Let $\alpha : \mathbb{C} \rightarrow M$ be an immersion and $M$ a $n$ dimensional complex manifold with complex structure $I$. Does then follow that $\alpha (\mathbb{C})$ is a one dimensional submanifold of $M$ ? If, yes, does the induced complex structure on $\alpha (\mathbb{C})$ by the immersion coincide with $I$ ?
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$\begingroup$ What if $M$ is an abelian variety of dimension $g\geq 2$ and $\mathbb C$ is an irrational diagonal line in the original $\mathbb C^g$? Isn't the induced map an immersion? $\endgroup$– Sándor KovácsCommented Jun 14, 2011 at 10:50
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$\begingroup$ Gregor, this is the second of your questions of this type. I think, it would be reasonable if you write a longer detailed question (say 5-10 times longer) that explains what you are thinking about and what really bothers you. $\endgroup$– Dmitri PanovCommented Jun 14, 2011 at 18:58
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If $\alpha$ is only required to be a smooth embedding, then we can set $M = \mathbb{C}^2$, and embed $\mathbb{C}$ as a totally real subspace, e.g., $x + iy \mapsto (x,y)$. This is not a complex submanifold in the usual sense of the word, and the complex structure induced by $\alpha$ on its image is not compatible with $I$.
If $\alpha$ is required to be an immersion of complex analytic spaces but not an embedding, then the image may have singularities. This is even true if we require $\alpha$ to be a homeomorphism to its image, since the normalization map of the cuspidal cubic curve is a homeomorphism whose image is a topological manifold that is not a complex manifold.
If $\alpha$ is an analytic cover of its image, then the image is a complex submanifold of the ambient space, and the induced complex structure coincides with $I$.
answered Jun 14, 2011 at 7:52