# genus$=2$ implies Hyperelliptic.

In the book of Rick Miranda (Algebraic Curves and Riemann Surfaces), in Proposition 1.10 of Chapter VII (p. 198), the claim is that every compact Riemann Surfaces of genus $$2$$ is hyperelliptic. The proof is:

Let $$K_X$$ the canonical divisor has degree $$2g-2=2$$, and by Riemann-Roch Theorem we have $$\dim L(K_X)=2$$, therefore we may suppose that $$K_X>0$$, take $$f\in L(K_X)$$ nonconstant so when you look as a map to $$\overline{\mathbb{C}}$$ has degree $$=2$$ so it's Hyperelliptic.

It's a good proof but a just don't understand why we can suppose $$K>0$$.

I'm tried write $$K_X=P-N$$ where $$P,N\in \operatorname{Div}(X)$$ are non-negative divisors and use Riemann-Roch but I failed, thanks for any comments.

• By $K_X > 0$ you just mean that $K_X$ is a nonzero effective divisor class? Recall that a divisor class $D$ is effective iff $L(D) \neq 0$, and that $D$ is trivial iff $L(D) = 1$. Nov 13, 2020 at 21:32
• Yes, just that $K_X(p)$ is a positive integer for every $p \in supp(K_X)$. Nov 13, 2020 at 21:46
• Cool, then the rest of my comment should do the trick. Doesn't Miranda prove both of these fundamental facts? Nov 13, 2020 at 21:57
• I don't understand still, how i can find a divisor positive from $K_X$ of degree $=2$ and such that $\dim L(-)=2$? Nov 13, 2020 at 22:00
• If $K_X$ were not effective, we would have $L(K_X) = 0$. Thus $K_X \geq 0$. If $K_X$ were the zero divisor class, we would have $L(K_X) = 1$. Since $L(K_X)$ is known to be $2$, in must be the case that $K_X > 0$. Nov 13, 2020 at 22:07

Recall that $$\DeclareMathOperator{\div}{div} L(D) = \{f \in \mathbb{C}(X) : D + \div(f) \geq 0\} \cup \{0\}$$ where $$D$$ is a divisor on $$X$$. Since $$\ell(K_X) = 2 > 0$$, then there is a nonzero function $$f \in L(K_X)$$, i.e., a function $$f \in \mathbb{C}(X)$$ such that $$K_X + \div(f) \geq 0$$. Since $$K_X$$ is only defined up to linear equivalence we can replace $$K_X$$ by $$K_X' := K_X + \div(f)$$: we have $$K_X' \geq 0$$ by the definition of $$L(K_X)$$, and as pointed out in the comments, $$K_X' \neq 0$$ since $$\ell(0) = 1$$ and $$\ell(K_X') = \ell(K_X) = 2$$.