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Let $Q=[0,\infty)\times [0,\infty)\subset \mathbb C$ and $f: Q\times Q\to Q\times Q$ be a diffeomorphism. such that $f$ is holomorphic in the interior of $Q\times Q$. Can we extend this map analytically across the boundary.

Motivation: We have following proposition:

Let $U$ and $V$ are open subsets of $\mathbb R^n_k=[0,\infty)^k\times \mathbb R^{n-k}$ and $f:U\to V$ be diffeomorphism, then

(a). $x\notin \partial U \Leftrightarrow f(x)\notin \partial V$

(b). $f|Int(U)$, and $f|\partial U$ are diffeomorphism.

This proposition gives: If $f:Q\to Q$ is diffeomorphism and holomorphic in the interior. Then either

1- $f$ maps Y-axis to Y-axis and X-axis to X-axis origin goes to origin. OR

2-$f$ maps X-axis to Y-axis and Y-axis to X-axis origin goes to origin.

And using Schwarz reflection principle we have extension in both case. So for $f: Q\to Q$ we have extension across the boundary. I have doubt for $f:Q\times Q\to Q\times Q$.

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Not only $f$ admits an analytic continuation across boundary, in fact, $f$ is the restriction of a linear transformation. Indeed, the interior of $Q\times Q$ is the 2-dimensional polydisk (more precisely, it is biholomorphic to the standard polydisk by a product map). Biholomorhic automorphisms of polydisks are compositions of permutations of components and products of conformal automorphisms of factors (first proven by Poincare and could be found in any textbook on several CVs). Therefore, by composing $f$ with permutation of coordinates if necessary, $f$ is the product map $(f_1, f_2)$, where $f_k: Q\to Q, k=1,2$. Since you are assuming that $f$ is smooth on the boundary, each $f_k$ fixes points $0, \infty$ on the boundary of $Q$. Hence, each $f_k$ is a dilation and, thus, $f$ is linear.

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  • $\begingroup$ This argument is valid for map(diffeomorphism and in interior holomorphic) $f: Q\times H\to Q\times H$ where $H$ is upper half plane... Nothing special about $Q$.. am i right? $\endgroup$ – zapkm May 23 '12 at 16:12
  • $\begingroup$ @Pradip: Yes, in the case of $Q\times H$ the conclusion will be that the map $f$ is affine (same proof). $\endgroup$ – Misha May 23 '12 at 17:56

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