Let X be a compact Kähler 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 $U(M)\subseteq X$?
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
13
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!
answered Sep 23, 2011 at 21:33
-
1$\begingroup$ Dmitri, you are probably thinking on Van de Ven Theorem. See mathoverflow.net/questions/61596/… $\endgroup$ Commented Sep 25, 2011 at 0:51
-
1$\begingroup$ Yes Georges, this is correct, one can construct this splitting explicitly. Namely take two points on the conic -- $p$ and $q$ and for any point $x\in C$ chose two directions in $T_x\mathbb CP^2$: the directions of the lines $[xp]$ and $[xq]$. It is easy to see that this defines two transversal line sub-bundles. $\endgroup$ Commented Sep 25, 2011 at 12:49
-
1$\begingroup$ Dear Dmitri, this visual proof of the splitting is really heart-warming to a geometer. And the symmetry of your construction also shows that the two sub-bundles are isomorphic, hence isomorphic to $\mathcal O(3)$ since the degree of the restricted tangent bundle is $6$. Прекрасно, поздравляю! $\endgroup$ Commented Sep 25, 2011 at 16:57
-
1$\begingroup$ Dear YangMills, the case of $(-1,-1)$ curve is the "good" one indeed. But one needs some work to prove this. I just started to think how to do this. It is standard to say that if you blow down $(-1,-1)$ curve then you get an ordinary double point, i.e. singularity $(x_1^2+...+x_4^2)=0$. Of course, from this the existence of bundle follows. But how to prove that this is so? I never saw the proof, so I decided to go to arxiv.org/PS_cache/alg-geom/pdf/9602/9602006v1.pdf . There Miles Reid speaks (in chapter 4) about contracting curves on surfaces. For example a $\mathbb CP^1$ with normal $\endgroup$ Commented Sep 30, 2011 at 5:58
-
1$\begingroup$ Dear Dmitri, thanks a lot for your answer, which was very helpful. I found an explanation for why $M$ can be contracted to an ordinary double point in the paper of Friedman "On threefolds with trivial canonical bundle": basically he blows up $X$ along $M$ and on the blowup the exceptional divisor is $P^1\times P^1$ with normal bundle $O(-1,-1)$ so this can be contracted to an ordinary double point, and the resulting variety is also the contraction of $M$ inside $X$. $\endgroup$ Commented Sep 30, 2011 at 15:13
$\begingroup$
$\endgroup$
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
- $H^0 (M, TM) = 0$
- $\dim H^0(M, [X]) \geq 2$, and
- $H^1(M, [X]^{\vee} \otimes TM) = 0$,
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.
answered Sep 23, 2011 at 23:31