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Let $X$ be a non-singular projective complex variety and $\mathcal{L}$ be an invertible sheaf on $X$ with negative degree. Is it true that $\mathcal{L}$ has no global sections? If so, can someone suggest a reference?

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    $\begingroup$ Without further data, "degree" only makes sense for line bundles (=invertible sheaves) on curves. $\endgroup$ Commented Jun 21, 2015 at 21:39
  • $\begingroup$ I have edited the question. Use the definition of degree as in Ex. II. 6.2 of Hartshorne after identifying invertible sheaves with corresponding divisors as explained in chapter II.6 of Hartshorne. $\endgroup$
    – user54369
    Commented Jun 21, 2015 at 22:51
  • $\begingroup$ @user54369 : But Hartshorne Ex. II. 6.2 , isnt written for curves? $\endgroup$
    – Holonomia
    Commented Jun 22, 2015 at 6:15
  • $\begingroup$ @user54369 : it is Ex II. 6.12 the one for curves, sorry. But if you use Hartshorne definition of degree (i.e. Ex II. 6.2.) it seems to me that your claim follows from the 1-1 correspondence of Proposition 13 plus the fact that the zero set of a section is an effective divisor whose degree seems to be non negative. I think that the last follows also from Ex. II. 6.2. part c). $\endgroup$
    – Holonomia
    Commented Jun 22, 2015 at 6:32
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    $\begingroup$ Let me try to expand on my comments.That exercise in Hartshorne is talking about the degree of a divisor on a variety with a fixed embedding in projective space. That is the "extra data" I was referring to. This sounds like a petty point to argue over, but in fact it really matters: there are varieties $X$ which come with many different embeddings into projective space, and depending on which embedding you choose, you can get a different answer for the degree of a subvariety. In other words, this notion of degree is not intrinsic to $X$ (unlike the case for curves). $\endgroup$ Commented Jun 22, 2015 at 22:51

2 Answers 2

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I guess it's a bit late but it might be useful to someone.

As emphasized by @Relapsarian, if we work with a non singular curve $C$ (I will assume moreover that it's a projective curve), then we can talk about the degree of an invertible sheaf.

The correct formulation of the O.P. question should be then:

([Proposition) An invertible sheaf of negative degree has no non-zeros global sections.

Suppose we have a negative-degree invertible sheaf $\mathcal L$ on $C$ with a non-zero section $s$. Then $s$ has no poles and probably some zeros, so $\operatorname{deg} \mathcal L\ge 0$ which contradicts the assumption.

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That seems to be correct, at least if you have in mind compact manifolds. See the introduction of http://www.google.it/url?sa=t&rct=j&q=&esrc=s&source=web&cd=62&ved=0CCoQFjABODw&url=ftp%3A%2F%2Fftp.math.ethz.ch%2Fhg%2FEMIS%2Fjournals%2FNYJM%2Fjdg%2Farchive%2Fvol.50%2F1_4.ps.gz&ei=DiOHVae6LKOcygOuooGQAQ&usg=AFQjCNEBF-29nDprqrWQeOXRVDa2wZ6RQg&sig2=A87X-NRzKdVC4BzeUEf2yQ

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  • $\begingroup$ But isnt this written for curves? $\endgroup$
    – user54369
    Commented Jun 21, 2015 at 20:16
  • $\begingroup$ Yes, sorry. I am going to edit my answer. $\endgroup$
    – Holonomia
    Commented Jun 21, 2015 at 20:52
  • $\begingroup$ I am assuming that the variety is projective as well $\endgroup$
    – user54369
    Commented Jun 21, 2015 at 23:00

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