There are a few posts that are related to what I'm asking (e.g. this one) but aren't precisely what I had in mind. The question is fairly basic but is one that confuses me:

Does a holomorphic vector bundle $V$ over a complex manifold $X$ admit a zero section? If so, why is $H^0(X, V) = 0$ (for bundles in which the structure group acts non-trivially)? I'm sure there's a fairly straightforward answer to this but I'm not so well-informed on the subject.

  • $\begingroup$ All vector bundles have zero sections. This of course does not necessarily mean $H^0(X,V)=0$, it may have non-zero sections. $\endgroup$ – Mohan May 11 '18 at 0:28

Yes, there's always a zero section. $H^0(X,V)$ is the vector space of sections. Writing $H^0(X,V) = 0$ means that it is the zero vector space, which contains exactly one element, namely the zero section.

  • $\begingroup$ I think I understand now - because it is a single value (0) instead of, say, $\mathbb{R}$, it is zero-dimensional instead of being one-dimensional, correct? That is, $H^0(X, V) = 0 \rightarrow h^0(X, V) = 0 $ but $H^0(X, \mathcal{O}_X) = \mathbb{R} \rightarrow h^0(X, \mathcal{O}_X)) =1$? $\endgroup$ – nonreligious May 11 '18 at 4:19
  • $\begingroup$ Yes, that's correct $\endgroup$ – Nathaniel Mayer May 11 '18 at 4:21

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