Does there exist a complex manifold M which is a quasiprojective variety in two "essentially" different ways? Let me be more specific. I'm looking for a complex manifold M together with two projective varieties X and Y such that M is a locally closed Zariski dense subset of both X and Y. This gives two different notions of "algebraic" objects on M, and I want them to be different. For instance, can the resulting sheaves of regular function on M be different? Or the Picard groups? etc...
$\begingroup$
$\endgroup$
4
asked Oct 15, 2009 at 19:10
-
$\begingroup$ I'm afraid I don't understand the question. In abstract algebraic geometry, things like the sheaf of regular functions and the Picard group of a variety M are intrinsic, so they should not depend on the projective embedding of M. But one can have two independent ample line bundles, which give two projective embeddings. Is that what you mean: find M such that Pic(M) is large, and the ample cone is not just a ray? $\endgroup$– shenghaoCommented Oct 16, 2009 at 21:10
-
2$\begingroup$ The key point is that this is not exactly a question about abstract algebraic geometry. Rather, I am looking for (and, fact, found -- see Georges's comment below) a complex manifold M that can be "stiffened up" to be a quasiprojective variety in two different ways. On a naked complex manifold, there is no notation of things being "algebraic", only holomorphic. The example given by Georges says that knowing which things are "algebraic" is really new information that cannot be gleaned from the complex structure. $\endgroup$– Andy PutmanCommented Oct 17, 2009 at 4:44
-
1$\begingroup$ A fancier thing one could say is that the moduli space of algebraic structures on a complex manifold can contain more than one point. I wonder how large it can be? $\endgroup$– Andy PutmanCommented Oct 17, 2009 at 4:45
-
$\begingroup$ I forgot to mention one thing which should be kept in mind. Serre's GAGA paper says that if M is a COMPACT complex manifold, then M can be given an algebraic structure in at most one way. Thus the M that Georges supplied was by necessity noncompact. $\endgroup$– Andy PutmanCommented Oct 17, 2009 at 14:20
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
Dear Andy,
consider the punctured complex line C* and let M=C* x C*, seen as a holomorphic manifold.
You can embed M into P^1 X P^1 and get the standard algebraic affine structure on M, call it A.
On the other hand hand given an elliptic curve E there exists a P^1-bundle X over E such that, if you delete a section of this bundle X -->E, you get a Zariski open subset U of X which is algebraic, NOT AFFINE, but nevertheless biholomorphically isomorphic to M.
The Picard group of this U is isomorphic to that of E and so definitely not isomorphic to that of the affine variety A.
This construction is due to Serre and is mentioned in Hartshorne's "Algebraic Geometry" (Appendix B, Example 2.0.1, page 440).
Yours friendly,
Georges
answered Oct 15, 2009 at 20:52
$\begingroup$
$\endgroup$
This may be relevant, although I'm not sure if it answers your question directly. The Russell cubic x+x^2y+t^2+z^3=0 is diffeomorphic to affine space A^3. But, it is known not to be algebraically isomorphic to A^3. I'm afraid I don't know the references for this.
answered Oct 24, 2009 at 15:47