0
$\begingroup$

How to justify the following statement:

deg$(K-D)\le p-1 \Rightarrow $ deg(D)$\ge p-1$.

at where D is a divisors on a compact Riemann surface.

Thank You!

  • 2
    Is $p$ the genus (in which case $\deg K = 2p - 2$)? – 2017-01-09
  • 0
    $p$ is a genus! – 2017-01-09
  • 0
    This statement is in Jost's book: Compact Riemann Surface. – 2017-01-09
  • 1
    The degree map is additive on divisors, so the implication comes down to an elementary inequality. :) – 2017-01-09
  • 1
    then use that $\deg (A+B)=\deg A+\deg B$, for all divisors $A$ and $B$. – 2017-01-09
  • 0
    @AndrewD.Hwang , this map deg:Div($\Sigma$)$\rightarrow \mathbb{Z}$? – 2017-01-09
  • 0
    Yes. Maybe you should start with rational functions on the Riemann sphere (which is compact) and show $deg(D_f) = 0$ where $D_f$ is the divisor of any such rational/meromorphic function ($f(z)$ is meromorphic on the Riemann sphere if $f(z)$ and $f(1/z)$ are meromorphic on $\mathbb{C}$) – 2017-01-09
  • 1
    Manoel : I did downvoted your question. I think you should show what did you try instead of putting question and waiting for the answer. – 2017-01-10
  • 0
    @N.H. I'm reading Jost's book (compact riemann surface). I need to learn about Divisors and Riemann-Roch Theorem... On page 216 is written the implication that is in my question. I did not know how to justify it. Particularly it affirmation (my question) is important to me, because (affirmation) + riemann-roch + ($deg(D)<0\Rightarrow h^{0}(D)=0 $) And I can prove that: $h^{0}(D)=deg(D) - g+1$. For D divisor in S (compact r. surface ) with $deg (D)\ge 2p-1$. Thank you for your remark! – 2017-01-10

1 Answers 1

0

$\deg(K - D) = \deg(K) - \deg(D) = 2p - 2 - \deg(D) \leq p - 1$ by hypothesis, so it follows immediately that $\deg(D) \ \geq p - 1$.

I am assuming you know that $\deg(K) = 2p - 2$ : if not, first prove it for $\mathbb P^1$ and then you can prove it by pullback for any compact projective curve.