12 The Spectral Geometry of Riemann Surfaces 217ThenSis a Belyi surface.Our proof is based on the proof in [4]. The converse to this theorem is also
true, and proved by Belyi, but we will not need it.
Proof.:Letφ:S^2 →S^2 be a rational function. Iffis a holomorphic function
fromStoS^2 , then the critical values ofφ(f) are the image underφof the
critical values off, together with the critical values ofφ.
Suppose thatz 1 ,...,znarenpoints onS^2 lying inK∪{∞}, and letkbe
the maximal degree of thezi’s, which we may assume isz 1. Then there is a
polynomialPof degreekwith integer coeffcients such thatP(z 1 )=0.
Pdoes not raise the degree of any of the pointszi, and sendsz 1 to some-
thing of degree 1, namely the point 0. Furthermore,P introduces critical
points of degree≤k−1, namely the solutions ofP′(z) = 0. Hence the num-
ber of critical points of degree≥nreduces by at least one.
Arguing inductively, we may reduce to the case where all thezi’s have
degree 1, that is they are rational numbers. After adding and multiplying by
rational constants, we may assume that the critical values include 0, 1 ,∞,and
at least one value between 0 and 1, which we may then write asα/(α+β).
We now consider the map
P(z)zα(1−z)β.It sends 0 and 1 to 0,∞to∞, and has critical points at (at most) 0, 1, and
α/(α+β). So the total number of critical values decreases by at least one.
This concludes the proof.
As a simple consequence, we haveCorollary 1.LetSbe a Riemann surface. Then for everyε, there exists a
surfaceSεwithinεofS, such that
S=SC(G,O)
for some(G,O).
Proof.According to Riemann–Roch, for any Riemann surfaceSthere is a
holoporphic function
f:S→S^2.We do not specify what metric we use to measure , because they are all
equivalent. We could use any of the standard notions of distance in moduli
space in the proof.
Take a small diffeomorphism ofS^2 taking the critical points offto points
lying in some number fieldK, for instance the fieldQ[i], which is already
dense inS^2. Then lift this conformal structure toSto obtainS.