Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am stuck on this problem. Let $E$ be a holomorphic vector bundle on a complex manifold $M$, and $N$ a complex submanifold of codimension at least two. Prove that every section of $E$ on $M\backslash N$ extends to a section of $E$ over $M$. If $F$ is another holomorphic vector bundle over $M$ that is isomorphic to $E$ over $M\backslash N$, then prove that $F$ is isomorphic to $E$ over $M$.

I appreciate any suggestion.

Thank you.

share|cite|improve this question

The extension
As Kevin very aptly said, you should use the following amazing theorem of Hartogs:
Every holomorphic function on $M\setminus N$ extends to a holomorphic function defined on all of $M$.
In other word the restriction morphism of rings $\mathcal O(M) \to \mathcal O(M \setminus N)$ is surjective and thus actually an isomorphism .
If now $s\in \Gamma (M\setminus N,E)$ is a section and $n\in N$ is a point where it is not defined, take a trivializing neighbourhood $U$ of $n$ in $M$ and consider a trivializing isomorphism of vector bundles $\phi:E|U\to U\times \mathbb C^r$. Then $\phi \circ s$ is a section on $U\setminus N$ of the trivial bundle $U\times \mathbb C^r$, hence must be of the form $U\setminus N\to U \times \mathbb C^r: u\mapsto (u, (s_1(u),s_2(u),...,s_r(u)))$ with $s_i\in \mathcal O(U\setminus N)$.
Now each of the $s_i$'s is just a holomorphic function to which Hartogs's theorem applies.
So it can be extended to $\hat s_i\in \mathcal O(U)$, this yields a section $(\hat s_1,\hat s_2,...,\hat s_r)$ defined on the whole of U of the trivial bundle on U and by applying $\phi^{-1}$ you get a section of $E|U$.
Since this can be done at every $n\in N$, you get the desired extension of $s$.

The isomorphism
You are given an isomorphism of vector bundles $u:E|(M\setminus N) \to F|(M\setminus N)$.
To reduce to the above, the trick is to consider the holomorphic vector bundle $HOM(E,F)$ (note the capital letters in $HOM$ !), whose fiber at $m\in M$ is $Hom(E_m,F_m)=\mathcal L (E_m,F_m)$, the linear maps between two good old vector spaces.
You can then consider that $u\in \Gamma(M\setminus N, HOM(E,F))$ and extend that section holomorphically to $\Gamma(M, HOM(E,F))$, obtaining a morphism defined on all of $M$ of our bundles $\hat u:E\to F$.
End of story? No! You must show that $\hat u$ is an ISOmorphism also on$N$.
Here the trick is to look at the induced morphism on maximal exterior powers $\Lambda ^r \hat u:\Lambda ^r E\to \Lambda ^r F$. It can be considered as a section of a line bundle , namely $s= {\Lambda ^r \hat u} \in \Gamma(M,(\Lambda ^r E)^*\otimes \Lambda ^r F)$.
Now the points where $\hat u$ is not an isomorphism correspond to the points where the section $s$ is zero: this is of codimension one or empty.
But since we know that $u$ is an isomorphism outside of $N$, that zero-set must be a subset of $ N$, hence not of codimension one, hence empty.
So $\hat u:E\to F$ is indeed an isomorphism (everywhere!) of the given bundles.
(All of this second section can also be done by more explicit local calculations which translate the above)

Hartogs's theorem is false in the real analytic case: the analytic function $\frac {1}{x^2+y^2}$ on $\mathbb R^2 \setminus \lbrace (0,0) \rbrace$ cannot be extended continuously, let alone analytically, across $(0,0)$.

share|cite|improve this answer

Look up Hartogs's theorem (probably the book of Griffiths-Harris is a good place to look); use the fact that vector bundles are locally trivial.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.