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How can we define a smooth domain in $\mathbb{C}^2$ and a smooth boundary of bounded domain in $\mathbb{C}^2$?

{where $\mathbb{C}^2$ := Cartesian product of complex plane }

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The same as you define any other smooth manifolds, I'd say. In particular, it depends on whether you want them to be complex-smooth or real-smooth. –  tomasz Dec 26 '12 at 12:21
    
Well, to expand a little, a domain is usually understood in analysis to be open and connected, so the adjective "smooth" seems rather superfluous in this case, unless you meant just a smooth manifold. –  tomasz Dec 26 '12 at 12:23
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1 Answer

Every open subset of $\mathbb{C}^2$ will be smooth as a manifold, so the term "smooth domain" is redundant, unless you have a different notion of smoothness in mind.

The boundary $\partial U$ of a bounded domain $U\subset\mathbb{C}^2$ will be smooth as a manifold if and only if for every point $p\in \partial U$ there exists smooth coordinates $\varphi = (x_1,x_2,x_3,x_4)\colon V\to \mathbb{R}^4$, defined on some neighborhood $V$ of $p$ in $\mathbb{C}^2$, such that $V\cap \partial U = \{x_4 = 0\}$. To put in simpler terms, $\partial U$ is smooth if, locally in some smooth coordinates, it looks like the inclusion $\mathbb{R}^3\subset \mathbb{R}^4$.

There are other, sometimes preferable, ways of saying that $\partial U$ is smooth. I think $\partial U$ is smooth if and only if there is a smooth function $\rho\colon \mathbb{C}^2\to \mathbb{R}$ such that

  1. the differential $d\rho$ does not vanish on $\partial U$.
  2. $\rho(z)<0$ if and only if $z\in U$
  3. $\rho(z) = 0$ if and only if $z\in \partial U$.

Such a $\rho$ is sometimes called a defining function for $U$.

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