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Let $C$ be a smooth projective curve over $\mathbb{C}$. Let $A$ be a degree $d$ line bundle on $C$, and $M$ be a degree 0 line bundle on $C$ such that $M^2=\mathcal{O}_C$, that is, it is a 2-torsion line bundle. Therefore, we have that $deg(A)=deg(A\otimes M)$. Is it in general true that $h^0(C,A)=h^0(C,A\otimes M)$? If we assume that $deg(A)>2g(C)-2$, we get that $h^0(C,A)=h^0(C,A\otimes M)$. But otherwise can we say anything?

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We have the subvariety $W^r_d(C)=\{A\in Pic^d(C)|h^0(C,A)\geq r+1\}$ of $Pic^d(C)$. A 2-torsion line $M$ acts on $Pic^d(C)$. Assume $0< d<2g-2$, and that $W^r_d(C)\setminus W^{r+1}_d(C)=\{A\in Pic^d(C)|h^0(C,A)=r+1\}$ is nonempty. Then will the action of $M$ on $Pic^d(C)$ take some open subset in $W^r_d(C)\setminus W^{r+1}_d(C)$ to itself?

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    $\begingroup$ What do you want to say? If $A=\mathcal{O}_C$, clearly one of them is 1 and the other is zero. In general degree of $A$ has to be sufficiently large. $\endgroup$
    – Mohan
    Sep 30, 2015 at 1:05
  • $\begingroup$ For a general choice of $A$ in $Pic^d(C)$ with $d\leq 2g-2$, do we get $h^0(C,A\otimes M)=h^0(C,A)$? $\endgroup$ Sep 30, 2015 at 1:48

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No, in general we don't have $h^0(C,A)=h^0(C,A\otimes M)$.
Let $M$ be a non-trivial line bundle of degree $0$ such that $M^2=\mathcal O_C$: such bundles exist as soon as $C$ is not isomorphic to $\mathbb P^1$.
Then by taking $A=M$ (so that $d=0\leq 2g-2$) we get $$0=h^0(C,A)\neq h^0(C,A\otimes M)= h^0(C,M^2)=h^0(C,\mathcal O)=1 $$

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  • $\begingroup$ Thanks @Georges! Can we say anything for a general element? That is, a 2-torsion line bundle $M$ acts on $Pic^d(C)$. We have the subvariety $W^r_d(C)=\{A∈Pic^d(C)|h^0(C,A)≥r+1\}$. Will $ M$ take some open subset of $W^r_d(C)\setminus W^{r+1}_d (C)$ to itself? I will perhaps add edits to the question. $\endgroup$ Sep 30, 2015 at 14:54

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