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 $f(U)$ ?
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1$\begingroup$ It seems to me there is a confusion in you question. Namely, biholomorphism can be defined only between two complex manifolds. So if you want to have a biholomorphism between $U$ and $f(U)$ you need to have a complex structure on $f(U)$. But you did not say how you define complex structure on $f(U)$. So, it is impossible to answer the question... $\endgroup$– Dmitri PanovCommented Jun 11, 2011 at 18:40
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$\begingroup$ yes you are right, i have now changed the question and i hope its right now ;). see above $\endgroup$– gregorCommented Jun 11, 2011 at 19:18
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1 Answer
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Provided I understand the question this time, the answer is no.
Example. 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 $\mathbb C^5$.
answered Jun 11, 2011 at 22:03