0
$\begingroup$

This may be a very basic question, so my apologies if that is the case.

But I was interested in having some examples of meromorphic (singular) maps into complex projective space from complex surfaces (or higher dimensional objects) which are in the Sobolev space $W^{1,2}$.

The prototype of the examples I want would be the map $(z_1,z_2) \mapsto [z_1:z_2]$ from $\mathbb{C}^2 \rightarrow \mathbb{C} P^1$. This is a rational map and has the desired regularity.

Furthermore I thought that I could also build a map by using some explicit examples of globally generated holomorphic line bundles on complex surfaces. For this I would want to take a subcollection of the generating set of sections, say $s_i$, which vanish at some point $p_0 \in M$ and consider the map $p \mapsto [s_1(p):s_2(p):...:s_m(p)]$. The regularity of this should just be governed by the regularity of the sections near the vanishing point.

Any help would be appreciated, or some specific cases to look at would be great.

$\endgroup$
  • $\begingroup$ Your map $\Bbb C^2\to\Bbb CP^2$ is not well-defined at the origin. Further, there are no nonconstant maps from $\Bbb P^n\to\Bbb P^m$ when $n>m$, which has been discussed on this site before. Remember that a map to $\Bbb CP^1$ is given by two sections of a line bundle, which cannot both be zero at the same time. $\endgroup$ – KReiser Apr 15 at 23:31
  • $\begingroup$ It is not defined at the origin you're right, but that is fine for me (I am only really interested in rational maps anyway). The question you link to is useful thanks. I am not sure I understand the reasoning why a rational map would have to be constant though as it suffices to define the map away from a subvariety of codimension 2? Maybe I am being a bit slow though! $\endgroup$ – ben Apr 16 at 14:21
  • $\begingroup$ Your post previously did not make it clear that you were looking for rational maps. Thank you for updating your post to clarify that that's what you are looking for, There are perfectly good rational maps $\Bbb P^n\to \Bbb P^m$ with $n>m$, like those that you have given. $\endgroup$ – KReiser Apr 16 at 18:47
  • $\begingroup$ Yes sorry about that. I see, that's good news then. Do you know of anywhere where with some written down? I am still a bit confused about the answers contained in the post you linked to. In particular the answer of "Tony" appears to be for rational maps but it seems to me that as long as the vanishing locus of the polynomials is at least codimension 2 then there is no contradiction ... $\endgroup$ – ben Apr 18 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.