Analytic extension across the boundary. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T07:35:54Z http://mathoverflow.net/feeds/question/97687 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97687/analytic-extension-across-the-boundary Analytic extension across the boundary. Pradip Mishra 2012-05-22T17:31:02Z 2012-05-23T04:09:41Z <p>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.</p> <p>Motivation: We have following proposition:</p> <p>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</p> <p>(a). $x\notin \partial U \Leftrightarrow f(x)\notin \partial V$</p> <p>(b). $f|Int(U)$, and $f|\partial U$ are diffeomorphism.</p> <p>This proposition gives: If $f:Q\to Q$ is diffeomorphism and holomorphic in the interior. Then either </p> <p>1- $f$ maps Y-axis to Y-axis and X-axis to X-axis origin goes to origin. OR</p> <p>2-$f$ maps X-axis to Y-axis and Y-axis to X-axis origin goes to origin.</p> <p>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 $f:Q\times Q\to Q\times Q$. </p> http://mathoverflow.net/questions/97687/analytic-extension-across-the-boundary/97717#97717 Answer by Misha for Analytic extension across the boundary. Misha 2012-05-23T04:09:41Z 2012-05-23T04:09:41Z <p>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. </p>