Polydisc is not biholomorphic to any strictly pseudoconvex domain

Question

I want to prove the poly disc $P=\left\{z\in \mathbb{C}^2 : |z_1|<1,|z_2|<1\right\}$ is not biholomorphic to any strictly pseudo convex domain in $\mathbb{C}^2.$ Can any one provide a hint?

Answer 1

This is not completely trivial, and some details are missing from the sketch below, but essentially the argument goes as follows.

Assume that $F : P\to D$ is biholomorphic, where $D\subset {\Bbb C}^2$ is strictly pseudoconvex, and consider what $F$ does to discs $\{z_1\}\times \Delta$ as $|z_1|\to 1$. Use a subsequence argument to show that you can find sequences $$F(z_1^{(k)},\cdot) :\Delta\to D$$ with $|z_1^{(k)}|\to 1$ which converge to some holomorphic $\phi : \Delta \to \overline D$. Argue that $\phi$ must map $\Delta$ into the boundary $\partial D$, and use the strict pseudoconvexity to show that $\phi$ must be constant. For fixed $z_2$, since $${\partial F\over\partial z_2}(z_1^{(k)},z_2)\to \phi'(z_2)=(0,0)$$ as $k\to\infty$, the map ${\partial F\over \partial z_2}(\cdot,z_2) : \Delta\to{\Bbb C}^2$ can be extended continuously to $\overline\Delta$ with boundary values zero. It follows that ${\partial F\over \partial z_2}(\cdot,z_2)\equiv 0$ on $\Delta$, which means that $F(z_1,z_2)$ is independent of $z_2$. This contradicts the injectivity of $F$.