Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Consider the affine space $\mathbb C^n$ and then, because of reasons, compactify it to obtain the projective space $\mathbb P^n$. One of the most basic axioms or propositions of geometry is that through two distinct points in $\mathbb P^n$ there passes a single projective line.

Somehow this seems to be very much a fact of projective space. Thus it seems natural to ask how this fact changes when we consider a different ambient space.

So, for simplicity, let's consider a compact complex surface $X$. Again, for simplicity, assume that it admits some rational curves, of degree 1 even. How many of those pass through two (or more?) points of $X$? (We do not assume that a line passes through any two points. We just want to know that given two points that lie on a line, could another line pass through those points?)

From what I understand, basic intersection theory was developed to answer exactly these problems, so one should be able to obtain the answer by intersecting relevant Chern classes on the right vector bundles. Any reasonable attempt at this kind of answer leads to the badly understood problem of calculating the Chow groups and intersection products of surfaces or manifolds. Thus we arrive at a perhaps answerable

Question: Is there an example of a surface (or manifold) $X$ on which we know that more than one projective line passes through two given points? In that case, do we know how many points are necessary to determine a line?

share|improve this question
    
If by line you mean geodesic, there are likely to be very bumpy manifolds where several geodesics share the same $\kappa$-many points for appropriate cardinals $\kappa$. Even with fewer bumps, considering hyperbolic planes and similar geometries may suggest examples to you. Gerhard "Ask Me About System Design" Paseman, 2012.01.24 –  Gerhard Paseman Jan 24 '12 at 18:19
    
The questions seems difficult to make sense of, because of the phrase "rational curves of degree 1". The degree of a subvariety is defined with respect to a given projective embedding of your surface (or complex manifold, or whatever). But given such an embedding, the fact that there is a unique line in projective space through two given points implies the same thing for the surface. –  Artie Prendergast-Smith Jan 24 '12 at 18:23
1  
The issue now is that your original motivation is no longer true: in projective space there can be lots of rational curves through a given set of points. For instance in P^2, there is a 3-dimensional family of conics through any pair of points, and any smooth member of that family is isomorphic to P^1. –  Artie Prendergast-Smith Jan 24 '12 at 18:41
2  
Anyway, all that is just to say that the question in your last paragraph has a rather trivial answer, for either interpretation of the term "projective line". On the other hand, the problem of enumerating the rational curves passing through certain sets of points (even in projective space) is a very famous and actively-researched one. I guess the keywords to search for are Gromov-Witten invariants and Kontsevich moduli spaces. –  Artie Prendergast-Smith Jan 24 '12 at 18:49
1  
I think Edison said he needed a hundred tries before coming up with one good idea. The same seems to be true for learning anything, and this process now seems to be public. –  Gunnar Magnusson Jan 25 '12 at 8:34

1 Answer 1

up vote 9 down vote accepted

As Artie points out, you can't do this with actual lines, but the next best thing is to ask the question with rational curves of minimal degree. If you have lines those will automatically be of minimal degree.

If $X$ is Fano (like $\mathbb P^n$), then a natural ample line bundle to use for degree is the anticanonical bundle. Mori's celebrated Bend & Break Theorem (II.5.4 in Rational curves on algebraic varieties by János Kollár) implies that what you are asking cannot happen for minimal rational curves (with respect to the anticanonical bundle), that is, no two of them will pass through two separate points. In other words, those minimal rational curves actually behave like lines. It also follows that the anticanonical degree of minimal rational curves on Fano varieties is at most $\dim X+1$ which is achieved by lines on projective spaces and by a recent theorem of Cho-Miyaoka-Shepherd-Barron that's the only case when that bound is achieved.

In fact, more generally, the following is true. Suppose there is a one dimensional compact family of rational curves that goes through two points in some ambient space (no assumption on singularities of this space), then by Bend and Break one of the members of the family is either non-reduced or reducible. In other words, if you consider minimal rational curves in the sense that they are such that they do not decompose as a linear combination of other (rational) curves, then it follows that two of these cannot pass through two fixed points (even if the two points are allowed to collide, but that's another, more recent theorem of Hwang-Mok).

The difference between this and the above is that here we're not talking about a given degree, although since the curves in question should be deformations of each other, they will have the same degree, but that degree does not have to be minimal among rational curves.

If you do not require the curves to be deformation equivalent, then all kind of things can happen. For instance a line and a conic in the projective plane will have two intersection points and this occurs quite a lot and similar intersections all over. Or you can take your favorite smooth surface that contains a rational curve with a simple node (say $\mathbb P^2$ with a nodal cubic) and blow up the node. The proper transform of the rational curve with the node will intersect the exceptional curve in two points.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.