Suppose that $L,L^{'}$ are a line bundle over a compact riemann surface $C$. Take $H^0(C,L\otimes L^{'})$.
Is it true that $h^0(C,L\otimes L^{'})=h^0(C,L)+h^0(C,L^{'})$ where $h^0(V)$ ,means the complex dimension?
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dario
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No. Have you tried some examples? – Alex Youcis Aug 21 '15 at 15:48
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@Alex Youcis is the statement negative even if i suppose that both of lines bundle satisfy the relation that the double tensor product is equivalent to the caninical bundle? – dario Aug 21 '15 at 15:53
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This is false: let $C = \mathbb{P}^1_{\mathbb{C}}$ and let $L = \mathcal{O}(-1)$ and $L' = \mathcal{O}(1)$. Then, $$2 = \underbrace{h^0(C,L)}_{=0} + \underbrace{h^0(C,L')}_{=2} \neq h^0(C, L \otimes L') = 1.$$
It might be more reasonable to expect the dimensions to multiply i.e. have $$ h^0(C,L) \cdot h^0(C,L') = h^0(C,L\otimes L'), $$ because there is a natural map $H^0(C,L) \otimes_{\mathbb{C}} H^0(C,L') \to H^0(C,L \otimes L')$. However, this will not always be true, since this map is not an isomorphism, in general. (If $C$ is an affine scheme over $\mathbb{C}$, then this will be true: see the excellent answer here by Georges Elencwajg.)
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I think you made a mistake. $h^0(C,L\otimes L')=0$. Maybe $L=\mathcal{O}(-1)$ and $L'=\mathcal{O}(1)$ – Alex Youcis Aug 21 '15 at 16:48
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