Let $X$ be a complex manifold, and $Y\subset X$ a compact submanifold. Is it true that the tangent bundle $TY$ may be extended (as a holomorphic vector bundle) to some open neighbourhood of $Y$ in $X$?

P.S. This looks like a very simple question, but somehow I am stuck.

  • $\begingroup$ Around each point of $Y$ there is a chart i.e. local holomorphic coordinates of $X$ say $z_1,....,z_n$ such that $Y$ is locally defined as $z_m =... =z_n= 0$. The union of the domains of such charts is an open neihbourhood of $Y$ in $X$. It seems to me that the distribution given by $dz_m = ... = dz_n = 0$ extends $TY$ as complex vector bundle on the above mentioned open neighbourhood. $\endgroup$
    – Holonomia
    Sep 5 '16 at 21:18
  • $\begingroup$ Holonomia, this method does not work. If you substitute equation $z_n=0$ with $z^{new}_n=g(z)z_n=0$, where $g(z)$ is a non-vanishing holomorphic function in the local chart, then distribution given $dz_m=\dots=dz_{n-1}=dz^{new}_n=0$ by will be different outside of the zero locus of $z_n$. $\endgroup$ Sep 6 '16 at 9:23
  • $\begingroup$ @Yuri Ustinovskiy: you are right, I was wrong. Perhaps it better I delete my comment so it do not bother anyone interested in the OP question. $\endgroup$
    – Holonomia
    Sep 6 '16 at 17:29
  • $\begingroup$ @Alex Gavrilov: Did you try with $Y$ an exceptional divisor ? e.g. $X$ is the blow up $P^2$ in one point and $Y=P^1$ is the exceptional divisor. Maybe it is obvious but I have no time to check it. $\endgroup$
    – Holonomia
    Sep 7 '16 at 6:55
  • $\begingroup$ @Holonomia: Yes, it is quite obvious. In this case, $TY$ is a linear bundle which extends to the whole $X$. (Not just to a neighbourhood.) When the exceptional divisor has higher dimension it may be more complicated, but I suspect that this is not a counterexample either. $\endgroup$ Sep 7 '16 at 11:49

Let us discuss first the differntial geometry case. Here, you can use the tubular neighbourhood theorem to extend the vector bundle. Explicitly, you have a neighbourhood $U$ of $Y$ in $X$ which is isomorphic to a neighbourhood of the zero section of the normal bundle. So you have a projection $$ \pi\colon U \to Y $$ and use it to pull-back $TY$ to $U$.

In complex geometry, the tubular neighbourhood theorem holds just locally, because you do not have partition of unity. However, under some apporpiate condition, this theorem holds, so you can perform the same construction. This is a classical result of Grauert. See for instance the survey


and references therein.


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