2
$\begingroup$

I have the following problem:

Let $S$ be a compact Riemann Surface of genus $g$, and let $p\in S$. Show that there is a meromorphic function $F$ on $S$ with pole of order $n$ at $p$, if $n\ge 2g$.

I made a rough draft of the solution:

Solution: Let $p \in S$ and consider the divisor $D=np$. For a canonical divisor $K$ on a compact Riemann surface of genus $g$, we have

$deg K=2g-2$,

and so $deg (K-D)=degK-degD$.

Like $D=np \Rightarrow degD=n \Rightarrow deg(K-D)=2g-n-2.$ But $n\ge 2g \Rightarrow 0\ge2g-n \Rightarrow deg(K-d)=2g-n-2 <0$, and thus, $h^{0}(K-D)=0$. The riemann-roch teorem therefore yields

$h^{0}(D)=deg(D) -g+1=n-g+1=n+1-g.$

But $n\ge 2g \Rightarrow n+1>2g>g \Rightarrow n+1-g>0$. Therefore, $h^{0}(D)>0$.

Therefore, we can find a nonconstant meromorphic function $F$ with $D+(F) ≥ 0$, i.e. with at most a pole of order $n$ at $p$, and as $F$ is nonconstant, it must have a pole somewhere, and so it does have a pole of order $n$ at $p$.

I need some help to make sure that it is correct and also for writing. The final part seems unclear (confused), in my view.

Please be welcome to any correction!

I noticed that in case the genre be $g=0$, this exercise shows that: $S$ is equivalent to the Riemann sphere.

$\endgroup$
1
$\begingroup$

To show that there exists a meromorphic function with a pole of order precisely $n$ at $p$, you should show that $h^0(np) > h^0((n-1)p)$, i.e. that there exists a section $\mathcal O_S(np)$ which fails to be a section of $\mathcal O_S((n-1)p)$. You can do this using the method in your post. Does this make sense?

$\endgroup$
  • $\begingroup$ This no make sense for me :..( And my solution, does this make sense? $\endgroup$ – Manoel Feb 26 '17 at 5:11
  • 1
    $\begingroup$ Your method of calculation is fine, but you really need to compute both $h^0(np)$ and $h^0((n-1)p)$. Let me explain like this. $h^0(np)$ is the dimension of the space of meromorphic functions there are with at most an order $n$ pole at $p$. $h^0((n-1)p)$ is the dimension of the space of meromorphic functions with at most an order $n-1$ pole at $p$. So $h^0(np) - h^0((n-1)p)$ is the dimension of the space of meromorphic functions with at most an order $n$ pole at $p$ quotiented by the space of meromorphic functions with at most an order $n-1$ pole. $\endgroup$ – Kenny Wong Feb 26 '17 at 5:25
  • 1
    $\begingroup$ Thus, if $h^0(np) - h^0((n-1)p) \geq 1$, then there must exist a meromorphic function with a pole at $p$ of order precisely equal to $n$. $\endgroup$ – Kenny Wong Feb 26 '17 at 5:25
  • 1
    $\begingroup$ I got to understand what you said! Thank you for your attention! $\endgroup$ – Manoel Mar 1 '17 at 1:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.