Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 5 / 4 / Solution

Let $p:\mathbb{D}\to D$ be a universal covering map. On a small evenly covered neighborhood in $D$, choose a local inverse $s$ of $p$ and setAny two such lifts differ by a deck transformation of the disc. Every automorphism of the unit disk preserves the disc density: apply the derivative Schwarz-Pick lemma to the automorphism and its inverse to get equality. Hence the definition is independent of the local inverse. Uniqueness of the universal cover up to a covering isomorphism gives independence of the chosen cover as well. This is the Poincare metric on a Riemann surface, normalized to curvature $-1$.

For a piecewise continuously differentiable path $\gamma$, define its length by $L_D(\gamma )=\int {\lambda }_D(\gamma )|{\gamma }^{\prime }|$, and the distance by the infimum of lengths of joining paths. Path reversal and concatenation prove symmetry and the triangle inequality. The distance between distinct points is positive: choose a small closed Euclidean disc around one point not containing the other, on which the positive continuous density is bounded below by $c>0$. Every joining path must leave that disc and has length at least $c$ times its radius before doing so. Joining paths of finite length exist because a plane domain is path connected and the density is locally bounded. Thus this construction gives an actual metric, not just a nonnegative path-length expression.

For the punctured unit disc, use $p(w)=e^{iw}$ from the complex upper half-plane. The half-plane line element is $|dw|/\mathrm{Im}w$, while $|dz|=|z||dw|$ and $\mathrm{Im}w=\log (1/|z|)$. ConsequentlyThe domain called an annulus in the paper has inner radius zero, so it is this punctured-disc metric, not the metric of an annulus with positive inner radius.

We now prove the Great Picard theorem. In its holomorphic form, an essential singularity takes every complex value, with at most one exception, infinitely often in every punctured neighborhood. It suffices to prove that a holomorphic function on a punctured disc omitting $0$ and $1$ extends meromorphically through the puncture: if two values occurred only finitely often, a smaller punctured disc would omit both, and an affine change of target would reduce them to $0,1$.

Use the universal covering by the modular lambda function $\lambda :\mathbb{H}\to Y=\mathbb{C}\setminus \{0,1\}$, whose construction is given in the following solution. Its deck transformation group is the projective level-two principal congruence subgroup $\Gamma (2)$. Rescale the source punctured disc to unit radius. Lift $f(e^{2\pi iw})$ to a holomorphic function $L:\mathbb{H}\to \mathbb{H}$. Uniqueness of lifts givesfor a fixed deck transformation $T\in \Gamma (2)$. The Schwarz-Pick lemma givesas $\mathrm{Im}w\to \infty$. A nonidentity deck transformation has no interior fixed point, so it is either a hyperbolic Möbius transformation or a parabolic Möbius transformation. The hyperbolic case is conjugate to a dilation $w\mapsto aw$ with $a>0$, $aeq1$. Since $|d\log \mathrm{Im}w|\le d{s}_{\mathbb{H}}$, its displacement is at least $|\log a|>0$, contradicting this inequality.

If $T$ is the identity, compose $L$ with the Cayley transform between the half-plane and disk. Its periodicity lets the resulting bounded function descend to the punctured disc and extend by the Riemann removable singularity theorem. A nonconstant extension takes its central value inside the disc by the maximum modulus principle. The covering map then gives an extension of $f$; a constant lift gives a constant $f$.

In the parabolic case, its fixed point is a rational modular cusp. Indeed, a parabolic integral matrix of trace $\pm 2$ has a rational eigenvector for its repeated eigenvalue. Conjugate by an element of the modular group sending this cusp to infinity. The level-two subgroup is normal, so the conjugated transformation is $\zeta \mapsto \zeta +2n$ for a nonzero integer $n$. The conjugated lift $P$ consequently satisfiesThe modular transformations permute the three target cusp values $0,1,\infty$, as seen from the lambda transformation formulas below. It is therefore enough to examine $\lambda (P(w))$.

To fix the sign of $n$, the function $\mathrm{Im}P$ is periodic in the real direction. Its mean $M(y)={\int }_0^1\mathrm{Im}P(x+iy)dx$ satisfies $M^{\prime }(y)=2n$ by the Cauchy-Riemann equations and the increment of $\mathrm{Re}P$. Positivity for all $y>0$ forces $n>0$. The functionis well-defined, nonvanishing and bounded by one on the punctured disc, so it extends holomorphically to zero. Its winding number along a circle around the puncture is $n$, because a unit increment in $w$ increases $\pi iP$ by $2\pi in$. The argument principle therefore shows that $Q$ has a zero of order $n$ at the origin.

At the infinite cusp, the modular lambda function has the convergent expansion $\lambda (\zeta )=16q+O(q^2)$ in $q=e^{\pi i\zeta }$. Thus $\lambda (P(w))=16Q(z)+O(Q(z{)}^2)$ extends and tends to zero. Undoing the target permutation gives a removable singularity or a pole for $f$. In every case the puncture is not essential, which proves the holomorphic Great Picard theorem. For a meromorphic function with an essential singularity, if three spherical values occurred only finitely often, a Möbius transformation would make the omitted values $0,1,\infty$. The same proof applies because omission of infinity makes the transformed function holomorphic. Hence the meromorphic version has at most two exceptional spherical values.