Past exam of the geometry course of the University of Oxford / 2008 / B / 8

A divisor on an algebraic curve is a finite formal sum $D=\sum n_{p}p$ with integer coefficients. Its degree is $\mathrm{deg}D=\sum n_p$. The principal divisor of a nonzero meromorphic function is $(f)=\sum {\mathrm{ord}}_p(f)p$. For a nonconstant $f$, zeros and poles are fibers over zero and infinity of the same map to the sphere, each of total multiplicity equal to its degree; hence $\mathrm{deg}(f)=0$. A nonzero constant has zero divisor.

Write $L(D)=\{f:(f)+D\ge 0\}\cup \{0\}$ and $\ell (D)=\dim L(D)$. The Riemann-Roch theorem stateswhere $K_C$ is a canonical divisor. Taking $D=p_1+\cdots +p_{g+1}$ gives $\ell (D)=2+\ell (K_C-D)\ge 2$. Constants span only one dimension, so there is a nonconstant $f\in L(D)$. It has no poles outside these points and at most a simple pole at each. Thus the required nonconstant function exists, with actual poles at some nonempty subset because a holomorphic function on a compact connected curve is constant.

Since every $p_i$ is fixed, $f\circ \phi$ has poles only among the first $g+1$ points, with order at most one: a biholomorphic change of coordinate preserves pole order. Thus $h=f-f\circ \phi$ also has at most $g+1$ poles, counted with multiplicity. At every $p_i$ with $i>g+1$, both terms are finite and equal, so $h(p_i)=0$.

If $h$ were not zero, its principal divisor would have degree zero. It would therefore have at most $g+1$ zeros counted with multiplicity. But it has at least $n-(g+1)$ distinct zeros. When $n>2g+2$ this number exceeds $g+1$, a contradiction. Consequently

Choose an actual pole $p_i$ of $f$; one exists by nonconstancy and compactness. In a local coordinate $z$ centered there, its simple pole has expansion $f(z)=a_{-1}/z+O(1)$ with $a_{-1}\ne 0$. HenceThe holomorphic inverse function theorem makes $1/f$ a local coordinate. Here the source notation $f^{-1}$ denotes the reciprocal, rather than an inverse map of the meromorphic function.

The chosen pole is among the fixed points, so $\phi$ maps a sufficiently small neighborhood of it into the coordinate neighborhood. If $f=f\circ \phi$, then $w=w\circ \phi$ there. Injectivity of this coordinate forces $\phi (x)=x$ throughout a nonempty open set. The stated identity principle then gives $\phi$ equal to the identity everywhere.

If a nonidentity holomorphic automorphism had more than $2g+2$ fixed points, choose that many distinct ones and apply the function constructed above. The zero-and-pole count would give $f=f\circ \phi$, and the local reciprocal coordinate would force $\phi$ to be the identity, a contradiction. Therefore