3
$\begingroup$

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.

$\endgroup$
7
  • 2
    $\begingroup$ 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 $L(D) = 1$. $\endgroup$ Nov 13, 2020 at 21:32
  • 1
    $\begingroup$ Yes, just that $K_X(p)$ is a positive integer for every $p \in supp(K_X)$. $\endgroup$
    – Zhooo
    Nov 13, 2020 at 21:46
  • $\begingroup$ Cool, then the rest of my comment should do the trick. Doesn't Miranda prove both of these fundamental facts? $\endgroup$ Nov 13, 2020 at 21:57
  • $\begingroup$ I don't understand still, how i can find a divisor positive from $K_X$ of degree $=2$ and such that $\dim L(-)=2$? $\endgroup$
    – Zhooo
    Nov 13, 2020 at 22:00
  • 1
    $\begingroup$ 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 $K_X > 0$. $\endgroup$ Nov 13, 2020 at 22:07

1 Answer 1

5
$\begingroup$

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 $\ell(K_X') = \ell(K_X) = 2$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .