Low-degree Riemann-Roch computations

By [F3] at $A=0$, $\mathrm{\ell }(K)=i(0)=g$. At $A=K$, duality gives $i(K)=\mathrm{\ell }(0)=1$, so Riemann–Roch gives $g-1=\mathrm{deg}K+1-g$, hence $\mathrm{deg}K=2g-2$. If $\mathrm{deg}D<0$, [F2] gives $\mathrm{\ell }(D)=0$ and [F3] gives $\chi ({\mathcal{O}}_X(D))=\mathrm{\ell }(D)-i(D)=-\mathrm{\ell }(K-D)$. If $\mathrm{deg}D>2g-2$, then $\mathrm{deg}(K-D)<0$ and [F2] gives $\mathrm{\ell }(K-D)=0$; [F3] therefore gives $i(D)=0$ and $\mathrm{\ell }(D)=\mathrm{deg}D+1-g$.

Suppose $\mathrm{deg}D=0$ and $0\mathrm{\ne }f\in L(D)$. The effective divisor $(f)+D$ has degree zero by [F2], so all its nonnegative coefficients vanish; thus $(f)=-D$ and $D\sim 0$. Conversely, if $D\sim 0$, [F2] and [F3] identify $L(D)$ with the one-dimensional space $L(0)=\mathbb{C}$. This proves the degree-zero equivalence and the dimension bound.

A degree-one proper nonconstant holomorphic map to the sphere is a biholomorphism: [F5] makes every fibre a single point of multiplicity $1$, so the map is bijective; [F6] provides local holomorphic inverses, which agree with its unique inverse and hence glue to a holomorphic inverse. It would force $g=0$ by [F7]. Now let $\mathrm{deg}D=1$, $g\ge 1$, and suppose $f_0,f_1\in L(D)$ are independent. Put $E_j=(f_j)+D$, effective of degree one by [F2], and $r=f_1\mathrm{/}{f}_0$. Independence makes $r$ nonconstant, and its pole divisor is bounded by $E_0$ since $(r)=E_1-E_0$. By [F4] and [F5], it is a proper map of degree at most one and at least one, contradicting the preceding genus consequence. Thus $\mathrm{\ell }(D)\le 1$. A nonzero section gives an effective degree-one divisor $(f)+D=[p]$, so $D\sim [p]$. Conversely $D\sim [p]$ identifies $L(D)$ with $L([p])$, which contains constants and has dimension at most one; hence equality holds. For effective degree-one $D=[p]$, this also proves $L(D)=\mathbb{C}$.

Let $\mathrm{deg}D=2$ and $g\ge 2$. Choosing any point $p$, [F8] and step 1.3 give $\mathrm{\ell }(D)\le \mathrm{\ell }(D-[p])+1\le 2$. Suppose $\mathrm{\ell }(D)=2$ and choose independent $f_0,f_1\in L(D)$, representing sections as in [F9]. Their effective zero divisors $E_j=(f_j)+D$ each have degree two. Let $B$ be their common effective divisor, with coefficient $B(q)=\min \{E_0(q),E_1(q)\}$. If $B\mathrm{\ne }0$, choose a point $q$ in its support; then both $f_j$ belong to $L(D-[q])$, contrary to the degree-one bound in step 1.3. Hence $B=0$. The nonconstant ratio $r=f_1\mathrm{/}{f}_0$ has divisor $E_1-E_0$; because the two effective divisors have disjoint support, its pole divisor is exactly $E_0$, of degree two. By [F4] and [F5], $r:X\to \widehat{\mathbb{C}}$ has degree two, proving hyperellipticity in the stated analytic sense. Its contrapositive and the bound $\mathrm{\ell }(D)\le 2$ give $\mathrm{\ell }(D)\le 1$ for every degree-two divisor on a nonhyperelliptic surface.