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Let X be a compact kahler manifold, $M\subseteq X$ be a complex submanifold, is there a biholomorphism between a neighborhood of the zero section of the normal bundle $NM$ of $M$ and a neighborhood $U(M)\subseteq X$?

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The existence of such a biholomorphism is rather rare. For example, as a simple exercise you can check that already for a conic in $\mathbb CP^2$ such a biholomorphism does not exist. At the same time, in the case $M$ is "exceptional" in $X$, for example, $X$ is obtained by a blow up from $X'$ at a point $x'\in X'$ and $M$ is the exceptional divisor, the situation is the often the one you want (I don't specify purposefully here what "exceptional" means).

Solution to the exercise. Everything follows from the fact that each holomorphic map from $\mathbb CP^1$ to itself has at least one fixed point. Indeed if a neighbourhood of a conic $C$ in $\mathbb CP^2$ were biholomorphic to a neighbourhood of $O(4)$ bundle over $\mathbb CP^1$ then the restriction of $T \mathbb CP^2$ to $C$ would holomorphically decompose as a sum $N_C\oplus TC$ where $N_C$ is a line subbundle of the restriction. Now consider a line $L_x$ through each point $x$ of $C$ tangent to the direction $N_C(x)$. Finally consider $y(x)=L_x\cap C$ the second point of the intersection of $L(x)$ with $C$. We got the map $C\to C$ without fixed points, this is a contractioction.

PS. I remember a theorem that states that the only smooth hypersurface in $\mathbb CP^n$ that has a neighbourhood isomorphic to a neighbourhood of a zero section in a line bundle is a linear $\mathbb CP^{n-1}$. Thanks to jvp for the link below!

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Dmitri, you are probably thinking on Van de Ven Theorem. See mathoverflow.net/questions/61596/… – jvp Sep 25 2011 at 0:51
Dear jvp,thanks a lot: your link is very interesting. I can't access Van de Ven's article nor Ionescu/Repetto's right now: can you (or someone else) help? – Georges Elencwajg Sep 25 2011 at 6:52
Dear Dmitri, thank you for posting your solution to the exercise.However I don't understand how the existence of the biholomorphic map gets you that family of lines in $\mathbb CP^2$ parametrized by the conic. – Georges Elencwajg Sep 25 2011 at 7:27
@Dmitri: First of all thank you for the answer. Let me see if i understand, Let X be a kahler manifold and M a submanifold, M could not have a neighborhood that retract holomorphically to it, but if i blow up X along M and call E the exceptional divisor, E has a neighborhood that retracts holomorphicaly to it because it has "enough" complex lines to retract? – Italo Sep 25 2011 at 8:14
Dear Georges, sorry I was too sloppy, hope now the "solution" reads well. Dear Italo, I have to apologize, I would not be able to state a correct general version of the theorem on exceptional divisors. I just have a collection of examples in my head when this works. I have to say, that I don't think, that if you take say a blow up of $\mathbb CP^3$ in a curve of high degree then the exceptional divisor will have a good neighbourhood. In fact, I think, it will not... On the other hand, if you blow up an isolated singularity, you have better chance (again I don't know a precise result..). – Dmitri Sep 25 2011 at 11:13
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The answer is "no" because in general there does not even exist a neighbourhood $U$ of your submanifold that holomorphically retracts to the submanifold.
In fact Finnur Lárusson has proved the following (warning: I have kept his notations, so as to faithfully quote him. In particular $[X]$ is $\mathcal O(X)$, the line bundle on $M$ associated to the divisor $X$)

Theorem Let $X$ be a connected hypersurface in a compact complex manifold M of dimension at least $2$. If
(1) $H^0 (M, TM) = 0$
(2) $dim H^0(M; [X]) \geq 2$, and
(3) $H^1(M; [X]^{-1} \otimes TM) = 0$,
then no neighbourhood of $X$ retracts holomorphically onto $X$.

Note that the hypotheses are very easy to satisfy: (1) holds for $K3$ surfaces for example and (2), (3) will follow from $X$ being sufficiently ample.

Here is a freely downloadable version of Lárusson's article, from his homepage.

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