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This question arised when I was studying Beauville's book 'Complex Algebraic Surfaces'.

Castelnuovo's theorem says that a smooth rational curve $E$ on an algebraic surface $S$ is an exceptional curve iff $E^2=-1$. The proof in Beauville's book is to find a very ample divisor $H$ satisfying $H^1(S,\mathcal{O}_S(H))=0$ first, and then set $H'=H+kE$ where $k=H\cdot E$. The linear system of $H'$ gives a projective morphism from $S$ to $\mathbb{P}^n$ which contracts $E$, and then some topological arguments implies that the image of $S$ is actually smooth.

Although this proof is not difficult to understand, I still want a proof based on complex manifolds but not algebraic geometry.

Question: Is there any holomorphic version of the tubular neighborhood theorem?

I have several reasons to raise this question:

  1. If we have some holomorphic tubular neighborhood theorem, we can identify some neighborhood $U$ of $E$ in $S$ with neighborhood $V$ of the zero section in $N_E$. Here $N_E$ is the holomorphic normal bundle of $E$. Then $E^2=-1$ easily implies $N_E\cong\mathcal{O}_{E}(-1)$, so $E$ can be contracted in $U$ directly. Thus we not only prove Castelnuovo's theorem but also generalize it to non-algebraic surfaces.

  2. There exists a symplectic version of the tubular neighborhood theorem, so I guess the holomorphic case is also true.

Any answers or comments are welcome. I'll really appreciate your help.

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1 Answer 1

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+100

I don't have a proof that this works, but according to Joe Harris, if $C \subset \mathbb{P}^2$ is a smooth conic curve, then there is no holomorphic map from a neighborhood of $C$ to $C$, something that would be doable if there was a holomorphic tubular neighborhood theorem. Additionally, he claimed that, if you let $X$ be the total space of the normal bundle of $C$, with $C \subset X$ as the zero section, then even the first order neighborhoods of $C$ in $X$ and in $\mathbb{P}^2$ are not isomorphic. Again, I lack proof, but the second statement is something that I assume can be checked algebraically.

Aside: there is another way that you could try to cook up a counterexample. If there was a tubular neighborhood theorem, with $X \subset Y$ complex manifolds, let $Z$ be the total space of the normal bundle of $X$ in $Y$. Then one has that $T_Z|_X = T_X \oplus N_{Y/X}$ in $Z$, and so if you follow through the supposed map in the holomorphic tubular neighborhood theorem, you get that $T_Y|_X = T_X \oplus N_{Y/X}$.

Edited to add: here is something that I can show is a counterexample. Let $\pi: X \rightarrow V$ be a non-isotrivial proper family of curves of genus greater than $2$. Choose a smooth fiber $C = \pi^{-1}(p)$. Then a neighborhood of $C$ contains $\pi^{-1}(U)$ for some neighborhood $U$ of $p$, and I have that $\pi^{-1}(q) \neq \pi^{-1}(p)$ for a dense open set of all $q \in U$. Then, any map from $\pi^{-1}(U) \rightarrow C$ must factor through $\pi$: any fiber not isomorphic to $C$ admits no nonconstant maps to $C$ and so any holomorphic map from $\pi^{-1}(U)$ factors through $\pi$ on a dense open and so it does so on all of $\pi^{-1}(U)$. But then, if the holomorphic tubular neighborhood theorem were true, there would be a holomorphic map from $\pi^{-1}(U) \rightarrow C$ that restricts to the identity on $C$, which is in contradiction with what was just remarked.

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Thanks, and actually I also ask this question on MathOverflow: mathoverflow.net/questions/114414/… –  Yuchen Liu Jul 2 '13 at 9:57

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