# Extending holomorphic functions

Suppose $K\subset \mathbb{C}^n$ is a compact subset and $f:\mathbb{C}^n\setminus K\to \mathbb{C}$ is a holomorphic function. Then, provided $n>1$, $f$ extends to a holomorphic function defined on the whole $\mathbb{C}^n$. This is the Hartogs' extension theorem, and a proof can be found e.g. somewhere in the very beginning of Griffiths-Harris.

This theorem is still true if one replaces $\mathbb{C}^n$ by a connected open subset $U$ of $\mathbb{C}^n$ (upd: and takes $K$ to be a compact subset of $U$ that does not disconnect it, or else there may be a problem even with a locally constant function), see e.g. theorem 2 in http://www.encyclopediaofmath.org/index.php/Hartogs_theorem

I would like to ask if this theorem can be generalized to other complex manifolds (e.g., to Stein manifolds). If there is an answer in the algebraic case, that would be particularly interesting.

• You left out an important hypothesis: $K$ does not disconnect the domain (${\mathbb C}^n$ or the connected open subset of it) Jun 16, 2014 at 7:27
• Dear Robert -- yes, indeed, thanks! Jun 16, 2014 at 13:25
• Connectedness of the complement of $K$ is necessary only if one wants uniqueness of the extended function! Jun 17, 2014 at 14:13
• diverietti -- yes and no: if say a function equals 1 in the interior of the unit ball and 0 in the exterior, then it does not extend to the whole space, although its restrictions to the connected components of its domain do. Jun 17, 2014 at 15:46

Theorem. Let $X$ be a connected, normal Stein space of dimension $\geq 2$. Let $K$ be a compact subset of $X$ and let $h$ be a holomorphic function on $X \setminus K$. Then there is a unique holomorphic function $H$ on $X$ such that $H=h$ on the unbounded component of $X \setminus K$.