Topology in Molecular Biology

12 The Spectral Geometry of Riemann Surfaces 223

Proof.Consider the function

g(z)=

{f(z)

z ,z=0

f′(0),z=0.

On|z|=r<1, we have

|g(z)|=

|f(z)|

|z|

≤

1

r

.

Hence|g(z)|≤ 1 /rfor all|z|≤r. Lettingr→1 complete the proof.

It was observed by Pick that the Schwarz lemma admits a geometric in-
terpretation: if ds^2 Hdenotes the hyperbolic metric onD,

ds^2 =

4(dx^2 +dy^2 )

(1−|z|^2 )^2

,

then any holomorphic mapD→Dis distance nonincreasing, with the distance
between two points preserved if and only iffis an isometry, that is to say a
M ̈obius transformation preservingD.
The proof follows from what we have already done by first observing that
the M ̈obius transformations preserving the disk are isometries of the metric
ds^2 H, and secondly by composingfwith such a M ̈obius transformation so that
it sends 0 to 0.
Ahlfors observed that the argument could be made even more geometric
by introducing the Gauss curvatureκof a metric. If the metric ds^2 is given
in conformal coordinates by ds

ds^2 =λ^2 (z)[dx^2 +dy^2 ],

then

κ(ds^2 )=

−∆(log(λ))

λ^2

,

where, since we are currently in “analyst’s mode,”

∆(f)=fxx+fyy.

The proof is an elementary calculation using Christoffel symbols.
We will give a number of different versions of the Ahlfors–Schwarz lemma.
The first is due to Scott Wolpert:

Lemma 2.LetS 1 andS 2 be two compact Riemann surfaces, with metricsds^21
andds^22 respectively, and

f:S 1 →S 2

a holomorphic map.
Suppose thatκ 2 (f(z))<κ 1 (z)< 0. Then f is distance decreasing.