A Riemann surface $X$ is hyperelliptic iff there is a holomorphic involution of $X$ that has exactly $2g(X)+2$ fixed points

algebraic-geometryriemann-surfaces

I’d like to show that a Riemann surface $X$ is hyperelliptic iff there is a holomorphic involution $\tau : X \rightarrow X$ such that $\tau$ has exactly $2g+2$ fixed points, where $g$ is the genus of $X$.

(I’m working with the following definition of a hyperelliptic Riemann surface: $X$ is hyperelliptic iff there is a holomorphic degree $2$ map $p:X \rightarrow \mathbb{CP}^1$).

My thoughts so far (please tell me if this is the correct idea):

Assuming $X$ is hyperelliptic, I was thinking of defining $\tau$ such that $p^{-1}(z)=\{x,\tau(x)\}$ for all $z \in \mathbb{CP}^1$. That way, $\tau$ would be an involution of $X$ that would fix the ramification points of $p$. By using Riemann-Hurwitz, I found $p$ would need to have $2g+2$ ramification points, so that seems to work. I’m not sure how to show that $\tau$ would have to be holomorphic, though.

For the other direction, I don’t really know how to proceed… any hint would be helpful.

Best Answer

Edit: thanks to Ted Shifrin who pointed out an error. Riemann-Hurwitz applied to a degree $2$ map from a genus $g$ surface to $\Bbb{P}^1$ gives:

$$ 2g - 2 = 2(-2) + \sum_{P\in X} (e_P - 1) $$

where $e_P$ is the ramification index at $P\in X$. So, $2g+2$ is the number of ramification points (since they all must have $e_P = 2$). Consequently, I think the problem statement should be modified to "$2g + 2$ ramification points". Note that this generalizes the statement for a genus $1$ surface, $E$, which admits a map $E\to \Bbb{P}^1$ ramified at $4$ points. See here.

Your idea is correct. Given a hyperelliptic curve with map $f:X\to \Bbb{P}^1$ as in the definition, you can define a map $\alpha:X\to X$ given by permuting the points in each fibre $f^{-1}(p) = \{q_1,q_2\}$ of cardinality $2$ and fixing each singleton fibre. $\alpha$ is easily seen to be a bijection of $X$ to itself and it is also clearly fibre preserving. To see that $\alpha$ is holomorphic, you can check locally around each point. There are basically two cases:

  1. $q\in X$ lies in a fibre of cardinality $2$: i.e. $q\in f^{-1}(p) = \{q,q'\}$. In this case, the holomorphic map $f$ is unramified at both $q$ and $q'$ so that there exists a small open neighborhood $U$ of $p$ such that $f|_{f^{-1}(U)}:f^{-1}(U)\to U$ is equivalent to the map $U\times \{1,2\}\to U$ given by $(x,i)\mapsto x$. $\alpha$ maps $f^{-1}(U)$ into itself and acts by $(x,1)\mapsto (x,2)$ and $(x,2)\mapsto (x,1)$. That is, $\alpha$ is (locally) the "switch sheets of the cover" map. So, $\alpha$ is holomorphic near such $q$.
  2. $q\in X$ lies in a fibre of cardinality $1$: i.e. $q$ is a ramification point for the map $f$. In local analytic coordinates, $f$ is given $z\mapsto z^2$. In this case, $\alpha$ acts by $z\mapsto -z.$ Hence, it is again holomorphic and we are done.

By definition, the ramification points of $f$ are the fixed points of $\alpha$. Next, given such an involution $\alpha$, we basically want to form the quotient of $X$ by the associated $\Bbb{Z}/2\Bbb{Z}$ action. Long story short is that you can, and the associated map $X\to X/\alpha$ is degree $2$ with ramification points at the $2g + 2$ fixed points. An application of Riemann-Hurwitz shows that $X/\alpha$ is genus $0$, i.e. biholomorphic to $\Bbb{P}^1$.

Here is a sketch proof that $X/\alpha$ is well-defined: You can first consider $X_0 = X\setminus \{p_1,\ldots, p_{2g+2}\}$, i.e. the complement of the fixed points. $\alpha$ still acts on this set and in particular it acts freely. Consequently, $X_0/\alpha$ inherits a Riemann surface structure in a unique way compatible with the map $\pi_0:X_0\to X_0/\alpha$. Now, there is a unique way to complete $X_0/\alpha$ to a compact Riemann surface $Y$ so that $\pi_0:X_0\to X_0/\alpha$ extends to a map $\pi:X\to Y = X/\alpha$. The idea is that you map "holes" in $X_0$ to "holes" in $X_0/\alpha$. More precisely, given a curve $\gamma(t)$ in $X$ such that $\lim_{t\to 0} \gamma(t) = p_i$ for one of the $p_i$, define a curve by $(\pi_0\circ \gamma)(t)$ in $X_0/\alpha$. Define $\pi(p_i)\in Y$ in such a way that $\lim_{t\to 0}(\pi_0\circ \gamma)(t) = \pi(p_i)$. This should be enough that you can fill it in yourself if you don't know this already.

An extra comment: if you learn the equivalence of categories between transcendence degree $1$ field extensions over $\Bbb{C}$ (i.e. finite extensions of $\Bbb{C}(t))$ and Riemann surfaces, you will know that a map $X\to \Bbb{P}^1$ corresponds functorially to a field extension $\mathcal{M}(X)$ of $\Bbb{C}(t) = \mathcal{M}(\Bbb{P}^1)$. The degree of the map corresponds to the degree of the extension of fields. So, a hyperelliptic curve $X\to \Bbb{P}^1$ is nothing more than a degree $2$ field extension $\mathcal{M}(X)\supset \Bbb{C}(t)$. Because $[\mathcal{M}(X):\Bbb{C}(t)] = 2$, $\mathrm{Aut}_{\Bbb{C}(t)}(\mathcal{M}(X)) \cong \Bbb{Z}/2\Bbb{Z}$, but then it follows that $\mathrm{Aut}_{\Bbb{P}^1}(X) = \Bbb{Z}/2\Bbb{Z}$. Consequently, there exists an order $2$ automorphism of $X$ commuting with the map $X\to \Bbb{P}^1$. Such a map takes ramification points to themselves, as needed.

Note, though, that in spite of the fancy "functorial" language, the methods from earlier on in the post are really needed to set up this correspondence. This perspective does result in quick answers to these sorts of questions, though.

Related Question