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Suppose I have a variety $X$ and suppose that I have three line bundles $L_1, L_2$ and $L_3$ over $X$ such that $L_1 \oplus L_2 \oplus (L_2 \otimes L_3)$ has a nonzero global section. Does it imply that we have a nonzero global section of the bundle $L_1 \oplus L_2 \oplus L_3$ ?

Michael Albanese
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jack
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  • Yes. Take $X$ to be a smooth projective curve, then there is a nontrivial line bundle $L$ whose square is trivial. Then $L$ has no nontrivial global sections. Take $L_1=L_2=L_3=L$. – Aphelli Feb 22 '23 at 08:42
  • @Aphelli: you say "Yes", but probably mean "No", right? – Sasha Feb 22 '23 at 09:12
  • Can we put some conditions on $L_i$ such that the answer is affirmative ? – jack Feb 22 '23 at 09:28
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    Hi, In my opinion, the condition that $L_2$ has a nonzero global section is enough to get an affirmative answer. – sti9111 Feb 22 '23 at 09:45
  • You can also take any non-trivial line bundle $L$, put $L_1=L_2=L$ and $L_3=L^$, the dual line bundle. Then $L_2]otimes L_3=L\otimes L^$ has a canonical nowhere vanishing section (corresponding to the identity map), but $L^*$ is non-trivial since $L$ non-trivial. – Andreas Cap Feb 22 '23 at 09:49
  • @Sasha: you’re right, thanks for pointing it out. – Aphelli Feb 22 '23 at 11:00
  • @sti9111: How do you use the nonzero global section of $L_2$ to construct a nonzero section of $L_1 \oplus L_2 \oplus L_3$ ? – jack Feb 22 '23 at 17:29
  • Since, $H^0(X,L_1\oplus L_2 \oplus L_3)\cong H^0(X,L_1)\oplus H^0(X,L_2)\oplus H^0(X,L_3)$, then if $H^0(X,L_2)$ is not trivial this implies that $L_1\oplus L_2 \oplus L_3$ has a nonzero global section. – sti9111 Feb 22 '23 at 17:45
  • Is this question in the context of algebraic geometry? Cause this question has the algebraic-topology and differential-geometry tags and you're not specifying in the post with what kinds of bundles and what kinds of sections you're working with. In the continuous or smooth categories, the answers become very different compared to the algebro- or complex-geometric settings. – Thorgott Feb 22 '23 at 23:12
  • Yes, it is from algebraic geometry. – jack Feb 23 '23 at 03:33
  • @Thorgott, I believe that the question is clear even with your suggestions, because the word Variety is in the context of algebraic geometry, and the bundles are lines bundles. – sti9111 Feb 23 '23 at 07:58
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    @sti9111 variety can also be used for a manifold in French. More importantly, tags should be clear and specific - if this question doesn't have a relation to algebraic topology or differential geometry, those tags shouldn't appear here. – KReiser Feb 23 '23 at 15:22

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I think the answer is No. It is because you can take $X$ to be an Enriques surface; for this type of surface $K_X$ is not trivial, but $H^0(X, K_X)=\{0\}$ and $K_X\otimes K_X= 2K_X\sim \mathcal{O}_X$. Then, taking $L_i=K_X$, the expression $L_1\oplus L_2 \oplus (L_2\otimes L_3)= K_X\oplus K_X \oplus \mathcal{O}_X$, but $L_1\oplus L_2\oplus L_3= K_X\oplus K_X \oplus K_X$. This is basically the @Aphelli answer.

sti9111
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