Is there any holomorphic version of the tubular neighborhood theorem?

Question

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.

Answer 1

Unfortunately, the tubular neighborhood theorem is not true in general in the holomorphic context. To see the obstruction, consider the exact sequence of holomorphic vector bundles over $E$: $0\to TE\to TS|_E \to NE \to 0$. If we were to have a holomorphic embedding of a neighborhood of the zero section of $NE$ into $S$, in particular we would obtain a splitting of this exact sequence. The obstruction to doing so is a class in $Ext^1(NE,TE)$, as can be seen by applying the left-exact functor $Hom(NE,\cdot)$ and extending on the right to an exact sequence.

To be more explicit, let $q:TS|_E\to NE$ denote the quotient map, and consider the following terms of the long exact sequence $Hom(NE,TS|_E) \to Hom(NE,NE) \to Ext^1(NE,TE)$. The first map takes $f:NE\to TS|_E$ and composes it with $q$ to get a map from $NE$ to itself, and the second is the coboundary map $\delta :Hom(NE,NE)\to Ext^1(NE,TE)$. By definition, $f$ is a splitting of the short exact sequence above if and only if $qf$ is the identity map on $NE$. By exactness, such an $f$ exists if and only if $\delta(id_N)=0$, so this class is the obstruction; it may or may not be easy to compute it in any given example. (in the last equation, "N" means "NE"... I couldn't get it to typeset properly)

Also, although I'm not an expert, it's my understanding that there's a construction known as "deformation to the normal cone", which allows one to get around the failure of the holomorphic tubular neighborhood theorem in my situations. In particular, I've heard it's often useful in intersection theory, so it may have some bearing on the problem you describe in your question.

Answer 2

This answer is rather late, but one version of a holomorphic tubular neighbourhood theorem is proved in: "H. Grauert, Uber Modifikationen und exzeptionelle analytische Mengen, Math. Ann. 146 (1962), 331–368". Specifically, on page 363, the corollary to theorem 7 states that if $A\subset X$ is a compact complex submanifold for which $H^1(A,TA\otimes (N^\ast)^\nu)=H^1(A,(N^\ast)^\nu)=0$ for all $\nu\geq1$, then a neighbourhood of $A$ in $X$ is biholomorphic to a neighbourhood of $A$ in the total space of its normal bundle.

Answer 3

To start with something positive, it is indeed true that whenever you have a $\mathbb CP^1$ with negative self-intersection in a complex surface it has a standard holomorphic neighbourhood.

On the other hand the comparison in 2) between symplectic and complex geometry is not correct unless you consider $2$-spheres with negative self-intersection in symplectic $4$-manifolds.

Namely in symplectic geometry whenever we have a symplectic submanifold $N^{2k}$ in a symplectic manifold $M^{2n}$ a symplectic neighbourhood of $N^{2k}$ is always standard. I.e. such a neighbourhood depends only on type of almost complex structure on the normal bundle to $N^{2k}$. On the other hand in complex geometry the existence of a standard neighbourhood is extremely rare especially if the normal bundle of $N^{2k}$ in $M^{2n}$ is not negative.

Example. Consider a smooth quadric $Q$ in $\mathbb CP^n$ and let us show that its neighbourhood is not byholomorphic to a neighbourhood of the zero section of the $O(4)$ bundle over $Q$. Indeed if this were the case, one would be able to realise the normal bundle to $Q$ in $\mathbb CP^n$ as holomorphic sub-bunlde $N$ of $T\mathbb CP^n$ restricted on $Q$. But the later is impossible. Indeed, in this where possible through each point of $Q$ one would be able to draw a line that would be tangent to this normal sub-bundle $N$. Such a family of lines would generate an involution of $Q$ without fixed point, which is absurd.

Answer 4

Tubular neighborhoods can fail to exist in a strong sense: it can happen that, given a complete smooth subvariety $X$ of a smooth complex variety $V$, there is no neighborhood (in the classical topology) of $X$ in $V$ that possesses a holomorphic retraction onto $X$. For example, take $V$ to be the projective plane and $X$ to be a smooth curve of degree at least $3$; in any neighborhood $U$ of $X$ in $V$ there are perturbations $X'$ of $X$ that are not isomorphic to $X$ (in the case of degree $3$ they will have unequal $j$-invariant), while any retraction of $U$ onto $X$ would give a map $X'\to X$ of degree $1$, so an isomorphism.

Answer 5

The existence of tubular neighborhoods is a remarkably strong constraint on submanifolds of projective space. A theorem of Morrow and Rossi (Math. Ann., 1978, if I can find a free link I'll attach it) says the following: Suppose $X$ is a connected holomorphic submanifold of ${\Bbb P}^n$ that has a holomorphic tubular neighborhood. Then $X$ is a linear subspace. (They also prove a similar statement for submanifolds of complex tori.)

Answer 6

Perhaps it is worth saying that while in general a complex submanifold of a complex manifold does not admit a holomorphic tubular neighborhood, it does admit a special smooth tubular neighborhood; the leaves that get contracted are complex manifolds, but they do not vary holomorphically.