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I have seen the following two definitions of a domain of holomorphy; I was wondering if they are actually equivalent:

1) A domain $\Omega$ is a domain of holomorphy if there exists a holomorphic function on $\Omega$ that does not extend to a larger domain $\Omega' \supset \Omega$;

2) A domain $\Omega$ is a domain of holomorphy if for every domain $\Omega'$ which intersects the boundary of $\Omega$, and for every connected component $U$ of $\Omega \cap \Omega'$, there exists a holomorphic function $\varphi$ on $\Omega$ such that $\varphi|_U$ does not extend to $\Omega'$.

I think that at least the second condition implies the first; but does the first imply the second?

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The first is not a good definition. Define $\Omega \subset {\mathbb C}^2$ in the following way: $$ \Omega = \{ (z,w) \mid z e^{i|w|^2} \text{is not a nonnegative real number} \} $$ For each fixed $w$, the set is simply a branch cut of the plane. It is not difficult to show that for the complex line given by $z = -r$ for a negative real number $-r$ hits the boundary of $\Omega$ at a single point. That is the boundary of omega except for the set $z=0$ is actually a strictly pseudoconvex hypersurface. But it is $\Omega$ on both sides of it.

Take any $f$ holomorphic on $\Omega$. You can see that according to the second (correct) definition, $\Omega$ is not a domain of holomorphy. The $f$ will extend from the pseudoconcave side. You should also note how the "every connected component" plays into this definition.

On the other hand, there clearly is not a large domain $\Omega'$ for the first definition. You can just take a branch of $\log z$ so that it is holomorphic in $\Omega$. That will be discontinuous at the branch cuts and so cannot extend.

The issue is connected to the idea of there not being a "largest domain of holomorphy containing a domain", just like there isn't a largest domain onto which a specific holomorphic function extends. In one dimension, this leads to branched Riemann surfaces. In several dimensions you also get a branched domain if you attempt to extend analytic functions . These branched domains are called the "envelopes of holomorphy".

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  • $\begingroup$ Why does the complex line $z = -r$ intersecting the boundary of $\Omega$ at a single point imply that $\partial\Omega$ is a strictly pseudoconvex hypersurface? $\endgroup$
    – Kurosu
    Jul 1 at 16:16
  • $\begingroup$ Well to be perfectly honest the line only shows that you can fit a hartogs figure to extend every function across the boundary from the pseudoconcave side. Or perhaps use Kontinuitatsaatz (if I spelled that correctly). That is not quite strict pseudoconvexity from the Levi-point of view, but the domain is strictly pseudoconvex in the Levi sense at those points. To see that change coordinates to make the complex line one of the coordinate lines. $\endgroup$
    – Jiri Lebl
    Jul 2 at 18:35

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