most recent 30 from http://mathoverflow.net 2013-05-22T18:38:24Z http://mathoverflow.net/feeds/question/67735 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67735/immersion-submanifold-of-complex-manifold gregor 2011-06-14T06:43:22Z 2011-06-14T07:52:34Z

<p>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 $I$ ? </p>

http://mathoverflow.net/questions/67735/immersion-submanifold-of-complex-manifold/67737#67737 Hsueh-Yung Lin 2011-06-14T06:56:37Z 2011-06-14T07:04:32Z

<p>If $\alpha$ is an immersion, $\alpha$ needs to be a homeomorphism to guarantee that $\alpha(\mathbb{C})$ is a one dimensional submanifold of $M$.</p>

http://mathoverflow.net/questions/67735/immersion-submanifold-of-complex-manifold/67741#67741 S. Carnahan 2011-06-14T07:52:34Z 2011-06-14T07:52:34Z

<p>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 $I$.</p> <p>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.</p> <p>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 $I$.</p>