# Positive vector bundles

In the case of a line bundle over M, positivity of such a bundle (one whose curvature form which is Kahler) gives rise to an embeddings of M into the projective space.

Now I have in mind (more or less) the following definition. Let E be a holomorphic vector bundle over M with a hermitian metric. Moreover let D be a connection on E. Then we can define D^2 so that the curvature matrix of 2-forms. Such a curvature matrix (tensor) gives rise to a Hermitian form O_E on the bundle TM\otimes E. We can say that E is positive if such a hermitian form O_E is positive on all the tensors in TM\otimes E.

Then, What is the geometric meaning (if any) of the positivity in a vector bundle? rank>1.

I think, there are several definitions that generalize the concept of positive line bundle. Can you say which is the more standard one and why?

I edited the previous question since it was ambiguous.

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I don't understand what sense of positivity you're talking about here. –  Scott Morrison Nov 6 '09 at 7:21
I believe there is a definition of positivity of a vector bundle in Lazarsfeld's book "Positivity in Algebraic Geometry." That said, I would also like to know what defnition Csar is using. –  David Speyer Nov 6 '09 at 11:22
I'm not sure if this helps, but when I hear 'positive vector bundle', I think of something along the lines of eom.springer.de/p/p073980.htm. Anyway, it would be good to know what Csar has in mind exactly. –  1-- Nov 6 '09 at 15:22
Let's say that a vector bundle E is positive if its curvature form is positive everywhere. What is the geometric info about our manifold that this fact carries? –  Csar Lozano Huerta Nov 6 '09 at 16:58
Perhaps your question isn't specific enough. What do you mean by "curvature" for a general vector bundle? –  Ryan Budney Nov 7 '09 at 1:05

The positivity you are talking about is named Nakano positivity. It implies Griffiths positivity (you only require positivity of the hermitian form on $T_M\otimes E$ on rank-one tensors) which implies ampleness.

To finish with, an ample vector bundle $E$ is such that there exists a $k_0$ such that the symmetric powers $S^kE$, for $k\ge k_0$ are very ample. In particular, the natural maps $$\psi_{H^0(M,S^kE)}\colon M\to G_r(H^0(M,S^kE))$$ define an embedding (here $G_r(H^0(M,S^kE))$ is the Grassmannian of $r$-codimensional linear subspaces of $H^0(M,S^kE)$.

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I think there might be some confusion between the following notions:

1. A complex vector bundle on a manifold (yields a map to BGLn(C)).

2. A holomorphic vector bundle on a complex manifold (gives an embedding to projective space when "positive" in a suitable sense).

This distinction confused me for a while. A holomorphic structure on a bundle is not a trivial thing. It roughly amounts to half of the data of a connection, by forcing holomorphicity on local sections.

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Vamsi, the answer to your question "Is positivity sufficient to ensure that the bundle is generated by finitely many sections?" is yes. If E is positive, E is in particular globally generated which means that the evaluation map H°(X,E)->E_x is onto for every x in X. But it is (well?)-known that in such a case, thanks to Baire theorem applied in those Frechet spaces, there exists V a finite dimensional subspace of H°(X,E) such that V generates all fibers E_x. Moreover, we can choose V such that dim V is less or equal to dim X+dim E.

More precisions on Demailly's online book, Complex analytic and algebraic geometry, chapter VII, prop. 11.2, avalaible at http://www-fourier.ujf-grenoble.fr/~demailly/books.html

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Mmm... I know plenty of positive vector (even line) bundles with no non-zero holomorphic global section... I would say rather the contrary... A globally generated vector bundle carries a hermitian metric such that its curvature is Griffiths positive (and hence the dual bundle is Nakano negative). –  diverietti Jan 25 '11 at 23:31
You are perfectly right! I had in mind the case of high powers of line bundles, and I got confused! –  Henri Jan 26 '11 at 8:35

I think if you impose the condition of "positivity" on a holomorphic vector bundle (meaning that the curvature is positive definite on vector-valued forms), you can find global holomorphic sections that generate all the fibres. Thus, the bundle is very ample and gives you an embedding into a Grassmannian (see Griffiths and Harris' section on Grassmanians). But, I am not sure of whether positivity is sufficient to ensure that the bundle is generated by finitely many sections. One might use Hormander's theorem to prove this (but I haven't gone over the details myself).

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