Let be a nonsingular projective variety over , and let be a one-dimensional closed subscheme. We have , the ideal sheaf of , and we assume that is a local complete intersection, or what is equivalent, is a locally free sheaf of -modules of rank . For example, this holds if is a nodal curve.
Definition. The normal bundle of the curve in is
We have that
Example. In the case where , with is a smooth genus curve, and having at worst simple branchings, then there are no obstructions to deformation and
A map of moduli spaces
Let be a nonconstant morphism with a nonsingular projective variety over , and let be a closed subscheme which is a smooth curve of genus , such that the ramification of is simple. In a (formal) neighborhood of , the spaces and the are the same. The map
Corollary. If is contained in the smooth locus of , and is ``sufficiently positive'', then the morphism
One argues that the Hilbert scheme is smooth at the point since one can twist by a small number of points and keep that the vanishes. In particular, the corollary implies that the morphism is surjective.
We are now ready to prove:
Theorem. [G, Harris, Starr] If , , then any rationally connected variety over for a curve has a rational point.
Step . Take a general complete intersection ; it will be smooth, irreducible, of say genus and degree . The condition implies that is in the smooth locus and (by Bertini) has at worst simple branching.
Step . Choose a large integer and choose general points , and rational curves such that:
Now let . The basic property is that . Moreover, with colength and assumption (iv) gives that this is ``general''. This gives that the sheaf on is sufficiently positive.
Now deform this curve to a simply branched curve, and this gives the result; conclude by the corollary.
We must deal with the case when fails. Suppose we have a family of varieties with fibres at irreducible of multiplicity . Since the curve must intersect these fibres transversally, this must be preserved in any deformation, meaning that the ramification index at will be divisible by .
In this case, the problem is: cannot dominate. Instead, we consider consider the subset consisting of stable maps , of genus , of degree , such that all ramification indices above are equal to .
Now we have the additional problems: Which reducible curves are in ? And perhaps is too small? To resolve both problems, enlarge the genus (but not ) by adding loops to : join two points with a good rational curve. This allows you to break off a component even in this case.
This is work with Jason Starr. What will guarantee the existence of a rational point on a variety over a function field in two variables? Is there a geometric condition which would be like rational connectedness in this case? This is too much to hope for, there are many surfaces with a nontrivial Brauer-Severi variety . Maybe there are geometric restrictions on the fibres such that one obtains a rational section.
A good guess for this condition: demand that certain moduli spaces of rational curves on the fibers are themselves rationally connected. For example, Starr and Harris proved that for hypersurfaces of degree in with , the moduli spaces of rational curves of fixed degree on are themselves rationally connected.
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