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If C is a quadric hyperelliptic curve ($g(C)=3$ and the canonical line bundle is very ample) contained in the two dimensional complex projective space and $K_C$ is the canonical line bundle of $\mathbb{P}^2$ restrected to my curve.
I take four distinct points on $C$ named $p_1,\ p_2,\ p_3,\ p_4 $ so $p_1+ p_2+p_3+p_4 \in K_C= |\mathcal{O}_{\mathbb{P^2}}(1)| $. Than i take another points different from the previous, named $q$ and the divisor on $C$ $\ D=p_1+ p_2+p_3+q$.
I want to compute
$h^o(D)$.
So using Riemann Roch theorem i can write:
$h^o(D)=h^1(D)+deg(D)-g+1=h^1(D)+4-3+1=h^1(D)+2$.
Now i've used Serre Duality and i get:
$h^1(D)=h^0(K_C-D)$. Is there a way to say that $h^0(K_C-D)=0$ for example using some vanishing theorem?

Jyrki Lahtonen
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dario
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  • there is an error $K_C=(|\mathcal{O}_{\mathbb{P^2}}(1)|_C)$ – dario Sep 20 '14 at 15:11
  • sorry C is a QUARTIC curve NON hyperelliptic – dario Sep 20 '14 at 15:41
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    $h^0(K_C-D)=h^0(p_4-q)$, and this is the dimension of the vector space of all rational functions $f$ on $C$ such that $\mbox{div}(f)+p_4-q\geq0$ (or $f=0$). This means that $f$ can only have one pole (at $p_4$) and must have a 0 at $q$ which can't be; therefore $f$ must be 0. – rfauffar Sep 20 '14 at 20:55
  • The function $z-q \fract{z-p_4} doesn't work? – dario Sep 20 '14 at 23:43
  • Sorry in this case Why f would be a rational function? – dario Sep 20 '14 at 23:46
  • i mean the function $\frac{z-q}{z-p_4}$. This function have $q$ as zero and $p_4$ as pole. So why doesn't it work? – dario Sep 21 '14 at 11:37
  • Notice that that function is an isomorphism of $\mathbb{P}^1$ with itself, so it wouldn't work for any other curve (note that when writing $z$, you're specifying a local coordinate). – rfauffar Sep 21 '14 at 21:54
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    $f$ is a rational function because $H^0(C,\mathcal{O}_C(D))$ can be identified with the vector space of all rational functions $f$ that satisfy $\mbox{div}(f)+D\geq0$, along with 0. This is in Hartshorne, for example. – rfauffar Sep 21 '14 at 21:56
  • Related: https://math.stackexchange.com/questions/2470645 – Watson Nov 23 '18 at 08:41

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