5
$\begingroup$

I am using Rick Miranda's book "Algebraic curves and Riemann Surfaces" to try and check some things about hyperelliptic curves. I have completed almost all of one of the exercises, but there is one part of my proof I am not sure of. The question (question R) 2), page 245), is as follows:

Let $X$ be an algebraic curve of genus $g\geq 2$. Show that $2\notin G_P(|K|)$ if and only if $X$ is hyperelliptic and $P$ is a ramification point.

In this case, $2\notin G_P(|K|)$ simply means that there does not exist a meromorphic function with a pole order exactly 2 at $P$ and no other poles (I think this is the right definition to be using). Technically, $G_P(|K|)$ is the set of gap numbers.

I have shown the if part fine. The other direction I think is true because there will be two points in the preimage of the projection to $\mathbb P^1$ if $P$ is not ramified. Hence, I want to say that if there is a pole at $P$, there will be a pole at the other point in the preimage. But this can't in general be true, because by Riemann-Roch applied to the divisor $nP$ for sufficiently large $n$, we get a function with a pole only at $P$. So can anyone help point out where I am going wrong?

$\endgroup$
0

1 Answer 1

4
$\begingroup$

You don't have a projection to $\mathbf{P}^1$. You have to "construct" one. The fact that $2$ is not a gap number for $P$ means that there is a rational function $f$ such that the divisor of poles $(f)_{\infty}$ equals $2\cdot P$. Consider the morphism $\overline{f}:X\to \mathbf{P}^1$ given by this rational function $f$. It is of degree $2$ (because $\deg \overline{f} = \deg (f)_{\infty} = \deg (f)_{0}$). So you have a hyperelliptic map on $X$. Moreover, you clearly see that $P$ is a ramification point because it has multiplicity two in the divisor of $f$.

So this proves that if $2$ is not a gap number for $P$, then $X$ has a hyperelliptic map for which $P$ is ramification points.

Let us prove the other implication. It is just as easy.

Now, suppose that $X$ is hyperelliptic and let $P$ be a ramification point. This means that there is a finite morphism $\pi:X\to \mathbf{P}^1$ of degree $2$ such that $P$ ramifies. Now, we may and do assume that $P$ maps to $\infty$. In fact, composing with an automorphism that sends $\pi(P)$ to $\infty$ will do the trick. Now, consider the divisor of poles of $\pi$. This is the divisor $(f)_{\infty} = 2\cdot [P]$. In fact, we already know that $P$ is in the support of $(f)_{\infty}$ because $P$ maps to $\infty$ and we also know that $P$ ramifies. So it must have multiplicity 2. So we see that there is a meromorphic function $f$ on $X$ such that $(f)_{\infty} =2 [P]$. But this means precisely that $2$ is not a gap number of $P$.

You can show that, if $X$ is hyperelliptic of genus $g\geq 2$, the hyperelliptic map $X\to \mathbf{P}^1$ has precisely $2g+2$ ramification points, and for each ramification point $P$, the gap-sequence $\Gamma(P)$ at $P$ equals $$\{1,3,5,\ldots,2g-1\}.$$ Hence, each $P$ has weight $g(g-1)/2$, and the ramification points are exactly the Weierstrass points of $X$.

$\endgroup$
3
  • $\begingroup$ Thank you very much for clarifying, I clearly wasn't thinking about it properly. Do you have any idea about the other direction i.e. if $P$ is not ramified then 2 is a gap number? $\endgroup$
    – Joe Tait
    Commented May 2, 2012 at 23:12
  • 1
    $\begingroup$ See my edited answer. $\endgroup$
    – Harry
    Commented May 3, 2012 at 7:06
  • $\begingroup$ That's brilliant, thanks very much, makes complete sense now. And in believe that you meant {1,3,5...,2g-1}. $\endgroup$
    – Joe Tait
    Commented May 3, 2012 at 9:46

You must log in to answer this question.

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