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There is a mistake in the following argument but I cannot see where. Can someone help me, please?

Let $C$ be any smooth curve of genus $g\geq 1$ and $D$ a general effective divisor of degree $g+2$. Then given any pair of points $p,q\in C$, we have by Riemann-Roch $h^0(K+p+q)=g+1$ so we get a morphism $$\phi:C\rightarrow \mathbb P^g$$ given by the sections of $K+p+q$, where $\phi(C)$ is non degenerate. Then since a general set of $g+2$ points are not contained in a hyperplane of $\mathbb P^g$ and $\phi(C)$ is non degenerate, $h^0(K+p+q-D)=0$ so that, $h^0(D-(p+q))=h^0(D)-2=1$ which means that $D$ separates points and tangent directions so it gives an embedding of $C$ in $\mathbb P^2$.

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    $\begingroup$ I think you just got wrong when you claim that $h^0(K+p+q-D)=0$, because you need this holds true for all pairs of points $p, q$. However, what you showed is that if you fix $p,q$, then $h^0(K+p+q-D)=0$ for a general $D$. The generality of $D$ depends on $p,q$, so you did not show that there is a $D$ such that $h^0(K+p+q-D)=0$ for all $p,q$. $\endgroup$
    – Chen Jiang
    Commented May 18, 2018 at 11:58
  • $\begingroup$ Thank you very much for your answer. Is it right then, that the locus in $C^{(d)}$ of $D$ whose image in $|K+p+q|$ (for a given pair $p,q$) is contained in a hyperplane is a determinantal variety? $\endgroup$
    – pi_1
    Commented May 23, 2018 at 11:32

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