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This question is related to another question of mine. Let $X$ be a kahler manifold with $\dim_{\mathbb{C}}(X)=n$, let $\pi:E\rightarrow M$ be a holomorphic vector bundle of $rank_{\mathbb{C}}(E)=n-k$ over a kahler manifold $M$ with $\dim(M)=k$. Suppose there is a holomorphic embedding $\Phi$ of a neighborhood $N$ of the zero section of $E$ in $X$ that is $$\Phi:N\rightarrow X$$

What can be said about $X$? If $M$ is a point nothing, but what if $M$ is a higher dimensional manifold?

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Dear Italo, it would be nice if you can give a hint on the "type" of statement that you would like to have, i.e. give a slightly more specific question. At least, do you encounter such situations naturally? –  Dmitri Sep 26 '11 at 15:26
I know that the question is rather vague, i imagined a situation like this thinking about my last question and your answer, in particular van de ven theorem. I was wondering if situations like the one described above are "natural" or conditions above are quite restrictive (as i suspect). And as consequence of the rigidity there is a statement like this: if you have a kahler manifold $X$ a vector bundle $E$ as above necessarily X is isomorphic $\mathbb{P}(E\osum \mathcal{L})$ with $\mathcal{L}$ a line bundle. –  Italo Sep 26 '11 at 16:02

2 Answers 2

If your $X$ is projective and $E$ is an ample vector bundle over $M$ then the field of meromorphic functions of $X$ is a finite extension of the field of meromorphic functions of $\mathbb P(E \oplus \mathcal O_M )$. This follows from Corollary 6.8 of Hartshorne's Cohomological dimension of algebraic varieties.

There is also an analytic version by Andreotti, which precedes Hartshorne's paper, implying the same result.

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I would like to make one guess concerning one particular situation.

Correction In fact, the following guess is wrong (thanks to jvp)! But I don't want to delete this answer, since jvp left a comment to it with a cool link, that shows why the guess is wrong

Guess. Suppose that $M\cong \mathbb CP^1$, and $E=O(1)\oplus...\oplus O(1)$. In this case $X$ is rational, i.e. birational to $\mathbb CP^n$.

I have to say that I don't know how to prove this guess and even not 100% sure that this guess is correct. But first of all, this statement is obviously true if $n=2$ and I rather belive it should be possible to prove this for $n=3$ using twistor theory. The reason to make this guess is that if you have such a line in $X$ you can start to deform it and it deforms (at least locally) in exactly the same way as a line in $\mathbb CP^n$. Also, it is obvious that $X$ is rationally connected in this case.

Of course, this guess is a pure speculation. But maybe someone can prove it? :)

Remark 1. You can always blow up something in $X$ that does not touch the image of the zero section of $N$, so one can only say something (if one can at all) about birational type of $X$.

Remark 2. Note that if you don't assume $X$ to be Kaehler, but only ask it to be complex, then the guess is completely wrong. There exist huge amount of complex 3-folds containing lines with neighborhoods biholomorphic to a neighbourhood of a line in $\mathbb CP^3$. Such examples come from twistor theory -- one should take the twistor space of a conformally flat 4-manifold.

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Related paper arxiv.org/PS_cache/math/pdf/0304/0304006v1.pdf –  jvp Sep 26 '11 at 17:28
Thank you for this link, this is an interesting paper! –  Dmitri Sep 26 '11 at 18:26
You are welcome Dmitri. –  jvp Sep 26 '11 at 20:02

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