Is complex analytic extension of real-analytic diffeomorphism a diffeomorphism ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T10:47:47Z http://mathoverflow.net/feeds/question/69510 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69510/is-complex-analytic-extension-of-real-analytic-diffeomorphism-a-diffeomorphism Is complex analytic extension of real-analytic diffeomorphism a diffeomorphism ? Analysis Now 2011-07-05T03:26:09Z 2011-07-06T05:22:59Z <p>Hi, my question is :</p> <p>Let $D$ be the open unit disk in $\mathbb{C}=R^2$ and $f:D\to D$ be a real-analytic diffeomorphism. Let us think of the canonical embedding : $\mathbb{C}=R^2\subset \mathbb{C^2}.$ Does there exist a complex analytic diffeomorphism $F$ ( analytic in two complex variables ) whose domain is either $D^2\subset \mathbb{C^2}$ or the complex 2-dimensional unit open polydisk $\Omega ={{(z,w): |z|^2+|w|^2 &lt; 1}}$ in $\mathbb{C^2}$ such that its restriction to $D\subset D^2$ or $D\subset \Omega$ is $f$ ? By restriction , I mean $F(z,0) = f(z)$ in the case of $D \subset D^2 \subset \mathbb {C}^2$ .</p> <p>The range of $F$ does not necessarily have to be $D^2$ or $\Omega$, but it would be even better if they are !</p> <p>If this is a very well-known result, you can cite a reference. </p> <p>Is the same result true in 1-dimension as well , i.e. replacing $D$ by $I\subset R$ and changing the complex-analytic/conformal diffeomorphism $F$ accordingly , i.e. asking that domain of $F$ is $I^2$ or $D$ with restriction $f$ ?</p> <p>Thank you .</p> http://mathoverflow.net/questions/69510/is-complex-analytic-extension-of-real-analytic-diffeomorphism-a-diffeomorphism/69511#69511 Answer by Alex Gavrilov for Is complex analytic extension of real-analytic diffeomorphism a diffeomorphism ? Alex Gavrilov 2011-07-05T04:52:06Z 2011-07-05T04:52:06Z <p>Indeed, a real-analytic map can always be extended to some complex neighbourhood. The problem is, the neighbourhood may be very small. Consider, for example, the map $$f:I\to I,\, I=[-1,1],$$ defined by $$f:x\mapsto x+\frac{a^3x(x-1)^2}{x^2+a}.$$ For small $a>0$ this is a diffeomorphism of $I$, but it cannot be extended very much due to the poles near $x=0$. The similar map (though a bit more contrived) can be designed for a disc. So, the answer to the question is no.</p> <p>Of course, all this is well known but what is a proper reference I cannot say.</p> <p>P.S. This is an answer to the question as I understand it. There are some points I do not understand. $\Omega$, as it is defined, is a sphere. And, I hope, the restriction is not defined by $F(z,0)=(f(z),0)$: if it is, you can always take $F(z,w)=(f(z),w)$.</p> http://mathoverflow.net/questions/69510/is-complex-analytic-extension-of-real-analytic-diffeomorphism-a-diffeomorphism/69531#69531 Answer by Thomas K for Is complex analytic extension of real-analytic diffeomorphism a diffeomorphism ? Thomas K 2011-07-05T12:38:43Z 2011-07-05T12:38:43Z <p>I'm not 100% sure I understand your question, so pardon me if I'm saying something unrelated.</p> <p>It seems to me that you are asking whether a given real-analytic function $f \colon \mathbb{R}^2 \to \mathbb{R}^2$ (perhaps only defined on suitable open subsets), given by</p> <p>$$f(x,y) = [ f_1(x,y), f_2(x,y) ]$$</p> <p>is in fact complex-analytic, i.e. there is $F \colon \mathbb{C} \to \mathbb{C}$ such that</p> <p>$$f(x,y) = F(x+iy) = u(x,y) + iv(x,y)\text{ .}$$</p> <p>This is patently the case if and only if $u$ and $v$, hence $f_1$ and $f_2$, satisfy the Cauchy-Riemann equations, and thus any pair of real-analytic functions $f_i \colon \mathbb{R}^2 \to \mathbb{R}$, $i=1,2$, which do not satisfy CR furnish a counter-example.</p> <p>On the other hand, what you may have wanted to ask, is whether a real-analytic function <strong>extends</strong> to a complex-analytic function, e.g. if for a real $f(x)$ there is a complex $F(z)$ such that $F(\Re{z}) = f(\Re{z})$. This can indeed always be done on <em>some</em> Stein neighbourhood, but you don't in general have control over the size of the neighbourhood.</p>