Extending the tangent bundle of a submanifold

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.

• 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. Sep 5 '16 at 21:18
• 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 $z_n$. Sep 6 '16 at 9:23
• @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. Sep 6 '16 at 17:29
• @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. Sep 7 '16 at 6:55
• @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. 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 $U$.