Equivalent definitions of hyperelliptic Riemann surfaces

Question

Let $X$ be a connected Riemann surface. I’m trying to understand the equivalence about these two definitions (I’m still new to this subject):

1. $X$ is hyperelliptic iff there is a holomorphic degree $2$ map $p:X \rightarrow \mathbb{CP}^1$.

2. $X$ is hyperelliptic iff there is a degree $2$ divisor $D$ such that $\dim(L(D)) \geq 2$ (here $L$ denotes the Riemann-Roch space).

I think I can see that $2)$ implies $1)$: if $\dim(L(D)) \geq 2$, then there must be a function $f \in \dim(L(D))$ such that either $f$ has two single poles or $f$ has a double pole at some point $p\in X$. In either case, $\deg(\tilde{f})=\sum_{p\in \tilde{f}^{-1}({\infty})}mult_p(\tilde{f})=2$, where $\tilde{f}:X \rightarrow \mathbb{CP}^1$ is the holomorphic function obtained by “extending” the meromorphic function $f:X \rightarrow \mathbb{C}$ to the point at infinity. (Is this correct?)

However, I don’t see how to prove that $1)$ implies $2)$. Do I need to find a divisor $D$ that satisfies the conditions? Any help would be appreciated.

Answer 1

Your argument for $2\implies 1$ is correct, but I don't know what definition of degree you are using. The one I will assume is: given a map of Riemann surfaces $f:X\to Y$, $\deg(f) := \#f^{-1}(p)$ for any $p\in Y$, counted with multiplicity. It is of course a claim that this definition of degree is well-defined.

For $1\implies 2$: given such a holomorphic map $f:X\to \Bbb{P}^1$, consider the divisor $$f^*(\infty):= \sum_{P\in f^{-1}(\infty)} \mathrm{mult}_P(f)\cdot P.$$

$f$ is a meromorphic function and by definition $f\in L(f^*(\infty))$ . Indeed, $\mathrm{div}(f) + f^*(\infty) = 0$. Consequently, $\dim L(f^*(\infty)) \ge 2$ since it contains the constants and $f$, which is nonconstant by hypothesis.

Answer 2

I don't think your argument for $2 \implies 1$ is correct. In principle your (non-constant) $f \in L(D)$ could have more than 2 poles or only 1 pole. In the latter case, we are biholomorphic to $\mathbb{CP^1}$ so we are done. Otherwise, we need that $D>0$ to conclude with your argument. But this can be easily achieved by replacing D with $\tilde{D} := D + (f)$. Then notice that $\deg(D)=\deg(\tilde{D})$ and by Riemann-Roch $\dim L(D)=\dim L(\tilde{D})$.

For the other direction, since $p$ must have two poles counted with multiplicity, we can take the divisor $D = (x_1)+(x_2)$, where $x_1$ and $x_2$ are the (possibly identical) poles of $p$ and conclude that $\deg (D) =2$ and $\dim(L(D)) \geq 2$ because $p$ is not constant.