0
$\begingroup$

The usual convex set is the real linear convex set, if we change the real linear map into complex linear map, we can get the complex convex set. A system way to do this is in the several complex analysis, at wiki here: Holomorphically convex hull, changing the holomorphic functions into complex linear functions.

Now my questions is what is the complex convex set looks like.

First, the complex convex set must be convex set, but does every convex set must be complex convex?

If not, at least in the one complex dimension case, the complex convex is complex polynormally convex, if it is a compact, its complement must be connected. Can you say something more?

The same question is at here.

Improve this question
$\endgroup$
2
  • $\begingroup$ Are you defining a complex-convex set in a finite-dimensional complex vector space as a set which admits a support complex hyperplane at every boundary point? By "support", I mean here that the complex hyperplanes intersects the convex set, but does not intersect its interior. $\endgroup$
    – alvarezpaiva
    Commented Apr 18, 2013 at 12:02
  • $\begingroup$ It is also a possible definition, I think it is equivalence to the definition here, defining the complex convex hull first, then it is complex convex iff its hull is itself. $\endgroup$
    – Strongart
    Commented Apr 19, 2013 at 11:33

1 Answer 1

Reset to default
2
$\begingroup$

It is hard to understand the formulation of your post, but let me attempt an answer anyway.

I will use the notion of a domain of holomorphy: a domain $\Omega \subset \mathbb{C}^n$, $n \geq 1$, is a domain of holomorphy if there exists a function $f$ holomorphic in $\Omega$ which does not extend holomorphically to any larger domain. Cartan-Thullen theorem says that for a domain in $\mathbb{C}^n$, $n \geq 1$, being a domain of holomorphy is equivalent to being holomorphically convex. It can be proved (using Hahn-Banach theorem) that every convex domain is a domain of holomorphy, so it is holomorphically convex. This is really interesting when $n \geq 2$, because when $n=1$ any domain is a domain of holomorphy. A visualization of a strange-looking domain of holomorphy with $n >1$, aptly called a worm, first described in the paper Diederich, Klas; Fornaess, John Erik A strange bounded smooth domain of holomorphy. Bull. Amer. Math. Soc. 82 (1976), no. 1, 74–76, can be seen here (in a note by Harold Boas)

http://www.ams.org/notices/200305/what-is.pdf

Also, when $n \geq 2$, polynomial convexity is no longer equivalent to simple connectedness. Counterexamples (going both ways) can be found in the book MR1818167 Nishino, Toshio Function theory in several complex variables. (English summary) Translated from the 1996 Japanese original by Norman Levenberg and Hiroshi Yamaguchi. Translations of Mathematical Monographs, 193. American Mathematical Society, Providence, RI, 2001. xiv+366 pp. ISBN: 0-8218-0816-8

Edit: To those with editing power: It took me a longer while to realize that OP is (probably) asking what happens when in the definition of a hull the class of holomorphic functions is replaced by the class of $\mathbb{C}$-linear function, not quite the question I have just answered. It would benefit from editing.

Improve this answer
$\endgroup$
1
  • $\begingroup$ A good background for my question, thank you all the same. $\endgroup$
    – Strongart
    Commented Apr 19, 2013 at 14:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .