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Let $X$ be a smooth projective variety with $L$ a line bundle.

Assume $L$ has global sections and it is primitive (i.e. it is not $L=M^{\otimes k}$ for some $k>1$ and some other line bundle $M$).

For any global section $s\in H^0(L)$, its powers $s^k\in H^0(L^{\otimes k})$, for $k>0$. More generally, any homogeneous polynomial of degree $k$ in the global sections of $L$ will be a global section of $L^{\otimes k}$.

Are all the sections of $L^{\otimes k}$ obtained like this?

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  • $\begingroup$ What does primitive mean? $\endgroup$
    – Mohan
    Mar 1, 2016 at 19:00
  • $\begingroup$ Dear @Mohan, I mean that it has no $k$-th roots. I will edit to clarify $\endgroup$ Mar 1, 2016 at 19:05

1 Answer 1

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Almost never. For example, take $X$ to be a smooth projective curve of positive genus and $L=\mathcal{O}(P)$, for a point $P$. Then $L$ has exactly one dimensional global sections, so taking powers, you only get a 1-dimensional subspace of $H^0(L^k)$. But, as $k$ increases, this vector space is approximately $k-g+1$ dimensional.

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  • $\begingroup$ What if the original line bundle is ample? $\endgroup$
    – Pierto
    Nov 7, 2022 at 14:18
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    $\begingroup$ @Pierto My line bundle above is ample. $\endgroup$
    – Mohan
    Nov 7, 2022 at 15:14

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