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Let $C$ be a smooth projective curve of genus $g\geq 1$ over an algebraically closed field. Let $\mathcal{M}$ be a line bundle with $deg \mathcal{M}\geq 2g -1$. Let $T$ be torsion and denote by $U$ the kernel of the map $\mathcal{O}^n\to T$ ( Edit: where $n=H^0(T)$ ). $\mathcal{M}$ is non-special, so $H^1(\mathcal{M})=0$.

I am trying to understand that $H^1(\mathcal{M})=0$ implies $H^1(M\otimes U)$ vanishes. Any help is appreciated,

I got this from the proof of lemma 7 in 'Orlov:Remarks on Generators and Dimension of Triangulated Categories, http://arxiv.org/abs/0804.1163

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    With the new hypothesis on $n$, you can reduce to the case $T=O_C/O_C(-p)$. Then it works when $\deg\mathcal M\ge 2g$ as in your answer below.2011-12-18

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This is false in general. Consider the case $T=O_C/O_C(-(d+1)p)$ where $p$ is a given point in $C$, $d\ge 2g-1$ and $M=O_C(dp)$. Then we have a natural surjection $O_C\to T$ (so $n=1$) whose kernel is $U=O_C(-(d+1)p)$. So $M\otimes U=O_C(-p)$. But if $K_C$ is a canonical divisor on $C$, then $H^1(C, O_C(-p))\simeq H^0(C, O_C(K_C+p))\supseteq H^0(C, O_C(K_C)) \ne \{ 0 \}.$

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    thanks! I do not really understand how to reconcile this with my answer below, though.2011-12-18
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In the above setup, let $\mathcal{M}$ be a line bundle of degree $d$, $d\geq 2g$. Take any point $p$ on $C$. There is an exact sequence $0\to \mathcal{O}(-p)\to\mathcal{O}\to\mathcal{O}_p\to 0.$ The degree of $\mathcal{O}(-p)\otimes\mathcal{M}$ is $deg \mathcal{O}(-p)+ deg \mathcal{M}\geq -1+2g$ and thus $\mathcal{O}(-p)\otimes\mathcal{M}$ is nonspecial, whence $H^1(\mathcal{O}(-p)\otimes\mathcal{M})=0$.

The decomposition of a torsion sheaf $T$ into a direct sum of skyscrapers then yields the result for the lower bound $d\geq 2g$.

Agreed?

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    Thanks again for your input QiL. I edited the question- I set $n=H^0(T)$. I'll try to proof that "new" statement later.2011-12-18