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I have the following question:

I know that every holomorphic function $f$ defined on a closed complex submanifold $M$ of the space $\mathbb C^d$ can be extended to a holomorphic function on the total space $\mathbb C^d$.

Now, what about functions which are only defined on an open subset $U$ of $M$? Is it possible to extend them to an open subset of $\mathbb C^d$ containing $U$ ?

I know the corresponding statements are true for real smooth functions since we may une smooth partitions of unity but for holomorphic functions this obviously does not work...

I would be very thankful for any ideas on this subject.

Thanks in advance, Tom

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  • $\begingroup$ You can surely do that if $M$ is nonsingular. Write $M$ locally as the range of a biholomorphic map $h : V\subset(\mathbb C^n\times\mathbb C^{d-n}) \to\mathbb C^d$ for some $h$ holomorphic in such a way that $h(X,0)\subset U$. Define simply $\hat f(p):=f(h(X,0))$ for all $p=h(X,Y)$. $\endgroup$
    – Loïc Teyssier
    Commented Feb 2, 2015 at 13:55
  • $\begingroup$ Thanks, but as far as I understood it, this just gives you a local extension. But if I want this extension to be defined on a big set containing my set U, I still have to glue the local extensions together? Or am I missing something? $\endgroup$
    – Tom
    Commented Feb 2, 2015 at 13:58
  • $\begingroup$ How can you extend the holomorphic function $1/z$ defined on $U=\mathbb{C}-{0}$ to the whole of $\mathbb{C}$? $\endgroup$
    – Francesco Polizzi
    Commented Feb 2, 2015 at 16:17
  • $\begingroup$ Maybe my question was too unclear. I have an open subset of the submanifold and I want to extend the function defined on that subset to an open subset of the whole space. I do not need to extend it to the whole space. In that way, I do not have to extend 1/z to a bigger set since it is already defined on an open subset of the whole space. $\endgroup$
    – Tom
    Commented Feb 2, 2015 at 16:47
  • $\begingroup$ @Tom: The glueing together should not be a problemas. Actually you can cover the whole $U$ by the range of a single analytic function $h$ as in this question. $\endgroup$
    – Loïc Teyssier
    Commented Feb 2, 2015 at 19:04

1 Answer 1

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The construction as suggested by Loïc Teyssier can be globalized, with the help of a Stein neighborhood basis. By Corollary 1 in ''Every Stein subvariety admits a Stein neighborhood'' by Siu, every Stein manifold $M$, which is a submanifold of some manifold $N$ has a neighborhood $W$ in $N$ such that there exist a holomorphic retraction $r : W \to M$. In particular, when $N$ is Stein (as is the case here, with $N = \mathbb{C}^d$), then also $M$ is Stein, and the theorem applies.

Thus, if $U \subseteq M$ is open, $f : U \to \mathbb{C}$ is holomorphic, and one lets $V$ be the open set $V := r^{-1}(U) \subseteq W \subseteq \mathbb{C}^d$, then $F = f \circ r : V \to \mathbb{C}$ is an extension of $f$.

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  • $\begingroup$ Wonderful, this was exactly what I was hoping for. Thank you! $\endgroup$
    – Tom
    Commented Feb 3, 2015 at 8:02
  • $\begingroup$ The existence of a holomorphic retraction in the situation considered by OP is a Theorem of Docquier and Grauert (1960) Math Ann vol 140 see for example the book of Gunning and Rossi page 257 $\endgroup$
    – Mohan Ramachandran
    Commented Feb 4, 2015 at 16:14
  • $\begingroup$ Thank you for this clarification of the history, which I was not aware of. $\endgroup$
    – Richard Lärkäng
    Commented Feb 5, 2015 at 16:23

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