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Let $\Omega\subset\mathbb{C}^n$ be a domain of holomorphy, $m<n$. Is $\Omega\cap\mathbb{C}^m$ necessarily a domain of holomorphy?

If true, it seems that it will be useful in inductive arguments. However, I could neither prove nor disprove this. It is true in some nice cases, e.g., if $\Omega$ is Reinhardt, but I have no idea in general.

Thanks in advance!

Yuxiao Xie
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1 Answers1

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It need not be a domain to begin with, that is, the intersection could be disconnected. Furthermore, even if the original has smooth boundary, the intersection may not. But other than that, using the Hartogs-pseudoconvex definition (there exists a continuous plurisubharmonic exhaustion function), the intersection is Hartogs-pseudoconvex if the original was. (That a restriction of a plurisubharmonic function to a subspace is plurisubharmonic is obvious from the definition).

Jiri Lebl
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  • So every connected component of the intersection is a domain of holomorphy? – Yuxiao Xie Aug 22 '20 at 02:45
  • Yeah, but note that it could have weird behavior anyway. Many results in several complex variables are for domains with some sort of regularity on the boundary, and by intersection with a subspace, you are going to lose any regularity. You might get really weird boundaries in fact. – Jiri Lebl Aug 23 '20 at 03:07