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How to prove that $\mathbb {C^2\setminus R^2}$ is not a domain of holomorphy while $\{(z_1,z_2)\in \mathbb C^2;z_2\not=0\}$ is a domain of holomorphy?

My initial proof: If $f\in \mathcal O(\mathbb {C^2\setminus R^2})$. Then W.L.O.G we can consider the "shift" function $f(z_1+i,z_2+i)$ and hence this new $f$ is holomorphic in the polydisc $\mathbb D^2$??

The Hartogs figure $H=\{(z_1+i,z_2+i); |z_1+i|>1/2\}\cup\{(z_1+i,z_2+i); |z_2+i|<1/2\}\subset\mathbb C^2\setminus\mathbb R^2$. Thus by Hartogs extension theorem the $f$ can be extended to $\mathbb D^2$??

How would we also conclude that $f$ is everywhere holomorphic?

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  • $\begingroup$ I suppose it is really $\mathbb{R}^2=\{(z,w)\in\mathbb{C}^2\mid z,w\in \mathbb{R}\}$, note that the domain is $\mathbb{C}^\color{red}{2}$. $\endgroup$
    – b00n heT
    Aug 3 '16 at 20:51
  • $\begingroup$ Is $\mathbb{C^2\setminus R^2}$ connected? $\endgroup$
    – Ronald
    Aug 3 '16 at 21:16
  • $\begingroup$ it is path connected, so yes, in particular connected. $\endgroup$
    – b00n heT
    Aug 3 '16 at 21:27
  • $\begingroup$ ok i will write my attempt so if you like you can write your comments about it $\endgroup$
    – Ronald
    Aug 3 '16 at 21:35
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In the second example the function $(z_1,z_2)\mapsto 1/z_2$ is a holomorphic on the given domain which can not be extended to a strictly bigger domain.

The first case is non-trivial. In Steven Krantz (Several complex variables) you may find a theorem (p 160 in my version) about tube domains. Let me multiply your domain by $i$ (a biholomorphic map) to fit with Krantz' notation. Let $T_\omega={\Bbb C}^2 \setminus (i {\Bbb R}) \times (i {\Bbb R})= \{ z\in {\Bbb C}^2 : \mbox{Re } z \in \omega \}$ with $ \omega = {\Bbb R}^2\setminus \{(0,0)\} . $ The theorem states that $T_\omega$ is a domain of holomorphy if and only if $\omega$ is (geometrically) convex which is clearly not the case here. I don't know of an easy proof but that might exist.

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  • $\begingroup$ I have written an attempt solution $\endgroup$
    – Ronald
    Aug 3 '16 at 22:09
  • $\begingroup$ When you translate the variables you should also translate the domain. So the argument is at least not that simple. $\endgroup$
    – H. H. Rugh
    Aug 3 '16 at 23:00
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The first case is not as difficult as it seems: First notice that the missing points are all in the closure of the domain, so if you prove that the function extends continuously to the closure, and if this extension which is then unique, is holomorphic, you are done.

The trick is to use a different transformation than a linear one. By a local biholomorphic transformation you can change the domain near the origin to ${\mathbb C}^2 \setminus \{ (z,w) | \Im w = (\Re z)^2, \Im z = 0 \}$ (exercise is to find this local biholomorphism, it is a quadratic polynomial). Then for small $|w|$ and $|z| < 1$ write $$ F(z,w) = \frac{1}{2\pi i} \int_{|z| = 1} \frac{f(\xi,w)}{\xi-z} d \xi $$ If $\Im w < 0$, then clearly $F(z,w) = f(z,w)$ by Cauchy. $F(z,w)$ is holomorphic simply by differentiating under the integral. So $F$ and $f$ are equal for all small enough $w$ and $|z| < 1$ where $f$ is defined. In particular $f$ extends near the origin (it extends continuously to the missing points and this extension is holomorphic).

You can also "fit" a Hartogs figure in there, but the proof is just using Cauchy, so I usually find it easier to just look for those "discs" used by Cauchy for the extension. Instead of "bending space" as we did above, you could also "bend" the Hartogs figure, but for this you must bend something, a linear version does not do the job here.

As an exercise try to use this idea to also prove that every holomorphic function on ${\mathbb C}^2 \setminus \{ (z,w) | \Im w \geq 0, \Im z = 0 \}$ extends to all of ${\mathbb C}^2$.

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I'm not sure if this works, but this is how I would have approached the problem.

Let $S_1^{n}$ and $S_2^{1/n}$ be two smooth $(2n-1)$-dimensional hypersurfaces which respectively bound domains $G_1^n \supset \mathbb{R}^2$and $G_2^{1/n}$. The purpose of having the superscript $n$ is to indicate that as we let $n \to 0$, $G_1^n$ shrinks to $\mathbb{R}^2$ and $G_2^{1/n}$ extends to $\mathbb{C}^2$. Now if $f \in \mathscr{O}(\mathbb{C}^2 \backslash \mathbb{R}^2)$, then $f \in \mathscr{O}(G)$, where $G: = G_2 \backslash G_1$. Therefore, for $z \in G$, we may expand $f$ into a Martinelli-Bochner integral $$f(z) = \int_{S_2^{1/n}} f(\zeta) \omega_{\text{MB}}(\zeta -z) - \int_{S_1^{n}} f(\zeta) \omega_{\text{MB}}(\zeta -z).$$ Using the fact that $z \not \in G$, the second integral above vanishes (this is a particular property of the Martinelli-Bochner integral), and so we are left with $$f(z) = \int_{S_2^{1/n}} f(\zeta) \omega_{\text{MB}}(\zeta -z).$$ The right hand side of the above equation represents a function which is holomorphic on all of $S_2^{1/n}$ and so, by taking uniform limits, we see that $f$ extends to a function which is holomorphic on all of $\mathbb{C}^2$.

Note that this technique can be used to prove the Hartog's extension theorem.

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