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In the correspondence between projective and Kaehler geometry an ample line bundle corresponds to a positive line bundle, where the latter requires that the curvature of the Chern connection is a positive $(1,1)$-form. A very ample is a strengthening of ample (no need to take powers). Is there a corresponding Kaehler notion, and in this case, is there a differential geometric proof that such a line bundle is generated by holomorphic sections?

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J. P. Demailly, Multiplier ideal sheaves and analytic methods, p. 26, (3.15) A holomorphic line bundle $F$ over a compact complex manifold $X$ is

a) very ample if the map $\Phi_{|F|}\colon X \to \mathbb{P}^{N−1}$ associated to the complete linear system $|F| = \mathbb{P}(H^0(X,F))$ is a regular embedding (by this we mean in particular that the base locus is empty, i.e. $B_{|F|} = \emptyset$).

b) ample if some multiple $mF$, $m > 0$, is very ample.

By Example 3.14, every ample line bundle $F$ has a smooth hermitian metric with positive definite curvature form; indeed, if the linear system $|mF|$ gives an embedding in projective space, then we get a smooth hermitian metric on $F^{\otimes m}$, and the $m$-th root yields a metric on $F$ such that $(i/2\pi)Θ(F )=(1/m)\Phi^*_{|mF}\omega_{FS}$. Conversely, the Kodaira embedding theorem tells us that every positive line bundle $F$ is ample (see Exercise 5.14 for a straightforward analytic proof of the Kodaira embedding theorem).

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  • $\begingroup$ So it's basically the same as in the algebraic case? $\endgroup$ – Pierre Dubois Nov 5 '19 at 20:25
  • $\begingroup$ Yes. It is algebraic, because the Kodaira embedding theorem says that $X$ must be a smooth complex projective variety. $\endgroup$ – Ben McKay Nov 5 '19 at 21:26
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As Ben says, the complex-geometric version of an ample line bundle is a positive line bundle: A line bundle is ample if and only if it admits a positively curved Hermitian metric.

However, I'm not aware of a complex-geometric version of the notion of a very ample line bundle. That is, I don't know of a condition one can put on a positively curved metric on a line bundle that ensures that the morphism the global sections of the bundle define to projective space is an embedding.

Some such thing presumably exists, and can be dug out of some of the proofs of the Kodaira embedding theorem. One needs the annihilation of cohomology groups that obstruct the morphism to projective space from being an embedding; possibly one can find such a condition locally and use compactness to conclude that the condition can be satisfied globally on a manifold.

In conclusion, consider an elliptic curve $E$ and a point $p \in E$. The line bundle $\mathcal O_E(p)$ is ample, and thus positive. However, it has no nonzero global sections, as any such section would lift to a doubly-periodic entire function on $\mathbb C$ with a single pole, and no such thing exists by integration over the boundary of a fundamental parallelogram that contains the pole. How can we tell that this positive bundle does not define an embedding to projective space from just a positive metric on the bundle?

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  • $\begingroup$ $\mathscr{O}_E(p)$ has a nonzero global section, which vanishes exactly at $p$ — with your identification, it corresponds to a constant function on $\mathbb{C}$. $\endgroup$ – abx Sep 1 '20 at 9:39
  • $\begingroup$ Hmm... it's been a while so I may be confused. The point of talking about that line bundle was supposed to be that it's ample but not very ample, which I think is still the case modulo my confusion. $\endgroup$ – Gunnar Þór Magnússon Sep 1 '20 at 10:31
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    $\begingroup$ Yes, of course. The rest of your answer goes through. I was just pointing out a small mistake. $\endgroup$ – abx Sep 1 '20 at 12:11

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