most recent 30 from http://mathoverflow.net 2013-06-20T06:50:39Z http://mathoverflow.net/feeds/question/67528 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67528/biholomorphism-complex-manifold-induced-structure gregor 2011-06-11T18:10:14Z 2011-06-11T22:03:12Z

<p>Let $X$ be a $n$ dimensional complex manifold with complex structure $I$ and assume one has a diffeomorphism $f : \mathbb{C} \rightarrow X$ of some open set $U$ in $\mathbb{C}$ into its image $f(U)$. Also $f$ is holomorphic (from this it follows that $f$ is a biholomorphism). Is then $f(U)$ a complex $1$ dimensional submanifold of $X$ ? And if so, does the induced complex structure from $\mathbb{C}$ via $f$ coincide with the complex structure $I$ in the tangent space of $f(U)$ ?</p>

http://mathoverflow.net/questions/67528/biholomorphism-complex-manifold-induced-structure/67546#67546 Dmitri 2011-06-11T22:03:12Z 2011-06-11T22:03:12Z

<p>Provided I understand the question this time, the answer is no.</p> <p><strong>Example.</strong> There are plenty of contrexamples. Let us construct one, so that the closer of $F(U)$ is a complex manifold of complex dimension $2$. To do this, take a complex algebraic torus $T^2$ of dimension $2$ and consider a linear map from $\mathbb C^1$ that has an everywhere dense image in $T^2$. Now, embed $T^2$ in $\mathbb CP^5$ and throw away a hyperplane from $\mathbb CP^5$. This will give you a map from an open subset of $\mathbb C^1$ to $\mathbb C^5$ such that the closer of the image equals the part of $T^2$ contained in $\mathbb C^5$.</p>