MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $f(U)$ ?

share|cite|improve this question
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... – Dmitri Jun 11 '11 at 18:40
yes you are right, i have now changed the question and i hope its right now ;). see above – gregor Jun 11 '11 at 19:18
up vote 4 down vote accepted

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 $\mathbb C^5$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.