3
$\begingroup$

$X$ is a compact Riemann surface of genus $g$. I'm starting to wonder whether this statement (in my title) is true at all or whether it's a typo...it seems impossible to prove using just the Riemann-Roch theorem, unless I'm not seeing something simple. Do I need to use more than just Riemann-Roch? I can only do the case $\deg(D) = -1$ right now.

I'll just explain my notation here, I'm using Otto Forster's notation from his book Lectures on Riemann Surfaces. $D$ is a divisor on $X$. $H^0(X, \mathcal{O}_D)$ is obviously the 0-th cohomology group of X with respect to the sheaf $\mathcal{O}_D$ where $\mathcal{O}_D (U) = \{ f \in \mathcal{M}(U)| \mathrm{ord}_x(f) \geq -D(x) \, \forall x\in U \}$ and $\mathcal{M}$ is the sheaf of meromorphic functions.

The version of Riemann-Roch with which I'm familiar is: for a compact Riemann surface $X$ of genus $g$ and a divisor $D$ on it, $\dim H^0(X, \mathcal{O}_D) - \dim H^1(X, \mathcal{O}_D) = 1 - g + \deg D$.

I also tried using the consequence of the Serre Duality which asserts that $\dim H^1(\mathcal{O}_D) = \dim H^0(X, \Omega_{-D})$ where $\Omega_{-D}$ is the sheaf of meromorphic 1-forms that are multiples of $-D$. But I just cannot get the above inequality!

Could I please get a hint?

$\endgroup$
1
$\begingroup$

You don't need Riemann-Roch. Let $D$ be a divisor with $\dim H^0(X,D) \neq 0$. Then, there is a non-identically zero section $s \in H^0(X,D)$, say $s(p) \neq 0$. Evaluation at $p$ gives an exact sequence $$ 0 \to H^0(X, D- p) \to H^0(X,D) \to k \to 0$$ and you are done by induction since you know that there are no sections for a divisor with negative degree.

$\endgroup$
  • $\begingroup$ Actually I'm a little confused on what we're doing induction on? $\endgroup$ – Acton Feb 19 '18 at 2:08
  • $\begingroup$ Never mind, I figured it out. Thanks for your explanation. $\endgroup$ – Acton Feb 19 '18 at 2:50
  • $\begingroup$ @ActonGrey : perfect, you're welcome ! $\endgroup$ – Nicolas Hemelsoet Feb 19 '18 at 6:50
  • $\begingroup$ Hi. Sorry for the very late comment. May I know if there is a way to do so using Riemann Roch? $\endgroup$ – thedilated Jul 31 at 5:32

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.